18.S997 Notes
Sam Elder
December 12, 2012
1
Wednesday, September 5, 2012
At the beginning of the class, he wrote on the board: Polynomial Method. Larry Guth, lguth@math.mit.edu,
Website: math.mit.edu/ lguth/PolynomialMethod.html. As we waited, the classroom completely filled.
As a new professor, Professor Guth got to teach a class about whatever he wanted. He’ll be teaching about
the polynomial method, which is a new method in the last 5 years. It came originally from computer science
and then people in pure math picked it up and used it in problems seemingly unrelated to polynomials. We’ll
start with talking about the method and then address how the class will be organized.
Let F be a field and x1 , . . . , xs ∈ Fn . First we’ll examine this problem: Find a nonzero polynomial p with
p(xi ) = 0 for all i, and deg p minimal.
Example. Consider the points (j, 2j ) ∈ R2 , for j = 1, . . . , 106. What polynomials vanish at these points?
You can do something clever like −y + 1 + x1 + x2 + · · · + 10x6 , but you could also do something dumb like
(x − 1)(x − 2) · · · (x − 106 ), and both of those have degree 106 . You could also do (y − 2)(y − 4) · · · (y − 106 ).
Or you could take 5 · 105 lines through pairs of points and take their product. You can do better in other
ways.
This question can be solved with linear algebra. Let V (d) be a vector space of polynomials of degree ≤ d
(over F in n variables). Then the evaluation map E(p) = (p(x1 ), . . . , p(xs )), E : V (d) → Fs is a linear map,
and its kernel is the polynomials of degree ≤ d that vanish at xi .
Corollary. There exists a polynomial-time algorithm that finds p of minimal degree.
Corollary. If dim V (d) > s, then there exists a non-trivial kernel.
Now, V (d) has a P
basis of monomials: if our variables are (t1 , . . . , tn ), then the monomials are of the
form td11 · · · tndn with
dn di ≤ d. When n = 2, there are a triangular number of these, and in general,
dim V (d) = d+n
≥ n! .
n
Corollary. For any s points in Fn , there exists a nonzero p such that p = 0 at all the points and deg p ≤
ns1/n .
So in our example, there is a polynomial of degree 2000, much less than the ones we were able to find
off the top of our head. In general, you can either combine or create polynomials or use abstract dimension
counting arguments that will tell you that they exist.
In general, the polynomial method will have a structure like this: You have some problem involving some
finite set of points in a vector space; you find a polynomial of small degree that is a solution to a problem,
and you use that to solve your problem. Now we’ll see some possible applications.
Example. Coding theory. Let F be a finite field of q elements, and p : F → F a polynomial of degree
√
≤ q. This is our information, and now it’ll be corrupted. They’ll receive some function F : F → F, where
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F (x) = p(x) for ≥ 100
q values of x. The question is, is it possible to recover F ? In this situation, it is
theoretically possible to recover p, and it follows from a famous lemma.
1
Lemma. If p : F → F has degree ≤ d and p = 0 at > d points, then p is the zero polynomial.
That immediately implies that this problem cannot have more than one solution:
Corollary. For all F : F → F, there is at most one polynomial p of degree ≤ q 1/2 , so that F (x) = p(x) 51%
of the time.
Proof. Indeed, if two such polynomials p1 , p2 exist, then p1 (x) = p2 (x) at least 2% of the time. Therefore,
2
q zeros, which dominates the degree ≤ q 1/2 for q > 104 . Therefore, by the lemma,
p1 − p2 has at least 100
p1 − p2 = 0.
This gives us an algorithm: Keep guessing until you get the right one. But that’s really inefficient.
There’s a superior algorithm called the Berlekamp-Welch algorithm (1986).
The big idea is to find the lowest degree polynomial R(x, y) that vanishes at all of the points. After a
while, they decided to look at polynomials of the form R0 (x) + R1 (x) · y. And if you do that, then you’ll get
the graph of p plus a vertical line going through all of the errors. This can be computed in polynomial time,
so this problem can be solved quickly.
This was a cool story in computer science, and then it found its way into math through combinatorics.
Example. N ⊂ [0, 1]n is called Nikodym if ∀x ∈ [0, 1]n , there exists a line L(x) such that x ∈ L(x), and
L(x) \ {x} ⊂ N . For example, we can take [0, 1]2 and remove the line y = 1/2. One might feel that a
Nikodym set should be pretty big, so this was surprising:
Theorem (Nikodym). There exist Nikodym sets of measure 0.
Still, they’re pretty big in other ways:
Conjecture. Every N-set has full Hausdorff dimension.
This is still a major open problem with applications to PDE and analysis. Being faced with a big problem
we can’t solve, we’ll imagine various alternatives. So someone suggested looking at the finite field version of
a Nikodym set.
Definition. Finite field Nikodym sets: N ⊂ Fn , with F a field with q elements, is said to be Nikodym if
∀x ∈ Rn , there exists an affine line L(x) such that x ∈ L(x) and L(x) \ {x} ⊂ N .
There was a corresponding conjecture:
Conjecture. Every finite field Nikodym set has ≥ cn · q n points.
Similar partial results had been found for each of these conjectures. Then this one was proven, with a
proof that was about a page and a half and which we’ll see right now.
Proof. Suppose |N | < (2n)−n q n . Lok at the lowest degree polynomial p that vanishes on N . So by construction, deg p ≤ 2q by the formulas we worked out above. Fix x. Then Lx \ {x} is in N , so p = 0 on the q − 1
points of Lx . Then the restruction of p to Lx is equivalently zero, which means that p(x) = 0. Since x was
arbitrary, p(x) = 0 for all x ∈ Fn . There’s a little work to do here, but you can then conclude that p is the
zero polynomial.
This was a surprising result, since the finite field Nikodym set definition doesn’t refer to polynomials.
The intuitive thing to do is to try to build up a large set from a bunch of lines, but apparently that’s not
the way to start. So this had a big impression on Professor Guth, and he’d like to investigate this method
in more detail.
That’s an introduction to what the polynomial method is about. Then he talked about the plan of the
course:
(1) The fundamental examples of the polynomial method: BW algorithm, finite field Nikodym, etc. We’ll
do this for about a week.
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(2) Give some context of these problems in combinatorics (incidence geometry).
(3) The distinct distance problem, which is a much more substantial proof using this method. This will take
up a big part of the class.
(4) Connections to other areas besides computer science and combinatorics. We’ll see connections to geometry, analysis, topology, and number theory. These things we’ve seen so far are recent, but are similar
(in hindsight) to proofs in number theory about 100 years ago in the field of Diophantine equations.
There will be homework for people who are interested in homework and taking the class for credit. There
will be four assignments every three weeks or so, with a range of difficulties between easy and open problems.
As far as prerequisites, nothing fancy is required. Any math or CS graduate student should be fine. Having
said that, we will spend a while on long proofs, so it’s still a serious course.
Professor Guth is thinking of trying to write a book on this topic. He’ll be writing some notes himself,
but he might try to get students to get involved. Sometime soon, an assignment will require that you e-mail
him, and he’ll create an e-mail list from that.
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Friday, September 7, 2012
The class filled up again, and we started at 2:04. We’ll talk about two other applications of the polynomial
method starting today.
2.1
Distinct distance problem
Given a finite set P ⊂ R2 , |P | = N , we can talk about d(P ) := {|p − q|}p,q∈P,p6=q (we exclude 0).
Example (Example 0). For generic points, |d(P )| = N2 ∼ N 2 . That’s clearly the maximum; we’re more
interested in the minimum.
Example (Example 1). An evenly spaced line produces |d(P )| = N − 1 ∼ N .
Example
(Example 2). A square grid gives a slight improvement: Erdos compute that there are |d(P )| ∼
√
N/ log n, and conjectured that this was tight.
Theorem (Guth-Katz). For every P , |d(P )| ≥ cN/ log N , which was better than previous estimates, which
couldn’t reach the power N 1 .
There have also been connections to Diophantine equations in number theory. One of the newest results,
due to Thue, was that a lot of Diophantine equations only have finitely many solutions, like y 3 − 2x3 = 1 or
y 4 − 7xy 3 + 16x4 = 107, and so on. This was nice because previous efforts had focused on very particular
setups, while his was general.
Theorem (Thue). If P (x, y) is irreducible and homogeneous and A ∈ Z, then P (x, y) = A has only finitely
many integer solutions.
In this theorem, irreducible isn’t really necessary because if it reduces, that’s fine. Homogeneous is a big
restriction, and the fact that we can only use two variables is also key. Arbitrary Diophantine equations are
undecideable, and that might happen as soon as 3 variables, but Guth wasn’t sure.
These are two results that will be down the road and take a while in this course. But for now we’ll
return to the Berlekamp-Welch algorithm. Recall what we were doing: If F is a field with q elements, p(x)
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a polynomial of degree < q/100, and then the data gets corrupted into f (x) = p(x) for ≥ 100
q values of x,
then there is an efficient algorithm to recover p from f . We used
Lemma (Vanishing lemma). If p(x) has degree ≤ D and p vanishes at D + 1 distinct points, then p is the
zero polynomial.
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The key to the efficient algorithm is to look at polynomials that vanish on the graph {(x, y) : y = f (x)},
and it’s nice to only consider polynomials of the form R(x, y) = R0 (x) + R1 (x)y. Define W (d) to be the
vector space of such polynomials with deg R0 , deg R1 ≤ d. This has dimension 2d + 2, and we have q points.
Therefore, as long as 2d + 2 ≥ q, there is at least one polynomial that vanishes.
Lemma. There exists R(x, y) ∈ W (d) with d <
q
2
which vanishes on the graph.
Now we have the crucial claim, that this polynomial of two variables also vanishes on the graph of p. In
fact, something even stronger is true: R(x, P (x)), which is a polynomial in x, is the zero polynomial.
Proof. We know that R(x, f (x)) = 0 for all x ∈ F, and we know that R(x, p(x)) is true for at least 51% of
the values of x. This is a polynomial in x with the form R0 (x) + R1 (x)p(x). We know that R0 and R1 have
degree at most q/2, and p(x) has degree at most q/100, so the whole thing has degree at most (51/100)q.
0 (x)
So then R0 (x) + R1 (x)p(x) = 0, implying R1 (x) | R0 (x), and we can recover p(x) = −R
R1 (x) .
We said last time that the zero set of R(x) is the graph of p(x) along with vertical lines through every
error. Define the error set E = {e ∈ F : f (e) 6= p(e)}.
Q
Claim (Claim 2). R(x) = c(y − p(x)) e∈E (x − e).
The vanishing lemma is so important to what we’re doing that we’ll prove it, even if most people here
have seen it before. It relies on another lemma.
Lemma (Divisibility Lemma). If p(x) is any polynomial with p(x1 ) = 0, then p(x) = (x − x1 )p1 (x).
PD
Proof. If we have any polynomial p(x) = j=0 aj xj , then we can always write p(x) = (x − x1 )(bD−1 xD−1 +
· · · + b0 ) + r. This is just the division algorithm: Match the coefficients from top to bottom, leaving a
remainder. Plugging in x = x1 yields p(x1 ) = r, so the result follows.
We can then prove the vanishing lemma by applying the divisibility lemma repeatedly at each of the
roots, assuming it is not the zero polynomial. We’ll also need the following variant:
Lemma (Divisibility Lemma 2). If R(x, y) and P (x) are polynomials, and R(x, P (x)) = 0, then R(x, y) =
(y − P (x))R1 (x) for some polynomial R1 (x).
PD
j
Proof. Expand R(x, y) =
j=0 aj (x)y as a polynomial in y. Then again we can write R(x, y) = (y −
P (x))(bD−1 y D−1 + · · · + b0 (x)) + r(x). Again we choose the expressions sequentially from highest to lowest
degree. Then we know that R(x, p(x)) = r(x) and the claim follows.
Later we’ll talk about how to see if something is divisible by a more complicated expression than y −P (x).
We can use these to prove the claims.
Proof of Claim 2. The Divisibility Lemma 2 implies that R(x, y) = (y − P (x))R1 (x, y). On the other hand,
since R is linear in y, R1 can’t have any dependence on y, so we write R(x, y) = (y−P (x))R1 (x). Additionally,
Q
since e ∈ E, 0 = R(e, F (e)) = (F (e) − p(e))R1 (e), so R1 (e) = 0. Therefore, we see that R1 (x) = e∈E (x −
e)R2 (x) by the Divisibility Lemma 1. All such polynomials vanish on the graph of f , so the minimal one
must have R2 (x) = c ∈ F, and we’ve proven the claim.
Let’s come back to Divisibility Lemma 2. We want to describe what it should say if you have polynomials
of two variables. It is closely related to Bezout’s theorem, which we’ll discuss now.
Theorem (Bezout). If P (x, y) and Q(x, y) are polynomials in two variables x and y, define Z(P, Q) =
{(x, y) : P (x, y) = Q(x, y) = 0}. Then either |Z(P, Q)| ≤ deg P deg Q or P and Q have a nontrivial common
factor (a polynomial of positive degree).
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We can see that Bezout’s theorem implies the divisibility lemma where one of them has the special form
y − P (x). We’ll prove Bezout’s theorem later on in the course.
Let’s talk some more about coding theory. We had corrupted data which agreed with the polynomial
51% of the time. Of course, that was chosen to be slightly greater than a half. There’s an obvious reason: If
you chose two low degree polynomials and picked some data that agreed with one of them half the time and
the other one the other half of the time, and you’d have no idea how to distinguish them. And you might
imagine that if it only agreed, say, 1% of the time with the polynomial there’s not much you could hope for.
In these cases, there are multiple possibilities, but the surprising thing is that there is an efficient algorithm
to come up with a list of all of them.
Theorem (Sudin 1997). If F is a field with q elements, and F : F → F is any function, then there is an
efficient algorithm that finds all polynomials P such that (i) deg P < q 1/2 /200 and (ii) P (x) = F (x) on at
least q/100 varlues of x.
In the end, this proof will also produce a bound on the total number of such polynomials, as we’ll see.
Proof. Look at the graph of F , and find a low-degree polynomial
that is zero on it. Define V (d) to be
1 2
polynomials in 2 variables of degree ≤ d with dim V (d) = d+2
∼
2 d . There are q points, so we can find
√ 2
R(x, y) 6≡ 0 that is zero on the graph of F with deg R ≤ 2 q. And we want to claim that R(x, P (x)) = 0.
Proof. We know that R(x, F (x)) = 0, so R(x, P (x)) = 0 for at least q/100 values of x. On the other hand,
R(x, P (x)) is a polynomial of degree at most deg R · deg p. (This is because when
we substitute p(x) for y,
√ √q
q
and the worst case would have been if we had y deg R .) So deg R(x, P (x)) < 2 q 200 = 100
, and the number
of zeros exceeds the degree, so it must be the zero polynomial, as desired.
Q So y − P (x) divides R(x, y) by DL2. Now note that F[x, y] has unique factorization, so R(x, y) =
j Rj (x, y) where Rj (x, y) are irreducible. Since y − P (x) is irreducible, it must be one of these. Here we
rely on another algorithm as a black box that factors any polynomial into irreducible factors in polynomial
time. Therefore, the only possibilities for the polynomial must each have their own factors, so they are
bounded by the degree of R, which is q 1/2 .
We might be able to improve this, since it’s clear how you’d get 100 different polynomials that it agrees
with 1% of the time each, but how you’d approach q 1/2 is unclear.
That was the end of the notes for the class, but as a bonus, we got to see a sketch of an extension to
two variables. If P (x, y) is a polynomial in 2 variables and F (x, y) agrees with P half the time, then find
a low degree polynomial R(x, y, z) = 0 on the graph of F . Then dim V (d) ∼ d3 and there are q 2 elements
in F2 so deg R ≤ q 2/3 . Then R(x, y, F (x, y)) = 0 for all x, y and R(x, y, P (x, y)) = 0 for at least 21 q 2 times.
The degree of this is at most deg R deg P ≤ q 2/3 deg P . We run into the general question: How many zeros
can a nonzero polynomial in two variables with a fixed degree have? If we can answer that question, we can
similarly bound deg P and get (z − P (x, y)) | R.
A quick summary: Polynomials have been around for a long time, but something new has been brought
to them from computer scientists asking other kinds of questions. We’ll take those methods and apply them
to pure mathematics.
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Monday, September 10, 2012
Today we’ll see how to prove the finite field Nikodym theorem and the finite field Kakeya theorem. The
theorems are short but these were important examples in showing the power of this.
These are both related to still-open conjectures in Euclidean space.
Definition. A set K ⊂ Rn is called Kakeya if it contains a unit line segment in each direction.
Example. For instance, B n (1/2) is a Kakeya set.
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You might think this was the smallest, but Besicovitch showed in the 20s that there exist arbitrarily
small Kakeya sets, and therefore by a limiting procedure, one of measure 0, at least for n ≥ 2. If you look
at other notions of size like Hausdorff measure, they all have dimension n, which is a conjecture.
Also recall the definition of Nikodym sets.
Definition. N ⊂ [0, 1]n is a Nikodym set if ∀x there exists a line L(x) so that x ∈ L(x) and L(x) \ {x} ⊂ N .
Theorem (Nikodym). There are Nikodym sets with measure 0.
Conjecture (Nikodym). Every N set has Hausdorff dimension n.
Are these known to be equivalent? Guth isn’t sure, but he’ll look it up for the next class. These problems
have been open for so many years that people have asked variations on each of them, and we’ll be looking
at finite field variants. Let F be a field with q elements.
Definition. Let K ⊂ Fn . If ∀v 6= 0 in Fn , ∃a such that a + tv ∈ N for every t ∈ F. Define La,v = {a + tv :
t ∈ F}. A finite field Kakeya set has the property that ∀α, there exists some line Lα ⊂ K.
Now if λ ∈ F∗ , La,v = Lλv , so really there are only (q n − 1)/(q − 1) directions. We can think of them as
n
F∗ . The corresponding conjecture is
forming a projective space P Fn−1 = F /{0}
/
Conjecture (Finite field Kakeya conjecture). There is a positive constant cn such that every Kakeya set
contains at least cn q n elements.
Before we prove this, let’s see some straightforward methods and how well they do.
Proposition (Bush method). |K| ≥ 21 q (n+1)/2 .
Proof. There are q n−1 lines each with q points in K. From Pigeonhole, there is an x with at least q n / |K|
lines Lα through it. Suppose |K| < 21 q (n+1)/2 , so there are at least 2q (n−1)/2 lines here. The picture of all
the lines going through x looks something like a bush. The lines must be disjoint except for at x. Then we
have 2q (n−1)/2 (q − 1) > q (n+1)/2 points in our Kakeya set, a contradiction.
Proposition (L2 -method). ∀K ⊂ Fnq Kakeya, |K| ≥ 12 q 2 (for all n ≥ 2).
First let’s compare these bounds:
Dimension n
2
3
4
|K| & with bush
q 3/2
q2
q 5/2
So the bush method is better for dimension at least 4, but this method is better for dimension 2.
S
n
Proof. Let K = α Lα . Then L has at least q n−1
P lines. If α 6= β, |Lα ∩ Lβ | ≤ 1. Then define χα : F → F
by χα (x) = 1 if x ∈ Lα and 0 otherwise. Then x∈Fn χα (x)χβ (x) ≤ 1 if α 6= β and = q if α = β.
P
P
2
2
2
Claim.
x∈Fn (
α χα (x)) ≤ L + Lq ≤ 2L .
Proof. We have
LHS =
X X
X
(
χ)(
χ)
x∈Fn
α
β
!
=
X X
α,β
x
and then we can use the bounds we established.
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χα (x)χβ (x) ,
Now we use the Cauchy-Schwarz inequality to relate the L2 norm of this to its L1 norm. We have
L·q =
x∈K
X
χα (x)
!2 1/2 "

!
X
·1≤
α
X
x∈K
X
χα (x)
α
#1/2
X

1
≤ (2L2 )1/2 |K|
1/2
.
x∈K
So then L2 q 2 /2 ≤ 2L2 |K|, and we’re done.
Jacob interjected that there’s a simpler proof of this bound: Add the lines one by one. The first one adds
q new points; the second q − 1 new points, and so on until the qth which produces at least 1 new point. In
2
all, this gives us q+1
points, which is at least q2 .
2
We’re not really getting close to the actual bound. In fact, we’ve only used the fact that the lines only
are different and intersect in at most one point. Using only that assumption, we can’t actually do better
than q 2 .
Example. In F2q there are q 2 + q lines. If we think of a plane within F3q , it’ll have q 2 different lines, so it’s
really no good to only know that there are q 2 different lines. It was a big step to increase that exponent in
F3q .
Now we’ll explain how to solve these two problems with the polynomial method quickly. We sketched
the finite field Nikodym theorem in the first lecture but this time we’ll be more careful.
Theorem (Dvir). If N ⊂ Fn such that ∀x there is a line L(x) such that x ∈ L(x) and L(x) \ {x} ⊂ N , then
|N | ≥ (2n)−n q n .
Proof. Let V (d) be the vector space of all polynomials in n dimensions of degree at most ≤ d.
dn
Lemma. dim V (d) = d+n
≥ n! .
n
Proof Sketch. This has a basis of z1d1 · · · zndn with d1 + · · · + dn ≤ n. There’s a quick way of encoding these
exponents with the “stars and bars” format, which we’ll illustrate: For z1 z22 with d = 4 and n = 2, we get
∗| ∗ ∗|∗. We can see there’s a bijection there and that there are d stars and n bars, so there are d+n
basis
n
elements.
Now suppose that |N | < (2n)−n q n .
Corollary. There exists a nonzero P such that P = 0 on N and deg P ≤ 2q .
In fact, you don’t need to see the constant of (2n)−n in front, just know that we rigged it to be small
enough. Now this polynomial is going to interact in interesting ways with these lines.
Fix x. Then P = 0 on L(x) \ {x}, which contains at least q − 1 points. We had our vanishing lemma.
Lemma (Vanishing lemma). If P : Fn → F has degree D, L is a line, and P = 0 at > D distinct points on
L, then P = 0 on L.
Essentially we want to parameterize our line to make it only one dimensional.
Proof. Parameterize L by γ(t) = a + tv. Then P (γ(t)) is a polynomial in t with degree at most D and > D
zeroes, so P (γ(t)) ≡ 0.
So in this case, we must have P (x) = 0, so it’s zero everywhere. We’re almost done, but not quite. We
can’t immediately and directly contradict it being zero. We need to know what kind of degree a polynomial
in several variables can be to have a lot of zeros. There’s a corresponding generalization of the result from
one variabe.
Lemma (Vanishing lemma 2). If P is a nonzero polynomial in n variables over Fq with degree D, then P
has at most Dq n−1 zeroes in Fnq .
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Proof. Induct on n; we know it to be true for n = 1. Let z ∈ Fn and z = (~z, zn ) with ~z ∈ Fn−1 . Then we
PE
know that we can write P (z) = j=0 Pj (~z)znj , where E ≤ D and PE is nonzero. We know that for all ~z,
either P (~z, zn ) has at most E zeroes, or PE (~z) = PE−1 (~z) = · · · = 0 (so all the coefficients are zero). Divide
the space into type 1 and type 2 based on this. We know that there will be at most Eq n−1 zeros in type
1, and by induction, PE (~z) has ≤ (D − E)q n−2 zeroes ~z ∈ Fn−1 . So in all, the total number of zeroes is at
most Eq n−1 + (D − E)q n−1 = Dq n−1 , as desired.
In particular, in this context if we have a polynomial with degree less than q that vanishes at every point,
it must be the zero polynomial, finishing the proof.
Note: Just the application of the second vanishing lemma to this problem is a little simpler.
We’ll start next class with the Kakeya conjecture proof. You can go home and try to imitate this proof
between now and then.
A couple announcements:
• Class will be cancelled for the Jewish holidays on September 17 and September 26.
• The first problem set will be ready by Wednesday, due in two to two and a half weeks after that.
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Wednesday, September 12, 2012
Last time we had proven the finite field Nikodym theorem and were about to prove the finite field Kakeya
theorem (Dvir 07).
Theorem (Finite field Kakeya). If K ⊂ Fnq is a Kakeya set, then |K| ≥ (2n)−n q n .
A lot of the details of the proof are about the same, but we’ll have to something tricky at one point.
Proof. Suppose |K| < (2n)−n q n . So there exists some polynomial P such that P = 0 on K and deg P ≤
q/2 < q. We know that there’s a line in every direction, so ∀v 6= 0, ∃a such that a + tv ∈ K for all t ∈ F.
Fix a and v, and define R(t) =: P (a + tv), which is a polynomial in t of degree at most deg P < q. So by
the vanishing lemma, R is the zero polynomial.
What does this mean for the coefficients of P ? When we expand out P (a+tv), its coefficients will depend
on a and v in an annoying way, but we can focus on the largest degree term. Let P = Pd + P 0 , where Pd is
homogeneous of degree d and deg P 0 < d. So then P (a + tv) = Pd (v)td + lower order terms, so Pd (v) = 0.
Now Pd (0) = 0 since Pd is homogeneous of degree > 1. Therefore Pd has too many zeros for its degree, and
it must be the zero polynomial, a contradiction.
This proof is a little bit more complicated than the finite field Nikodym theorem. The sketch of finite
field Nikodym says that there is a polynomial that vanishes at a bunch of points, so it must vanish at some
more points, which become too many, a contradiction. The Kakeya one is a little more complicated, so here’s
another way to look at it:
Think in terms of projective planes. Fn ⊂ P Fn = Fn ∪ {points at ∞}, where we have one point at ∞
for each direction. Any polynomial P has roots Z(P ) ⊂ Fn which is an affine variety. If P is homogeneous
(which we can make it by sticking on copies of another variable), these are contained in the Zproj (P ) ⊆ P Fn ,
a projective variety. The Kakeya polynomial must vanish at all of the points at ∞ since it is zero on one
line in each direction. And you can translate the previous method into this language.
8
4.1
The joints problem
We’ll spend the rest of today looking at the Joints problem, which is another problem using this method.
Definition. Let L be the set of lines in R3 . A joint is a point x ∈ R3 that lies in 3 non-coplanar lines.
The Joints problem asks: What is the maximum number of joints that you can make with L lines.
Let’s think of some examples.
Example.
1. Use vertical, horizontal, and inward lines in a cube. If we have a S × S × S grid, we have
3S 2 lines and S 3 ∼ L3/2 lines.
2. A tetrahedron has 6 edges (or lines) and 4 joints.
Can this generalize? We could think about taking
S planes in general position, and make their S2 intersections our lines and their triple intersections
give S3 joints. Again, the number of joints is L3/2 , but our constant is a little better. This is the best
known example in fact.
What can we do for upper bounds? There’s a trivial upper bound of L2 ∼ L2 since each joint must
be the intersection of some two lines (in fact, three). We can improve this slightly by using the point that
you need three lines. But not much. Indeed, if you just looked in a plane, you could get order of L2 triple
intersections: Make a grid, and draw in the diagonals as well. But these wouldn’t be joints since the lines
are coplanar, so it’s clear that condition is important.
Theorem. For any L lines, there are at most 10L3/2 joints.
Proof. We’ll walk through this proof by audience participation and be opimistic and write “proof.” (Laughter.)
What do we do first? Find a nonzero polynomial P of minimum degree that vanishes at all J joints. Its
degree is at most 3J 1/3 . Consider P restricted to a line l ∈ L that goes through a bunch of joints (we’ll
clarify this later). If l contains at least 3J 1/3 points, then P ≡ 0 on l. This is equivalent to saying that
either (a) l has at most 3J 1/3 joints or (b) P ≡ 0 on l.
We could take out lines that fall into case (a) and do a sort of induction on what’s left. Thus we know
that J ≤ 3J 1/3 + 10(L − 1)3/2 , by the induction hypothesis. There’s a bit of a calculation to be done here,
so he just showed us what to do. If J ≤ 10L3/2 , we’d be happy, so assume otherwise that J > 10L3/2 ,
J
J
. Thus we can write J < L
+ 10(L − 1)3/2 , or
which implies (with some loss) that 3L < J 2/3 , or 3J 1/3 < L
L−1
3/2
1/2
3/2
, so J < 10L(L − 1)
≤ 10L . This was a bit of an annoying calculation but you
L J < 10(L − 1)
can do it.
So case (a) was fine, and what do we do in case (b)? Some ideas:
• Look at a joint. Does the polynomial being zero on all those lines imply that it’s zero in general? No,
with counterexamples the origin and the axes with the polynomial xyz, or just xy.
• Both of those involve planes... Does it imply that it is zero on a plane? No, with a clever example
xy + yz + zx.
• All of the terms are degree at least 2. This seems likely...
• Maybe the derivative vanishes at a joint: ∇P (x) = 0. Indeed, its derivative in three independent
directions is zero, so ∇P (x) = 0 at all the joints.
But ∂1 P , ∂2 P and ∂3 P are all smaller degree than P , and zero at all of the joints. Therefore, ∇P = 0 and
hence P is constant, which means it must be zero.
9
4.2
Axis-parallel joints
There’s a special case of this which is easier to prove. Suppose all of the lines are in the directions of the
axes: Let Lj ⊂ L be the lines parallel to the xj axis, j = 1, 2, 3.
Theorem (Loomis-Whitney, 51). If |Lj | ≤ L for each j, there are at most L3/2 joints.
He’ll prove this at the beginning of next class, but you can think about it (and possibly prove it with
your own bare hands) next time.
Another similar problem: “nearly axis parallel joints.” Lj are within 1 degree of parallel to the x-axis,
and again define the joints to be those that are on one line of each type. And you’d get the same bound.
A couple bureaucratic points: First, the homework is posted and is due two weeks from Friday. Included
is an assignment to send an e-mail to him introducing yourself. He’d prefer you do that sooner than when
the homework is due. Also, he’d like to add another lecture on September 21, a student holiday on a Friday,
to make up for the Jewish holidays when he’ll have to skip class. E-mail him if you don’t want him to do
that, and if he gets enough e-mails he’ll cancel class. Finally, he’d like to ask people enrolled in the class to
take detailed notes one or two days throughout the semester so he can save himself some time.
5
Friday, September 14, 2012
Today we’re going to try to understand the strength of the polynomial method. Our motivation is that we’d
like to come up with a ratio of lengths of the proofs of the joints theorem using the polynomial method and
not using the polynomial method. If the proof needs to be much longer without the method, why is that?
To approach this, we’ll recap the joints method.
5.1
Recap: Joints
Let’s recap the lemma from class.
Lemma (Main). If there are J joints, then one line contains ≤ 3J 1/3 joints.
Theorem. L lines implies at most 10L3/2 joints.
To prove the theorem using the lemma, we removed lines one at a time, which meant that J ≤ L · 3J 1/3 ,
or J 3/2 ≤ 3L, or J ≤ 10L3/2 . And yes, we could use 33/2 instead of 10, but Guth isn’t very careful about
those kinds of things.
To prove the main lemma, we let P be a nonzero minimal degree polynomial vanishing at all the joints.
Then deg P ≤ 3J 1/3 . We supposed that the main lemma was false and each line has > 3J 1/3 joints on it.
Then P = 0 on each line, so we concluded that ∇P = 0 at each joint. By minimality, ∇P ≡ 0 so P is
constant.
5.2
Axis-Parallel Joints
S
There’s an analog of the problem where the lines are all parallel to the axes: L = j∈{1,2,3} Lj where Lj is
the set of lines parallel to the xj axis. This case is actually a lot easier to prove from elementary methods.
We’ll just prove the analog of the main lemma.
Proof. Assume each line has > 3J 1/3 joints. Take a line in the x1 direction. Then through each of the 3J 1/3
joints there are lines in the x2 direction with at least 9J 2/3 joints, which then must lie on at least 9J 2/3
different lines in the x3 direction. Then there are at least 27J different joints, a contradiction.
10
5.3
Nearly Axis-Parallel Joints
Set some parameter α > 0 and let Lj be a disjoint set of lines whose angle with the xj axis is less than α.
Where does the previous proof fail? Well, we might not get 9L2/3 lines in the penultimate step. He drew a
potential example of this.
S
We could also allow the lines to curve rather than just being lines. Suppose T = j∈{1,2,3} Tj where
TJ is a disjoint set of curves whose tangent vector is always < α from the xj axis. The axis-parallel curves
question is an open question: How many joints can we make with L α-nearly parallel curves? For instance,
if α = 1/100, is it less than CL3/2 or even, for that matter, CL1.99 ?
Of course, the polynomial method solves the nearly axis parallel question, since they’re still lines. But it
doesn’t say anything about the curves, which is why it’s still an open question.
5.4
Lines Intersecting in R3
3
We have several questions we could ask. First, if there are L lines in R
, how many intersection points can
L
there be? We could just put them all in a plane and there would be 2 . So we could restrict oursevles away
from that by requiring that at most 10 lines are in any plane. How many intersection points can there be?
The answer is still ∼ L2 ! The construction uses a surface S where z = x · y. S contains many lines:
“horizontal lines” t 7→ (t, y0 , ty0 ) and “vertical lines” t 7→ (x0 , t, tx0 ). If we take L/2 horizontal lines and
L/2 vertical lines, then we get L2 /4 intersection points.
Does every plane contain at most 10 of these lines? In fact, no three of them are coplanar. One way to
see this is that the intersection of a plane with this surface is a conic, which could be degenerately two lines,
but won’t contain more than that. Another way is that the
So we’ve seen that one way that we can arrange for a lot of lines to intersect is to have them all lie in a
plane, but you can also arrange for them to lie in any sort of low-degree algebraic surface.
Our next question then: If there are L lines in R3 and ≤ 10 in any algebraic surface of degree at most 2.
How many intersection points can there be? This is actually an open question.
Well, there’s a theorem that says this is bounded by CL3/2 . Let’s see what sort of lower bounds we can
come up with.
We could just add the lines one by one to intersect with the previous line to get L − 1 lines. You might
also be able to connect the last one up to the beginning and get L, a slight improvement.
We messed around with picking a fourth line starting with three lines that we’d like to hit. Eventually it
seemed like parameter fitting, so we asked how many parameters there are in the set of all lines in R3 . We
concluded from various arguments that the answer was 4.
If we want to hit a given line, there will be 3 free parameters. And if we want to go through two lines,
there will be 2. In fact, that’s easy to see because you just pick the two points on the lines. And if you
want to hit 3 lines, there’s a 1-parameter family. These lines will sweep out an algebraic surface of degree 2
(which we’ll prove in a week or so) called a regulus. Generically, you can also find a finite number of lines
that will intersect any 4 lines.
So this suggests another method: We could start with
a number L0 of general lines and then add one
that will intersect any 4 of them. So then L = L0 + L40 . Thus the number of intersections is almost 4L.
This is the best possible known example, and Guth believes it’s close to that. He does know that it doesn’t
help to look along degree 3 or 4 surfaces.
A short list of bureaucratic things: No class on Monday. We will have class a week from today to make
up for it. And Adam will TeX up notes for next Wednesday.
6
Wednesday, September 19, 2012
The next section of the class is to get some background in the fields of math that these problems come from.
We’ll focus on one field called incidence geometry within combinatorics.
11
Let L be a set of L lines in R2 . Let Pk (L) = {x s.t. x lies in ≥ k lines}, called a k-fold intersection. A
simple but not necessarily easy question: What is the maximum of |Pk (L)| for some k, L?
Example (1). An easy example is to get L/k k-fold intersections by picking L/k points and drawing k lines
through each of them.
We’re going to think of this by imagining where the points are and then arranging for enough lines to go
through them. Naturally, we’re going to want to have a lot of points on some lines. A good way to do this
is to use a grid:
Example (2). Consider a N × N grid of points, and let L = {lines that contain ∼ R points of the grid},
where by that we mean it contains between R/2 and 2R points. We claim that there are ∼ (N/R)2 lines
through each point. The idea is that we consider a point x, and a line in L going through x, then there must
be another point on L within a L × L grid with x at a corner. Conversely, given such another point y, we
can draw the line through x and y, and if it goes “towards the middle”, we’re good.
There’s the problem that we’ll be overcounting the line through x and x + (1, 1) and the line through x
and x + (2, 2). So we need a lemma:
Lemma. For every B, there exist at least
1
2
100 B
integer pairs (x, y) with
B
2
≤ x, y ≤ B with gcd(x, y) = 1.
Proof Sketch. There are B 2 /4 total points. 1/4 of them are both divisible by 2; 1/9 of them are both divisible
by 3, and so on, and 1/4 + 1/9 + 1/16 + . . . < 4/5.
Now let the smallest degree of any grid point be K, so |PK | ≥ N 2 . Now by the above argument we have
K ∼ (N/R)2 , so L ∼ |PK | · K/R ∼ N 4 /R3 = L. Therefore |PK | ∼ L2 K −3 . We’ve sketched how to prove:
√
Proposition. For every K < L, there is a configuration such that |PK | ≥ cL2 K −3 .
√
Now the L/K bound might be better if K > L. There’s an important theorem which says that
whichever of these is better is the bound.
√
L2
L
+K
. If K > L, the first term dominates,
Theorem (Szemeredi-Trotter, early 1980s). |PK | ≤ C K
3
√
and Example 1 is sharp up to C. If K ≤ L, the second term dominates, and Example 2 is sharp up to C.
We’ll focus on proving this today and Friday. There are some basic upper bounds you can get through
counting:
2
2
Proposition (1). |PK | ≤ L2 / K
2 ≤ 2L /K .
Proof. There are L2 pairs of lines, and at each x ∈ PK , K
2 pairs must intersect at x, so this is trivial.
Proposition (2). If
K2
4
> L, |PK | <
K
2 .
Proof. Suppose |PK | ≥ K/2, and take P ⊆ PK with |P | = K/2. For each x ∈ P , there are at least K/2 lines
K2
through x that don’t contain any other point of P . Thus L ≥ |P | K
2 = 4 , as desired.
Proposition (2B). If K 2 /4 > L, then |PK | < 2L/K.
Proof. By proposition 2, since |PK | ≤ K/2, ∀x ∈ PK there are K/2 lines through x that don’t pass through
any others, and as above, L ≥ |PK | K/2, as desired.
Let’s take stock of what we’ve proved so far. We drew a graph of K versus |PK | in terms of L. Proposition
1 gives us a line between (L, 1) and (1, L2 ). Proposition 2 has a lower line from (L, 1), and the theorem
claims that this is a bound, and the other bound is the line from the end of this one to (1, L2 ). It’s kind of
hard to describe the drawing.
So far we’ve only used the fact that there’s only one intersection point between any two lines. But if we
look over finite fields, we get a different answer. There’s an important example:
12
Example. Just take all lines in F2q . Then L = q 2 + q, and there are K = q + 1 lines through each point. To
make it simpler, we can ignore vertical lines and then |PK | = q 2 with L = q 2 and k = q.
This is a point above the region that we want to prove with Szemeredi-Trotter. It means that over finite
fields there’s essentially a phase transition when you get to K = L1/2 . Again, this is hard to describe, but
the point is that we need to use something further about the plane, like its topology. We’re going to end up
using the Euler formula:
Consider a disc with a bunch of line segments through it. Then let V = Vint + 2L be the number of
intersections, plus the vertices at the edges of the lines, and E the sections of the line segments as divided
by other lines, also E = Eint + 2L. The Euler formula says that V − E + F = 1, or Vint − Eint + F = 1.
The vertices and edges are closely related to what we want to measure, but the faces aren’t, so let’s
remove them. By looking at the faces on the interior, they each have at least 3 edges, so 3F ≤ 2Eint + 2L.
or Eint ≤ 3Vint + 2L.
Plugging this in, 1 ≤ V − E + 23 Eint + 23 L = Vint − 13 Eint + 23 L,P
Looking locally at each vertex, we can rewrite this formula as v∈Vint ( 12 deg(v)−3) ≤ L. Here we sent the
contribution of each interior edge half to each of its endpoints, and it became L because we subtract off the
edges that aren’t interior. We notice that 21 deg(v) is the number of lines going through v. Unfortunately,
we see that the 2-fold intersections count negatively in this; 3-fold intersections don’t count at all; and
higher intersections count positively. This is a little unfortunate because the negative contributions can’t be
bounded by this. But if we ignore them, we get
Proposition. If every intersection has multiplicity ≥ 3, then |PK | ≤
2L
K−3 .
We see that this rules out the sort of behavior we found in the finite field case, but it still doesn’t improve
our bounds considerably in general.
Guth was thinking about this more, and realized that the inequality we had was sharp if the faces are
triangles. He drew out the grid example on a sheet of paper to see if it would be possible, and while he saw
a lot of triangles, there were also plenty of quadrilaterals and pentagons. He passed his drawing around the
room.
This method can, tangentially, be used to prove that K5 , the complete graph on 5 vertices, cannot be
embedded into the plane, or equivalently, S 2 .
Proof. Suppose it could be, and applying the Euler formula on S 2 , V − E + F = 2. Then V = 5 and E = 10,
and 3F ≤ 2E = 20 so F ≤ 6. Then 2 = V − E + F ≤ 5 + 10 − 6 = 1, a contradiction.
So we’ll get into some graph theory, because it has to do with proofs of these.
6.1
Crossing Numbers of Graphs
Let G be a graph and consider a legal map F of this into the plane. We say that vertices must go to distinct
points, and edges must go to curves from one edge to the other. Then the crossing number of F is the
number of pairs of edges (E1 , E2 ) with F (E1 ) ∩ F (E2 ) 6= ∅, and the crossing number of a graph G is the
minimum over all legal maps. Of course, the crossing number is zero iff it is planar.
A classic question:
What is the crossing number of Kn for large n? For K5 , it’s 1. And we can get a
trivial bound of E2 using generic points and straight lines. Laszlo asks: Is it the same for the curves to be
straight lines? Apparently not, says Nate. There are some really weird examples.
7
Friday, September 21, 2012
Today we’ll do the crossing number proof of the Szemeredi-Trotter theorem, which Guth learned as a postdoc.
He gave a talk about it then and brought his notes about it. At the end, someone said that Dvir had solved
the finite field Kakeya problem, which he also wrote on his notes to look up and got him started on this
whole field.
At the end of last class, we had this proposition:
13
Proposition. If G is a graph with E edges and V vertices and E is planar, then E ≤ 3V .
Proof. Embed G into S 2 . Then it cuts the plane into some number F faces. The Euler formula gives
2 = V − E + F . Last time we realized that each face needs to have at least 3 sides, and therefore, 3F ≥ 2E
since each edge counts for two faces, or F ≤ 32 E. Plugging that in, 2 ≤ V − 13 E, which gives E ≤ 3V − 6.
Wait, is this true? If our graph only has two vertices and an edge, then 3V − 6 = 0 < 1 = E. Well, in
this case, we can’t really say that the face has at least three edges. But if we have at least 3 vertices, we can
add more edges, which will only hurt us, to make all of the faces triangles. Moreover, if it’s disconnected, we
can do the same thing to make it connected. And if there are 1 or 2 vertices, we can just check hte cases.
Proposition. Let G be a graph with E edges and V vertices, then the crossing number k(G) ≥ E − 3V .
Proof. Map it into S 2 with k(G) crossings. Remove an edge for each crossing to get a graph G0 ⊂ G which
is planar. Then E 0 = E − k and V 0 = V , so by the first proposition, E − k ≤ 3V , as desired.
Example. What bound does this give us for the crossing number of the complete graph? We had E = n2
and V = n, so the bound is k(Kn ) & n2 . Of course, k(G) ≤ E2 , and we’ll see that actually, we need to get
k(Kn ) . n4 is the closer bound.
How could we do better than this? Well, the crossing number might be bigger if there are edges that go
through many other edges. We’ll see a probabilistic argument which does this in a little different way.
Theorem. If E ≥ 4V then k(G) ≥
1
3 −2
.
64 E V
Let’s see this for the complete graph: k(Kn ) & (n2 )−3 n−2 ∼ n4 .
Proof. Let 0 < p ≤ 1. Pick a subgraph G0 ⊂ G where we random include each vertex with probability p
and include an edge iff its endpoints are included. Then E(V 0 ) = pV and E(E 0 ) = p2 E. Then E(k(G0 )) ≥
p2 E − 3pV from k(G0 ) ≥ E 0 − 3V 0 . That’s a lower bound on the crossing number, and we also have an upper
bound from the original graph:
Claim. E(k(G0 )) ≤ p4 k(G).
In fact, if we use the same map into S 2 with k(G) crossings, we expect a crossing to be preserved with
probability p4 .
But what if two of those vertices coincided, so we had something looking like ∝. Well, we want to prove
that if our mapping was minimal, then we can’t have a crossing like this. What we do is look at the paths
going to that vertex, and then pick the one that crosses fewer other edges. Send both along that one (slightly
shifted) so they don’t run into each other and you’re done.
Anyways, continuing, we have p2 E − 3pV ≤ E(k(G0 )) ≤ p4 k(G). So k(G) ≥ p−2 E − 3p−3 V . It turns
out to be best to let p = 4V
E , and since we’re assuming E ≥ 4V , this is allowed. Then this bound is
1 E3
3 E3
1 E3
−
=
.
16 V 2
64 V 2
64 V 2
So we see these different regimes: from E = 3V to E = 4V it’ll increase linearly (and we can see an
example with E = 4V of a grid with diagonals). Then we’re sort of saturated and it’ll increase at the E 3 /V 2
rate, and these bounds are tight up to constant factors.
7.1
Proving Szemeredi-Trotter
That theorem had been known before, but it wasn’t applied to prove this until later.
Theorem (Szemeredi-Trotter). For any L lines in the plane, if Pk is the set of points on at least k lines,
then |Pk | ≤ C(L/K + L2 /K 3 ).
14
Proof. We’d like to correspond this to a graph G. For its vertices, we’ll choose the points of Pk , and put
edges between two of these
if they are consecutive points on a line. Then the crossing number of this graph
is at most L2 (really, L2 since we require two different lines). So by the crossing number theorem, either
E < 4V or
−2
L2 ≥ k(G) ≥ cE 3 |Pk | .
How many edges are there? (L + P ) − 2L: The points divide the lines up into at least L + kP segments, but
we subtract the two ends of each line to get E ≥ kP − L. Substituting in, we get L2 ≥ ck 3 P , so P ≤ L2 /k 3 .
Otherwise, we have E < 4V , so P < 2L/k, as desired.
This is one of the most important problems in incidence geometry, so we’ve actually gotten quite far in
our brief survey. There are several open problems related to this:
• The unit distance problem: Given N points in the plane, how many unit distances can be have?
• The distinct distance problem: What is the smallest possible number of distinct distances between N
points?
• S-T for circles: Given N circles in the plane, how big can Pk be?
Let’s try to solve the unit distance problem by class participation for the last few minutes of class. Let’s
start with some examples.
Example. A square grid. What is the most common distance in a square grid? Well, it becomes a number
theory problem of counting how many solutions (x, y) there are to R2 = x2 + y 2 . And we’d want R < s/2 or
something like that to actually realize this. Now how does that number of solutions scale? It’ll be something
bigger than s, or ω(N ), but also o(N 1+ ).
Guth thinks it would be good to have office hours. He’ll next have them on Tuesday afternoon at 4pm.
Laszlo will type up notes on Monday (but I still will as well). On Monday, we’ll talk about how to coordinate
with the math retreat next Friday (the buses are supposed to leave at 2:45).
8
Monday, September 24, 2012
It seems tough to have class this Friday, if the class leaves at 2:45. We talked and laughed about several
options: class on the bus, class at the retreat. So we’re not going to have class on Friday. To turn in the
homework, there will be an envelope on Guth’s door.
We’ll now use the crossing number technique to try to approach the distance problems. This is the
biggest single approach that has been around.
Suppose that we have N points in the plane with t distinct distances between them. We can draw N t
circles around each point so that each point is contained in N − 1 circles (the ones centered at all the other
points). We drew a possible picture.
Think of this picture as a graph G, with vertices the points and edges the arcs of circles between
consecutive points. Then each point lies on N − 1 circles, so there are N (N − 1) ∼ N 2 edges in all. We
can also measure
the crossing number of this graph, since two circles only cross in two points each. So
k(G) ≤ 2 N2t ∼ (N t)2 .
Apply the crossing number theorem from last time. Then N 2 t2 & κ(G) & E 3 V −2 ∼ (N 2 )3 N −2 ∼ N 4 ,
so t & N . How do we feel about that proof?
So technically, this is a multigraph, and that’s a problem. We could have several circles going through
the same point. And in fact, if we tried to extend it to multigraphs, we could just add a ton of multiple
edges and not produce any more crossings. Where did this come up in the proof? We had to use that the
faces were triangles, which isn’t true in this case because we have a lot of bigons. So this proof
√ is wrong, and
the result is incidentally also wrong. The best example known is that there are order of N/ log N distinct
15
distances, and the best known lower bound is actualy N/ log N , the proof of which uses the polynomial
method and which we’ll be spending a while on later.
But for now, let’s try to salvage this result. We might think that it doesn’t have many multiple edges.
Say that if G is a multigraph has multiplicity M ult(G) ≤ M if between any two vertices there are ≤ M
edges. What can we say about the crossing number in terms of the multiplicity?
Proposition. If M ult(G) ≤ M and E ≥ 4M V , then k(G) ≥ E 3 V −2 M −3 .
Proof. Let G0 ⊂ G be an honest graph, which has at least E 0 ≥ E/M edges, and apply the previous
result.
What about loops? We could have loops in our examples, but when we actually fix this up, we’ll see that
we can just delete those circles.
To have multiple edges, we’d need the other points to lie on the perpendicular bisector. So if we rule this
out by fiat, then we’re done.
Theorem. If N points in the plane have at most 100 on any line, then the number of distinct distances is
at least cN .
Proof. First let’s deal with loops. Remove all the circles with only one point. We removed at most N t
edges and still have at least 12 N 2 edges. Then the multigraph has multiplicity at most 101 and use the
proposition.
This is a result that Guth isn’t really sure if anyone found interesting enough to write down. We’ll come
back to the distinct distance problem but for now segue to the unit distance problem.
We were examining a square grid with N points, and found that the number of unit distances U was
bounded by ω(N ) ≤ U ≤ O(n1+ ). There’s a conjecture that this is the bound for all sets.
Theorem. Given a set S of N points in the plane, U ≤ CN 4/3 .
Proof. Draw the unit circles with centers in S that contain at least 2 points of S. This defines a multigraph
with the N points and arcs between consecutive points on a given circle. Assume U ≥ 10N , and we have
E ≥ 12 U . Then the multiplicity of this graph is at most 4 since there are at most two unit circles through
any two given points. Then 2N 2 ≥ k(G) & E 3 V −2 & U 3 N −2 , so U . N 4/3 .
This is still the best known bound. Why is this so hard? Well, it’s hard to distinguish unit circles from
shapes like “unit parabolas:” polynomials of the form y = x2 + ax + b. Taking a = 1, . . . , s, and b = 1, . . . , s2 ,
we get s3 parabolas. Consider 3s3 points [1, . . . , s] × [1, . . . , 3s2 ], each parabola contains s points. The “unit
distances” would correspond to the number of pairs (p, Γ) of a point p and a parabola Γ with p ∈ Γ, and
there are s4 pairs here.
To talk about this better, we need some helpful definitions:
Definition. Let S be a set of points and L a set of curves. Then an incidence I(S, L) = {(x, l) ∈ S × L :
x ∈ l}.
Note that if we could come up with an example like the previous one with unit circles, we could just add
in the centers and still have a counterexample to the unit distance conjecture. We also have examples like
this using a different norm on R2 .
Let’s return to the problem of bounding crossing numbers for multigraphs with multiplicity ≤ M . Let’s
start with the example of K5M , taking K5 with M copies of each edge. We think there should be at least
M 2 crossings, if we just replace a given edge of K5 with M closely spaced edges. Can we do better? No. If
we randomly choose one representative edge from each multiedge of K5M to get a graph G0 ' K5 , then we
get at least one crossing since K5 is not planar, so 1 ≤ E(k(G0 )) ≤ M12 k(G).
Let’s generalize this notion.
16
Lemma. If G is a multigraph with multiplicity ≤ M , E ≥ 4M V , and each edge has multiplicity ≥ M/2,
then k(G) ≥ cE 3 V −1 M −1 .
Proof. Let G0 ⊂ G be an honest graph with one edge for each set of parallel edges of G at random. Then it
1
E ≥ 4V edges, and M42 k(G) ≥ E(k(G0 )) ≥ (E 0 )3 V −2 ∼ M −3 E 3 V −2 .
has E 0 ≥ M
In this proof, we needed the multiplicities to be big, which is weird. So can we remove that condition?
Proposition. If G is a multigraph with multiplicity M and E ≥ 100M V , then k(G) ≥ c0 E 3 V 2 M −1 .
Proof. The idea is to look at the edges with multiplicity at least M/2. If that’s a large portion of the graph,
we use the previous argument, and if it’s not, then we remove those and use induction.
1
E ≥ 10M V , by our lemma, k(G0 ) ≥
Let G0 ⊂ G be the edges with multiplicity > M/2. If E 0 ≥ 10
1
0 3 −2
−1
0 3 −2
−1
0
c(E ) V M
≥ c E V M , as desired. Otherwise, E < 0 E and we induct on E. Say G1 ⊂ G is
9
the set of edges of multiplicity ≤ M/2. By induction, k(G1 ) ≥ C0 E13 V −2 (M/2)−1 . Then E1 ≥ 10
E, so
3
3 −2
−1
3 −2
−1
k(G1 ) ≥ (9/10) · 2c0 E V M ≥ E V M .
Next time we’ll try to use this to get a bound for the distinct distance problem. There is no class on
Wednesday or Friday, unfortunately, so the next class will be next Monday. Office hours are on Tuesday and
the homework is due on Friday afternoon before the retreat.
9
Monday, October 1, 2012
We got our homework back. The next homework will be a little longer.
There are several versions of Szemeredi-Trotter:
1. If S is a set of S points and L a set of L lines in the plane, then |I(S, L)| ≤ C(S 2/3 L2/3 + S + L).
2. Given a set L of L lines in the plane, let Pk be the set of points on k lines. Then |Pk | ≤ C(L2 K −3 +
LK −1 ).
3. Given a set S of S lines in the plane, let Lr be the lines containing at least r points of S. Then
|Lr | ≤ C(S 2 r−3 + Sr−1 ).
You can see that versions 2 and 3 are dual. It’s fairly easy to see how to go between these. It might be
tedious to have multiple versions, but we’d like to be able to use these different versions.
We were talking about the distinct distance problem. Today we’ll prove an actual result on it.
Theorem (Szekely). Given a set S of N points in the plane, there are at least cN 4/5 distinct distances
between them. In fact, for p ∈ S, {d(p, q)}q∈S has at least cN 4/5 distances.
Proof. Assume that for every p, {d(p, q)} has at most cN 4/5 points. Then let t = maxp |{d(p, q)}|. Draw
N t circles, and each point lies in (N − 1) circles. Keep circles with ≥ 2 points on them. This gives us a
multigraph G with vertices the points and edges which are arcs on these circles between consecutive edges.
Then E ≥ N 2 − N − N t ≥ 21 N 2 (the N t is for subtracting entire circles). Then the crossing number
k(G) ≤ 2 N2t ≤ N 2 t2 because each pair of circles can cross at at most 2 points.
To use this, we’d like to compute the multiplicity. If two points lie on many circular arcs, then the centers
of these arcs lie on a line. Picking the last one of these points, there can be at most t other points on the
line, so the crossing number is bounded by Ct for some constant C around 2. Recall our theorem:
Theorem. If G is a multigraph with V vertices and E edges with multiplicity ≤ M and E ≥ 100M V , then
k(G) & E 3 V −2 M −1 .
We can apply this because N 2 ≥ 100tN . Applying this, we have N 2 t2 & k(G) & N 6 N −2 t−1 , or t & N 2/3 .
We’d like to improve on this. The idea is that we could have up to N 2/3 points on a line, but we couldn’t
have too many lines like that. We’ll have to be careful.
17
Lemma. In G, the number of edges with multiplicity > M is at most (blank).
We’ll fill in the bound later after deriving it.
Proof. In our graph, map edge arcs to the perpendicular bisector of its endpoints. If in this map, a line l
contains A vertices in d is the image of at most 2At edges, each with multiplicity ≤ A. Then let Lj be the
j−1
lines
< A ≤ 2j points of S on them. Then the number of edges of multiplicity M is bounded by
P with 2
j
j,2j ≥M |Lj | 2 · 2 t. By S-T, this is at most
X
C(N 2 2−3j + N 2−j )2j t . N 2 M −2 t + tN log N.
j,2j ≥M
So now our lemma says:
Lemma. In G, the number of edges with multiplicity > M is at most N 2 M −2 t + tN log N .
To see that this is useful, see that when M ∼ t, we get much less than the ∼ N 2 edges. We’d like to
1
prune out the edges of high multiplicity. Choose M so that this expression is at most 10
N 2 . We can do this
because we’re assuming t . N 4/5 . So we can choose M = αt1/2 for α a large constant. Then choose G0 ⊂ G
to be the edges with multiplicity at most M . (Note here that multiplicity is a little unusual; each of the
multiple edges are said to have that multiplicity.) Then the number of remaining edges is at least 14 N 2 , and
using the crossing number result,
N 2 t2 & k(G0 ) & E 3 V −2 M −1 & N 4 t−1/2 =⇒ t5/2 & N 2 =⇒ t & N 4/5 .
That’s enough incidence geometry in the plane. What could we ask in 3 dimensions? Say we want to
generalize Szemeredi-Trotter.
1. Suppose we were given S points and P planes in R3 . How many incidences could we have between
these points and planes?
2. We could also ask about lines and planes, where we define incidence as containment.
3. We could do triples of points, lines, and planes.
Problem 1 is not that interesting, because we could have all the points on one line and all the planes
containing that line. So we can modify these to 1a: There are at most B points on a given line, or 1b: at
most B planes containing any given line, or 1ab: both.
Another question: (4) Just the intersections of points and lines. We can define I3 (S, L) to be the maximum
number of intersections, and let I2 (S, L) be the bound that we got in 2 dimensions (following S-T). In fact,
I3 ≥ I2 by picking a plane in R3 and I3 ≤ I2 by projecting onto a generic plane, so I2 = I3 . And we can see
that this is dual to (2), so (2) should have the same answer.
We can then modify this to (4a): at most B points in a plane or (4b): at most B lines in the plane.
Recapping, we see that when we go to three dimensions, there’s a proliferation in the number of questions
you can ask, and then some of the questions turn out to be trivial or equivalent to two dimensions.
10
Monday, October 3, 2012
We chatted at the beginning of class about what we could talk about at the beginning of class. Then it came
to 2:05 and we started.
We’ll start to work on this question that we discussed earlier, from incidence geometry in 3 dimensions:
Given L, a set of L lines in R3 with at most 10 lines in any plane or degree-2 surface, how many intersecting
points can there be? (Here we only require that 2 lines intersect, though we can generalize this to k.)
We’ll see a useful tool in this situation.
18
Proposition. Any 3 lines in R3 are contained in the zero set of a degree 2 polynomial.
This seems to have been known for a long time, but Guth couldn’t find a proof. He’d like to approach
this in a way that’s a theme for a class.
Proof. Let V (2) be the vector space of polynomials of degree ≤ 2 in R3 . dim V (2) = 10 since it has a basis
of x2 , y 2 , z 2 , xy, xz, yz, x, y, z, 1. Choose three points on each line, so we can define a nonzero polynomial
that is zero at all nine points. Then restricting to each line, we get a degree 2 polynomial with 3 roots, so
it must be zero on each line.
Corollary. If `1 , `2 , `3 are pairwise skew (not parallel, not intersecting), then there is an irreducible degree
2 algebraic surface R(`1 , `2 , `3 ) called a regulus such that every line intersecting all three lines lies on R.
We saw this for the example of the surface z = xy.
Proof. Let R be the zero set of the three lines. We must check that it contains any line that intersects all
three lines. Given a line intersecting all three, it must intersect them in three different points. Then the
polynomial defining the surface restricted to that line is a degree 2 polynomial with 3 roots, so the polynomial
must be zero and the surface contains the line.
Additionally, we claimed that it was irreducible and degree 2. If not, it is just 1 or 2 planes, but no two of
the lines lie in the same plane, so this is impossible and it is indeed irreducible and degree 2, as desired.
Are there lines going through every point?
Proposition. If x ∈ R, there exists a line ` ∈ x intersecting `1 , `2 , `3 .
Proof Idea. Find ` 3 x which intersects `1 and `2 . We didn’t finish this.
Seven mathematicians were able to use reguli to prove a weaker bound on the joints problem than we
have, but that wouldn’t be very informative to see. So we’ll prove something else new using the same
methods.
Theorem. If L lines in R3 have at most 10 on any plane or degree 2 surface, then the number of intersection
points is . L5/3 .
Proof. Define an L × L matrix A but Aij = 1 if `i ∩ `j 6= ∅ and 0 otherwise. Then |A| is the number of
intersection points. For any t, let (At )ij be 1 iff `i ∩ `j = p and p lies onP> 2t−1 and ≤ 2t (we write ∼ 2t )
lines of L, and 0 otherwise. Then the number of intersection points is ∼ t |At | 2−2t .
Lemma. At has no 3 × 20 · 2t minor of all 1’s.
Proof. Indeed, consider any three rows, corresponding to three lines `1 , `2 , `3 . The main case is when they
are pairwise skew, in which case there are 20 · 2t lines in R(`1 , `2 , `3 ).
Otherwise, some two `1 , `2 lie in a plane P . If they are parallel, there are 10 · 2t lines in plane P , and if
they intersect in p, then there are 10 · 2t points lines through p, which contradicts the definition of At .
Now this situation where you don’t have these minors of all 1’s in your matrix has been analyzed in
combinatorics, with this result:
Theorem (Kővári-Sós-Turán). If a M ×N 0-1 matrix with no V ×W minor of all 1’s, then the total number
of 1’s is at most C(V )W 1/V M N (V −1)/V .
P
Applying this to our situation, we get . t L5/3 2t/2 2−2t . L5/3 , as desired.
Now we’ll switch gears and focus on this KST theorem. Let’s prove it.
19
Proof of KST Theorem. The hardest part is remembering what all the variables mean, so he drew a picture.
Let cj be the jth column for j = 1, . . . , N , where we regard them as cj ⊂ [M ]. Then cvj is the subsets of
P
cj with cardinality v. We know each V -subset must show up fewer than W times, so j |CVj | < W M
V .
That’s all, now we
to do some algebra. Here we regard constants depending on V Pas constant.
P just have
V
So this becomes j |Cj | . W M V . We can then bound the number of 1’s using Hölder:
j |Cj | · 1 ≤
P
1/V
N
V
N (V −1)/V . W 1/V M N (V −1)/V .
j=1 |Cj |
How close is this bound to the truth? Let’s consider a simple example.
Example. (1) N × N matrices with no 2 × 2 minor of all 1’s. KST gives us a bound of . N 3/2 . What can
we do? Look at a finite field projective plane’s incidence matrix. Let’s take a while to explain this.
Incidence matrices are pretty easy to understand. Given a set S of points and L of lines, the incidence
matrix is Aij = 1 if pi ∈ lj and 0 otherwise. We can notice that there are no 2 × 2 minors because no
two lines have more than one intersection. So in general, KST gives the first two bounds we have on
incidence, S 1/2 L and SL1/2 . We didn’t actually notice these, since we went to further. But this is the
best you can do in F2q or P F2q . The second is the sharpest, but it’s a little more work so we’ll look at the
first.
Consider a finite field projective plane’s incidence matrix. There are q 2 = N points in F2q and we can
pick q 2 = N lines in F2q . Each line has q points on it, so the number of ones is q 3 = N 3/2 .
(2) N × N matrices with no 3 × 3 minor of all 1’s. The bound is N 5/3 and there is a matching example due
to Brown in the early 60s which will be on the problem set.
(3) N × N matrices with no 4 × 4 minor of all 1’s. Here there is no tight example of N 7/4 ; in fact, the best
example known is Brown’s.
(4) We can also address large V . Again excluding V × V minors, KST gives us . N 2−1/V points. There’s a
random construction that will also be on the problem set reaching N 2−2/(V +1) 1’s. That’s a fairly large
gap that’s been stable for 40-50 years.
All of this corresponds to bipartite graphs, and you can talk about this problem as the exclusion number
of the complete bipartite graph. It’s so easy to go between these languages that it doesn’t help much, but
some questions are more natural on each side. For instance, you can look at other bipartite graph exclusion
numbers which are more awkward with matrices.
11
Friday, October 5, 2012
Today’s the last background lecture in incidence geometry. We’ll discuss one of the latest methods to
approaching the distinct distance problem, which has a cool connection to incidence geometry.
Before going into that, let’s review how we were thinking about the distinct distance problem. Suppose
we have N points and t N distances. Draw tN circles around each point, and consider the circles as arcs.
But we never used the fact that the radii at each of the points must be the same.
Let P ⊂ R2 be the set of points and d(P ) the set of nonzero distances. Let Q(P ) = {(p1 , q1 , p2 , q2 ) ∈
4
P : |p1 − q1 | = |p2 − q2 | =
6 0}. We’d expect |Q(P )| to be large.
Lemma (1). |d(P )| |Q(P )| ≥ (N 2 − N )2 & N 4 .
2
Proof.
P Let d(P ) 2= {d1 , . . . , ds } where s =P|d(P2)|. Then we can just count nj = {(p, q) ∈ p : |p − q| = dj }
and j nj = N − N . Then |Q(P )| = j nj (pick the distance first and then choose two pairs at that
Ps
P 2 1/2 1/2
1/2
1/2
distance). Then we just use Cauchy-Schwarz: N 2 − N = j=1 nj · 1 ≤
nj
s
= |Q(P )| |d(P )| ,
as desired.
20
This is not at all surprising: If there are few distances, then there should be a lot of quadruples. So we’d
also like to count these quadruples in another way, and figuring out a way to do this was their key insight.
Let G be the group of orientation-preserving rigid motions of the plane.
Lemma (2). |p1 − q1 | = |p2 − q2 | =
6 0 iff ∃!g ∈ G with g(p1 ) = p2 and g(q1 ) = q2 .
That got people thinking about which rigid motions take p1 to p2 . Let Sp1 ,p2 = {g ∈ G : g(p1 ) = p2 }.
This is a 1-dimensional curve in G, which is a 3-dimensional Lie group.
Lemma (3). Assume p1 6= q1 . Then |p1 − q1 | = |p2 − q2 | iff |Sp1 ,p2 ∩ Sq1 ,q2 | = 1, and |p1 − q1 | 6= |p2 − q2 |
iff |Sp1 ,p2 ∩ Sq1 ,q2 | = 0.
So we can look at the incidence geometry of these curves in G. If a point lies on two curves, it corresponds
to two quadruples, and if a point lies on three curves, it corresponds to six quadruples. Let G=k := {g ∈ G :
g lies in exactly k curves of S}.
S
Let E : Q(P ) → g be given by lemma 3. Then the image of E is contained in k≥2 G=k . We have
E −1 (g) = 2 k2 , since we can take the quadruples in either order. Therefore, we have
PN
Lemma (4). |Q(P )| = k=2 |G=k | 2 k2 .
We usually calculated things wiht Gk = {g ∈ G : g lies in ≥ k curves}. So writing in terms of these, we
have
PN
Lemma (5). Q(P ) ∼ k=2 |Gk | k.
Proof. We have
k
|Q(P )| = |G=k | 2
2
=
N
X
[|Gk | − |Gk+1 |](k 2 − k)
k=2
=
X
|Gl | [(l2 − l) − ((l − 1)2 − (l − 1))]
l
=2
X
l |Gl | .
l
We also have this other characterization of Gk :
Lemma. Gk = {g ∈ G : |gP ∩ P | ≥ k}.
This is sort of a generalization of symmetries, where we’d require gP = P . So Guth likes to call these
partial symmetries.
Example.
Suppose
our set of points is an s × s square grid with N = s2 . Then |Gs2 | = 4. What about things
1
3 −2
1 2 ? Well, it takes a while to explain, but |Gk | ∼ N k
N.
like G 10
for all 2 ≤ k ≤ 10
s
That this is the best you can do was a conjecture:
Conjecture (ES1). If P ⊂ R2 with |P | = N and 2 ≤ k ≤ n, then |Gk | . N 3 k −2 .
This has since been proven, and we’ll prove
in this class P
using the polynomial method.
Pit
N
N
3 −1
Let’s see the consequences. |Q(P )| ≤
. N 3 log N . Then |d(P )| &
k=2 N k
k=2 |Gk | k .
3
N / |Q(P )| & N/ log N . We’ll prove this whole chain of implications using the polynomial method. We
see that we’ve claimed that the conjecture itself is sharp for the square grid, so we know that the square
grid indeed does have that many quadruples. But we could have lost some at the last step because we used
21
√
Cauchy-Schwarz, and indeed, we checked earlier that there are N/ log N distinct distances in the large
square grid.
Let’s get a better feel for these rigid motions. We first have the translations T , which are congruent to
the plane R2 .
Lemma. |T ∩ Gk | . N 3 k −2 .
Proof. Consider the number of translation quadruples QT ⊂ Q(P ) = {(p1 , q1 , p2 , q2 ) ∈ P 4 such that p1 −q1 =
p2 −q2 6= 0}. Then #QT ≤ N 3 because
∀p 1 , q1 , p2 , there is at most one choice
of p2 . Then define E : QT → T
similar to before, and if g ∈ Gk , E −1 (g) ∼ k 2 . So |QT | ≥ |Gk ∩ T | · 2 k2 , and we’re done.
Now we’d like to “straighten” G0 := G/T . There’s a way to do this to make it correspond to the incidence
geometry of points and lines. G0 is a rotation around a fixed point (x, y) ∈ R2 by angle θ ∈ (0, 2π). Then
we define ρ : G → R3 by ρ(x, y, θ) = (x, y, cot θ/2).
Proposition. ρ(Spq ∩ G0 ) is a line `pq .
Proof. Indeed, if we rotate from p to q, the point of rotation must be on the perpendicular bisector of p and
q. It’s just trigonometry from here.
In fact, we have the following:
2 q1 −p1
Proposition. Let v = p2 −q
be a vector perpendicular to p − q with length
2 ,
2
Then ρ(Spq ∩ G0 ) is a line parameterized by `pq : t 7→ (a + tv, t).
1
2
|p − q| and a =
p+q
2 .
Let L = {`pq }p,q∈P , N 2 lines. Then |G0k | is the number of points in ≥ k lines of L. Remember that the
incidence geometry depended on whether they were all in the plane.
Lemma. If q 6= r then `pq and `qr are skew.
Proof. Spq = {g ∈ G : g(p) = q}, so Spq ∩ Spr = ∅. This shows that `pq ∩ `pr = ∅, so we also have to show
they aren’t parallel. The “slope” ((dx/dz, dy/dz)) of `pq is v(p, q) and these slopes are different.
We also realized that there was a problem if too many lines lay in some regulus. We won’t prove this in
class today but defer this proof to a while later, but there are ≤ N lines of L in any degree 2 surface.
Conjecture (ES2A). If L is a set of L lines with at most L1/2 in any plane or degree 2 surface, then
|P2 | . L3/2 .
Conjecture (ES2B). If L is a set of L lines with at most L1/2 in any plane and 3 ≤ k ≤ L1/2 , then
|Pk | . L3/2 k −2 .
Finally, Guth drew another log-log graph of the bounds we had. We had the S-T bound for any L lines
that was piecewise linear with two regions, from 2 to L1/2 and L1/2 to L. Then if we assume the number of
lines in any plane or degree 2 surface is small, then we lower the first line.
This finishes our background on incidence geometry. On Wednesday, we’ll pick up with the polynomial
method. The next problem set is almost ready and will be due two weeks from Monday.
12
Wednesday, October 10, 2012
Definition. For S ⊆ Fn , define the degree of S deg(S) as the minimum degree of a nonzero polynomial that
vanishes on S.
1/n
We proved that deg(S) ≤ n |S| . Now we’ll try to investigate the relationship between the degree of S
and its combinatorics.
As a warmup, let S be the union of L lines in F3 . What can we say about deg(S)?
22
• deg(S) ≤ L by picking a plane containing each line.
• deg(S) ≤ 2L/3 by picking a degree 2 surface containing any three of the lines.
√
• Can we get a L bound? Pick A points on each line, totalling LA points. We can find a polynomial
of degree 3L1/3 A1/3 that is zero on all of those, and we want to conclude that there are more points
on each line, so it must be zero on each line. We can do this if A > 3L1/3 A1/3 , so A > 10L1/2 suffices.
We’ve proven
S
L
1/2
Proposition. deg
if `i are lines in F3 .
i=1 `i ≤ 10L
What if our lines have lots of intersections?
Proposition. If L is a set of L lines in F3 , each line containing ≥ A intersecting points with other lines of
L (distinct points, not intersecting lines). Then the degree of the union of the lines is at most cL/A.
Of course, this is only interesting if A L1/2 .
Example. With L lines in R3 , if the number of intersecting points is L3/2 , we want to show that a lot of
lines lie in a plane or regulus. If the intersecting points are evenly distributed, then this should give us an
algebraic structure.
We’ll use a dimension-counting argument. Let’s be more careful with our proof.
1 2
d lines, so there exists
Proof. Let d be a parameter to be chosen later. Randomly choose L0 ⊂ L with 10
a nonzero polynomial P of degree ≤ d that is zero on L0 . If another line has d + 1 intersections with lines
in L0 , the polynomial must vanish on that line as well. So we should estimate how many times a line will
intersect L0 .
Let’s do this heuristically first. Fix some `, with ≥ A intersection points. A given one will have another
d2
line meeting it, which will be on L0 with probability ≥ |L0 | / |L| = 10L
(since there could be other lines
1 2
meeting that point). The expected number of intersection points is therefore E = 10
d AL−1 . We want to
arrange that E > 100d, or equivalently, d > 1000L/A. So choose some 1000L/A < d ≤ 1001L/A.
Now here’s the heuristic part: If on average, E > 100d, then P = 0 on most of the lines of L. Let’s be
more rigorous now. We’ve got to be more careful with our probabilities.
Lemma (Probability Lemma). If S is a set of N elements, X ⊂ S a random subset where each element
1
is included with probability p independently. Then E |X| = pN , and P[|X| > 2pN ] ≤ exp(− 100
pN ) and
1
1
P[|X| < 2 pN ] ≤ exp(− 100 pN ).
Guth wrote a proof of this, but we’ll just assume it for now and maybe come back to the proof later.
This gives us essentially what we want in this problem, except there’s a slight problem that the intersections
are not quite independent. So we’ve got to slightly adjust our construction. Let’s start over.
1 d2
Pick L0 ⊂ L by including each line with probability p = 20
L . This means that by the probability
1 2
lemma, P[|L0 | > 10
d ] is very low, and we can essentially ignore it. Therefore, there exists a polynomial of
degree ≤ d where P = 0 on L0 .
Fix some ` ∈ L with ≥ A intersection points. For each intersection point x, P[x ∈ L0 \ `] ≥ p, and the
events are independent for different x ∈ `. So by the probability lemma, the probabilty of less than 12 Ap
points on ` being chosen is at most exp(−cAp).
1 d2
Now we want to choose d so that 21 Ap > 2d, or 12 A 20
L > 2d, so d > 100L/A suffices. Choose d ∈
1
1
[100L/A, 101L/A]. Then the probability that P does not vanish on ` is at most exp(− 100
E) ≤ exp(− 100
d) ≤
1 L
−10
exp(− 100 A ). Therefore, if L/A > 1000 log L, this probability is at most L . So usually P = 0 on all lines.
There’s the annoying case where A is close to L, where our conclusion gets stronger. To solve this, we
can take d = 106 L/A, so the probability is less than 1000. Therefore, P = 0 on some set of at least 99.9% of
the lines. Any bad line must have a quite small number of intersecting points with the good lines, so the bad
lines form a set with a lot of intersections like our original setup. We can iterate with the smaller system
and multiply the polynomials to finish.
23
We have two more examples of degree reduction which we’ll see now. The methods are essentially the
same.
Example (2). Let X ⊂ F3q be a union of at set of L lines in X. Suppose X has the special property that
every point lies in at least two lines. (Some examples: a plane, all the lines, a regulus, a union of planes and
reguli.) Then deg X ≤ c(log q) |X| q −2 .
In this one, the log q term is annoying, like our previous case.
In a previous class, we asked how big the size of a zero set of a polynomial could be. In F3q , we had
1/3
|Z(P )| ≤ (deg P )q 2 . Therefore, we know that |X| q −2 ≤ deg q ≤ |X| . The lower bound means they have
a lot of algebraic structure, and the upper bound means they have not a lot of algebraic structure. The
examples we discussed are those with a lot of algebraic structure, and there are conjectures that they are
the only ones.
This proof is fairly similar to the previous one. Suppose each point lies in at most 10 lines of L. Then
|L| ≤ 10 |X| /q =: L, and each line ` ∈ L has A ≥ q intersection points with other lines of L. The previous
proposition gives us deg X ≤ CL/A = C |X| q −2 . And then you have to deal with the annoying case where
points lie on lots of lines on L, and that’s where you get the log q term.
There’s a way to do this without degree reduction, and that’s probably the case for a lot of these results.
But here’s one that doesn’t have an easy proof without degree reduction:
S
S
Proposition. Given some lines {`i }i∈I , `i ⊂ F3q . Take Si ⊂ `i with |Si | = 2q . Let X = Si and Y = `i .
Then |Y | ≤ c(log q) |X|.
1/3
Let’s try to prove this with the straight up polynomial method. Let P = 0 on X, so deg P . |X| .
We can assume that deg P < q/2. Then P = 0 on al lines, so P = 0 on Y . The next thing we could do
1/3
17/6
is |Y | ≤ (deg P )q 2 . |X| q 2 . That isn’t the best estimate; if |X| = q 5/2 , then |Y | . |X|
. There’s
considerable loss there, and somehow we’ve got to get within a log factor. We’ll see next time how to do
this.
13
Friday, October 12, 2012
Let’s finish what we were doing last time.
Proposition. Let `i be a set of lines in F3q , and Si ⊂ `i with |Si | ≥ q/2. Let X :=
Then |Y | ≤ C(log q) |X|.
1/3
S
i∈I
Si ⊆
S
i∈I `i
=: Y .
At the end of last class, we used a straight polynomial method to get |Y | ≤ |X| q 2 . So when |X| = q 5/2 ,
|Y | . q 17/6 , which is lessthan q 3 but not at |X| yet.
S
Let Imin := {i ∈ I : Si ∩ j6=i Sj ≤ q/4}, the set of lines that contain at most q/4 points not in any
S
other Sj . Thus |Imin | 4q ≤ X, so i∈Imin `i ≤ |Imin | q ≤ 4 |X|. So the heart of the matter is Imaj = I \Imin .
We’d like to do degree reduction onSthose lines.
Degree reduction says that deg( i∈Imaj `i ) ≤ C |X| q −2 (log q). This immediately implies a volume bound
S
i∈Imaj ≤ C(log q) |X|.
Define a subset of the lines I1 as follows: Look at the lines one at a time S1 , S2 , . . . . We decide whether
or not to include i by this rule: include it if Sj has ≥ q/4 points not already included in the previous Si , i.e.
for i < j. That is, we always include S1 .
q
This
setShas two good features: |I1 | 4 ≤ |X|, so |I1 | . |X| /q. The other good feature is that if j ∈ I \ I1
then Sj ∩ i∈I1 Si ≥ q/4.
At this point, we can pretty much do what we did before by choosing a random subset. For some d
1 2
which we’ll pick later, randomly choose I0 ⊂ I1 with |I0 | ∼ 10
d (choose them independently so that the
probability is rigged to get that many). Then let P be a nonzero degree ≤ d polynomial which is zero on `i
for i ∈ I0 .
24
S
S
Now fix some j ∈ I \ I1 and look at E[Sj ∩ i∈I0 Si ]. There will be q/4 points where Sj ∩ i∈I1 Si , so
2
2
this expected number is at least & 4q |Id1 | ∼ dq 2 |X|.
We want E > 2d, and E > 1000 log q so the probability of missing is small enough that the probability of
not missing for every j is positive. So choose d . |X| q −2 log q, so P = 0 on `j for all j ∈ I \ I1 .
Now fix j ∈ Imaj . If Sj intersects many Si for i ∈ I1 , then P = 0 on `j as before. Otherwise, Sj intersects
many Si for i ∈ I \ I1 , so P = 0 on `j by the vanishing lemma, and we’re done.
There’s a fundamental fact about the degrees of polynomials in general. This is known as Bezout’s
theorem, which we’ll prove today in a manner that’s very much in the spirit of this course.
Theorem. If P, Q ∈ F[X, Y ] have no common factor. Then the number of points where they both vanish
|Z(P, Q)| ≤ deg P deg Q.
Proof. Let I¯ be the ideal generated by P, Q, and S = F[X, Y ]/I.
Lemma. |Z(P, Q)| ≤ dim S (since S is a vector space over F).
Proof. For X ⊂ F2 , we have the evaluation map EX taking polynomials to functions from X to F. An
elementary lemma says that if X is finite, EX is surjective, i.e. we can write any function as a polynomial.
That’s pretty simple to see, but we’ll come back to it.
If X ⊂ Z(P, Q), I¯ ⊂ ker EX , so the result follows.
The rest of the proof is more complicated, so we’ll go one step at a time. Start with I = hP i ⊂ F[X, Y ].
We’d like to keep track of the dimensions along the way, but this is tricky since these are infinite dimensional.
So let Vd be the polynomials of degree at most d, Id = I ∩ Vd , and RD = Vd /Id . These are finite dimensional
spaces and we can figure out their degree.
Let’s start with Vd , and we found before that dim Vd = d+2
2 , since we’re in two variables. Then to
compute dim Id , let D = deg P and E = deg Q. Then the multiplication map by P : Vd−D → Id is an
isomorphism, so dim Id = d−D+2
.
2
Then dim Rd = dim Vd − dim Id = d+2
− d−D+2
= Dd + 23 D − 12 D2 . To simplify things, let
2
2
c(D) = 23 D − 12 D2 .
Now let J ⊂ R be the ideal generated by Q and R/J = S. Similarly define Jd = I ∩ Rd and Sd = Rd /Jd .
How can we compute deg Jd ? Multiplication by Q takes Rd−E → Jd . (Here Rd−E is the equivalence class of
a polynomial of degree at most Rd−E .) Is this still an isomorphism? P and Q have no common factors; how
can we use that? Suppose r1 ∈ Rd−E , so suppose it comes from P1 ∈ Vd−E . Then r1 Q = 0 implies P1 Q ∈ I,
so P1 Q = P P2 , hence P | P1 and r1 = 0. This proves that this map is injective.
What about surjective? If we had some r1 ∈ J2 then r1 = Qr2 in R, so r1 comes from some P1 of degree
d. r2 also comes from some P2 , which we’d like to have small degree, but it doesn’t necessarily. We have
P2 Q + P P3 = P1 , but there might be some high-degree cancellation in the left side, so P1 can have a small
degree without P2 having one as well.
But we don’t actually need that; this just shows that dim Jd ≥ dim Rd−E . Then we have dim Sd =
dim Rd − dim Jd ≤ dim Rd − dim Rd−E = (D · d = c(D)) − (D(d − E) + c(D)) = DE. This is independent
of d, so dim S ≤ DE, as desired.
Now we’ll see some more cultural results. An important part of the proof was determining the dimension
of Rd . We can pose this in general: If I ⊂ F[x1 , . . . , xn ] is an ideal, we can take R = F[x1 , . . . , xn ]/I and ask
for dim Rd . We’ve done this for a couple examples:
(1) In the case when n = 2 and I = hP i with deg P = D, we found dim Rd = Dd + c(D) for d ≥ D.
1 n
(2) When I = 0, dim Rd = d+n
n , a polynomial with leading term n! d .
There’s a theorem probably due to Hilbert that for every I, there exists a polynomial hI (d) such that
dim Rd = hI (d) for all d ≥ d0 (I). Moreover, the leading term is the transcendence degree of R, or alternatively, the dimension of the algebraic variety Z(I). This is a general notion that can be defined as the
25
dimension. The leading coefficient is also the “degree” of Z(I) divided by m!. We haven’t defined those in
this class so they’re in quotes. They describe exactly how the space of polynomials are growing.
There’s Bezout in higher dimensions as well. There’s one other variation that we’ll need and will do next
time. Guth is interested in whether it’s possible to do this in general.
A little logistics: There will be office hours next Wednesday after class. On the website, there is the next
problem set and a list of further projects along the lines of this class.
14
Monday, October 15, 2012
Office hours are Wednesday after class. Problem set 2, problem 6 has been slightly updated.
The theme of today is “special points in surfaces.”
Definition. Let P ∈ R[x1 , . . . , xn ], then Z(P ) := {x ∈ Rn : P (x) = 0}. We say that x is a critical point if
∇P (x) = 0.
Theorem (Implicit Function Theorem). If x ∈ Z(P ) is not a critical point, then Z(P ) is a smooth manifold
in a neighborhood of x.
We’d like to ask how big the set of critical points can be. There’s an unfortunate example: P = x21 ,
which has critical points along the entire x1 = 0 plane. To avoid this, we define
Definition. P is squarefree if whenever Q2 | P then deg Q = 0.
Lemma (1). P is squarefree iff P, ∂1 P, . . . , ∂n P have no common factor.
Q
Proof. Let P ∼ Q if P = cQ for some c ∈ R\{0}. Then we can write P as a product of irreducibles P = j Pj ,
which are distinct if P is squarefree, and if an irreducible Q | P then Q ∼ Pj1 for some P
j1 . We needQto check
that Q does not divide all of these derivatives, so suppose it does. We have ∂i P = j0 (∂i Pj0 ) j6=j0 Pj .
If Pj1 | ∂i P , then Pj1 | ∂i Pj1 . But ∂i Pj1 is smaller degree, so it must be zero and Pj1 is constant, a
contradiction.
Proposition (2). If n = 2 and P is a nonzero degree d squarefree polynomial, then the number of critical
points of Z(P ) is ≤ 2d2 .
Q
Proof. x is a critical point of Z(P ) iff P (x) = 0 = ∂1 P (x) = ∂2 P (x). Let P =
Pj be the irreducible
factorization. There are two types of critical points: (a) x ∈ Z(Pj ) for at least two j, and (b) x is a critical
point in Z(Pj ) for some j. He drew a picture of what these looked like: (a) were the intersections of the
curves Z(Pj ) and (b) were the critical points of those curves as P
well.
We can bound the number of critical points of type (a) by j1 ,j2 (deg Pj1 )(deg Pj2 ) ≤ d2 . Since the Pj
are irreducible, ∃i such that ∂i Pj 6= 0. Then Pj and ∂i Pj have no common factor, so P
the number of critical
points in Z(Pj ) is at most (deg Pj )2 . Hence the number of type (b) critical points is ≤ j (deg Pj )2 ≤ d2 .
We aren’t too concerned with two dimensions, but we are interested in three dimensions, so we’ll do the
following variation:
Theorem (Bezout for lines in 3 dimensions). If P, Q ∈ R[x1 , x2 , x3 ] have no common factor, then Z(P, Q)
contains at most ≤ (deg P )(deg Q) lines.
There was a proof of this in last class, though we didn’t recognize it as such. Here’s the proof heuristic:
Let π be a plane in R3 and restrict P, Q to π, getting P̃ and Q̃. Then |Z(P, Q) ∩ Π| ≥ L for almost every
π (not those parallel to or containing the lines). We could also consider
Z(P̃ , Q̃), and (here’s the sketchy
part) for almost every π, P̃ and Q̃ have no common factor. Then Z(P̃ , Q̃) ≤ deg P̃ deg Q̃ ≤ deg P deg Q,
and we’re done, modulo that unproven claim.
26
Proposition. If P ∈ R[x1 , x2 , x3 ] is degree d and squarefree, then Z(P ) contains ≤ 2d2 critical lines (lines
of critical points).
In summary, algebraic points have critical points, but they can’t have too many. In R3 , the critical
points will be arranged in one-dimensional lines, but we can control the number of lines. Could it have more
isolated points, and do we know a bound on those? Yes, it could. This is an algebraic structure given by
P = 0, ∂1 P = 0, ∂2 P = 0 and ∂3 P = 0. Morally, lines come from intersections of two of them where the
others happens to be zero and points come from intersections of three of them where the fourth happens to
be zero.
Now we’ll move to an outline of the original proof of the joints theorem, which is less pretty than the
polynomial method one we proved.
Lemma (Main lemma). Given L lines in R2 , one has ≤ 1000L1/2 joints on it.
Then you just have to delete lines with few joints on them, and the proof isn’t hard from here.
Proof. By contradiction. First, we use degree reduction. ∃P nonzero with P = 0 on the lines and deg P ≤
1 1/2
. We can also assume P is squarefree, since the minimal polynomial which is zero on all of these lines
10 L
will be squarefree.
The second step is that the gradient of P must vanish on all of the joints. This is because the directional
derivative vanishes in three independent directions. So the joints are all critical points of this algebraic
surface. The moral of the proof is that the algebraic surface is too low degree to have this many joints.
That’s not quite true, but we can get it to work. Once the gradient vanishes on the lines, it must vanish on
all the lines since ∇P has too low a degree. Terefore, Z(P ) has L > 2(deg P )2 critical lines, contradicting
the previous proposition.
Maybe that’s not as exciting, but we can prove various things about varieties now. We’ll stay in n = 3
dimensions.
Definition. Let x ∈ Z(P ) be a non-critical point. Translate and rotate the surface so that x = 0 and
Tx Z(P ) is {x3 = 0}. Then this is locally the graph of x3 = Q(x1 , x2 ) + O(x1 , x2 )3 where Q is homogeneous
of degree 2. x is flat if Q = 0.
We’d like to prove that there aren’t many flat points. Being flat by this definition, it isn’t clear that this
is where some polynomial that is zero at the flat points, though, so we need to massage this definition.
Let N be the unit normal vector of Z(P ). We can tell whether the surface is curved by looking at how
∇P
the normal vector is changing. Then x is flat iff ∂v N (x) = 0 for every v ∈ Tx Z(P ). We have N (x) = |∇P
|,
and v ∈ Tx Z(P ) iff v · ∇P = 0. We still haven’t quite written this as a polynomial. We need some tricks.
1. ∂v N = 0 iff ∂v∇P is parallel to ∇P . This is equivalent to saying that ∂v ∇P × ∇P = 0. This is nice
because if v is any particular vector, this is a polynomial.
2. The next problem is that we have a list of infinitely many vectors to check. So the next trick is to look
at {ej × ∇P }j=1,2,3 ⊂ Tx Z(P ), which span Tx Z(P ). So it’s sufficent to check it for just these three,
and we’re done.
Let’s summarize what we’ve done. Let SP (x) := {∂ej ×∇P ∇P × ∇P }j=1,2,3 . This gives three vectors in R3 ,
so nine polynomials of degree 3d − 4 ≤ 3d (you can work it out). Succinctly, then,
Proposition. Let x ∈ Z(P ). Then SP (x) = 0 iff ∇P = 0 or x is flat.
Proof. It’s clear if ∇P = 0 then this vanishes. We also know that if it’s flat, that all of those cross products
will vanish. Conversely, if we know that ∇P 6= 0 and this always equals zero, by the previous logic, we know
that x is flat.
27
Of course, it’s still possible for every point to be flat, and we call that a plane. So we need to rule out
that possibility in counting the number of flat points.
Definition. ` ⊂ Z(P ) is special if SP = 0 on `.
Proposition. If P is irreducible, degree d, not linear. Then Z(P ) contains ≤ 3d2 special lines.
The proof of this is very similar to before.
Proof. Case 1: P does not divide every component of SP . Then pick one which it doesn’t divide and use
Bezout. Case 2: P | SP . Then SP = 0 on all of Z(P ). We’d like to argue that in this case, it must be a
plane, which is actually rather annoying.
Suppose there is a non-critical piont x ∈ Z(P ). Then we have a smooth neighborhood of x which is
flat, and hence a relatively open subset of a plane. By the vanishing lemma it must contain every line
through that set, hence the whole plane. Thus P must be divisible by the linear polynomial defined by P ,
a contradiction.
Otherwise, all the points are critical, so every line is critical, so there are at most 2d2 lines at all, hence
fewer special lines.
We’ll see what our target is now and shoot at it on Wednesday:
Theorem (Special case of ES conjectures). Suppose we have L lines in R3 and ≤ B bounds in any plane
(mainly, B ≥ L1/2 ). Then the number of points in any three lines |P3 | ≤ CBL.
The ES conjectures said that |Pk | . L3/2 k −2 . (The conjectures are proved, but we’ll start with the easier
one.)
15
Wednesday, October 17, 2012
We’ve proved some tools in the polynomial method and now we’ll try to prove things in combinatorics using
them.
Our next goal in this class is to prove the distinct distance problem. First, we’ll prove this conjecture:
Theorem (Katz-Guth). Given L lines in R3 with at most L1/2 lines in any plane, and 3 ≤ k ≤ L1/2 , then
|Pk | . L3/2 k −2 .
In class last time, we mentioned a theorem that covers some cases of this:
Theorem. Given L lines in R3 , ≤ B lines in any plane, B ≥ L1/2 , then |P3 | . LB.
We’ve spent about three days gathering tools to prove this, and today we’ll try to prove this theorem by
group participation. Recall that Pk is the points that lie on at least k lines.
What should we do first? Suggestion: Let P be a nonzero polynomial that vanishes on these lines of
minimum degree d. What do we get for d?
Well, the way you can do this is take d + 1 points on each line, and if a polynomial vanishes on (d + 1)L
points, we need L(d + 1) < 16 d3 , and in this case, we can arrange that d ∼ L1/2 . When we did this in class,
we get d ≤ 3L1/2 .
How can we relate this to the plane condition?
Well, we’d like to know whether the planes are (linear)
Q
factors of our polynomial. So let P =
Pj irreducible. Some of the factors are linear, so let them be
π1 , . . . , πt with t ≤ d.
Let’s throw out the lines without many points on them. We can use induction to remove lines that have
. B points in P3 . We could prove a little lemma there. So we can suppose that every line has & B points.
Maybe we could have instead looked a polynomial that vanishes on the points of P3 . We can suppose
by contradiction that |P3 | & LB, but in the end, this doesn’t help because we’re trying to bound |P3 | from
above.
28
Well, some of these points are joints, and we had a previous bound. Call J ⊂ P3 the joints. Here, ∇P = 0
on J, so we can use our previous bounds on J. What about P3 \ J? Well, there’s a plane π(x) naturally
associated to each of these points, containing these three lines. Does it mean then that ∇P = 0? No, it
could be perpendicular.
What can we say about the derivatives of P ? Well, it will be perpendicular, but we’d know that for any
x ∈ P2 . Hmmm, maybe it’s flat there. Recall that flat means that the surface is approximated by a plane
to degree 2. Can we prove this? Let’s restrict the polynomial to P̄ = P |π(x) , a polynomial in two variables.
Then P̄ , ∂ P̄ , ∂ 2 P̄ = 0 at x since there are at least three linear factors that all vanish at x.
Our definition of flatness was that if we translate it to the origin so that the tangent plane is {x3 = 0}
then x3 = Q(x1 , x2 ) + O(|x1 | , |x2 |)3 and Q = 0. This isn’t quite the same. How can we translate this
problem? Translate and rotate so that π(x) is indeed {x3 = 0}. Suggestion: Consider a map F : R3 → R3
with F : (x, y, z) 7→ (x, y, P (x, y, z)). There was a lot of discussion involving inverse function theorems
about whether this was still true. Eventually, we sorted this out F −1 (x, y, 0) = (x, y, h(x, y)), the height
function that we needed to show was zero. Since a bunch of derivatives of p vanish, so do the corresponding
derivatives of F , and therefore, h, as we want.
This about works, but we won’t go through those computations any longer. We’ll bracket it and move
on.
Now, we had a proposition last time that a polynomial was zero iff x was critical or flat.
Proposition. Let x ∈ Z(P ). Then SP (x) = 0 iff ∇P = 0 or x is flat.
The polynomial was SP (x) = {∂ej ×∇P ∇P × ∇P }j=1,2,3 , which has degree ≤ 3d. And this vanishes on
all of P3 , prety nice. The degree is kind of unfortunate, though, since it isn’t smaller than d. So it’s not as
simple as the joints problem.
We also had results last time related to whether SP = 0 on the lines. Can we conclude this? Well, we
had d ≤ 3L1/2 , and each line contains ≥ 100B ≥ 100L1/2 points of P3 on it, so indeed, it vanishes on all the
lines, so they’re special, with the terminology from last time.
Last time, we also had this proposition: If P is irreducible, deg p > 1, then SP = 0 on ≤ 3(deg P )2 lines.
If our P in this problem is actually irreducible,
this is concerning, since it only gives us Z(P ) ≤ 27L special
Q
lines, which is too many. What if P = Pj ? If SP = 0 on `, could we have SPj = 0 on ` for some j? No,
not quite. For example,
we could have ` ⊂ Z(P1 ) ∩ Z(P2 ) so ∇P = 0 on `. We do have a bound on the
P
number such `, as j1 ,j2 (deg Pj1 )(deg Pj2 ) ≤ d2 ≤ (3L1/2 )2 = 9L. That’s also unfortunately large.
So it seems like we would be helped if d was just a bit smaller. Maybe we can do degree reduction?
Unfortnately, we’re out of time for today. Next class, Guth will recap what we’ve said so far in the first 10
minutes, then we’ll pick it up from there.
16
Friday, October 19, 2012
Last time, we started proving a theorem by group participation. It was going pretty well, but we didn’t
finish the proof. We’ll start by recapping what we did achieve, and then take it to the finish line.
Here’s what we were trying to prove:
Theorem. If we have L lines in R3 , with at most B in any plane, where B ≥ L1/2 , then |P3 | . LB.
Our first step was to remove the lines that didn’t have many 3-rich points on them. We formulated this
as a lemma which would suffice:
Lemma. If L lines in R3 have at most B lines in any plane, then one of the lines has . B points of P3 .
We started the proof of this lemma by contradiction, so assume each line contains ≥ 1000B points.
(1) Let P be a minimum degree nonzero polynomial that is zero on the lines. Then d := deg P ≤ 3L1/2 by
the number of lines.
29
(2) We proved this lemma:
Lemma. x ∈ P3 is a critical point or a flat point of Z(P ), and therefore, SP (x) = 0.
Proof. If x ∈ `1 , `2 , `3 ∈ L. We had two cases: First, if `1 , `2 , `3 are not coplanar, then ∇P (x) = 0 as we
saw in connection with the joints problem. Second, if `1 , `2 , `3 are coplanar and ∇P (x) 6= 0, we claimed
x was flat. WLOG, x = 0 and Tx Z(P ) = 0. Locall, Z is defined by x3 = h(x1 , x2 ) = Q(x1 , x2 )+higher
order terms. The shortest proof involves looking at the highest order terms, Q = 0 on `1 , `2 , `3 , so
we must have Q = 0 by the vanishing lemma. You can also probably look at the definition of SP (x)
directly.
(3) The lines are special, i.e. SP = 0 on all the lines. This is because deg SP ≤ 3 deg P ≤ 9L1/2 , but there
were ≥ 1000B ≥ 1000L1/2 points on each line, so SP (x) must be zero on the lines.
(4) We recalled this lemma from the previous class:
Lemma. If Q is irreducible and not linear, then SQ = 0 on at most ≤ 3(deg Q)2 lines in Z(Q).
We noticed then that in the special case that P is irreducible, L ≤ 3d2 ≤ 3(3L1/2 )2 = 27L, which is
true and therefore useless. So we noticed that if the degree were just a bit lower, this would give us a
contradiction in this case. So we wanted to reduce the degree of P .
That was the recap. What can we do next? We’d like to get a lower degree. One observation we made is
that we can assume that each line has 1000B lines in P3 − P4 . Well, if they’re all on exactly three lines,
that won’t help us, so that’s basically the hardest case. Our method was to pick a set of points on each line.
Well, if we had a point that was on several lines, we could possibly do better. Let’s recall a theorem we
proved regarding degree reduction:
Theorem. If L lines in R3 have ≥ A points of P2 on each line, then the degree of the union of the lines is
. L/A.
L
In this case, we would get d ≤ c 1000B
≤ cc0 L1/2 , where we can adjust the constant c0 (formerly known
1
L
1/2
. Great!
as 1000). So in fact, we can let d ≤ Q
100B ≤ 100 L
Now in full generality, let P =
Pj be a factorization into irreducibles. Then Z(P ) contains planes
L
1
π1 , . . . (at most d ≤ 100B
of them). Each plane contains ≤ B lines, which accounts for ≤ 100
L lines in those
planes.
Well, can we divide up the special lines of P according to its factors? If ` is a special line in Z(P ), is ` a
special line in Z(Pj ) for some j? No, the intersection of two planes is a critical line, so a special line. But
what if a line is not contained in two of Z(Pj )? Do we conclude it then?
Well, first, we could count the number of lines that
Bezout’s
theorem
P are contained in the intersection.
P
P
does this: The number of lines in two Z(Pj )’s is ≤ j1 ,j2 (deg Pj1 )(deg Pj2 ) = j1 deg Pj1 j−2 deg Pj2 =
d2 ≤ 10−4 L.
What about the question then? Yes, it’s true. Let’s write this a little more formally:
Lemma. If ` is a special line of Z(P ) then either ` is contained in two of the Z(Pj )’s, or ` is a special line
of Z(Pj ) for some j.
As a sketch, this line only meets finitely many points of Z(Pk ) for k 6= j, and elsewhere it must be special,
so it must be special everywhere.
P
1
L.
So let Uj be the number of special lines in Z(Pj ). Then Uj ≤ 3 j (deg Pj )2 ≤ 3d2 < 1000
1
1
Are we done now? Well, at most 100 L lines lie in the planes, at most 10000 L lines lie in two Z(Pj )’s,
1
and at most 1000
L lines are special lines of Z(Pj ). These categorized all the lines, so we get a contradiction
since L is much bigger than the sum of these (1.11%L). Yay!
Let’s outline this proof:
30
1. Degree reduction.
2. The points of P3 are special (geometric).
3. The lines are special.
4. Categorizing special lines. They come in three types: Lines in two components, and special lines in one
component, which can be not
Pa plane or a plane. The first type were bounded by Bezout’s theorem,
and the second bounded by j 3d2j ≤ 3d2 . Finally, we used the assumption that not many lines lie in
each plane to bound the last number.
We can think of what we’ve done as proving that there must be a lot of lines on a plane:
Lemma. If L lines are in R3 and ≥ 108 B points of P3 on each line, with B ≥ L1/2 , then ∃ a plane with
> B lines.
We can walk through this proof and see that it generates these lines.
This feels high-tech. Does it? The definition of SP and flat points seems quite complicated. Does it
follow from the incidence axioms of points, lines, and planes? We talked about what this would mean a bit.
Hopefully this problem motivates you to look at degree reduction again.
17
Monday, October 22, 2012
The homework is now due on Friday, so after class on Wednesday there will be office hours. It’ll last until
at least 4 if you have a class then.
Last class, we finished this theorem about how many 3-rich points there were. Putting together everything, it was the longest proof we’ve done. We needed three lectures of build-up (degree reduction, Bezout,
and flat points) and two lectures to finish it. We’ll start by putting it in context and then look a little
broader than that at the main techniques and what they’re good for.
Here was that theorem:
Theorem. Given L lines in R3 with ≤ B in any plane (B ≥ L1/2 ), then |P3 | . BL.
We’ll first see that this example is sharp:
Example. If you put B lines in the plane, you can get ∼ B 2 3-rich points points. Just make a grid with
diagonals. Then just pick L/B planes and B lines in each plane. Each plane contributes ∼ B 2 3-rich points,
so |P3 | ∼ LB. Well, technically you have to rule out there being some other plane containing B + 1 lines,
but if you do things generically, it works.
Now we’ll see why this theorem is so difficult. We’ll see that it implies another theorem which is also
hard.
Proposition. Given L lines in R3 with |P3 | ≥ L1.99 . Then there exists a plane containing three of those
lines.
Let’s think about how we might try to find such a plane. We could consider the incidence matrix,
considering points and lines. If we found a triangle, i.e. three lines each pair of which intersect, that would
give us a plane containing those three lines. We’d want to prove that the incidence matrix contains something
like this.
What does the incidence matrix look like? It’s L × |P3 | ∼ L × L1.99 and each point is contained in at
least 3 lines, and any two lines intersect in at most one point. We could try to conclude directly from this
information that it contains a triangle, but actually there’s a cool example that shows it doesn’t.
Proposition (Suk-Solymosi). Fix > 0. There exists L a set of L lines in R2 ⊂ R3 where P ⊂ P3 (L) with
|P | ≥ L2− but which contains no triangles.
31
This is based on a really cool example in combinatorics from Behrend in 1946. He was interested in
arithmetic progressions a, a + d, a + 2d, . . . . He was interested in if you look at a subset of the numbers from
1, . . . , N if it contains no 3-term arithmetic progression, how big can that subset be? He constructed some
suprisingly large examples. Here was what he proved:
Theorem (Behrend 46). For > 0, then for arbitrarily large N , ∃P0 ⊂ {1, . . . , N } such that |P0 | ≥ N 1−
and P0 has no 3-term AP.
Using that example, we’ll build lines in a grid. There will be three kinds of lines: Horizontal lines of the
form y = b, b = 1, . . . , S; vertical lines x = a, a = 1, . . . , S, and diagonal lines x − y = c for c = −S, . . . , S.
Then L = 4s + 1 ∼ s. There are s2 triple points, i.e. |P3 (L)| = S 2 . But it has a lot of triangles. He drew
this on the board to illustrate a typical triangle. Our goal is to remove some of these points to have few
triangles.
Let P0 be a Behrend subset of [S/2, . . . , 3S/2] with no 3-AP and |P0 | ∼ S 1− . Take the points (a, b) ∈ [S]2
so that a + b ∈ P0 . That is, we take some of the diagonal lines in the opposite direction. The beautiful
thing is that if you take any triangle in our original grid, you get a 3-term AP. Indeed, our triangles must be
isosceles right triangles, and projected down, that gives a 3-term AP. Then |P | ∼ S 2− ∼ L2− , as desired.
This example illustrates that the proposition is hard, that it can’t just follow from incidence geometry
(it might show that it doesn’t follow from the axioms of incidence geometry, but we haven’t checked that
precisely). So to see that sort of definition, let’s see the axioms of incidence geometry:
• Lines and planes are identified with subsets of points.
• For eery two points, there is a unique line containing them.
• For every three points that don’t all lie on a line, there is a unique plane containing them.
• And so on (there are some more).
Now we haven’t proven anything yet, but this example suggests:
Conjecture. In the axioms of incidence theory, we can have L lines, |P3 | = L1.99 , but each plane contains
at most two lines.
But we won’t go into these details. You can search for the axioms of incidence geometry online.
There is another proof of this proposition using reguli. On the problem set, you prove that there exists
a regulus or plane containing ≥ L.99 lines. We’d just have to rule out the case that it’s a regulus. The basic
fact about a regulus is that the lines in a regulus have no 3-rich points. So each point on the regulus must
have another line going through it. And reguli are a structure that is not part of the axioms of incidence
geometry. Including reguli is almost the same as including all of the degree 2 algebraic surfaces. We should
note that reguli aren’t enough to prove our theorem; you get stuck at a larger lower bound of B, like B ≥ L2/3
or B ≥ L3/4 . There’s some limit below which the 3 × 3 grid doesn’t occur, and that surface implies you are
in a regulus or plane. In the polynomial method, we basically did the analog of extending this to algebraic
surfaces of all degrees.
Can this bound on B be lowered further? Well, that’s not really clear. The polynomial method gets
stuck.
Let’s take a step back. We’ve seen that usually incidence geometry isn’t enough to get us the bounds
we want. The first technique we discovered was to use the topological structure of R2 , which involved the
crossing number lemma and Szemeredi-Trotter. Secondly, we looked at the algebraic structure of Rn , which
we saw with the joints problem and the P3 estimate. Let’s look at what these were good for.
How did we know to use the topological structure? Well, S-T was false over F2q (in particular, to take all
the lines in F2q ). So a statement being false over finite fields indicates that topology is important to look at.
We also noticed that S-T was true for pseudolines which topologically behaved the same but didn’t have an
algebraic definition.
32
What’s the corresponding indication that it’s a good idea to use the polynomial method? Well, it’d
probably have to be true over finite fields as well. We don’t know that yet, but maybe it’s true. Perhaps
you’d need a lower bound on q. This question is harder to answer. The easiest clue we have is if the problem
looks like other problems we’ve solved with the polynomial method. Also if you are able to get started with
e.g. degree reduction, that’s a good sign.
These methods are somewhat complementary. If you try to use the polynomial method to attack
Szemeredi-Trotter, for instance, you might take the lowest degree polynomial that vanishes on the lines,
but we know exactly what that is; it’s the product of the linear factors for each line. Similarly, you can try
to attack the joints problem using crossing numbers, but crossing number is a very 2D notion. You need to
project down and it’s pretty awkward.
Let’s look at our next theorem that we want to prove.
Theorem. Given L lines in R3 , ≤ L1/2 in any plane and 3 ≤ k ≤ L1/2 . Then |Pk | . L3/2 k −2 .
This is similar to the theorem we’ve proved, in that the case k = 3 in this one and B = L1/2 in that
one are the same. The thing about this theorem is that it smells like both the topological and algebraic
structure.
For instance, it’s false over Fq . Let L be all the lines in F3q , so L ∼ q 4 lines and we have ≤ q 2 in any
plane. Each point lies in ≥ q 2 lines and let’s take k = q 2 . Then q 2 = |Pk | = L3/2 k −3/2 . So it satisfies our
test telling us to prove with the topological structure. On the other hand, it’s similar to the other theorem
we proved with the polynomial method. So we need to unify these two methods. We’ll see this one ring to
rule them all on Wednesday.
18
Wednesday, October 24, 2012
Our goal in this section is to combine the Szemeredi-Trotter technology and the algebraic structure. The
crossing number method doesn’t combine very well, but we’ll look at another method in incidence geometry
called the cellular method from the 1990s.
To prove, for instance, Szemeredi-Trotter, we imagine a divide-and-conquer strategy where we split our
space into cells and bound what’s going on in each. We introduce a bunch of auxiliary lines that split up
our space into sections, and just look at the lines going into each piece.
Proposition (Counting estimate). If L lines in R2 , L ≤ 14 k 2 , then |Pk | ≤ 2L/k.
We proved this in the very first lecture of incidence geometry. We’d like to apply this in each cell. Of
course, this wouldn’t be useful if all of the points and lines were in one of the cells. Let’s make this precise.
What makes a good cell decomposition?
First note that if we have around d lines chopping up our space, we have ∼ d2 connected components.
This is because generically, the kth line crosses all the previous lines and adds ∼ k new cells. So now
any given line ` can only enter ≤ d + 1 connected components. If either the lines or the points are evenly
distributed, that would be good. There are several options for what we could mean:
A. The points are evenly distributed. So no k-rich points are in the dividing lines, and ≤ 10 |Pk | /d2 lie in
each cell.
B. The lines are evenly distributed. Again, no k-rich points are in the dividing lines, and ≤ 10 |L| /d lines
intersect each cell.
It’s tricky to prove there exist good decompositions, so we’ll save that for later. We’ll just try to prove the
Szemeredi-Trotter theorem by applying this elementary
counting P
estimate in each of the cells. Let the cells
P
be Oi , with Ni points and Li lines in Oi . Then
Li < 2Ld and
Ni = |Pk |.
Lemma. If d >(blank), then |Pk | ≤(blank).
33
Proof. First, Ni ≤ 10 |Pk | d−2 , and if Li ≤ 14 k 2 , then |Ni | ≤ 2L/k.
about
P So we have to care
P which cells have
Ni = B · 10 |Pk | d−2 + l 2Li /k where B is
a lot of lines. Let Oi be a big cell if Li > 14 k 2 . Then |Pk | =
P
2
the number of big cells. So how many big cells can there be? Clearly B k4 ≤
Li ≤ 2Ld, so B ≤ 8k − 2Ld.
Plugging that in, we get
|Pk | ≤ 80k −2 Ld−1 |Pk | + 4Ldk −1 .
If 80k −2 Ld−1 > 1, this is useless. But if it’s, say, less than 1/2, then it’s a good estimate: |Pk | ≤ 8Ldk −1 .
That’s our lemma, so let’s plug in what it says.
Lemma. If d > 160Lk −2 , then |Pk | ≤ 8Ldk −1 .
Corollary. If there exists a cell decomposition obeying A, with d = 160Lk −2 lines, then |Pk | . Ldk −1 ∼
L2 k −3 (S-T).
It turns out that the same is true for type B cell decompositions. In fact, you get the same d, up to a
constant factor.
Can we always do this? Let P ⊂ R2 be a set of N points, and d some number. Do there exist d lines
obeying A (that is, each cell has ≤ 10N d−2 points)? We’ll assume d ≤ N 1/2 , which if you trace it through
this proof of S-T, is fine.
Well, you can always split it in half, and if you mess with the angle and position of the second line, you
can split it into quarters. This reminds us for the Ham Sandwich theorem:
Theorem (30’s and 40’s). If Uj ⊂ Rn are sets of finite positive volume for j = 1, . . . , n. Then there exists
a hyperplane that bisects all of them.
Does this extend? Well, when we draw the next line, it can only enter three of these regions. Even
optimistically, it might bisect all three, but there would be more in the third set. It’s not clear that this
could reasonably continue in this way.
Could we possibly consider pseudolines, curves that pairwise intersect once? Well, there could be problems
with lines in the S-T theorem could intersect these cell walls. Since we’ve gotten a little stuck, let’s play the
other side and look for examples of sets of points we can’t cut up evenly.
Example. Let’s just think of examples: A square grid. Points on a line. Hmmm, that one is interesting. If
we want to split up with lines, those lines will only intersect that line once, splitting it into L + 1 points.
Well, maybe lines are special. But if you take the points on a circle, you have at most two intersections, so
it’s the same difficulty. Well, is it just low degree curves? Well then you could take a strictly convex curve.
Things are getting messy.
So maybe we can salvage B, evenly distributing the lines. This is the main idea of the cellular method:
Pick d random lines in L, with d ∼ Lk −2 . This is disobeying our rule that they not intersect the points, but
we only get ∼ |Pk | /k of them, since each point lies on k lines.
The key observation is that a typical edge intersects at most L/d lines. Heuristically, if we imagine the
line where this edge is on, it will intersect the d lines we chose, splitting into L/d edges, and then each cell is
a triangle and so has ≤ 3L/d lines. But while this is typically true, there are rare exceptions, and you need
to do some hard probability in order to fix this.
Let’s try to carry over this to our problem in R3 . What does it look like? We divide our space with
d planes. Then each line intersects ∼ d cells, and there are ∼ d3 cells in all. (Adding a new plane adds
∼ d2 regions because its intersection has lines and we apply the previous argument.) So our good cell
decompositions of part B (lines evenly distributed) will have ≤ 10L/d2 lines intersecting each cell. Suppose
we have a good cell decomposition of type B.
Lemma. If d2 > 40L/k 2 , then |Pk | . Ldk −1 .
P
P
Proof. If Li < k 2 /4, then Ni ≤ 2Li /k, so |Pk | = Ni ≤ ( Li )/k ≤ Ldk −1 . To do this, we need Li ≤ k 2 /4,
and we know that Li ≤ 10L/d2 . Therefore, we need d2 > 40L/k 2 .
34
So if condition B holds, d ∼ L1/2 k −1 , then |Pk | . L3/2 k −2 . This is the bound in the ES conjecture.
Unfortunately, it isn’t possible to find a good cell decomposition. You have the same problem with the points
in the plane when you look at lines in R3 .
Our idea is that instead of using d planes, we use a degree d algebraic surface. It will cut the space into
roughly d3 pieces still, and there are a lot more of those. That’s the big idea. We can also do the same thing
in the plane to prove Szemeredi-Trotter again.
19
Friday, October 26, 2012
Today we’ll talk about polynomial cell decompositions. We saw that it would have been useful to do this
with planes in general, but the idea is that we’d like to use algebraic surfaces to split them up. This gives
us a lot more degrees of freedom.
Theorem. Suppose S ⊆ Rn is a finite set, d ≥ 1. Then there is a nonzero polynomial P of degree ≤ d such
that each component of Z(P )C has . sd−n points of S.
The caveat here is that some (or all!) of S may lie in Z(P ). We’ll see as we go along how that may
happen. That might be a downer, but it’s actually more like a dichotomy. If a lot of the set lies in Z(P ),
then that’s saying that that subset is small degree. Otherwise it can be nicely subdivided by a degree d
variety.
This is similar to what we did early in the class bounding the degree, but we’re going to use topology
now. It’ll be the same sort of topology as the Ham Sandwich theorem which we saw last time:
Theorem (Ham Sandwich). If U1 , . . . , Un ⊂ Rn are finite volume open sets, then there exists a hyperplane
π that bisects all Ui .
This is reasonable from parameter counting, because we’d imagine that each bisection loses one parameter,
and we have n parameters for our hyperplane. There’s a vast generalization of this that is quite useful and
confims that this naive parameter counting idea is correct.
Let V be a vector space of continuous funcions on Rn . For f ∈ V , Z(f ) := {z ∈ Rn : f (x) = 0}. We say
that f bisects U if the volume of {x ∈ U : f > 0} is the same as the volume of {x ∈ U : f < 0} and is half
of the volume of U . The Ham Sandwich theorem looks at V = {a1 x1 + · · · + an xn + b} and Z(f ) planes.
We’d like to bisect dim V − 1 sets. Here’s the theorem that says we can do that under mild conditions on
our space of functions:
Theorem (Stone-Tukey, 42). If V is a vector space of continuous functions on Rn and U1 , . . . , UN ⊆ Rn
finite volume and open, where N < dim V . For every f ∈ V \ {0}, assume that Z(f ) has 0 Lebesgue measure.
Then we conclude that f ∈ V \ {0} that bisects all Ui .
We’ll prove the Ham Sandwich theorem and see that the same result generalizes to Stone-Tukey. Both
use the Borsuk-Ulum theorem from topology.
Theorem (Borsuk-Ulam). If F : S N → RN is continuous and antipodal, i.e. F (−x) = −F (x), then 0 is in
the image of F .
We won’t prove this, but it’s widely useful in topology and even combinatorics.
We’re looking for a function that can bisect all of these sets, and we can measure how far we are. For
i = 1, . . . , N and f ∈ V \ {0}, let Φi (f ) = Vol{x ∈ Ui : f > 0} − Vol{x ∈ Ui : f < 0}. Then Φi (f ) = 0 iff
f bisects Ui . Define the big function Φ : V \ {0} → Rn , and WOLOG dim V = N + 1 (otherwise look at
a subspace). Then S N ⊆ V \ {0}, so we can think of Φ : S N → RN . Now Φi (−f ) = −Φi (f ) is clear. The
last thing to check is continuity, and that’s where we use the condition that the functions vanish on measure
zero sets. We can see why that would be a problem, because slightly perturbing f would cause a jump in
the volume on which it is positive. We’ll write this as a lemma:
35
Lemma. IF m(Z(f )) = 0, for all open U , Vol(U ∩ {f > 0}) is continuous in f over f ∈ V \ {0}.
Alright, we had a polynomial existence lemma:
Lemma. If x1 , . . . , xN ∈ Rn with N < dim V (d) =
that P = 0 on xi for all i.
d+n
n
, then there exists a nonzero P of degree ≤ d such
We have an analogous polynomial ham sandwich theorem then:
Lemma. If U1 , . . . , UN ⊂ Rn of finite volume, and N < dim V (d) =
polynomial P of degree ≤ d which bisects each Ui .
d+n
n
. Then there exists a nonzero
Now this is a little bit of a different thing, because we’re talking about bisecting finite sets, not volumes.
How can we relate those? Well, we could take small balls around each point. Let S1 , . . . , SN be finite sets
in Rn , where N < dim V (d). Then let Nδ (Si ) be a δ-neighborhood of Si . By the polynomial ham sandwich,
there exists Pδ bisecting eahc Nδ (Si ) with deg Pδ ≤ d.
What could happen? We talked through pathological examples. It seemed necessary that we let δ → 0.
Let’s precisely say what happens.
Corollary. If S1 , . . . , SN are finite sets, and N < dim V (d). Then there exists a nonzero polynomial P of
degree ≤ d with #{x ∈ Si : P > 0} ≤ 12 |Si | and #{x ∈ Si : P < 0} ≤ 21 |Si |.
Can we do better and get that these are actually equal? We talked through an example with two sets in
the plane that we were trying to simultaneously bisect.
It’s clear what we need to do to prove this, taking that limit limδ→0 Pδ . Well, can we do this? Scale Pδ
so that it lies in a sphere, which is compact and take a subsequence which converges. Then compare the
limit to some Pδ which is close and you get a contradiction to having more than half the points on one side.
Let’s relate this to cell decomposition now, by proving our theorem:
Theorem. If S ⊂ Rn is finite then there exists a nonzero polynomial P of degree ≤ d so that each component
of Z(P )C has . |S| d−n points.
Proof of Cell Decomposition Theorem. Let S be a finite set in Rn . In step 0, choose a degree 1 polynomial
P0 which bisects S, outputting S+ and S− (we don’t label S0 ), with |S+ | , |S− | ≤ 12 S. Then bisect them
both with a low degree surface, creating 4 sets. We continue this for a while and eventually stop at step J
having used some low degree polynomials (we’ll figure out the bound later) that bisects all 2J sets from step
J − 1.
C
Our polynomial is P0 P1 · · · PJ . Each component of |Z| has ≤ |S| 2− J points. We need to choose J and
figure out our degree bound. That degree bound comes from the corollary above, dnj ∼ dim Vdj > 2j , so we
PJ
PJ
can arrange that deg Pj = dj . 2j/n . Therefore, d = deg P = j=0 deg Pj . j=0 2j/n . 2J/n , so indeed
each component has . |S| 2−J .
Is this tight? Well, in one dimension it is. In general, there’s probably a bound on the number of
components of Z(P )C on the order of dn . And then you have to worry a little about the points not lying on
the polynomial, but if they’re in general position of some sort, that’s not a problem.
Next time, we’ll use this new tool to go back and prove Szemeredi-Trotter and more.
20
Wednesday, October 31, 2012
We got our homework back. It’s very convenient, actually, to have a hurricane when you have a big stack of
problem sets to grade. Guth was quite impressed. The average level of the rigor in the solutions was higher
than in the questions.
S
We decided to go over the notoriously vague 6c. Let’s remind ourselves of the setup: X = `∈L ` ⊆ F4q .
Every point of X is in ≥ 2 lines, and |X| q 4 . P was the minimal degree nonzero polynomial with P = 0
1/2
on X, and deg P . |X| q −1 . What Guth wanted us to prove was:
36
S
P
Proposition. X ⊂ Z(Pk , Qk ) such that Pk , Qk have no common factor, and
deg Pj deg Qj / |X| q −2 ,
p
where A / B if A ≤ C(log q) B for some constant p.
Q
We broke P into irreducibles P = Pj . And we can assume |X| is small enough that deg P < 2q . Then
many of us proved that each line of L ⊂ Z(Pj ) for some j. We wanted to look at {` ⊂ Z(PJ )}, but this
didn’t quite have the properties we wanted.
Instead, look at Lj := {` ⊂ Z(Pj ), ` 6⊂ Z(Pj 0 )},Pand Lj,k = {` ⊂ Z(Pj ) ∩ Z(Pk )}, and maybe more
2
−2
similar things. Lines of the second type are good:
. So from the
j,k deg Pj deg Pk ≤ (deg P ) . |X| q
point of view of the proposition, we only have to look at the first type.
Then
for ` ∈ Lj , let `uni ⊂ ` be the set of points not in any other Z(Pj ). Then since deg P < q/2,
uni
` ≥ q/2. Now we just use those half-lines in the same way that we used lines before.
Where do the logs come in? When we try to do the argument of 6d again using `uni , we don’t have a
bound on their size, so we need to have a log there.
If you liked that problem, you could try to do 5 dimensions as well!
20.1
Polynomial Cell Decompositions Continued
Now we’ll go back to polynomial cell decompositions and prove Szemeredi-Trotter, then go back to that
theorem we’ve been talking about and prove it. Recall what Szemeredi-Trotter says; it has several different
formulations, but here’s what we’ll use:
Theorem. Given S points S and L lines L in R2 , the number of incidences is
I(S, L) ≤ C0 [S 2/3 L2/3 + S + L].
Proof. Recall the elementary counting bounds that we proved:
Lemma. I(S, L) ≤ L + S 2 and I(S, L) ≤ S + L2 .
theP
lines
that
Proof. For x ∈ S, let Lk be the lines of L containing x, and Luni
x
uni
contain x and no other point
. Summing over x, I(S, L) ≤ S 2 +
Lx ≤ S 2 + L. The other one is
of S. Then |Lk | ≤ S + Luni
k
x
similar.
Now we’ll divide up our space into cells and apply this estimate in each.
Recall polynomial cells: ∃P of degree
d, each component Oi of Z(P )C having . Sd−2 points Si of S,
P
and Li is the lines meeting Oi , with
Li ≤ (d + 1)L . dL. Then our estimate gives I(Si , Li ) ≤ Li + Si2 , so
defining Scell to be the sets that are in the interiors of the cells,
X
X
X
I(Scell , L) =
I(Si , Li ) ≤
(Li + Si2 ) . Ld + Sd−1
Si ≤ Ld + S 2 d−2 .
i
i
We’ll want to optimize this with a particular d, but first, let’s take care of the points on the boundary. Let
Salg = S ∩ Z(P ), and L = Lcell ∪ Lalg , where Lalg is the lines in Z(P ). We have I(Salg , Lcell ) ≤ Ld since
Z(P ) is a degree d surface. Therefore, it remains to bound I(Salg , Lalg ).
We’re in the plane, so a line is a factor, and there are at most d lines in Lalg , which would give us a S + d2
bound. But that isn’t actually a much better bound: If d L, then we could just handle this last term by
induction. So let’s do that, choosing d < L/2. Under that constraint, we’d like to minimize Ld + S 2 d−2 . So
we’d like d ∼ S 2/3 L−1/3 , in which case this becomes L2/3 S 2/3 . But we have to check that S 2/3 L−1/3 is in
the range 1 ≤ d ≤ L/2.
√
1 2
1 2
Well, because of the lemma, if L > 10
S or S > 10
L , we’re done anyways. Otherwise, 10s1/2 ≤ L ≤
1 2
2/3 −1/3
L
. L, and we can choose it.
10 S , so 1 . S
Then we have to work out our induction. Then I(S, L) ≤ CL2/3 S 2/3 + C0 [S 2/3 (L/2)2/3 + S + L/2]. Then
we can just make the constant C0 big enough and we’re done.
37
Not too bad, eh? Let’s go up a dimension.
Theorem. For S points S and L lines L in R3 with at most B on any plane,
I(S, L) ≤ C0 [S 1/2 L3/4 + B 1/3 L1/3 S 2/3 + S + L].
Let’s look at this a bit. The last two terms are easy to understand. The second one is what happens
when you put L/B planes with B lines in each. And the first one corresponds to a 3-dimensional grid.
Proof Outline. We still have the usual L + S 2 and S + L2 bounds, and for some range of L and S, this will
be good enough. But we also have Szemeredi-Trotter, which is stronger, and which we can take as a lemma.
Lemma. I(S, L) ≤ C1 [S 2/3 L2/3 + S + L].
(You do the usual thing of projecting onto a plane here.)
Now use a degree d cell decomposition and use S-T on each cell. Again split up S = Scell ∪ Salg and
L = Lcell ∪ Lalg . Then I(Scell , L) ≤ some bound depending on d. Additionally, I(Salg , Lcell ) ≤ dL, so we’re
done there. The tricky part is I(Salg , Lalg ), because our surface is now 2-dimensional, and can have infinitely
many lines. We talked about several examples of how this might happen.
The other problem is that we haven’t used the B bound at all. To handle these, define Lplanar ⊂ Lalg
to be the lines in a plane in Z(P ). There are at most d planes in Z(P ), and each of those have at most B
lines of L, so I(Salg , Lplanar ) is also bounded, and it remains to look at I(Salg , Lalg \ Lplanar ). We’ll break
these up based on whether the lines and points are special. Recall that x ∈ Z(P ) is special if it is critical or
flat, and SP (x) = 0 with deg SP ≤ 3d. Lines are special iff every point on them is special.
(1) The number of lines in Z(P ) which are not planar are bounded by 10d2 .
(2) If x is in three lines, then x is special, so we can bound the number of intersections involving non-special
lines.
(3) A non-special line has at most 3d special points.
We’ll see this complete argument with all the algebra on Friday.
21
Friday, November 2, 2012
We’ll prove the 3D incidence geometry estimate that we stated last class. If there’s some time, we’ll ask the
question, “What’s special about polynomials from the point of view of geometry and topology?”
Theorem. Let S be a set of S points, L lines in R3 with at most B lines in any plane, with B ≥ L1/2 .
Then
I(S, L) ≤ C0 [S 1/2 L3/4 + B 1/3 L1/3 S 2/3 + L + S].
Proof. We’ll first write down the things that we already know:
• Counting bounds. I ≤ S 2 + L, L2 + S.
• Szemeredi-Trotter. I ≤ C[L2/3 S 2/3 + L + S].
If it so happens that these bounds are smaller than what we want to prove, then we’re done. So we can
assume S 1/2 . L . S 2 . We can also do induction on L.
Now let’s get rolling. Let d ≥ 1, which we will choose later. So there is a polynomial cell decomposition
P of degree d. We obtain an estimate from these cells:
Lemma. I(S, L) ≤ C[d−1/3 S 2/3 L2/3 + dL + Scell ] + I(Salg , Lalg ).
Recall: Algebraic points and lines are those in Z(P ).
38
Proof.
Let Oi be the cells, Si the points of S P
in Oi , and Li the lines of L that intersect Oi . We know that
P
Si = Scell ≤ S, and Si . d−3 S. Moreover,
Li . dL. Now we’ll count the incidences in each cell, using
the best thing that we already know, Szemeredi-Trotter, in each cell.
P
P 2/3 2/3
Then I(Scell , L) = i I(Si , Li ) . i Si Li + Li + SiP
. We can obviouslyP
sum the last two terms, but
what do we do with the other one? We have a bound on
Si , max Si , and
Li . This is a Hölder-type
1/3
inequality, so what we do is bound Si by its maximum, getting
X
X
(d−1 S)1/3 (
Si )1/3 (
Li )2/3 = d−1/3 S 2/3 L2/3 .
The other sums give dL + Scell . And there’s one more term we can tackle. Lcell is the lines that intersect
some of the cells, so I(Salg , Lcell ) ≤ dL. Therefore, the lemma is proved.
Now let’s talk about general strategy. We’ve been counting this big number of incidences, and have
broken it up into a bunch of smaller terms. If they’re especially small, we might be able to use induction.
Next we’ll examine the lines that are in planes, because we have this bound B on the number of lines
that we’d like to bound. Define Lplanar to be the lines in at least one plane of Z(P ), and Splanar the points
in at least one plane of Z(P ). Moreover, define Lmultiplane to be those lines in more than one plane of Z(P ),
and Luniplane to be those in exactly one, and define Smultiplane and Suniplane similarly.
With all of this, we can describe the planar estimate:
Lemma. I(Salg , Lplanar ) ≤ C[B 1/3 L1/3 S 2/3 + dL + Suniplane ] + I(Smultiplane , Lmultiplane ).
This is nice, it fits into the bound we proved for the other intersections, but we have that annoying term.
Well, we can also bound it from |Lmultiplane | ≤ d2 . We’re going to pick d to be small enough that this term
can be handled by induction.
Proof. Multiplanar lines can only meet multiplanar points, so we just needPto bound I(Salg , Luniplane ). This
is straightforward: Divide the lines up by planes {π} in Z(P ). So this is π I(Sπ , Luniπ ). We have to be a
little careful with Smultiplane , but we can do this. Each line can hit at most d other planes, so this is at most
P
P 2/3 2/3
π dLuniπ + I(Suniπ , Luniπ ). Then this is ≤ dL +
π Suniπ Luniπ + Luniπ + Suniπ by Szemeredi-Trotter.
The later terms will do what we want, and the first one is bounded by the same computation we did earlier.
Using Luniπ ≤ B and Hölder, we get
I(Salg , Luniplane ) . dL + B 1/3 L1/3 S 2/3 + Suniplane ].
We’re left with the algebraic lines that are not planar. This gives us the third structure, the geometry of
an algebraic variety, in particular, its flat points and special points.
Recall that a special point is a flat or critical point, and a special line is a line consisting of only special
points. We can make some corresponding vocabulary: Sspec is the set of special points in Salg and Snotspec
is the set of non-special points in Salg (still in Z(P )). Correspondingly define Lspec and Lnotspec . We have
one last estimate:
Lemma (Algebraic Estimate). I(Salg , Lalg \ Lplanar ) ≤ C[dL + Snotspec ] + I(Sspec , Lspec \ Lplanar ).
This doesn’t seem like we’ve done much, but |Lspec \ Lplanar | ≤ 10d2 , so we can choose d small enough
to do induction here, too.
Proof. Recall that:
(1) If x is on three lines in Z(P ), then x is special.
(2) x is special iff SP (x) = 0, with deg SP ≤ 3d.
(3) The number of lines which are special but not planar is at most 10d2 .
39
We want to bound I(Salg , Lalg \ Lplanar ). This splits into three terms: I(Snotspec , Lalg ) + I(Sspec , Lnotspec ) +
I(Sspec , Lspec \ Lplanar ). The first term is bounded by 2Snotspec by fact (1), and the second by 3dL by fact
(2), so this proves the estimate.
Now let’s do some bookkeeping to put these three estimates together. Let’s just distinguish between lines
we’ve estimated and lines we still have to estimate by induction. The latter we’ll call L2 = Lmultiplane ∪
(Lspec \ Lplanar ). There are at most L2 ≤ 11d2 of these. We do the same with the points: We’ve already
estimated all of S1 = Scell ∪ Suniplane ∪ Snotspec and haven’t estimated S2 = Smultiplane ∪ Sspec . Putting
everything together, we’ve proved that
I(S, L) ≤ C[d−1/3 L2/3 S 2/3 + dL + B 1/3 L1/3 S 2/3 + S1 ] + I(S2 , L2 ).
Now, we want to choose d to be small enough that 11d2 L, so choose some 1 ≤ d ≤
some expression that we’ll write down later.
1 1/2
10 L
minimizing
Lemma. The minimum of this expression is ∼ S 1/2 L3/4 + B 1/3 L1/3 S 2/3 + S1 .
The proof is just high school algebra, and uses that B ≥ L1/2 and see its results. We have S2 ≤ S and
L2 ≤ L/2. Then
I(S, L) ≤ C[S 1/2 L3/4 + B 1/3 L1/3 S 2/3 + S1 ] + C0 [S 1/2 (L/2)3/4 + B 1/3 (L/2)1/3 S 2/3 + L/2 + S2 ].
Now choose C0 > 100C so that this is at most C0 [S 1/2 L3/4 + B 1/3 L1/3 S 2/3 + L + S].
We’ve talked about the importance of polynomials in computer science and over finite fields. What was
interesting was that you could mess with them a little bit and you’d know where they came from. But what
about geometry and topology?
21.1
Efficiency of polynomials
Let’s do an example. We’ll start by talking about C polynomials.
Theorem. If P : C → C is a polynomial, let F : C → C be a smooth function. Suppose that F = P outside
Ω, a bounded domain. Morally, the number of zeros of P is at most the number of zeros of F . However, we
have to worry about multiplicity, so we’ll fix this by assuming that 0 is a regular value of P and F . Then
#Z(P ) ≤ #Z(F ).
What is a regular value? A function F : M m → N n , with M, N manifolds. x ∈ M is a critical point if
dFx is not surjective, and a regular points is not critical. Then y ∈ N is regular if all preimages are regular.
Proof Sketch. If x ∈ Z(F ), 0 is a regular value if dFx : R2 → R2 is an isomorphism. There are two kinds
of isomorphisms, orientation-reversing and orientation-preserving, so we have a sign σ(x) = +1 if dFx is
orientation-preserving and −1 if it is orientation-reversing.
Complex polynomials, however, are always orientation preserving.
topological signifP These signs have aP
icance, that the winding number of a f : ∂Ω → C \ {0} is equal to x∈Z(F )∩Ω σF (x) = x∈Z(P )∩Ω σP (x).
But the latter are all +1, so there need to be at least as many terms in the first sum.
This is the first of a series of results we’ll sketch to see how polynomials are efficient in geometry and
topology.
40
22
Monday, November 5, 2012
Today we’ll talk about polynomials in geometry and topology, partly to take a bit of a break from other
stuff and partly because we’ll see things like the ham sandwich theorem again in another context.
The big idea here is that polynomials are efficient. Last time we saw this theorem:
Theorem (1). If P : C → C is a polynomial and F : C → C is C 1 and agrees with P outside D and 0 is a
regular value of F, P , then |Z(P )| ≤ |Z(F )|.
Here we didn’t really use that P was a polynomial but that it was holomorphic in a neighborhood of D.
If 0 is not a regular value, we need a multiplicity definition that can work for general functions. Also, 0 isn’t
special, we’re just looking at the preimage.
Let’s go up a little bit and look at polynomials in several variables.
Theorem (2). If P : Cn → C is a polynomial and F : Cn → C is smooth (thinking of it as R2n → R) and
P = F outside B 2n . Then Vol2n−2 [Z(P ) ∩ B] ≤ Vol2n−2 [Z(F ) ∩ B].
This was proved by DeRahm and Federer in the 1950s. A related observation is that Z(P ) is a minimal
surface, i.e. given that it intersects the ball in certain points, it has the smallest surface area. This is a
pretty hard theorem.
Proof Sketch. How can we get a handle on this? Restrict P and F to some complex line L ⊂ Cn . Then
P, F : L → C agree outide B 2n ∩ L. It doesn’t quite follow that
0 will be a regular value, but this will
be usually true. Then by applying theorem 1, Z(P ) ∩ L ∩ B 2n ≤ Z(F ) ∩ L ∩ B 2n . This is some local
information, and we have to assemble it into a sort of global statement. This is the role of integral geometry,
which we won’t go into. Later today, we’ll see a similar proof, but for now, trust that this conclusion implies
the volume bound in the theorem.
We’ve looked at this from a geometric perspective; there’s also a topological perspective over which
polynomials are efficient.
Theorem (Kronheimer-Mrowka, 1990s). If P : C2 → C is a polynomial and F : C2 → C agree outside B 4 ,
0 is a regular value of each, and Z(P ) and Z(F ) are conneced 2-dimensional surfaces. Then the genus of
Z(P ) is at most the genus of Z(F ).
We drew a potential picture there, but it didn’t have enough dimensions. We thought that maybe we
could fill in the boundary without holes, but the boundary is in S 3 and might be knotted. So this theorem
is interesting in that polynomials provide the most efficient way to unknot things. Unfortunately, it’s far
outside the scope of this course.
One thing we might take issue with this whole presentation is that these things have been true of
holomorphic functions. We might ask if anything like this is true for real polynomials. Well, Theorem 1
is pretty clearly false; we can just take some easy interpolation. But there’s a recent fix to this, a new
statement that the polynomials are efficient. The key is to look at the vector space of polynomials of small
degree, and arguing that this space is efficient rather than any individual polynomial.
Theorem (1’). If W is a vector space of functions R → R with dim Vd = dim W . Then ∃F ∈ W \ {0} such
that |Z(F )| ≥ d, and ∀P ∈ Vd \ {0}, |Z(P )| ≤ d.
That is, if you stack up the space of polynomials against any other space (with the same dimension to
be fair), it is the most efficient in terms of not having many zeroes. But this has other higher-dimension
analogs:
Theorem (A). If P : Rn → R is a nonzero polynomial of degree ≤ d, then Voln−1 Z(P ) ∩ B n . d.
Theorem (B). If W is a vector space of functions B n → R with dim W = dim Vd , then ∃F ∈ W \ {0} such
that Voln−1 Z(F ) ∩ B & d.
41
Theorem A was proved by Crofton in the early 20th Century, and Theorem B by Gramov in the 2000’s.
These say that up to a constant, real polynomials are most efficient in terms of having small zero sets.
Proof of Theorem A. Let L be a line in Rn . Then either |Z(P ) ∩ L| ≤ d or L ⊂ Z(P ), and the second one
isn’t going to happen too often. We have to compare the intersections with lines to the overall intersections.
Let AG(1, n) := {lines in Rn }, G the group of rigid transformations, which acts on AG(1, n). Then there
is a measure dµ on AG(1, n), the Haar measure on the group of quotients, which is G-invariant up to scaling.
R
Proposition (Crofton formula). Voln−1 (X) = αn AG(1,n) |L ∩ X| dµ(L) where αn is a universal constant.
This is like the needle theorem, where if you have a curve in the plane and drop a needle onto it
“randomly,” the probability of intersecting it is just dependent on the length of the curve. This is analogous,
except we’re taking infinitely long needles and counting intersections with multiplicity.
Now we’d like to compute this for X = Z(P ) ∩ B. Several observations:
• If some line ` intersects X, then it intersects the sphere twice.
• dµ{` ⊂ Z(P )} = 0, so we can ignore it.
R R
• AG |` ∩ Z(P ) ∩ B| dµ ≤ d2 AG ` ∩ S n−1 dµ =
spheres as Z(P ), we get that bound.
d
2
Vol S n−1 . This is even sharp in that if we take d/2
Now we’ll sketch a proof of Crofton’s formula. Let C(x) be the right hand side. Instead of starting with one
and getting the other, we’ll show that these have so many common properties that they must be the same.
• They respect disjoint unions: X = X1 t X2 . In both cases, the volume and C sum.
• They are invariant under rigid motions.
Now choose αn such that Vol[0, 1]n−1 = C([0, 1]n−1 ) = 1. So Vol(X) = C(X) for unions of unit cubes.
We can also do this for 1/N -cubes for N ∈ N because we can stack them together. Then the last step is a
limiting argument by cubical approximation for general X. Then we have to prove that the volume of the
approximation is close to the volume of the surface, and that the Crofton of the approximation is close to
the Crofton of the surface.
Now we’ll go back and prove Theorem B.
Proof. Recall that dim Vd ∼ dn ∼ dim W . We had the general ham sandwich theorem, which said that with
U1 , . . . , UN where N < dim W in the unit ball, ∃F ∈ W \ {0} such that F bisects Ui . The trick is that you
take Ui to be disjoint balls with radius ∼ d−1 . So then we have to argue that to bisect each ball, the surface
needs to be a certain size. Basically, we need this lemma from geometry:
Lemma. IF F bisects B n (1), then Z(F ) ∩ B has Voln−1 ≥ c(n).
This can be proved from the isoparametric inequality for the ball, and the equatorial disk is the best thing
to do, but it’s even easier than that. From this lemma, Voln−1 (Z(F )∩Ui ) & d−(n−1) , so Voln−1 (Z(F )∩B n ) &
dn d−(n−1) ∼ d, as desired.
Some morals to this story:
1. This was an early application of using ham sandwiches to prove estimates, and later provided another
inspiration to use polynomials in combinatorics.
2. Polynomials do a pretty good job of this. Guth tried constructing a family of polynomials by hand
that solved this problem, but it was hard to compete with them. He could do it, but it was hard. He
couldn’t do the analogs for complex polynomials. This is kind of analogous to the other applications
of the polynomial method in that we used them to come in and do things we couldn’t do ourselves.
42
23
Wednesday, November 7, 2012
Recall, we recently proved this theorem:
Theorem. Given L lines L in R3 , ≤ B in any regulus or plane, B ≥ L1/2 , then |P2 (L)| . BL.
This was harder to prove than 3-fold intersections. That was because if we had 3-rich points, we had a
way to detect planes with critical points and flat points:
Lemma (Plane Detection). If P is a polynomial, then SP is another polynomial with
(1) deg SP ≤ 3 deg P .
(2) If x is on three lines in Z(P ), then SP (x) = 0.
(3) If P is irreducible and SP = 0 on Z(P ), then Z(P ) is a plane.
The problem with applying this to 2-fold intersections is that we can’t conclude (2). But we also can’t
detect reguli, so we’ll build a corresponding regulus-plane detection lemma:
Lemma (Regulus Detection). If P is a polynomial, then RP is a finite list of polynomials, with
(1) deg RP ≤ C deg P .
(2) If x is in two lines in Z(P ), then RP (x) = 0.
(3) If P is irreducible, RP = 0 on Z(P ), and there exists one point x on two lines of Z(P ) which is not
critical or flat, then Z(P ) is a regulus.
We tried to make this look a little nicer, but there didn’t seem to be any guarantee that Z(P ) contains
any lines. When we want to use this in incidence geometry, we’ll have that condition pretty easily.
At this point, we’ll take a brief digression and talk about ruled surfaces, where a similar detection lemma
arises. There might also be complex lines that don’t arise in Rn .
23.1
Ruled Surfaces
We’ll work in C3 for the rest of the class. Question: If P is an irreducible polynomial, how many lines can
Z(P ) have?
There can certainly be infinitely many, for instance for planes, reguli, cones, and cylinders. There’s a
more general example: Consider Φ : C2 → C3 , (s, t) 7→ Φ1 (s)t + Φ0 (s). This contains infinitely many lines.
Moreover, these examples all have this property:
Definition. Z(P ) is ruled if every point is on some line in Z(P ).
In all other cases, though, there can be only finitely many lines:
Theorem. If P is irreducible, Z(P ) not ruled, the number of lines in Z(P ) is . (deg P )2 .
The contrapositive amounts to this:
Lemma (Ruled surface detection). If P is a polynomial in C3 , then ∃F P with
(1) deg F P ≤ C deg P .
(2) If x lies on one line in Z(P ), F P (x) = 0.
(3) If P is irreducible, F P = 0 on Z(P ), then Z(P ) is ruled.
There’s a formula for F P in terms of around 11 different derivatives of P which we won’t try to write
down, but we want to think about why it exists.
43
Definition. z ∈ C3 is flecnodal iff ∃0 6= v ∈ C3 such that 0∇v P (z) = ∇2v P (z) = ∇3v P (z).
The job of the flecnodal polynomial F P is to determine whether or not a point is flecnodal: F P (z) = 0
iff z is flecnodal.
How do we understand these derivatives? If I = (i1 , i2 , i3 ) is a multiindex, ∂I P = ∂z∂i ∂z∂i ∂z∂i P and
1
1
1
P
v I = v1i1 · · · , then ∇sv P = |I|=s I!v I ∂I P .
P
I
M
Lemma (1). Consider
:
|I|=s v aI = 0 for s = 1, 2, 3, which we’ll call (1). Then the set {a ∈ C
(1) has a nonzero solution v} is algebraic, so we’ll call it the zero set of a finite list of polynomials G.
In this case, G can be taken to be a single polynomial, but as we’ll see later, in a more general setup, G
could be multiple polynomails. This lemma implies that F P obeys:
(1) deg F P ≤ (deg G)(deg P ) ≤ C deg P .
(2) F P = 0 if and only if there is a nonzero v such that ∇sv P (z) for s = 1, 2, 3, iff z is flecnodal. So if z is
on a line in Z(P ), z if flecnodal, so F P (x) = 0.
The last property is harder to prove.
We’ll need some projection theory from algebraic geometry:
Let F be a field, and Z ⊆ Fm × Fn an algebraic set. What happens when we project onto Fm ?
Let’s see some examples.
Example. If Z is a sphere in R2 × R1 , then π(Z) is a closed disc, not an algebraic set. If Z is a hyperbola
like {xy = 1} ⊂ R × R, then π(Z) is R \ {0}.
Can we do anything to save this? Working over the complex numbers gets rid of the sphere example,
but the hyperbola is still a problem. We talked about whether these things were dense for a bit.
Can we fix the hyperbola? Well, it has a “solution” at infinity, so maybe we should count that as well.
Let’s work over projective space then. So now we can state:
Theorem (Fundamental Theorem of Projective Theory, Version 1). If Z ⊆ Fm × P Fn is an algebraic set
and F is algebraically closed, then the projection of Z to Fm is an algebraic set.
We haven’t talked about how to define an algebraic set in projective space. It amounts essentially to
only looking at homogeneous polynomials, and we can translate the statement over like this:
Theorem (Fundamental Theorem of Projective Theory, Version 2). If Q(x, y) is a finite list of polynomials
which are homogeneous in y and F is algebraically closed, then {x ∈ Fm : ∃0 6= y ∈ Fn , Q(x, y) = 0} is an
algebraic set.
We can think of the previous discussion as motivation, and we can see that this is a more general case
of what we wanted to prove. v I were homogeneous and we wanted to know when there would be a nonzero
solution.
Proof. We’ll start proving this today, and do the rest on Friday. We’d like to have some way to tell when
there is a nonzero homogeneous solution.
Write the list of polynomials as Qj (x, y), a list of polynomials. Let Qj,x (y) as a polynomial just in y,
and let I(x) = hQj,x (y)i ⊆ F[y]. This is a homogeneous ideal: if q ∈ F[y], q=d the degree d part of q, then I
is homogeneous iff ∀q ∈ I, ∀d, q=d ∈ I.
Proposition (1). If I is any homogeneous ideal in F[y], F algebraically closed, then ∃ a nonzero y ∈ Z(I)
iff F[y]/I is infinite-dimensional as an F-vector space.
44
Proof. The forward direction is easy. If y is a nonzero solution, then the line from 0 to y is in Z(I) and there
is a map from F[y]/I to the polynomials on this line. That is an infinite-dimensional space, so F[y]/I is as
well.
Now, if Z(I) = 0, then by Nullstellensatz, rad I = I{0} = (y1 , . . . , yn ). Then F[y]/ rad I is finite
dimensional, and in general if I = I0 ⊂ I1 ⊂ · · · ⊂ IJ = (y1 , . . . , yn ) where Ij is Ij−1 with a radical element
adjoined. Then let Rj = F[y]/Ij , so Rj−1 is Rj with a nilpotent element adjoined. This maintains the finite
dimensionality, so since RJ is finite dimensional, so is R1 , as desired.
We’ll finish the proof of the fundamental theorem on Friday.
24
Friday, November 9, 2012
Let’s remind ourselves where we are. We were talking about the Fundamental Theoreom of Projection
Theory:
Theorem (FTPT). If Qj (x, y) are polynomials over an algebraically closed field F, homogeneous in y of
degree dj , then SOL = {x ∈ Fm : ∃y 6= 0, Qj (x, y) = 0∀j} is an algebraic set, so it is the zero set of a finite
list of polynomials G.
We can consider Qj (x, y) as a polynomial in y, and let I(x) be the ideal generated by these. We proved
last time:
Proposition. If F is algebraically closed, I ⊂ F[y] a homogeneous ideal, then Z(I) contains y 6= 0 iff F[y]/I
is infinite-dimensional (as a F-vector space).
This is reminiscient of Bezout-type degree arguments. Let H=d ⊂ F[y] be homogeneous degree d polynomials, and I(x)=d = I(x) ∩ H=d . We can see that F[y]/I[x] is infinite-dimensional iff I(x)=d ( H=d for
all d. This is easy to see: If these are all strict, we get at least one degree for each dimension, and if some
I(x)=d = H=d , then all higher ones are equal as well, so the whole thing is finite-dimensional. Now we can
write SOL = {x : dim I(x)=d ≤ dim H=d − 1∀d}.
Lemma (2). If d, B ≥ 0 are integers, then {x ∈ Fm : dim I(x)=d ≤ B} is an algebraic set. (Here we don’t
need F to be algebraically closed.)
If we assume this for a moment, then SOL is a countable intersection of these algebraic sets, so by the
Noetherian property of rings, it is algebraic as well.
P
Proof of Lemma 2. Let M : ⊕j H=d−dj → I(x)=d be the linear map M (f1 , . . . , fJ ) = j fj Qj,x . This works
because Qj,x is homogeneous of P
degree j. The important thing that happens here: M is surjective. If
f ∈ I(x)=d . By
definition,
f
=
fj Qj,x for some polynomials fj ∈ F[y]. Since Qj,x is homogeneous of
P
degree dj , f =
fj,=d−dj Qj,x ∈ im M .
So dim I(x)=d = rank M . The coefficients of M are the coefficients of Qj (x, y), so if we compute the
determinants of the (d + 1) × (d + 1) blocks, that’s a finite number of polynomials, so the set is algebraic as
desired.
n
Let P ∈ C[zP
Cn such that ∇sv P (z) = 0 for s = 1, . . . , t.
1 , . . . , zn ]. Then z ∈ C is t-flecnodal iff ∃0 6= v ∈ P
Now, ∇sv = |I|=s I!v I ∂I P (z) = 0. So by the theorem, {aI : |I|=s V I aI = 0, s = 1, . . . , t} = Z(G(aI ))
for some polynomials G by the FTPT. Call this the tth flecnodal polynomial Ft P (z) = 0 with deg Ft P ≤
(deg G)(deg P ).
This ends our discussion of projection theory.
A couple general comments: There was a problem with one of the problems on the problem set. We took
a vote on whether Wednesday or Friday next week would work for office hours, and it was close enough that
we’ll do it both days.
The last thing we need related to flecnodes is the following theorem:
45
Theorem (Cayley). If P ∈ C[z1 , z2 , z3 ] and F P = 0 on Z(P ). Then Z(P ) is ruled.
Recall that ruled means there are lines in the surface through every point. Here F P = F3 P .
So for every z ∈ Z(P ), there is a 0 6= v ∈ C3 with 0 = ∇sv P (z) for s = 1, 2, 3 (∗). This is a local-to-global
sort of transition. Let’s assume that we can assume v(s) smoothly obeying this equation. Look at the
integral curves of v.
There’s a problem with this strategy: If Z(P ) = C2 and V (Z) is any vector field in the plane, then it
will obey (∗) but the integral curves won’t be lines. This is essentially the only problem, though, so we’ll
essentially assume it’s not a plane. There are annoying cases of what happens if Z(P ) is no a plane. We’ll
tackle one of these.
Proposition. If P ∈ R[x1 , x2 , x3 ], O ⊂ Z(P ), V (x) is a smooth vector field on O obeying (∗), O is regular
(∇P 6= 0 on O) and ∇2 P : T Z × T Z → R which is non-degenerate on O. Then all integral curves are
straight lines.
P
That is, ∇2V,W F (z) = i,j Vi Wj ∂i ∂j F , which is a map ∇2 F (x) : R3 × R3 → R is a linear map. Nondegenerate means that ∀V 6= 0 ∈ T Z there exists a W ∈ T Z such that ∇2V,W P 6= 0. This is related to the
second fundamental form.
Proof. If we have a vector field, how can we check that the integral curves are straight lines? It suffices
to check that ∇V V is a multiple of V at every point. That is, a change in the tangent vector can only be
acceleration or deceleration.
From here, it’s a matter of being clear about second derivatives. There is another derivative called the
Lie derivative. We have LV F = ∇V F but LV (LV (F )) 6= ∇2V F . In fact, LV (LV F ) = ∇2V F + ∇∇V V F .
Now LV P = 0 on O, so 0 = LV (LV P ) = ∇2V P + ∇∇V V P = ∇∇V V P , so ∇V V ∈ T Z (in the tangent
space). So we’ve eliminated the possibility that the vector is turning into or out of the surface, but it could
be rotating in the surface, like those integral curves in the plane. Now we’ve got to use the third order
condition.
Let’s compute LV (∇2V,V F ). We took a while to do this, but eventually got LV (∇2V,V P ) = ∇3V P +
2∇2V,∇V V P . So from this, we have ∇2V,∇V V P . Since ∇2 P is nondegenerate and T Z is 2-dimensional and
∇2V,V P = 0, we must have ∇V V a multiple of V , as desired.
25
Wednesday, November 14, 2012
We have office hours after class today and Friday for the problem set.
Today is the last lecture on lines in 3-dimensional space, before we go on to talking about number theory.
The goal for today is to prove the regulus detection lemma. After that, we’ll take some time to package
everything together.
Lemma. If P ∈ R[x1 , x2 , x3 ], then ∃ a finite set of polynomials RP such that
(1) deg RP ≤ C deg P .
(2) If x lies in two lines in Z(P ), then RP (x) = 0.
(3) If P is irreducible and RP = 0 on Z(P ) and ∃x0 ∈ Z(P ) such that x0 is regular and not flat and lies in
two lines of Z(P ), then Z(P ) is a regulus.
The key feature of a regulus is that through every point pass two lines, i.e. it’s doubly ruled. The only
two doubly ruled surfaces in R3 are planes and reguli. The idea is that our polynomials RP will test whether
there are two different directions along which P is still zero. The technical condition says that at one point,
Z(P ) looks like a regulus, and it says that this germ of a regulus crystallizes in the entire surface.
Let’s find RP . If Qs (x) is a homogeneous degree s polynomial in v = (v1 , v2 , v3 ) ∈ R3 . Then let the
ideal I := hQ1 , Q2 , Q3 i and recall we defined I=d to be the homogeneous degree d polynomials in I, and H=d
those in R[v].
46
Lemma. The set {(Q1 , Q2 , Q3 ) : dim I=3 ≤ 8} = Z(R) where R is a finite list of polynomials in coefficients
of Qj .
Then define Qs,x (v) := ∇sv P (x). This is indeed a homogeneous polynomial in v of degree s. Then
RP := R(Q1,x , Q2,x , Q3,x ). So deg RP ≤ deg R deg P ≤ C deg P , so the first condition holds. We see that if
I(x) is the ideal generated by Q1,x , Q2,x , Q3,x so RP (x) = 0 iff dim I(x)=3 ≤ 8 (∗).
Now we claim that the point of defining RP is to detect when two directions are lines in the plane.
Lemma (1). If x ∈ Z(P ) is a regular point, ∇2 P : T Z × T Z → R has signature (1, 1) (i.e. has one
positive and one negative eigenvalue, corresponding to a saddle shape), so there exist two linearly independent
vectors ν1 , ν2 which vanish to second order: ∇2ν1 P (x) = ∇2ν2 P (x) = 0. Supposing that, RP (x) = 0 iff
∇3ν1 P (x) = ∇3ν2 P (x) = 0.
We’ll prove this first, then explain why we need that weird assumption.
Proof. Let I˜ = (Q1,x , Q2,x ). We claim that this is exactly ideal of polynomials that vanish on the line
through 0 and ν1 and the line through 0 and ν2 . It’s easiest to do this in coordinates, so linearly change
the coordinates so that x = 0 and Tx z is the v1 , v2 plane. So ∇P (x) = Q1,x (v) = cv3 for c 6= 0. Then
consider the map Q2,x → R[[v]]/(Q1,x ) = R[v1 , v2 ]. We know that Q2,x (ν1 ) = 0 and Q2,x (ν2 ) = 0, so choose
coordinates so that ν1 = (1, 0, 0) and ν2 = (0, 1, 0). That implies that Q2,x (v) ∈ (v1 v2 , v3 ), so I˜ = (v3 , v1 v2 ),
which is exactly the ideal of polynomials vanishing on these two lines.
So then we examine I˜=d , the set of polynomials that vanish at ν1 and ν2 . Because of this nice description,
we can read off its dimension easily as dim I˜=d = dim H=d − 2, and in particular, dim I˜=3 = 8. So all of
˜ So now RP (x) = 0 if and
that information is what comes out of this setup and the way we chose I and I.
only if dim I=3 ≤ 8. Remember, I was the dimension when we throw in Q3,x , so this is true if and only if
Q3,x ∈ I˜ if and only if Q3,x (ν1 ) = Q3,x (ν2 ) = 0, and this was just an abbreviation for ∇3ν1 P (x) = 0 and
∇3ν2 P (x) = 0.
Let’s take a step back. Any generic saddle shape has two lines which are tangent to second order, but
what we’ve found here is that the third order condition is relevant.
In general in projection theory, it’s hard to have a polynomial definition that tells us that there are two
different solutions. We can’t algebraically distinguish a double solution from two distinct solutions. The
classic example is a cylinder, which only has one line through each point but RP will still vanish.
So that’s why we need that ugly assumption. But once we have it, then we know where to look, since
we only have two directions where it vanishes. So then we just need to check whether the third-order part
vanishes in those directions as well. We encoded this in this dimension-counting condition, which is algebraic.
It is helpful to know a little bit about what happens in the opposite situation, where x is not a regular
point with that signature. Here are a couple facts:
• If x is a critical point, RP (x) = 0.
• If x is a flat point, RP (x) = 0.
These are in the notes, but are boring enough for us not to show in lecture. Now we can prove our next
lemma.
Lemma (2). If x is in two lines of Z(P ), then RP (x) = 0.
Proof. By the above points, we can assume that x is not critical and not flat. Let `1 , `2 be the two lines,
and ν1 , ν2 be the two tangent vectors. Then ∇sνi P (x) = 0 for all s, in particular for s = 2. Then ∇2 P :
T Z × T Z → R is nonzero since it is not flat, and has two independent null directions, so it must have
signature (1,1). So RP (x) = 0 iff these derivatives vanish to third order, which they do.
We’ve now checked the first two properties of the regulus detection lemma, and we’d like to prove the
last part.
47
Lemma (3). If P is irreducible, RP = 0 on Z(P ), and x0 ∈ Z(P ) is regular, not flat, and lies in two lines
of Z(P ), then Z(P ) is a regulus.
Proof. The first observation is that x0 satisfies the conditions of lemma 1. These are also open conditions,
so we can choose a small neighborhood O ⊂ Z(P ) where all x ∈ O are regular and ∇2 P (x) has signature
(1,1). So there are vector fields v1 , v2 so ∇2v1 P (x) = ∇2v2 P (x) = 0 where v1 , v2 are independent on O. Then
RP (x) = 0 so by Lemma 1, they are tangent to third order: ∇3v1 P (x) = ∇3v2 P (x) = 0 for all x ∈ O.
Last time in our class on local-to-global transitions, we proved:
Proposition. If O ⊂ Z(P ) is an open subset satisfing ∀x ∈ O, x is regular with ∇2 P (x) nondegenerate
(e.g. signature (1,1)) and V a vector field such that ∇sV P (x) = 0 for s = 1, 2, 3, then the integral curves of
V are straight line segments.
So the integral cuves of v1 , v2 are line segments. We’re very close to showing that this is regulus. Call
the integral curves of V1 line “horizontal lines” and integral curves of V2 “vertical lines.”
Claim. There exists a neighborhood of x0 where every horizontal line inersects every vertical lines and the
vertical lines don’t intersect each other.
In this case, three vertical lines determine a regulus, so this is looking suspicious. Choose three vertical
lines; if they’re all skew, then we have infinitely many horizontal lines in a regulus, which implies that Z(P )
is a regulus. If two are in a plane, then there are infinitely many horizontal lines in the plane (since the
two vertical lines don’t intersect in this neighborhood), and that would imply that Z(P ) is a plane, which is
impossible because of x0 . So it suffices to prove the claim.
Proof of Claim. Let L1 be the horizontal line through x0 and L2 the vertical line through x0 and change
coordinates so that L1 = {(t, 0, 0)} and L2 = {(0, t, 0)}. Then let L2 (t) be the vertical line through (t, 0, 0)
and L1 (u) be the horizontal line through (0, u, 0). We’d like to prove that these intersect.
Choose O small enough so that Z(P ) ∩ O is a graph x3 = h(x1 , x2 ), so L1 (t) is the restriction of the graph
to a line `1 (t) in the plane x1 , x2 plane, and similarly L2 (u) is the restriction of the graph to a line `2 (u).
Then `1 (t) is nearly horizontal, and `2 (u) is nearly vertical, so they must intersect and then the original lines
must intersect, as desired.
What are the counterexamples if you don’t include that x0 condition? There’s obviously the plane, but
you can also consider any surface with Gaussian curvature 0 everywhere. Cylinders are a typical example,
but there are others. Think of bending a piece of paper around.
26
Friday, November 16, 2012
We’ll start by briefly wrapping up incidence geometry. Let’s remind ourselves what we’ve set out to prove
and where we are.
Theorem. If L is a set of L lines in R3 with at most B in any plane or regulus and B ≥ L1/2 , then
|P2 (L)| . BL.
We didn’t really mention this result, but we’d proved this regulus detection lemma. On the last problem
set, we’ll prove this theorem from the regulus detection lemma.
Is this tight? Well, we can obviously put L/B planes with B lines each, and ∼ B 2 intersection points in
each plane. Is B ≥ L1/2 necessary? That’s not as clear, Guth doesn’t know a counterexample.
We also talked about what we can say when we have a lot of lines, which we can derive from the bound
on I(S, L).
Theorem. For 3 ≤ k ≤ L1/2 , |Pk (L)| . BLk −2 , and . L/k for k ≥ L1/2 .
48
26.1
Number Theory
We’ll spend the next couple of weeks on number theory, the first week being on a Diophantine equation
problem where a similar method was invented. We’ll start by getting background.
Diophantine equations are equations for which we want integer solutions. We’ll start with the famous
Fermat equation xd + y d = z d . When d = 2, there are many Z solutions, but when d ≥ 3 there are no Z+
solutions.
Let’s ask about the intuition behind this. Why might we expect this sort of behavior? Well, there are a
lot of squares; they are quite dense, but there aren’t a lot of cubes or higher powers. Let’s try to understand
this.
Let P be a polynomial in n variables which is homogeneous of degree d. We want to know how many
solutions there are to P (x) = A where A ∈ Z and x ∈ Zn . From an analysis perspective, let’s try to estimate
how big {x ∈ Zn : P (x) − A = 0, |x| ∼ 2s } is. Well, by the degree of P , |P (x)| . 2sd .
Naively, let’s model the polynomial with P̃ (x) an i.i.d. random integer in [−2sd , 2sd ]. Then we can easily
estimate E {x ∈ Zn : P̃ (x) − A = 0, |X| ∼ 2s } ∼ 2ns−ds .
This suggests some naive conjectures:
Conjecture.
1. If n > d, there are lots of solutions to P (x) = A, meaning infinitely many solutions or
they might grow like that.
2. If n < d, there are only finitely many solutions to P (x) = A.
Unfortunately, these are false. What are some counterexamples?
1. x21 + x22 + · · · = −1 has no solutions, but has degree 2.
2. x21 x22 · · · = 0 has infinitely many solutions, but degree 2n > n.
Can we fix these setups? What if we require having real solutions? The first doesn’t have any. Well, we
could take 4x2 + 4y 2 + 4z 2 = 3, which has no solutions. For the second one, what if we didn’t let A = 0? We
had several ideas, one of which was xn xd−1
+ P (x1 , . . . , xn−1 ) = 1. To get a bunch of solutions, let x1 = 1,
1
x2 , x3 , . . . , xn−1 arbitrary, and xn be whatever it needs to be.
So these naive conjectures aren’t true, but something similar actually is. For the first one, the HardyLittlewood circle method is the analog, and we won’t discuss it in this class. The other side is what we will
discuss:
Theorem (Thue). If n = 2, d ≥ 3, P irreducible. Then there are only finitely many solutions.
We can see that our counterexamples needed more than 2 variables, or were reducible. Another reducible
example would have been (x1 − x2 )1 0 = 1, which has infinitely many solutions.
Thue’s theorem is connected to Diophantine approximation, and is a corollary of an even stronger theorem
there.
Here’s a typical example: x3 − 2y 3 = 7. This problem is actually almost as hard as the general case. If
3
−3
we divide by y 3 and imagine x and y are rather large, then (x/y)
− 2 = 7y , which is small, so x/y is a
√
1/3 x −3
3
good approximation to 2. In fact, this easily implies 2 − y ≤ C |y| .
−2−
Proposition. For a.e. irrational number β ∈ R, ∀ > 0, β − xy ≤ |y|
has only finitely many Zsolutions.
This is a classic problem in measure theory. Guth quickly went over the proof. However, we’re approximating algebraic numbers, and they’re countable so this doesn’t actually say anything we care about. But
it’s kind of like the naive conjectures in that it would surprise us to find an infinite number of solutions, and
see that something special is going on here.
49
Proposition (Liouville). If β is an irrational algebraic number, Q(β0 = 0 for some Q ∈ Z[x], then β − xy ≥
c |y|
− deg Q
.
Proof. We can assume deg Q is minimal, so Q(x/y) 6= 0 (since otherwise we could take a linear factor out).
We can also assume Q0 (β) 6= 0 since otherwise Q0 would be minimal. Then we look at Q(x/y), and find
− deg Q
that |Q(x/y)| ≥ |y|
. So then the Taylor series gives us Q(x/y) ≈ Q(β) + Q0 (β)(β − x/y). Omitting
some details,
we can assume x/y is close to β so we can ignore the higher order terms. So then we have
− deg Q
x
0
|Q (β)| β − y ≥ |y|
.
If it’s a quadratic root, this says
√ about the same thing as the proposition. We talked through that
equivalence. But for something like 3 2, there’s a bit of a gap in its approximation. We can ask whether it’s
a typical number, i.e. matches the earlier proposition, or if you can approximate it better than that.
Theorem (Thue).
If β is algebraic and irrational, γ > deg(β)/2 + 1, then there are only finitely many Z
−γ
x
solutions to β − y ≤ |y| .
Notice that for deg β = 2, γ = 2 again, but for deg β = 3, γ = 5/2. This gets us halfway to the
proposition for general numbers. In fact, Roth proved in the 1950s that the original proposition holds for
algebraic numbers. His proof had significant new ideas, but built off of the same ideas that Thue had.
−3
Let’s go back to our Diophantine equation. We had a bound of 21/3 − xy ≤ C |y| . Now Liouville
−3
proved that 21/2 − xy ≥ c |y| , which doesn’t rule out solutions in between. But if we can improve the
bound slightly, even to y −2.99 , this rules out all arbitrarily large solutions. This is what Thue’s theorem
does, and why it implies the theorem we stated earlier.
Let’s sketch this curve x3 − 2y 3 = 7. It will do something possibly complicated in the middle, but will
asymptote out to the line x/y = 21/3 . This will happen in general, and the slopes will be the solutions to
P (β, 1) = 0.
Lastly, we’ll sketch the proof of Thue’s theorem, in analogy to the cleanest, nicest polynomial method
proof, that of the finite field Nikodym problem.
Assume that N ⊂ Fnq is a small Nikodym set.
(1) Find a nonzero P which vanishes on N , and bound deg P by counting parameters.
(2) Then P = 0 lots of other places (using the Nikodym set property).
(3) So P = 0 too many places, a contradiction.
Thue’s proof has the same structure. It’s also a proof by contradiction. Assume (r1 , r2 ) are very good
rational approximations of β.
(1) Find nonzero P ∈ Z[x, y] such that P (β, β) = 0 to high order. Get a bound on P and the size of the
coefficients of P by counting parameters.
(2) Then P (r1 , r2 ) also to high order.
(3) Then P vanishes to too high order at (r1 , r2 ).
The tricky part is going to two variables. You can’t do this with only one rational number. You can also
do this with higher numbers of variables, and that was one of the main things Roth did when he improved
Thue’s theorem. People knew how to do the first two steps for more variables, but the third step is somewhat
different and takes more work.
50
27
Monday, November 19, 2012
We’ll start by investigating Thue’s theorem. Let P be a polynomial with Z coefficients and r a rational
point. Then there’s a relationship between P (r) = 0 to high order and deg P and |P |, the maximum of its
coefficients.
Let’s do this in one variable. If ∂j P (r) = 0 for j = 0, . . . , l − 1, how small can we make its coefficients?
l
l
Well, our first guess is P = (qx − p)l , which has |P | ∼ |q| , |p| . This is related to the complexity of the
l
rational number r, defined as krk := max{|p| , |q|}, so |P | ∼ krk . Is this the best we can do?
Proposition (Gauss). If P ∈ Z[x] with ∂ j P (r) = 0 for j = 0, . . . , l −1, then P = (qx−p)l P1 with P1 ∈ Z[x].
Proof. We have P = (qx − p)l P2 with P2 ∈ R[x]. In fact, P2 ∈ Q[x] since we can solve for them. Moreover,
1
P̃2 with P̃2 ∈ Z with M not having any factor
we can pull out a common factor M , so P = (qx − p)l M
dividing all of the coefficients of P̃2 . Then M P = (qx − p)l P̃2 . Now if we let s be a prime divisor of M ,
working modulo s gives us an easy contradiction.
Well, we can go down in degree by multiplying:
Okay,
what light does this bear on our
original question?
(x − 1)2 = 2 but (x − 1)2 (x + 1) = (x − 1)(x2 − 1) = 1. However, q l divides the top coefficient, and pl
l
divides the least (nonzero) coefficient, so we still can’t do better than krk .
Let’s move on to two variables. Let P ∈ Z[x1 , x2 ] and r = (r1 , r2 ) ∈ Q2 , with ∂j P (r) = 0 for j ∈ J a list
of pairs (possibly ∀j, |j| ≤ l − 1). Can we say anything similar?
Let’s ask a simpler question: Suppose krk = max kr1 k , kr2 k is enormous, and P (r) = 0. Must |P | be
large?
No. We could have r1 = r2 and just P = x2 − x1 , as simple as can be. Well, what if we require that
∂j P (r) = 0 for j with 0 ≤ j ≤ l − 1? We could just do P = (x2 − x1 )l then. So that isn’t very helpful.
We need some sort of other definition of the complexity here that rules out them being equal. Say, the
coefficients of one as a rational function of the other.
Let’s take another approach: Only allow polynomials of the form P (x1 , x2 ) = P1 (x1 )x2 +P0 (x1 ). Suppose
∂1j P (r) = 0. What can we say?
We can plug in r2 and get ∂ j (q2 P1 + p2 P0 )(r1 ) = 0 for j = 0, . . . , l − 1, making this into Z coefficients.
l
l
1k
Then |p2 P1 + q2 P0 | ≥ kr1 k by what we already found, and this expression is about kr2 k |P |, so |P | & kr
kr2 k .
This is kind of a weird bound.
What example polynomials can we come up with given j, l? First, take P = q2 x2 − p2 , with |P | = kr2 k.
l
then ∂1j P (r) = 0 for all j (and any l). We can also just ignore P1 , giving us P = (q1 x1 −p1 )l , with |P | ≥ kr1 k .
We’ve done the two obvious reductions to one variable.
As a flashback, this reminds us of a problem we did on the first day of class, where we wanted to find
a polynomial that goes through the points (j, 2j ) for j = 1, . . . , 106 . We fumbled around with some ideas
that seem similar to restricting ourselves to one variable until just counting the dimensions. We think there
might be something similar at work here.
The fundamental thing we need to understand is integer solutions to linear systems.
Proposition. If L : Zm → Zn is linear and M ⊃ N , then ∃0 6= X ∈ Zm with LX = 0.
It is obviously true over a field, but the proof over the integers is even more elementary. Define QM
s :=
{X ∈ Zm : |Xi | ≤ S, i = 1, . . . , M }.
M
N
Sketch of proof.
M TheMmain idea
N is thatNL :NQS → QCS where C only depends
on
L. Then by the Pigeonhole
> QN . Therefore, there exist
principle, QS ∼ S and QCS ∼ C S . We can choose S so that QM
S
CS
X1 6= X2 with L(X1 ) = L(X2 ) and L(X1 − X2 ) = 0.
We’d like to figure out how big X needs to be, so let’s do this proof a little more carefully. First, C is the
operator norm of L, C = |L| = max|X|∞ =1,X∈RM |LX|∞ . In particular, we know that |L| is at most M times
the maximum of the coefficients of L. Now what can we choose for S? We need (2S + 1)N > (2 |L| S + 1)N ,
51
N/(M −N )
N
but let’s be careful. It suffices to have (2S + 1)M > |L| (2S + 1)N , so 2S + 1 > |L|
. Notice
N/(M −N )
that 2S + 1 is about what comes out for |X1 − X2 |, so we can say that |X|∞ ≤ |L|
. This is the
proposition we have.
So strategically, we see that we’ll do better when M − N is a significant fraction (e.g. 1%) of N , than if,
say, M = N + 1.
Let’s go back and compare parameter counting methods to the 1-dimensional case, where we came up
with (qx − p)l , even if it sounds weird.
Let r = p/q and we want to look for a polynomial P of degree D with ∂j P (r) = 0 for j = 0, . . . , l − 1.
PD
We have P = i=0 ai xi where the ai are the variables we get to play with. Each of these is an equation:
PD
PD
i!
i!
ri−j = 0. Clearing denominators, i=0 ai (i−j)!
q D−(i−j) p(i−j) .
∂j P (r) = i=0 ai (i−j)!
We have D + 1 variables, and l equations with Z coefficients. What can we say about the size of these
D
coefficients? We can bound the q and p component by krk . There’s a trick of the trade where we notice that
D
i!
j! | (i−j)!
, and that ji ≤ 2i . Then divide out by j! in each term and get a new bound of 2D krk = (2 krk)D .
l
So we can find a polynomial P of degree D with ∂j P (r) = 0 for j = 0, . . . , l − 1 and |P | ≤ ([2 krk]D ) R−l ∼
l D
1.01l
l
krk D−l . So for D = 100l and |P | ∼ krk
. The optimal formula of (qx − p)l gives D = l and |P | ∼ krk .
This is a general philosophy where someone might come up with a formula for somehting and then you
can ask, well, given the number of possible formulas taht could arise, is it surprising that we could come up
with one? We’ll tackle multiple dimensions next time.
28
Wednesday, November 21, 2012
Let’s recall the situation: r ∈ Q2 is very complicated, and we want to find P ∈ Z[x1 , x2 ] with ∂1j P (r) = 0
for j = 0, . . . , l − 1. We had taken deg2 P ≤ 1, i.e. P (x1 , x2 ) = P1 (x1 )x2 + P0 (x1 ). We had two examples:
l
(q1 x1 − p1 )l with |P | ∼ kr1 k and q2 x2 − p2 with |P | ∼ kr2 k.
Then we wanted to count parameters. We had a theorem that L : Zm → Zn has a nonzero solution X
N/(M −N )
with |X| ≤ |Lop |
≤ [M |Lcoef f |]N/(M −N ) .
PD
PD
Write P1 = i=0 bi xi and P0 = i=0 ai xi with D to be chosen later. Then we know that
X i i−j D−i+j
1
bi
p q
p2 + (similar expression with the ai ’s).
0 = q1D q2 ∂1j P (r) =
j!
j 1 1
i
D
This gives us l equations in 2D variables, with |Lcoef f | ≤ 2D kr1 k kr2 k, so there is a solution P with
il/(2D−l)
h
lD/(2D−l)
l/(2D−l)
D
= C l kr1 k
kr2 k
.
|P | ≤ C D kr1 k kr2 k
Now we can choose D so that the second term is small and the first term is close to l/2. That is, we have
Proposition. For any > 0, there exists P ∈ Z[x1 , x2 ] with deg2 P ≤ 1, partialij P (r) = 0 for 0 ≤ j ≤ l − 1
l/2+
.
and |P | ≤ C()l kr1 k
l/2
So we can approximately summarize our parameter counting example as this: |P | ∼ min(kr1 k
In the other direction, the best we can do is:
, kr2 k).
Proposition (Schneider).
If P ∈ Z[x1 , x2 ] with deg2 P ≤ 1 and ∂1j P (r) = 0 for j = 0, . . . , l − 1 and l ≥ 2,
(l−1)/2
then |P | ≥ min kr21 kdeg P , kr2 k .
So our upper and lower bounds are somewhat similar.
Proof. If ∂ j P1 (r1 ) · r2 + ∂ j P0 (r1 ) = 0 for j = 0, . . . , l − 1 we can write this using V (x) = (P1 (x), P0 (x)) as
∂ j V (r1 ) · (r2 , 1) = 0. So these are all vectors in the same direction. If V, W are vectors, let [V, W ] be the
matrix with columns V and W . Then we have det[∂ j1 V, ∂ j2 V ](r1 ) = 0 with j1 , j2 = 0, . . . , l − 1.
52
Claim. ∂ j det[V, ∂V ](r1 ) = 0 for j = 0, . . . , l − 2.
Proof. det[V, W ] is linear in V, W , and there’s a Leibniz rule: ∂ det[V, W ] = det[∂V, W ] + det[V, ∂W ]. So we
can expand ∂ j det[V, ∂V ] to get expressions of the form we know is zero.
l−1
Now, det[V, ∂V ] is a polynomial with integer coefficients, so by the Gauss lemma, |det[V, ∂V ]| ≥ kr1 k
or det[V, ∂V ] ≡ 0.
l−1
2
In the first case, kr1 k
≤ |det[V, ∂V ]| = |∂P1 · P0 − ∂P0 · P1 | ≤ 4(deg P )2 |P | (just multiply them out;
differentiating may introduce a factor of deg P , and multiplying might involve adding up to deg P terms).
This is just the first inequality on |P |.
In the second case, ∂P1 P0 − ∂P0 P1 = 0. In the first case, P1 ≡ 0 so P = P0 (x1 ) and this is just the
l
one-variable case, so |P | ≥ kr1 k . Otherwise, we can divide to get P0 /P1 which has zero derivative and is
therefore a constant, −r2 . So our polynomial has the form P = (x2 − r2 )P1 = (q2 x2 − p2 )P̃1 which must
have rational coefficients by an argument we saw earlier. So P is divisible by q2 x2 − p2 , so |P | ≥ kr2 k.
So in summary, we had these two trivial examples, and imagined that there were some better examples
of this form. Indeed, we found some and then bounded all the examples. We found that they were either of
these trivial forms or were at least this complicated.
We talked a bit about the analogy to finding the degree of a set of points. There’s a range in both cases
between very ordered rational points and, say, a plane, and the worst-case scenario.
Let’s go on and talk about polynomials that vanish at algebraic points rather than rationals. We can
rationally approximate those, but we’ll see that that won’t help us with them.
Proposition. If β is algebraic, given l, > 0, then ∃P ∈ Z[x1 , x2 ] with deg2 P ≤ 1 such that
• ∂1j P (β, β) = 0, j = 0, . . . , l − 1.
• |P | ≤ C(β, )l .
• deg P ≤ (1 + ) deg2 β l.
The proof is similar to the rational number one, but there’s one key idea about how to deal with algebraic
numbers. So we’ll just do a sketch.
P
P
Proof Ideas. Let D = deg P be chosen later. Then let P1 =
bi xi , P2 =
a1 xi , so we have 2D Z-variables
to choose. Let’s write down our equations to solve. We have
D
D
∂1j P (β, β) X
i i−j+1 X
i i−j
0=
=
bi
β
+
ai
β . (∗)j
j!
j
j
i=0
i=0
The difference with before is that we have equations with β’s in them, rather than integer coefficients. The
point is that (∗)j is equivalent to deg(β) equations with Z coefficients. That’s because 1, β, . . . , β deg(β)−1 is
a basis for Q[β] over Q. In particular, any β d can be written as a rational combination of 1, β, . . . , β deg(β)−1 .
So we can substitute this into the original formula, which is a little messy: (∗)j implies
"
#
deg(β)−1
X
X
X
bi Bik +
ai Aik β k
k=0
i
i
for some A, B ∈ Q. So this is equivalent to deg β equations with Z coefficients. If you go through and track
the bound we get, it’s |A| , |B| ≤ C(β)D . This bound is somewhat unpleasant to write out. You induct on
D, writing β D in terms of lower degree terms and adding the factor of a single constant each time you go
up a degree.
We have 2D variables, and l deg β equations, and we’d prefer to give this difference a little nudge:
deg P = D ≈ (1 + ) deg2 β l equations. Then |P | ≤ [C(β)D ]1/ and l ∝ D so this becomes C(β, )l as
desired.
53
Lastly, we’ll outline the proof of Thue’s theorem in this light.
Outline of Thue’s Proof. If β is an algebraic number, r1 , r2 very good rational approximations.
(1) Find P with integer coefficients and so on such that ∂1j P (β, β) = 0 for many j and |P | is pretty small.
(2) Because r is a good approximation to (β, β), ∂1j P (r) = 0 for many j. (This is like the vanishing lemma,
we haven’t discussed yet.)
l
(3) P vanishes for too many r. (The coefficients would need to be around krk but aren’t.)
A quick explanation of the second point: It divides into two steps. First, since P vanishes to high order and
r is a good approximation, by Taylor’s theorem, ∂1j P (r) needs to be very small. Second, these are rational
numbers and we can bound their denominators, so they can’t be too small without being zero.
29
Monday, November 26, 2012
Today we’ll finish proving Thue’s theorem. Let’s remind ourselves where we were before Thanksgiving.
Theorem. If β is an algebraic number, γ >
−γ
krk .
d(β)+2
2
then only finitely many r ∈ Q exist such that |β − r| ≤
Proof. Here’s our strategy: We’ll prove this by the polynomial method, by contradiction. So suppose there
are infinitely many solutions. Call two of them r1 and r2 , which have kr1 k , kr2 k very big.
(1) Find P that vanishes a lot at (β, β).
(2) Show that then P vanishes a lot at (r1 , r2 ).
(3) Show that P vanishes too much at (r1 , r2 ).
We did step 1 before the break. We saw that for any m ≥ 1 and > 0, ∃P ∈ Z[x1 , x2 ] such that
• ∂1j P (β, β) = 0 for 0 ≤ j < m.
• deg2 P ≤ 1 and deg P1 ≤ (1 + ) d(β)
2 m.
• |P | ≤ C(β, )m .
Now let’s do step 2.
m
m+1
Proposition. If kr1 k ≤ kr2 k ≤ kr1 k
∂1j P (r1 , r2 ) = 0 for 0 ≤ j ≤ c(β, γ)m.
, is sufficiently small, kr1 k big enough, and m big enough, then
The sketch
of this
is that by Taylor’s theorem, the jth derivative is very small near (β, β), which will
j
imply that ∂1 P (r) is very small. But it’s a rational number with a controlled denominator, so we’ll know
that it’s actually zero. Let’s quickly review Taylor’s theorem:
Pm−1 Pm−1 1 (i)
(x)hi + E, where
Theorem. If f : R → R is a smooth function, then f (x + h) =
i=0 i! f
i=0
1 (m)
m
|E| ≤ maxy∈[x,x+h] m! f (y) h .
So if a function vanishes to high order, then the sum there is zero, and the error term is an upper bound.
m
Corollary. If Q(x) ∈ R[x] and Q = 0 at x0 to order m with |h| ≤ 1, then |Q(x + h)| ≤ C(x)deg Q |Q| |h| .
1 m
Proof. We just need to check that the bound m!
∂ Q(y) ≤ |deg Q| 2deg Q |Q|)(|x| + 1)deg Q as desired.
54
Now let’s apply that here: P (j) =
1 j
j! ∂1 P
∈ Z[x1 , x2 ] has P (j) ≤ 2deg P |P | ≤ C(β, )m . Then we have
(j)
P (r1 , r2 ) ≤ P (j) (r1 , β) + P (j) (r1 , r2 ) − P (j) (r1 , β) .
−γ(m−j)
m−j
Letting Q(x) = P (j) (x, β), we can bound this by C(β, )m [|β − r1 |
+ |β − r2 |] ≤ C(β, )m (kr1 k
+
−γ
kr2 k ). In words, when we move the first coordinate, it’s dominated by the first nonvanishing term in the
m−j
Taylor expansion, which is |β − r1 |
, and when we move the second coordinate, it’s dominated by the
first nonzero term in that Taylor expansion, which is just |β − r2 |.
m
Now, which term is more significant? Well, we’ve chosen kr2 k to be bigger than kr1 k , so that term is
smaller and we can ignore it. What about the constant and the kr1 k? Well, we can take kr1 k to be as large
C(γ,β)m
as we want: so kr1 k ≥ C(β, ). Let’s make some notation: A . B means A ≤ kr1 k
B. So we can
(j)
−γ(m−j)
. That’s our first step.
write our upper bound as P (r1 , r2 ) . kr1 k
deg P deg P
But P (j) ∈ Z[x1 , x2 ] so P (j) (r) is a rational number with denominator q1 1 q2 2 . If P (j) (r) 6= 0, then
d(β)
d(β)+2
(j) P (r) ≥ kr1 k− deg1 P kr2 k−1 & kr1 k− 2 m−m . So we must have kr1 k− 2 m . kr1 k−γ(m−j) . Extracting
m ≤ −γ(m − j) + am. Manipulating this a bit, we see that γj ≥ [γ − d(β)+2
]m − am
exponents, − d(β)+2
2
2
and so we are assuming that is a positive fraction. We can choose small enough that it is still positive, and
therefore, we’ve proved the proposition.
Now this polynomial vanishes too much at (r1 , r2 ). Recall that if P ∈ Z[x1 , x2 ] with integer coefficients
(l−1)/2
1
with deg2 P ≤ 1 and ∂1j P (r) = 0 for j = 0, . . . , l − 1 with l ≥ 2, then |P | ≥ 2 deg
, kr2 k). In
P min(kr1 k
(cm−1)/2
m
our case, this implies that C(β, )m ≥ min(kr1 k
, kr1 k ). The second term will dominate, and then
if m is big enough, we conclude that kr1 k ≤ C 0 (β, ) as desired.
We had 15 minutes left, so what could we talk about with this? Where can we get estimates out of this?
−γ
Corollary. Let β, γ be as in Thue’s theorem. Then |β − r| ≥ c krk
for every r ∈ Q.
The most relevant quantitative question is, what is c? Let’s say for instance, β = 21/3 and γ = 2.9. This
relates to Diophantine equations like 2x3 − y 3 = 15. We might like to know some estimate of how big the
solutions could be because then we could just check on a computer everything up to that constant.
But our proof gives no estimates. That’s because we have to rely on two solutions, one of which is much
bigger than the other. So there could be some isolated very good approximation with no approximations
much bigger than it. There could even be a range of good approximations that are close enough. You might
be able to bound the number of these, but you can’t get any estimate on their size. There isn’t any evidence
of such an island, but it’s an artifact of this clever proof.
This method was very influential.√ In transcendental number theory, it was used by Gelfand and Schneider
to prove that numbers like log2 3, 2 2 and others were transcendental. Their proof shared in common some
auxiliary function P (not a polynomial in this case). You do some parameter counting to conclude that it
vanishes at a bunch of points. This was a potential unit in this course, but there wasn’t time.
Some bureaucratic things to mention: The last problem set has been graded and was handed back. The
last problem set is up and will be due in a week and a half, so it’s shorter than usual. There will be no
class on Friday since Larry will be away for a couple of days. For the “summarize the proof” problems on
the problem set, the last problem set asks you to imagine you’re in a coffee shop and describing the proof of
Thue’s theorem to a friend.
30
Wednesday, November 28, 2012
Today we’re going to look at how combinatorics and analysis interact. We’ll be studying the Loomis-Whitney
Inequality.
55
Consider a unit cubical lattice in Rn . Let X be a set of these unit cubes, and consider the projections
πj onto the hyperplane perpendicular to xj . We want to know: How big can |X| be if we know the size of
|πj X|.
Theorem (LW 50s). If |πj X| ≤ A, then |X| . An/(n−1) .
The original proof uses Hölder a lot, so you can practice that if you want. The constant is 1, achieved
for the cube.
Proof. Define a column to be a line in the xj direction from any cube.
P
Lemma. If
|πj X| ≤ B, then there exists a column of cubes with between 1 and B 1/n−1 cubes of X.
Proof. Suppose not, so every column has > B 1/(n−1) cubes. Look in a second direction from each of the
B 1/(n−1) cubes in a given column, and continue this until there are > B cubes in a given x1 , . . . , xn−1 -plane,
a contradiction.
P
Now as a corollary, we claim that |X| ≤ B n/(n−1) . Indeed, remove a small column, so
|πj X 0 | ≤ B − 1.
0
n/(n−1)
1/(n−1)
0
By induction, |X | ≤ (B − 1)
. Then |X| ≤ B
+ |X |, so the result follows.
Theorem (LW). If U is an open set in Rn , |πj U | ≤ A for each j, then |U | . An/(n−1) .
Proof. Approximate U by U which is a union of -cubes in an -lattice. Then |U | ≤ An/(n−1) , and we can
take |U | → |U |.
Corollary (Isoperimetric inequality). If U is a bounded open set in Rn , then Voln U . (Voln−1 ∂U )n/(n−1) .
Proof. Indeed, if we project, the image will be at least as big as the boundary. (This is where we use that
it’s bounded.) So |πj U | ≤ Voln−1 (∂U ), and the result follows.
R
1
Next we’ll look at the Sobolev inequality. Question: Suppose u ∈ Ccomp
(Rn ) with |∇u| = 1. How big
can u be? We can’t make u short or tall, because there are examples that would contradict any bounds we
could make.
Theorem. kukLn/(n−1) . k∇ukL1 .
Recall the Lp norms:
R p 1/p
Definition. kukLp =
|u|
.
The natural example to look at is hχA , and then khχA k = h |A|
Given u, we’ll let S(h) = {x ∈ Rn : |u(x)| > h}.
1/p
.
Proposition. If kukp ≤ M , then |S(h)| ≤ M p h−p .
R p
Proof. We have M p = |u| ≥ hp |S(h)|.
It isn’t equivalent, but a bound on kukp is related to a bound on |S(h)| for all h.
Proof of Sobolev inequality. First, we’ll prove a lemma.
1
Lemma. If u ∈ Ccomp
(Rn ), |πj S(h)| ≤ h−1 k∇ukL1 .
R
Proof. Let x ∈ S(h), and consider the line through x in the xj direction. Then ` |∇u| ≥ h. Indeed, since
the function is compactly supported, u = 0 somewhere on this line, and then the integral of the gradient
along the line will be at least this difference |u(x)| > h. So
Z
Z
Z
k∇ukL1 ≥
|∇u| =
|∇u| dxj dx−j ≥ |πj S(h)| h,
πj S(h)×R
πj S(h)
as desired.
56
R
n/(n−1)
By Loomis-Whitney, |S(h)| . h−n/(n−1) k∇uk
. So this is almost the same as bounding the Lp
norm of u, but it’s not and we have to fix things. These are the main ideas, though.
Why can we hope to get a little more leeway? Well, we’ve combined everything of a given height
h first, but we could first do some integrating without that. Let’s change our definitions slightly: Let
Sk := {x ∈ Rn : 2k−1 ≤ |u(x)| ≤ 2k }. We have a slightly different lemma:
R
1
Lemma. If u ∈ Ccomp
, |πj Sk | 2−k Sk−1 |∇u|.
Proof. Considering x ∈ Sk−1 , again consider a line ` in the xj direction, and the function reaches u = 0 at
some point.
to x within Sk−1 , the function has to change for at least 2k−2 (from 2k−2 to
R So in getting1 up
k
k−1
2
), so Sk−1 ∩` |∇u| ≥ 4 2 .
n
Corollary. |Sk | . 2−k n−1
n/(n−1)
|∇u|
.
Sk−1
R
So the proof of Sobolev’s lemma is that
Z
|u|
n/(n−1)
∼
∞
X
k=−∞
kn/(n−1)
|Sk | 2
.
X Z
n/(n−1) Z
≤
Sk−1 |∇u|
n/(n−1)
,
|∇u|
Rn
k
as desired.
In summary, the Sobolev inequality corresponds to this statement about sizes of sets with small projections, and that statement came down to this combinatorial cube counting problem that we could solve in a
similar way to the stuff we’ve been doing.
The Kakeya problem relates to some topics in analysis related to Lp estimates of linear operators.
Let’s describe some things we might do with functions:R
Convolution: f, g ∈ Rn → R (or C), then f ∗ g(x) := Rn f (y)g(x − y)dy. Let’s hear a story describing
this. Let’s suppose we have a factory that produces a cloud of pollution g(−y). In fact, there are many
factories that produce the same cloud of pollution, and you could have many factories you might want to
approximate with something continuous. So if f (x) is the density of factories generating pollution, f ∗ g(x)
is the shape of the pollution.
R
−α
−α
Example. Let Tα f := f ∗ |X| . So Tα f (X) = f (y) |x − y| dy. If 0 < α < n and f is integrable with
compact support, this integral will converge. This is a linear operator, and it’s related to questions in PDE
like the evolution of the wave equation. Those aren’t technically convolutions, but they’re similar enough
that we can study these.
So let’s try some f . Try χB1 , the characteristic function on the unit ball. So if |x| ≤ 1, then this will
−α
be about constant, and if |x| > 1, |x − y| ≈ |x|, so this falls off as |x| . What about a bigger ball: χBr .
−α
Again, if |x| ≤ r, we’ll get approximately rn r−α , and if |x| > r, it will fal of as rn |x| .
Any other examples? A delta function might be reasonable, but it’s just going to be similar to the limit
for small r. Yeah, there aren’t many more examples to think about.
Now we can start to pose math questions about these things, and see whether they capture all of the
behavior. This is where Lp estimates come in.
Question: Fix α, n. For which p, q is there an inequality kTα f kq . kf kp for all f . Let’s figure these
things out in our examples first, though that’s really easy. Then we can ask if these examples show all that
there is.
R
q
In the first example, kχB1 kp ∼ 1, while kTα χB1 kq q ∼ Rn (1 + |x|)−αq dx. This could be infinite, and is
finite if αq > n. If it’s finite, the inequality is true since it’s bounded by a constant, so the question works
for αq > n. In fact, let’s assume this from now on.
Then let’s do example 2. Now kχBr kp ∼ r1/p , and we need to estimate kTα χBr kq . If you think about it,
if αq > n, it means the tail doesn’t really matter. So this is basically the same as krn−α χBr kq ∼ rn−α+n/q .
So this holds in example 2 if and only if n/p < n − α + n/q and αq > n. So we need the exponents to be
57
equal, because otherwise we could look at small and large r. Therefore, n/p = n − α + n/q. So now the
question we can ask is whether this inequality holds for all functions. We’ll answer that question next time
on Monday.
31
Monday, December 3, 2012
We started about 8 minutes late today. Last time, we were studying a particular operator, Tα f := f ∗ Kα ,
−α
where Kα (x) = |x|
for 0 < α < n. We looked at f = χBr as an example to understand the behavior. We
then proved:
Proposition. kTα χBr kq . kχBr kp iff n − α +
n
q
=
n
p
and αq > n (or equivalently p > 1).
So we wondered whether balls tell the whole story, and in this case, they are.
Theorem (Hardy-Littlewood-Soboler). If n − α +
n
q
=
n
p
and p > 1, kTα f kq . kf kp .
P
Our goal today will be to prove this. For now we’ll consider a slightly different case of f = j χBj
where Bj are some balls. So it helps to think about the geometry and combinatorics of overlapping balls to
think about general functions. So we’re going to take a break and study the geometry and combinatorics of
overlapping balls. The main result there is:
Lemma (Vitali S
Covering Lemma).
If {Bi }i∈I is a finite collection of balls. Then ∃J ⊂ I such that {Bj }j∈J
S
are disjoint but i∈I Bi ⊂ j∈J 3Bj where 3Bj is a ball with the same center and three times the radius.
The idea is that you take a greedy algorithm, taking the biggest ball that’s still eligible, and it works
out. What happens if you take infinite I?
Here’s a bad example: B(0, r) for r ∈ Z+ . So no two of them are disjoint but you can’t cover them all
with only one.SSo how would we handle this? Analysis textbooks
soften the conclusion by saying for any
S
compact K ⊂ i∈I Bi , we can choose some J(K) with K ⊂ j∈J(K) 3Bj .
We’ll use this to understand some geometry and analysis of how balls overlap.
S S
Lemma (Ball doubling). If {Bi }i∈I is a finite collection of balls, | 2Bi | ≤ 6n Bi .
Proof. By the
covering lemma
(which we didn’t see), Bi ⊂ 3Bj for some j, so 2Bi ⊂ 6Bj .
S Vitali S
P proof S
Therefore | 2Bi | ≤ | 6Bj | ≤ 6n |Bj | ≤ | Bi |.
Is this sharp? It seems it should be 2n , but that’s at least hard to prove. These results are morally
involved, but not literally involved in the proof we’ll give now.
H
R
1
Definition. Let A f := Vol(A)
f , the average value. The Hardy-Littlewood maximal function is M f (x) :=
A
H
supr B(x,r) |f |.
Also let Sg (h) := {x ∈ Rn : |g| > h}.
Lemma. |SM f (h)| . h−1 kf k1 .
H
R
Proof. Suppose x ∈ SM f (h). Then ∃r(x) such that B(x,r(x)) f ≥ h, or B(x,r) |f | ≥ h |B(x, r)|. The idea is
R
that these balls cover SM f (h) so by the Vitali covering lemma, there are Bj disjoint with Bj |f | ≥ h |Bj |.
R
P
Then |SM f (h)| . j |Bj | ≤ h−1 S Bj |f | ≤ h−1 kf k1 .
Proposition. kM f kp . kf k1 for all 1 < p < ∞.
This seems relevant, and we can connect them in the following way:
(2k ) = {x ∈ Rn : 2k <
R Let pTM fP
∞
k+1
k
|M f | ≤ 2
} ⊂ SM f (2 ). Then using the estimate we just proved, |M f | ∼ l=−∞ TM f (2k ) 2kp .
P −k kp
2 kf k1 which unfortunately diverges. But there’s a way to tweak it so that this theorem pops out.
k2
58
Lemma. |SM f (h)| . h−1
R
Sf (h/2)
Proof. Start the same as before:
h
2
|f |.
R
Bj
|f | ≥ h |Bj |. But
R
Bj \Sf (h/2)
|f | ≤
h
2
|Bj |. So indeed,
R
Bj ∩Sf (h/2)
≥
|Bj |. Then everything else is the same.
If we plug this in and do the same computation, we’ll get the result we want.
R
R
P∞
P∞
p
Proof of Proposition. We have |M f | ∼ k=−∞ SM f (2k ) 2kp . k=−∞ 2kp 2−k Sf (2k−1 ) |f |. We’d like
R
R
R
P
p−1
p
to switch the sum and the integral here, and get Rn |f | 2k−1 ≤|f | 2k(p−1) ∼ Rn |f | |f |
= Rn |f | .
Let’s quickly review what we’ve done so far. We defined this Hardy-Littlewood maximal function and
then looked
at when it could be big. You use that bound on the maximal function and turn a 2k -crank,
R
p
getting |Mf | . kf kp . Now we’ll prove the Hardy-Littlewood theorem, Guth’s best attempt to tame the
number of minor computations.
Proof. Here’s the three-step summary:
H
1. Tα f (x) can be written in terms of B(x,r) f . That is,
H
R∞
Lemma. Tα f (x) = 0 rn−α−1 B(x,r) f dr.
Proof. This is just a computation. Just apply Fubini to the right-hand side and it works out, because
Kα is spherically symmetric.
2. Bound above
H
B(x,r)
f . How could we do that? We could just do
H
f ∼ r−n
B(x,r)
−n
R
B(x,r)
n(p−1)/p
|f | ≤ r−n
R
|f |.
But what if p > 1? We could use Jensen/Hölder instead to get . r kf kp r
= r−n/p kf kp .
Alternatively, we can bound by M f (x). We’d like to use both upper bounds. How do we know which
one to use where? You might think you want to look at where r > 1 and r < 1, say, but actually the
best thing to do is to look at whichever one of these is smaller.
We’ll pick rcrit (x) the critical radius. Use M f (x) for r ≤ rcrit and the other bound for r ≥ rcrit . Then
B
you can do the calculation and get |Tα f (x)| . M f A kf kp for some A, B with A + B = 1. It doesn’t
depend on what you get there, though, so we won’t worry about that.
R
q
Bq R
Bq
Aq
β
α
M f Aq . rcrit
M f + rcrit
3. |Tα f | . kf kp
kf kq . kf kp kf kAq . (The second step is where we plug
in the rcrit which minimizes that expression. α, β, A, B are some exponents that aren’t especially
q
important. Well, that’s interesting, but it would be useful if Aq = p, in which case we get kf kp . So
we need A = p/q and B = (q − p)/q. These become exactly the first condition that we needed in the
hypothesis. You might even be able to make a rigorous proof without checking.
So we see that analyzing this operator amounts to knowing how balls intersect. Later we’ll look at long
narrow tubes and see that those intersect in a similar way to lines intersecting, so we can apply things like
Szemeredi-Trotter.
59
32
Wednesday, December 5, 2012
Today we’re looking at Fourier analysis, but it’ll be bracketable.
R
0
If f : Rn → R or C, the Fourier transform is F̂ (ω) := Rn f (x)e−2πiωx dx. This is well-defined if f ∈ Ccomp
.
This captures the plane waves in different directions.
R
Theorem (Fourier Inverse Theorem). f (x) = Rn fˆ(ω)e2πiωx dω.
∞
If f ∈ Ccomp
, then fˆ decays strongly, so this integral is defined.
R
A natural thing to look at is MR f (x) := B n (R) fˆ(ω)e2πiωx dω. We can then ask if f = limR→∞ MR f for
0
f ∈ Ccomp
. These things were discovered about this:
• MR f (x) does not converge at all for a particular x ∈ Rn .
• MR f → f in L2 .
• MR f → f in LP for all 1 < p < ∞ in dim n = 1 (Riesz).
We’d like to understand better to what extent this is true in dimension n > 1. Now, if you understand well
what happens when you restrict to the unit ball B n (1), then you can sort of rescale to understand other
balls, so we’d like to know when kM1 f kp . kf kp .
Why is this? Well, suppose we had ∀B, some function fB satisfying kM1 fB kp ≥ B kf kp . Then we
could make this into functions fB,R with kMR fB,R kp ≥ B kfB,R kp . We could construct a bad function from
1
1
f100,2 + 100
f10000,3 + · · · , so it’s clear we’d like this to be true. The other direction
something like f10,1 + 10
is a complicated functional analysis argument saying that this is all that can go wrong. So at least modulo
these explanations, this is something we’re interested in.
−α
Last time, we studied Kα (x) := kxk
and defined Tα f = f ∗ Kα . We can modify these a little bit
by introducing some oscillations, and instead look at K̃α (x) := [1 + |x|]−α cos |x| and T̃α f := f ∗ K̃α . In
this sense, M1 f is similar to T̃α . Unfortunately, T̃α is not very well-understood and it’s a well-known open
problem to characterize it completely like we did for Tα .
Let’s try to understand it anyways. Let’s start with some examples.
R
Example.
1. f = χBr . Then T̃α f = Br [1 + |x − y|]−α dy. What happens? Well, the hard part is the
cosine term, which introduces some oscillation. There are positive and negative regions. Let’s try to
understand some limiting behavior.
p
When r < 1/100, there isn’t really any cancellation. Then T̃α χBr ∼ cK̃α . So if kf kp ∼ 1, T̃α f ∼
p
R
[1 + |x|]−αp < ∞ only if αp > n.
R
2. We could also try to make T̃α f (0) = f (y)[1 + |y|]−α cos |y| dy large. So take f (y) = χBr cos |y|, for
R
r ≥ 1. Then T̃α f (0) ∼ Br [1 + |y|]−α dy ∼ rn−α . How far away do we have to go to make this not
true? Something like π/2 because that’s when thecosines
will be out of phase. So we can safely say
that it’s about this big for |x| < 1/100. So in all T̃α f & rn−α . Therefore, since kf kp ∼ rn/p , we
p
need n/p ≥ n − α.
n
<p≤
So puttting together what we’ve discovered from these examples, we need α
we got to this point, we raised the question of whether these estimates are enough.
n
n−α .
Last time, when
First, why do these give opposite bounds? Well, in the first example, we had a small ball, so our incoming
function was localized and the outgoing function was rather spread out. In the second, our incoming function
is spread out over a large ball, and our outgoing function is rather localized.
This reminded Guth of something Terry Tao wrote about PDEs. In some, we have initial data u at
time 0 and final data T u at time 1. In many examples, we have kT uk2 = kuk2 . You can ask whether
60
kT ukp . kukp for various p. This can be phrased as whether energy is concentrating or diffusing, where the
2
energy is e(u) = |u| .
Diffusion of energy is like going from a tall rectangle (e(u)) to a flat rectangle (e(T u)), and concentration
is the reverse. Suppose we had exactly that situation, where e(u) has height h20 and width A0 , and similarly
p
p
p
define h21 and A1 . So kukp ∼ hp0 A0 . Then in diffusion, kT ukp increases for p < 2 and in concentration, kT ukp
increases for p > 2.
Back to our examples, the first one was an example of diffusion and the second was an example of
concentation.
We’ve just looked at spherically symmetric functions so far... do we have to? We talked about this for
a while. We could split up an arbitrary function into a radial part and an asymmetric part f0 that’s 0 on
every sphere centered at the origin. Then T̃α f0 (0) = 0, but we only know at the origin.
Then we talked about how a few small balls might “interfere.” It isn’t quite analogous to a dipole moment
setup, because there could be strong positive interference. It’s more analogous to the double slit experiment.
Example. There’s an example which gives a large value, the long thin tube. Let T be a cylinder of length L
1
L1/2 . Define fT = χT eivT α and T + = T2 vT L, a second shifted cylinder along the axis.
and radius 1000
Proposition. T̃α fT (x) & L(n+1)/2−α on T + .
eivT y cos |x − y| [1 + |x − y|]−α dy. Now |x − y| is almost |vT · x− vT · y|:
2
|x − y| = |vT x − vT y| + s2 , where s ≤ .002L1/2 , so s2 ≤ ·10−5 L. So |vT x − vT y| − |x − y| ≤ 10−5
Proof. The left side equals
2
R
T
2
because |x − y| ≥ L. We can then take this out of the integral, or up to a small error jsut write it as
cos(vT x − vT y). Then write cos(a) = 21 (eia + e−ia ), so we get one term with eivT ·x and another term with a
R
similar oscillation. That first term gives us T [1 + |x − y|]−α dy, which goes as L−α Vol T ∼ L(n+1)/2−α , as
desired.
We can think of this example as a way to recover our ball from one dimension which had perfect cancellation in multiple dimensions. If our cylinder is small enough, it’s a legit example.
33
Friday, December 7, 2012
Today, we get to see the punchline in the unit about analysis. Let’s remind ourselves of what was going on.
We were trying to understand the Lp norms of some operators. The operator we were looking at was
based on K̃α (x) = (1 + |x|)−α cos |x|, T̃α f := f ∗ K̃α . We were plugging in example functions to this, and
had originally only tried radially symmetric functions. Then we looked at a tube T with length L ≥ 1 and
width L1/2 /1000. Letting vT be the unit vector along the tube, we defined fT = eivT ·x χT . We then proved:
n+1
Proposition. T̃α fT (x) & L 2 −α on T + .
Corollary. If α <
n+1
2 ,
However, if α <
case is α = n+1
2 .
n+1
2
then T̃α f . kf kp is false for all p.
p
wouldn’t tell us anything, because we’d get something smaller. So the interesting
Theorem (Fefferman 71). Bounds T̃ n+1 f . kf kp is false for all p 6= 2.
2
p
P
Proof. We’ll prove it for p > 2. The idea is to look at many tubes Ti , and take f i fTi . Suppose the Ti are
+
all disjoint, but maybe the Ti are not. This raises a question: Can you put the tubes in such a way that
these translated tubes all intersect in some good way? Besicowitch came up with a cool arrangement:
61
Theorem (Besicowitch,
For
1, there exists a collection of tubes
S 20s).
S any L ≥P
S Ti (as above) such that the
1
1
| Ti |, and
χT + ∼ K on a set of size ∼ K
| Ti |, where K & (log L).99 .
Ti are disjoint, but Ti+ ≤ K
i
It isn’t quite enough to just intersect all of the Ti+ . Suppose x lies in K tubes Ti+ . Then T̃ n+1 f (x) has
2
about K contributions of size around 1. However, not all of these are necessarily positive, so there could be
contributions. As
√ a heuristic, take K values of ±1 with equal probability and independent, then the absolute
value is around K with high probability. So it isn’t likely that we get ∼ K.
For any particular set of tubes, it’s tricky to figure out how these things cancel out. But something else
is easier:
P
Proposition. If gi are some
take gran =
±gi where the signs are taken randomly. Then for
functions,
1/2 P
2
with high probability.
|gi |
all 1 ≤ p ≤ ∞, kgran kp ∼ p
We won’t give the detailed proof of this. It’s similar to the point in the course when we wanted with
high probability some value to be
P close to its expected value.
S
So apply this to get fran =
±fTi . Then fran ∼ 1 on a set of size | Ti | since the Ti are disjoint. But
R
R
S
S
p
p
1
T̃ n+1 fran ∼ K 1/2 on a set of size ∼ K
| Ti |. So taking the Lp norms, |fran | = | Ti | and T̃ n+1 fran ∼
2
S
p
K 2 K −1 | Ti |. If p > 2,
2
T̃ n+1 fran 2
p
→ ∞, as desired. It remains to find the tubes.
What happens when p = 2? You can prove that T̃ n+1 f . kf k2 . This has some relationship with
2
2
Fourier analysis.
How do we make the tubes? Let’s first make some lines `j (x) = Nj x + H(j). Let Rj be the 1/N neighborhood of `j , for j = 0, . . . , n − 1.
PA
Let N = AA , where A is an integer and work in base A. So Nj = a=1 j(A)A−a . Then define the heights
PA a
j(a)A−a . We’ll work out the proof algebraically, then try to understand what’s going on
H(j) = − a=1 A
S
log N
.99
.
geometrically. The theorem says that | Rj | ≤ 10
A . Then A log A = log N , so A ≥ log log N ≥ (log N )
j
J Lemma. If j(A) = J(a) for a = 1, . . . , b − 1. Then N − N
≤ A−(b−1) .
PA
Proof. The first part is easy because the base-A representations coincide. Then Nj x = i=1 xj(a)A−a . We
P
PA a
A
a
j(a)A−a . When we subtract these,
defined H(j) = − a=1 A j(a)A−a . Therefore, `j (x) = a=1 x − A
the first b − 1 terms disappear, and if we plug in x = Ab , the bth term disappears. The remaining terms have
size ≤ A−b , and we have two sets of them, so the lemma follows.
kfran kp
b
Corollary. If j(a) = J(a) for a = 1, . . . , b − 1, |`j (x) − `J (x)| ≤ 4A−b for x ∈ [ b−1
A , A ].
Proof. They are within 2A−b at b/A, and their slope difference is at most A−(b−1) over the interval A−1 , so
there is a change of at most A−b , as desired.
S
b
(b−1)
Corollary.
Rj ∩ ([ b−1
horizontal strips (really, parallelograms) of width
A , A ] × R) is covered by A
−b
6A .
Proof. There are Ab−1 choices for j(1), . . . , j(b − 1). Then there is a strip within these of this width that
covers all of the possible lines.
Corollary. The area of the Rj is at most 10A−1 .
Proof. Just add up the area within each strip according to the last corollary.
62
That’s the proof, but we also saw a picture of what this looks like. We took A = 2, so N = 22 , pretty
much the only manageable case. There were some pretty pictures with colored chalk.
There are some more bells and whistles to check: that when we slide the rectangles to the right to [2, 3],
they become disjoint, and that there are a lot of points in this construction contained in many rectangles.
We’ve thought through T̃ n+1 , but what about T̃α for α > n+1
2 ?
2
Conjecture (Bochner-Riesz). If α > n+1
2 , then T̃α f . kf kp for
p
n
α
<p<
n
n−α .
The Besicovitch construction would seem to contradict this, but its compression is only something like
log L, while when α > n+1
2 , we would get some power of L. In other words, this conjecture would imply
Conjecture
(Tube
S + S translation conjecture). If Ti are tubes of length L as above, > 0, the Ti disjoint, then
T ≥ c L− | Ti |.
i
Next time, we’ll look at the questions that people care about involving tubes and how those relate to
lines and the Kakeya problem.
34
Monday, December 10, 2012
Now we finally get to write down the Kakeya problem. We’ve been trying to motivate it from analysis up
to this point.
Let Ti ⊂ Rn are cylinders of length N and radius 1. Then {Ti } is a Kakeya set of tubes if v(Ti ) are
1
2
n−1
, where v(Ti ) are unit vectors in the “direction of the tube.”
N -separated and N -dense in S
The main question: How small can a Kakeya set be?
Last time, we gave a construction of Besicovitch which started with a bunch of rectangles and compressed
them by a factor of ∼ 1/ log N . We did this in two dimensions, but you could do it in higher dimensions.
We want to know whether or not this is the right compression factor, or if you can get polynomial
compression. As an aside, to get evenly spaced directions, you need ∼ N n−1 tubes, so the way we constructed
these, the total volume if they didn’t intersect would be N n .
S
Conjecture (Kakeya). For every Kakeya set of tubes in Rn and , | Ti | ≥ c N n− .
We can see the connection with the finite field Kakeya problem we mentioned earlier in the class. Technically, it implies:
Proposition (Segment version of the Kakeya problem). K ⊂ Rn is a set containing a unit line segment in
each direction. (1) There exists a Ki with m(Ki ) → 0. (2) There exists K with m(K) = 0.
Besikovitch’s construction, scaled down, implies the first of these. The second requires some slight
modifications to just taking the intersections, because they aren’t nested. This is a bit different from
Conjecture (Hausdorff dimension Kakeya conjecture). For every > 0, the Hausdorff dimension of K is
at least n − .
This implies our version of the Kakeya conjecture, which Guth prefers because it’s more combinatorial
and doesn’t require you to define Hausdorff dimension. Anyways, let’s continue investigating our Kakeya
problem.
Proposition. The Kakeya conjecture is true in 2 dimensions.
Proof. Given two tubes with angles θ1 , θ2 , the overlap is .
1
|θ1 −θ2 | .
Plugging in this estimate, we get
Z X
2 X X Z
XX
χ
χTi χTj =
|Ti ∩ Tj | . (log N )2 N 2 .
Ti =
i
j
i
63
j
On the other hand, by Cauchy-Schwarz,
2
N =
Z X
χ Ti
Z X
[ 1/2
2 1/2 [ 1/2
. (log N )N Ti .
χTi ·1≤
Ti S
Unwinding this, we get | Ti | & N 2 (log N )−p for some power p that we don’t want to work out. This implies
the Kakeya conjecture.
We see that this is related to Lp estimates on the Kakeya sets, so there’s even a stronger statement.
R P
p
Conjecture (Lp version). | χTi | . (N ) · (what happens if all of them are centered at 0).
These conjectures imply the Kakeya conjecture volume version by the same argument.
There are then versions of the bush and hairbrush arguments for tubes:
• Bush argument: For finite fields, we got & q
n+1
2
, and for the Rn version, we get & N
• Hairbrush argument: For finite fields, we got & q
n+2
2
n+1
2
.
, and for the Rn -version, we get & N
n+2
2
.
The bush argument is easier to see. Pick some point in a lot of cylinders, and while the cylinders might
overlap, move out a distance of N/10 from that point, and after that, you get at a size of at least N times
the number of tubes in the bush.
For the hairbrush argument, we’d like to pick some tube and look at those tubes that intersect it. Again,
we need to go out a distance of at least N/10. We run into the problem that when you do this, some tubes
might not even make it out that distance. It’s hard, but you can rule out that case; Tom Wolff did that in
the mid-90s.
It’s been remarkably hard to improve on this. Kats-Laba-Tao produced a very tiny improvement under
some minor assumptions. Being stuck at this point, Tom Wolff proposed a bunch of toy problems related to
this. One of these was the finite field Kakeya problem that we looked at at the beginning of this class.
Another problem he looked at was replacing lines in different directions with circles of different radii in
the plane. So we look at annuli with thickness 1/N and radii of distance between 1 and 2. That problem is
still 2-dimensional, but it’s harder than the 3-dimensional problem. In order to solve that, he used incience
geometry, stuff related to Szemeredi-Trotter. That was cool because it was brought a whole different set of
techniques to this area, so people in harmonic analysis learned about this area of mathematics.
Of course, when Dvir solved the finite field Kakeya problem with the polynomial method, that raised the
question of whether we could say anything about tubes with this method.
Let’s recall the main ideas of the finite field Kakeya proof. If K ⊂ Fnq is a Kakeya set with |K| = q n−γ ,
look at a nonzero polynomial P with P = 0 on K and deg P q. That implies that the polynomial vanishes
at some other places, and the ball is rolling.
Let’s try to adapt that. We suppose that K ⊂ Rn is a Kakeya set of 1 × N tubes, and |K| = N n−γ .
What could we do that’s analogous to picking polynomial P ? We opened this question to the class.
1. P = 0 on the center lines. But these could be disjoint; this condition is stable and we can take a
random perturbation.
2. P = 0 on ∂Ti , so P = 0 on the infinite surfaces.
3. Ham sandwich idea: P = 0 roughly bisects the tubes. We would like it if they did it the “hard way,”
which means along the axis, not the “easy way.”
4. Maybe we can say Z(P ) intersects each cross-section of the tube. Or stronger than that, bisects each
cross-section. The problem here is that this consists of infinitely many conditions
on the polynomial.
S
Maybe we can modify this to say it bisects every 1 × 1 × 1 · · · × 1 cube in Ti . Well, there are still
infinitely many, so let’s restrict to lattice cubes, say, on something like a 2−n lattice.
64
When we plug in the polynomial ham sandwich theorem, we get the same bound on the degree:
γ
deg P . N 1− n . Okay, now can we do the last step? Can we conclude that it bisects some other cubes,
in analogy? Maybe we can get it to intersect a line a bunch of times...
5. P is small on the tubes, but some normalization like the maximum coefficient is small.
We’ll see what happens in the last class on Wednesday.
35
Wednesday, December 12, 2012
Last time, we had a Kakeya set K of 1 × N tubes in Rn . The volume of K is ∼ N n−γ . Look at all unit
cubes in the unit cube lattice which intersect K. Then there are N n−γ cubes, and by the polynomial ham
sandwich theorem, there’s a polynomial P which bisects all of those cubes and has deg P . N 1−γ/n ≪ N .
Pick a random line in T parallel to the axis of T . We could construct situations where these lines didn’t
intersect Z(P ), but if |{` ∩ Z(P )}| > deg P , then ` ⊂ Z(P ). So we know for every line `, |` ∩ Z(P )| ≤
deg P . N 1−γ/n .
So our more accurate picture is that the polynomial slices most of the cubes at an oblique angle, roughly
slicing it parallel to the tube, because there isn’t much intersection with lines parallel to the tube. We can
wonder whether this zero surface reaches out to more cubes outside of the tube. But it might curve sharply
away, and we can’t really do that.
One thing we can observe is that other tubes might pass through the same cubes, but the polynomial is
going to have a surface that’s one hyperplane. We can’t have many tubes perpendicular to this. This is a
property known as planiness, which Katz-Laba-Tao discovered: For most q ∈ K, and for most Ti ⊃ q, the
Ti are near a hyperplane. They used this seemingly strong property to only improve the three-dimensional
bound slightly, though.
We might also notice that, speaking informally, these tubes don’t have many joints. This observation
was used to prove the following result that will be our last for the class:
Theorem (Joints theorem for long thin tubes). Let Tj,a ⊂ Rn be a cylinder of radius 1, infinite length, for
j = 1, . . . , n and a = 1, . . . , A, such that Tj,a axis is an angle < (100n)−1 to the xj axis. Let I := {x ∈ Rn :
∀j∃a such that x ∈ Tj,a }. Then Vol I . An/(n−1) .
If the tubes were exactly parallel to the axes, this would follow from the Loomis-Whitney theorem, since
you could project onto a hyperplane.
Our method will involve taking random lines through the surface, so we’ll need some definitions to work
with that.
R
Definition. If v is a unit vector and S is a hypersurface with normal vector N , Vs (v) = S |N · v| d VolS .
Example. The unit circle in a plane has directed volumes of 4 in every direction. The integral on any small
projection is the horizontal length, but there are two halves to the circle. This generalizes.
R Lemma. Let π : Rn → v ⊥ be the orthogonal projection. Then VS (v) = v⊥ π −1 (y) ∩ S dy.
Corollary (Cylinder estimate). If T is an infinite cylinder of radius r, core unit vector v, then VZ(P )∩T (v) .
rn−1 deg(P ).
The last cool thing about directed volumes is that if you know a lot of directed volumes, you can recover
the ordinary volume.
Pn
1
, then Voln−1 S ≤ 2 j=1 VS (vj ).
Lemma. If ej are the coordinate vectors, |vj − ej | < 100n
Moreover, Voln−1 S is at least as big as any individual directed volume, so these are comparable.
P
P
Proof. It suffices to show that 1 ≤ 2 j |N · vj | for each unit vector N . Indeed, j |N · ej | ≥ 1 (CauchyP
1
Schwarz exercise), and the error |ej − vj | ≤ |vj − ej | ≤ 100
so this follows.
65
Proof of Theorem. Let Q1 , . . . , Qv be unit cubes (in the unit lattice) that intersect I, so it suffices to prove
V . An/(n−1) . Let P be a polynomial such that Z(P ) bisects each Qi , deg P . V 1/n . Then for all i,
Voln−1 Z(P ) ∩ Qi & 1. This is similar to the isoparametric inequality.
Now for each i,
Pchoose Tj (Qi ) ∈ {Tj,a } such that Tj (Qi ) intersects Qi . Let vj,i be the unit vector of Tj (Qi ).
Then for each i, j vZ(P )∩Qi (vj,i ) & 1. Assign each cube Qi to one tube Tj (Qi ) and VZ(P )∩Qi (vj,i ) & 1. By
Pigeonhole, one tube T has & V /A cubes Q̄i . Then VZ∩Q̄i (v) & 1 and Q̄i intersects T .
Let T̃ be a concentric cylinder of radius n + 1. Then T̃ contains all of the Q̄i (so it’s big enough).
Therefore, V /A . VZ(P )∩T̃ (v) . V 1/n . Rearranging, V (n−1)/n . A, so V . An/(n−1) , as desired.
We used at several points in this course that having a polynomial vanish at a lot of points on a line, it
has to vanish on that line. We tried to replace points with cubes, this time that were bisecting, but that
doesn’t propagate as well. In some directions, it will propagate well, but we can’t guarantee that it does
that. In the joints problem, we were able to guarantee that it propagated in one of the possible directions,
so get an improvement there. But in other problems, it’s hard to get it to do what you want.
66