Geometry: Old and New
Will Traves
Department of Mathematics
United States Naval Academy
St. Mary’s College of Maryland
Natural Science and Math Colloquium
21 OCT 2014
Traves (USNA)
St. Mary’s
21 OCT 2014
Ancient Geometry
Line Arrangement due to Pappus of Alexandria (Synagogue; c. 340)
Richter-Gebert: 9 proofs in Perspectives on Projective Geometry, 2011
Traves (USNA)
St. Mary’s
21 OCT 2014
Pascal’s Mystic Hexagon Theorem
Pascal: placed the 6 intersection points on a conic (1639)
Traves (USNA)
St. Mary’s
21 OCT 2014
Why Mystic?
Traves (USNA)
St. Mary’s
21 OCT 2014
The Projective Plane
P2 is a compactification of R2
P2 = R2 ∪ line at ∞
Parallel lines meet at infinity - one point at ∞ for each slope.
Line at infinity wraps twice around R2 .
Traves (USNA)
St. Mary’s
21 OCT 2014
The linking numbers of triangles 124 and 356 is 0 because they are not linked;
apart from each other without being caught like a chain link.
Topology of the Projective Plane
Thicken line at infinity:
P2 = disk ∪ Möbius band
The linking numbers of triangles 246 and 135 is 1 because they are linked toge
P2 can’t be embedded in R3
(Conway, Gordon, Sachs (1983):
linked triangles in K6 )
If you add all of the linking numbers of all sets of triangles in this partic
find the sum to be 1 (246 and 135 are the only linked triangles). This is consist
Gordon and Sachs who claimed that as long as we are working in R3 then we w
linking number.
Traves (USNA)
St. Mary’s
21 OCT 2014
Bézout’s Theorem
Compactness of P2 allows us to count solutions:
Theorem (Bézout)
Any two curves, without common components, defined by the
vanishing of polynomials of degrees d1 and d2 meet in d1 d2 points in
P2 , suitably interpreted.
Lines meeting an
y=3
y=2
y=1
y=0
4x2+9y2=36
Traves (USNA)
St. Mary’s
Line m
points
Doubl
tangen
4x2 +
21 OCT 2014
Projective Coordinates: Möbius’s model for P2
(x, y) ∈ R2 ↔ (x, y, 1) ∈ R3
(x, y) ∈ R2 ↔ lines containing both (0, 0, 0) and (x, y , 1)
= {(x : y : z) ∈ R3 }/ ∼
where (x : y : z) ∼ (λx : λy : λz) for λ 6= 0
Traves (USNA)
St. Mary’s
21 OCT 2014
Bashelor’s Work
Enumerative Geometry of Conics: How many conics pass through p
NA
Project
pointsTrident
and are tangent to
` lines and c conics in general position?
work with
and my
Ksir.
Our work grew outSt.ofMary’s
Bashelor’s Trident
Traves (USNA)
21 OCT 2014
Conics through 5 points
ax 2 + bxy + cy 2 + dx + ey + f = 0
ax 2 + bxy + cy 2 + dxz + eyz + fz 2 = 0
(a : b : c : d : e : f ) ∈ R6 / ∼= P5
Point conditions force (a : b : c : d : e : f ) to lie on a hyperplane.
(x0 : y0 : z0 ) on C ⇐⇒ ax02 + bx0 y0 + cy02 + dx0 z0 + ey0 z0 + fz02 = 0.
Intersecting 5 hyperplanes in P5 gives a single point corresponding to
the one conic through all 5 points.
Traves (USNA)
St. Mary’s
21 OCT 2014
Determinants
Given three points in P2 we can list them as columns of a 3x3 matrix,


ax bx cx
 ay by cy  .
az bz cz
The determinant [abc] of this matrix measures six times the volume of
the tetrahedron with edges a, b and c.
[abc] = 0 ⇐⇒ points a, b, c are collinear.
Can recover coordinates from full knowledge of determinants.
Traves (USNA)
St. Mary’s
21 OCT 2014
Plücker Relations
The determinants of the 3x3 submatrices of a larger matrix satisfy
quadratic relations.
For example, given five points in P2 we form the matrix


ax bx cx dx ex
 ay by cy dy ey  .
az bz cz dz ez
Cramer’s Theorem implies that
[abc][ade] − [abd][ace] + [abe][acd] = 0.
In particular, if [abc] = 0 then [abe][acd] = [abd][ace].
Traves (USNA)
St. Mary’s
21 OCT 2014
Conics and Pascal’s Theorem
Six points lie on a conic
⇐⇒ [abc][aef ][dbf ][dec]
= [def ][dbc][aec][abf ].
conic:
[159] = 0
[168] = 0
[249] = 0
[267] = 0
[348] = 0
[357] = 0
[257][987] = 0
=⇒ [125][136][246][345] = +[126][135][245][346]
=⇒
[157][259] = −[125][597]
=⇒
[126][368] = +[136][268]
=⇒
[245][297] = −[247][259]
=⇒
[247][268] = −[246][287]
=⇒
[346][358] = +[345][368]
=⇒
[135][587] = −[157][358]
⇐=
[297][587] = +[287][597]
If points 2, 5, and 7 are not collinear, [257] 6= 0 so [987] = 0.
Traves (USNA)
St. Mary’s
21 OCT 2014
Computer Assistance I: Conjecture + Proof
Conjecture (Kepler)
√
No packing of spheres covers more than π/3 2 (approximately 74%)
of the filled volume.
History: Kepler and Harriot and Sir Walter Raleigh (1611)
Fejes Tóth (1953)
Hales and Ferguson (1992-2006)
FlysPecK (Formal Proof of Kepler; 2014)
Traves (USNA)
St. Mary’s
21 OCT 2014
Computer Assistance II
Prediction: Computers will become our mathematical assistants, vastly
raising the level of our mathematical reasoning (c.f. advanced chess
and CAD/origami).
Traves (USNA)
St. Mary’s
21 OCT 2014
Invariant Theory
Can change basis in P2 by multiplying each tuple of homogeneous
coordinates by an invertible 3x3 matrix.
Theorem (Fundamental Theorem of Invariant Theory)
The only invariant quantities are quotients of polynomials in the
brackets. The relations among invariants are all derived from the
Plücker relations.
The change of basis preserves incidence structures but allows us to
move any four non-collinear points to any other 4 non-collinear points.
This suggests that the only interesting questions in Projective
Geometry involve incidence structures. In particular, the action
suggests that distances and angles aren’t invariant quantities.
Traves (USNA)
St. Mary’s
21 OCT 2014
Homogenization
Can’t talk about the parabola y = x 2 :
If we scale the coordinates (x : y : z) = (λx : λy : λz) then we’d
require
λy = λ2 x 2
for all λ (true only for the origin (x, y) = (0, 0)).
Curves in P2 are defined by the vanishing of homogeneous
polynomials:
y − x 2 = 0 ←→ yz − x 2 = 0
Traves (USNA)
St. Mary’s
21 OCT 2014
Circles
Circle with center at (a,b) and radius r has equation
(x − a)2 + (y − b)2 = r 2
which homogenizes to
(x − az)2 + (y − bz)2 = r 2 z 2 .
Where does the circle hit the line at infinity? Bézout’s Thm ⇒ 2 points.
z = 0 and x 2 + y 2 = 0 ⇒ (x : y : z) = I(1 : i : 0) or J(1 : −i : 0)
Every circle passes through I(1 : i : 0) and J(1 : −i : 0).
General conic determined by 5 points but circles determined by 3
points (plus I and J).
Traves (USNA)
St. Mary’s
21 OCT 2014
Cross-Ratios
Given four lines coincident at point 0, any other line hits these lines in
four points which always satisfy an invariant condition.
Cross-ratio:
[0ac][0bd]
(a, b; c, d) =
[0ad][0bc]
(a, b; c, d) = (a0 , b0 ; c 0 , d 0 )
Traves (USNA)
St. Mary’s
21 OCT 2014
Measuring Angles
Theorem (Laguerre’s Theorem)
The angle between two lines meeting the line at infinity at L and M is
given by
1
ln(L, M; I, J) mod π.
2i
Traves (USNA)
St. Mary’s
21 OCT 2014
My current work
There is at least one degree 3 curve through every set of 9 points.
Question
When do 10 points lie on a plane curve of degree 3?
Smooth curves of degree 3 are called elliptic curves and play a role in
both elliptic curve cryptography and in Wiles’s proof of Fermat’s
Last Theorem.
There is a non-zero solution when the determinant of a 10 × 10 matrix
is zero (25 million terms).
Reiss (1842) wrote out a 20 term polynomial of degree 10 in the 3 × 3
brackets that computes the determinant faster.
Traves (USNA)
St. Mary’s
21 OCT 2014
Ruler and Compass Construction
David Wehlau and I created a straightedge-and-compass construction
that checks whether 10 points lie on a cubic.
Such constructive questions are now back in vogue since they
leverage all sorts of ideas in computational geometry.
Traves (USNA)
St. Mary’s
21 OCT 2014
The Key Idea: Cayley-Bacharach
10 points on a cubic precisely when 6 auxiliary points on a conic.
Traves (USNA)
St. Mary’s
21 OCT 2014
Straightedge Construction
Meet and Join Algebra: Allows algebraic formulation of
straightedge-only constructions
Theorem (after Sturmfels and Whiteley)
There exists a straightedge construction to determine if 10 points lie on
a cubic (with about 100 million lines).
Traves (USNA)
St. Mary’s
21 OCT 2014