QUALIFICATION EXAM
UMUT VAROLGUNES
Abstract. Here are some notes from my preparation for the exam. My two
minors were finite dimensional Lie algebras, and characteristic classes. The
Lie algebras section is a standard linear exposition whereas the characteristic
classes one is random notes. My major is an assortment of things related to
symplectic geometry and mirror symmetry. The notes there are mostly about
the parts that I found more esoteric. I enjoyed writing these, someone else will
maybe enjoy reading them - very unlikely.
1. Lie Algebras
1.1. Lie algebras from Algebraic Groups.
• An algebraic group over a field F is a collection of polynomials over F in
the variables corresponding to entries of a matrix over F. Condition is that
the set of invertible solutions for every base extension (algebra over F) is
closed under multiplication and inversion in the group of matrices.
• Use the algebra of dual numbers F[]/(2 ) to define the corresponding Lie
Algebra, i.e. all X ∈ Matn (F) for which I + X is in G(F[]/(2 )).
1.2. Engel’s Theorem.
Theorem 1. Let g → gl(V ) be a representation. If all elements of g act nilpotently
on V , then there exists a vector in V which is killed by all of g.
Proof. Induction on dimg. Suffices to show for subalgebras. Take a maximal proper
subalgebra h of g. Key step: Consider the representation of h on g given by the
adjoint action. Clearly, h ⊂ g is a subrepresentation, so we can consider the quotient
representation. It is elementary to check that h acts nilpotently. Hence, there is
an element in g/h which is killed by h and let us lift that element to g, and call it
a. By construction [a, h] ⊂ h. Hence, h ⊕ F · a = g, by maximality. Now we finish
with the more obvious use of the induction hypothesis - take the subspace that is
killed by h, and show that it is invariant under the action of a.
Corollary 1. Let g → gl(V ) be a representation. If all elements of g act nilpotently
on V , then V has a basis in which the matrices that represent the elements of g are
all strictly upper-triangular.
Corollary 2. The adjoint representation is nilpotent if and only if the Lie algebra
is nilpotent.
1.3. Lie’s Theorem.
Lemma 1. Let g be a Lie algebra, and V its finite dimensional representation.
Furthermore, let h be an ideal and λ ∈ h∗ . Then the weight space Wλh of λ is
invariant under the action of g.
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Proof. We need to show that for any g ∈ g and h ∈ h, λ([g, h]) = 0, for Wλh
nonempty. Let v ∈ Wλh .
Let Wn be the span of v, g · v, . . . , g n · v. Notice that Wn get strictly bigger as
n increases up to some point (say N ), and then it stabilizes completely. We claim
that for every n, and for every a ∈ h, a is upper triangular with diagonal entries
λ(a) on Wn , with respect to its defining basis. We do this by induction on n. Base
case n = 1 is clear. Let us assume it for Wn−1 . We have
(1)
a · g n · v = [a, g] · g n−1 · v + g · (a · g n−1 (v)),
which finishes the proof. In particular, the trace of a as an operator on WN is equal
to N λ(a). Since WN is closed under the action of g as well, we have T rWN ([g, h]) =
0, this finishes the proof if N is coprime with the characteristic of our base field (so
we assume something in the beginning), by setting a = [g, h].
Theorem 2. Let g be a solvable Lie algebra and V its finite dimensional representation. Then there exists a common eigenvector to all elements of g.
Proof. We do induction on the dimension - base case by algebraically closedness.
Let h be a maximal subalgebra of g which contains [g, g]. It is easy to see that
h is also an ideal, and hence a solvable ideal. It is also of codimension one by
maximality. By the induction hypothesis, h has a common eigenvector. We take
the corresponding weight space, and apply the previous theorem. Using again that
the field is algebraically closed, this finishes the proof.
Corollary 3. Let V be a finite dimensional representation of a solvable Lie algebra
g. Then there exists a basis of V , with respect to which all elements of g are upper
triangular.
Corollary 4. A Lie algebra g is solvable if and only if its adjoint representation
can be made simultaneously upper-triangular.
Corollary 5. A Lie algebra g is solvable if and only if [g, g] is nilpotent.
1.4. Weight space decomposition.
Lemma 2. Let a ∈ g, α, β ∈ F. We define gaα , and Vβa to be the generalized
eigenspaces of ad(a) and a : V → V with eigenvalues α and β. We then have
a
.
gaα · Vβa ⊂ Vα+β
Proof. We need to show that for v ∈ Vβa , and g ∈ gaα , there exists N ∈ Z such
that (a − α − β)N · g · v = 0. Now do binomial expansion on the left hand side
according to the identity (left multiplication −α − β) = (adjoint action-α) + (right
multiplication−β), and choose large N
We can define the generalized weight spaces Vλ , and in particular gλ , in the
obvious way. What we have shown implies gλ · Vγ ⊂ Vλ+γ , and in particular
[gλ , gγ ] ⊂ gλ+γ .
Theorem 3. Let h be a nilpotent Lie algebra, and V its finite dimensional representation. Then V decomposes into its generalized weight spaces.
Proof. Let a be any element of h. Then by assumption ha0 = h, and by the lemma
h · Vαa ⊂ Vαa . Hence, if there exists any a with two different eigenvalues, then
by Jordan decomposition for a, we can do induction on dimV . Let us assume
QUALIFICATION EXAM
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otherwise, then by Lie’s theorem we can find a basis of V such that all elements
of g is upper triangular with the same diagonal entries. This proves that V is a
generalized weight space itself.
Corollary 6. We get very fundamental decompositions of representations and of
the Lie algebra itself, if we are given a nilpotent subalgebra inside.
1.5. Cartan subalgebras.
Definition 1. Let h ⊂ g be a nilpotent subalgebra. If it is maximal in either of the
following senses (which are equivalent):
• If [a, h] ⊂ h, then a ∈ h
• There doesn’t exist a nilpotent subalgebra strictly containing it.
we call it a Cartan subalgebra.
One direction of equality follows from Jacobi identity. For other side, consider
the commutator series of the larger nilpotent algebra and take the largest one which
is contained in the smaller nilpotent algebra.
Proposition 1. Let h be a Cartan of g. Then the zero part of the generalized
weight space decomposition is h itself.
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Proof. Let g = λ∈h∗ gλ . Trivially, h ⊂ g0 . Consider g0 /h as an h representation.
This is by definition a nilpotent action, and Engel’s theorem contradicts maximality.
Theorem 4. Let a ∈ g be a regular element. Then the 0-eigenspace of ad(a) is a
Cartan.
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Proof. Let g = λ∈F gaλ = ga0 ⊕V . ga0 is a subalgebra, and the second decomposition
is one of ga0 representations by the magic identity. Let’s assume that ga0 is not
nilpotent. Then, we can find an element b ∈ ga0 which acts nonsingularly on V , and
nonnilpotently on ga0 - as these are both non-empty open sets. This contradicts the
regularity of a. Maximality is easy.
Corollary 7. Let h ⊂ g is a Cartan, and a ∈ h be a regular element. Then ga0 = h.
Theorem 5. Any Cartan subalgebra is in fact of the form constructed in the above
theorem. Moreover, for any two Cartan there exists an automorphism of the Lie
algebra taking one to the other.
Proof. The key point is to find a regular element in h. Consider the generalized
weight space decomposition g = h ⊕ V . The important observation is that the pure
weight elements of V act nilpotently under adjoint representation. This lets us
exponentiate them and obtain isomorphisms of the Lie algebra. Heuristically, by
applying exponentials in all directions from h we can cover most of g, which will
make us hit a regular element as they are in abundance - also note that regular
elements are preserved under automorphisms.
Define φ : g → g, by φ(Σαi bi + h) = Exp(α1 ad(b1 )) . . . Exp(αn ad(bn ))h. It is
not hard to show that for a ∈ h,
(2)
dφa (b + h) = h + [b, a].
Hence, if we choose a outside the hyperplanes given by the roots, dφa is nonsingular.
Now we use that everything is defined by polynomial equations, which implies that
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UMUT VAROLGUNES
the image of φ contains a Zariski dense subset. This proves the first claim - note
this only used that the image contains an open set. Doing this for two Cartan
subalgebras and taking their intersection proves the second claim - this actually
uses Zariski open.
1.6. Killing form. If we are given a representation V of a Lie algebra g then we
can define a bilinear form < g, h >V = tr(π(g)π(h)). This is trivially symmetric
and invariant i.e. < [g, h], g 0 >V =< g, [h, g 0 ] >V . The Killing form is the case of
adjoint representation.
Lemma 3. Let h be a Cartan subalgebra, and V a representation. Let e ⊂ gα and
f ⊂ g−α for some root α ∈ h∗ . Then, for any λ ∈ h∗ such that Vλ 6= 0, we have
λ(h) = r · α(h), where r is a rational number depending only on α and λ.
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Proof. Let U be
n∈Z Vλ+n·α . Clearly, e and f both preserve this subspace so
tr(h |U ) = 0. On the other hand h preserves all the summands of U , and using
tr(h |Vλ+n·α ) = dim(Vλ+n·α )(λ(h) + n · α(h)) we conclude the result.
Theorem 6. Let g ⊂ gl(V ). Then the following are equivalent:
(1) < g, [g, g] >V = 0.
(2) < a, a >V = 0, for all a ∈ [g, g].
(3) g is solvable.
Proof. (1) implies (2) is obvious. For (3) implies (1), choose an upper-triangular
basis for V , and use that product of an upper-triangular and strictly upper triangular matrix is strictly upper triangular. The actual content is in (2) implies
(3).
Consider the derived series of g and call the subalgebra that it stabilizes to p.
Take a Cartan h of p. By construction p = [p, p]. Therefore, h has a basis {hαi }
such that hαi ∈ [gα , g−α ].
By assumption < hαi , hαi >V = 0. If we compute the LHS using the definition,
we get Σλ dim(Vλ ) · λ(hαi )2 . Now we use the previous lemma and get
(3)
2
Σλ dim(Vλ ) · (rαλ α(hαi ))2 = (α(hαi ))2 (Σλ dim(Vλ ) · rαλ
) = 0.
Hence, V = V0 - as the equation shows that λ(g) = 0 for all λ. This implies that
p = p0 . But, then since h is Cartan p = h, which implies p = 0.
Corollary 8. g is solvable if and only if K(g, [g, g]) = 0.
Remark 1. In some of these theorems we can drop the algebraically closed hypothesis. These can be deduced from the algebraically closed case by passing to the
algebraic closure, which preserves nilpotency, solvability, and abelianness.
1.7. Semi-simplicity. We call a Lie algebra semi-simple if it has no solvable (or
equivalently abelian) ideals.
Let R(g) be the maximal solvable ideal of g, which can be constructed as the
sum of all solvable ideals. Clearly, g is semisimple iff R(g) = 0.
Lemma 4. g/R(g) is semisimple.
Proof. Assume that a is an abelian ideal of g/R(g), and take its preimage in g,
which is also a solvable ideal containing R(g) - contradiction.
Yet stronger than this is the following theorem (Levi’s theorem), which will be
proven later.
QUALIFICATION EXAM
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Theorem 7. In fact there exists a semisimple subalgebra s such that g = R(g) ⊕ s
Theorem 8. Let g ⊂ glV such that V is irreducible. Then either g is semisimple,
or it is the direct sum of F · IV and a semisimple Lie algebra.
Proof. Consider V as a representation of R(g). By Lie’s theorem, there is at least
one weight space that is nonempty, and by Lie’s lemma this gives a subrepresentation of V as a rep. of g. Hence, it should be the whole of V . This implies that
R(g) is either empty or the multiples of identity, as desired.
Theorem 9. g is semisimple if and only if the Killing form is nondegenerate.
Proof. Assume that there is an abelian ideal a, and choose a complement V to it
inside g. Let a ∈ a and g ∈ g, and let us compute K(g, a) = tr(ad(g) ad(a)). Using
that a is an ideal we see that ad(g) is upper-triangular, and also using that a is
abelian that ad(a) is strictly upper triangular. Hence the Killing form is degenerate.
Conversely, if the Killing form is degenerate, we take its kernel a, which is an ideal
by the invariance property of the Killing form. Consider the adjoint representation
of a on g. This shows that a/center is solvable by Cartan’s criterion, which then
shows that a is solvable.
Theorem 10. If the Killing form is nondegenerate, then the center of g is trivial,
and all its derivations are inner.
Proof. Its obvious that the center is trivial by the definition of the Killing form.
We identify g with its image in glg , so we have inclusions of Lie algebras g ⊂
Der(g) ⊂ glg . Consider the trace form on glg . Its restriction to g is nondegenerate
by assumption. We can take the orthogonal complement g⊥ of g inside Der(g). It
is easy to see that g⊥ is an ideal, just as g. Hence [g, D] = 0, if D ∈ g⊥ ⊂ Der(g).
This shows that D is the zero derivation as the center is trivial, which finishes the
proof.
1.8. Jordan decompositions. One can characterize the standard Jordan decomposition as writing a matrix as a sum of a semisimple matrix and a nilpotent matrix
which commute, A = As + An . Some facts:
• The proof is actually not so hard.
• There is a polynomial P such that As = P (A) - do it first for a Jordan
block.
• This decomposition is unique - consider the differences, this is clearly nilpotent, it is semisimple by the previous item, and finish by noting that the
only semisimple nilpotent element is zero.
An abstract Jordan decomposition of an element a of g is obtained by writing
a = as + an such that ad(as ) is semisimple, ad(an ) nilpotent, and [an , as ] = 0. The
standard Jordan decomposition of a matrix is an example of this.
Lemma 5. If the center is trivial then there is at most one Jordan decomposition
of an element.
Proof. This follows immediately from the uniqueness of the standard Jordan decomposition.
Theorem 11. If all derivations are inner, and the center is trivial, then there
exists a Jordan decomposition.
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UMUT VAROLGUNES
Proof. We embed g in gl(g) by the adjoint representation. We need to show that the
summands in the standard Jordan decomposition ad(a) = An + As are derivations
of g. Generalized eigenspaces of ad(a) are the eigenspaces of As , and generalized
eigenspaces play nicely with respect to Lie bracket by the magic formula. That As
is a derivation can be checked on a basis, and finishes the proof.
1.9. Simplifications
for the root decomposition in the semisimple case.
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Let g =
gα be a root space decomposition for h. Let K be the Killing form.
• K(gα , gβ ) is zero unless α + β = 0. Notice that ad(a)ad(b) is a nilpotent
operator unless α + β = 0. This doesn’t use semisimplicity.
• K is nondegenerate on gα ⊕ g−α and h. Follows immediately from the
previous point, and semisimplicity.
• h is abelian. This follows from Cartan’s criterion and semisimplicity: h is
solvable, so K(h, [h, h]) = 0.
• h is semisimple. Use Jordan decomposition h = s + n. It follows from
the properties of the standard Jordan decomposition that anything that
commutes with h commutes also with s and n. By maximality they should
both lie inside h. Since n is nilpotent, and ad(n) commutes with the adjoint
action of all elements in h, it follows that n is in the kernel of K |h , which
shows that it is zero as desired.
• At this point, the generalized root space decomposition became much simpler. We make the obvious definition of a root (given a Cartan), and call
all of them ∆ ⊂ h∗ . K defines an isomorphism ν : h → h∗ , and a bilinear
form on h∗ .
• Let e ∈ gα and f ∈ g−α , then we can compute K([e, f ], h), for any h ∈ h
- use the invariance property of K - and we find K(e, f )α(h). This is
equivalent to [e, f ] = K(e, f )ν −1 (α).
• Let us take a root α, apply the previous point, and find an sl2 triple inside g.
If α(ν −1 (α)) = 0, then one can check that the Cartan criterion is satisfied,
and sl2 is solvable, which is a contradiction. So we proved that K(α, α) 6= 0.
Lemma 6. Let V be a finite dimensional representation of sl2 , i.e. < E, F, F >
with [E, F ] = H, [H, E] = 2E, and [H, F ] = −2F . Let v ∈ V be such that Ev = 0
and Hv = λv, where λ 6= 0. Then the following hold:
(1) HF n v = (λ − 2n)F n v
(2) EF n v = n(λ − n + 1)F n−1 v
(3) λ ∈ Z+ . Moreover, F λ+1 v = 0, and the ones before that are linearly
independent.
Proof. The first two are trivial inductions (pay attention to normalizations though).
For the last part note that since they are eigenvectors with different eigenvalues,
F n v’s are linearly independent as long as they are nonzero. By finite dimensionality
they should become zero at some point, let N be the smallest one. By the second
identity this means that λ = N − 1.
• gα is one dimensional. Let’s assume it is not and let us choose an sl2 triple
with E in gα . Then by nondegeneracy of Killing form g−α is at least two
dimensional too, and hence contains a vector v which is perpendicular to
E. This means that it commutes with E from the formulas above. We also
know that [H, v] = −2v, but this is a contradiction from the lemma above.
QUALIFICATION EXAM
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• The string property. Let α and β be two roots. Then the intersection of
{β + nα} with ∆ ∪ {0} is a string consisting of adjacent roots such that
the length of the positive and negative sides differ by 2K(α,β)
K(α,α) . Consider
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the sl2 triple corresponding to α, which acts on
gβ+nα . Let gβ+pα be
the last nonempty one (and q be the first). Then it follows that H acts
as β(H) + 2p on gβ+pα , and it also follows that the ones in between p and
p − β(H) − 2p are nonzero. Similarly, it follows that the ones between q
and q − β(H) − 2q are nonzero. Using the definitions of q and p, we get
q < β(H) − p and β(H) − q < p. When we sum this two we get an equality,
so we have that the two chains are in fact the same and that its length is
p + q = β(H) = 2K(α,β)
K(α,α) . This also shows, by going back to the numbers
that were supposed to be positive, that q < 0 < p.
• [gα , gβ ] = gα+β - if α + β is not a root this is obvious. This follows from
the proof above, as β, β + α is part of the string - so they are connected
with the sl2 action corresponding to α.
• If α is a root, nα is a root iff n = ±1. Assume n > 1. We have that
2K(nα,α)
2
K(nα,nα) = n is an integer. Then n = 2. This is impossible by the previous
point.
• ∆ spans h∗ . Say it doesn’t, so we can choose a nonzero element in the
K-orthogonal complement of span(∆). This shows by construction that
the K-dual of that vector in h commutes with everything.
• K(α, β) is rational, for α, β ∈ ∆. If we can show this for K(α, α) we would
be done by the string property. Let us find a formula for K(α, β) for all
α, β ∈ h∗ regardless. Dually, for h1 , h2 in the Cartan subalgebra,
(4)
K(h1 , h2 ) = tr(ad(h1 )ad(h2 )) = Σα∈∆ α(h1 )α(h2 ),
which implies K(α, β) = Σγ∈∆ K(α, γ)K(β, γ). Plug in β = α and divide both sides by K(α, β)2 , and use the rationality obtained by the string
property again.
• K is positive definite on the Q-linear span of the roots. This follows from
the previous point (and the formula derived there) directly. Note that the
rationals here is independent of the ground field.
1.10. Indecomposable root systems and simple Lie algebras.
Proposition 2. Let a be an ideal of semisimple g. Then g is isomorphic to a ⊕ a⊥ .
Proof. It follows that a ∩ a⊥ is empty from Cartan’s criterion. This also implies
that [a, a⊥ ] = 0. Dimension counting finishes the proof.
We define a simple Lie algebra to be a non-abelian one with no proper ideals.
Note that we could equivalently say non-solvable. By the proposition above, we
have that a Lie algebra is semisimple if and only if it is a direct sum of simple Lie
algebras.
Let us say that two roots α and β are connected if there are roots a = γ1 , . . . , γn =
b such that γi + γi+1 ’s are also roots. Relatedly, we call a root system indecomposable, if it can’t be written as the disjoint union of disconnected subsets of roots.
If the roots we get from a semisimple Lie algebra is decomposable, the decomposition induces in the only way possible a decomposition of the Lie algebra. Moreover,
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UMUT VAROLGUNES
if the root system is decomposable, the connected components actually lie on complementary orthogonal subspaces. This shows that the decomposition of the root
system into indecomposable root systems is unique.
1.11. Abstract root systems.
Definition 2. We call a pair (∆, V ), where V is a vector space with an inner product,
and ∆ a finite subset of vectors, an abstract root system if
(1) 0 ∈
/ ∆, and ∆ spans V .
(2) If α ∈ ∆, then nα ∈ ∆ if and only if n = ±1.
(3) The string property, i.e. if α, β ∈ ∆ then {α + nβ} ∩ (∆ ∪ {0}) is a
connected string with nonempty positive and negative sides differing in
length by 2(α,β)
(α,α) .
Note that there is no extra rationality assumption - except the one imposed by
the string property. There is an obvious notion of an isomorphism of abstract root
systems.
Lemma 7. The inner product in the definition is unique up to scaling.
Proof. Fix a root, which determines the scaling factor. Then for any other root
use the string property on one side to determine the scaling factor for the inner
product of the two roots, and then for the length of the root.
Here are some examples of abstract root systems.
• ∆An = {i − j | i 6= j, 1 ≤ i, j ≤ n}
• ∆Bn = {i − j , i + j , −i − j , i , −i | i 6= j, 1 ≤ i, j ≤ n}
• ∆Cn = {i − j , i + j , −i − j , 2i , −2i | i 6= j, 1 ≤ i, j ≤ n}
• ∆Dn = {i − j , i + j , −i − j | i 6= j, 1 ≤ i, j ≤ n}
Recall the definition of an even lattice: a lattice in a vector space with inner product
such that the the norms of the elements of the are all even - this is a weaker condition
than all inner products being integer. The most important example of such lattice
is Γr = {i ∈ Z + 21 , Σi ∈ 2Z} in the standard Rr , where r is a multiple of 8. It is
easy to see that the minimal norm nonzero elements of the lattice, assuming that
they span the vector space, form an abstract root system.
• Let us call the abstract root system we get for Γ8 above E8 .
• E7 is the set of roots perpendicular to vector (1/2, 1/2, . . . , 1/2) inside the
perpendicular vector space.
• E6 is the one perpendicular to (0, . . . , 0, 1, 1) inside E7 .
• F4 is the norm 1 or 2 elemensts inside the lattice of all half integer, or all
integer vectors in standard R4 .
• G2 lives inside the lattice of A2 and is the elements of the lattice with norm
2 or 6.
1.12. Cartan matrix and Dynkin diagrams. Let us now choose a functional f
on h∗ which doesn’t vanish on any of the roots. This divides the roots in two parts,
positive and negative. Define a simple root to be a positive root which can’t be
written as a sum of two positive roots. Let us also define a highest root be a root
with maximum value under f .
String property lets us prove a bunch of statements about the simple roots. Let
α and β denote two different simple roots, θ be a highest root vector, and γ be a
positive root.
QUALIFICATION EXAM
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• (α, β) ≤ 0. This follows from the fact that α − β can’t be a root by
simplicity.
• (θ, γ) ≥ 0. This from that θ + γ can’t be a root.
• Any positive root is a nonnegative integer linear combination of the simple
roots. This follows by induction using the filtration given by the value of
f.
• There is no linear relation between the simple roots. Assume that there is
a relation, which we write as an equality of two sums, we can assume that
the coefficients are positive integers. Then use the pairing with one side of
the equation, and get the result by the first point above and the positive
definiteness of the Killing form.
• We define the decomposition of the simple roots in the obvious way. This is
equivalent to the decomposition of the abstract root system in a functorial
way.
• If the root system is indecomposable, there is a unique highest root. Write
the highest root as a nonnegative integer linear combination of the simple
roots. If there is a coefficient which is zero, consider the pairing between
the two and deduce that the root system is decomposable using the first
two points above - a contradiction. Now assume that there are two highest
roots, then their difference or sum can’t be a root, so by the string property
they are orthogonal. Now notice that since a highest root can’t be simple,
it should pair nontrivially with at least one of the simple roots. We get
a contradiction to the orthogonality by expanding out one of the highest
roots in terms of the simple roots and using everything above.
• Let us define the Cartan matrix using the simple roots and the quantity
that appears in the string property. This has diagonal consisting of 2’s.
Other entries are all non-positive, and their negativity is symmetric along
the diagonal. Most nontrivially, Cartan matrix is a positive definite matrix.
This can be seen by writing it as the product of a diagonal matrix with
positive entries and the Gramm matrix of the Killing form with respect to
the basis given by the simple roots.
• Cartan matrix can be represented concisely by a quiver called the Dynkin
diagram. Symmetry described above along with Cauchy-Schwartz inequality simplifies the situation a lot. There can be only four kinds of connections between vertices, no connection, standard one edge connection, and
directed two and three edge connections.
• Finally note that if we also add the highest root to the simple roots and
form the Cartan matrix of one larger size, we get a positive semi-definite
matrix, which satisfies the other properties of the Cartan matrix.
1.13. Classification of simple Lie algebras. We first classify all positive definite
Cartan matrices. The strategy for that is to first create a large pool of positive semidefinite matrices, which are barely not positive definite. This is done by taking the
extended Dynkin diagrams obtained by adding the highest root to the simple roots.
Dynkin diagrams which give positive definite Cartan matrices should not contain
these Dynkin diagrams. It is possible to show that the matrices obtained this way
are positive definite of corank one, i.e. any principal minor is positive definite. This
can be done in a more or less systematic way by generalizing the definition of a
Cartan matrix in a suitable manner (see Kac’s book). A particularly delicate point
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is to show that Dynkin diagrams can’t contain multiply laced cycles. As a result we
get the A, B, C, D general families and E6 , E7 , E8 , F4 , G2 . Now, we have to perform
two tasks to finish the classification
(1) Show that if we start with a simple Lie algebra, make all the choices,
and get a Dynkin diagram, any other simple Lie algebra which potentially
induces this Dynkin diagram is isomorphic to the original Lie algebra. This
involves writing down the obvious generators and relations, and showing
that the seemingly undetermined Lie brackets are actually determined by
semisimplicity.
(2) Show that all Dynkin diagrams actually come from Lie algebras. This
involves actually constructing the undetermined Lie brackets.
Let us start with the first point above. Take generators {Ei , Fi , Hi } for each
2ν −1 (αi )
node of the Dynkin diagram such that [Ei , Fi ] = K(α
= Hi . This implies
i ,αi )
the following commutation relations: [Hi , Hj ] = 0, [Hi , Ej ] = aij Ej , [Hi , Fj ] =
−aij Fj , [Ei , Fj ] = 0. The last one is because the roots are simple. Then define the
decomposition g = n+ ⊕ h ⊕ n− of Lie algebras, by declaring the plus ones to be
the span of the root spaces of the positive roots. In Kac’s choices these become the
upper triangular and lower triangular matrices in all of the types A, B, C, D.
Lemma 8. All ideals of g intersect h nontrivially.
Proof. Let’s say I doesn’t. Then, I is a subrepresentation of h on g, and the weight
space decomposition of I is just obtained by intersecting the weight spaces of g with
I. Note that this fact about the decompositions of the representations of abelian
Lie algebras is true for infinite dimensional representations as well, we will use it
below. This implies that I contains gα for some α. But then it would have to
contain an element of h by the nondegeneracy of the Killing form.
Let the Lie algebra generated by {Ei , Fi , Hi } with only the relations above be
g̃. We will show that g is a uniquely defined quotient of g̃ proving uniqueness.
Proposition 3.
• Define ñ± in the obvious way. All pure commutators are
either inside ñ± or in h.
• g̃ = ñ+ ⊕ h ⊕ ñ−
• There exists a unique maximal ideal J in g̃ which doesn’t intersect h, and
this ideal is proper.
• The natural map g̃ → g factors through g̃/J and induces an isomorphism
of Lie algebras.
Proof. The first item is an easy induction. The second one is a consequence of the
first item. The third one is also easy. The only thing to note is by the above general
easy fact, if I doesn’t intersect h then I = I+ ⊕ I− , and hence the direct sum of all
ideals not intersecting h is a proper ideal, which doesn’t intersect h. Fourth item is
just by construction.
This finished the first point. Now we come to the second point. Before that note
the exceptional isomorphisms A1 = B1 = C1 , B2 = C2 , A3 = D3 , D2 = A1 ⊕ A1 ,
and there is no D1 (it is abelian). I should remember these statements in the Lie
group level, which we did in Vogan’s class as a homework.
Now we actually start from Cartan matrix. In the simply-laced case, there is a
nice reconstruction of all the roots from the simple roots, and you grind your way
QUALIFICATION EXAM
11
to define a Lie algebra structure by considering the relation [Eα , Eβ ] = (α, β)Eα+β
and what should satisfy. This seems to work only in the simply laced case, so
A, D, E. You can in general reconstruct all the roots using the string property but
it is not as nice (should look at this at some point though). We already know B, C
exists. To construct F4 and G2 , we go back to the previous tilde construction. It
suffices to show that the quotient is finite dimensional apparently (need to check
that Lie algebra has the given Cartan matrix and is semisimple) and for that we
construct maps from g̃(A) to Lie algebras we know E6 for F4 and D4 for G2 . These
maps are injective on the h’s inside, and hence the kernels can only get bigger when
we do the actual quotient, which proves finite dimensionality.
1.14. More on the Killing form.
Proposition 4. On simple Lie algebra there exists a unique nondegenerate symmetric invariant bilinear form up to a constant factor.
Proof. We could go back and do the proofs again with using any such bilinear form
and we would find that the string property is still satisfied. Hence we know that it
the statement is true on a Cartan subalgebra. Now we use the properties cleverly
and show it is correct everywhere.
In the simply laced case, we have a good construction of the Lie algebra from
the Cartan matrix. Similarly for the Killing form, we can construct it from the
Cartan matrix. We let it be what it was on the Cartan, and then declare (h, Eα ) =
0, (Eα , Eβ ) = −δα,−β . The previous proposition shows that this is the Killing form.
Armed with this description, we now take our base fields to be C, which we see
as the base extension of the one with the base field R. We define the automorphism
ω |hR = −Id, and ω(Eα ) = E−α , and extend to the complexification by antilinearity.
This is an automorphism of the complex Lie algebra, and its fixed point set is called
the compact form of it (because the corresponding Lie group inside is compact) and
the Killing form restricted there is negative definite.
1.15. Weyl group. Weyl group W ∈ O(V, <, >) is the group generated by the
reflections along the orthogonal hyperplanes of the roots - we work with an indecomposable abstract root system here. Here are some of its properties:
• W preserves the roots. This is a direct consequence of the string property,
we know that one side that is A more than the other, this is asking for the
Ath one in the longer side.
• If we assume the previous point, and also that 2<α,β>
<α,α> is an integer, the
string property follows. This is not trivial. Note that if a root is k times
the other then 2k and 2/k are both integers by integrality and k can only
be ±1, ±2. I don’t think that there is a way to exclude the 2, but this
is ok for the string property anyways. Assuming that the roots are not
multiples of each other: first we show that there is at least one root in
the side that should be longer right next to the existing root, and this is
by proving that at least one of 2<α,β>
<α,α> should be ±1. Then we show that
the string is connected, let’s assume it is not, then we write down what is
required to have that from the sentence below, and get a contradiction to
the positive definiteness. Finally, we prove the exact formula with 2<α,β>
<α,α>
by noting that the reflection along the hyperplane perpendicular to the
12
UMUT VAROLGUNES
string can send one end only to the other end - the formula gives exactly
what is needed.
• W (An ) = Sn , W (Bn ) = Sn n Zn2 , W (Cn ) = Sn n Zn2 and W (Dn ) =
Sn n Zn−1
2
Choose an f ∈ V ∗ and define positive roots, simple roots and stuff. Call the
reflections of the simple roots simple reflections.
• Reflection of a simple root can’t make any positive root other than itself
negative. We have that any positive root is a positive combination of simple
ones and just reflection can change only one of these.
• A positive but not simple root can be simply reflected such that the sum of
the coefficients in its representation with simple roots decreases. Assume
the contrary, which means that it pairs with all simple roots negatively,
which is impossible, as we have shown before - or you can just look at its
norm and expand one side only.
• Hence we can apply simple reflections and bring any positive root to the
position of a simple root. This shows that W is generated by simple reflections because if we want to make a reflection at a positive root, we
first change coordinates by the reflection obtained in the previous sentence
(conjugation) and do the simple reflection instead.
Now we define the chambers to be the closures of the connected components
of the complement of all the orthogonal hyperplanes to the roots. Define the fundamental chamber as the closure of the portion that pairs positively with all the
simple roots.
Lemma 9. The fundamental chamber is a chamber.
Proof. First, the fundamental chamber can’t intersect with the hyperplanes. Because if they do we find a vector which pairs positively with all the simple roots,
but pairs trivially with a positive root which is impossible. Since the fundamental
chamber is connected, it should lie completely in one of the chambers. Finally, if
we look at the map from that chamber to the reals given by pairing with any of
the simple roots we see by continuity and connectedness that all of that chamber
should be in the fundamental chamber.
Theorem 12.
(1) W acts transitively on the chambers.
(2) If we choose another f 0 and get other positive roots, there is a reflection
that takes one set of positive roots to the other.
Proof. First one is quite good. Find points in the two chambers such that the
straight line connecting them doesn’t hit any of the points that are in the intersection of two (or more) of the hyperplanes. We walk along that segment, and once we
hit a hyperplane, we make the corresponding reflection (do not change the straight
line).
For the second one, we find the fundamental chambers from the two functionals.
By the previous point, there is a reflection sending one to the other. This implies
that that one reflection sends one set of simple roots to the other, which finishes
the proof.
Corollary 9. Cartan matrix is independent of the choice of the functional.
QUALIFICATION EXAM
13
Lemma 10. Let sit−1 , . . . , si1 be simple reflections and αit be a simple root such
that si1 . . . sit−1 (αit ) is a negative root. Then si1 . . . sit−1 sit can be reduced to the
word si1 . . . sˆir sit−1 .
Proof. There will be a first time when a positive root becomes a negative one. In
that step, the root we have should be the same as the reflection we are applying
by a previous point. Let w be the reflection that took αit to αir which is that
root. Then sir . . . sit−1 sit is the same as sir+1 . . . sit−1 by the change of basis trick,
proving the result.
Theorem 13. W acts simply transitively on the chambers.
Proof. We do this by showing that if a reflection fixes the fundamental chamber,
then it is the identity. Indeed, in that case, it has to fix the set of simple roots. Let
us write the reflection as a reduced word si1 . . . sit−1 sit in the simple reflections,
then si1 . . . sit−1 (αit ) should be a negative root, and by the lemma, this gives a
contradiction to the word being reduced.
1.16. A bit more genera theory.
Definition 3. We define the universal enveloping algebra U (g) of a Lie algebra to
be smallest (in the categorical sense) unital associative algebra with a Lie algebra
map from g with respect to its commutator Lie algebra structure. It is unique up to
unique isomorphism by abstract nonsense. It exists, because we can choose a basis
of the Lie algebra, and take the algebra with the commutation relations. Notice
that representations of g is the same as the representations of U (g).
Theorem 14. Poincare-Birkhoff-Witt theorem. Consider the explicit construction
above. Denote the basis by a1 , . . . , an . Then the ordered monomials form a basis of
U (g).
Theorem 15. It is easy to see that they span U () - unordered monomials does by
construction, then we use commutation relations. For the linear independence, we
define the following map U (g) → B, where B is the vector space generated by the
ordered monomials in some set of n generators. We send the ordered monomials
to unordered monomials, and define the map for unordered polynomials by making
some random choices of commutations but in the most effective way (there is a
number only depending on the arrangment of the numbers), we get down to a linear
sum of ordered monomials, and define the map by linearity. We need to show
that this map is well defined, i.e. it doesn’t depend on the choices. For this we
make induction on the number of inversions on a word. We end up with two cases
(reminiscent of braid relations), and when the commutations are far apart, it is
easy, when they are next to each other, we need Jacobi identity.
Let us take a symmetric invariant nondegenerate pairing (·, ·) on g. Then take
a basis (u1 , . . . un ) of g, and let (v1 , . . . vn ) be the dual basis. Then the element
Σui vi doesn’t depend on the choice of basis, and it is called the Casimir element Ω.
(do this computation!) The matrix of an adjoint action of an element a ∈ g with
respect to these two bases are negative transposes of each other.
Given a representation V of g, define the Lie algebra cohomology as follows The
cochains are given by alternating k-multilinear maps Λl g → V , and the differential
is
(5)
dφ(g1 , . . . , gl ) =
14
UMUT VAROLGUNES
(6) Σ(−1)i gi φ(g1 , . . . , ĝi , . . . , gl ) + Σi<j (−1)i+j φ([gi , gj ], g1 , . . . , ĝi , . . . , ĝj , . . . , gl )
The cohomology of this complex is called the Lie algebra cohomology. Zero cycles
are the elements killed by all of g. One coboundaries are just action on a fixed
element. One cocycles satisfy
(7)
φ([g1 , g2 ]) = g1 φ(g1 ) − g2 φ(g2 ).
If we take a simply connected Lie group, its cohomology should be computable
from the data of its Lie algebra, and this is what Lie algebra cohomology does when
the module is itself I believe. It should be thought of as the equivariant version of
the deRham complex. Need to think more about this, it’s cool.
The main theorem on cohomology will tell us that the first cohomology will
vanish for all representations if g is semisimple. This is a generalization of the
fact that all derivations are inner in a semisimple Lie algebra - take the adjoint
representation - which was used to prove that the abstract Jordan decomposition
is unique, so we do need to prove that seperately. Before that note that Lie algebra
cohomology is additive under direct sums of representations. Also:
Lemma 11. If φ is a one cocycle for V , then we have the following Ωφ(a) =
aΣui φ(vi ). As a corollary, multiplication by Ω gives a representation automorphism
of V .
Proof. The corollary follows from the first statement when φ is a cocycle. Ωφ(a) =
Σui vi φ(a) = Σui aφ(vi ) + Σui φ([a, vi ]) = Σaui φ(vi ) + Σ[ui , a]φ(vi ) − Σui φ([a, vi ]).
Moreover, Σ[a, ui ]φ(vi ) + Σui φ([a, vi ]) = Σaij uj φ(vi ) + Σui φ(−aji vj ) = 0 by some
previous computation.
Theorem 16. When g is semisimple, H 1 (g, V ) = 0.
Proof. First we can reduce to the case, when the representation is faithful. We mod
out by the kernel, check that the semisimplicity is preserved, that the theorem for
the quotient implies the original one, which are both easy.
If we apply Cartan’s criterion for solvability, we get that the invariant symmetric
bilinear form (·, ·)V defined through taking trace inside V is also nondegenerate.
Now take Ω with respect to (·, ·)V . As for every operator V splits into a direct
sum of Ω’s generalized eigenspaces and these are invariant under g by what we have
discussed before. So, we can assume that Ω is either nilpotent or invertible. If it is
nilpotent, we compute the trace of Ω by expanding it using its definition, and find
that the dimension of g has to be zero. If it is invertible, we use the main statement
of the lemma above, where pass omega to the other side, and commute it with a,
which says explicitly that φ is a coboundary.
The following theorem is called Weyl reducibility theorem.
Theorem 17. If g is semisimple, then U () is semisimple.
Proof. Let V be a representation of U (g), and hence of g, and let U be a subrepresentation. We want to show that there exists a complementary subrepresentation
U 0 such that V is isomorphic to U ⊕ U 0 as a representation. Take an arbitrary
complementary subspace W and we will modify it. Note that this lets us define a
projection P0 = V → U - the subset of all projections to U form an affine space
over the subspace M of End(V ) which is zero on U , and its image lies inside U ,
QUALIFICATION EXAM
15
and we just fixed an origin for it. Note that M as a subrepresentation of End(V ),
where we define the action of g by taking commutators with endomorphisms.
Now define a one cocycle g → M , by φ(a) = aP0 − P0 a. By the main theorem on
cohomology, this cocycle is in fact of the form φ(a) = ab − ba for some b ∈ M . Now
we can check easily that P0 − b gives a g-invariant splitting, since it is a projector
to U and also commutes with the actions of all elements of g.
Now we also give a proof of Levi’s theorem which was promised before.
Theorem 18. Let Rad(g) be the solvable radical of g. Then, there exists a semisimple s such that g = Rad(g) + s.
Proof. We will do induction the dimension of g. We divide the argument into
two cases. If Rad(g) is not abelian. Then, we consider g0 = g/[Rad(g), Rad(g)]
(because what the hell, why not). This has smaller dimension, so we use our
induction hypothesis, and write g0 = Rad(g0 ) + s0 . Let us call g1 the preimage of
s0 inside g. We have g = g1 + Rad(g). Note that g1 6= g, because the image of
Rad(g) is a non-empty solvable ideal in the quotient, so g0 can’t be semisimple.
Now we apply our induction hypothesis again to g1 , and write g1 = Rad(g1 ) + s1 .
By elementary but kind of confusing computations we see that Rad(g) + Rad(g1 )
is again a solvable ideal. Hence, it should be that it is equal to Rad(g), and we are
done.
Now assume that Rad(g) is abelian. Do the trick in the proof of Weyl complete
reducibility for V = g the adjoint representation, and U is the subspace given by
Rad(g) ∈ End(g). So we get M̃ . Note that we naturally have Rad(g) ∈ M̃ because
Rad(g) is an abelian ideal. We take quotient s = g/Rad(g), and also the quotient
M̃ /Rad(g), which is an s-module because Rad(g) acts trivially on M̃ (so s has an
action there), and then we use again that Rad(g) is an ideal to say that it is a
subrepresentation and we can pass to the quotient representation.
Now we take a projector P0 : g → Rad(g), and define φ : s → M by φ(s) =
s̄P0 − P0 s̄, where s̄ ∈ g is a lift of s. This is apparently a one cocycle. Then we
find a primitive m ∈ M , and want to use the projection P0 − m, but this time we
don’t directly get what we want because we have to lift everything up to g again,
we have the formula
(8)
ad(ã)(P0 − m̃) − (P0 − m̃)ad(ã) = ad(ra ),
where ra is some mystery element in the radical. If ad(ra )’s are all zero, then we
are done - in fact we have such a decomposition to ideals. If not, then we consider
g1 = {a ∈ g | P ad(a) = ad(a)P , which is a proper subalgebra, and we can put the
equation above into the form P (ad(a) − ra ) = (ad(a) − ra )P , which shows that any
element of g can be put in g1 by subtracting an element which is in the radical.
Hence, we have g = g1 + Rad(g), so we do the same thing that we did in the above
nonabelian case.
There is also the following uniqueness theorem for the Levi decomposition, which
is called Malcev’s theorem. I may write down a proof later, it follows a similar
strategy to Levi’s theorem - it’s easier.
Theorem 19. Any two Levi decompositions are related by an inner automorphism
(exponential of an inner derivation) of the Lie algebra.
16
UMUT VAROLGUNES
1.17. Representation Theory of Semisimple Lie algebras. In addition to all
the choices we have been making, choose an ordering of all positive roots, this will
be used in conjunction with PBW theorem.
The key definition is of a highest weight g-module. A highest weight module
V with highest weight Λ ∈ h∗ is defined by the property that it contains a vector
which is in the weight space VΛ , which is killed by n+ , and which generates the
whole of V under the action of the universal enveloping algebra. Notice that for
the generation it is equivalent to require that it is generated by the part of the
universal enveloping algebra generated by n− .
A singular weight vector is one that is described in the above paragraph without
the generation requirement, and such weight is called a singular weight. Let us also
define D(Λ) for a weight Λ, the negative quadrant of the lattice generated by the
simple roots with origin taken to be Λ. Let V be a highest weight g-module with
highest weight Λ.
L
• V = λ∈D(Λ) Vλ . This follows from the PBW theorem and a small computation of how the eigenvalues of elements in h change when a highest
weight vector is hit by E−β1 . . . E−βn - it becomes Λ − β1 − . . . − βn . Note
that some of these weight spaces can be empty.
• VΛ is one dimensional and all weight spaces are finite dimensional. This
follows from the fact there are only finitely many ways to split a nonnegative
integer vector into other nonnegative (nonzero) vectors. First statement is
a degenerate case of this too.
• V is irreducible if and only if Λ is its only singular weight. If there is another
singular weight, then the part that it generates is a proper submodule.
Conversely, if there
L is a submodule U , then its weight space decomposition
is given by U = λ∈D(Λ) (Vλ ∩ U ). Let us take one of the minimal λ’s such
that Vλ ∩ U is nonempty. Then this weight should be singular.
• V contains
a unique proper maximal submodule. This also follows from
L
U = λ∈D(Λ) (Vλ ∩U ), because if we take the sum of all proper submodules,
it will not contain the highest weight vector.
• Let us compute Ω(v) where v is a singular vector with weight λ. We
choose a basis of g which looks like E±α , α ∈ ∆+ and Hi a basis of h, such
that K(Eα , E−α ) = 1. Then Ω(v) = (ΣEα E−α + ΣE−α Eα + ΣH i Hi )v =
2ΣE−α Eα v+Σν −1 (α)v+Σν −1 (Hi )λ(Hi )v = λ(ν −1 (Σα+Σλ(Hi )Hi ))v =<
λ, λ + 2ρ > v, where 2ρ = Σα. Note that if v is the highest weight vector,
this implies the same result for the whole of V . Also if λ is another singular weight, then we have < λ, λ + 2ρ >=< Λ, Λ + 2ρ >, which can also be
rewritten as < λ + ρ, λ + ρ >=< Λ + ρ, Λ + ρ >.
• If the highest vector happens to be a real valued functional (as opposed
to for example complex valued), then this shows that there can be only
finitely many singular weigths, because we know that they should lie in
some lattice and the previous points shows that they also lie in a compact
subset.
Let us define M (Λ) to be the universal example of a highest weight module with
highest weight Λ. This can be defined as the obvious quotient of the universal
enveloping algebra, and clearly satisfies the universal property (any such module is
a quotient of M (Λ)). This comes with a basis once a functional is chosen in h∗ . By
one of the points above it has a unique maximal proper submodule, the quotient of
QUALIFICATION EXAM
17
M (Λ) by that submodule is called L(Λ). Note that Λ determines M (Λ) and L(Λ),
and more surprisingly the converse is also true.
We make the last set of important definitions now. Let b = n+ ⊕ h be called
−1
(α)
a Borel subalgebra. We have [b, b] = n+ . Let us define the coroots to be 2ν(α,α)
,
where α is a root. Then define the lattice P of functionals in h∗ which evaluate to
an integer for all coroots to be the weight lattice. If all the integers are positive,
then we call such a weight dominant, of which set we represent by P+ . Notice that
the lattice of roots lies inside P by the string property. Also let ρ = 1/2Σα as
before.
Theorem 20. The g-modules {L(Λ}Λ∈P+ are all finite dimensional g-modules.
Proof. We know that L(Λ)’s are irreducible, and that they are finite dimensional
follow from Weyl’s dimension formula. For the other direction, note that b is
solvable (its commutator subalgebra is nilpotent). By Lie’s theorem, there is a
λ ∈ b∗ such that bv = λ(b)v, for every b ∈ b. If n ∈ n+ , then n = [b, b”], and it
follows that nv = 0. Since the representation is irreducible v should be a generator.
To see that the weight is dominant integral, we use the sl2 -triples inside, and of
course the main lemma on sl2 .
Theorem 21.
• dim L(Λ) = Πα∈∆ (Λ+ρ,α)
(ρ,α) , for Λ ∈ P+
• Let us define the formal character of a representation by ch(L(Λ)) = Σλ∈h∗ dim Vλ eλ .
Then, for Λ ∈ P+ , we have eρ Rch(L(Λ)) = Σw∈W (det w)ew(Λ+ρ) , where
R = Πα∈∆+ (1 − e−α ).
Proof. The dimension formula follows from the character formula as follows. We
can define the map Fτ from the formal characters (with finitely many terms) to
real numbers by eλ → e(λ,τ )t for all τ ∈ h∗ . Now let us apply Fρ to both sides of
the Weyl character formula. We find:
(9)
et(ρ,ρ) Π(1 − e−t(ρ,α) )(Σ dim L(Λ)λ et(λ,ρ) ) = Fρ+Λ (Σw∈W (det w)ew(ρ) ).
By putting Λ = 0 in the character formula we derive the Weyl denominotor formula
(10)
eρ R = Σw∈W (det w)ew(ρ) ,
which we can substitute in the previous equation.
1 − e−t(Λ+ρ,α)
.
1 − e−t(ρ,α)
We finally do something cool and let t → 0. Using L’Hospital’s rule for the RHS
we get exactly what we want.
Now we start proving the Weyl character formula. We do it step by step:
(1) Weyl group acts on formal characters by acting on the exponents. We first
show that ch(L(Λ) is invariant under this action. This will be by reduction
to sl2 , in which case the main lemma shows the claim. Let α be a root
and let < E, F, H > be the corresponding sl2 triple, call it s. We will show
that the reflection corresponding to α fixes the character. If we restrict the
representation to s, we can compute its character simply by projecting the
exponents into the line generated by α. Moreover, if V is an irreducible
subrepresentation then by the main lemma, we know that it should be
generated by one vector by applying F 0 s and this shows that the orthogonal
complement of < α > acts exactly the same way on all of V . Now the only
(11)
e(t(ρ,ρ) (Σ dim L(Λ)λ et(λ,ρ) ) = et(Λ+ρ,ρ) (Πα∈∆+
18
UMUT VAROLGUNES
(2)
(3)
(4)
(5)
(6)
(12)
things to show is that this splits into irreducible s representations. Since
Λ(H) is integral, we can turn the highest weight vector into another singular
vector by applying F enough times, and that should be zero. Now take the
direct sum of all irreducible s subrepresentations of L(Λ). It is not hard to
see that this is a g-submodule and hence it should be everything.
w(eρ R) = det(w)eρ R. It suffices to show this for w = si . Here we note
that ρ is also equal to ω1 + . . . + ωn , where ωi are the fundamental roots.
This is because the set of positive roots except αi is invariant under the
reflection si , and writing down the actualy formula for what the reflection
does on ρ gives the desired claim. Using the same argument, w(eρ R) =
eρ−αi (1 − eαi )Πα6=α1 (1 − e−α ), which proves the statement.
We can compute the characters of the Verma modules directly. RchM (Λ) =
eΛ .
We can write the character of a finite dimensional highest weight representation of weight Λ in as a positive integer linear combination of the
characters of L(λ)’s where λ satisfies < λ + ρ, λ + ρ >=< Λ + ρ, Λ + ρ >,
and the coefficient of L(Λ) is 1. This is by induction, using the fact that all
irreducible subrepresentations are of the form L(λ) for such λ’s, when we
take quotient by such subrepresentation all the properties are still satisfied,
and finally that the character splits under short exact sequences.
The same with M (λ)’s instead of L(λ)’s. For this we use the previous
statement for V = M (λ). This will give us some matrix with positive
integer entries aµν , where µ and ν are weights satisfying < λ+ρ, λ+ρ >=<
Λ + ρ, Λ + ρ >. I don’t quite understand this step. Let me just skip it for
now.
We use the previous step to decompose the character into characters of
M (λ)’s. We multiply both sides with eρ R, and use the formula for characters of M (λ)’s. Then we use the fact that the LHS of the equation
eρ RchL(Λ) = Σλ∈B(Λ) aλ eλ+ρ is antilinear with respect to the Weyl group
action, so should be the RHS. We have the terms obtained from the 1 · eΛ+ρ
and we want to show that there is no other. If there is one such aλ 6= 0, we
can make it so that λ + ρ is dominant integral weight by reflections. Also,
Λ − λ = α pairs nonnegatively with al Hi ’s.
< λ + ρ, λ + ρ > − < Λ + ρ, Λ + ρ >=< Λ + 2ρ, α > + < α, λ > .
Λ and λ + ρ pairs nonnegatively α, and ρ pairs positively, which finishes
the proof.
2. Characteristic Classes
2.1. Rather arbitrary notes.
• One can consider Pn as the set of symmetric rank 1 projections with trace
one by means of exterior product (as physicists do in bra-ket notation).
This generalizes in the obvious way to other Grassmanians.
• For the naturality axiom of characteristic classes, it is better to say it for
all vector bundle maps (require tit to be an isomorphism of the fibre vector
spaces) covering the given map.
QUALIFICATION EXAM
19
• Two real manifolds are cobordant if and only if their Stiefel-Whitney numbers are the same. One side is elementary, but the other is a theorem of
Thom.
• For Stiefel-Whitney classes you really work with Z2 coefficients, which
makes the computations with real projective spaces very pleasant.
• To relate the tangent bundle of projective spaces to the tautological line
bundle one either uses Euler sequence or its identification with the bundle
Hom(γ, γ ⊥ ) - this is easy to see when one passes to the sphere above the
projective space.
• If the total space of a principal G-bundle is weakly contractible (aspherical)
then it is a model for a universal principal G-bundle (or its base for a
classifying space).
• One can construct an EG → BG explicitly by either Milnor’s join construction or as the geometric realization of a simplicial set constructed
using group operations. One can also deduce the existence of EG → BG
using Brown representability theorem. In the case of vector bundles (i.e.
G = GLn ) there is a more concrete description via infinite Grassmanians.
• A central problem is to compute the cohomology of BG. For example G =
GLn (R) is hard. This is where characteristic classes come from essentially.
The method I know is to work with the whole family BGn , and do induction
using SSS. One needs to identify what is Gn /Gn−1 - in the cases one usually
considers for characteristic classes this is a sphere.
• To embed a vector bundle into a trivial bundle there are two approaches.
One can first choose sections which generate everything, i.e. a projection
from a trivial bundle, and take an orthogonal complement to the kernel, or
one can cover the base with trivializations, pick cutoff functions subordinate
to the covering, and map to (Rn )×N , where N is the number of open sets.
When the base is paracompact (CW complexes, metric spaces, countable
unions of compact regular spaces) then the second procedure lands you in
R∞ after a refinement of the covering - by the local finiteness condition.
Also note that a bundle map to the tautological bundle over the Grassmanian (infinite or not) is the same thing as a vector bundle embedding as
above.
• A very good toy case of classifying space theory: P rinG (S n ) = πn−1 (G),
heuristically this is clear by clutching function construction.
• Homotopy covering theorem: We have the following diagram of fiber bundles where the upper map is a fibrewise homeomorphism and X is nice:
(1)
E
/F
X
/Y
If we are given a homotopy X × I → Y , then there is a covering homotopy
(again a fibrewise homeomorphism) E ×I → F making the obvious diagram
commute.
• Homotopy covering theorem can be used to prove that homotopic maps
induce isomorphic bundles. For principal G-bundles over smooth manifolds this can also be done using connections. In MS, they never use this
somehow.
20
UMUT VAROLGUNES
• MS does prove that any vector bundle over X gives rise to a well defined
homotopy class of maps X → Grn (R∞ ) via some trickery.
• Schubert cells provide a cell decomposition for Grassmannians. These are
cells indexed by partitions of m into n numbers, with easily computed
dimensions.
• Thom isomorphism for orientable bundles. This is very easy to prove using
the Serre Spectral sequence. The only tricky point is to use SSS for fiber as
the pair (vectors, non-zero vectors), and realize that the orientablity is the
same as the local system on the base being trivial. Note that the isomorphism is given by taking cup product with the Thom class which restricts
to the given orientation of each fiber. This class can also be constructed
using differential geometry.
• The Euler class is a characteristic class in H n (M, Z) for oriented real vector
bundles. It satisfies all the possible axioms it could satisfy. It is two-torsion
when n is odd. Its mod−2 reduction is the top dimensional Stiefel-Whitney
class. It can be defined using Chern-Weil theory by taking the invariant
polynomial as the Pfaffian - note that this exists only when the bundle is
orientable. It can also be defined by extending the Thom class from (E, E0 )
to E, this correspoonds to pulling back the square of Thom class via the
Thom isomorphism.
• Steenrod squares are usually used in the construction of Stiefel-Whitney
classes.
• Obstruction theory: Let F → E → B be a fibration, where B is a CWcomplex, and let B [k] be its k−skeleton. Assume that we are given a section
σ on B [k] , then we can extend it to a section on B [k+1] only if a certain
k+1
class in HCW
(B, πk (F )) ' H k+1 (Homπ1 (B) (C∗cell (B̃) → πk (F ))) vanishes.
This class is constructed as follows: For each k + 1-cell α : Dk+1 → B, we
send α̃ to an element in πk (F ) represented by σ |δα : S k → α∗ E under the
isomorphism πk (α∗ E) ' πk (Fα̃(0) ) ' πk (F ), where the second isomorphism
can be chosed consistently for all cells of the universal cover. It is not trivial
to see that this is a cocycle. Once that is established one can also show that
the cohomology class of the cocycle is independent of the section chosed on
the k-skeleton. The point is that if the section extends to k + 1-skeleton,
the obstrcution class vanishes.This construction is natural with respect to
pullbacks. See Hutching’s notes.
• Consider the Stiefel bundle of k−frames associated to a rank−n real vector
bundle. One can check that Vk (n) is n − k − 1 connected - consider the
bundle GL(k) → Vk (n) → Grk (n), note that Grk (n) → Grk (∞) is n − k − 1
connected and the map ΩGrk (∞) → GL(k) is a homotopy equivalence.
Also there is a well defined mod−2 reduction of the obstruction class for
extending section along the n−k+1-skeleton, which gives rise to an element
H n−k+1 (B, Z/2Z), which is precisely wn−k (E). In the case of k = 1, and
orientable E the local system is trivial so we get a class in H n (B, Z), and
this class is the Euler class.
• The way MS defines Chern classes is actually very similar to Grothendieck’s
method. You define them inductively You take a rank-n complex vector
bundle E → B, define a rank-n − 1 vector bundle over E0 by taking orthogonal complements (either by picking a Riemannian metric or by taking
QUALIFICATION EXAM
•
•
•
•
•
21
quotient). Then write down the Gysin sequence for E0 → B - note that E
has canonical orientation - and define the characteristic classes except the
top dimensional one. Finally, the top Chern class to be the Euler class of
the underlying oriented vector bundle.
Pontrjagin classes are defined by complexifying the real vector bundle.
They exist at degrees which are multiples of four. They are directly related to the cohomology of the Grassmanian of real oriented planes with
coefficients in a ring where 2 is inverted. If one takes bundle E which is
already complex and complexifies it, he gets E ⊕ Ē, and correspondingly
there is a formula relating Pontrjagin and Chern classes of E. If we have
an oriented even rank real vector bundle, its top degree Pontrjagin class is
the square of its Euler class.
Chern and Pontrjagin numbers - defined in the obvious way. Again these
are cobordism invariants. Note that the definition can be modified to use
any basis of nth cohomology of the classifying space (do the pairing there).
We will use the one given by taking symmetric sums of monomials (powers
equal to only 1 or 0 gives the Chern classes, one variable monomials are
the Newton sums). Notice how a partition defines directly a class this
way. There is a product formula for Whitney sums. The nth Newton sum
element is crucial for the case of the tangent bundle of an n dimensional
manifold - this vanishes for product manifolds for example, deduce that
projective space is not a product. A nice theorem is that these numbers
are linearly independent - proved very ingeniously by taking manifolds with
dimensions 1, . . . , n with the crucial numbers nonzero, and arranging all the
possible numbers we get to a triangular matrix with nonzero entries in the
diagonal.
To prove splitting principle one should take the projectivization - use LerayHirsch. Universal instance of that and stuff it’s all very simple and nice.
This is the way to prove Cartan formula also. Never forget the functorial
meaning of BG it gets many things for free.
Boundary maps in the long exact sequence of cohomology is actually not so
weird. When we think about the Puppe sequence we see that they in fact are
just some actual pullback maps composed with the suspension map. For
example this shows that Steenrod squares commute with the connecting
boundary maps. This in turn shows that transgression commutes with
Steenrod squares.
Transgression. We say that a class in H n (F ) transgresses to a class in
H n+1 (B) if they map to the same element under the maps δ : H n−1 (F ) →
H n (E, F ) and H n (B) → H n (E, F ). Note that this is not map or something, just a relation - it is a map from a subspace to a quotient though.
A better way to understand trasgression is through the Serre spectral sequence, which identifies what this subspace and quotient are in terms of
the edge homomorphism. Consider the edges of the spectral sequence,
the mystery subspace and quotient and the well defined map is precisely
dn : En0,n−1 → Enn,0 . There is also a homological version of this and the
nice point there is that the spherical classes always transgress. This transgression explains the relationship between the cohomologies of U and BU
for example.
22
UMUT VAROLGUNES
• Do not get confused between the universal vector bundles and universal
principal G-bundles over the infinite Grassmanians. Also keep in my mind
the frame bundles and the Stiefel manifolds.
• One can compute the cohomologies of the Stiefel manifolds by brute force
with rational coefficients say. You first compute the low degree homotopy
groups - this I talked about a little bit above. You first compute for the
2-frame bundles using the spherical fibration. Then you use two different
kinds of fibrations for the induction, when the euler class should be there
you use the spherical one, otherwise you take two vectors and use one with
fiber a 2 frame bundle. In particular, you get the cohomology of O(n) and
SO(n).
• The cohomology of Grassmannians is harder. You get pretty combinatorial
formulas for the ring structure over integers involving Young tableux stuff
like Littlewood-Richardson numbers, well that precisely gives the structure
coefficients.
2.2. Grothendieck’s method for defining characteristic classes. The key
point here is the classical Leray-Hirsch theorem.
Theorem 22. Let F → E → B be a fibration such that H ∗ (E) → H ∗ (F ) is
surjective, and also assume that it has a section q : H ∗ (E) → H ∗ (F ). Let us also
assume that B is connected for safety. Then, the induced map H ∗ (F ) ⊗ H ∗ (B) →
H ∗ (E) is an H ∗ (B)-module isomorphism. If q is an isomorphism of rings, then we
get ring isomorphism.
Proof. This all follows from the edge homomorphism for the fibre edge, which means
that the map H ∗ (E) → H ∗ (F ) factors through the fibre edge of the E∞ page. By
construction of the Serre spectral sequence, the fibre edge of the E2 page is the
π1 invariants of the cohomology of the fibre. Since, H ∗ (E) → H ∗ (F ) is surjective,
these should actually be all of H ∗ (F ), and the differentials starting there should all
be zero after this point. Hence, the local system is trivial, and the spectral sequence
collapses at the E 2 page. To finish we notice that the map H ∗ (F )⊗H ∗ (B) → H ∗ (E)
respects the filtration of the antidiagonal and the filtration on H ∗ (E) given by the
construction of the spectral sequence. Since this map is an isomorphism at the level
of associated graded’s, we finish the proof.
Now the construction of Grothendieck goes as follows. Construct c1 or w1 in
some way for line bundles (take sections, or I guess more generally use the Thom
isomorphism). Take the projectivization of the given vector bundle, there is a
canonical line bundle over that, and the naturality of c1 shows that the conditions
of Leray-Hirsch theorem hold, with a preferred section. Then consider the image
of c1 ⊗ 1 in H ∗ E. The n + 1th power of it is not zero but it is in the image of
H ∗ (F )⊗H ∗ (B), and hence its preimage in H ∗ (F )⊗H ∗ (B) can be written uniquely
as Σni=0 ci ⊗ ci1 , the ci ’s are defined to be required classes. Their naturality follows
from among everything in the construction, the naturality of SSS.
For the Cartan formula one needs to grind a little bit. Take the natural embedding j : P(E) → P(E ⊕ E 0 ). There is a projection p : P(E ⊕ E 0 ) − j(P(E)) → P(F ).
Let us also denote the inclusion P(E ⊕ E 0 ) − j(P(E)) → P(E ⊕ E 0 ) by ρ. Define
E ∗
i
0
α = Σrank
πP(E⊕E 0 ) (ci (E))ξE⊕E
0 and let β be the same for E . What we are trying
i=0
∗
∗
to show is α ∪ β = 0. By naturality of c1 , both j α and ρ β are zero. Since ρ∗ β is
QUALIFICATION EXAM
23
zero, β is in the image of H ∗ (P(E ⊕ E 0 ), P(E ⊕ E 0 ) − j(P(E))), and α is in the image
of H ∗ (P(E ⊕ E 0 ), j(P(E))). Now by some generalities on cup product show what
we are trying to show (essentially the classes are supported in disjoint regions).
2.3. Computation of the cohomologies of BU, BO, BSO with convenient
coefficients. Cohomology of BU. We have U (n − 1) → U (n) → S 2n−1 . This
can be extended to the right by S 2n−1 → BU (n − 1) → BU (n). This is because
for any Lie group inclusion H → G, EG → EG/H is a model for the universal
principal H-bundle, and the natural map EG/H → EG/G is the natural map
BH → BG given by functoriality. For this it is enough to show that EG pulls back
to EG ×H G under the map.
Note that a model for BU (1) is CP ∞ , of which cohomology ring is just graded
polynomial ring in one variable with degree 2. I here want to remind myself that
it is actually note so easy to get the ring structure of the cohomology of projective
spaces. On that note I also want to remind myself how to construct Poincare duals
of submanifolds using the Thom isomorphism (also recall more generally the Gysin
map), and how it relates the cup product to geometric intersection.
We write the SSS for the fiber sequence above, and do induction. First define
cn - the image of the generator of the cohomology of sphere, which should map to
something non-zero, since it has to die. Show that the cohomology should lie in even
dimensions. Using the Leibniz rule show that all differentials are multiplication by
cn . Everything in the upper row should die, so we have an injective map which
is multiplication by cn and the quotient is a polynomial ring. We can choose a
section of this exact sequence, which is a ring homomorphism. Hence, our ring
is a direct sum of the image of cn and a polynomial ring. There is an obvious
homomorphism from a polynomial algebra, which is clearly surjective, and a bit
less clearly injective.
We can use the same proof for computing the cohomologies of BSU (n), BSp(n).
For U (n), SU (n), Sp(n), we write down the fibration and the spectral sequence.
This time the spectral sequence collapses by the Leibniz rule as all generators in
the first column are sent to zero. The short exact sequences split and we deduce
the ring structure from the spectral sequence.
Cohomology of BO. We want to do induction again: S n−1 → BO(n − 1) →
BO(n). We know that H ∗ (BO(1), F2 ) is a polynomial ring in one variable. We run
the Serre spectral sequence. We see that the generators of H ∗ (BO(n − 1), F2 ) are
in the image of the induced map H ∗ (BO(n), F2 ) → H ∗ (BO(n − 1), F2 ), hence this
map is surjective. Therefore the Gysin exact sequence splits, and shows that the
critical differential is injective. This differential is multiplication by some element,
which is the new generator.
Cohomology of BSO. For this we use a coefficient ring in which 2 is inverted.
I use rational coefficients. H ∗ (BSO(2), Q) is a polynomial ring on one variable of
degree two, and H ∗ (BSO(3), Q) is a polynomial ring on one variable of degree four.
Now we would like to do induction but we see that it doesn’t work because we lost
the surjectivity of the BO(n) case. Hence, we do something else. This works for a
general compact Lie group in fact.
Let T be a maximal torus in G, and let N be its normalizer. The Weyl group
is W = N/T . We factor the map T → G as T → N → G, which induces BT →
BN → BG. Using the trick for finding a model of the classifying space of a
subgroup, we see that BT = EG/T , and BN = EG/N . Let us first analyze
24
UMUT VAROLGUNES
r : H ∗ (BN, Q) → H ∗ (BT, Q) using the W → BT → BN = BT /W . The important
point is that there is also a map H ∗ (BT, Q) → H ∗ (BN, Q) called the transfer
map. These two maps compose to |W | times the identity in one direction showing
injectivity of r. The image of r is invariant under the action of W - one can show
easily that the map BT → BN is invariant under the action of W . Final point to
make is that this map is surjective onto the invariant classes, but this is also trivial
using the fact the transfer map is defined on chains first.
Now we consider the fibration (G/T )/W → BN → BG. First we show that
H ∗ (G/T, Q) is concentrated in even degrees and G/T has Euler characteristic equal
to |W |. This is a direct consequence of Bruhat decomposition. There should also be
some explanation using Morse theory, but I don’t exactly know how to do that. But,
after we mod out by W , the Euler characteristic becomes 1, and hence (G/T )/W has
trivial homology. Hence the SSS tells us that BN → BG induces an isomorphism
of rings on cohomology.
We just need to find the invariants in the polynomial algebra corresponding
to the cohomology of BT under the Weyl group action. In the case of SO(n),
this action is described by permuting and changing signs of variables. In the case
n is odd, we can change signs whichever way we want, so we get the symmetric
polynomials in the squares of the variables (which originally had degree 2). When
n is even, we can change signs only even times. So we also have the product of the
variables, which is the Euler class. Note that above I worked with rationals but
the only prime that is actually relevant is 2 it turns out. I am a bit confused about
this.
There are also the following two very cool theorems, which tells us that the
cohomology of BG is in even degrees directly for example. The first one is Hopf’s
theorem, and was the reason why Hopf algebras was invented in the first place.
The second one is called Borel’s transgression theorem. If we already know that the
generators transgress, then the rest is actually not so hard to prove by constructing
some model and using the comparison theorem for spectral sequences (see Zeeman’s
paper).
Theorem 23. The rational (or maybe real actually) cohomology of a compact connected Lie group is an exterior algebra on odd generators.
Theorem 24. If we have a spectral sequence that looks like the E 2 page of a Serre
spectral sequence where the local coefficients is trivial, and the total space has no
cohomology, and we further have that the fibre edge is an exterior algebra on odd
generators, then:
• We can find homogenous generators of the exterior algebra which transgress.
• The cohomology of the base is a polynomial algebra generated by the transgressions of those generators - in particular, these generators are in even
degree.
3. Major
3.1. Classical Integrable Systems.
3.1.1. Duistermaat’s paper.
QUALIFICATION EXAM
25
What is here?
• Construction of action-angle coordinates locally
• The monodromy and gluing invariants of a Lagrangian fibration, and how
they completely determine the symplectic type of the fibration.
Let (X 2n , ω) be a symplectic manifold. A classical integrable system can be
described as either:
• A Lagrangian fibration π : X → B n possibly with singularities.
• A surjective proper map π : X → B n such that the Poisson bracket of any
two functions pulled back from B is zero, and that π is a submersion on an
open dense subset of X.
Initially we assume that the map π : X → B n doesn’t have singularities, i.e.
it is a Lagrangian fibration on the nose, or it is a submersion everywhere. Their
equivalence can be seen using Ehressmann’s theorem plus some manipulations with
the Lagrangian condition.
Notice that there is a fiberwise action of T ∗ B on X - take a covector at b,
its pullback defines a covector on π −1 (x), which in turn defines a vector field on
π −1 (x), which is tangent to the fibre by the Lagrangian condition, and finally take
the time-1 map of this vector field. Since the fibres are compact this implies that
they should be disjoint union of n-tori - the orbits have to be open subsets, so the
orbits are precisely the connected components, result follows from the compactness
of the fibres and properties of discrete subgroups in Rn . This also gives a smoothly
varying full rank lattice P in T ∗ B assuming that the fibers are connected (or one
can think about the ramified finite covering of B having the smooth connected
components of the fibers as its points). The fact that this is an integral affine
structure follows from the existence of action coordinates.
Theorem 25. Around each of these tori, there are coordinates I1 , . . . , In on the
base (around the point it projects to) and θ1 , . . . , θn in the fibers such that ω =
dI∧dθ. In other words, there is a fibre preserving symplectomorphism with a product
neighborhood of the tori and T ∗ T n → Rn - Weinstein neighborhood theorem would
imply the existence of only a symplectomorphism.
Proof. Notice first that the key point is really to produce the action coordinates.
We want to produce them in such a way that they are coordinates on the base, and if
we lift them to functions on the total space and run the Hamiltonian flow, their time
one flows are all identity. Once we do that we can first locally construct a section
over the open subset in the base by locally (in the total space) completing the action
variables to a symplectic coordinate system - which we do by a generating functions
argument. Then we can just move the points around using the Hamiltonian flows
of the functions and parametrize the torus using that.
In the first proof as one of the steps is tricky in my opinion. Choose a trivialized
chart U and coordinates x1 , . . . , xn in the base. Also choose a basis of P over this
chart, which we see as one-forms over U . We are trying to show that these sections
(i)
are closed (and hence exact). More explicitly, we have functions Tj (x1 , . . . xn ),
(i)
(i)
and we want to show that ∂k Tj = ∂j Tk . We have a fiberwise action of U × Rn
on π −1 (U ) as described before, hence a map π −1 (U ) ×U (U × Rn ) → π −1 (U ).
Restricting this action to the graph of T (i) inside U × Rn gives the identity map,
so we write the identity map as a composition of the inclusion and the action map,
26
UMUT VAROLGUNES
∂
and see what happens to the vector field ∂x
using the chain rule. This is essentially
l
equivalent to computing how the action changes when we change xl along the graph,
which changes both xl and T ( i) - Duistermaat does this by heart. I would have to
do it by computing the Jacobians.
In the second proof, we start with a product neighborhood which is also a Weinstein neighborhood of a fibre. In this neighborhood, the symplectic form is exact.
We choose a primitive, and also a smoothly varying basis for the first integral homology of the fibres. We will integrate the primitive along the cycles to find the
action coordinates. Note that changing the basis would change the coordinates by
the action of GL(n, Z), and changing the primitive by the translation action of Rn
- and this is all the freedom there is when the lattice is given.
Let y, θ be the Weinstein coordinates. We have a Lagrangian foliation by the
Liouvile tori, and all the leaves are of the form (θ, y(θ, I)), i.e. they are graphical
over the center leaf, and we use I to parametrize the set of leaves. Also fix some θ0 .
Define the multivalued generating function S(θ, I) by integrating ydθ along paths
lying inside the leaves and starting at θ0 . Now consider the canonical transformation
defined by S and denote the canonical momentum of I by φ, which is a multivalued
function. Let’s see what happens to this functions when we go along the n linearly
independent vector fields a full turn (the basis is the same basis we used for H1 ).
Change in the ith φ coordinate after a full turn in jth basis element is easily
computed to be 2πδij (just change the places of difference with derivative, also
there is a normalization mistake here but anyways) which proves the claim.
We can ask when X → B is isomorphic to T ∗ B/P (as a bundle). Clearly, they
are smoothly isomorphic if and only if there is a section of X → B. Morever, this
isomorphism is also a symplectomorphism if and only if the image of the section is
Lagrangian. But there is a better answer.
Cover X with action-angle coordinates, and choose arbitrary Lagrangian sections
in each chart. In the intersections of charts the sections differ by a Lagrangian
section (over the intersection) of T ∗ B defined up to translation by P . These satisfy
the cocycle condition, and hence define a Cech cocyle µ and a Cech cohomology
class, [µ] ∈ H 1 (B, ΛL (T ∗ B/P )), where ΛL (T ∗ B/P ) denotes the sheaf of Lagrangian
sections. Clearly, the vanishing of [µ] implies the existence of a Lagrangian section.
If we instead see µ as a 1-Cech cocycle of Λ(T ∗ B/P ), then the vanishing of the
cohomology class implies the existence of a smooth section. Consider the following
diagram of short exact sequences.
(1)
0
/P
0
/P
=
/ ΛL (T ∗ B)
/ ΛL (T ∗ B/P )
/0
/ Λ(T ∗ B)
/ Λ(T ∗ B/P )
/0
Identifying the sheaf of Lagrangian sections of T ∗ B with the sheaf of closed oneforms on B, the following short exact sequence (of sheaves) also comes in handy
(always does).
(2)
0
/R
/ C ∞ (B, R)
/ C ∞ (B, T ∗ B)
clsd
/0
QUALIFICATION EXAM
27
Being vector bundles (and hence O-modules), Λ(T ∗ B) and C ∞ (B, R) have no
∞
higher cohomology. So we get isomorphisms H k (Cclsd
(B, T ∗ B)) → H k+1 (B, R).
It is also nice to describe them explicitly. Take a Cech cover (contractible intersections) of B and take sections on k + 1-fold intersections which define a cocycle,
and pick primitives for the closed forms. The coboundary operator applied to the
primitives gives a constant in the k + 2-fold intersections. This is a cocycle because
in the Cech complex is a complex. This construction of course works in general
for the connecting map in the long exact sequence of sheaf cohomologies. For the
case k = 1, there is another description which will be used below. Take a one
cocycle, this is a coboundary in Λ(T ∗ B), so choose a 0-cochain lift there (we will
have a given lift below). Now apply exterior derivative to this, which glues to a
well-defined 2-form, whose deRham cohomology class is what we need.
Using the second row in (1) above, we can push the obstruction class to H 2 (B, P ).
If this class vanishes, then we get a section B → X, since Λ(T ∗ B) has no cohomology. This doesn’t give a Lagrangian section yet. Let us now assume that for some
section s : B → X, s∗ ω is exact. Take a cover of B by action angle coordinates,
s defines local one-forms over the action coordinate charts, and their coboundary
∞
defines an element of H 1 (Cclsd
(B, T ∗ B)) (modulo P ). This class maps to [µ] under
the map in the first row of (1), because if we instead of si ’s took a section of P we
would get the same cocycle. Finally, note that dsi = s∗i ω, and hence the deRham
2-cohomology class we constructed above is really the class of ω. Hence, if ω is
exact [µ] = 0.
Having answered the question when X is the same as T ∗ B/P , we now turn to
the question of when T ∗ B/P is a trivial bundle.
The local system P is trivial if and only if its monodramy representation is
trivial. First notice that by passing to a covering space of the base, we can make
the local system P trivial - we can transfer to whole data to a covering space by
pullback, and the local system there, which is also just the pullback local system,
will be trivial.
In case we have P trivial, we have a trivialization of the cotangent bundle with
closed one forms. If those one forms happened to be exact, they would define a
local diffeomorphism to Rn , and the integral affine structure on B would be pulled
back from the standard one on Rn .
Passing to the universal cover (maybe less) we can kill all the H 1 , and make
these forms exact but they are also automatically
exact if ω is exact. We will show
R
this by showing that for any loop γ in B, γ a, where a is one of these forms. Let
us form a two dimensional cycle β over γ by spinning a lift of γ capped off to a
loop within the fiber over the basepoint using the vector field associated to a. By
a local computation
R
R(use partitions of unity and cover by action-angle coordinates)
we can show γ a = β ω, which finishes the proof.
3.1.2. Almost toric manifolds.
What is here?
• Integral affine structures associated to Lagrangian fibrations
• Toric manifolds and symplectic boundary reduction
• Singularities of classical integrable systems, in particular the local model
for the focus-focus singularity in 4 dimensions
• Base diagrams, and operations on the almost toric manifold that can be
described through them
28
UMUT VAROLGUNES
• Classification of almost toric manifolds
In this paper, we tend to see the integral affine structure in the tangent bundle
instead of the cotangent bundle. Here one point that confuses me is that a lattice
in the cotangent bundle defines an integral affine structure if locally there is a basis
given by differentials of functions but in the tangent bundle we should have the Lie
brackets of the vector fields to vanish. To see the equivalence of the two (for dual
lattices) one has to make a computation, that I still did not do.
A more direct way to construct the lattice in the tangent bundle is to glue the
locally defined integral affine structures (via the action-angle coordinates) together.
More precisely, we have
Lemma 12. Fiber-preserving symplectomorphisms of (Rn × T n , dpdq) are (p, q) 7→
(A−T p + c, Aq + f (p)), where A ∈ GL(n, Z), and A− 1 ∂f
∂p is symmetric.
Proof. Such map Rn × T n → Rn × T n is equivalent to a fiber preserving map
T ∗ Rn → T ∗ Rn which sends every translate of the standard lattice to a translate of
the standard lattice - apparantly I get confused about maps T n → T n . Hence if we
write the map as (p, q) 7→ (φ(p), A(p)q + f (p)) and write down what it means to be
a symplectomorphism, we are done - A(p) has to be constant.
Note that we didn’t assume any integrality anywhere. This is kinda surprising
actually.
We can define affine subspaces, and affine length.
There are two different notions of monodromy when one is given a Lagrangian
fibration. One is the monodromy of the first homology bundle and the second is the
monodromy of the local system. If we take the local system in the cotangent bundle, then these are the same under the induced identification of Tb∗ with H1 (Fb , R),
hence they are transposed inverses of each other if we use the tangent bundle.
Interlude: Atiyah-Guillemin-Sternberg convexity
Lemma 13. Let M be a compact connected smooth manifold, and f be a MorseBott function such that all the critical manifolds have index and coindex not equal
to one, i.e. the ascending and descending manifolds are not of codimension one.
Then the level sets are connected.
Proof. The key point here is that the complement of a codimension not equal to one
submanifold of a connected manifold is also connected, which is in stark contrast
with the Jordan-Brouwer seperation theorem for example. To prove this for high
codimension we use the Sard Theorem, which implies that any two submanifolds
can be perturbed so that they become transverse
We first show that there is exactly one index zero critical manifold - same holds
for coindex of course. We know that M is the union of all the ascending manifolds,
and if we use the previous paragraph, we find that the union of all codimension zero
ascending manifolds is connected, as desired. Hence we have unique local minimum
and maximum.
If c is just a bit above the minimum: then take two points in the level set, connect
them by going down to the minimal value critical manifold, move the path to miss
the critical manifold, and push them up with gradient flow. It continues like this.
I skip it for now.
QUALIFICATION EXAM
29
One of the ideas is to notice that if we choose a point in the Lie algebra of the
torus, which defines a one parameter subgroup of the torus, and also a function on
the manifold, then this function is Morse-Bott, and its critical manifolds are fixed
points of that one parameter subgroup. Moreover one can show that dimensions,
indices, and coindices of all its critical manifolds are even, moreover the critical
submanifolds are symplectic. The key point here seems to be to choose a torus
invariant almost complex structure.
Now by combining the two discussions, and doing symplectic reduction one dimension at a time, we can do induction and show that the preimages of the regular
values of the moment map are connected. Then by induction again, we want to
show that the image is convex. This clearly holds for circle actions. For the induction step we use codimension one tori which get denser and denser. The moment
map for such torus is a projection to some hypersurface of the original moment
map. For any two points lying over the same point in that hypersurface, we look
at the preimage of the moment map (for the small torus of course) at this point
which is also assumed to be regular and we know that it is connected, so we connect
them and their image under the big moment map should lie on a line. For any two
points, we approximate them by such pairs and prove the claim. With a bit more
work, we can show that this convex image is the convex hull of the images of the
fixed submanifolds of this action.
We finish the discussion of Atiyah-Guillemin-Sternberg convexity here.
In this paper, we want to work things like toric manifolds, but we only want
to think about the image of the moment map and recover the symplectic manifold
from the image of the moment map using extra date and forget about the action
of the torus. More precisely, if we have a toric manifold (not necessarily compact),
we have the projection to the quotient space X → B, and an immersion B → Rn
obtained by the moment map.
The observation is that for a toric manifold, the points on the toric divisors
have normal forms for their neighborhoods. Since outside of the divisor a toric
manifold is boring - torus bundle over an open convex polytope in euclidean space we can maybe just take the closed polytope, take the trivial torus bundle with the
standard symplectic form, and do something on the boundaries and reconstruct the
toric manifold. It turns out that in order to do this boundary reduction it is very
crucial that we keep track of the normal directions of the faces which are integral
- we have a boundary which seems like there was an S 1 action and someone took
µ−1 (0) but didn’t do the S 1 quotient, in order to do that we should have a foliation
by circles, which should actually be the characteristic foliation (so that the quotient
is symplectic), for this we should take the normal to the face, and take the integral
curves of the corresponding vertical vector field in the fibres.
Hence it is clear that we should keep some kind of integrality to recover the
toric manifold from its base and that’s how the integral affine structure enters the
picture. We have it for all Lagrangian fibrations, and it is a structure on the base,
so we might as well use it. In the non-closed case, we should actually do this
procedure on the quotient space (not the image of the moment map) and there how
will we keep track of this integrality, well, by the integral affine structure that is
pulled back from the map to Rn , which is of course the same as the one induced
by the Lagrangian fibration on the regular points - in particular the integral affine
structure extends to the critical values of the fibration.
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UMUT VAROLGUNES
Let me describe the normal form now. Let η be a point in a k-dimensional
face and a neighborhood of a point in its preimage (this neighborhood can be
quite large, i.e. the whole preimage of the star of the face) looks like the following. Take Rn , and take the portion of it which evaluates to a nonnegative
number under the first k-coordinates. Take the trivial T n bundle over it with the
standard symplectic structure, on this we have coordinates (p1 , . . . , pn , q1 , . . . , qn ),
where the first k momentum coordinates are delimited to [0, ∞) and qi ’s are angular coordinates. Now consider the manifold R2k × (S 1 × R)n−k with coordinates
(x1 , y1 , . . . , xk√, yk , pk+1 , qk+1 , . . . , p√n , qn ). We have a map from the first to the second by xi = 2pi cos qi and yi = 2pi sin qi , which is symplectic, bijective outside
the boundary of the first, and the image of the boundary is a symplectic submanifold. This is the normal form. In the general case one should make linear change of
variables for the angular coordinates according to the normals of the faces meeting
at the point.
A base manifold B, a stratification S, and an integral affine structure A on B
satisfying the conditions we kind of talked about above indeed defines uniquely a
toric manifold - and such data comes from a toric manifold if and only if those
conditions are satisfied.
If we can glue two such structures, then the glued data defines a toric manifold
which is glued from the original toric manifolds - CP2 is a good example. I should
think more about Hirzebruch surfaces. There is also a cool example of an exotic
R4 constructed with this methods.
In this paper we want to have more than toric manifolds. The first idea that
comes to mind is of course to add other kinds singularities that don’t occur in toric
manifolds. So we go through the list of stable singularities that happen in integrable
systems. Note that an integrable system is sometimes defined as a Lagrangian
fibration with the base immersing to Rn .
A singularity of an integrable system is nondegenerate of rank k for this papers
purposes if and only if around each point in X there are symplectic coordinates
(x1 , y1 , . . . , xn , yn ) and also coordinates in the base such that the first k components
of the projection just kills the y part, and for the others it is one of the following
three:
• elliptic: x2j + yj2 - exactly the ones that come up in toric manifolds.
• hyperpolic: x2j − yj2
• nodal-nodal: if we define zi = xi + Jyi just for notation, then the i, j
components of the projection combined looks like zi · z¯j , or more clumsily
(xi xj +yi yj , xj yi −xi yj ). Note that this is the same singularity that happens
in Lefschetz fibrations.
Remark 2. If we take R2n and consider the Lie algebra of homogenous quadratic
functions with respect to the standard Poisson bracket, we get sp(2n, R. You see
this by observing that there is a map from the above Lie algebra to the Lie algebra
of the linear symplectomorphisms by taking the Hamiltonian vector field.
In the more general definition, we look at the symplectic vector space ker DF (x)/ <
HFi > and see if the quadratic parts of Fi form a Cartan subalgebra for this Lie
algebra. They do form an abelian subalgebra, and of the right expected dimension,
so being Cartan is generic. We kind of ignore the higher order terms in Fi ’s for this
definition but it seems like this is ok.
QUALIFICATION EXAM
31
Let us restrict to four dimensions from now on. We want to allow some of these
singularities but we still want to reconstruct everything from the base and this
prevents us from including hyperbolic singularities. For example in two dimensions
(which of course affects the higher dimensions as well by crossing with nonsingular
stuff) these produce crosses in the base, but when two crosses intersect it is not
clear if there is a singular point above that point or not, it is even worse if the
intersection points can also be in the centers of the crosses. Elliptic singularities all
have neighborhoods as in the toric case, so they look like boundary points of convex
polytopes in the base, and in the total space they can be reconstructed from that.
For the nodal singularities, we have the neighborhood of such a singular point (in
the total space) fibred by cylinders and a transverse pair of discs, mapping down
to a disc in the base. If we assume that there is no other singular fibre in that disc,
the cylinders should be a part of a torus, this shows that the singular fiber will be
a torus pinched at different circles belonging to the same homology class, because
it is determined by the fixed subspace of the monodromy.
Note that in two dimensions if we don’t allow hyperbolic singularities, the only
closed manifolds we get are the sphere and the torus, by the classification of 1d
manifolds (and also the mapping class group of the torus). If we do allow hyperbolic
singularities we get everything.
Definition 4. If we have a non-degenerate Lagrangian fibration with nondegenerate
singularities which does not have any hyperbolic singularities, we call it an almost
toric fibration. If a triple (B, S, A) can be obtained this way we call it an almost
toric base. Note that A here is only defined outside of the nodal singularities, and
we declare the nodal singularities on the base to be of top stratum. Also note that
this fails to be toric manifold in two ways, one the presence of nodal fibres, and two
the non-existence of an immersion of the base to Rn .
Now we want to understand the neighborhood of the nodal fibre. If we are
interested only in the symplectic type of this neighborhood (which we are, the other
possibility would be to be interested in its Lagrangian fibration type) then there
is unique model for this, constructed as a self plumbing of the cotangent bundle
of the sphere. It is easy to see that this is unique by pulling back the symplectic
structure to the cotangent bundle of a two sphere (there is such immersion) and
using Weinstein neighborhood theorem. It is not hard to check in this model
that the monodromy is given by the unipotent matrices depending on the number
of singular points in the fibre. Note that the general form of such matrix (once
the number is given) depends only on two integers, hence it is enough to give its
eigenvector, or what it does to a given vector to determine it - one caviat: if we are
only given the eigenline then also the number of singular points should be provided.
Moreover something quite nice happens, the vanishing cycle is precisely the unique
eigenvector of the monodromy operator (this was alluded to above). This can also
be seen easily from the model above.
Secondly, we want to understand (find good models) for the integral affine structure around the nodes in the base. We will show that they can always be embedded
in R2 once we make a cut starting at the node (make the base simply connected so
the integral affine structure becomes trivial). We define the eigenline and eigenray
in the obvious way. The model is constructed by a developing map - the rays are
not chosen to be straight. Kindly note what happens in the case when we choose
32
UMUT VAROLGUNES
the eigenray, we don’t seem to be doing anything but we are regluing the tangent
spaces along the cut, which is very confusing.
In order to prove that this is really the only model, we need to show something.
Namely, we make the cut at an eigenray and then we can define an affine map to
the plane but we need to show that it is injective. This uses later results in the
paper and I will skip it for now.
Theorem 26. If (B, S, A) is such that (B, A) is an integral affine manifold with
nodes, then it is an almost toric base if and only if (B − nodes, S, A) is locally toric.
Note that we don’t get uniqueness (the Chern class doesn’t need to vanish) except
when B has boundary or it is open. Even in that case, we need to use Zung’s
theorem that says that two non-degenerate Lagrangian fibrations over the same
(B, S, A) are the same (as a Lagrangian fibration) if and only if they are roughly
symplectically equivalent and the Lagrangian Chern class vanishes. Duistermaat
stuff is not enough here because of the singularities.
We start working with base diagrams. Stuff we can do:
• Branch moves. These don’t change the actual stuff but they change the
way the base is immersed in the plane. Usually best seen by cutting the
base by a cut involving the branch locus and regluing the pieces in different
ways.
• Blow-up and down. There is the usual toric blow-up at a corner. In addition
to that we can also create a node by removing a perpendicular isosceles
triangle with one of the edges on a boundary edge and introducing a node at
the interior edge (the vanishing covector/class is the same as the collapsing
covector/class).
• Nodal slides. We can move one of the nodes through its eigenline and the
resulting symplectic manifolds are symplectomorphic - this only changes a
neighborhood of the preimage of the eigenline above and apparently we can
work in exact neighborhoods of that and use Moser trick to do this, though
I don’t quite see these exact neighborhoods.
• Nodal trades. If we have an eigenline hitting an edge such that the vanishing
and and collapsing covectors form basis of the lattice, we can get rid of the
node.
Note one cool point from the classification: Say we have a almost toric fibration
over a closed surface (so no elliptic singularities), then there the monodramy around
a small circle that doesn’t contain any node is the product of all single monodromies
around all the nodes. But all these monodromies are unipotent and it is known that
one needs at least 12 of those to get identity. Hence the number of nodes should be
divisible by 12. I will sketch here the proof of the fact that it is actually 12χ from
KoSo.
We first define an explicit homomorphism from the universal cover of SL(2, Z)
1
Z. Then we take the circle bundle over the affine manifold, which has as
to 12
its fibers the straight rays starting at that point, this is weird at the singularities.
Then we trivialize the tangent bundle, seeing it as a complex line bundle, outside of
some balls around the singularities and another fixed point. But we do it in such a
way that all the zeros are pushed away from the hole which doesn’t come from the
singularity. Now we consider the SL2 (2, R bundle and lift that to a universal cover
of SL2 (2, R bundle. This has a connection induced by the integral affine structure.
QUALIFICATION EXAM
33
Now we compute the monodromy around the not singularity point, by first doing
some kind of Gauss-Bonnet formula using this connection and the section we chose,
and we get χ(B)u, where u is the translation generator, and second as minus the
sum of the monodromies around the singular points. If the singular points are
focus-focus, then the homomorphism sends the monodromies to 1/12 - this is also
a proof of the divisibility by 12 that is mentioned above by the way. Since u is sent
to 1, this gives the desired number.
Also note that the discriminant of an elliptic curve has degree 12 in its coefficients, and hence we are from that perspective looking at the solutions of a degree
24 equation in the projective line.
For the case where there is boundary they use Gauss-Bonnet to show that in
that case as well the Euler characteristic is nonnegative, so there are not that many
choices for the base: disc, Moebius band, cylinder. The full list of almost toric
manifolds is as follows:
• S 2 × S 2 (toric, if we do blow up we can cancel the node by nodal trade, so
we will eventually end up with something toric)
• CP 2 and its blow-ups (all toric)
• S 2 × T 2 and its blow-ups (non are toric)
˜ 2 this glued by the antipodal map in one of the monodromies (the
• S 2 ×T
blow-ups of this turn out to be the same as the blow-ups of S 2 × T 2 , which
is very cool)
• Torus bundles over torus with (unipotent, identity) monodromy
• Base Z2 quotients along the trivial monodromy circle of the previous one,
so torus bundles over the Klein bottle
• K3 surface or its base Z2 quotient which is called Enriques surface (so over
the projective plane)
3.1.3. Kontsevich-Soibelman paper.
What is here?
• Basics of Lagrangian fibrations and examples
• Families of integrable systems and PL actions on the base
There is a cohomology class associated to integral affine structures, but I don’t
∂
⊗ dxi , where xi are integral
completely understand it. It is obtained by gluing Σ ∂x
i
affine coordinates. You try to define a global Euler vector field as an affine vector
field. You take random Euler vector fields along charts, then because of the form
of our transition maps, in the overlaps they differ by constant vector fields, which
defines a one Cech cocycle in the tangent bundle sheaf, in fact in the sheaf of parallel
sections with respect to the affine connection. Also note that we have a short exact
sequence:
(3)
0
/R
/ Af f
/ C∞
parallel (B, T
∗
B)
/0
∞
Hence Af f defines an element of Ext1 (Cparallel
(B, T ∗ B), R) = H 1 (B, T M ). The
last equivalence can be seen by noting that they are the derived functors of Γ(T M ⊗
·) = Hom(T ∗ M, ·), or by resolving R using the twisted deRham complex, which
are actually injective.
They define a Lagrangian fibration on K3 surface. They do it by taking a generic
bidegree (3, 2) hypersurface in CP 2 × CP 1 , and projecting it down to the second
factor. The symplectic form is not the standard symplectic form. You first take
34
UMUT VAROLGUNES
the standard holomorphic form on C3 × C2 and divide it by the defining equation,
and take its residue along the affine hypersurface, then you contract by the Euler
vector fields of the two factors. I checked in my head that this indeed descends to
a form on CP 2 × CP 1 . The fact that it is nonzero follows from the smoothness of
the hypersurface. This is a holomorphic 2-form (in particular not a Kahler form)
and hence the fiber elliptic curves are obviously complex Lagrangian. Taking the
real part of this gives us an integrable system.
Note that we had a lot of parameters in the above construction (the coefficients of
the defining polynomial), let us consider them all in a family. First of all the moduli
space of parameters have real dimension 2 × (3 × 10 − 3 − 8) = 38. All of these
integrable systems has some monodromy representation π1 (S 2 − {24 points}) →
SL(2, Z)nR2 , which is so that monodromy around each puncture is conjugate to the
standard unipotent matrix. The moduli space of all such representations up to conjugacy has 20 dimensions. I claim that for a given cohomology invariant above there
is only a discrete set of such representations because it determines the translational
part of the monodromy completely - and clearly all invariants are represented - build
them up one chart each time. Hence, the dimension will be equal to the dimension of H 1 (S 2 − {24 points}, T (S 2 − {24 points})with the given connection). This
I compute using the Serre spectral sequence of the fibration over S 2 − {24 points}
2
that we are thinking about. We need the E1,1
term. It is easy to check that the
sum of dimensions of the stuff in the second antidiagonal is equal to the second
Betti number of what is left of the K3 and I think I find 1 + 20 + 1 = 22. So
we have a foliation of rank 18 = 38 − 20 of the parameter space. Now by nodal
slides (or moving worms) we can change the integrable system without changing
the conjugacy class of the monodromy representation. It turns out that we can go
everywhere in a fixed leaf by such moves. So if we are given a point in a leaf and
a loop based at that point, we can move the S 2 with an integral affine structure
around (i.e. this explicit realization of the deformation gives us something like a
connection of this bundle of S 2 ’s over the leaf) and get some P L action of the
fundamental group of the leaf on S 2 .
3.2. Symplectic cohomology.
What is here?
• The heuristics behind the construction
• The general construction
• The properties of symplectic cohomology related to symplectic embeddings
and action cutoffs
• Maximum principle derivation
• Monotone homotopies in the presence of boundary
• Computational strategies
Hamiltonian Floer homology is a kind of Floer homology theory designed initially
to get information about periodic orbits of Hamiltonian flows.
I want to note some analysis results here.
• Arzela-Ascoli theorem: A uniformly bounded equicontinuous sequence of
functions has subsequence which is convergent in L∞ norm. Note that
equicontinuity follows from uniformly bounded derivatives.
• Banach-Alaoglu theorem: The unit ball in the dual of a Banach space (with
respect to the sup norm) is compact in the weak ? topology. This topology
QUALIFICATION EXAM
35
is usually not metrizable so this doesn’t imply sequential compactness. In
the seperable case, it does. Also note that a Hilbert space is its own dual.
• Garding inequality: Let D be an elliptic operator of order m on a compact
manifold, then the inequality
kukL2k ≤ C(kDukL2k−m + kukL2k−m ).
(4)
•
•
•
•
There is also Lp versions of this, where you replace the rightmost term with
the L1 norm of u. This is what lets you do elliptic bootstrapping along with
Sobolev embeddings - i.e. what happens when u is a solution.
Sobolev embedding, and also compactness of embeddings when you stricly
lose regularity. Note specifically the embeddings into Holder spaces, i.e.
once you have that something is in a Sobolev space of high regularity, then
depending on p and k, you actually have that it is continously differentiable
to some degree.
Inverse function theorem for Banach spaces: If f : U ⊂ B1 → B2 is a continuously differentiable operator and dfx is an isomorphism then there exists
a neighborhood V around x such that the map V → f (V ) is invertible.
Implicit function theorem for Banach spaces: If f : U ⊂ B1 → B2 is a
continuously differentiable operator and dfx is a surjection, and also need
to assume that the kernel of dfx admits a closed complement. Then it goes
as in the finite dimensional case.
The zero set of a Fredholm operator between Banach spaces is locally modeled on the zero set of a smooth map between two finite dimensional vector
spaces. If the cokernel is zero, then the range of this map is zero, and hence
the zero set is finite dimensional manifold.
I also want to make sketch of how to go about understanding the moduli space
of J- holomorphic things by listing facts.
• Symplectic area of a J-holomorphic curve is positive and it is equal to half
the energy of the surface. J-holomorphic discs are minimal surfaces - minimal surfaces are the critical points of the energy functional. To define the
energy of a parametrized surface, we only need a almost complex structure
on the surface, i.e. energy is conformally invariant.
• Localy there are as many J holomorphic curves as i holomorphic curves
- so there is one passing through every point. This uses the existence of
a 2d Green’s operator for the usual Cauchy-Riemann equation for maps
D 2 → Cn .
• Similar to the previous one, if we have two J-holomorphic curves agreeing up to infinite order at a point, then they agree everywhere (analytic
continuation).
• If a point on the domain (of a J-hol. curve) does not have a full rank
Jacobian (i.e. it is a critical point), then the differential vanishes completely
(since J doesn’t have real eigenvalues). Such points are isolated - pick the
largest order to which the curve is zero, approximate the almost complex
structure up to this order by i in a chart, and show that up to that order
the curve is holomorphic, and deduce that the critical point is isolated by
writing down the Taylor expansion. Hence if the domain is compact we
have finitely many critical points.
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UMUT VAROLGUNES
• A simple curve is a curve that is not multiply covered. The set of injective
points of a simple curve is open and dense (and nonempty). Factor the map
through the normalization of the image curve outside of the critical points.
Here we use some fact that the branch points which are not critical values
are isolated - not sure why this is correct (some normal form maybe?)
• To study the moduli space of J-holomorphic curves, we need to use some
functional analysis. For this it becomes handy to represent the ∂¯J operator
as a section of some Banach space bundle over a Banach manifold. For
proving transversality and stuff we work with Sobolev spaces, but for compactifications we need to use local Sobolev norms, because that is the kind
of convergence we get.
• We define Sobolev spaces as the abstract completions of smooth things with
respect to the Sobolev norms. This completion can be realized inside the
space of distributions, but in general we work with p, k so that we land in
continuous functions using Sobolev embeddings.
• Elliptic estimates kukLpk ≤ C(kDukLpk−1 + kukL1 . So when the domain is
not compact, this is not enough to show the Fredholmness of the linearized
operator.
• In order to get the semi-Fredholmness of this section at the zeros, we need
an inequality that is a little bit better than the elliptic estimates, namely
the norm of u that we use to bound should be replaced with the norm of
K for some compact operator. Also to show finiteness of kernel, it is best
to show that the unit ball is compact. For the other semi, we should work
with the adjoint.
• The missing piece for this is to use the fact that we are trying to connect
nondegenerate orbits. When we approach the two infinities the linearized
operator kind of converges to a special kind of s independent operators,
and if we write down the linearized operator for them, we see that they are
invertible operators. I don’t know if it follows from this but in the course of
the proof we also prove some stronger elliptic inequality for such operators,
which doesn’t have the L1 term at all in the right hand side, so we cut
our section into two parts (supported in a closed interval, and zero in the
same closed interval), and get the result using the two different estimates
we have.
• When we want to think about compactness, what we will be able to deduce
for the broken pieces of our limits will only be that they have finite energy,
but luckily that implies that we are converging to the right places, in fact
exponentially, which are the generator orbits. This is not trivial but true.
• The energy of bubbles is bounded below so the bubbling has to stop.
• Convergence modulo bubbling. The topology used here is quite weird, and
I still don’t have a good feeling for it.
• Infinite dimensional Morse theory doesn’t work for two reasons. One, when
we write down the gradient flow equation, we see that the tangent direction
to the path doesn’t lie in the space that it should lie (this doesn’t happen
in finite dimensions, because in finite dimensions we can work with smooth
target directly) since it loses regularity. Two is seen best for the problem
of the Morse theory for the loop space of Cn with the action functional
given by integration of the tautological one form. The gradient operator
QUALIFICATION EXAM
•
•
•
•
•
37
we get has non-negative spectrum, and once we Fourier expand it we see
that the solutions do not converge even for small time, so we do not get
local solvability.
Index computation: for this, at least in the Morse theory case, you look
at the number of crossings of the eigenvalues of the Hessian with the zero
axis. The point is that because we want solutions that are L2 or something
like that, after decoupling the system to n different one variable ODE’s, the
eigenvalues should do the right thing at both infinities - crossing the axis
once is right in only one direction, not crossing is never right, the other
possibilities are covered because index doesn’t change by compact . For
every eigenvalue that does the right thing we get one solution. So for every
eigenvalue that became negative we get one contribution to the index, and
this number is exactly the difference of the Morse indices.
Index for Gromov-Witten invariants - closed string. Here the idea is to
perturb the delbar operator so that we are actually looking for holomorphic
sections of a holomorphic vector bundle over our Riemann surface with
Chern number equal to the Chern class of the tangent bundle of the target
paired with the homology class we are working with.
Learn the linearization of the delbar operator
In Morse theory finiteness of energy for gradient flow lines follow directly
from Morse condition, gradient flow lines converge exponentially automatically.
Showing generic transversality. We define sections of the bundle parametrized
by the extra auxilary data that we are choosing, and apply the infinite dimensional version of the following very useful lemma:
Lemma 14. Let E → B be a vector bundle, say we have sections parametrized
by another manifold M , i.e. a map S : B × M → E. Assume that this map
is surjective to the vertical tangent space at the zero section. Then a section
given by b ∈ B is transverse to the zero section if and only if the restriction
of the projection map S −1 (0) → B has b as a regular value. Prove this
once.
• Great examples of bubbling and convergence modulo bubbling. Consider a
family of maps depending on t ∈ R, P1 → P2 given by [x : y] 7→ [x2 : y 2 :
txy] as t → ∞. The image is a conic xy = z 2 /t2 . If we parametrize the
sequence by [x : y] 7→ [x2 : y 2 /t2 : xy], then the image of the parametrized
limit looks like y = 0, whereas for [x : y] 7→ [x2 /t2 : y 2 : xy] it looks like
x = 0. Also consider a sequ P1 → P1 × P1 given by z 7→ (z, 1/n2 z). Away
from z = 0 on any compact set the first derivatives are uniformly bounded,
so a limit exists. If we look at the images of |z| = 1/n we see that they are
collapsing to a point but the energy lying in their interior stays the same,
and the image of the limit is the union of two coordinate axes.
• There is also the removal of singularities, which lets us close up the missing
the points using finite energy.
In Symplectic homology we want to use Hamiltonian Floer homology in order to
define symplectic invariants of open subsets of symplectic manifolds. The motivation to define this invariant is at least partly to study symplectic embeddings of
simple shapes into each other. A heuristic argument trying to improve a Gromov
non-squeezing type embedding suggests that the obstructions have something to
38
UMUT VAROLGUNES
do with closed characteristics in the boundary. The idea is to choose Hamiltonians
which are negative and approximately flat in the interior and very steeply increasing near the boundary - it should be infinitely steep in some sense. In order to
make sense of this steepness we will make use of a direct limit construction.
The general construction uses all functions seperating the open set in the sense
above and then takes the direct limit with respect to the monotone ordering of
hamiltonian, almost complex structure pairs. Here the key point is that since we
are in the non-compact setting, moduli spaces that we consider can have undesired
degenerations. As long as we have some kind of maximum principle, we can almost
act as if we are in the compact setting. The maximum principle is established for
certain special kinds of paths between ordered pairs of choices - defines a unique
homotopy class of chain maps as usual. This is one reason why continuation maps
are only defined in this restriced setting. There is also the formula relating the
energy of a pseudoholomorphic cylinder to the actions of the periodic orbits at the
endpoints with the error term easily controlled when the path increases the Hamiltonians in the infinite direction. The monotonicity is only crucial at infinity. People
use this in the construction of the more standard (now) symplectic cohomology of
Liouville domains.
The definition is very general but we would need to be in a nice situation to
actually compute it. The strategy for this is usually to find a well ordered set
of autonomous Hamiltonians depending on a parameter controlling the steepness
of the neck, and constructed by taking into account only one special coordinate
exploiting the convexity of the open set. Then one breaks the S 1 symmetry causing
nondegeneracy (Morse-Bott instead of Morse, circles in the loop space turn into two
critical points) by applying very controlled perturbations. This should be done in
a way that goes deep in the big filtered directed system. Also note that one usually
wants to avoid having periodic orbits at infinity, and also make everything standard
and autonomous there.
The general definition also allows one to make action cut-off’s and get welldefined homology groups. This gives exact triangles, and obvious inclusions, which
play nicely with the functoriality of symplectic homology under symplectic embeddings. Also note the symplectic isotopy invariance of the map induced on symplectic
homology (the maps are restriction maps, as in sheaves, not extension maps).
3.2.1. Maximum principle for Floer cylinders in Cn . .
Note first that what is well known is that there is a very well known Hopf
maximum principle, which doesn’t quite fit to the situation, because in the end we
get some elliptic second order partial differential inequality but the linear term has
negative coefficient.
Here is the key point: If we have a short cylinder S 1 × [0, δ] and a function f
satisfying −∆f − λf < a, f |∂ < b, where a and b are any real numbers and λ is
such that π / δ 2 > λ then there exists a constant depending on all the constants,
which is an upper bound for f . The inequality is necessary for avoiding solutions
−∆f − λf = 0, f |∂ = b, which would definitely violate the maximum principle.
Now I know how to show this, thanks to Peter.
The next step is to show that if we are given the continuation map equation
where we assume monotonicity and also a bound on the gradient of the s-dependent
Hamiltonian, then the actions of all s = constant loops in a solution are in a
QUALIFICATION EXAM
39
universal interval given that they are bounded above and below. Here we find a
formula for the s derivative of actions of the loops as well - it’s a simple formula.
We assume more now, we put some bound on the s derivative of the gradient
of L, and also make the bound on the gradient finer. This makes introduce a new
data A. But the bounds don’t depend on that. We can make the L21 norm of a loop
as small as we want by just bounding that derivative appeared above. Note that
this over a circle so L21 embeds compactly inside continuous functions.
Now we bring the above two paragraphs together and prove the following. If we
are given that the actions of a solution are bounded above and below, then for any
given δ we can divide the s coordinate of the cylinder into small chunks so that at
the loops dividing the chunks, we have some bound cδ on the absolute value of the
solution, depending on the given data J, L and δ. This is the first approximation
to a maximum principle, now we will use the key point.
Now it becomes absolutely crucial that J is standard outside some ball - we used
it before too. We simply throw away that ball. Write down the equations, and
derive a more Laplacian looking operator in the standard way from the Hamiltonian
deformed CR equations. Now it is in some shape that we can apply the key point,
which we can because we can divide up the cylinder into small parts in a nice way
as we proved before, and just get the uniform bound for all of them seperately.
Note that this bound on actions of all loops along with the maximum principle
implies bounded energy, and the rest of the argument goes in the way that is
supposed to be standard.
3.3. Mirror symmetry.
3.3.1. Paul’s paper.
What is here?
•
•
•
•
Hypersurfaces inside an algebraic torus associated to a polygon
Construction of the mirror as a toric degeneration
Generalization to hypersurfaces in abelian varieties
Examples
3.3.2. Dennis’ paper.
What is here?
• Classical SYZ construction
• The modification of the construction via considering moduli spaces of objects in the Fukaya category, and how this translates into the wall-crossing
formalism
• Toric Fano case
• C2 − conic and the blow up of C × C at (0, 1)
• Higher dimensional analogues of above
• Hirzebruch surfaces
• Eigenvalues of the cap product action of c1 (X) on quantum cohomology
• Different strategies for defining the self Floer cohomology of a Lagrangian,
and how they are related
• Pathologies when one doesn’t use Novikov field as the ground field
• Mirror symmetry for the anti-canonical divisor itself
40
UMUT VAROLGUNES
3.3.3. Eigenvalues of the quantum cap product action of c1 (X) on Lagrangian Floer
homology. Recall that quantum cohomology is something like a deformation of the
cup product structure on singular cohomology - let me just talk about the even
part of cohomology and avoid super-stuff. It is defined by counting J-holomorphic
spheres passing through the given cycles. Here we will define the cap product action
of QH ∗ (X) on HF ∗ (L, L), and describe it explicitly for the toric Fano case. We
will take the model of HF ∗ (L, L) based on singular chains of L.
We assume that the moduli spaces we consider have a fundamental class of the
expected dimension, and also that the classes we are fixing are transversal to the
evaluation maps. We define mk , k ≥ 0, for CF ∗ (L, L) = C ∗ (L, C) by taking k
input and 1 output points in the boundary, and pulling back the cycles to Mk (L, β),
taking their cup product, and then pushing forward to L. This really does what it
is supposed to (if we take a basis of cohomology then the structure constants of the
product structures would be as close as you get to the number of discs representing
β with marked points lying in the cycles).
The main point in all of this checking that things are well defined business is
that when we assume that all holomorphic discs that L bounds have Maslov index
at least 2, we get that the m0 term that is introduced in the equations don’t
change anything. More precisely, it becomes a multiple of [L], and because putting
[L] as a coincidence condition is the same as not putting any condition at all, by
degree considerations this element turns out to be a multiple of the unit for the
A∞ structure.
The key observation for what we are trying to prove is the following:
Proposition 5. c1 (X) ∩ [L] = m0 (L, ∇)[L], for L ∩ D empty.
Proof. We want to compute [D] ∩ [L] with the way we set things up. Since there
is no actual intersection, the constant disc doesn’t contribute to this action, and
because of our assumptions we only need to care about Maslov index 2 discs, which
means that they pass through D exactly once.
Now recall that when we defined the quantum cap action, we did not quotient
out the moduli space by reparamatrizations. We are counting discs where 0 ∈ D2
maps to [D] and 1 maps to [L], but this is exactly the same thing as just counting
Maslov index 2 discs up to reparametrization (output marked points are at −1 in
both cases). This finishes the proof.
Now because HF (L, L) is a module over QH(X), this shows that m0 (L, ∇) is
an eigenvalue of c1 (X)? : QH(X) → QH(X), whenever HF ∗ (L, L) 6= 0.
Let us define m1 (L, ∇) = Σµ(β)=2 nβ zβ [∂β]. A very surprising fact is that if
HF (L, L) 6= 0, then m1 (L, ∇) = 0. First of all, HF (L, L) 6= 0 is equivalent to
[L] being zero. Now let us compute δ[C] for a codimension 1 cycle in H(L). The
only contribution comes from the discs where the boundary meets [C], and this is
essentially saying that δ[C] = ([C] · m1 (L, ∇)[L]. Hence, if m1 is nonzero, then [L]
is actually a coboundary, and hence zero in cohomology. In the toric Fano case, it
turns out that the converse is also true.
The last observation is that if we compute the differential of W : M → C, we find
that it is equal to pairing the tangent vectors to m1 (L, ∇) - follows from the computation of dlogzβ . Hence we have proved that in the Fano case that HF (L, L) 6= 0
if and only if (L, ∇) is such that m0 (L, ∇) is an eigenvalue of the quantum multiplication with c1 (X) as an automorphism of the quantum cohomology.
QUALIFICATION EXAM
41
3.3.4. Hirzebruch surfaces. Hirzebruch surfaces are P1 bundles over P1 . Their holomorphic type is indexed by an integer, and they are toric varieties. More specifically
they are constructed as the projectivization of the line bundles O(n), so P(O(n)⊕O).
For n 6= 0, they contain a rational curve with self intersection −n < 0, so these
curves are rigid. They are also toric manifolds, and their polytopes can be drawn
as a right trapesoids with the slope of the slanted edge is 1/n for some integer n.
For n = 0, this is P1 × P1 , for n = 1 it is CP 2 blown up at a point. Symplectically
any Hirzebruch surface is isomorphic to these two smallest examples.
The strategy that Auroux uses to compute the mirrors of Hirzebruch surfaces
with their toric fibrations is to deform them to the toric fibration of F0 or F1
through nontoric fibrations of the latter. This is essentially a nodal slide, branch
move, and nodal slide trick. Note that the mirror of non-toric F0 was studied
before - compactify C2 to a CP 1 × CP 1 . The superpotential has the terms that
you would normally expect when you see the polygon plus a term that is caused
by the presence of the exceptional curve which causes bubbling and wall crossing
phenomenon to some extent.
3.3.5. Mirror symmetry for the anti-canonical divisor itself. Recall the classical
notion of the residue of a differential form. If we are given a holomorphic k-form ω
with poles along an effective divisor D. Let h be a locally defining equation for D,
then there exists locally defined regular forms η, ξ such that
dh
∧ ξ + η.
(5)
ω=
h
Moreover, the restriction of ξ to D is independent of the choices, and it is defined
to be the residue. Note that this choice of h is kind of like choosing a trivialization
of the normal bundle, and ξ is essentially the integral of ω along the circles of the
circle bundle with respect to the that normal neighborhood. We have the following
exact sequence.
(6)
0
/ Ωq (X)
/ Ωq (X, D)
res
/ Ωq−1 (D)
/0
There is a conjecture that near the boundary of the base the fibers of the Lagrangian fibration are contained near the divisor, and the smooth ones are actually
circle bundles over special Lagrangian tori inside the divisor. Hence, Auroux expects that special Lagrangian tori in X to accumulate into special Lagrangian tori
in D. This way we would get an SYZ fibration for D over ∂B, and its mirror would
be something that lies in some kind of boundary of the mirror of M - but not all
of it, need to get rid of one dimension from the flat connection part as well. This
boundary can be detected using functions zmer corresponding to the holomorphic
discs that correspond to the collapsing circle of a fibre near the boundary. In the
superpotential this term is supposed to dominate and by definition these functions
have absolute value close to 1. Also considering the arguments of these functions
we can define map to S 1 , and this map records exactly the holonomy around the
collapsed cycle. Since zmer ≈ W , fibres of this are close to fibres of W . Hence, we
expect the fibres of W to be SYZ mirrors to D. Note that this fibration over S 1 is
not trivial. A trivialization is given by choosing a rank n − 1 frame for homology of
the fibres nearby the boundary that together with the collapsing cycle form a basis.
If we go around a loop in the boundary, any of the members of this n − 1 element
list is going to undergo some monodromy which will add a multiple of the collapsing
42
UMUT VAROLGUNES
cycle to it. This after being careful and smart shows that the monodromy of the
fibration over S 1 gives us a symplectomorphism of the mirror of the divisor. If you
look at the mirror action on the derived category of coherent sheaves, that should
be given by tensoring with (KX ) |D .
Now we completely change the roles of the two sides and write down an HMS
conjecture:
(7)
Db (Coh(X))
/ Db (Coh(D))
Dπ F (M, MD )
/ Dπ (F (MD ))
3.4. Non-archimedean geometry of Lagrangian fibrations. Let K be a field
with a non-archimedean valuation ν : K → R∪{∞} satisfying the following axioms:
• ν(a) = ∞ if and only if a = 0
• ν(ab) = ν(a) + ν(b)
• ν(a + b) = min (ν(a), ν(b))
We can equivalently take |a| = e−ν(a) , which is called an absolute value, and list
the corresponding axioms for that. We call an absolute value or valuation discrete if
the image of the invertible elements inside R>0 is discrete. Absolute value function
gives rise to a distance function in the standard way. This satisfies the ultrametric
triangle inequality. Consequently, we have a topology on K. We call K complete
if every Cauchy sequence converges as usual.
Lemma 15. If |a| =
6 |b|, then |a + b| = max (|a|, |b|).
Proof. Wlog assume: |a| > |b|. Assume the contrary, then |a| = |a + b + (−b)| <
max (|a + b|, |b|) < |a|. Contradiction.
Lemma 16. The partial sums of Σai form a Cauchy sequence if and only if
limi→∞ |ai | = 0.
Proof. If they form a Cauchy sequence, then for every there is an N such that
+m
|ΣM
ai | < , for every M > N . In particular, |aM | < .
M
Conversely, let N be a number such that |aM | < , for every M > N . Then
+m
+m−1
|ΣM
ai | ≤ max (|ΣM
ai |, |aM +m |). Hence, induction on m finishes the proof.
M
M
Corollary 10. If K is complete, then a sum converges if and only if the terms
have absolute values going to zero.
Fun facts:
(1) Any triangle in K is isosceles
(2) Any point contained in a disc can be taken as its center
(3) Any two intersecting discs are concentric.
(4) The ”open” disc is both open and closed.
(5) The ”closed” disc is both open and closed.
(6) The circle around a point is both open and closed.
(7) Any subset of K which contains at least two points is disconnected.
(8) There is no continuous map from the unit interval to K other than the
constant ones.
QUALIFICATION EXAM
43
We want to define special functions over this field, but it turns out that differentiability is not a very good notion, so we will stick to analyticity. Because of the
above properties local conditions are too local to get anything reasonably global.
For example any disc is a disjoint union of other discs which do not see each other.
Hence we will always assume global convergence for functions.
Here we note a fact. If K is a complete non-archimedean field, its valuation extends uniquely to its algebraic closure K̄ which may not be complete anymore. But,
for any finite extension between K and K̄, it is complete. Also, inclusions are continuous. Moreover, the completion of the algebraic closure is also still algebraically
closed.
Let Bn (K̄) be the unit closed cube in K̄ n . The closed interval R in K is called
the valuation ring of K. It has exactly one maximal ideal m, which is the open
interval. Their quotient k is called the residue field as usual.
Lemma 17. A formal power series Σv∈Nn cv ξ1v1 . . . ξnvn is convergent in Bn (K̄) if
and only if lim|v|→∞ |ai | = 0.
Proof. One direction can be seen directly by evalauating at point (1, . . . , 1). For
the other direction, choose a point in Bn (K̄), since it has finitely many components,
there exists a finite extension K 0 which contains all its components, and convergence
can be checked inside K 0 . This then follows immediately from the convergence
criterion.
We define the Tate algebra Tn to be the algebra of the elements as in the lemma
- the convergent series. Define the Gauss norm k · k of f ∈ Tn to be maxv |cv |. This
makes Tn into a k-Banach algebra where again the triangle inequality is ultrametric.
This in particular says that it is complete with respect to this norm. Moreover the
Gauss norm is multiplicative, not just submultiplicative.
Let f ∈ Tn such that all its coefficients are in R. Then we define a epimorphism
π : Tn → k[ξ1 , . . . , ξn ] by f = Σv∈Nn cv ξ1v1 . . . ξnvn 7→ f˜ = Σv∈Nn c˜v ξ1v1 . . . ξnvn , where
tilde represents the residue of the element. Note that kf k < 1 if and only if its
reduction f˜ = 0. This shows that the Gauss norm is multiplicative since polynomial
rings are integral domains.
Another cool point is that kf k = max |f (x)|, where x is in the closed unit cube
in K̄ n - in particular the maximum is attained. We can again reduce to the case
when f has coefficients in R. It is clear that |f (x)| ≤ kf k, we cut off f (x) at a
finite time so that the remainder is smaller in absolute value then |f (x)|, then split
the sum to two parts, and use the triangle inequality two times. To show that the
maximum is attained, we pass to the reduction of f , which is non-zero for some
vector in k̄ n , where k̄ is the algebraic closure of the residue field. By some basic
field theory, k̄ is the residue field of K̄, and hence we can lift this vector to R̄n .
Then since the residue of f applied to this element is nonzero, it has to be in R̄, as
desired.
Theorem 27.
• Tn is Noetherian
• Tn is UFD
• Tn is regular
• For every maximal ideal m of Tn , Tn /m is a finite extension of K. So they
have uniquely defined complete non-archimedean valuations.
44
UMUT VAROLGUNES
• Every prime ideal of Tn is the intersection of the maximal ideals containing
it. This implies that for any ideal I of Tn , an element of Tn /I is nilpotent
if and only if it lies in all maximal ideals of Tn /I.
• Every ideal is closed with respect to the Gauss norm.
• Let K 0 /K be a finite extension. Then all continuous k-algebra morphisms
Tn → K 0 are given by evaluating polynomials at the closed unit cube inside
K 0n .
Now we start to consider the geometry of these things. Consider MaxSpec(Tn ),
we will denote elements of this by x and their residue class fields by k(x). For
every f ∈ Tn we have f (x) ∈ k(x). Combining previous points, we can now prove
kf k = max |x|, where x ∈ MaxSpec(Tn ). The last item in the theorem shows that
the maximum can’t get larger. If we take the vector (x1 , . . . , xn ) for which the
maximum is attained, we can find a surjective map Tn → K(x1 , . . . , xn ), which
defines a maximal ideal, and for some reason the values of f agree, I don’t know
why. This shows that the Gauss norm is independent of coordinates, more precisely,
it doesn’t change under a K-algebra automorphism.
We now define the local pieces of this stuff, which are called K-affinoid algebras.
These are just K algebras isomorphis to quotients of Tn by some ideal. We denote
its maximal ideals by M (A). The Theorem about the algebraic properties of Tn
translate to such algebras. M is functorial here, because the residue fields are finite
dimensional vector spaces over K - an integral domain, which is a finite dimensional
K-algebra is a field.
We want to define a K-Banach algebra structure on affinoid domains. We do
this by the residue norm construction (take infimum among representations) after
choosing coordinates, which is not necessarily multiplicative anymore. Note that
these norms are not exactly coordinate invariant, but they define equivalent norms
in the sense that the identity map provides a bounded map in each direction. In
fact, apparently any two k-Banach algebra norms on A are equivalent in this sense.
For example if we take a ∈ A the boundedness of the power sequence {an }n≥1 is
well defined. The theorems of standard functional analysis carry over to K-Banach
spaces setting.
There is a Noether normalization theorem for affinoid domains. The maximal
principle for functions (elements of A evaluated at M (A)) hold again, but the norms
are different in this case. This gives a canonical K-Banach algebra structure on A
when A is reduced. Reducedness is relevant because the Jacobson property tells us
that a ∈ A is nilpotent if and only if it vanishes at every point, making its norm 0.
This norm is also only submultiplicative, but is clearly power multiplicative. We
can use this to show that a ∈ A is power bounded if and only if a(x) < 1 for all
x ∈ M (A). This is appareantly true for not necessarily reduced affinoid domains
as well.
There is a universal property for the Tate algebras Tn in the category of KBanach algebras. Define the R-subalgebra A0 of power bounded bounded elements
in A a K-Banach algebras. This is obviously functorial with respect to continuous
maps. In particular if Tn → A is a continuous map, it sends ξi ’s to elements of
A0 . The universal property is that the corresponding map Hom(Tn , A) → (A0 )n is
bijective. This in particular shows a version (for any dimension) of the discs being
concentric point before. Namely let (c1 , . . . , cn ) be in the unit cube in K n , then by
QUALIFICATION EXAM
45
the universal property, there is a unique Tn → Tn continuous K-Banach algebra
automorphism ξi 7→ ξi − ci - it is obviously invertible.
Now we want to think about the topology on M (A). We somehow want to use
inequalities like |f | <≤ |g| not ones like f = g as we do in algebraic geometry. The
first attempt is what is called the canonical topology, which is apparently useless.
For each point x ∈ M (A), we have a map A → K(x), and we choose an embedding K(x) → K̄ and get a map A → K̄ of which image lies in a finite subextension.
Denote by A(K̄) all such maps. There is an Aut(K̄/K) action on this. Choosing
different embeddings of the residue field can be taken care of by this action. In fact
one can show that the above map describes a bijection M (A) → A(K̄)/Aut(K̄/K)
- this is a sort of Nullstallensatz but what sort, not now. It is also functorial. We
define a topology on A(K̄) using the sets of the following form as a basis:
(8)
{x ∈ A(K̄) | |fi (x)| ≤ i , |gj (x)| ≥ ηj , 1 ≤ i ≤ n, 1 ≤ j ≤ m},
where f1 , . . . , fn and g1 , . . . , gm are functions (i.e. elements of A) and η’s and ’s
are positive numbers. M (A) is topologized by the quotient topology. With this
topology A → A0 induces continuous maps M (A0 ) → M (A). This topology is
Hausdorff and totally disconnected.
We can define relative Tate algebras, kind of like polynomial rings over rings
instead of a fixed field. Here we change K to an arbitrary K-Banach algebra A,
for example affinoid algebras. The definition is exactly the same (convergence)
and we can define the Gauss norm also exactly analogously. If we change the
norm in A, the Gauss norm of A < Y > also changes but again they are all
equivalent. In fact, relative Tate algebras are affinoid domains too. If we choose
isomorphism Tn /I → A, then we have map Tn → A, which we can extend to a
map Tn+m → A < Y >. It is clear that this map is surjective, because Tn → A is
continuous and surjective. There is a universal mapping property for the relative
Tate algebras as well. Using this we can show (A < Y >) < X >= A < Y , X >.
Let A be an K-affinoid algebra. We introduce Laurent domains as some of the
future open subsets of M (A). Define A < a, a0−1 >= A < X, Y 0 > /(Xi − ai , a0i Yi −
1). There are not exactly the same as in the standard commutative algebra (maps
should be continous, which puts restrictions) and some care must be taken when
dealing with them. If we consider A < a >= A < X > /(X − a), then we will see
that this is not the same as A, because a ∈ A is not necessarily power-bounded,
whereas it is power bounded in A < a >, because X is power bounded in A < X >.
But, we will see that the induced map M (A < a >) → M (A) is injective and onto
the set where |a(x)| ≤ 1.
The UMP for relative Tate algebras says that a map φ : A → B factors through
A < a, a0−1 > if and only if the images of a’s are power-bounded and the images
of a’ are invertible and their inverses are power-bounded. If B is a K-affinoid
algebra, this can also be phrased in terms of the underlying spaces, namely, such
factorization exists if and only if the image of the induced map M (B) → M (A) has
in its image (set theoretic) only the points for which |ai (x)| ≤ 1 and |a0i (x)| ≥ 1. If
the factorization holds, then take a point x : B → K̄, then φ(a)(x) = a(x ◦ φ) ≤ 1,
similarly for a’s, and if the set theoretic condition holds for all points of B then the
canonical norm proves the converse. This shows that the map A → A < a, a0 >
gives rise to a bijection from M (A < a, a0 >) to the points which are delimited
by |ai (x)| ≤ 1 and |a0i (x)| ≥ 1 in M (A). Note that points are maps A → K̄ For
46
UMUT VAROLGUNES
injectivity, use the uniqueness in the universal property for such maps, and for
surjectivity the existence.
Another important class of open domains are the rational domains Now we are
given a0 , a1 , . . . , an which do not have a common zero. We define A < aa10 , . . . , aan >=
A < X1 , . . . , Xn > /(a0 Xi − ai ). These also have a universal property like the
Laurent domains. Now the related set theoretic condition is |ai (x)| ≤ |a0 (x)|.
For these two examples one of the main points if that the set determines the
algebra by the universal property. We now make this a definition.
3.5. One side of the homotopy perturbation lemma. Let (V, d) and (W, D)
be chain complexes, f : V → W , g : W → V are chain maps such that there exists
a chain homotopy g : W → W [1] such that 1 − f g = Dh + hD. Then, assume that
(W, D) also has an A∞ -structure. We can then define an A∞ structure on (V, d),
and also extend f and g to A∞ maps F and G with respect to that structure, and
the homotopy to an A∞ homotopy H between F G and 1.
In KS, they give formulas for the A∞ structure and F . These are both defined
by summing over certain decorated trees. We start with f , for the former we finish
with g, and for the latter with h. KS puts the homotopy operators also in vertices
by simply adding new vertices to the internal edges. This simplifies the notation in
their proof.
For the proof that what is defined is an A∞ structure, we add an extra vertex to
all the trees in all possible edges which is decorated by D. Then they sum over all
such trees. You seperate the terms where this vertex is added to an exterior edge,
and an interior edge. For the latter, we can use the homotopy identity to change
those to 1 − f g. If we could cancel the ones with 1 with the interior D’s, we would
be done. This indeed happens because of the A∞ relations in W . This can be seen
by locally modifying the trees and matching the terms with the new D adjacent to
a given a vertex, and all possible quadratic insertions in that vertex, with a vertex
labelled with 1 connecting to µ’s.