A *characteristic class* with values in a cohomology theory is a rule which assigns to each bundle a cohomology class such that (i) depends only on the isomorphism class of ; (ii) commutes with base change: if is any morphism, then .

In order to illustrate, let us take the prototypical example of -bundles. If is a topological group, a -bundle on a topological space is a map of topological spaces equipped with a continuous action of , such that, locally over , is isomorphic to a product with its natural action. It can be thought of as a continously varying family of -homogeneous spaces.

Notice that -bundles can be base changed: if is a continuous map, the fibre product is naturally a -bundle.

Let denote the functor which takes . It takes a continuous map to the map induced by base change; it is therefore a contravariant functor. Let be a cohomology theory. For our purposes, this need only be a contravariant functor which is homotopy invariant. Then, by definition, a characteristic class with values in is simply a natural transformation .

*Example of a characteristic class.* In this example we take denote the first Cech cohomology pointed set with coefficients in the (non-abelian) sheaf of continuous functions . Then any -bundle determines a class in in the following manner: pick a trivializing cover of , with -bundle isomorphisms . Over the intersection , the map is an automorphism of the -bundle , or, what is the same, a continuous map , i.e. a section of the sheaf over . The collection is a Cech -cocycle, and therefore it determines a cohomology class . Different choices of trivializations give cocycles which differ by coboundaries, so the cohomology class is in fact well-defined.

The characteristic class constructed above actually determines the bundle up to isomorphism, so that the pointed cohomology set actually classifies isomorphism classes of -bundles on . There is, however, another way to classify -bundles, using classifying spaces. Given a topological group , one can construct a topological space which represents the functor in the homotopy category of topological spaces. This means means that there is a canonical isomorphism

where denotes the set of homotopy classes of maps .

Thanks to the homotopy invariance of cohomology, we have the following proposition:

Proposition 1Given a cohomology theory , there exists a canonical bijection between the set of characteristic classes with values in , and the cohomology set of the classifying space .

*Proof:* Indeed, this is nothing but the Yoneda lemma:

This allows us to think of characteristic classes in two equivalent ways: either as rules which transform bundles into cohomology classes on the base, or as cohomology classes on a classifying space. The same thing will happen with modular forms.

An *elliptic curve over a scheme * (also called a *relative elliptic curve* over ) is defined as a smooth and proper morphism , equipped with a section , whose geometric fibres are smooth curves of genus one.

It can be shown that there is a unique structure of -group scheme on , for which is the identity, and which induces the group structure on each fibre (for which is the identity).

Elliptic curves can be based change: If is an elliptic curve and is a morphism, then the base change is an elliptic curve over . We think of as a *family of elliptic curves*, parametrized by the geometric points of .

Here are some examples:

- Let , where is an algebraically closed field. Then an elliptic curve over is an elliptic curve over in the classical sense.
- An elliptic curve over is an elliptic curve over having good reduction at .
- There are no elliptic curves over . Indeed, such a curve would provide an elliptic curve over having good reduction everywhere. However, by a result of Tate, this is impossible. (In fact, there are no abelian varieties over , of any dimension.)
In general, if is a ring which admits a morphism , then there is no elliptic curve over .

- Let and . Then an elliptic curve over is an elliptic curve over whose minimal discriminant divides .
- Weierstrass’ theory of elliptic functions shows that the equation
defines an elliptic curve over , where and are the formal power series

The Tate curve is therefore an elliptic curve over the formal punctured disc. It comes with the canonical differential , which is independent of . We write for the Tate curve and its canonical differential.

The Tate curve cannot be extended to the whole disc because the ring admits a morphism to , given by .

Given an elliptic curve , the sheaf of algebraic -forms is free of rank on . Since is proper, is free on . For example, if then can be identified with , the vector space of global differentials on (a one-dimensional vector space over ).

Now we define the notion of an *algebraic modular form*.

Definition 2Analgebraic modular form of weightis a rule which, to any elliptic curve over any scheme , assigns a section such that: (i) depends only on the isomorphism class of ; (ii) commutes with base change: if is any morphism, then .

Next time, I’ll expand on the definition and relate it with classical modular forms.

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We will follow the outline in Hartshorne (III.3 Problems 1 & 2 and III.4 Problems 1 & 2).

Theorem 1Let be an affine morphism of noetherian schemes. Then for any coherent sheaf on , there are natural isomorphisms for all ,

*Proof:* According to (II, Ex. 5.17), when is affine, the direct image functor induces an equivalence from the category of coherent -modules to the category of coherent -modules. Moreover, an equivalence of abelian categories (i.e. an additive functor which is also an equivalence) is exact. Therefore, if is a left additive functor, by the uniqueness of the -functor extending a given left additive functor, it follows that there exists a natural isomorphism for each .

Theorem 2Let be a noetherian scheme. Then is affine if and only if is.

*Proof:* Clearly is affine if is affine.

Conversely, suppose is affine. We prove that has cohomological dimension , hence it is affine by Serre’s theorem (III.3.7). Let be a quasi-coherent sheaf on . As indicated in the hint, we let denote the sheaf of nilpotents of and we consider the filtration

of . Since is noetherian, there exists an such that , so the filtration is finite.

We prove by descending induction on that is acyclic. For , it is trivial. Now consider the exact sequence of quasi-coherent sheaves on ,

The quasi-coherent sheaf is naturally a quasi-coherent -module, and its cohomology can be calculated either as an -module or as an module by Theorem 1 (using the fact that the reduction morphism is affine). Therefore, it is acyclic, since is affine by assumption. The sheaf is acyclic by the inductive hypothesis. By the long exact sequence of cohomology, we see that is also acyclic.

Theorem 3Let be a reduced scheme. Then is affine if and only if each irreducible component of is affine.

*Proof:* The irreducible components of are closed subschemes of , hence they are affine if is affine. Conversely, suppose that every irreducible component of is affine. We prove that has cohomological dimension .

We proceed by induction on the number of irreducible components of . If is irreducible, then the statement is vacuously true. Now suppose it holds for noetherian schemes with irreducible components. Suppose that has irreducible components, and write it as where is irreducible. Let be a quasi-coherent sheaf on . Denote the inclusion and the inclusion , where each closed subscheme is given the canonical reduced closed subscheme structure. Since is Noetherian, is also a quasi-coherent sheaf on , supported on . There is a canonical morphism , and . (Each of these two morphisms is a unit of the “inverse image – direct image” adjunction). Let

be their sum. It is easy to see that this morphism is surjective, and an isomorphism away from the intersection. Let . Then is quasi-coherent and supported in . Therefore we have an exact sequence

Since is affine by the induction hypothesis, is affine, being a closed subscheme of an affine scheme. Now, since , the cohomology of can be calculated either as an -module or as an -module, and therefore it vanishes. Similarily the sheaf is acyclic because and are affine. Therefore, by the long exact sequence of cohomology, is also acyclic.

Lemma 4Let be a finite surjective morphism of integral noetherian schemes. Then there is a coherent sheaf on , and a morphism of sheaves for some , such that is an isomorphism at the generic point of .

*Proof:* Let be the function field of and be the function field of . Then the morphism gives rise to a field homomorphism . Since is finite and surjective, is finite over , say of degree . Let be a basis for over . Each can be represented as a section of over an open set . Let be the inclusion. Let be the sheaf on . Obviously is coherent (in fact free of rank ). Let . Then is quasi-coherent on since is noetherian; since is finite, is in fact coherent. Let . Define the morphism by the global sections of (using the fact that represents the global sections functor ). Then, by construction, is an isomorphism of -vector spaces at the generic point of .

Lemma 5Let be a finite surjective morphism of integral noetherian schemes. Then for any coherent sheaf on , there exists a coherent sheaf on , and a a morphism which is an isomorphism at the generic point of .

*Proof:* We take , where is the sheaf and is the morphism of Lemma 4:

Remark that . Moreover, the sheaf naturally has a structure of -module. By (II, Ex. 5.17), when is an affine morphism, induces an equivalence between the category of coherent -modules and the category of coherent -modules. Therefore is isomorphic to an -module of the form , where is a coherent -module. Thus has the form .

Moreover, it follows from the fact that a coherent sheaf on a noetherian scheme is finitely presented that on such a scheme, taking sheaf commutes with taking stalks of morphisms; therefore is also an isomorphism at the generic point of .

Now we are ready to prove Chevalley’s theorem.

Theorem 6(Chevalley’s theorem). Let be a finite surjective morphism of noetherian separated schemes, where is affine. Then is affine.

*Proof:* By Theorems 2 and 3, we may suppose that and are reduced and irreducible. We prove by contradiction that is affine. Let be the collection of closed subschemes of which are not affine. Suppose it not empty; then it contains a minimal element , which we may view as having the reduced induced subscheme structure. Since finite morphisms are stable under base change, we may in fact suppose that (what this means is that we are replacing by its restriction to if necessary). Therefore, we suppose that every proper closed subscheme of is affine.

Let be a coherent sheaf on . By Lemma , there exists a coherent sheaf on and a morphism which is generically an isomorphism (and which is therefore surjective, since is irreducible). Thus, if , we have an exact sequence of sheaves on

Now, as in the proof of Theorem 3, we view as a quasi-coherent sheaf on the proper closed subscheme . By the minimality of , is affine and therefore is acyclic. Moreover, since a finite morphism is affine, we can apply Theorem 1 to see that is also acyclic. Therefore, by the long exact sequence of cohomology, is acyclic, so is acyclic.

Thefore, has cohomological dimension , which contradicts the assumption that it is not affine.

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Lemma 1(Lebesgue Number Lemma). Let be a compact metric space and let be an open cover of . Then there exists a real number such that every open ball of radius is contained in some .

*Proof:* First, remark that if any refinement of the cover satisfies this property, then also satisfies this property; thus, since is compact, we may replace the cover by a finite cover by open balls .

Define on by

Then is continous and its support is . Let . Then is continous, and because covers . Since is compact, attains its minimum, which is ; we call it . Now, if , the statement means precisely that for some , , so the ball of radius around is contained in .

Theorem 2Let be a continous function on the compact metric space . Then is uniformly continous.

*Proof:* Let . For each , let

Then is an open cover of . Let be a Lebesgue Number for the cover . Then, if are such that , there exists a such that ; therefore, since , we have

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It is -invariant, meromorphic, and has a double pole at each lattice point and no other poles. Its Laurent expansion at the origin is

where . Its derivative is

The functions and satisfy

in terms of the quantities and . Now if we take , where is in the upper half-plane, then , where is the weight Eisenstein series. These Eisenstein series, or rather the normalized Eisenstein series , satisfy certain relations, such as: (wikipedia)

In most basic texts on modular forms, these identities are derived by proving and exploiting the fact that the space of modular forms of weight is finite-dimensional. For instance, the fact that and have the same value at , combined with , implies . However, there is another, more “hands on” way to derive these identities.

Let us prove that . Substituting Laurent series at the origin in the equation , we see after some rearranging that the function

is identically , and therefore

Now

Thus, for instance, . Making these substitutions in the first equation and factoring out , we see that

Since , we have .

In fact, we have a bijection between and the set of infinituples of complex numbers satisfying the infinite system of equations :

Luckily, the values of and determine all of the others, and the ring is generated by and , and is in fact isomorphic to . This means precisely that is in bijection with the closed points of .

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*Proof:* Let be such a space, and suppose that it is infinite. Let be the collection of infinite closed subsets of . It is nonempty since , and therefore has a minimal member by the Noetherian assumption. Let be points of , and be disjoint open neighborhoods of and respectively (such and exist by the Hausdorff assumption). Then since and are disjoint, so . Now each of and is closed in , and is properly contained in (the first one doesn’t contain , and the second one doesn’t contain ). Therefore, by minimality of , each must be finite, and therefore is also finite, which is a contradiction.

**Corollary**: in any infinite Hausdorff space, there exists a strictly descending infinite chain of closed subsets . The proof above can be easily adapted to construct such a sequence.

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In what follows, is a topological space and are sheaves of abelian groups on . Let be a closed subset of . We let denote the global sections of with support in . The functor is a left-exact additive functor from sheaves on to abelian groups, and its right derived functors, denoted , is the -th cohomology of with support in . If is a sheaf, the presheaf is also a sheaf on , denoted and called the “subsheaf of with support in “.

The Mayer-Vietoris sequence, for a sheaf and for a pair of closed subsets , is the long exact sequence of cohomology with supports

We will prove the existence of this sequence in several steps.

Lemma 1Let be a flasque sheaf, a closed subset of , and . Then the sequence

is exact.

*Proof:* Trivial.

Lemma 2Let be a flasque sheaf, and let be closed subsets of . Then the sequence

is exact (where the first map is the diagonal embedding, and the second map is ).

*Proof:* Exactness is clear except possibly on the right. Let , and Let be the short exact sequences

and

where the maps are defined similarily as in the statement of the Lemma. There is an obvious morphism of short exact sequences . Since is flasque, this morphism is surjective onto each term of . By the snake lemma, and using Lemma 1, we get the desired short exact sequence.

Now we are ready to prove the existence of the Mayer-Vietoris sequence for . Let

be a flasque resolution of . By the lemma, we have a short exact sequence of complexes

The long exact sequence of cohomology associated to this short exact sequences of complexes is precisely the Mayer-Vietoris sequence.

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Exercise 1Let be the vector space of continuous real-valued functions on the interval . Then, for any ,

*Proof:* Let be the measure on . Then is a probability space, is Lebesgue-integrable on and is a convex function . By Jensen’s inequality,

Multiplying throughout by we get the claimed inequality.

Exercise 2Let be a linear operator on a finite-dimensional vector space . (a) Prove that if every one-dimensional subspace of is -invariant, then is a scalar multiple of the identity operator. (b) Prove that if every codimension-one subspace of is -invariant, then is a scalar multiple of the identity operator.

*Proof:* (a) The hypothesis means that every nonzero vector of is an eigenvector of . Suppose are eigenvectors of with eigenvalues , . Since, by assumption is also an eigenvector, and and are independent, we can read off the eigenvalue of off of either coefficient in the equation , and therefore . Therefore is a multiple of the identity operator.

(b) Let be the dual operator on . We claim that satisfies the condition of . First, we have the following:

Lemma 1Two functionals (where is the ground field) have the same kernel if and only if they are multiples of each other.

*Proof:* Indeed, it is trivial if either of or is (in which case both are zero), so suppose neither is . Recall that if and we define , then we have a canonical isomorphism , which in particular implies . If we apply this to , we have, under assumption,

which has codimension since . Therefore has dimension , and and are scalar multiples of each other.

Now, back to . S suppose that . Then has codimension in , and therefore, under the hypothesis of (b), . This implies ; indeed, if , then since . Since is codimension , we either have equality, or . If there is equality, then and have the same kernel and therefore they are proportional, i.e. is an eigenvector of . If then is trivially an eigenvector of . In every case, we see that is an eigenvector of . By (a), , and therefore , is a multiple of the identity operator.

Exercise 3Let be a linear operator on a finite-dimensional inner product space .

- (a) Define what is meant by the adjoint of .
- (b) Prove that .
- (c) If is normal, prove that . Give an example when the equality fails (and, of course, is not normal).

*Proof:*

- (a) It is the unique linear operator on such that for every .
- (b) Indeed,
- (c) A normal operator is one which commutes with its adjoint, i.e. . Thus,
An example where the equality fails is supplied by the operator acting on in the standard way. The vector is in the kernel of but not of .

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Here are three problems to start.

**Problem**: Suppose that is an real matrix with distinct real eigenvalues. Show that can be written in the form where each is a real number and the are real matrices with , and if . Give a real matrix for which such a decomposition is not possible and justify your answer.

*Solution*: for each , let denote the matrix with a on the entry and zeroes everywhere else. Then and when . Since has distinct real eigenvalues , it is diagonalizable over , so there is a real matrix such that , where . Let . Then

Moreover, for we have .

For the second part, notice that if the matrix is decomposed in the manner described above, the numbers are necessarily eigenvalues of . Indeed, multiplying the equality by and using that when , we find that . Hence, let be any nonzero vector. Since , at least one of the terms in the sum is nonzero, say . Then

and therefore is an eigenvector of with eigenvalue . Thus, it is impossible for the matrix to have such a decomposition if, say, it has no real eigenvalues, for example

**Problem**: Let be a continuous function on . Suppose there exists such that for all ,

Prove that .

*Solution*: (With Juan Ignacio Restrepo.) By iterated integration, we find by induction that, for every integer ,

Since the integrand is uniformly bounded and as , it follows that .

**Problem**: Suppose that is a real sequence with and that is a nonnegative sequence with . Prove that

*Solution*: For notational simplicity, let us write . Let be so large that for all . Then, for , we have, by the triangle inequality,

where . Let be so large that for all , which exists since . Then for , we have

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I’d love to see a more elementary proof if you can come up with one.

First we need the following:

Lemma 1Let be the cyclic group with elements. Let be a positive divisor of , and consider as a subgroup of . Then the number of automorphisms of which fix pointwise is equal to (which, in particular, is an integer).

**Proof of the Lemma:** Note that any automorphism of fixes , though not necessarily pointwise: indeed has a unique subgroup of order , and thus any automorphism of must take this subgroup to itself. Thus we have a group homomorphism which is easily seen to be surjective; its kernel is precisely the subgroup consisting of those automorphisms of which fix pointwise. The statement follows by comparing orders.

Now to prove the initial claim, consider the field extension . Basic Galois theory tells that this is a Galois extension of degree . Consider the canonical homomorphism

which restricts an -automorphism to the group of units of . Clearly it is an injective homomorphism since is completely determined by where it sends the units. Moreover for any , lies in the subgroup of of those automorphisms fixing pointwise the cyclic subgroup of order , because the Galois group consists of -homomorphisms. By the lemma the subgroup of these automorphisms has order , whereas has order . This does it.

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Our main tool will be Burnside’s lemma, which states that if a finite group acts on a finite set , the average number of fixed points of the elements of is equal to the number of orbits of the action of on :

where is the set of fixed points of .

We let , the symmetric group on letters, act on component-wise. The elements of are -tuples consisting of integers between and . Now you may easily convince yourself that it is possible to send a tuple to a tuple if and only if whenever , we also have , and vice versa. In other words we view each -tuple as a function ; its fibres partition , and composition with a permutation preserves the fibres of . It is immediate that and are in the same orbit of if and only if they have the same collection of fibres. For instance, can be sent to by the cycle but there is no way to send to , because any permutation will send to a 3-tuple of the form .

Thus has orbits on . On the other hand since our permutations act component-wise, we have

i.e. the fixed points of acting on are the tuples consisting of fixed points of acting on . Therefore, by Burnside’s lemma, we have

In fact the same argument shows that for any , we have

In particular, the identity of fixes all of , so we have

In fact, by using the fact that a permutation of consists of a subset of (the subset of fixed points), and a derangement of the remaining elements, we easily obtain the formula

where is the number of derangements of a set of elements.

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