# Algebraic modular forms and characteristic classes

This post is the first of a series dedicated to algebraic modular forms and ${p}$-adic modular forms. I will be mostly following the foundational paper of Katz in the Antwerp proceedings.

A characteristic class with values in a cohomology theory ${H^\bullet}$ is a rule which assigns to each bundle ${E/S}$ a cohomology class ${f(E/S) \in H^\bullet(S)}$ such that (i) ${f(E/S)}$ depends only on the isomorphism class of ${E/S}$; (ii) ${f}$ commutes with base change: if ${g:S' \rightarrow S}$ is any morphism, then ${f(E_{S'}/S') = g^*f(E/S)}$.

In order to illustrate, let us take the prototypical example of ${G}$-bundles. If ${G}$ is a topological group, a ${G}$-bundle on a topological space ${S}$ is a map of topological spaces ${p: E \rightarrow S}$ equipped with a continuous action of ${G}$, such that, locally over ${S}$, ${E}$ is isomorphic to a product ${U \times G}$ with its natural action. It can be thought of as a continously varying family of ${G}$-homogeneous spaces.

Notice that ${G}$-bundles can be base changed: if ${S' \rightarrow S}$ is a continuous map, the fibre product ${E \times_S S' \rightarrow S'}$ is naturally a ${G}$-bundle.

Let ${b : \text{Top} \rightarrow \text{Set}}$ denote the functor which takes ${S \mapsto \{\text{iso. classes of } G \text{-bundles on } S}\}$. It takes a continuous map ${S'\rightarrow S}$ to the map ${b(S) \rightarrow b(S')}$ induced by base change; it is therefore a contravariant functor. Let ${H^\bullet : \text{Top} \rightarrow \text{Set}}$ 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 ${H^\bullet}$ is simply a natural transformation ${b \rightarrow H^\bullet}$.

Example of a characteristic class. In this example we take ${H^\bullet(S):=H^1(S, \text{Aut}(G))}$ denote the first Cech cohomology pointed set with coefficients in the (non-abelian) sheaf of continuous functions ${U \mapsto {\text{Aut}(G)}}$. Then any ${G}$-bundle ${p: E\rightarrow S}$ determines a class in ${H^1(S, {\text{Aut}(G)})}$ in the following manner: pick a trivializing cover ${\{U_i\}}$ of ${S}$, with ${G}$-bundle isomorphisms ${\varphi_i : E|_{U_i} \rightarrow U_i \times G}$. Over the intersection ${U_{ij}}$, the map ${\varphi_j \varphi_i^{-1}}$ is an automorphism of the ${G}$-bundle ${U_{ij} \times G}$, or, what is the same, a continuous map ${U_{ij} \rightarrow \text{Aut}(G)}$, i.e. a section of the sheaf ${ {\text{Aut}(G)}}$ over ${U_{ij}}$. The collection ${\{\varphi_j \varphi_i^{-1}\}}$ is a Cech ${1}$-cocycle, and therefore it determines a cohomology class ${\sigma_{E/S} \in H^1(S, {\text{Aut}(G)})}$. Different choices of trivializations give cocycles which differ by coboundaries, so the cohomology class ${\sigma_{E/S}}$ is in fact well-defined.

The characteristic class ${\sigma_{E/S}}$ constructed above actually determines the bundle ${E/S}$ up to isomorphism, so that the pointed cohomology set ${H^1(S, {\text{Aut}(G)})}$ actually classifies isomorphism classes of ${G}$-bundles on ${S}$. There is, however, another way to classify ${G}$-bundles, using classifying spaces. Given a topological group ${G}$, one can construct a topological space ${BG}$ which represents the functor ${b}$ in the homotopy category of topological spaces. This means means that there is a canonical isomorphism

$\displaystyle b(S) \cong [S, BG]$

where ${[S,BG]}$ denotes the set of homotopy classes of maps ${S \rightarrow BG}$.

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

Proposition 1 Given a cohomology theory ${H^\bullet}$, there exists a canonical bijection between the set ${\{f : b \rightarrow H^\bullet\}}$ of characteristic classes with values in ${H^\bullet}$, and the cohomology set ${H^\bullet(BG)}$ of the classifying space ${BG}$.

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

$\displaystyle \text{Nat}(b, H^\bullet) = \text{Nat}([-, BG], H^\bullet) = H^\bullet(BG).$

$\Box$

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 ${S}$ (also called a relative elliptic curve over ${S}$) is defined as a smooth and proper morphism ${E \xrightarrow{p} S}$, equipped with a section ${S \xrightarrow{e} E}$, whose geometric fibres are smooth curves of genus one.

It can be shown that there is a unique structure of ${S}$-group scheme on ${E}$, for which ${e}$ is the identity, and which induces the group structure on each fibre ${E_x}$ (for which ${e_x}$ is the identity).

Elliptic curves can be based change: If ${p: E \rightarrow S}$ is an elliptic curve and ${T \rightarrow S}$ is a morphism, then the base change ${E\times_S T = E_T \rightarrow T}$ is an elliptic curve over ${T}$. We think of ${E_{/S}}$ as a family of elliptic curves, parametrized by the geometric points of ${S}$.

Here are some examples:

1. Let ${S = \text{Spec }K}$, where ${K}$ is an algebraically closed field. Then an elliptic curve over ${S}$ is an elliptic curve over ${K}$ in the classical sense.
2. An elliptic curve over ${\mathbf Z_p}$ is an elliptic curve over ${\mathbf Q_p}$ having good reduction at ${p}$.

3. There are no elliptic curves over ${\mathbf Z}$. Indeed, such a curve would provide an elliptic curve over ${\mathbf Q}$ having good reduction everywhere. However, by a result of Tate, this is impossible. (In fact, there are no abelian varieties over ${\mathbf Z}$, of any dimension.)

In general, if ${R}$ is a ring which admits a morphism ${R \rightarrow \mathbf Z}$, then there is no elliptic curve over ${R}$.

4. Let ${\Delta \in \mathbf Z}$ and ${S=\text{Spec } \mathbf Z[1/\Delta]}$. Then an elliptic curve over ${S}$ is an elliptic curve over ${\mathbf Q}$ whose minimal discriminant divides ${\Delta}$.

5. Weierstrass’ theory of elliptic functions shows that the equation

$\displaystyle y^2+xy=x^3+B(q)x + C(q)$

defines an elliptic curve over ${\mathbf Z((q))}$, where ${B}$ and ${C}$ are the formal power series

$\displaystyle B(q) = -5 \sum_{n \geq 1} \sigma_3 (n) q^n, \qquad C(q) = \sum_{n \geq 1} \frac{-5\sigma_3(n) - 7 \sigma_5(n)}{12} q^n \qquad (\sigma_k(n) = \sum_{d \mid n}d^k).$

The Tate curve is therefore an elliptic curve over the formal punctured disc. It comes with the canonical differential ${dx/(2y+x)}$, which is independent of ${q}$. We write ${(\text{Tate}(q), \omega_{can})}$ for the Tate curve and its canonical differential.

The Tate curve cannot be extended to the whole disc ${\mathbf Z[[q]]}$ because the ring ${\mathbf Z[[q]]}$ admits a morphism to ${\mathbf Z}$, given by ${q \mapsto 0}$.

Given an elliptic curve ${E/S}$, the sheaf of algebraic ${1}$-forms ${\Omega^1_{E/S}}$ is free of rank ${1}$ on ${E}$. Since ${p : E\rightarrow S}$ is proper, ${ {\omega}_{E/S} : = p_*\Omega^1_{E/S}}$ is free on ${S}$. For example, if ${S=\text{Spec } K}$ then ${ {\omega}_{E/S}}$ can be identified with ${H^0(E, \Omega^1_{E/K})}$, the vector space of global differentials on ${E}$ (a one-dimensional vector space over ${K}$).

Now we define the notion of an algebraic modular form.

Definition 2 An algebraic modular form of weight ${k \in \mathbf Z}$ is a rule ${f}$ which, to any elliptic curve over any scheme ${S}$, assigns a section ${f(E/S) \in H^0(S, {\omega}_{E/S}^{\otimes k})}$ such that: (i) ${f(E/S)}$ depends only on the isomorphism class of ${E/S}$; (ii) ${f}$ commutes with base change: if ${g:S' \rightarrow S}$ is any morphism, then ${f(E_{S'}/S') = g^*f(E/S)}$.

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

# Chevalley’s theorem

The purpose of this post is to prove Chevalley’s theorem: If ${f: X \rightarrow Y}$ is a finite surjective morphism of noetherian separated schemes, with ${X}$ affine, then ${Y}$ is affine.

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

Theorem 1 Let ${f: X \rightarrow Y}$ be an affine morphism of noetherian schemes. Then for any coherent sheaf ${\mathcal F}$ on ${X}$, there are natural isomorphisms for all ${i \geq 0}$,

$\displaystyle H^i(X, \mathcal F) \simeq H^i(Y, f_* \mathcal F).$

Proof: According to (II, Ex. 5.17), when ${f}$ is affine, the direct image functor ${f_*}$ induces an equivalence from the category of coherent ${\mathcal O_X}$-modules to the category of coherent ${f_*\mathcal O_X}$-modules. Moreover, an equivalence ${\tau : A \rightarrow B}$ of abelian categories (i.e. an additive functor which is also an equivalence) is exact. Therefore, if ${F: B \rightarrow \text{Ab}}$ is a left additive functor, by the uniqueness of the ${\delta}$-functor extending a given left additive functor, it follows that there exists a natural isomorphism ${R^i(F \circ \tau) \simeq R^i F \circ \tau}$ for each ${i}$. $\Box$

Theorem 2 Let ${X}$ be a noetherian scheme. Then ${X}$ is affine if and only if ${X_{\text{red}}}$ is.

Proof: Clearly ${X_\text{red}}$ is affine if ${X}$ is affine.

Conversely, suppose ${X_{\text{red}}}$ is affine. We prove that ${X}$ has cohomological dimension ${0}$, hence it is affine by Serre’s theorem (III.3.7). Let ${\mathcal F}$ be a quasi-coherent sheaf on ${X}$. As indicated in the hint, we let ${\mathcal N}$ denote the sheaf of nilpotents of ${X}$ and we consider the filtration

$\displaystyle \mathcal F \supseteq \mathcal N \cdot \mathcal F \supseteq \mathcal N^2 \cdot \mathcal F \supseteq \dots$

of ${\mathcal F}$. Since ${X}$ is noetherian, there exists an ${n>0}$ such that ${\mathcal N^n = 0}$, so the filtration is finite.

We prove by descending induction on ${j}$ that ${ \mathcal N^j \cdot \mathcal F}$ is acyclic. For ${j=n}$, it is trivial. Now consider the exact sequence of quasi-coherent sheaves on ${X}$,

$\displaystyle 0 \rightarrow \mathcal N^j \cdot \mathcal F\rightarrow \mathcal N^{j-1} \cdot \mathcal F \rightarrow (\mathcal N^{j-1} \cdot \mathcal F) / (\mathcal N^j \cdot \mathcal F) \rightarrow 0.$

The quasi-coherent sheaf ${(\mathcal N^{j-1} \cdot \mathcal F) / (\mathcal N^j \cdot \mathcal F)}$ is naturally a quasi-coherent ${\mathcal O_X / \mathcal N \simeq \mathcal O_{X_{\text{red}}}}$-module, and its cohomology can be calculated either as an ${\mathcal O_X}$-module or as an ${\mathcal O_{X_{\text{red}}}}$ module by Theorem 1 (using the fact that the reduction morphism ${X_{\text{red}} \to X}$ is affine). Therefore, it is acyclic, since ${X_{\text{red}}}$ is affine by assumption. The sheaf ${\mathcal N^j \cdot \mathcal F}$ is acyclic by the inductive hypothesis. By the long exact sequence of cohomology, we see that ${\mathcal N^{j-1} \cdot \mathcal F}$ is also acyclic. $\Box$

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

Proof: The irreducible components of ${X}$ are closed subschemes of ${X}$, hence they are affine if ${X}$ is affine. Conversely, suppose that every irreducible component of ${X}$ is affine. We prove that ${X}$ has cohomological dimension ${0}$.

We proceed by induction on the number of irreducible components of ${X}$. If ${X}$ is irreducible, then the statement is vacuously true. Now suppose it holds for noetherian schemes with ${n-1}$ irreducible components. Suppose that ${X}$ has ${n}$ irreducible components, and write it as ${X=Y \cup X'}$ where ${Y}$ is irreducible. Let ${\mathcal F}$ be a quasi-coherent sheaf on ${X}$. Denote ${\tau}$ the inclusion ${Y \hookrightarrow X}$ and ${\iota}$ the inclusion ${X' \hookrightarrow X}$, where each closed subscheme is given the canonical reduced closed subscheme structure. Since ${Y}$ is Noetherian, ${ \tau_* \tau^* \mathcal F}$ is also a quasi-coherent sheaf on ${X}$, supported on ${Y}$. There is a canonical morphism ${\mathcal F \rightarrow \tau_* \tau^* \mathcal F}$, and ${\mathcal F \rightarrow \iota_* \iota^* \mathcal F }$. (Each of these two morphisms is a unit of the “inverse image – direct image” adjunction). Let

$\displaystyle g : \mathcal F \rightarrow \tau_* \tau^* \mathcal F \oplus \iota_* \iota^* \mathcal F$

be their sum. It is easy to see that this morphism is surjective, and an isomorphism away from the intersection. Let ${\mathcal G= \ker g}$. Then ${\mathcal G}$ is quasi-coherent and supported in ${Y \cap X'}$. Therefore we have an exact sequence

$\displaystyle 0 \rightarrow \mathcal G \rightarrow \mathcal F \rightarrow \tau_* \tau^* \mathcal F \oplus \iota_* \iota^* \mathcal F \rightarrow 0$

Since ${X'}$ is affine by the induction hypothesis, ${Y \cap X'}$ is affine, being a closed subscheme of an affine scheme. Now, since ${\text{Supp }\mathcal G \subseteq Y \cap X'}$, the cohomology of ${\mathcal G}$ can be calculated either as an ${\mathcal O_{(Y \cap X')}}$-module or as an ${\mathcal O_X}$-module, and therefore it vanishes. Similarily the sheaf ${\tau_* \tau^* \mathcal F \oplus \iota_* \iota^* \mathcal F}$ is acyclic because ${Y}$ and ${X'}$ are affine. Therefore, by the long exact sequence of cohomology, ${\mathcal F}$ is also acyclic. $\Box$

Lemma 4 Let ${f: X \rightarrow Y}$ be a finite surjective morphism of integral noetherian schemes. Then there is a coherent sheaf ${\mathcal M}$ on ${X}$, and a morphism of sheaves ${\alpha : \mathcal O_Y^r \rightarrow f_* \mathcal M}$ for some ${r>0}$, such that ${\alpha}$ is an isomorphism at the generic point of ${Y}$.

Proof: Let ${L}$ be the function field of ${X}$ and ${K}$ be the function field of ${Y}$. Then the morphism ${f}$ gives rise to a field homomorphism ${K \hookrightarrow L}$. Since ${f}$ is finite and surjective, ${L}$ is finite over ${K}$, say of degree ${r}$. Let ${\{x_1, \dots, x_r\}}$ be a basis for ${L}$ over ${K}$. Each ${x_j}$ can be represented as a section ${s_j}$ of ${\mathcal O_X}$ over an open set ${U_j}$. Let ${\tau_j : U_j \hookrightarrow X}$ be the inclusion. Let ${\mathcal E_j}$ be the sheaf ${\mathcal E_j = s_j \cdot \mathcal O_{U_j}}$ on ${U_j}$. Obviously ${\mathcal E_j}$ is coherent (in fact free of rank ${1}$). Let ${\mathcal F_j = (\tau_j)_*(\mathcal E_j)}$. Then ${\mathcal F_j}$ is quasi-coherent on ${X}$ since ${U_j}$ is noetherian; since ${f}$ is finite, ${\mathcal F_j}$ is in fact coherent. Let ${\mathcal M = \bigoplus_j \mathcal F_j}$. Define the morphism ${\alpha : \mathcal O^r_Y \rightarrow f_*\mathcal M}$ by the global sections ${x_j}$ of ${f_*\mathcal M}$ (using the fact that ${\mathcal O_Y}$ represents the global sections functor ${\Gamma(Y, -)}$). Then, by construction, ${\alpha}$ is an isomorphism of ${K}$-vector spaces ${K^r \cong L}$ at the generic point of ${Y}$. $\Box$

Lemma 5 Let ${f: X \rightarrow Y}$ be a finite surjective morphism of integral noetherian schemes. Then for any coherent sheaf ${\mathcal F}$ on ${Y}$, there exists a coherent sheaf ${\mathcal G}$ on ${X}$, and a a morphism ${\beta : f_* \mathcal G \rightarrow \mathcal F^r}$ which is an isomorphism at the generic point of ${Y}$.

Proof: We take ${\beta = \mathcal{H}\text{om}(\alpha, \mathcal F)}$, where ${\mathcal{H}\text{om}}$ is the sheaf ${\mathcal{H}\text{om}}$ and ${\alpha}$ is the morphism of Lemma 4:

$\displaystyle \beta: \mathcal{H}\text{om}(f_*\mathcal M, \mathcal F) \rightarrow \mathcal{H}\text{om}(\mathcal O_Y^r, \mathcal F).$

Remark that ${\mathcal{H}\text{om}(\mathcal O_Y^r, \mathcal F) \simeq \mathcal F^r}$. Moreover, the sheaf ${\mathcal{H}\text{om}(f_*\mathcal M, \mathcal F)}$ naturally has a structure of ${f_*\mathcal O_X}$-module. By (II, Ex. 5.17), when ${f}$ is an affine morphism, ${f_*}$ induces an equivalence between the category of coherent ${\mathcal O_Y}$-modules and the category of coherent ${f_*\mathcal O_X}$-modules. Therefore ${\mathcal{H}\text{om}(f_*\mathcal M, \mathcal F)}$ is isomorphic to an ${\mathcal O_Y}$-module of the form ${f_*\mathcal G}$, where ${\mathcal G}$ is a coherent ${\mathcal O_X}$-module. Thus ${\beta}$ has the form ${f_* \mathcal G \rightarrow \mathcal F^r}$.

Moreover, it follows from the fact that a coherent sheaf on a noetherian scheme is finitely presented that on such a scheme, taking sheaf ${\mathcal{H}\text{om}}$ commutes with taking stalks of morphisms; therefore ${\beta}$ is also an isomorphism at the generic point of ${Y}$. $\Box$

Now we are ready to prove Chevalley’s theorem.

Theorem 6 (Chevalley’s theorem). Let ${f: X \rightarrow Y}$ be a finite surjective morphism of noetherian separated schemes, where ${X}$ is affine. Then ${Y}$ is affine.

Proof: By Theorems 2 and 3, we may suppose that ${X}$ and ${Y}$ are reduced and irreducible. We prove by contradiction that ${Y}$ is affine. Let ${\Sigma}$ be the collection of closed subschemes of ${Y}$ which are not affine. Suppose it not empty; then it contains a minimal element ${Z \hookrightarrow X}$, 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 ${Z=Y}$ (what this means is that we are replacing ${f}$ by its restriction to ${f^{-1}(Z)}$ if necessary). Therefore, we suppose that every proper closed subscheme of ${Y}$ is affine.

Let ${\mathcal F}$ be a coherent sheaf on ${X}$. By Lemma ${5}$, there exists a coherent sheaf ${\mathcal G}$ on ${X}$ and a morphism ${\beta: f_* \mathcal G \rightarrow \mathcal F^r}$ which is generically an isomorphism (and which is therefore surjective, since ${Y}$ is irreducible). Thus, if ${\mathcal D = \ker \beta}$, we have an exact sequence of sheaves on ${Y}$

$\displaystyle 0 \rightarrow \mathcal D \rightarrow f_* \mathcal G \rightarrow \mathcal F^r \rightarrow 0.$

Now, as in the proof of Theorem 3, we view ${\mathcal D}$ as a quasi-coherent sheaf on the proper closed subscheme ${\text{Supp }\mathcal D}$. By the minimality of ${Y}$, ${\text{Supp }\mathcal D}$ is affine and therefore ${\mathcal D}$ is acyclic. Moreover, since a finite morphism is affine, we can apply Theorem 1 to see that ${f_* \mathcal G}$ is also acyclic. Therefore, by the long exact sequence of cohomology, ${\mathcal F^r}$ is acyclic, so ${\mathcal F}$ is acyclic. $\Box$

Thefore, ${Y}$ has cohomological dimension ${0}$, which contradicts the assumption that it is not affine.

# The Lebesgue Number Lemma and uniform continuity

In this post, I’ll prove the Lebesgue Number Lemma and use it to prove that a continous function on a compact metric space is uniformly continuous.

Lemma 1 (Lebesgue Number Lemma). Let ${(X,d)}$ be a compact metric space and let ${\{V_i\}_{i \in I}}$ be an open cover of ${X}$. Then there exists a real number ${\delta > 0}$ such that every open ball of radius ${\delta}$ is contained in some ${V_i}$.

Proof: First, remark that if any refinement of the cover ${\{V_i\}_{i \in I}}$ satisfies this property, then ${\{V_i\}}$ also satisfies this property; thus, since ${X}$ is compact, we may replace the cover ${\{V_i\}}$ by a finite cover by open balls ${\{B(x_i, r_i)\}_{i=1}^n}$.

Define ${f_i}$ on ${X}$ by

$\displaystyle f_i(x) = \min(0, r_i - d(x, x_i)).$

Then ${f_i}$ is continous and its support is ${B(x_i, r_i)}$. Let ${f= \min(f_1, \dots, f_n)}$. Then ${f}$ is continous, and ${f>0}$ because ${\{B(x_i, r_i)\}}$ covers ${X}$. Since ${X}$ is compact, ${f}$ attains its minimum, which is ${>0}$; we call it ${\delta}$. Now, if ${x \in X}$, the statement ${f(x)\leq \delta}$ means precisely that for some ${i}$, ${d(x, x_i) \leq r_i - \delta}$, so the ball of radius ${\delta}$ around ${x}$ is contained in ${B(x_i, r_i)}$. $\Box$

Theorem 2 Let ${f}$ be a continous function on the compact metric space ${(X, d)}$. Then ${f}$ is uniformly continous.

Proof: Let ${\epsilon>0}$. For each ${w \in X}$, let

$\displaystyle V_w = \{y \in X : |f(w)-f(y)|<\epsilon/2\}.$

Then ${\{V_w\}_{w \in X}}$ is an open cover of ${X}$. Let ${\delta}$ be a Lebesgue Number for the cover ${\{V_w\}}$. Then, if ${x, y\in X}$ are such that ${d(x, y)< \delta}$, there exists a ${w \in X}$ such that ${B(x, \delta) \subseteq V_w}$; therefore, since ${x, y \in V_w}$, we have

$\displaystyle |f(x)-f(y)|\leq |f(x)-f(w)| + |f(w)-f(y)| < \epsilon.$

$\Box$

# Eisenstein series identities, directly

Let ${\Omega \subseteq \mathbf C}$ be a lattice. The Weierstrass ${\wp}$-function is defined as

$\displaystyle \wp(z) = \frac{1}{z^2} + \sum_{\omega \in \Omega^*} \left(\frac{1}{(z-\omega)^{2}} - \frac{1}{\omega^{2}}\right).$

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

$\displaystyle \wp(z) = \frac{1}{z^2} + c_2z^2 + c_4z^4 + c_6z^6 + \dots,$

where ${c_{2m} = (2m+1)\sum \frac{1}{\omega^{(2m+2)}}}$. Its derivative is

$\displaystyle \wp'(z) = \sum_{\omega \in \Omega} \frac{1}{(z-\omega)^3} = \frac{-2}{z^3} + 2c_2z + 4c_4z^3 + 6c_6z^5 +\dots.$

The functions ${\wp}$ and ${\wp'}$ satisfy

$\displaystyle \wp'(z)^2 = 4\wp(z)^2 - g_2\wp(z) - g_3$

in terms of the quantities ${g_2 = 20c_2}$ and ${g_3 = 28c_4}$. Now if we take ${\Lambda = \left<1, \tau\right>}$, where ${\tau}$ is in the upper half-plane, then ${\sum \frac{1}{\omega^{2m}} = G_{2k}(\tau)}$, where ${G_{2k}}$ is the weight ${2k}$ Eisenstein series. These Eisenstein series, or rather the normalized Eisenstein series ${E_{2k} =\frac{G_{2k}}{G_{2k}(i\infty)}= \frac{G_{2k}}{2\zeta(2k)}}$, satisfy certain relations, such as: (wikipedia)

$\displaystyle \begin{array}{rcl} E_{8} &=& E_4^2 \\ E_{10} &=& E_4\cdot E_6 \\ 691 \cdot E_{12} &=& 441\cdot E_4^3+ 250\cdot E_6^2 \\ E_{14} &=& E_4^2\cdot E_6 \\ 3617\cdot E_{16} &=& 1617\cdot E_4^4+ 2000\cdot E_4 \cdot E_6^2 \\ 43867 \cdot E_{18} &=& 38367\cdot E_4^3\cdot E_6+5500\cdot E_6^3 \\ 174611 \cdot E_{20} &=& 53361\cdot E_4^5+ 121250\cdot E_4^2\cdot E_6^2 \\ 77683 \cdot E_{22} &=& 57183\cdot E_4^4\cdot E_6+20500\cdot E_4\cdot E_6^3 \\ 236364091 \cdot E_{24} &=& 49679091\cdot E_4^6+ 176400000\cdot E_4^3\cdot E_6^2 + 10285000\cdot E_6^4. \end{array}$

In most basic texts on modular forms, these identities are derived by proving and exploiting the fact that the space ${M_{2k}}$ of modular forms of weight ${2k}$ is finite-dimensional. For instance, the fact that ${E_8}$ and ${E_4^2}$ have the same value at ${i\infty}$, combined with ${\dim M_{8} = 1}$, implies ${E_8=E_4^2}$. However, there is another, more “hands on” way to derive these identities.

Let us prove that ${E_8 = E_4^2}$. Substituting Laurent series at the origin in the equation ${\wp'(z)^2 - 4\wp(z)^2 + g_2\wp(z) + g_3=0}$, we see after some rearranging that the function

$\displaystyle (12c_2^2 - 36c_6)z^2 + (12c_2c_4 - 44c_8)z^4 + (-4c_2^3 + 4c_4^2 + 20c_2c_6 - 52c_10)z^6 + ...$

is identically ${0}$, and therefore

$\displaystyle \begin{array}{rcl} 0 &= &12c_2^2 - 36c_6\\ 0 &=& 12c_2c_4 - 44c_8 \\ 0 &=& -4c_2^3 + 4c_4^2 + 20c_2c_6 - 52c_{10}\\ & \dots & \end{array}$

Now

$\displaystyle c_{2m} = (2m+1)G_{2m+2} = (2m+1)2\zeta(2m+2)E_{2m+2} = (-1)^{m+1}(2m+1)\frac{(2\pi)^{2m+2}}{(2m+2)!}B_{2m+2}.$

Thus, for instance, ${c_6 = 14 \zeta(8)E_8 = \frac{14}{9450}\pi^8E_8}$. Making these substitutions in the first equation and factoring out ${\pi^8}$, we see that

$\displaystyle 0= \frac{12 \cdot 36}{90^2}E_4^2 - \frac{14 \cdot 36}{9450}E_8.$

Since ${\frac{12 \cdot 36}{90^2} = \frac{14 \cdot 36}{9450}}$, we have ${E_4^2 = E_8}$.

In fact, we have a bijection between ${\{(\wp_\Omega(z), \wp_\Omega'(z)) : \Omega \subseteq C\}}$ and the set of infinituples ${(c_2, c_4, c_6, \dots)}$ of complex numbers satisfying the infinite system of equations ${I}$:

$\displaystyle \begin{array}{rcl} 0 &= &12c_2^2 - 36c_6\\ 0 &=& 12c_2c_4 - 44c_8 \\ 0 &=& -4c_2^3 + 4c_4^2 + 20c_2c_6 - 52c_{10}\\ & \dots & \end{array}$

Luckily, the values of ${c_2}$ and ${c_4}$ determine all of the others, and the ring ${\mathbf C[c_2, c_4, c_6, \dots]/I}$ is generated by ${c_2}$ and ${c_4}$, and is in fact isomorphic to ${\mathbf C[c_2, c_3] = \mathbf C[G_4, G_6]}$. This means precisely that ${\{(\wp_\Omega(z), \wp_\Omega'(z)) : \Omega \subseteq C\}}$ is in bijection with the closed points of ${\text{Proj} (\mathbf C[G_4, G_6]) = \mathbf P^1_{\mathbf C}}$.

# A Noetherian and Hausdorff space is finite

In this post, I will prove that a Noetherian and Hausdorff topological space is finite (and therefore has the discrete topology, being Hausdorff). The proof is very short and pleasant.

Proof: Let ${X}$ be such a space, and suppose that it is infinite. Let ${\Sigma}$ be the collection of infinite closed subsets of ${X}$. It is nonempty since ${X \in \Sigma}$, and therefore has a minimal member ${Z}$ by the Noetherian assumption. Let ${p,q}$ be points of ${Z}$, and ${U,V}$ be disjoint open neighborhoods of ${p}$ and ${q}$ respectively (such ${U}$ and ${V}$ exist by the Hausdorff assumption). Then ${X = (X-U) \cup (X-V)}$ since ${U}$ and ${V}$ are disjoint, so ${Z = (Z \cap (X-U)) \cup (Z \cap (X-V))}$. Now each of ${Z \cap (X-U)}$ and ${Z \cap (X-V)}$ is closed in ${X}$, and is properly contained in ${Z}$ (the first one doesn’t contain ${p}$, and the second one doesn’t contain ${q}$). Therefore, by minimality of ${Z}$, each must be finite, and therefore ${Z}$ is also finite, which is a contradiction. $\Box$

Corollary: in any infinite Hausdorff space, there exists a strictly descending infinite chain of closed subsets $Z_1 \supset Z_2 \supset Z_3 \dots$. The proof above can be easily adapted to construct such a sequence.

# The Mayer-Vietoris sequence in sheaf cohomology

In this post, I will prove the Mayer-Vietoris Sequence for sheaf cohomology.

In what follows, ${X}$ is a topological space and ${\mathcal F, \mathcal G, \mathcal H}$ are sheaves of abelian groups on ${X}$. Let ${Z}$ be a closed subset of ${X}$. We let ${\Gamma_Z(X,\mathcal F)}$ denote the global sections of ${\mathcal F}$ with support in ${Z}$. The functor ${\Gamma_Z(X, -)}$ is a left-exact additive functor from sheaves on ${X}$ to abelian groups, and its right derived functors, denoted ${H^i_Z(X, -)}$, is the ${i}$-th cohomology of ${X}$ with support in ${Z}$. If ${\mathcal F}$ is a sheaf, the presheaf ${U \mapsto \Gamma_{Y \cap U}(U, \mathcal F)}$ is also a sheaf on ${X}$, denoted ${\mathcal H^0_Y(\mathcal F)}$ and called the “subsheaf of ${\mathcal F}$ with support in ${Y}$“.

The Mayer-Vietoris sequence, for a sheaf ${\mathcal F}$ and for a pair of closed subsets ${Y,Z \subseteq X}$, is the long exact sequence of cohomology with supports

$\displaystyle \dots \rightarrow H^i_{Y \cap Z}(X, \mathcal F) \rightarrow H^i_Y(X, \mathcal F) \oplus H^i_Z(X, \mathcal F) \rightarrow H^i_{Y \cup Z}(X, \mathcal F) \rightarrow H^{i+1}_{Y \cap Z}(X, \mathcal F) \rightarrow \dots$

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

Lemma 1 Let ${\mathcal E}$ be a flasque sheaf, ${Y}$ a closed subset of ${X}$, and ${U=X-Y}$. Then the sequence

$\displaystyle 0 \rightarrow \Gamma_Y(X, \mathcal E) \rightarrow \Gamma(X, \mathcal E) \rightarrow \Gamma(U, \mathcal E) \rightarrow 0$

is exact.

Proof: Trivial. $\Box$

Lemma 2 Let ${\mathcal E}$ be a flasque sheaf, and let ${Y, Z}$ be closed subsets of ${X}$. Then the sequence

$\displaystyle 0 \rightarrow \Gamma_{Y \cap Z}(X, \mathcal E) \rightarrow \Gamma_Y(X, \mathcal E) \oplus \Gamma_Z(X, \mathcal E) \rightarrow \Gamma_{Y \cup Z}(X, \mathcal E) \rightarrow 0$

is exact (where the first map is the diagonal embedding, and the second map is ${(s, t) \mapsto s-t}$).

Proof: Exactness is clear except possibly on the right. Let $U=X-Y, V=X-Z$, and Let ${D, E}$ be the short exact sequences

$\displaystyle 0 \rightarrow \Gamma(X, \mathcal E) \rightarrow \Gamma(X, \mathcal E) \oplus \Gamma(X, \mathcal E) \rightarrow \Gamma(X, \mathcal E) \rightarrow 0$

and

$\displaystyle 0 \rightarrow \Gamma(U \cup V, \mathcal E) \rightarrow \Gamma(U, \mathcal E) \oplus \Gamma(V, \mathcal E) \rightarrow \Gamma(U \cap V, \mathcal E) \rightarrow 0$

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

Now we are ready to prove the existence of the Mayer-Vietoris sequence for ${\mathcal F}$. Let

$\displaystyle 0 \rightarrow \mathcal F \rightarrow \mathcal E^0 \rightarrow \mathcal E^1 \rightarrow \dots$

be a flasque resolution of ${\mathcal F}$. By the lemma, we have a short exact sequence of complexes

$\displaystyle 0 \rightarrow \Gamma_{Y \cap Z}(X, \mathcal E^\bullet) \rightarrow \Gamma_Y(X, \mathcal E^\bullet) \oplus \Gamma_Z(X, \mathcal E^\bullet) \rightarrow \Gamma_{Y \cup Z}(X, \mathcal E^\bullet) \rightarrow 0.$

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

# Ph.D. Comprehensive exam practice problems, Round 2

Exercise 1 Let ${V}$ be the vector space of continuous real-valued functions on the interval ${[0,\pi]}$. Then, for any ${f \in V}$,

$\displaystyle 2 \int_0^\pi f(x)^2 \sin x dx \geq \left(\int_0^\pi f(x) \sin x dx\right)^2.$

Proof: Let ${d\mu}$ be the measure ${\frac{\sin x dx}{2}}$ on ${[0,\pi] = X}$. Then ${(X, d\mu)}$ is a probability space, ${f}$ is Lebesgue-integrable on ${X}$ and ${t \mapsto t^2}$ is a convex function ${\mathbf R \rightarrow \mathbf R}$. By Jensen’s inequality,

$\displaystyle \int_0^\pi f(x)^2 d\mu \geq \left(\int_0^\pi f(x) d\mu\right)^2.$

Multiplying throughout by ${4}$ we get the claimed inequality.
$\Box$

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

Proof: (a) The hypothesis means that every nonzero vector of ${V}$ is an eigenvector of ${T}$. Suppose ${v_1, v_2}$ are eigenvectors of ${T}$ with eigenvalues ${\lambda_1}$, ${\lambda_2}$. Since, by assumption ${v_1+v_2}$ is also an eigenvector, and ${v_1}$ and ${v_2}$ are independent, we can read off the eigenvalue of ${v_1 + v_2}$ off of either coefficient in the equation ${T(v_1+v+2)= \lambda_1 v_1 + \lambda_2 v_2}$, and therefore ${\lambda_1 = \lambda _2}$. Therefore ${T}$ is a multiple of the identity operator.

(b) Let ${T^\vee}$ be the dual operator on ${V^\vee}$. We claim that ${T^\vee}$ satisfies the condition of ${(a)}$. First, we have the following:

Lemma 1 Two functionals ${f, g : V \rightarrow k}$ (where ${k}$ 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 ${f}$ or ${g}$ is ${0}$ (in which case both are zero), so suppose neither is ${0}$. Recall that if ${W \subseteq V^\vee}$ and we define ${\mathrm{Ann}(W) = \{v \in V : f(v) = 0\: \: \forall w \in W\}}$, then we have a canonical isomorphism ${\mathrm{Ann}(W) \cong (V^\vee/W)^\vee}$, which in particular implies ${\dim \mathrm{Ann}(W) = \mathrm{codim}(W\subseteq V^\vee)}$. If we apply this to ${W=\left}$, we have, under assumption,

$\displaystyle \mathrm{Ann}(W) = \ker f \cap \ker g = \ker f = \ker g$

which has codimension ${1}$ since ${f,g \neq 0}$. Therefore ${W}$ has dimension ${1}$, and ${f}$ and ${g}$ are scalar multiples of each other. $\Box$

Now, back to ${(b)}$. S suppose that ${0 \neq f \in V^\vee}$. Then ${\ker f}$ has codimension ${1}$ in ${V}$, and therefore, under the hypothesis of (b), ${T(\ker f) \subseteq \ker f}$. This implies ${\ker T^\vee(f) \supseteq \ker f}$; indeed, if ${v \in \ker f}$, then ${T^\vee(f)(v) = f(Tv) = 0 }$ since ${Tv \in \ker f}$. Since ${\ker f}$ is codimension ${1}$, we either have equality, or ${T^\vee(f) = 0}$. If there is equality, then ${T^\vee(f)}$ and ${\ker f}$ have the same kernel and therefore they are proportional, i.e. ${f}$ is an eigenvector of ${T^\vee}$. If ${T^\vee(f)=0}$ then ${f}$ is trivially an eigenvector of ${T^\vee}$. In every case, we see that ${f}$ is an eigenvector of ${T^\vee}$. By (a), ${T^\vee}$, and therefore ${T}$, is a multiple of the identity operator. $\Box$

Exercise 3 Let ${T}$ be a linear operator on a finite-dimensional inner product space ${V}$.

• (a) Define what is meant by the adjoint ${T^*}$ of ${T}$.
• (b) Prove that ${\ker T^* = \mathrm{im}(T)^\perp}$.
• (c) If ${T}$ is normal, prove that ${\ker T = \ker T^*}$. Give an example when the equality fails (and, of course, ${T}$ is not normal).

Proof:

• (a) It is the unique linear operator ${T^*}$ on ${V}$ such that ${\left = \left}$ for every ${v, w \in V}$.
• (b) Indeed,

$\displaystyle \begin{array}{rcl} v \in \ker T^* &\Leftrightarrow& \left = 0 \: \forall w \in W \\ &\Leftrightarrow& \left = 0 \: \forall w \in W \\ &\Leftrightarrow& v \perp T(w)\: \forall w \in W. \end{array}$

• (c) A normal operator is one which commutes with its adjoint, i.e. ${TT^* = T^*T}$. Thus,

$\displaystyle \begin{array}{rcl} v \in \ker T^* &\Leftrightarrow& \left = 0\\ &\Leftrightarrow& \left = 0 \\ &\Leftrightarrow& \left = 0 \\ &\Leftrightarrow& \left = 0\\ &\Leftrightarrow& Tv=0. \end{array}$

An example where the equality fails is supplied by the operator ${T=\left(\begin{array}{ll} 1 & 1 \\ 0 & 1 \end{array}\right)}$ acting on ${(\mathbf R^2, \bullet)}$ in the standard way. The vector ${(1,-1)}$ is in the kernel of ${T}$ but not of ${T^*}$.

$\Box$