# The Cohomology Computation

Alright, I’m in a sort of tough spot. Yesterday I started typing this up, but I just don’t have the motivation. There are lots of tedious details that no one is going to read and will not come up in our study after this. It is all incredibly standard chasing Fourier coefficients around, so I’m not going to do it. This post will be an outline in how one would go about doing it, and I even may provide quick ideas behind it, but if it will be weeks before I continue on if I don’t just get through this. Someday, if it seems important, I’ll come back and fill it in. Or if someone comments and really wants to see one particular part, at least I’ll have motivation that someone is going to read it.

Here goes. Recall some of my conventions. $X$ is a compact complex Lie group. We let $A=\Gamma(X, \mathcal{C})$ the global sections of the sheaf of $C^\infty$-functions on $X$. This was important to the Dolbeaut resolution. $\overline{T}$ are the $\mathbb{C}$-antilinear functionals. We use $\bigwedge^q$ to mean $\bigwedge^q(\overline{T})$.

We’ve shown that $A\otimes_\mathbb{C}\bigwedge^q$ is isomorphic to $\Gamma(X, \mathcal{C}^{0,q})$ as complexes and hence we have the iso $H^q(X, \mathcal{O}_X)\simeq H^q(A\otimes_\mathbb{C}\bigwedge)$.

Now we want to show that we actually have an induced isomorphism on cohomology from the inclusion map $i: \bigwedge \hookrightarrow A\otimes \bigwedge$. We’ll do this by comparing Fourier series. So we set up a normalized measure on $X$ say $\mu$. By integrating functions against this measure we get linear function $\mu_{\wedge}$ (I called this something different last time).

Now we need the lemma that for all $\omega\in A\otimes \bigwedge^q$ we have $\mu_\wedge (\overline{\partial} \omega)=0$. This follows from periodicity and translation invariance of the vector field that comes up when you go to write it down.

The next step is to define the “Fourier coefficients” of a function. Now if we choose any integer valued function from the lattice defining $X$, say $\lambda$. Then it extends to an $\mathbb{R}$-linear function on the tangent space at the identity, $V$. We then have the exponentiation map $\displaystyle v\mapsto e^{2\pi i \lambda(v)}$. This respects the lattice and hence descends to $\latex X$. Call this map $c_\lambda$.

Define the $\mathbb{C}$-linear function $A\to \mathbb{C}$ by $Q_\lambda(f)=\int_X c_{-\lambda}fd\mu$. Don’t be intimidated here. This is just the standard Fourier coefficient when your in a familiar situation. i.e. $f=\sum e_\lambda \otimes Q_\lambda(f)$.

Now choose a Hermitian inner product, which gives us a norm to work with. Here things will become less detailed. One can next prove a lemma that the map $f\to \{Q_\lambda(f)\}_\lambda$ is an isomorphism $A\to$ the space of maps decreasing at infinity faster than $\|\lambda \|^{-n}$ for all $n$.

Then do a computation to see that $Q_\lambda (\overline{\partial}\omega)=(-1)^p 2\pi i \left(Q_\lambda(\omega)\wedge \overline{C}(\lambda)\right)$.

Lastly we want to get back to showing that the inclusion is a homotopy equivalence. Thus use the Hermitian inner product to define a map $\lambda^*\in Hom_\mathbb{C}(\overline{T}, \mathbb{C})$ for every $\lambda$ by $\displaystyle \lambda^*(x)=\frac{\langle x, \overline{C}(\lambda)\rangle}{2\pi i \|\overline{C}(\lambda)\|^2}$.

We need to define for $\omega\in A\otimes \bigwedge^p$ a map $k(\omega)$ which will only be defined in terms of its Fourier coefficients. We define $Q_\lambda(k(\omega))=(-1)^p \lambda^* \neg Q_\lambda(\omega)$ if $\lambda\neq 0$ and the coefficient is 0 if lambda is 0. Now it has all been set up so that comparing Fourier coefficients on $\overline{\partial} k + k \overline{\partial}$ we get exactly the same ones as in $id_{A\otimes\wedge} - i \mu_\wedge$. Thus we are done by the magic of Fourier coefficients being unique.

That last computation I left out requires use of things such as “Cartan’s magic formula” and breaking it into two cases. Anyway, for not doing any details, I think this is a pretty thorough outline and filling any of the details you don’t believe or would like to know shouldn’t be too hard.

# Cohomology of Abelian Varieties II

Two posts in the same week! Before we get started today, we need to introduce one more piece of new notation. Let ${\overline{T}}$ be the ${\mathbb{C}}$-antilinear maps ${V\rightarrow \mathbb{C}}$. Our goal is to prove that ${H^q(X, \mathcal{O}_X)\simeq \bigwedge^q\overline{T}}$ and ${H^q(X, \Omega^p)\simeq \bigwedge^pT\otimes \bigwedge^q\overline{T}}$.

To do the calculation we will use the Dolbeault resolution: ${0\rightarrow \mathcal{O}_X\rightarrow \mathcal{C}^{0,0}\rightarrow \mathcal{C}^{0,1}\rightarrow \mathcal{C}^{0,2}\rightarrow\cdots}$. This is an acyclic resolution of the structure sheaf, and so is fine to use for the calculation of cohomology. The first upper index of ${\mathcal{C}}$ refers to the degree of the ${\mathbb{C}}$-linear part and the second upper index refers to the degree of the ${\mathbb{C}}$-antilinear part. The map of the chain complex is ${\overline{\partial}}$.

Let’s examine the complex a little more closely. Define ${\phi_{p,q}: \mathcal{C}\otimes (\bigwedge^pT\otimes \bigwedge^q\overline{T})\rightarrow \mathcal{C}^{p,q}}$ by ${\sum f_i\otimes \alpha_i\mapsto \sum f_i\omega_{\alpha_i}}$. Where we define ${\omega_{\alpha}}$ to be the natural translation invariant ${(p,q)}$-form associated to ${\alpha\in \wedge^pT\otimes \wedge^q\overline{T}}$ by left-invariantizing.

Note that ${\omega_{\alpha\wedge\beta}=\omega_\alpha\wedge \omega_\beta}$. Thus to prove that all ${\omega_\alpha}$ are ${\overline{\partial}}$-closed (which we’ll denote ${d}$ from now on for simplicity), we only need to check this for ${(1,0)}$ and ${(0,1)}$ forms. Now ${exp: V\rightarrow X}$ is a local iso, so we also only need to check ${d(exp^*(\omega_\alpha))=0}$. But ${exp^*(\omega_\alpha)=d\alpha}$, so ${d}$ of this expression, is ${(d\circ d)(\alpha)=0}$.

This gives us that our map ${\phi_{0,q}}$ is an iso ${\Gamma(X, \mathcal{C})\otimes_\mathbb{C} \bigwedge^q\overline{T} \rightarrow \Gamma(X, \mathcal{C}^{0,q})}$. The map ${\overline{\partial}}$ is defined to be ${\overline{\partial}(f\otimes \alpha)=\overline{\partial}(f)\otimes \alpha}$. So since the ${\omega_\alpha}$ are closed, the iso commutes with the differential and we get these are actually isomorphic as chain complexes. Thus computing cohomology of one is equivalent to computing cohomology of the other.

Explicitly, we know that ${H^q(X, \mathcal{O}_X)\simeq H^q(X, \Gamma(X,\mathcal{C})\otimes_\mathbb{C} \bigwedge^q\overline{T})}$.

Since this notation is cumbersome, let ${A=\Gamma(X, \mathcal{C})}$ and ${\bigwedge^*=\bigwedge^*\overline{T}}$.

Let ${i: \bigwedge \rightarrow A\otimes_\mathbb{C} \bigwedge}$ be the inclusion. We want to show this gives an iso ${\bigwedge^q\stackrel{\sim}{\rightarrow} H^q(X, A\otimes \bigwedge^*)}$. This is precisely the goal given at the start of the post.

Now ${V}$ is a vector space so we have a natural Euclidean measure. Let ${\mu}$ be the measure on ${X}$ induced from this that is normalized so that ${\mu(X)=1}$. (For those familiar, this is just the unique translation invariant Haar measure, and ${X}$ is compact, so is finite and can be normalized). Define the linear map represented by ${\mu}$ by ${S: A\rightarrow \mathbb{C}}$. It is just ${S(f)=\int_X fd\mu}$. If ${W}$ is any ${\mathbb{C}}$-vector space, denote by ${S_W}$ the map ${A\otimes_\mathbb{C} W\rightarrow W}$. In particular, we have ${S_\wedge: A\otimes \bigwedge^*\rightarrow \bigwedge^*}$ so ${S_\wedge\circ i=id_\wedge}$.

We have a good ways to go yet, so next time we’ll pick up with the lemma that ${S_\wedge(\overline{\partial}\omega)=0}$ for all ${\omega\in A\otimes \bigwedge}$.

# Cohomology of Abelian Varieties

Hopefully I’ll start updating more than once a month. Since it’s been awhile and the previous post was tangent to what we’re actually doing, I’ll recap some notation. ${X}$ will be a compact complex (connected) Lie group of dimension ${g}$. We showed that we have an analytic isomorphism ${X\simeq (S^1)^{2g}\simeq (\mathbb{R}/\mathbb{Z})^{2g}}$. Let ${V=T_0X}$ (note that I’ll assume ${0}$ is the identity).

Under the exponential map ${exp: V\rightarrow X}$ (which we showed was a local isomorphism), we have that ${V}$ is the universal covering space of ${X}$. We showed that ${ker(exp)=U}$ is a lattice. Now the title of this post will seem a little silly to experts out there in cohomology, since we know that topologically these things are all tori. We’ll go through the details anyway.

First, we’ll show that ${H^r(X, \mathbb{Z})\simeq }$ the group of alternating ${r}$-forms ${U\times \cdots \times U\rightarrow \mathbb{Z}}$. We proceed by induction.

By general covering space theory ${exp^{-1}(0)=U=\pi_1(X, 0)}$. Thus we get the base case ${H^1(X,\mathbb{Z})\simeq Hom(U=\pi_1(X), \mathbb{Z})}$. Now we’ll want to show that the cup product induces the isomorphism ${\bigwedge^r\left(H^1(X,\mathbb{Z})\right)\rightarrow H^r(X,\mathbb{Z})}$. By the ${r=1}$ case this proves the statement.

We first reduce to the case of showing it is true for ${S^1}$. Since ${X}$ is just a product of ${S^1}$‘s, if we show that if the statement is true for ${X_1}$ and ${X_2}$, then it is also true for the product we can make the reduction. (For simplicitly, coefficients are in ${\mathbb{Z}}$, but I’ll omit that). Since we only need to apply this in the case where ${X_1}$ or ${X_2}$ is finite product of ${S^1}$‘s, we can also assume that the cohomologies are finitely generated and that they two spaces are connected for simplicity.

First, by the K\”{u}nneth formula: ${H^1(X_1\times X_2)\simeq \left(H^1(X_1)\otimes_\mathbb{Z}H^0(X_2)\right)\bigoplus \left(H^0(X_1)\otimes_\mathbb{Z}H^1(X_2)\right)}$, but the spaces are connected, so ${H^0(X_i)=\mathbb{Z}}$. Thus ${\bigwedge^r(H^1(X_1\times X_2))\simeq \bigwedge^r(H^1(X_1)\oplus H^1(X_2))}$
${\simeq \displaystyle\sum_{p+q=r}\bigwedge^p(H^1(X_1))\otimes \bigwedge^q(H^1(X_2))}$.

But now our inductive hypothesis is that for all $p$ less than $r$, ${\bigwedge^p(H^1(X_i))\simeq H^p(X_i)}$. Thus we get ${\displaystyle \sum_{p+q=r}\bigwedge^p(H^1(X_1))\otimes \bigwedge^q(H^1(X_2))}$
${\displaystyle \simeq \sum_{p+q=r} H^p(X_1)\otimes H^q(X_2)}$
${\simeq H^r(X_1\times X_2)}$. In other words, stringing all these isos together we get the iso we wanted. So we’ve reduced to the case of showing the statement for ${S^1}$, which follows immediately since ${H^n(S^1)=0}$ for ${n>1}$.

Note that if you have basic facts about singular cohomology at your disposal, this isn’t at all surprising. But let’s look at sheaf cohomology instead. This will require us to look at the Hodge structure which could be interesting. We won’t go very far today, but let’s at least get a few things out of the way.

Let ${\Omega^p}$ be the sheaf of holomorphic ${p}$-forms on ${X}$. We’d like to compute ${H^r(X, \Omega^p)}$. Let ${T=Hom(V, \mathbb{C})}$, i.e. the (complex) cotangent space at the identity to ${X}$. As with vectors and vector fields, every ${p}$-covector, i.e. element of ${\bigwedge^pT}$ can be extended uniquely to a left invariant ${p}$-form by pulling back along the left multiplication by ${-x}$ map. We’ll denote the correspondence ${\alpha\mapsto \omega_\alpha}$. This map defines an isomorphism of sheaves ${\mathcal{O}_X\otimes_\mathbb{C} \bigwedge^pT\stackrel{\sim}{\rightarrow} \Omega^p}$.

This says that ${\Omega^p}$ is a free sheaf of ${\mathcal{O}_X}$-modules. Now take global sections to get that ${\Gamma(X, \Omega^p)\simeq \Gamma(X, \mathcal{O}_X\otimes \bigwedge^pT)\simeq \bigwedge^pT}$, since the global sections of ${\mathcal{O}_X}$ are constants. Thus the only global sections of ${\Omega^p}$ are the ${p}$-forms that are invariant under left translation. Thus this isomorphism reduces our calculation to ${H^r(X, \Omega^p)\simeq H^r(X, \mathcal{O}_X)\otimes_\mathbb{C} \bigwedge^pT}$. So we’ll start in on that next time.