# 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.