# 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.