Dolbeault cohomology on torus

Question

Let $T=\mathbb{C}/\Gamma$ where $\Gamma$ is a lattice of $\mathbb C$. Given that $H_{dR}^1(T)=\mathbb{C}^2$. Prove that $H^{1,0}_\bar{\partial}(T)=\mathbb{C}$. I have no idea what to do. Can someone point me in the correct direction?

Answer 1

Here are a couple of ideas for doing this.

(1) A complex torus is a Kahler manifold, since any flat metric on Euclidean space is invariant under the action of the lattice that defines the torus. Hodge theory then says that $$ H^1(X,\mathbb C) = H^{1,0}(X) \oplus H^{0,1}(X) $$ and that those two vector spaces are conjugate to one another.

(2) A more low-tech approach is this. Let $p : \mathbb C \to X$ be the projection morphism and let $\sigma$ be a holomorphic $1$-form on $X$. Then $p^* \sigma$ is a holomorphic $1$-form on $\mathbb C$ and can thus be written as $p^* \sigma(z) = f(z) dz$ for some holomorphic function $f$. What's special about this function?

(3) A torus is a group, and the translations $\tau_v$, $x \mapsto x + v$, for $v \in X$ fixed are transitive. Given any smooth closed differential form $\sigma$ on $X$, consider the average $$ \tilde \sigma(x) := \int_{v \in X} \tau_v^* \sigma(x). $$ Prove that $\sigma$ and $\tilde \sigma$ are homotopic to each other, so they define the same cohomology class. Then the pullback $p^* \tilde \sigma$ is a constant form on $\mathbb C$, because we integrated all variation out of it. Picking a little more at this thread proves that $H^{p,q}(X) = \bigwedge^{p,q} \mathbb C^n $ for any $p,q$ and $n$-dimensional torus $X$.