# Dolbeault cohomology of complex tori.

Let $T=\mathbb C^n/\Lambda$ a complex torus. It is completely elementary to prove that the de Rham cohomology of $T$ in degree $q$ is isomorphic to the $q$-th exterior power of the dual of $\mathbb C^n$.

Indeed, if $\omega$ is a $d$-closed $q$-form on $T$ then for every translation $\tau\colon T\to T$ the form $\tau^*\omega-\omega$ is $d$-exact: one can explicitly determine a $(q-1)$-form $\eta$ such that $d\eta=\tau^*\omega-\omega$ simply by integration of $\omega$ along the segment $[x,x+a]$ if $\tau$ is the translation by $a$ (mimicking the homotopy formula for proving the Poincaré lemma). This shows that the average $\widetilde\omega$ of $\omega$ on $T$ (which has constant coefficients) is again cohomologous to $\omega$. In this way we have a surjective map form the $q$-th exterior power of the dual of $\mathbb C^n$ to $H^q(T,\mathbb C)$. This map is then straightforwardly seen to be injective.

Of course, this proof has nothing to do with the fact that $T$ is complex, but

is there an analogous elementary argument for the Dolbeault cohomology?

Everything goes through, except for the statement

if $\omega$ is a $\overline\partial$-closed $(p,q)$-form on $T$ then $\tau^*\omega-\omega$ is $\overline\partial$-exact.

Of course, I am not looking for a proof which passes through harmonic theory or any other non elementary argument.

Thanks in advance.

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Look at [Birkenhake - Lange, Complex Abelian Varieties, Theorem 1.4.1]. There is an "elementary proof" of Hodge decomposition for complex tori, maybe it can be helpful. –  Francesco Polizzi Nov 3 '10 at 7:43
I'll take a look, grazie. –  diverietti Nov 3 '10 at 8:33
That proof appears to use both harmonic theory and references to Griffiths and Harris. Another reference is to Mumford's Abelian varieties, pages 4-13, which uses Fourier series and apparently owes a debt to Weil's varie'te's Kahleriennes. –  roy smith Nov 18 '10 at 2:37