11
$\begingroup$

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.

$\endgroup$
4
  • 1
    $\begingroup$ 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. $\endgroup$ Nov 3, 2010 at 7:43
  • $\begingroup$ I'll take a look, grazie. $\endgroup$
    – diverietti
    Nov 3, 2010 at 8:33
  • 3
    $\begingroup$ 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. $\endgroup$
    – roy smith
    Nov 18, 2010 at 2:37
  • 1
    $\begingroup$ There is a simple argument using Fourier series. The (DC) condition stated in the introduction of [1] always holds for complex structures and you can apply the computations from section 7 of the same paper. [1]: arxiv.org/pdf/2106.09182.pdf $\endgroup$ Jul 23, 2021 at 19:00

0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.