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The above text is from Griffiths and Harris. $M=V/\Lambda$ is a complex tori where $V$ is a $n$-dimensional complex vector space and $\Lambda$ is a lattice isomorphic to $\mathbb{Z}^{2n}$ in $V$. We are trying to see conditions required for existence for a Hodge form on $M$. Can someone please explain in the above text assuming $\tilde{\omega}$ a rational form, how does $\omega$ become an integral form on $M$ ? Also does this idea generalize for any Lie group (integrating w.r.t the Haar measure similarly)?

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As for your particular question, I think the authors assumed that $\tilde{\omega}$ is integral. They really mean to take a integral multiple of the original Hodge form in Kodaira's theorem.

In general, taking average does not change the cohomology class.

More precisely, let $G$ be a compact Lie group acting transitively on a smooth manifold $M$, then there is a chain map between complex of differential forms and complex of $G$-invariant differential forms $$I:\Omega^*(M)\to \Omega^*_G(M)$$ $$I(\omega)=\int_Gg^*\omega\ dg.$$ Here $dg$ is the Haar measure on $G$ (satisfying $\int_G 1dg=1$). Let $J:\Omega^*_G(M)\to \Omega^*(M)$ be inclusion. The following is classical by Chevalley:

Theorem: $J$ induces isomorphism of on the level of cohomology, and the inverse map is induced by $I$.

So in particular any cohomology class on $M$ is represented by a $G$-invariant form.

Now apply for $G$= complex torus acting on itself, the forms $dz_i$ and $d\bar{z}_j$ are all $G$-invariant, that's why $$I(\tilde{\omega})=I(\sum \tilde{h}_{\alpha\beta}(z)dz_i\wedge dz_j)=\sum\big (\int_G\tilde{h}_{\alpha\beta}(z)dg\big )dz_i\wedge dz_j=\sum h_{\alpha\beta}dz_i\wedge dz_j,$$ which does not change the cohomology class. So in particular if $\tilde{\omega}$ is integral of type $(1,1)$, the invariant form $I(\tilde{\omega})$ is still integral and of type $(1,1)$.

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  • $\begingroup$ Is it because $\omega$ and $g^* \omega$ are in the same cohomology class, so $I(\omega)= \omega\in H^*$ ? $\endgroup$
    – reuns
    Feb 23 '20 at 2:37
  • $\begingroup$ @reuns That's right. $I(\omega)$ is cohomologous to $\omega$ is essentially the fact that $g^*\omega$ is cohomologous to $\omega$. $\endgroup$
    – AG learner
    Feb 23 '20 at 2:55

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