# Doubt about obtaining integral cohomology form from a rational cohomology form

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)?

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 $$(1,1)$$.

• Is it because $\omega$ and $g^* \omega$ are in the same cohomology class, so $I(\omega)= \omega\in H^*$ ? Feb 23, 2020 at 2:37
• @reuns That's right. $I(\omega)$ is cohomologous to $\omega$ is essentially the fact that $g^*\omega$ is cohomologous to $\omega$. Feb 23, 2020 at 2:55