# How to prove that in a Kähler manifold without boundary $\Omega \wedge \cdots \wedge \Omega$ is closed but not exact?

Let $M$ be a compact Kähler manifold without boundary.

1. How can I show that the volume form $$\Omega^{m} = \Omega \wedge \cdots \wedge \Omega$$ where we have the wedge of $m$ $\Omega$s is closed but not exact for the case that the complex dimension of $M$ is $m$. The hint I am given is to use Stoke's theorem.
2. Also, what happens with the cohomology of this system?
• I want to use the hint, i.e. Stoke's theorem but I don't know how to integrate this wedge product in order to reach some conclusion. – Marion May 28 '15 at 13:20

Hint Suppose $\Omega^m$ is exact; what does Stokes' Theorem say about the integral $\int_M \Omega^m$?
• Indeed, this is from Nakahara's Geometry, Topology and Physics. Well, if $\Omega^m$ is exact then we can write it as $\Omega^m = dY^m$. Then Stoke's theorem would say $\int_M \Omega^m = \int_M dY^m = \int_{\partial M} Y$. But we said that $M$ has no boundary. So, ok, I think I am confused now since $\Omega^m$ is not supposed to be exact. – Marion May 28 '15 at 23:09
• Right, since $M$ has no boundary, that integral is zero. Now, what can we say about the integral $\int_M \Omega^m$ given that $\Omega^m$ is a volume form? – Travis May 29 '15 at 3:36
• I am not quite sure what we can say about it. This is where I am confused about. I do not follow your thought. A) I am not sure what this integration to zero means and B) I am not sure why I have to assume that I can write $\Omega = dY$ since the exercise clearly says that $\Omega$ is not exact. – Marion May 29 '15 at 12:39
If $M$ is a Kahler manifold, then if $N \subset M$ is a compact complex submanifold without boundary of complex dimension $k$, then the chomology class of $\omega^{k}$ of $H^{2k}(M, \mathbb{R})$ and the homology class of $N$ in $H_{2k}(M, \mathbb{R})$ are non-zero.