In our lecture of Advanced Geometry, we had an example of a manifold on which there exist harmonic forms whose wedge product is not again harmonic. I didn't completely understand the example.
Let $S$ be an oriented, closed surface of genus $g \geq 2$. Then we know that $H_{dR}^{1}(S)= \mathbb{R}^{2g}$.
- I didn't know that. If anyone has a reference, that would be great.
Now we put some Riemannian metric $g$ on $S$. Let $\omega_1, \omega_2$ be two harmonic 1-forms. Then $\omega_1 \wedge \omega_2$ defines a nontrivial element in $H_{dR}^2(S)$. We know that $\omega_1$ is a smooth 1-form, so a smooth section of $T^{*}S$. Since $\chi(S) = 2-2g \neq 0$ we know: $T^{*}(S)$ cannot have a (smooth) section without zero.Thus $\omega_1\wedge \omega_2$ has zeros.
I don't understand how that shows that the wedge product is not harmonic. Does the proof somehow use the Hodge theorem ?
I don't know why we can't have a smooth nonzero section of $T^{*}S$