Inner product of De Rham cohomology classes

Question

Is there a well-defined inner product between cohomology classes? In particular, is it possible to extend the Hodge inner product? If I try, I obtain this:

$$\int *(\omega + d\lambda)\wedge (\sigma + d\mu) = \int (*\omega\wedge\sigma + *d\lambda\wedge\sigma + *\omega\wedge d\mu + *d\lambda\wedge d\mu)$$

Is it possible to discard the last terms as an exact form, or to define anyway a product that does something similar?

Thanks

Answer 1

No, this is not possible because the integration depends heavily on the representative you choose.

Here's an illustration: take a circle $S^1$ and a bump form $b_1$ around $1$ (I am thinking of the circle realized as unit complex numbers). This generates $H^1(S^1)$. Similarly, $b_{-1}$ around $-1$ is another generator of the same class. You would expect the inner product of these two forms be non-trivial. But dualizing and taking products will leave you with zero form that integrates to zero.

Answer 2

No. If $\omega = \sigma = 0$ and $\lambda = \eta$ is smooth, then $$ \int_M \ast (\omega+d\lambda) \wedge (\sigma+d\eta) - \int_M \ast \omega \wedge \sigma = \int_M \ast d\lambda \wedge d\lambda = \|d\lambda\|^2, $$ which vanishes if and only if $d\lambda = 0$.

Answer 3

Assuming you're working on a compact Riemannian manifold, each cohomology class is represented by a unique harmonic form, so use the harmonic forms.