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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

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3 Answers 3

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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.

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This is such a good picture! Thank you. –  geodude Jul 25 '13 at 13:41

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

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This kind of misses the point. Of course you can endow every finite dimensional vector space with inner product. But OP asked for one that works for classes, not just randomly picked representatives. –  Marek Jul 21 '13 at 14:33
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@Marek: You are missing the point. This is far from a random choice. Given the Riemannian structure (which we need for all this discussion), it is canonical. –  Ted Shifrin Jul 21 '13 at 14:40
    
I am well aware of the role of harmonic forms in Hodge theory, thank you. I am just saying that you answered a different question than OP asked.. –  Marek Jul 21 '13 at 14:43
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..which, as far as I can see, asked whether the inner product on all forms descends to cohomology. The answer is it doesn't. –  Marek Jul 21 '13 at 14:47
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Wrong. As I said, there is a unique harmonic form (the element of least norm) in each class. What is your problem? –  Ted Shifrin Jul 21 '13 at 15:35

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 $d\lambda = 0$.

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