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Given a Riemannian metric $g$ on a smooth manifold $M$, one defines an $L^2$-inner product on the space $\bigwedge^\ast(M)$ of differential forms by

$$ \langle \alpha, \beta \rangle_g = \int_M \alpha \wedge \ast_g \beta, $$

where $\ast_g$ denotes the Hodge-star operator relative to $g$, and $\alpha, \beta$ are forms of the same degree.

Question: Does every inner product on $\bigwedge^\ast(M)$ as a graded vector space come from some metric $g$? How about inner products on $k$-forms $\bigwedge^k(M)$ for a single $k$, especially $0 < k < \dim M$?

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  • $\begingroup$ Restrict your inner product on $\wedge^\ast (M)$ to functions... is it a metric? $\endgroup$
    – David Roberts ♦
    Commented Sep 9, 2011 at 23:48
  • $\begingroup$ @David: It is the usual $L^2$ metric induced by $g$: since $*1 = \operatorname{vol}_g$, $\langle h, k \rangle_g = \int_M h k \, d\operatorname{vol}_g$. $\endgroup$
    – Brian Clarke
    Commented Sep 10, 2011 at 0:36
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    $\begingroup$ Sorry, I meant $L^2$ inner product. $\endgroup$
    – Brian Clarke
    Commented Sep 10, 2011 at 0:43

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Fixing a $k$ for simplicity, there are many inner products on $\bigwedge^k (M)$ (which I would usually denote $\Omega^k (M)$). Since $\bigwedge^k T^* M$ is a vector bundle, there are, for example, Sobolev $H^s$ inner products on its space of smooth sections for any natural number $s$. See, for example, Palais, Foundations of Global Non-Linear Analysis.

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  • $\begingroup$ Just to expand on your answer a little (for my own benefit?): If I understand the question asked, it is whether every inner product on $\Omega^k(M)$ is the $L^2$ inner product for some Riemannian metric on $M$. Since (for closed manifolds) the Sobolev $H^s$ norm for a fixed $s$ but two different Riemannian metrics are equivalent, but for different $s$ are inequivalent, then e.g. the $H^1$ inner product on $\Omega^k(M)$ can never be the $L^2=H^0$ inner product for any riemannian metric on $M$. $\endgroup$
    – Paul
    Commented Sep 11, 2011 at 3:13
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    $\begingroup$ Thanks, Brian and Paul - that's exactly what I wanted to know. Thanks also to José for an entirely different (at least for me) point of view. $\endgroup$
    – Slobodan Simić
    Commented Sep 23, 2011 at 19:14
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One can think of differential forms on a smooth manifold $M$ as spinors of the Clifford bundle constructed out of the vector bundle $TM \oplus T^*M$ with the split signature inner product induced from the dual pairing.

This identification does not respect the grading, though, only the parity (i.e., the reduction mod 2 of the usual grading).

Spinor modules have invariant inner products and in particular those of split signature Clifford algebras $Cl(n,n)$ have inner products which are either symmetric with split signature or symplectic.

This is explained in Chapter 12 (especially table 12.30) in Harvey's Spinors and calibrations and also in Marco Gualtieri's thesis, where this inner product on forms goes by the name of the Mukai pairing.

It does not require a metric on $M$ for it to be defined, so the answer to your question seems to be "No", at least if by "inner product" you allow any non-degenerate symmetric bilinear form.

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