Inner products on differential forms 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 $0 < k < \dim
M$?
 A: 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.
A: 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.
