Is the pairing induced by the wedge product and integration nondegenerate on de Rham forms? Let $M$ be a compact, oriented, smooth $n$-manifold and let $\Omega^*_{\mathrm{dR}}(M)$ be the commutative differential graded algebra of de Rham forms on $M$. We can define a pairing:
\begin{align}
\langle -,- \rangle : \Omega^k_{\mathrm{dR}}(M) \otimes \Omega^{n-k}_{\mathrm{dR}}(M) & \to \mathbb{R} \\
\alpha \otimes \beta & \mapsto \int_M \alpha \wedge \beta
\end{align}
Question. Is this pairing non-degenerate? In other words, is the map $\alpha \mapsto (\beta \mapsto \langle \alpha, \beta \rangle)$ an isomorphism $\Omega^k_{\mathrm{dR}}(M) \to \operatorname{Hom}_{\mathbb{R}}(\Omega^{n-k}_{\mathrm{dR}}(M), \mathbb{R})$?
This is true on the level of cohomology, a result known as Poincaré duality. Thus, given a closed $k$-form $\alpha$ which is not a coboundary, there exists a closed $(n-k)$-form $\beta$ with $\int_M \alpha \wedge \beta \neq 0$ (the cohomology groups of a compact manifold are finite dimensional so this is an equivalent characterization of nondegeneracy).
But I haven't been able to find anything on whether this is true on the level of de Rham forms directly; in fact I rather expect it to be false.
 A: The Hodge duality gives the non degeneracy of the pairing.
The Hodge star define a duality on de Rham forms, that is, if $\alpha$ is a non-zero $k$-form then $\star \alpha$ is a non-zero $(n-k)$-form.
The defining property of the $\star$ operator is indeed given by $\beta \wedge \star \alpha = \langle \alpha, \beta \rangle \operatorname{vol}$, where $\operatorname{vol}$ is a volume form ($M$ is oriented).

Edit (after sitting on it for a while): While the Hodge star gives us the isomorphism between $\Omega^k_{\mathrm{dR}}(M)$ and $\Omega^{n-k}_{\mathrm{dR}}(M)$ (which is what it is explained above), I don't think it gives us an isomorphism of $\Omega^k_{\mathrm{dR}}(M)$ with the dual of $\Omega^{n-k}_{\mathrm{dR}}(M)$, as $\Omega^{n-k}_{\mathrm{dR}}(M)$ is infinite dimensional. So you were correct in the comment, the fact that $\star \star= (-1)^{k(n-k)}$ only guarantees the isomomorphism mentioned above.
A: Actually, the answer is almost yes on the level of forms themselves. You've provided an inner product on $\Omega^k(M)$, and once you pass to its Hilbert space completion (the so-called space of $L^2$ forms) it is true, by the Riesz representation theorem, that any continuous linear functional $\Omega^k_{L^2}(M) \to \Bbb R$ is uniquely represented by $\alpha \mapsto \langle \alpha, \beta \rangle$; that is, it's uniquely represented by integration against $*\beta$; that is, it's uniquely represented by integration against an $(n-k)$-form. (An $L^2$ form, to be more careful.) To be more precise yet, the map $\Omega^k_{L^2}(M) \to \left(\Omega^k_{L^2}\right)^*$, given by $\alpha \mapsto \langle \cdot,\alpha\rangle$ is an isometry.
The problem with the example you gave, $f \mapsto f(0)$, is that it is not continuous in the $L^2$-topology, so you definitely should not expect it to be given by integration against any kind of form. But given an $L^2$-continuous functional $\Omega^k(M) \to \Bbb R$, you know it's given by integration against an $(n-k)$-form - but only necessarily an $L^2$ one, as an artifact of the non-completeness of $\Omega^k(M)$.
