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.

• I might be horribly wrong, so forgive me if I say something stupid, but doesn't that follow by the fact that the Hodge $*$ of a non-zero form is always non-zero (i.e. Hodge duality)? – Silvia Ghinassi Jan 18 '16 at 15:02
• @Silvia You may be right, Wikipedia seems to say that $\alpha \otimes \beta \mapsto \int_M \alpha \wedge \star \beta$ is an inner product for a Riemannian manifold. Maybe you can make that into an answer? – Najib Idrissi Jan 18 '16 at 15:10
• Wait, doesn't that actually only say that the map to the dual is injective though? What about surjectivity? – Najib Idrissi Jan 18 '16 at 15:13
• Hold on. Let me think about it, in case I'll delete. That's not really my area, just reminiscences... – Silvia Ghinassi Jan 18 '16 at 15:16
• $\star \star= (-1)^{k(n-k)}$, so it should be ok. – Silvia Ghinassi Jan 18 '16 at 15:17

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.

• Thanks! This at least answers my initial question before the edit. I just realized that the map is not surjective though - when $k=0$, the application $f \mapsto f(0)$ is not of the form $\langle \alpha, - \rangle$... I'll accept anyway because this was what I was most interested in. – Najib Idrissi Jan 18 '16 at 15:34
• I'm glad I was able to help! – Silvia Ghinassi Jan 18 '16 at 15:48

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 $\Omega^k(M)$.

• I think but am less sure that if your functional is continuous with respect to the $C^\infty$ topology on forms (which of course is not induced by an inner product at all) then the representing form $\beta$ will be smooth. – user98602 Jan 19 '16 at 3:21
• Ah, interesting, thanks. – Najib Idrissi Jan 19 '16 at 8:28
• @MikeMiller I have a stupid question. Once you take the $L^2$ completion and use Riesz, is it possible that $\beta$ does not belong to $\Omega^k$, but only to its completion. Am I missing something? – Silvia Ghinassi Jan 19 '16 at 18:14
• @SilviaGhinassi: I agree with you. Indeed there must be functionals of this form (just pick one that actually is given by integration against an $L^2$-form!). This is what I was trying to say with the parenthetical (an $L^2$ form...) - apologies for any lack of clarity. – user98602 Jan 19 '16 at 18:15
• Yeah. I just noticed that that's what you have written, I read it too fast. But yeah, taking the $L^2$ forms is a good way around it. – Silvia Ghinassi Jan 19 '16 at 18:19