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.

  • $\begingroup$ 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)? $\endgroup$ – Silvia Ghinassi Jan 18 '16 at 15:02
  • $\begingroup$ @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? $\endgroup$ – Najib Idrissi Jan 18 '16 at 15:10
  • $\begingroup$ Wait, doesn't that actually only say that the map to the dual is injective though? What about surjectivity? $\endgroup$ – Najib Idrissi Jan 18 '16 at 15:13
  • $\begingroup$ Hold on. Let me think about it, in case I'll delete. That's not really my area, just reminiscences... $\endgroup$ – Silvia Ghinassi Jan 18 '16 at 15:16
  • $\begingroup$ $\star \star= (-1)^{k(n-k)}$, so it should be ok. $\endgroup$ – 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.

  • $\begingroup$ 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. $\endgroup$ – Najib Idrissi Jan 18 '16 at 15:34
  • $\begingroup$ I'm glad I was able to help! $\endgroup$ – 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 $\Omega^k(M)$.

  • $\begingroup$ 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. $\endgroup$ – user98602 Jan 19 '16 at 3:21
  • $\begingroup$ Ah, interesting, thanks. $\endgroup$ – Najib Idrissi Jan 19 '16 at 8:28
  • 1
    $\begingroup$ @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? $\endgroup$ – Silvia Ghinassi Jan 19 '16 at 18:14
  • $\begingroup$ @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. $\endgroup$ – user98602 Jan 19 '16 at 18:15
  • $\begingroup$ 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. $\endgroup$ – Silvia Ghinassi Jan 19 '16 at 18:19

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.