# Equivalent singular chains and differential forms, as functionals on forms, on compact Riemannian manifolds

On a compact Riemannian oriented manifold $M$,for each singular $k$-chain $\sigma$ (with real coefficients), $\sigma$ induces a linear functional on the $\mathbb{R}$-vector space of differential k-forms, by integration of the form over $\sigma$. At the same time the metric induces an inner product on that space, by $<\alpha,\beta>=\displaystyle\int_{M}{\alpha\wedge *\beta}$. This product also gives, for each given form, a functional on the space of forms.

I'm just playing around here with possible relationships between basic stuff I was learning about, but it seemed to me like an obvious way to compare singular chains and forms is to compare the induced functionals (kind of in the spirit of Poincare duality and Stokes' theorem, which pair classes of closed forms with classes of cycles. However, the restriction to only considering singular chains may not make sense in this context...since you just need to integrate k-forms on them, not take boundaries or anything...). For example, when can a chain $\sigma$ and a form $\omega$ have the same functional?

I believe one can show that for any $\sigma$ there is a unique corresponding "dual" form so to speak, say $D(\sigma)$, with the same functional. Since $M$ is compact, there exists a countable orthonormal basis $e_j$ for the space of k-forms, and every element is determined by its inner product with the basis elements (its coordinates). So if we have some chain $\sigma$, we take $< D(\sigma),e_j>$ to be $\displaystyle\int_{\sigma}{e_j}$, and then $\displaystyle D(\sigma)= \sum_{j}{\left(\int_{\sigma}{e_j}\right)e_j}$, and one checks using the basis again that by construction this form has the same functional as $\sigma$. Furthermore, the functional of a form completely determines it, so a priori the dual form must be unique (and it doesn't matter that we chose a basis).

So my question is obviously first of all, does the above make sense? I don't recall seeing it yet. Another obvious question is the other direction - given a form, does it have a dual singular chain? (Or if one broadens from considering just singular chains?) If not, what can one say about the set of forms that do have duals, relative to the whole space of forms? (e.g. it might be dense.)

EDIT: Thanks, Petya, I guess I need to restrict to smooth singular chains. Also thanks Gonçalo for pointing out that I appear to really be talking about currents - I will take a look at the book! My remaining question: first of all, it seems to me like in the context of a compact Riemannian manifold, the space of k-currents is naturally identified with the space of k-forms via the inner product. So in this space, is the set of currents given by integration over a smooth k-submanifold a proper subspace? I understand that the point of currents is that in the general case they are broader, but in the compact case it seems like maybe that doesn't happen, and I'm having difficulty understanding the statements about the mass norm that seem to concern this question.

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@Petya's objection is that you need to restrict to smooth singular chains (which you 'can', as the inclusion of the complex of smooth chains into the full singular complex is an homotopy equivalence, so you do not loose much...) –  Mariano Suárez-Alvarez Mar 15 '10 at 3:41
It depends on what kind of monster objects you are willing to consider as "differential forms" and "chains". If M is one-dimensional, and chains and forms are 0-dimensional, the forms integrate the argument against a smooth density while chains integrate against measures concentrated at finitely many points. Of course you can define a larger space of functionals containing both types, is this what you want? –  Sergei Ivanov Mar 15 '10 at 14:18
You can not identify currents with differential forms; the space of currents is much larger. What 1-form do you associate to the current T defined by $T(\omega)=\omega_p(v)$ where $p\in M$ is some (fixed) point and $v\in T_p(M)$ is some (fixed) tangent vector? –  Sergei Ivanov Mar 18 '10 at 18:33

Let me make sure I understand you correctly. Given a compact Riemannian manifold of dimension $n$, any smooth $k$-chain induces a functional on the space of $k$-forms, and any $n-k$-form induces a functional on the space of $k$-forms. You're asking what the relationship is between these two functionals.

Answer: There is no overlap between these two kinds of functionals. Why? Suppose a $k$-chain $\sigma$ induces the same functional as a $n-k$-form $\omega$. Clearly, $\omega$ must be nonzero on some open set $U$ that doesn't intersect any part of $\sigma$. Then if $\alpha$ is a test $k$-form, then $\sigma(\alpha)$ doesn't depend on the values of $\alpha$ on $U$, but $\omega(\alpha)$ certainly does.

Now, if you look at these functionals at the level of homology/cohomology, then as you probably know, the relationship is Poincare duality. That is, $[\sigma]$ and $[\omega]$ induces the same functionals on the $k$-th cohomology iff they are Poincare dual.

Here's what currents have to do with the story. $k$-currents are essentially defined to be all linear functionals on the space of smooth $k$-forms (with appropriate topology). Then the two types of functionals you describe, the $\sigma$'s and the $\omega$'s are now specific examples of $k$-currents. If you know much about distributions, then you pretty much already know about currents--you just define everything via integration by parts. The boundary operator on currents generalizes both the boundary operator on chains and the $d^*$ operator on forms. Not surprisingly, the homology of currents gives you the (real) homology of $M$. The reason I immediately said that the $\sigma$'s and $\omega$'s have no overlap is that, in analogy with plain old distributions, the $\omega$'s are like smooth functions, while the $\sigma$'s are like singular measures supported on sets of measure zero.

I don't know much history, but I think that your idea was what lead de Rham to invent currents.

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I'll try to answer your question "does the above make sense?" It seems to me answer is "no". Is it possible to integrate say 1-form over the Peano curve?

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Ideas from your question are very close to the original de Rham approach. See the book "Differentiable Manifolds" by De Rham. –  Petya Mar 15 '10 at 3:30
@Petya: It is possible. By Stokes theorem integration does only depend on the (co)homology-classes of the singular chain and the differential form respectively. Every homology class has a smooth representative so integration over non-smooth singular chains is possible. The idea is well known: Arbitrary small changes to the curve do not change the integral so choose a smooth curve that is very close to the original and integrate over that curve instead. –  Johannes Hahn Mar 15 '10 at 13:25
@Johannes Hahn Why you are talking about homology classes? Chain is not a cycle! I do not know how to integrate a 1-form over Peano curve (which is not a cycle). So, it is impossible, there is no such a pairing between forms and singular chains. –  Petya Mar 15 '10 at 13:49
Of course you can integrate over chains to. Think at complex analysis: holomorphic functions can be integrated over arbitrary continouos paths. Every path from $x_0$ to $x_1$ defines a homology class in $H_\ast(X,A)$ with A:={$a,b$} and integration over paths does only depend on this homology class. –  Johannes Hahn Mar 16 '10 at 22:14
@Johannes Hahn Differential form is not supposed to be closed. Try to integrate $ydx$ over Peano curve and you will see very interesting phenomenon. –  Petya Mar 16 '10 at 22:39

I think that what you are getting at is called a "Current". See any book on geometric measure theory like the recent one by Krantz called "Geometric Integration Theory" or the older one by Federer called "Geometric Measure Theory"

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