Equivalent singular chains and differential forms, as functionals on forms, on compact Riemannian manifolds - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-26T04:41:28Z http://mathoverflow.net/feeds/question/18238 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/18238/equivalent-singular-chains-and-differential-forms-as-functionals-on-forms-on-co Equivalent singular chains and differential forms, as functionals on forms, on compact Riemannian manifolds Asaf Reich 2010-03-15T02:55:03Z 2010-03-21T02:10:57Z <p>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 $&lt;\alpha,\beta>=\displaystyle\int_{M}{\alpha\wedge *\beta}$. This product also gives, for each given form, a functional on the space of forms.</p> <p>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?</p> <p>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 $&lt; D(\sigma),e_j>$ 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).</p> <p>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.)</p> <p><b>EDIT</b>: 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.</p> http://mathoverflow.net/questions/18238/equivalent-singular-chains-and-differential-forms-as-functionals-on-forms-on-co/18240#18240 Answer by Petya for Equivalent singular chains and differential forms, as functionals on forms, on compact Riemannian manifolds Petya 2010-03-15T03:19:50Z 2010-03-15T03:19:50Z <p>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?</p> http://mathoverflow.net/questions/18238/equivalent-singular-chains-and-differential-forms-as-functionals-on-forms-on-co/18247#18247 Answer by Gonçalo Marques for Equivalent singular chains and differential forms, as functionals on forms, on compact Riemannian manifolds Gonçalo Marques 2010-03-15T06:20:59Z 2010-03-16T18:19:04Z <p>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"</p> http://mathoverflow.net/questions/18238/equivalent-singular-chains-and-differential-forms-as-functionals-on-forms-on-co/18896#18896 Answer by Dan Lee for Equivalent singular chains and differential forms, as functionals on forms, on compact Riemannian manifolds Dan Lee 2010-03-21T02:10:57Z 2010-03-21T02:10:57Z <p>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.</p> <p>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.</p> <p>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.</p> <p>Here's what currents have to do with the story. $k$-currents are essentially <em>defined</em> 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.</p> <p>I don't know much history, but I think that your idea was what lead de Rham to invent currents.</p>