Why do I need densities in order to integrate on a non-orientable manifold? - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-20T05:53:13Z http://mathoverflow.net/feeds/question/90455 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/90455/why-do-i-need-densities-in-order-to-integrate-on-a-non-orientable-manifold Why do I need densities in order to integrate on a non-orientable manifold? ISH 2012-03-07T13:52:32Z 2013-03-02T20:32:40Z <p>Integration on an orientable differentiable n-manifold is defined using a partition of unity and a global nowhere vanishing n-form called volume form. If the manifold is not orientable, no such form exists and the concept of a density is introduced, with which we can integrate both on orientable and non-orientable manifolds. My question is: On a non-orientable n-manifold, every n-form vanishes somewhere, but shouldn't I be able to chose an n-form with say a countable number of zeros, which would then constitute a set of measure zero and thus allow me to use n-forms (with zeros) for global integration also on non-orientable n-manifolds?</p> http://mathoverflow.net/questions/90455/why-do-i-need-densities-in-order-to-integrate-on-a-non-orientable-manifold/90457#90457 Answer by Liviu Nicolaescu for Why do I need densities in order to integrate on a non-orientable manifold? Liviu Nicolaescu 2012-03-07T14:19:19Z 2012-03-13T10:57:05Z <p>First of all, the things that you actually integrate are densities, which are the differential geometric counterparts of measures. No orientation is needed. </p> <p>A degree $n$ form on an $n$-dimensional manifold is almost a density, but not quite. We need an orientation to associate to the top degree form a density. This is what you ultimately integrate when you integrate a form. For more details see page 105 of <a href="http://www.nd.edu/~lnicolae/Lectures.pdf" rel="nofollow">these notes</a>.</p> http://mathoverflow.net/questions/90455/why-do-i-need-densities-in-order-to-integrate-on-a-non-orientable-manifold/90476#90476 Answer by Steven Gubkin for Why do I need densities in order to integrate on a non-orientable manifold? Steven Gubkin 2012-03-07T16:23:38Z 2012-03-12T06:13:16Z <p>The problem is that there is no way to figure out signs - It would be like trying to integrate a function from $\mathbb{R}$ to $\mathbb{R}$ without knowing whether you were moving forward or backward. </p> <p>What you CAN actually integrate are pseudo-differential forms. The whole point of choosing an orientation is to turn a differential form into a psuedo-differential form. For those, I recommend the wonderful short story by John Baez found here:</p> <p><a href="https://groups.google.com/group/sci.physics.research/msg/3c6a1a7237b66c8c?dmode=source&amp;pli=1" rel="nofollow">https://groups.google.com/group/sci.physics.research/msg/3c6a1a7237b66c8c?dmode=source&amp;pli=1</a></p> http://mathoverflow.net/questions/90455/why-do-i-need-densities-in-order-to-integrate-on-a-non-orientable-manifold/90478#90478 Answer by Johannes NordstrÃ¶m for Why do I need densities in order to integrate on a non-orientable manifold? Johannes NordstrÃ¶m 2012-03-07T16:38:44Z 2012-03-07T16:38:44Z <p>You would expect the zero set of an $n$-form to have codimension 1 rather than being countable. Your suggestion of choosing some $n$-form on a non-orientable manifold $M^n$ and defining integrals relative to that essentially amounts to cutting $M$ into two orientable pieces along a codimension 1 submanifold, choosing an orientation on each, and adding the integrals on the two pieces. You can certainly do that, but since the answer depends on the choice of $n$-form/cutting it is not very natural or interesting (whereas the integral on an oriented manifold only depends on the orientation and not on the choice of orientation form).</p> http://mathoverflow.net/questions/90455/why-do-i-need-densities-in-order-to-integrate-on-a-non-orientable-manifold/90714#90714 Answer by alvarezpaiva for Why do I need densities in order to integrate on a non-orientable manifold? alvarezpaiva 2012-03-09T15:46:26Z 2012-03-10T13:22:16Z <p>This is not an answer, but on reading the discussion I thought that it would be nice if someone gave the definition of a density so that no one would think that it is a complicated object. I learned the following (somewhat more general definition) from I.M. Gelfand:</p> <p><strong>Definition.</strong> A $k$-density on a manifold $M$ is a continuous real-valued function defined on the cone of simple (a.k.a. decomposable) tangent $k$-vectors on $M$ that is homogeneous of degree one. A $k$-density $\varphi$ is said to be smooth if for every $k$-tuple of smooth linearly independent vector fields $X_i$ $(1 \leq i \leq k)$ defined in some open set $U \subset M$, we have that the function $$y \mapsto \varphi(X_1(y)\wedge \cdots \wedge X_k(y))$$<br> is smooth in $U$.</p> <p>A densitiy is called <em>even</em> if $\varphi(-v) = \varphi(v)$ for every simple tangent $k$-vector $v$. Likewise, we have <em>odd $k$-densities</em> that generalize differential $k$-forms</p> <p><strong>Examples and context</strong></p> <p>If $\Omega$ is a volume form on a manifold of dimension $n$, then both $\Omega$ and $|\Omega|$ are $n$-densities. The arc-length element of a Riemannian or Finsler metric is a $1$-density. The $k$-area integrand of a Riemannian or Finsler manifold is a $k$-density.</p> <p>Parametric integrands (in the sense of Federer-Fleming) define $k$-densities when restricted to the cone of simple vectors, but densities are way more general. </p> <p>Varifolds of dimension $k$ are elements of the dual to the space of even $k$-densities. This is basically their definition: because of their homogeneity even $k$-densities can be seen as continuous functions on the bundle of tangent $k$-planes.</p> <p><strong>A message from our sponsor</strong></p> <p>For more examples and for neat applications to integral geometry (if I may say so myself ...), which becomes much easier if differential forms are replaced by densities, see the paper <a href="http://www.springerlink.com/content/p70613p157373r5v/" rel="nofollow">Gelfand transforms and Crofton formulas</a>.</p> http://mathoverflow.net/questions/90455/why-do-i-need-densities-in-order-to-integrate-on-a-non-orientable-manifold/90867#90867 Answer by Wei Luo for Why do I need densities in order to integrate on a non-orientable manifold? Wei Luo 2012-03-11T01:52:33Z 2012-03-11T01:52:33Z <p>I think one reason that integration of forms instead of densities is prefered is that one can use Stokes theorems.</p>