orientations for zero-dimensional manifolds - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T18:32:18Z http://mathoverflow.net/feeds/question/47596 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/47596/orientations-for-zero-dimensional-manifolds orientations for zero-dimensional manifolds Keivan Karai 2010-11-28T17:06:30Z 2010-11-29T07:33:48Z <p>I am teaching a course on manifolds, and soon I will have to prove the Stokes' theorem which, of course, involves defining oriented manifolds. There are many ways to define an oriented manifold. My favorite way is by the reduction of the structure group of the tangent bundle. But this definition and a couple of other that I know give just one orientation for the point: $GL(V)/GL^{+}(V)$ is ${\mathbb Z}/2{\mathbb Z}$ when $\dim V \ge 1$, but when $\dim V=0$ then $GL(V)$ has only one element. Of course, it is possible to define two orientations of a point by convention. I would like to know if is there any way to define the orientation for smooth manifolds in a uniform way that would also yield two orientations of a point.</p> http://mathoverflow.net/questions/47596/orientations-for-zero-dimensional-manifolds/47597#47597 Answer by S. Carnahan for orientations for zero-dimensional manifolds S. Carnahan 2010-11-28T17:25:50Z 2010-11-28T17:25:50Z <p>I think instead of reducing the structure group of the tangent bundle, you could consider reducing the structure group of its top exterior power, which is always a one-dimensional vector bundle, even in dimension zero. Then the set of orientations of each connected component is a torsor under <code>$GL_1(\mathbb{R})/GL_1^+(\mathbb{R}) \cong \{ \pm 1 \}$</code>.</p> http://mathoverflow.net/questions/47596/orientations-for-zero-dimensional-manifolds/47600#47600 Answer by Pietro Majer for orientations for zero-dimensional manifolds Pietro Majer 2010-11-28T18:17:26Z 2010-11-28T18:17:26Z <p>I think the source of small oddity around the idea that a point has two orientations, is that a point is indeed <em>canonically</em> oriented, which is not the case for manifolds of positive dimension. There's a short discussion on this point in the <a href="http://en.wikipedia.org/wiki/Orientation_%28mathematics%29#Zero-dimensional_case" rel="nofollow">wiki article</a> about orientation.</p> http://mathoverflow.net/questions/47596/orientations-for-zero-dimensional-manifolds/47606#47606 Answer by AndrĂ© Henriques for orientations for zero-dimensional manifolds AndrĂ© Henriques 2010-11-28T19:12:28Z 2010-11-29T07:33:48Z <p>Here is a definition that works for all dimensions, including zero.<br> Let $V$ be a real vector space equipped with a Riemannian metric.<br></p> <blockquote> <p>An orientation on $V$ is an isometric isomorphism between $\mathbb R$ and the top exterior power of $V$.</p> </blockquote> <p>If you want to get rid of the Riemannian metric, you can also say that an orientation is a ray inside the top exterior power of $V$.</p> <hr> <p>Here is a similarly well-behaved definition of spin structure. It also works all the way down to dimension zero. As above, let $V$ be a real vector space equipped with a Riemannian metric.<br></p> <blockquote> <p>A spin structure on $V$ is a Morita equivalence between <em>Cliff</em>($\mathbb R^n$) and <em>Cliff</em>($V$). In other words, it is a <em>Cliff</em>($\mathbb R^n$)-<em>Cliff</em>($V$)-bimodule that has the property that it induces an equivalence between the category of <em>Cliff</em>($\mathbb R^n$)-modules and that of <em>Cliff</em>($V$)-modules</p> </blockquote> <p>To see the analogy with the above definition of orientation, note that $\Lambda^{top}\mathbb R^n=\mathbb R$. And so requiring an isomorphism between $\mathbb R=\Lambda^{top}\mathbb R^n$ and $\Lambda^{top}V$ is formally similar to requiring a Morita equivalence between <em>Cliff</em>($\mathbb R^n$) and <em>Cliff</em>($V$). You just replace the functor $\Lambda^{top}$ by the functor <em>Cliff</em>, and replace the 1-category of vector spaces with the 2-category of algerbas and bimodules.</p>