orientations for zero-dimensional manifolds - MathOverflow

orientations for zero-dimensional manifolds

2010-11-28T17:06:30Z

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.

Answer by S. Carnahan for orientations for zero-dimensional manifolds

2010-11-28T17:25:50Z

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 $GL_1(\mathbb{R})/GL_1^+(\mathbb{R}) \cong \{ \pm 1 \}$ .

Answer by Pietro Majer for orientations for zero-dimensional manifolds

2010-11-28T18:17:26Z

I think the source of small oddity around the idea that a point has two orientations, is that a point is indeed canonically oriented, which is not the case for manifolds of positive dimension. There's a short discussion on this point in the wiki article about orientation.

Answer by André Henriques for orientations for zero-dimensional manifolds

2010-11-28T19:12:28Z

Here is a definition that works for all dimensions, including zero.
Let $V$ be a real vector space equipped with a Riemannian metric.

An orientation on $V$ is an isometric isomorphism between $\mathbb R$ and the top exterior power of $V$.

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 $V$.

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.

A spin structure on $V$ is a Morita equivalence between Cliff($\mathbb R^n$) and Cliff($V$). In other words, it is a Cliff($\mathbb R^n$)-Cliff($V$)-bimodule that has the property that it induces an equivalence between the category of Cliff($\mathbb R^n$)-modules and that of Cliff($V$)-modules

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 Cliff($\mathbb R^n$) and Cliff($V$). You just replace the functor $\Lambda^{top}$ by the functor Cliff, and replace the 1-category of vector spaces with the 2-category of algerbas and bimodules.