# orientations for zero-dimensional manifolds

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.

• One natural way to go about it is to define orientation for vector spaces that have non-empty basis first, then to give the subspace orientation convention: that if $V$ is a finite-dimensional vector space, and $W$ a subspace then an orientation of any two of the three $\{V, W, V/W\}$ induces an orientation of the other. So when $W=\{0\}$ this convention extends to give a notion of orientation of points. Nov 28 '10 at 17:17
• Regardless of whether it's easy to get a definition of orientation that lets the point have two orientations, we want the point to have two, to allow e.g. the statement "If $A = B\oplus C$, orienting two uniquely orients the third." Nov 29 '10 at 11:48

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

• Thank you for the answer. It's still a bit strange to me that the reduction of the structure group of the tangent bundle gives the "wrong" answer. I will try to better understand the connection between the two definitions. Nov 28 '10 at 19:54

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.

• Thanks you. I have accepted the a variation of your answer by Scott Carnahan. Nov 28 '10 at 19:57

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.

• (I'm not bold enough to claim that the orientation of a point seems unnatural because it is natural). Nov 28 '10 at 18:18
• Informally, an orientation on a connected 1-manifold is the choice of a direction, and an orientation on a point is the choice of a sign $+$ or $-$. As Pietro said, there is a canonical way to orient all points: just choose the sign $+$. Nov 28 '10 at 22:37
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