The Unapologetic Mathematician

Mathematics for the interested outsider

Oriented Manifolds

We recall that if V is an n-dimensional vector space than the space \Lambda^n(V) of “top forms” on V is one-dimensional. And since we’re working over \mathbb{R} this means that the zero form divides this space into two halves. Choosing one — by choosing a nonzero n-form on V or, equivalently, by choosing a basis for V — makes V into an “oriented” vector space.

We can do the same thing, of course, for the tangent space \mathcal{T}_pM of an n-dimensional manifold; For each point p\in M the stalk \Lambda^*_k(M)_p is isomorphic to \mathbb{R}, and so we can pick a nonzero value at each point. But of course this isn’t really what we want; we want to be able to choose this orientation “smoothly”. That is, we want a top form \omega\in\Omega^n(M) such that \omega(p)\neq0 for all p\in M.

If we can find such a form, we say that it “orients” M, and — along with the choice of orientation — M is “orientable”. This is not always possible; there are “non-orientable” manifolds for which there are no non-vanishing top forms.

It turns out that M is orientable if and only if the bundle \Lambda^*_k(M) is isomorphic to the product M\times\mathbb{R}. That is, if we can find a map f:\Lambda^*_k(M)\to M\times\mathbb{R} that plays nicely with the projections down to M, and so that the restriction to the stalk f_p:\Lambda^*_k(M)_p\to\{p\}\times\mathbb{R} is a linear isomorphism of one-dimensional real vector spaces.

In the one direction, if we have an orientation given by a top form \omega, then at each point we have a nonzero \omega(p)\in\Lambda^*_k(M)_p. Any other point in \Lambda^*_k(M)_p is some multiple of \omega(p), so we just define the real component of our transformation f to be this multiple, while the M component is the projection of the point from \Lambda^*_k(M)_p to M. The smoothness of \omega guarantees that this map will be smooth.

On the flip side, if we have such a map, then it’s invertible, giving a bundle map f^{-1}:M\times\mathbb{R}\to\Lambda^*_k(M). We can take the section of the product bundle sending each p\in M to 1\in\mathbb{R} and feed it through this inverse map: \omega(p)=f^{-1}_p(1)\in\Lambda^*_k(M)_p.

August 25, 2011 Posted by | Differential Topology, Topology | 14 Comments