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.

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August 25, 2011 - Posted by | Differential Topology, Topology

14 Comments »

  1. Do you mean, in your opening paragraph, that V is an n-dimensional vector space, rather than a manifold?

    Comment by David Roberts | August 26, 2011 | Reply

  2. Yeah, sorry. Fixing…

    Comment by John Armstrong | August 26, 2011 | Reply

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  11. Hope this is not too dumb: must the form be nowhere-zero on the tangent space basis at each point, or nowhere-zero on any tangent vector field? I’m pretty sure the first does not imply the latter. Thanks.

    Comment by Jerry P. | January 18, 2014 | Reply

    • Well, an orientation form doesn’t just take a single tangent vector field, it takes n of them. The requirement is that at each point p, there must be some n-tuple of vectors in \mathcal{T}_pM such that when you put them into the n-form at p you get a nonzero value out.

      Comment by John Armstrong | January 19, 2014 | Reply

  12. Yes, sorry, I meant a basis (X_1,..,X_n) , or may the more standard (Del/Delx_1, Del/Delx_2,…., Del/Delx_n), but you’re right, we just need _a_ basis (meaning an assignment of an n-ple of V.Fields at each point) where the form is nowhere-zero. Then we need to have a change- -of-basis matrix M between any two points , so that DetM =/ 0.

    Comment by Jerry P. | January 20, 2014 | Reply

  13. Sorry, that is the change-of-basis M between any two (X_p1, X_p2,..,X_pn); (X_q1,…,X_qn) with DetM non-zero, for a “coherent orientation”. AFAIK, if w(X_p1,…,X_pn) =/0 and DetM>0 , then w(M(x_p1,…,X_pn))=/0 .

    Comment by Jerry P. | January 20, 2014 | Reply


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