Today we’ll do a nice standard result. The tangent bundle of a smooth manifold is orientable as a manifold (regardless of whether or not the manifold itself is).

This could be done rather easily if I had built some theory first, but I’ll build the structures I need in this post. The first thing I’ll build is the tautological symplectic form on the cotangent bundle. Let a point in the tangent bundle by specifying a point in the manifold q and a covector . Thus we have the projection by .

Now the pullback at q is . So we’ll put a 1-form on the cotangent bundle by . Thus, given a tangent vector , we get .

That was a mouthful, and we only got a 1-form out of it not a symplectic form. The claim now is that is a symplectic form on the cotangent bundle.

Given the standard coordinates at a point say where the coordinate representation is , then we get a coordinate representation for the projection . Thus and we get a coordinate representation for our one form: . Thus our 1-form is smooth.

Now is closed since it is exact. We also have a coordinate form for it . Aha, so it is symplectic.

So if you haven’t seen this proof done this way, then you are probably in mass confusion why I just put a symplectic form on the cotangent bundle when what I really want is a nowhere vanishing 2n-form on the tangent bundle.

You can now simply check what happens when you wedge this form with itself n-times. You’ll get a nowhere vanishing 2n-form on the cotangent bundle. Thus the cotangent bundle is orientable. Now let be a Riemannian metric on the manifold. This gives us a nice isomorphism between the tangent and cotangent bundles by way of the raising and lowering of indices. This part of the proof sort of scares me, so this isomorphism is actually a bundle isomorphism. Does this imply diffeomorphic as manifolds? I think so, since it’s smooth, but if any reader can confirm, that would be great!

But since the tangent bundle is diffeomorphic to an orientable manifold, it is itself orientable.

As usual, let’s extrapolate a little now that the specific standard problem is done. We showed that the cotangent bundle was orientable, but the only fact we used was that it was symplectic. So this same proof (wedging the symplectic form with itself n times) will work, and all sympletic manifolds are orientable.

Another thing to note is that we need to be careful to continually specify “as smooth manifolds” when talking about orientability in this context. Another theorem says that the tangent bundle is orientable, as a vector bundle, if and only if the manifold itself is orientable.

Another quick question, is there a cleaner way to do this? Of course, this proof is a couple sentences with knowledge that the cotangent bundle has a tautological symplectic structure, and that all symplectic manifolds are orientable, and then just hit it with the tangent-cotangent iso. But I feel like there must be a proof working directly with the tangent bundle and some more elementary facts about orientation.

September 12, 2009 at 6:13 pm

I didn’t know what a symplectic manifold was until yesterday, but I think I’ve seen this result proved by more elementary means, and I’ll try to reconstruct the proof.

Let be a coordinate chart with coordinate functions and sections of the cotangent space over . Now can be viewed as functions on the tangent bundle by projection in the first case. Then I think the canonical orientation is given by , since the two change of variables matrices for the and the ‘s should be the same, leading to a postive determinant.

September 12, 2009 at 6:41 pm

Before someone else points this out, being isomorphic as total spaces of

smooth vector bundlesmeans that they are diffeomorphic as smooth manifolds, and better yet, that diffeomorphism preserves extra structure.The term “vector bundle” always throws me, since I think of being isomorphic in some sort of linear sense is weaker than preserving all the smooth structure. But I should think through the definition before jumping to that conclusion.

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