The Tangent Bundle is Orientable

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 $(q, \phi)\in T^*M$ a point in the tangent bundle by specifying a point in the manifold q and a covector $\phi\in T^*_qM$. Thus we have the projection $\pi: T^*M\to M$ by $\pi(q, \phi)=q$.

Now the pullback at q is $d\pi^*_{(q,\phi)}:T_q^*\to T^*_{(q, \phi)}(T^*M)$. So we’ll put a 1-form $\tau\in \Omega^1(T^*M)$ on the cotangent bundle by $t_(q, \phi)=d\pi^*_{(q, \phi)}\phi$. Thus, given a tangent vector $X\in T_{(q, \phi)}(T^*M)$, we get $\displaystyle \tau_{(q, \phi)}(X)=\phi(d\pi_{(q, \phi)}(X))$.

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

Given the standard coordinates at a point $(q, \phi)\in T^*M$ say $(x^i, \zeta_i)$ where the coordinate representation is $\phi=\sum \zeta_idx^i$, then we get a coordinate representation for the projection $\pi(x, \zeta)=x$. Thus $d\pi^*(dx^i)=dx^i$ and we get a coordinate representation for our one form: $\tau_{(x, \zeta)}=\sum\zeta_idx^i$. Thus our 1-form is smooth.

Now $\omega$ is closed since it is exact. We also have a coordinate form for it $\omega=-d\tau=\sum dx^i\wedge d\zeta_i$. 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 $g$ 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.

Let $U$ be a coordinate chart with $x_i$ coordinate functions and $v_i = dx_i$ sections of the cotangent space over $U$. Now $x_i, v_i$ can be viewed as functions on the tangent bundle $TU$ by projection in the first case. Then I think the canonical orientation is given by $\{ x_1, \dots, x_n, v_1, \dots, v_n \}$, since the two change of variables matrices for the $x$ and the $v$‘s should be the same, leading to a postive determinant.