26
$\begingroup$

Let $M$ be a smooth manifold. How do I show that the tangent bundle $TM$ of $M$ is orientable?

$\endgroup$
3
  • $\begingroup$ so sorry for my writing a question here, @ you for 2) in your answer, how do we know that "tau_M" and "xi" are orthogonal to each other, i.e. why each of the fibers of tau are orthogonal to those of xi? $\endgroup$
    – wqr
    May 12 '12 at 8:43
  • $\begingroup$ actually I read Characteristic Classes but I could not get a clue about the vectors in "fiber direction" and "manifold direction" $\endgroup$
    – wqr
    May 12 '12 at 8:51
  • 1
    $\begingroup$ @gIS You're seriously editing a nine-year old post to change one letter? Do we need to bring this post to the forefront now? $\endgroup$ Jul 10 '21 at 23:24
25
$\begingroup$

Sorry to resurrect, but we leave a $($detailed$)$ proof here that $TM$ has the structure of an oriented $2n$-manifold, even if the $n$-manifold $M$ is non-orientable.

Let $\{(U_\alpha, \phi_\alpha)\}_{\alpha \in A}$ be a smooth atlas of $M$, and let $V_\alpha = \phi_\alpha(U_\alpha) \subset \mathbb{R}^n$. Then $(\phi_\alpha)_*: TU_\alpha \to TV_\alpha = V_\alpha \times \mathbb{R}^n$ is a homomorphism $($having inverse $(\phi_\alpha^{-1})_*$$)$. Moreover, the sets $TU_\alpha$ cover $TM$ and the transition maps$$t_{\alpha\beta} = (\phi_\alpha)_* \circ \left(\phi_\beta^{-1}\right)_* = \left(\phi_\alpha \circ \phi_\beta^{-1}\right)_*: V_\beta \times \mathbb{R}^n \to V_\alpha \times \mathbb{R}^n$$are orientation preserving. Let $x_1, \dots, x_n$ be coordinates on $V_\beta$, $x_{n+1}, \dots, x_{2n}$ be coordinates on the left copy of $\mathbb{R}^n$, $y_1, \dots, y_n$ be coordinates on $V_\alpha$, and $y_{n+1}, \dots, y_{2n}$ be coordinates on the right copy of $\mathbb{R}^n$. Note that $(y_1, \dots, y_n) = \phi_\alpha\left(\phi_\beta^{-1}(x_1, \dots, x_n)\right)$ does not depend on $x_{n+1}, \dots, x_{2n}$, so the Jacobian matrix $\left({{\partial y_i}\over{\partial x_j}}\right)$ of $t_{\alpha\beta}$ has all zeros in the upper right quadrant. It follows that$$\det\left({{\partial y_i}\over{\partial x_j}}\right)_{i,j = 1}^{2n} = \det\left({{\partial y_i}\over{\partial x_j}}\right)_{i, j = 1}^n \det\left({{\partial y_i}\over{\partial x_j}}\right)_{i, j = n +1}^{2n}.$$The first of these submatrices is the usual Jacobian of $\phi_\alpha \circ \phi_\beta^{-1}$. The second is the Jacobian of the linear transformation $\left(\phi_\alpha \circ \phi_\beta^{-1}\right)_{*,\, (x_1, \dots, x_n)}$. Therefore$$\det\left({{\partial y_i}\over{\partial x_j}}\right)_{i, j = 1}^{2n} = \det\left(\left(\phi_\alpha \circ \phi_\beta^{-1}\right)_{*,\, (x_1, \dots, x_n)}\right)^2 > 0.$$This proves $\{(TU_\alpha, (\phi_\alpha)_*)\}_{\alpha \in A}$ is an oriented atlas for $TM$.

$\endgroup$
2
  • 2
    $\begingroup$ Why do you say in the first equation (about the transition maps) that they are orientation preserving? How can I see this triviality (since you don't prove it I assume it is trivial)? Also, why in the very last equation in the RHS inside the determinant you get square of the transition maps? You showed on Jacobian is of $\phi_a \circ \phi_b^{-1}$ and the other for its push forward. $\endgroup$
    – Marion
    Jun 14 '15 at 14:13
  • $\begingroup$ @Marion, That's because the coefficient matrix of the linear transformation of the last $n$ slots is exactly the Jacobian of the transition map of the atlases of $M$, c.f. Prop. 3.18 of [Introduction to Smooth Manifolds] by John M. Lee, ed2. $\endgroup$ Sep 15 '20 at 22:34
23
$\begingroup$

Start with an atlas for $M$ and construct the corresponding atlas on $TM$, each of whose charts is constructed from one on $M$. Check that the transition functions in the latter have Jacobian with positive determinant.

$\endgroup$
4
  • 1
    $\begingroup$ This is actually a nice computation to make! $\endgroup$ Apr 9 '12 at 3:40
  • 1
    $\begingroup$ that was my first approach to the question, I thought that there was a easier method... $\endgroup$
    – Jr.
    Apr 9 '12 at 3:52
  • 5
    $\begingroup$ You should have told us that... $\endgroup$ Apr 9 '12 at 4:11
  • 9
    $\begingroup$ (You can decompose $TTM$ as a vector bundle over $TM$ as the direct sum of $\pi^*(TM)$ (the pullback of $TM\to M$ along $\pi:TM\to M$) and the subbundle of $TTM\to TM$ of vertical vectors (those which are tangent to the fibers of $TM\to M$), and show that the determinant bundles of these two subbundles are isomorphic, so that the determinant bundle of $TTM$ is the tensor square of a line bundle. This implies orientability... This is just the same thing done complicatedly.) $\endgroup$ Apr 9 '12 at 4:13
22
$\begingroup$

There is also a way to see it with fibre bundles and characteristic classes. It's possible the original poster is not familar, but other people might be and more importantly I need practice. It is based on two relatively basic facts (at least I'm mostly sure I've seen them before):

1) A smooth $n$-manifold $M$ is orientable iff the first Stiefel-Whitney class of its tangent bundle $\tau_M$ vanishes,

and

2) If $\xi$ is a smooth $k$-plane bundle with base space $M^n$, total space $E^{n+k}$ (both smooth manifolds) and projection $\pi:E\rightarrow M$, then $$\tau_E=\pi^*(\tau_M)\oplus\pi^*(\xi)$$

Then, if $TM$ is the total space of the $n$-plane bundle $\tau_M$ with projection map $\pi\colon TM\rightarrow M$, it is a smooth manifold with its own tangent bundle $\tau_{TM}$. Since $\pi$ is the projection map of $\tau_M$ we have $$\tau_{TM}=\pi^*(\tau_M)\oplus\pi^*(\tau_M)$$ so by the Whitney product formula $$\omega_1(\tau_{TM})=(\pi^*\omega_0)(\tau_M)\cup(\pi^*\omega_1)(\tau_M)+(\pi^*\omega_1)(\tau_M)\cup(\pi^*\omega_0)(\tau_M)=2\pi^*\omega_1(\tau_{M})=0\in H^1(TM;\mathbb{Z/2})$$

Hence the manifold $TM$ is orientable.

(But for all intents and purposes, writing down charts is the easiest way to go)

$\endgroup$
3
  • $\begingroup$ This is the characteristic-class version of the comment to my answer :) Much preferable! $\endgroup$ Apr 9 '12 at 4:26
  • $\begingroup$ @you In which book should I find this content? thanks. $\endgroup$
    – Jr.
    Apr 9 '12 at 4:33
  • $\begingroup$ "Characteristic Classes" by John Milnor and James Stasheff. They treat characteristic classes axiomatically and develop some basic theory, and then later show that they exist. There's a lot of good content in there about characteristic classes and fibre bundles. If you want to learn more about the general theory of fibre bundles, check out Norman Steenrod's "Topology of Fibre Bundles." And it's probably also in Hatcher somewhere. $\endgroup$
    – William
    Apr 9 '12 at 4:35
5
$\begingroup$

Since the cotangent bundle $T^*M$ has a natural symplectic form $\omega,$ and, as for any symplectic form, $\omega^n$, for $2n=dimT^*M$ is a volume form, the cotangent bundle is orientable. Since the $TM$ is diffeomorphic to $T^*M$ via, say, any choice of a metric on $M$, the tangent bundle is orientable.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.