Why is the tangent bundle orientable?

Question

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

Answer 1

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 $TM$.

Answer 2

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.

Answer 3

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)

Answer 4

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.