Let $M$ be a smooth manifold. How do I show that the tangent bundle $TM$ of $M$ is orientable?
Differential Geometry – Why is the Tangent Bundle Orientable?
differential-geometryfiber-bundlessmooth-manifoldstangent-bundlevector-bundles
Related Solutions
I think the simplest motivation is that of vector fields. We want to be able to assign a tangent vector to each point in the manifold, giving us a "field" of vectors on the manifold. That is, $F$ should be some map such that
$$ F(p)\in T_pM $$
for $p\in M$. So, what's wrong with just saying that? Well, nothing really, if you're only interested in the value of vector fields at a point. If you ever want to look beyond singular points, you need some structure connecting your different tangent spaces. For example, to look at continuity of $F$, we need the space of outputs of $F$ to have a topological structure; to look at differentiability, we need a differentiable structure.
So, we need some space which
- contains all the tangent vectors of $M$
- has the same "level" of structure that $M$ has
To satisfy (1), we simply glue together all the tangent spaces by taking a union. Since the tangent spaces are completely disconnected from one another, we can exemplify this fact by using a disjoint union
$$ TM = \coprod\limits_{p\in M} T_pM $$
The rest of the bundle structure is just there to "lift" the structure of $M$ onto $TM$. With that, we can now define vector fields as functions in the usual way
$$ F: M\to TM $$
and we are able to discuss continuity and differentiability to the extent that $M$ admits such properties. However, this definition isn't "complete" because it allows for e.g. attaching a tangent vector from $T_qM$ to $p$, which doesn't fit our idea of a vector field. This gives us another requirement,
- we need a way to determine which point a vector is tangent to
This is the bundle projection map, $\pi: TM\to M$, and so we can add the requirement on $F$ that $\pi(F(p)) = p$ everywhere.
Here, we created a bundle with a base manifold and a vector space at each point, but we could imagine a more general concept of bundle which just has some kind of space $B$ with some other kind of space $F_p B$ at each point $p\in B$. Even in this general setting, we can see the utility of attaching to each point of the base space some element of its attached space. We define a function
$$ \sigma: B\to FB = E $$
with $\pi(\sigma(p)) = p$ as a cross-section (or just section) of the total space $E$.
Applying this terminology to our original example, we can then reconstruct the more terse definition:
$$ \text{a vector field on } M \text{ is a section }\sigma\text{ of the tangent bundle } TM$$
The answer is given by Kajelad in the comment
You have proved that $M$ is orientable $\Rightarrow TM$ is orientable. Now we prove the opposite direction.
Assume that $TM$ is orientable. Then there is a family of open cover $\{U_i\}_{i\in \Lambda}$ of $M$ and for each $i\in \Lambda$, a local trivialization $$\varphi_i : TU_i \to U_i \times \mathbb R^n$$ so that for all $i, j$ with $U_i \cap U_j \neq \emptyset$, the transition function $$ g_{ij} : U_i \cap U_j \to \operatorname{GL}_n(\mathbb R)$$ has $\det g_{ij} >0$.
By shrinking to smaller open sets if necessary, we assume that each $U_i$ is a coordinates neighborhood. That is, there is $\psi_i : U_i \to \psi (U_i) \subset \mathbb R^n$ which is a local chart. By composing with a reflection of $\mathbb R^n$ if necessary, we assume that in $U_i$, both $$\left\{ \frac{\partial }{\partial x_1}, \cdots, \frac{\partial }{\partial x_n}\right\}, \{ \varphi^{-1}_i e_1, \cdots, \varphi^{-1}_i e_n\}$$ have the same orientation. Here $\{e_1, \cdots, e_n\}$ are the standard basis of $\mathbb R^n$. This is the same as saying that for all $x\in U$, the linear map $L_i (x)$ defined by the composition $$ \mathbb R^n \cong T_{\psi(x)} \psi_i (U_i))\overset{(\psi^{-1}_i)_*}{\to} T_xU \overset{\varphi_i|_{T_xU_i}}{\to} \mathbb R^n$$ has positive determinant.
Now we check that $M$ is orientable: whenever $U_i \cap U_j$ is non-empty, let $x\in U_i$. Then one need to check $J_{ij}:= J(\psi_i \circ \psi_j):\mathbb R^n \to \mathbb R^n$ has positive determinant, but this is true since
\begin{align} J(\psi_i \circ \psi_j^{-1}) &= (\psi_i)_* \circ (\psi_j^{-1})_* \\ &= (L_i^{-1} \circ \varphi_i) \circ (\varphi_j^{-1} \circ L_j)\\ &= L_i^{-1} \circ g_{ji} \circ L_j. \end{align}
Best Answer
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 $TM$.