Tangent bundle is an oriented manifold

Question

Recall that the tangent bundle $TM$ of a manifold $M$ consists of all pairs $(x, \overrightarrow{v})$ where $x \in M$ and $\overrightarrow{v}$ is the tangent space $T_xM$ of $M$ at $x$. Show that $TM$ is an oriented manifold.

An orientable manifold is one such that there exists a continuous vector field. If $(\phi, U)$ is a chart for $M$ we can obtain a natural chart $(\Phi, TU)$ for the tangent bundle by defining $\Phi: TU\rightarrow \mathbb{R}^n \times \mathbb{R}^n$ by $\Phi(x, y) = \phi(x), (D\phi_a^{-1})^{-1}y)$ where $D$ is the derivative map. We just need to check the jacobian of the transition maps have positive determinant. If $(U, \Phi)$ and $(V, \Psi)$ where $\Psi(x, y) = (\psi(x), (D\psi_a^{-1})^{-1}(y))$ are two overlapping charts of $TM$ then the transition maps is for $(x, y) \in \mathbb{R}^{2n}$. \begin{equation} \Phi \circ \Psi^{-1}(x, y) = (\phi\circ\psi^{-1}(x), (D\phi_a^{-1})^{-1} \circ D\psi_a^{-1}(y)) \end{equation}

Thus, the Jacobian of the transition map $\mathcal{J}(\Phi \circ \Psi^{-1}(x, y))$ \begin{equation} \left\lvert\begin{matrix} \mathcal{J}(\phi\circ\psi^{-1}(x)) & 0\\ 0 & \mathcal{J}((D\phi_a^{-1})^{-1} \circ D\psi_a^{-1})(y)) \end{matrix}\right\rvert \end{equation}

From here I'm not sure how to prove that this determinant is positive. Some hints would be greatly appreciated.

Answer 1

It seems to me that you got the computation of the derivative wrong. As you have written (with a small simplification as observed in the comment by @guidoar and using the chain rule), you get $$ \Phi\circ\Psi^{-1}(x,v)=((\phi\circ\psi^{-1})(x), D(\phi\circ\psi^{-1})(x)(v)). $$ (I have changed the notion to emphasize that $v$ is a tangent vector, while $x$ is a point in $M$.) But when differentiating this, you have to distinguish wheter you differentiate in $x$-directions or in $v$-directions: Differentiating the second component in $x$ directions, you get a second derivative, whereas the fact that the derivative is linear in $v$ implies that you just get back the first derivative when differentiating in $v$-directions. More formally this means that $$ D(\Phi\circ\Psi^{-1})(x,v)(w_1,w_2)=(D(\phi\circ\psi^{-1})(x)(w_1),D^2(\phi\circ\psi^{-1})(x)(w_1,v)+D (\phi\circ\psi^{-1})(x)(w_2)). $$ In matrix notation, this leads to a block form as $$ \begin{pmatrix} D(\phi\circ\psi^{-1})(x) & 0\\ D^2(\phi\circ\psi^{-1})(x)(\_,v) & D(\phi\circ\psi^{-1})(x) \end{pmatrix} $$ Taking the determinant, the off-diagonal block does not play a role, so you conclude that $\mathcal J(\Phi\circ\Psi^{-1})=\mathcal J(\phi\circ\psi^{-1})^2>0$, which implies the claim.

Answer 2

Well, when you defined the map $\overline{\phi}: TU \to \phi(U) \times \mathbb{R}^{n}$, you didn't write it the right way, because it is given by $\overline{\phi}(p,v) = (\phi(p), D_p\phi(v))$ (you need to take into account the derivative of the second component with respect to the point on the manifold, and not just the vector in the tangent space), so when you take the jacobian of a change of coordinates $\overline{\phi}_{\alpha} \circ \overline{\phi}_{\beta} ^{-1} : V_{\beta} \times \mathbb{R}^n \to V_{\alpha}\times \mathbb{R}^n$, if you call $(x_1,...,x_n)$ the coordinates in $V_{\beta}$, and $(x_{n+1},...,x_{2n})$ the coordinates in the $\mathbb{R}^n$ of the domain and do the same with $y$'s for the co domain, to calculate the jacobian, we need to write $\frac{\partial y_j}{\partial x_i}$ for any $i,j = 1,...,2n$. Now, obviously, it is zero when $1\leq j \leq n$ and $n+1 \leq i \leq 2n$ so the upper right quadrant in the matrix representation is zero, but the rest of them won't in general, so it will actually be a lower triangular block matrix which will have determinant equal to the product of the determinants of $[\frac{\partial y_j}{\partial x_i}]_{i,j=1,...,n}$ and $[\frac{\partial y_j}{\partial x_i}]_{i,j=n+1,...,2n}$ as you can check here Determinant of a block lower triangular matrix.

Now, if you notice, the second matrix I pointed out in the product of the determinants, will be exactly $det (J(D_p \phi_{\beta}^{-1} \circ D_p \phi_{\alpha}))$, where $p \in U$ is now fixed because we're calculating partial derivatives with respect to the variables indexed by $n+1,...,2n$. Now, $D_p \phi_{\beta}^{-1} \circ D_p \phi_{\alpha}$ is a linear map from $\mathbb{R}^n$ to itself, and as we all know, in that case, the derivative of a linear map is itself, so the jacobian is the matrix representation of $D_p \phi_{\beta}^{-1} \circ D_p \phi_{\alpha}$ which in turn is exactly $J(\phi_{\beta}^{-1}\circ \phi_{\alpha})$. We have now concluded that the determinant of the jacobian of $\overline{\phi}_{\alpha} \circ \overline{\phi}_{\beta} ^{-1}$ is exactly $det(J(\phi_{\beta}^{-1}\circ \phi_{\alpha}))^2 >0$.