# Why are the transition functions of a tensor bundle invertible?

I am doing the following exercise in Lee's "Introduction to Smooth Manifolds."

Exercise 12.18. Show that $$T^kT^*M$$, $$T^kTM$$, and $$T^{(k,l)}TM$$ have natural structures as smooth vector bundles over $$M$$, and determine their ranks.

I am trying to use the following two theorems.

Lemma 10.6 (Vector Bundle Chart Lemma). Let $$M$$ be a smooth manifold with or without boundary, and suppose that for each $$p\in M$$ we are given a real vector space $$E_p$$ of some fixed dimension $$k$$. Let $$E=\bigsqcup_{p\in M}E_p$$, and let $$\pi:E\to M$$ be the map that takes each element of $$E_p$$ to the point $$p$$. Suppose furthermore that we are given the following data:
(i) an open cover $$\{U_\alpha\}_{\alpha\in A}$$ of $$M$$
(ii) for each $$\alpha\in A$$, a bijective map $$\phi_\alpha:\pi^{-1}(U_\alpha)\to U_\alpha\times\mathbb R^k$$ whose restriction to each $$E_p$$ is a vector space isomorphism from $$E_p$$ to $$\{p\}\times\mathbb R^k\cong\mathbb R^k$$
(iii) for each $$\alpha,\beta\in A$$ with $$U_\alpha\cap U_\beta\ne\emptyset$$, a smooth map $$\tau_{\alpha\beta}:U_\alpha\cap U_\beta\to\operatorname{GL}(k,\mathbb R)$$ (which will be called a transition function) such that the map $$\phi_\alpha\circ\phi_\beta^{-1}$$ from $$(U_\alpha\cap U_\beta)\times\mathbb R^k$$ to itself has the form $$\begin{equation} \phi_\alpha\circ\phi_\beta^{-1}(p,v)=(p,\tau_{\alpha\beta}(p)v) \end{equation}$$ Then $$E$$ has a unique topology and smooth structure making it into a smooth manifold with or without boundary and a smooth rank-$$k$$ vector bundle over $$M$$, with $$\pi$$ as projection and $$\{(U_\alpha,\phi_\alpha)\}$$ as smooth local trivializations.

A Generalization of Proposition 12.10 (Abstract vs. Concrete Tensor Products). Let $$F$$ be a field. Let $$k,l\in\mathbb N=\{0,1,2,\ldots\}$$. If $$V_1,\ldots,V_k,V_{k+1},\ldots,V_{k+l}$$ are finite-dimensional vector spaces over $$F$$, then there is a unique linear map $$\begin{equation} f:V_1\otimes\cdots\otimes V_k\otimes V_{k+1}^*\otimes\cdots\otimes V_{k+l}^* \to L(V_1^*,\ldots,V_k^*,V_{k+1},\ldots,V_{k+l};F) \end{equation}$$ such that $$\begin{equation} f(v_1\otimes\cdots\otimes v_k\otimes\omega^{k+1}\otimes\cdots\otimes\omega^{k+l}) = v_1^{**}\otimes\cdots\otimes v_k^{**}\otimes\omega^{k+1}\otimes\cdots\otimes\omega^{k+l}\text{.} \end{equation}$$ Furthermore, $$f$$ is a vector space isomorphism.

Here, $$v^{**}$$ is the image of $$v\in V$$ with respect to the natural map $$V\to V^{**}$$.

I want to use the vector bundle chart lemma to give $$T^{(k,l)}TM$$ a vector bundle structure over $$M$$. For each smooth chart $$(U_\alpha,\phi_\alpha)$$ for $$M$$, define $$\phi_\alpha:\pi^{-1}(U_\alpha)\to U_\alpha\times\mathbb R^{n^{k+l}}$$ by $$\begin{equation} \phi_\alpha(\alpha^{i_1\cdots i_k}_{i_{k+1}\cdots i_{k+l}} \frac{\partial}{\partial x^{i_1}}\Bigr|_p\otimes\cdots\otimes\frac{\partial}{\partial x^{i_k}}\Bigr|_p \otimes dx^{i_{k+1}}|_p\otimes\cdots\otimes dx^{i_{k+l}}|_p) = (p,(\alpha^{i_1\cdots i_k}_{i_{k+1}\cdots i_{k+l}}))\text{.} \end{equation}$$ Let $$(U_\alpha,x^1,\ldots,x^n)$$ and $$(U_\beta,y^1,\ldots,y^n)$$ be two smooth charts for $$M$$. Now I have to show that there is a transition function $$\tau$$ for $$\phi_\alpha\circ\phi_\beta^{-1}$$. Let $$p\in U_\alpha\cap U_\beta$$. Then with some calculations using Proposition 12.10, I found that $$\tau(p)$$ should be the $$n^{k+l}\times n^{k+l}$$ matrix $$\begin{equation} \left(\frac{\partial y^{j_1}}{\partial x^{i_1}}(p)\cdots\frac{\partial y^{j_k}}{\partial x^{i_k}}(p) \frac{\partial x^{i_{k+1}}}{\partial y^{j_{k+1}}}(p)\cdots\frac{\partial x^{i_{k+l}}}{\partial y^{j_{k+l}}}(p)\right) \end{equation}$$ (The above entry runs over all indices $$i_1,\ldots,i_{k+l}$$ as it goes down from top to bottom row, and $$j_1,\ldots,j_{k+l}$$ as it goes from left to right column). But the hypothesis of the vector bundle chart lemma requires that $$\tau(p)$$ should be invertible, but I do not know why this is invertible.

• Isn't it a requirement a priori that $\tau_{\alpha \beta}^{-1}\tau_{\alpha \beta}$ is an isomorphism?$Mar 25 at 8:37 • @Kevin If the vector bundle is already constructed, then$\tau(p)$will be invertible for all transition function$\tau$. But I am trying to construct a vector bundle, so I don't think it's guaranteed that$\tau(p)\$ is invertible.
– zxcv
Mar 25 at 8:57

I think you are missing the point of Lemma 10.6. Let $$p \in U_\alpha \cap U_\beta$$. from (ii) it is given that $$\phi_\alpha\circ \phi_\beta ^{-1}|_{\{p\} \times \mathbb R^k}: \{p\} \times \mathbb R^k \to \{p\} \times \mathbb R^k$$ is a vector space isomorphism. Thus $$\phi_\alpha\circ \phi_\beta ^{-1}(p, v) = (p, \tau _{\alpha\beta} (p) v)$$ for some $$\tau_{\alpha\beta}(p) \in GL(k,\mathbb R)$$. The main assumption in (iii) is that the map $$\tau_{\alpha \beta }: U_\alpha \cap U_\beta \to GL(k,\mathbb R)$$ is smooth. (Say, if $$\tau_{\alpha\beta}$$ is merely continuous, you get only a continuous vector bundle).
Thus to use Lemma 10.6 it is sufficient to construct $$\phi_\alpha$$ which satisfies (ii) and checks the smoothness of $$\tau_{\alpha\beta}$$, which should be obvious in this case.