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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.

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  • $\begingroup$ Isn't it a requirement a priori that $\tau_{\alpha \beta}^{-1}\tau_{\alpha \beta}$ is an isomorphism?$ $\endgroup$
    – user284001
    Mar 25, 2021 at 8:37
  • $\begingroup$ @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. $\endgroup$
    – zxcv
    Mar 25, 2021 at 8:57

1 Answer 1

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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.

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