I am reading the proof that the tangent bundle can be made into a smooth $2n$-manifold from Introduction to Smooth Manifolds by John M. Lee, and wanted to ask for clarification on something.
I'm on chapter 3 (tangent spaces), where tangent spaces are defined using derivations on spaces of smooth functions, reading Proposition 3.18. For reference, here is the proposition:
For any smooth $n$-manifold $M$, the tangent bundle $TM$ has a natural topology and smooth structure that make it into a $2n$-dimensional smooth manifold.
Let $(U,\phi)$ be a smooth chart for $M$ and let $(x^1,\cdots,x^n)$ be the coordinate functions of $\phi$. They certainly exist because $\phi$ is a map from $U$ to an open subset of the real $n$-space. Define the map $\tilde{\phi}:\pi ^{-1}(U) → \mathbb{R}^{2n}$ that sends the derivation $ v^i \left.\frac{\partial}{\partial x^i}\right|_{p}$ to the point $(x^1,\cdots,x^n,v^1,\cdots,v^n)$.
It is straightforward to show that $\tilde{\phi}$ is a bijection onto its image.
The step that is tripping me up is checking that the transition map is smooth. In particular, suppose $(U,\phi)$ and $(V,\psi)$ are smooth charts for $M$, then the corresponding smooth charts for the tangent bundle $TM$ are $(\pi ^{-1} (U), \tilde{\phi})$ and $(\pi ^{-1} (V), \tilde{\psi})$. We need to prove that the transition map $$ \tilde{\psi}\circ \tilde{\phi}^{-1}:\phi(U \cap V) × \mathbb{R}^{n}→\psi(U\cap V) × \mathbb{R}^{n}$$ (obviously assuming the intersections are not empty) is smooth.
Lee writes:
$$\tilde{\psi}\circ \tilde{\phi}^{-1} (x^1,\cdots,x^n, v^1, \cdots, v^n)= (\tilde{x}^1(x),\cdots,\tilde{x}^n(x),\frac{\partial \tilde{x}^1}{\partial x^j}(x)v^j, \cdots, \frac{\partial\tilde{x}^n}{\partial x^j}(x)v^j) $$
Here are my questions:
- What does each $\tilde{x}^i$ mean? I assume, by analogy with each $x^i$, that these are the coordinate functions of $\psi$, especially considering that the book did use this convention before.
- What is $x$? I assume $x=(x^1, \cdots, x^n)$, but this does not make sense. For one, the point $x$ is an element of $\phi(U \cap V)$, which is a subset of $\mathbb{R}^n$. On the other hand, the coordinate functions $(\tilde{x}^i)$ of the coordinate map $\psi$ are defined on an open susbet ($V$) of the smooth manifold $M$, so what does it mean to write $\tilde{x}^i(x)$?
If my assumptions so far are correct, then there is probably some identification going on. For example, by identifying the local coordinates $x \in \mathbb{R}^n$ with its preimage $\phi^{-1}(x)$, one could be justified in writing $\tilde{x}^i(x)$. Is this the correct interpretation of what is happening?
Untangling this identification, I would write:
$$\tilde{\psi}\circ \tilde{\phi}^{-1} (x^1,...,x^n, v^1, ..., v^n)= \tilde{\psi} \left(v^i \left. \frac{\partial}{\partial x^i}\right|_{\phi^{-1}(x)} \right) = \tilde{\psi}\left(v^i \frac{\partial \tilde{x}^j}{\partial x^i}(x) \left. \frac{\partial}{\partial \tilde{x}^j } \right|_{\phi^{-1}(x)} \right) = (\tilde{x}^1(\phi^{-1}(x)),\cdots,\tilde{x}^n(\phi^{-1}(x)),\frac{\partial \tilde{x}^1}{\partial x^j}(x)v^j, \cdots, \frac{\partial\tilde{x}^n}{\partial x^j}(x)v^j) $$
Is this the correct interpretation?
I'm finding it difficult to keep track of all the variables and identifications, which is probably the root of my problem.
Note: I am using the Einstein summation convention