4

Let $M$ be a smooth $n$ manifold and let $TM$ denote its tangent bundle

$$ TM = \bigsqcup_{x \in M} \{(x,T_x M)\}$$

I am trying to put a smooth structure (atlas) on $TM$ using the atlas on $M$. But I'm a bit confused and could do with some help:

Say, $(x,T_x M)$ is a given point in the tangent bundle. So my goal is to find an open set containing $(x,T_x M)$ and a smoth diffeomorphism $\psi$ from this open set to an open set in $\mathbb R^{2n}$.

Let $(U,\varphi)$ be a chart on $M$ such that $x \in U$. I want to use this chart to construct a chart $\psi$ on $TM$:

First I need to think about the domain of $\psi$. It seems to me that it should look like $U \times$ some open set $V$ where the elements of $V$ are tangent spaces $T_x M$.

And this is where I am confused: Where exactly would such an open set $V$ live? There seems to be no space consisting of points of the form $T_x M$. What am I doing wrong?

a student
  • 4,365
  • good news, you have not lost your mind. – James S. Cook Jul 25 '15 at 05:41
  • $(x,T_xM)$ is not a "given point in the tangent bundle." A point in the tangent bundle is $(x,v)$ where $x\in M$ and $v\in T_xM$. And $TM=\bigsqcup_{x\in M}{x}\times T_xM$. – anon Jul 25 '15 at 05:53
  • @JamesS.Cook I'm glad :-D – a student Jul 25 '15 at 06:08
  • @whacka Oh, I see. Thank you for your comment. But this still does not resolve the problem of where the open set lives. no? – a student Jul 25 '15 at 06:11
  • The set $TU$ described below is open in the bundle. I can't see using smaller sets than $T_pM$ since over a given point $p \in M$ you want to allow for all possible $v \in T_pM$ in $(p,v)$. Using an open set smaller than all of $T_pM$ would put some tangent vectors outside the chart of the tangent bundle. That would defeat much of the purpose of studying $TM$. So, we use adapted charts. – James S. Cook Jul 25 '15 at 15:18

1 Answers1

3

The usual adapted chart is formed by using $TU$ which is the disjoint union of $\{ x \} \times T_xM$ for $x \in U$. I assume $U$ is the domain of a chart on $M$ then the formula for $\Psi: TU \rightarrow \mathbb{R}^{2n}$ is simply formed by stringing together the coordinates of the point and the vector which form a bundle-point. That is: assuming $\phi: U \rightarrow \mathbb{R}^n$ is the chart on $M$, $$ \Psi(p,v) = (\phi(p), d_p\phi (v)) $$ where you can also write for $\phi = (x^1, \dots , x^n)$ simply $$ d_p\phi (v) = (v^1, \dots, v^n)$$ and $$ v = v^1 \frac{\partial}{\partial x^1} \bigg{|}_p+ \cdots + v^1 \frac{\partial}{\partial x^n} \bigg{|}_p. $$ So in this notation, $$ \Psi(p,v) = (x^1(p), \dots , x^n(p), v^1, \dots , v^n)$$ A very enjoyable exercise I remember from my coursework on this material, show $\Psi$ smoothly overlaps with $\bar{\Psi}$ constructed in the same way from another coordinate chart on $M$ which overlaps $\phi$.

James S. Cook
  • 16,755