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As I understand it, the tangent space $T_{p}(M)$ to a manifold is given a vector space structure by taking a chart $\varphi:U\rightarrow V\subset\mathbb{R}^{n}$ and making the identification via the induced map $d\varphi_{p}:T_{p}(U)\rightarrow T_{\varphi(p)}(V)$, which is isomorphic to $\mathbb{R}^{n}$ (and this identification is independent of chart). How do we know that we can find a smooth map $\varphi$ for which $d\varphi_{p}$ is bijective?

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What is your definition of $T_p(M)$? –  Chris Eagle Apr 19 '12 at 20:12
    
The set of tangency clases of smooth maps $\gamma:I\rightarrow M$ where $I$ is a closed interval containing 0 and $\gamma(0)=p$. –  LCL Apr 19 '12 at 20:24
    
Not only can you find a chart $\phi$ such that $d\phi_p$ is an isomorphism, but you will have an isomorphism for every chart whatsoever. Details in the answer below. –  Georges Elencwajg Apr 19 '12 at 22:47
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2 Answers 2

up vote 2 down vote accepted

There are some subtle points here.
Given a chart $\phi :U\to V$, the bijection $d\phi_p: T_p(M) \stackrel {\cong}{\to} \mathbb R^n= T_{\phi(p)}(V)$ is defined by sending the equivalence class of a curve $\gamma$ passing through $p$ at time $0$ to the velocity vector $(\phi\circ \gamma)'(0)= \vec u\in \mathbb R^n $.
It is injective by the definition of the equivalence relation on curves through $p$ and surjective by consideration of the curves $t\mapsto \phi^{-1}(\phi (p)+t\vec u) \;( u\in \mathbb R^n)$

The difficulties are that these bijections $d\phi_p$ are very, very dependent on the chart $\phi$ and that they are not isomorphisms because $T_p(M)$ is a priori a set and not a vector space ! So what is to be done ?
Here is what:

Suppose you have two elements $v,w\in T_p(M)$ . How do you add them?
The recipe is: choose a chart $\phi: U\to V$, compute $d\phi_p(v)=v'\in \mathbb R^n$ and $d\phi_p(w)=w'\in \mathbb R^n$, add these vectors in $\mathbb R^n$ and obtain $v'+w'$. Finally the required sum is defined as $$v+w\stackrel {def}{=}(d\phi_p)^{-1}(v'+w')$$
Yes, but what if you had chosen another chart $\psi: U\to W$?
You would have obtained $d\psi_p(v)=v''\in \mathbb R^n$, $d\psi_p(w)=w''\in \mathbb R^n$ and $v''+w''\in \mathbb R^n$ with:
$v'$ very, very different from $v''$,
$w'$ very, very different from $w''$
$v'+w'$ very, very different from $v''+w''$

However you would observe with delight that $(d\phi_p)^{-1}(v'+w')=(d\psi_p)^{-1}(v''+w'') $ (very,very equal !) so that the definition of the sum of $v, w \in T_p(M)$ $$v+w\stackrel {def}{=}(d\phi_p)^{-1}(v'+w')=(d\phi_p)^{-1}(d\phi_p(v)+d\phi_p(v))\in T_p(M)$$ does not depend on the choice of the chart $\phi$, despite strong appearances to the contrary !
An analogous observation will show you that the definition $r\cdot v\stackrel {def}{=}(d\phi_p)^{-1}(r\cdot d\phi_p(v))$ defines the product of a vector by a real scalar and this puts the final touch to the definition of the vector space structure on $T_p(M)$ .
And so, finally, the answer to your question is:

" For every chart $\phi$ the map $d\phi_p: T_p(M)\to T_{p}(\mathbb R^n)$ is an isomorphism because it is a bijection and because we defined the vector structure on $T_p(M)$ in order that it be linear ! "

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Recall that a chart about $p \in M$ is a pair $(U, \varphi)$, where $U \subset M$ is an open neighborhood of $p$ and $$\varphi: U \longrightarrow V \subset \mathbb{R}^n$$ is a diffeomorphism. Since $\varphi$ is a diffeomorphism, $$d\varphi_p : T_p U \longrightarrow T_{\varphi(p)} V$$ is a bijection (you can easily show that $d(\varphi^{-1})_{\varphi(p)}$ is its inverse). $\varphi$ exists because $M$ can be covered by charts.

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