Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
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
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 ! "

share|cite|improve this answer

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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.