Why is the definition of the element $rv$ of the tangent space chart-independent?

Question

Let's assume we have a $m$-dimensional smooth manifold $\mathscr{M}$. We can define the tangent space $T_p\mathscr{M}$ at a point $p \in \mathscr{M}$ as the set of equivalence classes of curves in $\mathscr{M}$ going through $p$. Then to show that $T_p\mathscr{M}$ is a vector space my textbook defines $r v$ as

$$ r v := [\phi^{-1} \circ (r\phi \circ \sigma)] \, ,$$ where $r \in \mathbb{R}$, $v = [\sigma]\in T_p\mathscr{M}$ and $\phi$ a chart around $p$ satisfying $\phi (p) = 0$.

I was able to prove this definition doesn't depend on the choice of representative $\sigma$, but how do you proceed to prove that it doesn't depend on the choice of chart $\phi$?

Answer 1

I will try to outline the main ideas.

First, note that $\phi(U)$ being an open subset of $\mathscr{M}$ it is itself a smooth manifold of dimension $m$. So, instead of looking at $T_p\mathscr{M}$ one can look at $T_0(\phi(U))$ and the main advantage there is that a chart is not required anymore to define the vector space structure on $T_0(\phi(U))$. Indeed, $\phi(U)$ already being a subset of $\mathbb{R}^m$ a chart is not needed to push the values taken by the curves into $\mathbb{R}^m$ to use the vector space structure thereof.

The vector space structure on $T_0(\phi(U))$ is defined as follows.

• $v_1 + v_2 = [\sigma_1] + [\sigma_2] = [\sigma_1 + \sigma_2]$
• $rv = r[\sigma] = [r\sigma]$

Then, one can check that the vector space $T_0(\phi(U))$ is isomorphic to $\mathbb{R}^m$ with the isomorphism mapping $v \in \mathbb{R}^m$ to $[\sigma_v]$, where $\sigma_v(t) := t v$.

Moreover, we have a bijection $\phi_*$ from $T_p\mathscr{M}$ to $T_0(\phi(U))$ mapping $v := [\sigma]$ to $[\phi \circ \sigma]$. One can use this bijection to transport the vector space structure on $T_0(\phi(U))$ to $T_p\mathscr{M}$ making $\phi_*$ into an isomorphism of vector spaces. The resulting vector space structure on $T_p\mathscr{M}$ is precisely the one defined partially in your post.

Finally, given another chart $(V, \psi)$ with $p \in V$ and $\psi(p) = 0$, one can define in the same way a vector space structure on $T_p\mathscr{M}$ using $\psi_*$. But what has been shown above is that both vector space structures are the same, i.e. they are isomorphic, both being isomorphic to $\mathbb{R}^m$ (through $T_0(\phi(U))$ and $T_0(\psi(V))$, respectively).