0
$\begingroup$

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 $\phi$?

$\endgroup$
1
$\begingroup$

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).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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