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Show that $ T_{p}M $ is an $ n $-dimensional vector space.

Hint: Given two tangent vectors $ v_1 $ and $ v_2 $ at $ p $ with corresponding curves $ \gamma_1 $ and $ \gamma_2 $, we can “add” the corresponding curves in $\mathbb{R}^n$ and then move back to $M$: $ \tilde{γ} = \varphi^{−1}\circ (\varphi\circ \gamma_1 + \varphi\circ\gamma_2 − \gamma (p)) $ defines a curve through $ p $ with tangent vector $ v_1+v_2 $. $ \dots\square $

Thank you.

amWhy
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alireza911
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  • Do you assume that $\phi(p) = 0$ in your chart? –  Oct 16 '13 at 04:01
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    Do there not seem like $n$-obvious vectors that should span? Also, what precisely is your definition? Curves up to having the same derivative after composition with the a chart? – Alex Youcis Oct 16 '13 at 04:17

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By definition, $T_pM$ is the set of all maps $\gamma :(-\epsilon, \epsilon) \to M$ such that $\gamma(0) = p$, with the equivalent relation that $\gamma_1 \sim \gamma_2$ if and only if

$(\phi\circ \gamma_1)'(0) = (\phi\circ \gamma_2)'(0)$

for some coordinate chart $\phi$ containing $p$. This defines a map $\Phi : T_pM \to \mathbb R^n$ by $[\gamma] \mapsto (\phi\circ \gamma_1)'(0)$. The above equivalent condition shows that $\Phi$ is injective. You can also show that $\Phi$ is surjective by considering $\phi^{-1}(tw): w\in \mathbb R^n$.

$\mathbb R^n$ obviously has a vector space structure. With the identification $\Phi$, what is the vector space structure on $T_pM$?

Let $v_1, v_2\in \mathbb R^n $ be represented by $\gamma_1$ and $\gamma_2$ (via $\Phi$) respectively. Define as in your question the curve

$\gamma = \phi^{-1}(\phi\circ \gamma_1 + \phi\circ \gamma_2)$.

Then $\gamma$ is a curve such that $\gamma(0) = p$ (We assume $\phi (p) = 0$) and

$(\phi\circ \gamma)'(0) = (\phi\circ \gamma_1)'(0) + (\phi\circ \gamma_2)'(0)= v_1 + v_2$.

Thus $\gamma$ corresponds to the vector $v_1+v_2$ (via $\Phi$).