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This minor problem popped up while I was reading the book "Modern Differential Geometry for Physicists" by Chris J. Isham. It deals with introducing vector space structure on a tangent space $T_pM$ to a manifold $M$ at a point $p \in M$ (a related problem has been solved here: Giving tangent space a vector space structure). Let me explain what's involved. Suppose $T_pM$ denotes the usual set of all equivalence classes $[\gamma]$ with respect to the equivalence $\sim$ on the set of smooth curves defined by the formula $\gamma_1 \sim \gamma_2 \iff (\phi \circ \gamma_1)^{\prime}(0) = (\phi \circ \gamma_2)^{\prime}(0)$ for some (and hence any) chart $(U, \phi)$ around $p$ with the additional property that $\phi(p) = 0$ (here the $\gamma$'s are standard smooth curves from $(-\epsilon, \epsilon)$ to $M$ such that $\gamma_i (0) = p.$) For any $v_1 = [\gamma_1], v_2 = [\gamma_2] \in T_pM$ and $a,b \in \mathbb{R}$ define $a v_1 + b v_2 := [\phi^{-1}(a (\phi \circ \gamma_1) + b(\phi \circ \gamma_2))].$ Now the reader is asked to show that "these operations are independent of the choice of chart $(U, \phi)$ and representatives $\gamma_1,$ $\gamma_2$ of the tangent vectors $v_1,$ $v_2.$"

My questions are: 1) exactly in what sense are the operations independent? To be more precise: if $(V, \psi)$ is another chart with $\psi(p) = 0,$ is it really true that $$\phi^{-1}(a (\phi \circ \gamma_1) + b(\phi \circ \gamma_2)) \sim \psi^{-1}(a (\psi \circ \gamma_1) + b(\psi \circ \gamma_2))$$ as the assertion seems to suggest and if so, how to prove it? 2) What is the connection between the other approach via the mapping $d\phi_p: T_p(M) \stackrel {\cong}{\to} \mathbb R^n= T_{\phi(p)}(V),$ $T_pM \ni v = [\gamma] \mapsto (\phi\circ \gamma)'(0)= \vec u\in \mathbb R^n$ as it is sketched in Giving tangent space a vector space structure? I mean: it would be very sad if the two approaches weren't equivalent in some sense. And if they are, then in what sense precisely? Thanks for your help!

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Regarding your first question, the answer is yes. And the way to prove it is to verify the definition of the equivalence relation you have written down in your problem statement. A strategy for how to do this is the following. First, let's keep the coordinate chart fixed, and consider two pairs of representatives $\gamma_1,$ $\gamma_2$, $\hat\gamma_1$, $\hat \gamma_2$ representing respectively v_1, and v_2. A curve representing $av_1 + bv_2$ is, as you say, $\phi^{-1}(a\phi\circ\gamma_1 + b \phi\circ\gamma_2),$ let's denote this curve for the moment by $\sigma(t)$. Let's denote by $\hat\sigma(t)$ the corresponding curve with $\hat\gamma_1$ and $\hat\gamma_2$. Now if $\gamma_1 \sim \hat \gamma_1$ and $\gamma_2 \sim \hat\gamma_2,$ then calculate $\phi\circ\sigma'(0)$ and $\phi\circ\hat\sigma'(0)$ and then verify that they are equivalent. Now that you know that $av_1 + bv_2$ is a well-defined tangent vector, if you belive your previous statment that equivalence is independent of the coordinate chart, then you are already finished with the problem, i.e. it is only necessary to verify coordinate chart independence for an arbitrary tangent vector $v$, and then it automatically holds for linear combinations of tangent vectors.

Regarding your second question, sure, they're the same. In the setup to your second question, you are already assuming that $T_pM$ is defined by equivalence classes of curves, so I'm not really sure what you're asking.

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  • $\begingroup$ I'm not sure I got it right. Of course, taking different representatives $\gamma_1,$ $\hat\gamma_1$ of $v_1$ and $\gamma_2,$ $\hat\gamma_2$ of $v_2$ shows that the result is independent on the choice of representatives. However, what do you mean by "it is only necessary to verify coordinate chart independence for an arbitrary tangent vector"? Thanks. $\endgroup$ – Jorge.Squared May 13 '14 at 9:02
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Ok, after some attempts, here is another suggestion of how I think the problem can be solved. Let the setup be as explained in the question above. Now we want to verify that really \begin{equation} \phi^{-1}(a \phi \circ \gamma_1 + b \phi \circ \gamma_2) \sim \psi^{-1}(a \psi \circ \gamma_1 + b \psi \circ \gamma_2). \end{equation} Since the relation "$\sim$" doesn't depend on the choice of a particular chart, it is sufficient to check that the above "equation" is true for some (and hence for any) chart. So let's take $\psi$ and compute: $$ \frac{\rm{d}}{\rm{d}t}\left(\psi \circ \phi^{-1} (a \phi \circ \gamma_1 + b \phi \circ \gamma_2) \right)\!\!\bigg|_{t=0} = $$ $$ =\frac{\rm{d}}{\rm{d}t} \left(\psi \circ \phi^{-1}(a \phi \circ \psi^{-1} \circ \psi \circ \gamma_1 + b \phi \circ \psi^{-1} \circ \psi \circ \gamma_2) \right)\!\!\bigg|_{t=0} = $$ $$ = \mathbf{D}(\psi \circ \phi^{-1})(0) \cdot \left(a \frac{\rm{d}}{\rm{d}t}(\phi \circ \psi^{-1} \circ \psi \circ \gamma_1)\bigg|_{t=0} + b \frac{\rm{d}}{\rm{d}t}(\phi \circ \psi^{-1} \circ \psi \circ \gamma_1)\bigg|_{t=0} \right) = $$ $$ = \mathbf{D}(\psi \circ \phi^{-1})(0) \cdot \mathbf{D}(\phi \circ \psi^{-1})(0) \cdot \left(a \frac{\rm{d}}{\rm{d}t}(\psi \circ \gamma_1) \bigg|_{t=0} + b\frac{\rm{d}}{\rm{d}t}(\psi \circ \gamma_1) \bigg|_{t=0} \right) = $$ $$ = \mathbf{D}(\psi \circ \psi^{-1})(0) \cdot \frac{\rm{d}}{\rm{d}t} \left(a \psi \circ \gamma_1 + b \psi \circ \gamma_2 \right)\!\bigg|_{t=0}= $$ $$ = \frac{\rm{d}}{\rm{d}t}\left(\psi \circ \psi^{-1} (a \psi \circ \gamma_1 + b \psi \circ \gamma_2) \right)\!\!\bigg|_{t=0}, $$ which was to be proved. As for part 2) of the question, it should be clear from the above proof that the two approaches are in fact the same taking into account the very definition of $\rm{d} \phi_p.$

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