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Let M be a differentiable manifold of dimension m and also let $\{\xi_1,\dots,\xi_m\}\subset \text{T}_pM$ be an linearly independent set of the tangent bundle of M at a certain point $p\in M$.

I have to proof that there is a chart $(\varphi=(u_1,\dots,u_m), U)$ with $\varphi(p)=0$ and:$$ \xi_i =\left( \frac{d}{du_i}\right)_p$$

-My approach: Let $(\varphi, U)$ be a chart with $\varphi(p)=0$, there is an unique linear endomorphism $F$ of $\text{T}_pM$ that sends (since both sets are bases), $$\left( \frac{d}{du_i}\right)_p\longrightarrow \xi_i$$ from this point I supose you should be able to define a diffeomorphism $f:M\rightarrow M$ with $df(p)=F$ wich would end the argument by taking the chart $(\varphi \circ f,U)$. I really don't know how to define such a map or if it even exists.

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  • $\begingroup$ @john thanks for your fast answer, sadly it contains some notions I am not supposed to use yet, like forms or the dual of the tangent space. $\endgroup$
    – Smurf
    Jul 20 '15 at 19:55
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    $\begingroup$ I'd suggest that, instead of trying to cook up such an $f$, instead, find a matrix $A:\mathbb{R}^n\rightarrow \mathbb{R}^n$ and consider $A\circ \varphi$. $\endgroup$ Jul 20 '15 at 19:55
  • $\begingroup$ I think I'm close to the answer but can't get there. There is a matrix $\{a_{ij}\}$ wich transforms the basis $\{ \left( \frac{d}{du_i} \right) _p\}$ in $\{\xi_i\}$, I think the trick is use the chart $(A^{-1}\circ \varphi,U)$ where $A=\{a_{ji}\}$, but I am not sure how to write it with precision. $\endgroup$
    – Smurf
    Jul 20 '15 at 21:34
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You've made an initial choice of a random chart $(\phi,U)$ such that $\phi(p)=0$. Let $V_1,\ldots,V_m \in T_0(\mathbb{R}^m)$ be the image of $\xi_1,\ldots,\xi_m$ under $d\phi_p$. The desired effect is that $V_1,\ldots,V_m$ is the standard basis for $T_0(\mathbb{R}^m)$, but it might not be.

Instead of doing a precomposition in $M$, do a postcomposition in $\mathbb{R}^m$.

Using the standard identification of $T_0(\mathbb{R}^m)$ with $\mathbb{R}^m$, let $F$ be the linear transformation of $\mathbb{R}^m$ taking $V_1,\ldots,V_m$ to the standard basis for $\mathbb{R}^m$.

Finally, let $\phi' = F \circ \phi$. An application of the chain rule together with linearity of $F$ will give the proof that $\phi'$ is the desired chart.

Some more details: Here's the final calculations to put this all together. Applying the chain rule, $$d\phi'_p = dF_{\phi(p)} \circ d\phi_p = dF_0 \circ d\phi_p $$ Applying linearity of $F$ we have $dF_0=F$ and so $$d\phi'_p = F \circ d\phi_p $$ Letting $e_1,\ldots,e_m$ be the standard basis, $F$ was chosen so that $F(V_i)=e_i$, and so $$d\phi'_p(\xi_i) = F(d\phi_p(\xi_i)) = F(V_i) = e_i $$ which is what was wanted to be shown.

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  • $\begingroup$ Thanks for your answer @Lee . I get what you've done there, but I still can't manage to apply linearity and the chain rule correctly to get the job done. I'm sorry if I am asking for something too obvious, I've been thinking about this too much I can't think straight anymore. $\endgroup$
    – Smurf
    Jul 20 '15 at 22:10
  • $\begingroup$ @Jorge: I added some more details. $\endgroup$
    – Lee Mosher
    Jul 21 '15 at 0:35

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