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Let $M$ be a differential manifold and $X$ a vector field over $M$ s.t. $X(p) = 0$ for some $p \in M$. Let be $\phi_p : T_p(M) \to T_p(M)$ given as $$\phi_p(v) := [Y,X](p),$$ being $Y$ another vector field of $M$ s.t. $Y(p) = v$.

For $X = \sum_{i=1}^nX_i\frac{\partial}{\partial x_i}$ calculate the associated matrix of $\phi_p$ with respect the basis $\{(\frac{\partial}{\partial x_i})\}_{1 \leq i \leq m}$.

Can you help me, please?

And another question: Since $\phi_p(v) = [Y,X](p) = Y(X(p)) - X(Y(p)) = Y(0) - X(v) = -X(v)$, can we assert that $\phi_p$ does not depend of $Y$?

Thanks

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Your last computation is wrong (it actually doesn't make any sense to write $Y(X(p))$) and leads me to believe that the definition of $[Y,X]$ is not clear to you. Here is how it goes. Basically, you need to test how vector fields act on smooth functions. Let $f\in C^\infty(M)$ and let $X=X^i\partial_i$, $Y=Y^j\partial_j$, where by $\partial_i$ I denote $\partial/\partial_{x_i}$ and I use Einstein summation convention for conciseness. Then $$ \begin{split} [Y,X](f) &= Y(X(f))-X(Y(f)) \\ &= Y^j\partial_j(X^i\partial_if)-X^i\partial_i(Y^j\partial_jf) \\ &= Y^j(\partial_jX^i)\partial_if-X^i(\partial_iY^j)\partial_jf , \end{split} $$ because the terms involving second order derivatives of $f$ cancel by Schwarz's theorem.

At point $p$, we have $X_p^i=0$ and $Y_p^j=v^j$, therefore only the first term remains $$ [Y,X]_p = v^j(\partial_jX^i)_p\partial_i . $$ In particular, this shows that $[Y,X]_p$ only depends on the value of $Y$ at point $p$, which is $Y_p=v$, and not on the actual vector field itself.

The matrix representation can be recovered from $\phi_p(v) = \phi_p(v)^i\partial_i$ from which $$ \phi_p(v)^i = (\partial_jX^i)_p v^j = M^i_j v^j $$ with $M^i_j=(\partial_jX^i)_p$. The transformation $\phi_p$ is linear and depends only on the derivatives of $X$ at $p$.

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  • $\begingroup$ So you get $[Y,X](f) = \sum_{j=1}^n\sum_{i=1}^n v_j\Bigl(\frac{\partial X_i}{\partial x_j} \frac{\partial}{\partial x_i}\Bigr)(p).$ And the matix $\phi_p(v) = \sum_{i=1}^n\phi_p(v)_i \frac{\partial}{\partial x_i}$, with $\phi_p(v)_i = \Bigl(\frac{\partial}{\partial x_j} X_i\Bigr)(p)v_j$? $\endgroup$ – LH8 Nov 19 '18 at 19:56
  • $\begingroup$ Yes. That is precisely what I wrote, apart from putting the indices in their most common position. Also, you are missing an $f$ in the r.h.s. of your first formula. $\endgroup$ – Federico Nov 19 '18 at 22:23
  • $\begingroup$ Ok. And $\phi_p (v)_i = \phi_p(v_i)$? I suppose, since $\phi_p$ is linear (since the partial derivations are linears). $\endgroup$ – LH8 Nov 20 '18 at 10:24
  • $\begingroup$ No. $v^i$ is the $i$-th component of $v$ written in the base $(\partial_i)_{i=1}^m$, that is, $v=v^i\partial_i$. It doesn't make sense to write $\phi_p(v_i)$. $\phi_p$ requires a vector, you cannot feed it just a single coordinate. By $\phi_p(v)^i$ i mean the $i$-th component of $\phi_p(v)$ written in the base $(\partial_i)_{i=1}^m$, that is, $\phi_p(v)=\phi_p(v)^i\partial_i$. $\endgroup$ – Federico Nov 20 '18 at 13:32

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