# Matrix associated of an application between tangent spaces

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 $$\{(\frac{\partial}{\partial x_i})\}_{1 \leq i \leq m}$$.

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

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 $$p$$.
• 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 $\phi_p(v)_i = \Bigl(\frac{\partial}{\partial x_j} X_i\Bigr)(p)v_j$? – LH8 Nov 19 '18 at 19:56
• 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. – Federico Nov 19 '18 at 22:23
• Ok. And $\phi_p (v)_i = \phi_p(v_i)$? I suppose, since $\phi_p$ is linear (since the partial derivations are linears). – LH8 Nov 20 '18 at 10:24
• 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, $\phi_p(v)=\phi_p(v)^i\partial_i$. – Federico Nov 20 '18 at 13:32