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Motivated by this computation, I was trying to compute this derivative: $$ \left.\frac{d}{dt}\right|_{t=0}X_{\Phi_t^Y(p)} $$ where $X$ is a vector field on a manifold $\mathcal{M}$ and $\Phi_t^Y$ is the flow of the vector field $Y$, i.e $\Phi_t^Y(p)=\gamma(t)$ for $\gamma$ being a curve on $\mathcal{M}$ with $\gamma(0)=p$, $\gamma'(t)=Y_p$.

I am using the definitions in Agricola & Friedrich's Global Analysis so that, if needed, I am allowed to consider that all manifolds are embedded in $\mathbb{R}^n$ for some $n$. Then I can identify a vector $v \in T_p\mathcal{M}$ with a curve $\gamma$ such that $\gamma'(0)=v$ and $\gamma(0)=p$. Also, the definition of differential of a function $f:\mathcal{M}\to \mathcal{N}$ is a linear map $f_{*,p}:T\mathcal{M}\to T\mathcal{N}$ such that: $$ f_{*,p}(v)=\left.\frac{d}{dt}\right|_{t=0}f\circ \gamma(t) $$ since $f\circ \gamma$ is a curve on $\mathcal{N}$.

I have struggled for hours to do the computations with all rigour and finally I am going to post an answer using coordinates, but somehow I feel this should be much more fundamental. Am I missing something? Is there a coordinate-free way to get to the result only from the definitions?

Miguel
  • 3,265

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Since $X$ can be considered a function from the manifold to the tangent bundle $X:\mathcal{M}\to T\mathcal{M}$, with some abuse of notation we can write: $$ \rho(s)=X\circ \Phi_t^Y(p) =X\circ \gamma(t) $$ for a curve $\gamma$ with $\gamma'(0)=Y_p$ and then $$ \left.\frac{d}{dt}\right|_{t=0}X_{\Phi_t^Y(p)}=\left.\frac{d}{dt}\right|_{t=0}X\circ \gamma(t)=dX\;Y_p $$ To understand what is really the differential of $X$ as a map with values on the tangent bundle, let us work in coordinates $x_i$, $i=1\dots n$. Then, for the $i$-th coordinate: $$ \left.\frac{d}{dt}\right|_{t=0}X^i\circ \gamma(t)=\sum_j \left.\frac{\partial X^i}{\partial x^j}\right|_p \; \left.\frac{d}{dt}\right|_{t=0}\gamma^j(t)=\sum_j \left.\frac{\partial X^i}{\partial x^j}\right|_p\; Y_p^j=Y_p(X^i) $$ where in the last equality we consider the vector $Y_p$ differentiating the scalar function $X^i$.

Considering now that a vector differentiates a vector field coordinate-wise, leads to: $$ \left.\frac{d}{dt}\right|_{t=0}X_{\Phi_t^Y(p)}=Y_p(X) $$

Miguel
  • 3,265