Proving the Lie derivative of a vector field with respect to another is a smooth vector field

Question

Suppose $M$ is a smooth manifold and $V,W$ are smooth vector fields on $M$. Let $L_V W$ denote the Lie derivative of $W$ with respect to $V$. Then $(L_V W)_p$ exists for every $p \in M$ and $(L_V W)$ is a smooth vector field.

The proof begins as follows:

Let $\theta$ be the flow of $V$. For arbitrary $p \in M$, let $(U,(x^i))$ be a smooth chart containing $p$. Choose an open interval $J_0$ containing 0 and an open subset $U_0 \subset U$ contatining $p$ such that $\theta$ maps $J_0 \times U_0$ into $U$. For $(t,x) \in J_0 \times U_0$, write the component functions of $\theta$ as $(\theta^1(t,x)...\theta^n(t,x))$. Then for any $(t,x) \in J_0 \times U_0$, the matrix of $d(\theta_{-t})_{\theta_{t}(x)}: T_{\theta_{t}(x)}M \rightarrow T_xM$ is

$\bigg(\frac{\partial \theta^i}{\partial x^j}(-t,\theta(t,x))\bigg)$

I am confused how to compute this matrix. In particular, I know that the matrix of a map is the Jacobian of the (coordinate representation) of the map. But in particular how do I arrive at it being evaluated at $(-t,\theta(t,x))$?

Help would be greatly appreciated !

Answer 1

As you stated in your question, for all $(i,j)\in\{1,\ldots,n\}^2$, the $(i,j)$ entry of the matrix will be given by: $$\frac{\partial{\theta_{-t}}^i}{\partial x^j}(\theta_t(x)).$$ However, one has ${\theta_{-t}}^i=\theta^i(-t,\cdot)$ (this is true without the upper $i$) so that, one gets: $$\frac{\partial{\theta_{-t}}^i}{\partial x^j}=\frac{\partial\theta^i}{\partial x^j}(-t,\cdot).$$ Whence specifying the previous identity at the point $\theta_t(x)=\theta(t,x)$, one has: $$\frac{\partial{\theta_{-t}}^i}{\partial x^j}(\theta_t(x))=\frac{\partial\theta^i}{\partial x^j}(-t,\theta(t,x)),$$ which is what you required.