Clarification the proof of lie bracket equals lie derivative I was reading the proof showing that Lie bracket equals Lie derivative ($ L_V W = [V, W]$, where $V, W$ are vector fields) in John Lee's Smooth Manifolds.
So, in the book here https://math.berkeley.edu/~jchaidez/materials/reu/lee_smooth_manifolds.pdf on page 230, in case 1, given a smooth manifold $M$, for any $p \in M$, it assumed that we can choose a smooth coordinate $(u^i)$ on a neighborhood of p
so that $V = \frac{\partial}{\partial u^1}$. If we denote $\theta$ as the flow associated to the Lie derivative, then the flow of $V$ is $\theta_t(u) = (u^1 + t, u^2,..., u^n)$ for each fixed $t$, so the matrix of $d(\theta_{-t})_{\theta_t(u)}$ is the identity matrix.
Then it computes
\begin{align}
d(\theta_{-t})_{\theta_t(u)}(W_{\theta_t}(u)) &= 
d(\theta_{-t})_{\theta_t(u)}(W^j(u^1 + t, u^2,.., u^n)(\frac{\partial}{\partial u^j})_{\theta_t(u)}\\
&= W^j(u^1 + t, u^2,.., u^n)(\frac{\partial}{\partial u^j})_{u}
\end{align}
So, I'm not completely sure why we can write  $(\frac{\partial}{\partial u^j})_{u}$ instead of $(\frac{\partial}{\partial u^j})_{\theta_t(u)}$. They are different vector fields, so why is this change valid.
Thanks ahead.
 A: I just had the same question reading Lee's textbook. You can first notice that
$$
\theta_{-t}'[\theta_t(u)] : T_{\theta_t(u)}M \longrightarrow T_uM 
$$
so $\theta_{-t}'[\theta_t(u)] \cdot W_{\theta_t(u)}$ should be in $T_uM$. To see it, write $(U,\varphi)$ a smooth chart such that $V$ writes $\partial / \partial x^1$ on $U$. Take $q$ near $p$ and $t$ small, write $u = \varphi(q)$, what we know is that $\varphi_*V \in \mathfrak{X}(\varphi(U))$ has the flow $\tilde{\theta}_t(u) = (u^1+t, u^2,...,u^n)$. It is fairly easy to show that
$$
\mathcal L_{(\varphi_*V)}(\varphi_*W) = \varphi_* (\mathcal L_VW)
$$
so you can assume without loss of generality that $M = \mathbb R^n$, $U \subset \mathbb R^n$ is open, $V \in \mathfrak X(U)$, $W \in \mathfrak X(\mathbb R^n)$ and $\theta_t(u) = (u^1+t, u^2,...,u^n)$. Forall $t$, $\theta_t : [\mathbb R^n]_t \rightarrow [\mathbb R^n]_{-t}$ has Jacobian matrix the identity. Then it is fairly easy to see that when $F : \Omega \subset \mathbb R^n \rightarrow \mathbb R^n $ is smooth and with Jacobian matrix at $p \in \Omega$ the identity,
$$
F'(p) \cdot X_p = \sum_{i=1}^n X^i_p \left. \frac{\partial}{\partial x^i} \right|_{F(p)}.
$$
Using this you have
$$
(\mathcal L_VW)_u= \left. \frac{d}{dt} \right|_{t=0} \sum_{i=1}^n W^i_{\theta_t(u)} \left. \frac{\partial}{\partial x^i} \right|_u
$$
and the remaining computations are stepforward.
