Lie derivative of a vector field in coordinates I'm reading the proof that $L_XY=[X, Y]$ on page 225 in this book and I believe it is not quite correct to due an error in equation 20.6. The author writes 
if $Y=\sum b^j \partial/\partial^j$, then 
$$\phi_{-t, *}(Y_{\phi_t(p)})=\sum_j b^j(\phi_t(p)) \phi_{-t, *}\left(\frac{\partial}{\partial x^j}|_{\phi_t(p)}\right)=
\sum_{i,j} b^j(\phi_t(p)) \frac{\partial\phi^i}{\partial x^j}(-t, p)\frac{\partial}{\partial x^i}|_p$$
It seems to me that the RHS should be $\sum_{i,j} b^j(\phi_t(p)) \frac{\partial\phi^i}{\partial x^j}(-t, \phi_t(p))\frac{\partial}{\partial x^i}|_p$ and that this error carries over to (20.7) as well as (20.8). However, since the final result of the author's calculation turns out to be $[X, Y]$ as expected, the error should have somehow disappeared. Any thoughts on this would be appreciated.
 A: I think you are correct: $\phi_{-t *}:T_{\phi_t(p)}M \to T_pM$ should satisfy $$ \phi_{-t *} \left(\frac{\partial}{\partial x^j}\rvert_{\phi_t(p)} \right) = \frac{\partial\phi^i_{-t}}{\partial x^j}(\phi_t(p)) \frac{\partial}{\partial x^i}\rvert_p.$$
The author does rewrite this a couple of times, but in the end it does not really matter because we are eventually setting $t = 0$, and $\phi_0(p) = p$.
A: I also stumbled across this error, and worked out the right derivative. I'll share the details of the computation, they might save someone some time.
We want to know the value of $\frac{\partial}{\partial t}|_{t=0} \left[ \frac{\partial{\phi_i}}{\partial x^j}(-t, \phi_t(p))\right]$. The function we are computing the derivative of is the composition of
$$
f : \mathbb{R} \to \mathbb{R}^{n+1},\ t \mapsto (-t, \phi_t(p)) 
$$
and $\frac{\partial\phi_i}{\partial x^j} : \mathbb{R}^{n+1} \to \mathbb{R}$.
We have that 
$$
df = \left(-1, \frac{\partial \phi^1}{\partial t}, \ldots, \frac{\partial \phi^n}
{\partial t} \right)^T
$$ 
and
$$
d\frac{\partial\phi_i}{\partial x^j} = \left ( \frac{\partial\phi_i}{\partial t \partial x^j}, \frac{\partial\phi_i}{\partial x^1 \partial x^j}, \ldots, \frac{\partial\phi_i}{\partial x^n \partial x^j} \right ).
$$
It follows from the chain rule that
\begin{align*}
\frac{\partial}{\partial t}|_{t=0} \left[ \frac{\partial{\phi_i}}{\partial x^j}(-t, \phi_t(p))\right] = \left(d\frac{\partial\phi_i}{\partial x^j}\right)_{f(0)} (df)_0 = - \frac{\partial\phi_i}{\partial t \partial x^j}(0, p) + \sum_k \frac{\partial \phi^k}{\partial t}(0, p)\frac{\partial\phi_i}{\partial x^k \partial x^j}(0, p).
\end{align*}
However, $\frac{\partial\phi_i}{\partial x^k \partial x^j}(0, p) = 0$ for all $k$, since $\phi_i(0, p) = p$. So in the end, we have
$$
\frac{\partial}{\partial t}|_{t=0} \left[ \frac{\partial{\phi_i}}{\partial x^j}(-t, \phi_t(p))\right] = - \frac{\partial\phi_i}{\partial t \partial x^j}(0, p),
$$
which is what we wanted.
