I'm following Tu's proof of the fact that Lie derivative of a (smooth) vector field with respect to another is actually the Lie bracket of the two vector fields in his book An Introduction to Manifolds. I go through his proof and everything is fine but something went wrong in my argument.
To illustrate, let me recall some terminologies.
Definition 1. Given a vector field $X\in \mathfrak{X}(M)$ of a smooth manifold $M$, the Lie derivative of $Y\in \mathfrak{X}(M)$ with respect to $X$ is defined as follows: let $p\in M$, and let $\varphi_{t}(q)$ be a local flow around $p$; then the Lie derivative is given by $$ L_{X}Y := \lim_{t\to 0} \frac{D\varphi_{-t}(Y_{\varphi_{t}(p)})-Y_{p}}{t} = \frac{d}{dt}\bigg|_{t=0} D\varphi_{-t}(Y_{\varphi_{t}(p)}). $$
To prove $L_{X} Y=[X,Y]$ by local coordinates, I worked out the local representation for $D\varphi_{-t}(Y_{\varphi_{t}(p)})$, but I received a tiny different result.
Put the smooth function $\varphi(t,q)=\varphi_{t}(q)$. Write, locally, $Y=\sum_{j} b_{j}\frac{\partial }{\partial x^{j}}$. I think we should obtain $$ D\varphi_{-t}(Y_{\varphi_{t}(p)})= \sum_{i,j} \frac{\partial \varphi^{i}}{\partial x^{j}}(-t,\varphi_{t}(p))b_{j}(\varphi_{t}(p))\frac{\partial}{\partial x^{i}}\bigg|_{p}. \tag{1} $$ However, Tu wrote $$ D\varphi_{-t}(Y_{\varphi_{t}(p)})= \sum_{i,j} \frac{\partial \varphi^{i}}{\partial x^{j}}(-t,\color{red}{p})b_{j}(\varphi_{t}(p))\frac{\partial}{\partial x^{i}}\bigg|_{p}, \tag{2} $$ which I felt a bit off since the entries of the matrix for $D\varphi_{-t}$ should be evaluated at $\varphi_{t}(p)$, but Tu's result (2) seems to indicate they should be evaluated at $\color{red}{p}$ instead.
Although I don't agree Tu's result, I found an obstacle showing $L_{X} Y = [X,Y]$ if I use my result (1). The trouble is when carrying out $\frac{d}{dt}$, we obtain an additional term: $$ \sum_{i,j,m} \frac{\partial^{2} \varphi^{i}}{\partial x^{m}\partial x^{j}}(-t,\varphi_{t}(p))a_{m}(\varphi_{t}(p)), $$ where locally, $X=\sum_{m} a_{m} \frac{\partial }{\partial x^{m}}$ due to the chain rule. Without this term, however, we can soon finish the proof of $L_{X}Y=[X,Y]$.
So, my question is: is my result (1) actually incorrect? If both (1) and (2) are correct, how can I get from (1) to (2) by any chance?
Since $ Y_p = \sum_{i} Y^i(p) \frac{\partial}{\partial x_i}\bigg|_{p}$, we have $Y_{\varphi_t(p)} = \sum_{i} Y^i(\varphi_t(p)) \frac{\partial}{\partial x_i}\bigg|_{\varphi_t(p)}$. Now, the differential $D\varphi_{-t}$ applied to the basis vector $\frac{\partial}{\partial x_i}\bigg|_{\varphi_t(p)}$ at $\varphi_t(p)$ is mapped to $\frac{\partial \varphi_{-t}(\varphi_t(p))^j}{\partial x_i}\frac{\partial}{\partial x_j}\bigg|_{p}$, meaning \begin{align} D\varphi_{-t} Y_{\varphi_t(p)} = \sum_{i,j} \frac{\partial \varphi_{-t}(\varphi_t(p))^j}{\partial x_i}Y^i(\varphi_t(p))\frac{\partial}{\partial x_j}\bigg|_{p}. \end{align} So you are definitely right, in that Tu's equation is not correct. But from your formula, we can prove now prove the desired Theorem: Applying $\frac{d}{dt}\bigg|_{t=0}$, we find \begin{align} \frac{d}{dt}\bigg|_{t=0} \big(D\varphi_{-t} Y_{\varphi_t(p)} \big) &= \sum_{i,j} \bigg(\frac{d}{dt}\bigg|_{t=0} \bigg(\frac{\partial \varphi(-t,\varphi_t(p))^j}{\partial x_i} \bigg) Y^i(p) + \frac{\partial \varphi(0,p)^j}{\partial x_i} \frac{d}{dt}\bigg|_{t=0} Y^i(\varphi_t(p)) \bigg)\frac{\partial}{\partial x_j} \bigg|_{p} \\ &= \sum_{i,j} \bigg( -1\cdot\frac{\partial \varphi(0,p)^j}{\partial t\,\partial x_i} Y^i(p)+ \sum_k \varphi(0,p)^k\frac{\partial \varphi(0,p)^j}{\partial x_k\partial x_i} Y^i(p) + \frac{\partial \varphi(0,p)^j}{\partial x_i} \frac{\partial\varphi(0,p)^k}{\partial t}\frac{\partial Y^i(p)}{\partial x^k} \bigg) \frac{\partial}{\partial x_j} \bigg|_{p} \\ \end{align} Since $\varphi_0(x)=x$, we have $\frac{\partial \varphi(0,p)^j}{\partial x_i} = \delta^j_i$ and the second spatial derivatives vanish at $t=0$. Furthermore, we have $\frac{\partial \varphi(x,0)}{\partial t} = X_x$. Plugging that in, we find \begin{align} \frac{d}{dt}\bigg|_{t=0} \big(D\varphi_{-t} Y_{\varphi_t(p)} \big) &= \sum_{i,j} \bigg( -1\cdot\frac{\partial X^j}{\partial x_i} Y^i(p) +\sum_k \delta^j_i \frac{\partial\varphi(0,p)^k}{\partial t}\frac{\partial Y^i(p)}{\partial x^k} \bigg) \frac{\partial}{\partial x_j} \bigg|_{p} \\ &=\sum_{i,j} \bigg( -1\cdot\frac{\partial X^j}{\partial x_i} Y^i(p)+\sum_k \frac{\partial\varphi(0,p)^k}{\partial t}\frac{\partial Y^j(p)}{\partial x^k} \bigg) \frac{\partial}{\partial x_j} \bigg|_{p} \\ &=\sum_{i,j} \bigg( -1\cdot\frac{\partial X^j}{\partial x_i} Y^i(p) \bigg)\frac{\partial}{\partial x_j} \bigg|_{p} + \sum_{j,k} \bigg( X^k\frac{\partial Y^j(p)}{\partial x^k} \bigg) \frac{\partial}{\partial x_j}\bigg|_{p}\\ &=\sum_{i,j} \bigg( X^k\frac{\partial Y^j(p)}{\partial x^k} -\frac{\partial X^j}{\partial x_i} Y^i(p) \bigg)\frac{\partial}{\partial x_j} \bigg|_{p}\\ &=[X,Y]_p.\\ \end{align} This finishes the proof.