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I started reading about Lie derivative on vector fields and its properties, found an exercise, but i have doubts about my solution.

Given are two vector fields $X_{1}=\frac{\partial }{\partial x_{1}} + x_{2}\frac{\partial }{\partial x_{3}} + x_{2}^{2}\frac{\partial }{\partial x_{4}}$ and $X_{2}=\frac{\partial }{\partial x_{2}}$. Calculate $\left [ X_{1},\left [ X_{1}, X_{2} \right ] \right ]$.

OK, i start with $\left [ X_{1},\left [ X_{1}, X_{2} \right ] \right ] = \left [ X_{1}, \left [ X_{1}X_{2}-X_{2} X_{1}\right ] \right ] = X_{1}X_{1}X_{2} - X_{1}X_{2}X_{1} - X_{1}X_{2}X_{1} + X_{2}X_{1}X_{1} = \left [ X_{1}, X_{1}X_{2} \right ]- \left [ X_{1}, X_{2}X_{1} \right ]$

The book with the answers says that the result of $\left [ X_{1},\left [ X_{1}, X_{2} \right ] \right ]$ should be $0$.

Is my idea correct? Can we replace $X_{1}X_{2}$ with $X_{2}X_{1}$? I am not sure about this. The other thing that bothers me is that i didn't use the definitions of $X_{1}$ and $X_{2}$ the way they were given in the exercise. Can somebody help me? Thank you very much

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up vote 5 down vote accepted

Your calculation is correct so far, but to proceed further you need to substitute in the specific $X_1$ and $X_2$ that were given. So far your calculation works for any $X_1$ and $X_2$. Remember that $\partial/\partial x_i$ and $\partial/\partial x_j$ commute.

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Thank you very much, Ted! – Lullaby Feb 3 '13 at 22:16
Is it true that for these specific $X_1$ and $X_2$, (writing $\partial_z$ for $\frac\partial{\partial z}$), we have $$ \begin{align} X_2X_1 &= 1\cdot \partial_{x_3}+2x_2\cdot\partial_{x_4} \\ X_1X_2 &= 0 & ? \end{align} $$ – Berci Feb 3 '13 at 22:18
@Berci That can't be right... $X_2 X_1$ is a second-order operator. It has terms like $\partial_{x_2} \partial_{x_1}$ in it, not $\partial_{x_3}$. – Ted Feb 3 '13 at 23:13

You must understand the definiton:

For smooth vector fields $X$ and $Y$, $XY:C^\infty(M)\to C^\infty(M)$ is the linear transformation (not necessarily a vector field) $$f\to X(Yf)$$ (remember that $Yf\in C^\infty(M)$).

Hence, as @Ted said, $\frac{\partial}{\partial x_i}\frac{\partial}{\partial x_j}=\frac{\partial}{\partial x_i}\frac{\partial}{\partial x_j}$.

The formula $$[X,Y] = \sum_{j=1}^n\sum_{i=1}^n\left(X_i\frac{\partial Y_j}{\partial x_i} - Y_i\frac{\partial X_j}{\partial x_i}\right)\frac{\partial}{\partial x_j},$$ for smooth vector fields $$X=\sum_{i=1}^nX_i\frac{\partial}{\partial x_i}\quad\text{ and }\quad\sum_{i=1}^nY_i\frac{\partial}{\partial x_i}$$ and smooth functions $X_i$ and $Y_i\in C^\infty(M)$, follows from this definition. In fact, for $f\in C^\infty(M)$, we have that $$\begin{array}{rcl} [X,Y]f & = & X(Yf) - Y(Xf) \\ & = & X\left(\sum_{j=1}^nY_j\frac{\partial f}{\partial x_j}\right) - Y\left(\sum_{j=1}^nX_j\frac{\partial f}{\partial x_j}\right) \\ & = & \sum_{i=1}^nX_i\frac{\partial}{\partial x_i}\left(\sum_{j=1}^nY_j\frac{\partial f}{\partial x_j}\right) - \sum_{i=1}^nY_i\frac{\partial}{\partial x_i}\left(\sum_{j=1}^nX_j\frac{\partial f}{\partial x_j}\right) \\ & = & \sum_{i=1}^nX_i\sum_{j=1}^n\left(\frac{\partial Y_j}{\partial x_i}\frac{\partial f}{\partial x_j} + Y_j\frac{\partial^2 f}{\partial x_i\partial x_j}\right) - \sum_{i=1}^nY_i\sum_{j=1}^n\left(\frac{\partial X_j}{\partial x_i}\frac{\partial f}{\partial x_j} + X_j\frac{\partial^2 f}{\partial x_i\partial x_j}\right) \\ & = & \sum_{i=1}^n\sum_{j=1}^nX_i\frac{\partial Y_j}{\partial x_i}\frac{\partial f}{\partial x_j} - \sum_{i=1}^n\sum_{j=1}^nY_i\frac{\partial X_j}{\partial x_i}\frac{\partial f}{\partial x_j} \\ & = & \sum_{j=1}^n\sum_{i=1}^n\left(X_i\frac{\partial Y_j}{\partial x_i} - Y_i\frac{\partial X_j}{\partial x_i}\right)\frac{\partial f}{\partial x_j} \\ & = & \left(\sum_{j=1}^n\sum_{i=1}^n\left(X_i\frac{\partial Y_j}{\partial x_i} - Y_i\frac{\partial X_j}{\partial x_i}\right)\frac{\partial}{\partial x_j}\right)f. \end{array}$$

Using this formula in your case, we get $[X_1,X_2] = 0$ and, hence, $[X_1,[X_1,X_2]]=0$.

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