How to calculate lie bracket in this case

Question

I am doing my final year bachelors course in differential geometry and I am not able to understand how exactly lie bracket should be calculated in the following case:

Definition is : If $X$ and $Y$ are vector fields on $M$, the lie brackets $[X,Y]$ of $X$ and $Y$ is the vector field on $M$ defined by $[X,Y]_p = X_p (Yf) - Y_p (Xf)$ , $p\in M$.

Compute the $X= z\frac{\partial{ }}{ \partial x} +\frac{\partial{ }}{ \partial z} $, $Y = \frac{\partial{ }}{ \partial y}+\frac{\partial{ }}{ \partial z}$ .

Prove that $[X,Y] = -\frac{\partial{ }}{ \partial x} $.

There is no $p$ here and also no $f$ in the question.

Can you please show your calculation of the the proof?

Answer 1

This is just a calculation. For the purpose of this computation you have to choose some arbitrary point $p\in M$ on your own, as well as some function $f$ - the formula is supposed to hold for any $p$ and any $f$. So you have to calculate, in some point $p$, $$(z\frac{\partial }{\partial x}+ \frac{\partial }{\partial z})(\frac{\partial }{\partial y}+\frac{\partial }{\partial z})f - (\frac{\partial }{\partial y}+\frac{\partial }{\partial z})(z\frac{\partial }{\partial x}+ \frac{\partial }{\partial z})f$$ If you expand all the terms you will note that - due to the fact that $\frac{\partial^2 }{\partial x\partial y }= \frac{\partial^2 }{\partial y\partial y }$ - and similarly for the other derivatives - only one term remains, namely $-\frac{\partial}{\partial x}f$.

Answer 2

Let $M$ be a differentiable manifold, and $X=\sum_{i=1}^nX^i\frac{\partial}{\partial x^i}$, $Y=\sum_{i=1}^nY^i\frac{\partial}{\partial x^i}$ the local expressions respect to a chart $(U,x^1,\ldots,x^n)$ of two smooth vector fields $X,Y$ over $M$. Then: $$[X,Y]=\sum_{j=1}^n(XY^j-YX^j)\frac{\partial}{\partial x^j}\:\:(\blacktriangle)$$ In fact $[X,Y]$ is a smooth vector field, so it is locally determined by its value on the open set $U$, i.e. $$([X,Y]f)|_U=[X,Y](f|_U)$$ So we can calculate the local expression on a single chart $(U,x^1,\ldots,x^n)$: \begin{align} [X,Y]f&=\sum_{i^1}^nX^i\frac{\partial}{\partial x^i}\left(\sum_{j=1}^nY^j\frac{\partial f}{\partial x^j}\right)-\sum_{j^1}^nY^j\frac{\partial}{\partial x^j}\left(\sum_{i=1}^nX^i\frac{\partial f}{\partial x^i}\right)\\&=\sum_{i,j=1}^n\left(X^i\frac{\partial Y^j}{\partial x^i}\frac{\partial f}{\partial x^j}+X^iY^j\frac{\partial^2f}{\partial x^i\partial x^j}-Y^j\frac{\partial X^i}{\partial x^j}\frac{\partial f}{\partial x^i}-Y^jX^i\frac{\partial^2f}{\partial x^j\partial x^i}\right)\\&=\sum_{i,j=1}^n \left(X^i\frac{\partial Y^j}{\partial x^i}\frac{\partial f}{\partial x^j}-Y^j\frac{\partial X^i}{\partial x^j}\frac{\partial f}{\partial x^i}\right) \end{align} If $X=z\frac{\partial}{\partial x}+\frac{\partial}{\partial z}$ and $Y=\frac{\partial}{\partial y}+\frac{\partial}{\partial z}$, using the formula $(\blacktriangle)$, we obtain that: \begin{align} [X,Y]&=\left(X(0)-Y(z)\right)\frac{\partial}{\partial x}+\left(X(1)-Y(0)\right)\frac{\partial}{\partial y}+\left(X(1)-Y(1)\right)\frac{\partial}{\partial z}\\&=-\frac{\partial}{\partial x} \end{align}