Applying Lie bracket to a smooth function.

203 Views Asked by At

Let $\textbf{X}$ and $\textbf{Y}$ be vector fields on an $n$-dimensional manifold, $M$. Let $f^{-1} : M \to \mathbb{R}^n$.

We can represent $$\textbf{X} = a^1 \frac{\partial}{\partial x^1} + \dots + a^n \frac{\partial}{\partial x^n}$$ and $$ \textbf{Y} = b^1 \frac{\partial}{\partial x^1} + \dots + b^n \frac{\partial}{\partial x^n}$$ where $(x^1, \dots, x^n)$ is some coordinate representation.

I am found the following property of the Lie bracket, $$[\textbf{X}, \textbf{Y}]f^{-1} = \textbf{X} \textbf{Y} f^{-1} - \textbf{Y} \textbf{X} f^{-1}$$

However, I am not sure how to interpret $$XY$$ I am looking for an example/explanation of how to use the above property.

1

There are 1 best solutions below

0
On BEST ANSWER

If an $X$ is multiplied on an arbitrary function (here: $f^{-1}$) then this function is derived in the direction of $X$. Hence $XYf^{-1}=X(Yf^{-1})$ is the derivative in the direction of $X$ applied on $Yf^{-1}$ and $Yf^{-1}$ is the derivative of $f^{-1}$ projected into the direction of the vector field $Y$.

Hence $XY(...)$ means: At first $(...)$ is derived in $Y$-direction and after that it is derived in $X$-direction.

Simple example: If $X= \frac{\partial}{\partial x}, Y= \frac{\partial}{\partial y}$ then it can be shown that $[X,Y]=XY-YX=0$. This means that $XY=YX$ and therefore two-fold derivatives of functions are Independent of the direction in which the function is derived at first (in the case when there are defined only the derivatives $X$ and $Y$ on the manifold).