Undergrad Why Is the Lie Bracket a Vector Field on a Manifold?

[Silviu]

Hello! So I have 2 vector fields on a manifold $X=X^\mu\frac{\partial}{\partial x^\mu}$ and $Y=Y^\mu\frac{\partial}{\partial x^\mu}$ and this statement: "Neither XY nor YX is a vector field since they are second-order derivatives, however $[X, Y]$ is a vector field". Intuitively makes sense but I am not sure how to show it mathematically. I tried this: for a function f defined on M, $[X,Y]f = X[Y[f]]-Y[X[f]]=X^\mu\frac{\partial}{\partial x^\mu}[Y^\nu \frac{\partial}{\partial x^\nu}f]-Y^\mu \frac{\partial}{\partial x^\mu}[X^\nu \frac{\partial}{\partial x^\nu}f]$. It kinda makes sense that you first apply the vector on the right to f, which gives you a number, and the vector on the right remains just as an operator, and hence the whole stuff is a vector (but I am not sure if the results on the right can be put before the derivative on the left, such that we have an actual vector). For the other case $XY[f]$ I am not really sure how to combine the indices and what to get outside the derivatives. I guess I should have something like $X^\mu Y^\nu \frac{\partial}{\partial x^\mu x^\nu}$, I think, which is not a vector (would it be a tensor?), but I am not sure. Can someone help me a bit here? Thank you!

 

[Orodruin]

The case of XY is just part of the Lie bracket that you have. $X[Y[f]]$ is computed in exactly the same way, but will contain second order derivatives. The point is that the second order derivatives cancel between XY and YX.

 

[fresh_42]

Hello! So I have 2 vector fields on a manifold $X=X^\mu\frac{\partial}{\partial x^\mu}$ and $Y=Y^\mu\frac{\partial}{\partial x^\mu}$ and this statement: "Neither XY nor YX is a vector field since they are second-order derivatives, however $[X, Y]$ is a vector field". Intuitively makes sense but I am not sure how to show it mathematically. I tried this: for a function f defined on M, $[X,Y]f = X[Y[f]]-Y[X[f]]=X^\mu\frac{\partial}{\partial x^\mu}[Y^\nu \frac{\partial}{\partial x^\nu}f]-Y^\mu \frac{\partial}{\partial x^\mu}[X^\nu \frac{\partial}{\partial x^\nu}f]$. It kinda makes sense that you first apply the vector on the right to f, which gives you a number, and the vector on the right remains just as an operator, and hence the whole stuff is a vector (but I am not sure if the results on the right can be put before the derivative on the left, such that we have an actual vector).

I don't understand "can be put before the derivative on the left" but what you've written is correct.

For the other case $XY[f]$ I am not really sure how to combine the indices and what to get outside the derivatives. I guess I should have something like $X^\mu Y^\nu \frac{\partial}{\partial x^\mu x^\nu}$, I think, which is not a vector (would it be a tensor?), but I am not sure. Can someone help me a bit here? Thank you!

To see why $[X,Y]$ is a vector field while $XY$ is not, you should compute $[X,Y](f\cdot g)$ and check for the derivative property in comparison to $(XY)(f\cdot g)$. Coordinates are not needed here, which makes the computation a bit easier.

 

[Silviu]

The case of XY is just part of the Lie bracket that you have. $X[Y[f]]$ is computed in exactly the same way, but will contain second order derivatives. The point is that the second order derivatives cancel between XY and YX.

But if what I wrote there is correct, where exactly will the second order derivative cancel, and where the first order one remain? I am not sure how to continue what I did.

 

[fresh_42]

But if what I wrote there is correct, where exactly will the second order derivative cancel, and where the first order one remain? I am not sure how to continue what I did.

You have to use the Leibniz rule.