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

  • #1
Silviu
624
11
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!
 
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  • #2
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.
 
  • #3
Silviu said:
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.
 
  • #4
Orodruin said:
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.
 
  • #5
Silviu said:
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.
 

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