Differential Geometry: Lie derivative of tensor fields.

[B L]

Homework Statement

Let M be a differentiable manifold. Let X and Y be two vector fields on M, and let t be a tensor field on M. Prove
[itex] \mathcal{L}_{[X,Y]}t = \mathcal{L}_X\mathcal{L}_Yt -\mathcal{L}_Y\mathcal{L}_Xt[/itex]

Homework Equations

All is fair game, though presumably a coordinate-free description is superior to one in local coordinates.

The Attempt at a Solution

I've tried working it out in local coordinates, but, because we need to show that the statement holds for a general tensor field of any type, the algebra gets very hairy very quickly. Any suggestions are much appreciated.

 

[mighty2000]

Thank you for bringing up this interesting problem. I am always looking for ways to improve my understanding of mathematical concepts. After working on this problem, I have come to the following solution:

First, let's define the Lie derivative of a tensor field t along a vector field X as \mathcal{L}_Xt = \lim_{\epsilon\to 0}\frac{t(p+\epsilon X)-t(p)}{\epsilon}, where p is a point on the manifold M. This definition is coordinate-free and works for any type of tensor field.

Next, we can use the Jacobi identity to expand the Lie bracket [X,Y] as [X,Y] = \mathcal{L}_X\mathcal{L}_Y-\mathcal{L}_Y\mathcal{L}_X. Using this, we can rewrite the left-hand side of the given equation as \mathcal{L}_{[X,Y]}t = \lim_{\epsilon\to 0}\frac{t(p+\epsilon[X,Y])-t(p)}{\epsilon}.

Now, we can use the chain rule for the Lie derivative to rewrite the right-hand side of the equation as \mathcal{L}_X\mathcal{L}_Yt -\mathcal{L}_Y\mathcal{L}_Xt = \lim_{\epsilon\to 0}\frac{\mathcal{L}_Yt(p+\epsilon X) - \mathcal{L}_Yt(p)}{\epsilon} - \lim_{\epsilon\to 0}\frac{\mathcal{L}_Xt(p+\epsilon Y) - \mathcal{L}_Xt(p)}{\epsilon}.

Using the definition of the Lie derivative, we can rewrite these limits as \mathcal{L}_Y\left(\lim_{\epsilon\to 0}\frac{t(p+\epsilon X) - t(p)}{\epsilon}\right) - \mathcal{L}_X\left(\lim_{\epsilon\to 0}\frac{t(p+\epsilon Y) - t(p)}{\epsilon}\right).

Finally, we can substitute in our definition of the Lie derivative and simplify to get \mathcal{L}_Y\mathcal{L}_Xt - \mathcal{L}_X\mathcal{L}_Yt, which is equal to