Lie derivatives: a simple idea behind a messy calculation

I want to write about Lie derivatives. Because finding good proofs for Lie derivatives in books and on the internet is a lost cause. Because they have caused me a world of pain. Because we could all do with less pain.

In all that is written below, we assume that all Lie derivatives are being found in the direction of the vector field $X$ , and that $\phi_t$ is the flow along this vector field.

What is a Lie Derivative? A Lie derivative is a derivative, but for things more complicated than functions. Basically, for a given vector field $X$ , $T(p+tX)=f(p)+tL_XT$ (roughly speaking). Here $T$ is a tensor, and a function is a special case of a tensor:

$f(p+tX)=f(p)+tL_Xf=f(p)+tD_Xf$ , where $D_Xf=L_Xf$ .

How do we then find a workable definition for a Lie derivative? Let us calculate the Lie derivative of a vector field.

Let me first try and explain what I will try and do. I know how functions and their derivatives behave. $f(p+tX)\approx f(p)+ t$ (derivative of f in the X direction). We will use this simple derivative rule of $f$ , wherever we can, to find out the what the Lie derivative of a vector field is.

For a vector field

$Y$ , we have $Y(p+tX)=Y(p)+t(L_XY)(p)$ .

This, however does not make sense, as $Y(p)+t(L_XY)(p)$ is a vector field at $p$ , while $Y(p+tX)$ is a vector field at $p+tX$ . Hence, the correct definition should be

$Y(p+tX)=(\phi^t)_*(Y(p))+t(\phi)^t_*(L_XY(p))$

This is equivalent to the following definition:

$L_XY(p)=\frac{(\phi)^{-t}_*(Y(p+tX))-Y(p)}{t}$

This, in turn, is equivalent to the formulation

$(L_XY(p))(f)=\frac{(\phi)^{-t}_*(Y(p+tX))(f)-Y(p)(f)}{t}$

We shall now try to simplify these terms.

$Y(p)(f)=Y(p+tX)(f(p+tX))+tD_{-X}(Y(p+tX)(f))$

to first order in $t$ . Similarly,

$(\phi)^{-t}_*(Y(p+tX))(f)=Y(p+tX)(f\circ \phi^{-t}_*)=Y(p+tX)(f(p+tX)-tX(f))$

Adding these two terms together, we get $(L_XY(p))(f)=X(Y(f))-Y(X(f))$

A minor technical point is that the left hand side is at $p$ , while the right hand side is at $p+tX$ . However, implicitly we have taken a limit $t\to 0$ . Hence, as the vector fields are continuous, we get

$(X(Y(f))-Y(X(f)))(p+tX)\to (X(Y(f))-Y(X(f)))(p)$

This expression can be generalized really simply to tensors. Let us find out what the Lie derivative of a $(0,n)$ tensor is:

$(L_XT)(Y_1,Y_2,\dots,Y_n)=\frac{(\phi)^{-t}_*(T(p+tX)((Y_1,Y_2,\dots,Y_n))-T(p)((Y_1,Y_2,\dots,Y_n))}{t}$

The terms can be simplified in a similar way as above: $T(p)(Y_1,\dots,Y_n)$ is a function. Hence, it is equal to

$T(p+tX)(Y_1,\dots,Y_n)-tD_X(T(p+tX)(Y_1,\dots,Y_n))$ . On the other hand,

$(\phi)^{-t}_*(T(p+tX)((Y_1,Y_2,\dots,Y_n))=T(Y_1\circ\phi^{-t}_*,\dots,Y_n\circ \phi^{-t})=T((Y_1-tL_XY_1)(p+tX),\dots,(Y_n-tL_XY_n)(p+tX))$

This is easily simplified to give

$(L_XT)(Y_1,\dots,Y_n)=D_X(T(Y_1,\dots,Y_n))-T(L_XY_1,\dots,Y_n)-\dots-T(Y_1,\dots,L_XY_n)$

Lie derivatives: a simple idea behind a messy calculation

Published by -

Leave a Reply Cancel reply

Share this:

Like this:

Related

Published by -

Leave a Reply Cancel reply