# Shutz's derivation of the Lie Derivative of a vector field

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I have a question about Bernard Shutz's derivation of the Lie derivative of a vector field in his book Geometrical Methods for Mathematical Physics.

I will try to reproduce part of his argument here.
Essentially, we have 2 vector fields V and U which are represented by $\frac{d}{d\lambda}$ and $\frac{d}{d\mu}$ respectively. The Lie derivative $L_{\vec{V}}\vec{U}$ is defined so that we Lie drag the vector field U from a point Q with parameter $\lambda_0+\Delta\lambda$ back to a point P with parameter $\lambda_0$ on V, creating a new vector field U* and then "compare" it with U which is at the point $\lambda_0$ (by "compare" I of course mean the usual thing you do in calculus which is to take the delta lambda to 0).

To do this, Shutz Taylor expanded U*(Q) and then rearranged terms to get U*(P) in terms of U*(Q) since U*(Q)=U(Q) which makes sense. The part of his derivation that doesn't make sense to me is that in the Taylor expansion of U*(Q), you obviously get a term like
$$\Delta\lambda\frac{d}{d\lambda}\frac{d}{d\mu^*}$$ (evaluated at P)

Shutz uses the Lie dragging condition to say that the Lie bracket between U* and V is 0, so he just reverses the order of that derivative. In the end you get that the Lie derivative of U with respect to V is the Lie bracket [V,U]. What I don't get is, even though Shutz COULD reverse the order of the derivatives in the above expression (perfectly valid due to the Lie Bracket being 0), there's no obvious a priori reason to me that he SHOULD.

If he DIDN'T reverse the order, and you carry out the calculations, then at the end you don't get the Lie Bracket, you get 0 (it seems to me). This seems very weird to me. Can somebody shed some insight? Why did he HAVE TO switch the order?

I didn't want to type out the full derivation, so I may have bungled something somewhere...Hopefully this much information will be enough for you guys to help me.

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George Jones
Staff Emeritus
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I don't have this book, but I have looked at page 77 using amazon.com. Do you mean the following?
$$\lim_{\Delta \lambda \rightarrow 0}\left( \frac{d}{d\lambda }\frac{d}{d\mu }f-\frac{d}{d\mu ^{\ast }}\frac{d}{d\lambda }f\right)$$
Now, the difference between $\mu^*$ and $\mu$ is clearly a term of first order $\Delta \lambda$, which means we can replace $\mu^*$ by $\mu$

If the order had not been reversed, then
$$\lim_{\Delta \lambda \rightarrow 0}\left( \frac{d}{d\lambda }\frac{d}{d\mu }f-\frac{d}{d\lambda }\frac{d}{d\mu ^{\ast }}f\right)$$
Now, the difference between $\mu^*$ and $\mu$ is clearly a term of first order $\Delta \lambda$, which means we can replace $\mu^*$ by $\mu$

would give zero.

Because of differentiation by $\lambda$ after replacement, I don't think
Now, the difference between $\mu^*$ and $\mu$ is clearly a term of first order $\Delta \lambda$, which means we can replace $\mu^*$ by $\mu$

applies in the second case.

Gold Member
Ok, can you explain why that statement doesn't work if the order is reversed?

I guess I didn't quite understand that statement like I thought I did. Thanks. =]

George Jones
Staff Emeritus
Gold Member
An example might help.

If $U = x \frac{\partial}{\partial y}$ and $V = y \frac{\partial}{\partial x}$, what is the vector field $U^*$? I think I have worked this out (maybe not), but I don't how well I can explain what I have done.

First, find the congruence of curves to which $V$ is tangent. This, I can explain.

Gold Member
Ok, so, for V we have the equation:

$$y\frac{\partial}{\partial x}x^i=\frac{dx^i}{d\lambda}$$

Assuming we just have 2 directions,

$$\frac{dx}{d\lambda}=y$$

and

$$\frac{dy}{d\lambda}=0$$

Giving:

$$y=A$$
$$x=y\lambda+B$$

For constants A and B. So, the congruence curve for V is simply the x-coordinate lines?

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George Jones
Staff Emeritus
Gold Member
Yes, with caveat that the coordinate lines are the images of the curves. Different curves (which are maps) can have the same image, but can traverse the image at different "speeds", i.e., have different tangent vectors at the same points of the image.

I forgot to say that my example is for vector fields on $\mathbb{R}^2$.

What is the the vector field $U^*$?

Gold Member
Ok, so U has congruence curves, in analogy with V:

$$x=C$$
$$y=x\mu+D$$

These are just the y-coordinate lines, and they "move" at speed "x".

So, dragging the x=const lines a parameter $\Delta\lambda$ along the congruence of V, should give me...lines of slope 1, right?

Congruence of U* being:

$$y=x+E$$

I did that last step intuitively...the math was making me confused...

How would I put it in a parametric form? That part is confusing me...>.>