Thanks much for the response. I'm still working on this one. I'll look
at Nakahara's book, too -- learning a little more differential geometry is
a long-term project of mine (started 25 years ago :-) ) and another book
can't hurt...


On Thu, 15 Jul 2004 19:08:34 +0100, davidoff404 wrote:

sal wrote:
On Thu, 15 Jul 2004 12:20:57 -0400, sal wrote:

The other question I'm looking at here is how to prove, simply and
intuitively, that the form [XY - YX]

Brain bubbles -- I meant to say "the form [X,Y] = XY - YX".


really matches the definition of
the Lie derivative. The proof I'm looking at (p70 in Warner, Foundations
of Differentiable Manifolds and Lie Groups) certainly proves it but
doesn't do anything for my intuition at all. The definition is nice
because it's intuitively reasonable, whereas the bracket notation is
nice because it's easy to work with.


I haven't read that book so I'm not sure what the proof in it is like. I
usually turn to Nakahara's book "Geometry, Topology and Physics" when I
need a quick fix for differential geometry, so I'll give a brief outline
of a proof in that. Let me swap notation a bit and give my definition of
the Lie bracket: given a manifold M such that dim(M)=m, two vector fields
X,Y, and a smooth function f I say

[X,Y]f = X[Y[f]] - Y[X[f]].

This is the definition of the Lie bracket I have seen (and it's what I
attempted to write, above). I believe I've also seen the Lie derivative
defined this way. The Lie derivative definition given in Warner is nice
because it is much easier to see the motivation (for me, at least).


We know that we can write the vectors X and Y as

X = X^i @/@x^i,
Y = Y^i @/@y^i,

where @ represents partial differentiation. It's trivial to show that
[X,Y] is actually a vector field given by

(X^i @_i Y^j - Y^i @_i X^j) e_j

Right, because the partials commute, and the second partial terms cancel
when you apply it to a function and expand it out.


for a basis e_j. This is equivalent to demonstrating that the Lie
derivative of Y along X is actually the Lie bracket of the two vector
fields:

L_X Y = [X,Y].

Sorry, I'm probably being horribly dense here, but I didn't follow this.

Are you using the definition of the Lie derivative I gave to start with?
And how does showing that the Lie bracket [X,Y] is a vector field show
that it's identical to that derivative?

The definition I was using, to reiterate in more detail, was

L_X Y (f) = lim(t-0) { (1/t) * ((dX_-t(Y_(X_t(m)))(f)) - Y_m (f)) }

where X_t is the mapping defined by integrating the vector field X, with t
determining how far we go. dX_-t is perhaps better written as d(X_-t),
and it's the -- ah, nuts, I can't recall which of 17 names for the
derivative is appropriate here, so I'll just spell it out. It's the
linear map from the tangent space at X_t(m) to the tangent space at m
induced by the inverse diffeomorphism defined by integrating X. Make
sense? (I hope?)

In plain English, the Lie derivative is the rate at which Y changes
relative to X, along the path determined by integrating X. As such, it's
obviously a sensible derivative, and obviously related to the covariant
derivative -- and I still can't make the connection to the Lie bracket
seem "obvious".

When X and Y are nearly parallel, it's also clear from this definition
that L_X Y = -L_Y X, but when they're not nearly parallel it's not nearly
so obvious (to me, at least).


The interesting thing to note is that, contrary to what we might
initially suspect, [X,Y] is a vector field since it is first-order in
derivatives. This may be surprising at first glance since XY and YX are
second-order.

Yes, but as observed, the second partial terms cancel since the partials
commute.


The intuitive notion of the Lie bracket centers around the fact that it
is essentially a measure of the noncommutativity of two flows. Given
vector fields X and Y and flows a(s,x) and b(s,x) generated by the
vector fields, the Lie bracket measures the failure of the closure of
arbitrarily small parallelipipeds formed by the flows. The Lie bracket
is zero only when transport around parallelipipeds defined on these
flows closes. It might be helpful to remember that this is essentially
the basis of parallel transport that you're familiar with from general
relativity.

I'm going to have to think about this some more -- my mental picture still
isn't quite in focus.


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