Intuitively, it appears that the covariant derivative with respect to a
particular vector V is just the Lie derivative taken along a geodesic to
which V is tangent.

But I haven't gotten the algebra to work out, so I'm not quite sure.

Comments? Am I barking at the wrong tree here?


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

Intuitively, it appears that the covariant derivative with respect to a
particular vector V is just the Lie derivative taken along a geodesic to
which V is tangent.

But I haven't gotten the algebra to work out, so I'm not quite sure.

Comments? Am I barking at the wrong tree here?

I just gave it one last nudge and realized that a simple not-too-rigorous
proof can be had trivially.

I had been working directly from the definition of the Lie derivative,

L_X Y (f) = (d/dt)|_(t=0)((dX_-t(Y_(X_t(m)))(f))

which looks a whole lot worse in flat Ascii than on paper :-)

But turning the whole problem around, if we use the more common view of
the Lie derivative,

L_X Y = [XY - YX]

and fiddle a little, it's obvious that it's covariant. Then if we
evaluate it in flat space with e_a in place of X, it comes out to a simple
directional derivative along e_a, which is the same as the "covariant
derivative" wrt e_a in flat space.

So, since both derivatives covary, they must both be equal in all
coordinate systems (as long as the integral curve of X is a geodesic).

.... right...?

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


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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] 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]].

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

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].

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.

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.

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.


--
I can be contacted through http://www.physicsinsights.org

a little clarrification - you don't take the covariant derivative
w.r.t. a vector, but of a vector w.r.t. a neighbourhood. you do take
the Lie derivative w.r.t. a one parameter group of diffeo's and this
is related to the vector field. all this can be found in wald's book,
or nakahara as mentioned before.

a counter to the statemtent:

"Intuitively, it appears that the covariant derivative with respect to
a
particular vector V is just the Lie derivative taken along a geodesic
to
which V is tangent."

is simple one! consider the covariant derivative of the metric and the
Lie derivative of the metric and you get the point.

or, have a a look at equation 10.9.10 in weinberg's book and you'll
see that the Lie only reduces to the covariant in the special case
where certain terms vanish.

hope that helps.

paul

On Fri, 16 Jul 2004 03:29:33 -0700, paul wrote:

a little clarification - you don't take the covariant derivative w.r.t. a
vector, but of a vector w.r.t. a neighbourhood.

Ah -- I think my sloppiness in terminology has bitten me here.

When I talked about the covariant derivative of a vector, Y, with respect
to a vector, V, I meant

D_V Y^a = (@Y^a/@x^b)*V^b + C^a_bc * V^b * Y^c

which I would have described as the covariant derivative of Y with
respect to the vector V, taken at some point P (which doesn't appear in
the expression).

Of course, you need to know the metric on the neighborhood to evaluate
that, but 'V' can be defined at just one point -- there's no implication
that it's a member of a vector field.


you do take the Lie
derivative w.r.t. a one parameter group of diffeo's and this is related
to the vector field.

Right -- the vector field determines the "line" upon which you evaluate
it, as well as the linear backmap you use to pull the two tangent vectors
into the same space so you can take their difference.

all this can be found in wald's book, or nakahara as
mentioned before.

Duh -- I had totally forgotten that Wald covers this; his book was just
sitting on a shelf here! Thanks for the reminder.

As a rule, I've found it's easier to learn math from physics books, if any
happen to cover it. As Bilge has pointed out, physics books invariably
provide some sort of physical motivation for the math; OTOH math books
frequently don't.


a counter to the statemtent:

"Intuitively, it appears that the covariant derivative with respect to a
particular vector V is just the Lie derivative taken along a geodesic to
which V is tangent."

is simple one! consider the covariant derivative of the metric and the
Lie derivative of the metric and you get the point.

I dunno -- I'll have to think about that... Of course the covariant
derivative of the metric is zero.

But, consider the Lie derivative of the metric in a locally flat
coordinate system, taken along a path through the origin which is
parallel to a basis vector. Isn't that derivative necessarily zero, also,
as a result of the local flatness? The second derivatives shouldn't enter
into it, and the firsts are all zero.

That would seem to imply that whenever we take the Lie derivative of the
metric WRT a one-parameter family of diffeo's defined such that the path
we're evaluating it along is a geodesic, the result should be zero.

No...?


or, have a a look at equation 10.9.10 in weinberg's book and you'll see
that the Lie only reduces to the covariant in the special case where
certain terms vanish.

Well this is a first! I can't recall anyone ever posting a reference to
Weinberg in this newsgroup before.

His approach is a bit different -- I'll have to stare at this for a while
to make sense of it.


hope that helps.

Thanks. The picture is gradually getting clearer.


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I can be contacted through http://www.physicsinsights.org