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Lie vs. covariant derivative

by wandering.the.cosmos
Tags: covariant, derivative
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wandering.the.cosmos
#1
Jan5-07, 09:48 PM
P: 23
Is there any relationship between the Lie ([itex]\pounds[/itex]) and covariant derivative ([itex]\nabla[/itex])?

Say I have 2 vector fields V, W and a metric g, the Lie and covariant derivative of W along V are:

[tex]\pounds_{V}W = [V,W][/tex]
[tex]V^\alpha \nabla_\alpha W^\mu = V^\alpha \partial_\alpha W^\mu + V^\alpha \Gamma^\mu_{\alpha \nu} W^\nu[/tex]

which appear rather different.

But conceptually I thought both derivatives help us to define what parallel transporting a vector in a general manifold means? Is there a good place to read about such issues?

Thanks!!
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Ratzinger
#2
Jan6-07, 12:47 PM
P: 302
To take covariant derivatives you need a metric structure on your manifold, which is not needed when taking Lie derivatives.

I like Schutz book on diffgeometry.
coalquay404
#3
Jan6-07, 01:35 PM
P: 218
Quote Quote by Ratzinger View Post
To take covariant derivatives you need a metric structure on your manifold, which is not needed when taking Lie derivatives.
Essentially, this is the right answer. In order to construct a derivative you generally need to have a sufficient amount of extra structure on your manifold; the more sophisticated the derivative, the more sophisticated this extra structure needs to be.

For example, the simplest possible derivative operator on a manifold is the coordinate partial derivative [itex]\partial/\partial x^i[/itex]. In order to construct this derivative all one needs is a coordinate chart in an open neighbourhood of some point.

At the next level of complexity is the Lie derivative. Since the Lie derivative of, for example, one vector field along another is simply equal to their Lie bracket you need to have both a coordinate chart and some means to define the idea of the 'flow' of a vector field. The generalization to Lie derivatives of higher order tensors along the flow of a vector field is then a straightforward generalization of this idea.

At an even higher level of complexity is the idea of the exterior derivative. Since this acts on differential forms, one needs to have not only a coordinate chart, but also a definition of the vector space of differential forms at a point and, at a further level of generalization, a means to define the space of differential [itex]p[/itex]-forms over a manifold.

Finally, one arrives at the covariant derivative. As stated above, one needs a metric structure on the manifold in order to define the covariant derivative (depending on your point of view and your upbringing, you may prefer to speak of the need to have a linear connection instead of a metric structure; these are, however, equivalent requirements).


Quote Quote by Ratzinger View Post
I like Schutz book on diffgeometry.
Schutz's book is good, but I think Nakahara's book contains possibly an even better discussion of the different derivative operators.

Hurkyl
#4
Jan6-07, 02:05 PM
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Lie vs. covariant derivative

Quote Quote by coalquay404 View Post
At the next level of complexity is the Lie derivative. Since the Lie derivative of, for example, one vector field along another is simply equal to their Lie bracket you need to have both a coordinate chart and some means to define the idea of the 'flow' of a vector field.
[X, Y] does not depend on a choice of coordinate chart.
Chris Hillman
#5
Jan6-07, 02:36 PM
Sci Advisor
P: 2,340
Hi, Wandering,

Quote Quote by wandering.the.cosmos View Post
Is there any relationship between the Lie ([itex]\pounds[/itex]) and covariant derivative ([itex]\nabla[/itex])?
Yes:

[tex] {\mathcal L}_{\vec{X}} \vec{Y} = \left[ \vec{X}, \vec{Y} \right] = \nabla_{\vec{X}} \vec{Y} - \nabla_{\vec{Y}} \vec{X} [/tex]

(In differential geometry books, you'll see this mentioned as one of the definining properties of a Riemannian connection [itex]\nabla[/itex].)

Quote Quote by wandering.the.cosmos View Post
But conceptually I thought both derivatives help us to define what parallel transporting a vector in a general manifold means? Is there a good place to read about such issues?
Yes, indeed. Frankel, Geometry of Physics, Cambridge University Press, 1997 (there is now a second edition) is readable and well-illustrated and a gold-mine of information and insight both geometrical and physical.

Following are some elaborations on what coalquay told you:

Quote Quote by coalquay404 View Post
For example, the simplest possible derivative operator on a manifold is the coordinate partial derivative [itex]\partial/\partial x^i[/itex]. In order to construct this derivative all one needs is a coordinate chart in an open neighbourhood of some point.
And in differential geometry textbooks (and in fact wherever vector fields are used in modern mathematics), you'll find vector fields described as first order linear partial differential operators acting on functions, for example on ordinary three-dimensional Euclidean space we might write
[itex] \vec{W} = W^x\, \partial_x + W^y \, \partial_y + W^z \, \partial_z [/itex]
where [itex]W^x, \; W^y, \; W^z[/itex] are functions and where [itex]x, \, y, \, z[/itex] are Cartesian coordinates.

Differential geometry is, if you like, very broadly speaking the study of rates of change on general smooth manifolds. Note surprisingly, derivatives of various kinds occur everywhere in this subject! And since manifolds occur everywhere in modern physics, this means that derivatives of various kinds occur everywhere in physics!

Here's a short list off the top of my head:

1. Darboux derivatives (act on maps between manifolds),

2. Vector fields (act on functions),

3. Lie derivatives (wrt some vector field; act on vector fields, or even on tensor fields),

4. Exterior derivatives (act on exterior forms),

5. Covariant derivatives (wrt some vector field; act on vector fields, or even on tensor fields).

Exterior forms also have a differential character, e.g. the exterior derivative of a function is a one-form dual to the gradient from undergraduate vector calculus.

As coalquay said, the notion of levels of structure is crucial here: Riemannian manifolds are defined by adding additional "geometric structure" to smooth manifolds, and so on.

There are various relations between these which are important. In particular, if [itex]\omega[/itex] is a one-form, then [itex]d\omega[/itex] is a two-form which can be defined by giving its value on a pair of vector fields:

[tex] d\omega \left( \vec{X}, \vec{Y} \right) = \vec{X} \left( \omega \left( \vec{Y} \right) \right) - \vec{Y} \, \left( \omega \left( \vec{X} \right) \right) - \omega \left( \left[ \vec{X},\vec{Y} \right] \right) [/tex]

In the case of left-invariant vector fields on a Lie group, we obtain a very simple relationship:

[tex] \nabla_{\vec{X}} \vec{Y} = 1/2 \, {\mathcal L}_{\vec{X}} \vec{Y} [/tex]

Going back to the Lie derivative, this has to do with a certain kind of asymmetry in the covariant derivative, between the vector field doing the Lie dragging and the vector field which is being dragged. One could also ask about possible failure of covariant derivatives to commute, and then (by definition)

[tex]R \left( \vec{X}, \vec{Y}; \vec{Z}) = \left( \nabla_{\vec{X}} \nabla_{\vec{Y}} - \nabla_{\vec{Y}} \nabla_{\vec{X}} - \nabla_{\left[ \vec{X}, \vec{Y} \right]} \right) \, \vec{Z} [/tex]

where [itex]R[/itex] is the Riemann curvature tensor, and where in the equation above, both sides define the same vector field. Here, the combination

[tex] \nabla_{\vec{X}} \nabla_{\vec{Y}} - \nabla_{\vec{Y}} \nabla_{\vec{X}} - \nabla_{\left[ \vec{X}, \vec{Y} \right]} [/tex]

is sometimes called the curvature operator. The commutator or Lie bracket is needed, in general, in order to "close up the quadrilateral"; this bracket vanishes if [itex]\vec{X}, \, \vec{Y}[/itex] are two of the coordinate vector fields in some chart.

Quote Quote by coalquay404 View Post
Schutz's book is good, but I think Nakahara's book contains possibly an even better discussion of the different derivative operators.
The second book is probably

Nakahara, Geometry, Topology, and Physics, IOP, 1990.

I'd consider this too sketchy to meet your needs. Frankel should be much better for you, IMO.

Widely used differential geometry textbooks include:

Boothby, An Introduction to Differentiable Manifolds and Riemannian Geometry, 2nd Ed., Academic Press, 1986.

Dubrovin, Fomenko, Novikov, Modern Geometry-- Methods and Applications (two volumes), 2nd Ed., Springer, 1992.

This thread should probably be moved to the differential geometry board, BTW, since I think this is really more about mathematics than about gtr. All of the above is useful in areas of physics far removed from gravitation physics.
Chris Hillman
#6
Jan8-07, 10:27 PM
Sci Advisor
P: 2,340
Good grief, I forgot to mention the torsion tensor at the same time that I mentioned the Riemann tensor:
[tex] T \left( \vec{X}, \vec{Y} \right) = \nabla_{\vec{X}} \vec{Y} - \nabla_{\vec{Y}} \vec{X} - \left[ \vec{X}, \vec{Y} \right] [/tex]

The curvature and torsion tensors defined in terms of vector fields as above are related to slightly more abstract notions of curvature and torsion two-forms which are defined for any affine connection. For a general affine connection, both the curvature and torsion might be nonvanishing. For a "Levi-Civita connection" (really due to Christoffel; it is probably best to think of this notion as the notion of a "(semi-)Riemannian connection"), the curvature is generally nonzero but the torsion vanishes. (Which is why, in the previous post, I mentioned that I was using a Levi-Civita connection.) For a "Weitzenboeck connection" (used in teleparallel theories of gravitation), the torsion is generally nonzero but the curvature vanishes. In gauge theories, in the first case the connection plays the role of a "potential" and the curvature plays the role of a "field strength". This is a solid and very useful analogy, but from other points of view, or in other contexts, it can also be somewhat misleading, as for example in discussing gravitoelectromagnetism (GEM).
wandering.the.cosmos
#7
Jan22-07, 03:16 PM
P: 23
Thanks for the replies so far. I have T. Frankel's Geometry of Physics, and will be looking at it in due time.

I'd like to ask a more specific question regarding this Lie vs. covariant derivatives. Say I have 2 vector fields V, W and a metric g. I also have the integral curve of V passing through some point P; i.e. there is a parameter [itex]\lambda[/itex] such that [itex]dX^\mu/d\lambda\vert_p = V[p][/itex]. Suppose I wish to find the components of W some parameter [itex]\xi[/itex] down from P along the integral curve of V. Do I do

[tex]
\exp\left[\xi V^\alpha \nabla_\alpha\right] W^\beta
[/tex]

or do I do

[tex]
\left( \exp\left[\xi \pounds_V\right] W\right)^\beta\?
[/tex]

and how are these related, if at all?
zenmaster99
#8
Mar12-07, 11:16 PM
P: 20
I have a related question:

I was not happy with most physics books' implicit definition of a covariant derivative so I started looking at lifts of curves in vector bundles... which became associated to principal fibre bundles (because Morita taught me that one may not have the lift of a curve defined all the way to the fibre over the endpoint of the curve unless the fibre is compact)... which took me into Lie groups and algebras... and Adjoint (and adjoint) representations...

And now I'm looking for the path from a connection form in a principal fibre bundle back down to the Christoffel symbols defined on the associated vector bundle which is compatible with the metric.

Oh how I wish I had just listened to Wald and his algebraic argument.

Is there any book (or books) that do this without too much more pain?

Much thanks,

ZM
bchui
#9
Jul21-07, 11:02 PM
P: 42
Quote Quote by Ratzinger View Post
To take covariant derivatives you need a metric structure on your manifold, which is not needed when taking Lie derivatives.
.
So, if there is a metric structure [tex]g(X,Y)[/tex] on the manifold, would the two derivatives be the same?
The definiteion of connection [tex]\nabla[/tex] does not require a metric structure. what is the relationship between a connection, covariant derivative and Lie derivative?
nickj1
#10
Dec6-11, 09:01 AM
P: 2
It's been half a decade since anyone's posted on this thread, so I don't know if it's appropriate for me to. But this seems to be the go-to thread for Lie vs. covariant for everyone at pf and even across the Web. So, since I'm struggling with trying to understand these issues myself, I hope it's alright if I ask some questions I still have about some of these posts. Maybe if a physicist helps me, we can improve the thread even more for all those people who are linked here.

Quote Quote by coalquay404
Quote Quote by Ratzinger
To take covariant derivatives you need a metric structure on your manifold, which is not needed when taking Lie derivatives.
Essentially, this is the right answer. In order to construct a derivative you generally need to have a sufficient amount of extra structure on your manifold; the more sophisticated the derivative, the more sophisticated this extra structure needs to be.

For example, the simplest possible derivative operator on a manifold is the coordinate partial derivative [itex]\partial/\partial x^i[/itex]. In order to construct this derivative all one needs is a coordinate chart in an open neighbourhood of some point.

At the next level of complexity is the Lie derivative. Since the Lie derivative of, for example, one vector field along another is simply equal to their Lie bracket you need to have both a coordinate chart and some means to define the idea of the 'flow' of a vector field. The generalization to Lie derivatives of higher order tensors along the flow of a vector field is then a straightforward generalization of this idea.
Quote Quote by Hurkyl
Quote Quote by coalquay404
At the next level of complexity is the Lie derivative. Since the Lie derivative of, for example, one vector field along another is simply equal to their Lie bracket you need to have both a coordinate chart and some means to define the idea of the 'flow' of a vector field.
[X, Y] does not depend on a choice of coordinate chart.
There's an ambiguity here. Indeed, the Lie derivative does not depend on a choice of a particular coordinate chart. For that matter, neither does the "simplest" derivative (which should really have been listed as the directional derivative, with the "partial" derivative just a particular choice thereof--in this case, a somewhat conceptually distracting one). What all derivatives do depend on, however, is the generic ability to create such a coordinate chart about any given point--manifold structure, in other words. So I see what coalquay is saying. There's been a much larger problem here, however, as brought out especially by what comes next:

Quote Quote by coalquay404
At an even higher level of complexity is the idea of the exterior derivative. Since this acts on differential forms, one needs to have not only a coordinate chart, but also a definition of the vector space of differential forms at a point and, at a further level of generalization, a means to define the space of differential [itex]p[/itex]-forms over a manifold.
I don't get this whole levels of "complexity" thing. I know I can sort directional, Lie, and exterior derivatives by (for example) what sort of objects they can operate on, and by that yardstick maybe directionals could be considered the least "complex" because they only work on (0,0)s. But that's the only thing I can think of. How it's put here--where e.g. one needs to have such-and-such for the Lie derivative, but for the exterior derivative you need everything you need for the Lie, plus some other stuff--is just baffling to me. I'm sure coalquay is right, but I'm not understanding it.

Note, in fact, that what is needed was explicitly described as structure. Recall that first bit:

Quote Quote by coalquay404
In order to construct a derivative you generally need to have a sufficient amount of extra structure on your manifold; the more sophisticated the derivative, the more sophisticated this extra structure needs to be.
But isn't it true that for all of these derivatives, not one less than the others, one needs in the way of structure nothing more than the (sufficiently differentiable) manifold itself? It's only the covariant that needs a little extra something-something. Speaking of which...

Quote Quote by coalquay404
Finally, one arrives at the covariant derivative. As stated above, one needs a metric structure on the manifold in order to define the covariant derivative (depending on your point of view and your upbringing, you may prefer to speak of the need to have a linear connection instead of a metric structure; these are, however, equivalent requirements).
Does one really need a metric to have an appropriate connection? I was under the impression that that was a stronger requirement. (Would appreciate some guidance here too; GR textbooks tend to gloss over this matter, since a metric is required anyway.)

Finally:

Quote Quote by Chis Hillman
...And since manifolds occur everywhere in modern physics, this means that derivatives of various kinds occur everywhere in physics!

Here's a short list off the top of my head:
...
3. Lie derivatives (wrt some vector field; act on vector fields, or even on tensor fields),
...
5. Covariant derivatives (wrt some vector field; act on vector fields, or even on tensor fields).
Isn't the whole point, so to speak, of covariant derivatives that they aren't "wrt some vector field"? It does have a "slot" for "direction," so that would specify a vector at that given point, but that's about it. It's an inherent artefact of the geometric structure of the manifold (ie that extra structure imposed on it), not something relative to any particular field living on it.


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