Covariant derivative vs. Lie derivative

  • Thread starter Quchen
  • Start date
13
0
Hey there,

For quite some time I've been wondering now whether there's a well-understandable difference between the Lie and the covariant derivative. Although they're defined in fundamentally different ways, they're both (in a special case, at least) standing for the directional derivative of a vector. However, the Lie derivative doesn't make any assumptions about whether there's a connection on the manifold or not.
So where's the difference in between the two derivatives, where's the structure of the manifold playing into what the result of the derivative will be?

With best regards,
Quchen
 
1,444
4
Lie derivative depends on the derivatives of the vector vector field along which you take it, the covariant derivative does not:

[tex]\nabla_{fX} = f\nabla_X [/tex]

for any function [tex]f[/tex]

Lie derivative does not have this nice property.
 
13
0
Ah that makes sense. The vector the covariant derivative operates with respect to merely gives a direction on the manifold, whereas in the Lie derivative, the extension of the field is considered. What the metric/connection does is provide such an extension, but that extension is a property of the space, not a field defined on it ... correct?
 
1,444
4
Well, I would not go that far as to attribute some property to "space" itself. This is the additional geometrical structure (connection in this case) that is being used here. The same "space" may carry a whole bunch of different connections. (but any two connections differ by a tensor, so one connection is enough).
 

lavinia

Science Advisor
Gold Member
3,076
535
Ah that makes sense. The vector the covariant derivative operates with respect to merely gives a direction on the manifold, whereas in the Lie derivative, the extension of the field is considered. What the metric/connection does is provide such an extension, but that extension is a property of the space, not a field defined on it ... correct?
A connection may be thought of as a field but not as a vector field. For instance, on an orientable surface, a connection may be thought of as a field of 2 planes on the unit circle bundle of the manifold. This field must be rotation invariant in the sense that the differential of the action of rotation on the fiber circles should preserve the planes.
 
13
0
It's becoming clearer and clearer. I guess it would have been more obvious if almost all fibre bundles I've encountered hadn't been vector bundles so far.
Thanks to both of you.
 

lavinia

Science Advisor
Gold Member
3,076
535
It's becoming clearer and clearer. I guess it would have been more obvious if almost all fibre bundles I've encountered hadn't been vector bundles so far.
Thanks to both of you.
The unit circle bundle is just the vectors of length 1 in the vector bundle. So really you have already encountered it.
 
Last edited:
13
0
>So really you have already encountered it.
Ah, it's been called the "unit tangent bundle", so yes.
 

lavinia

Science Advisor
Gold Member
3,076
535
BTW: A vector field along a curve may be thought of as a curve in the vector bundle itself. For a surface, if the vector field has length 1 then it is a curve in the unit circle bundle. The derivative of this curve lies in the tangent bundle of the unit circle bundle. If the derivative actually lies in the field of 2 planes of the connection then the vector field is said to be parallel along the curve.
 

lavinia

Science Advisor
Gold Member
3,076
535
Ah that makes sense. The vector the covariant derivative operates with respect to merely gives a direction on the manifold, whereas in the Lie derivative, the extension of the field is considered. What the metric/connection does is provide such an extension, but that extension is a property of the space, not a field defined on it ... correct?
This is correct. The covariant derivative needs only to know a direction and a vector field along a curve that fits the direction. Parallel transport of a vector is the solution of an ODE along the curve whose initial condition is the vector you want to parallel transport.
 

lavinia

Science Advisor
Gold Member
3,076
535
On any vector bundle over a smooth manifold, one can think of the covariant derivative of a vector field, Y, i.e. a section of the vector bundle, as a 1 form that takes values in other sections of the same bundle. Given a tangent vector field,X, DY(X) is another section of the vector bundle. It is a 1 form because it is smooth and acts linearlly at each point of the tangent bundle i.e. DY(aX + bZ) = aDY(X) + bDY(Z) pointwise.

This definition highlights the property that covariant derivatives do no depend on the flow of the tangent fields but only on their point values.

Notice also that this definition does not require the two vector fields to both come from the tangent bundle. So there is no notion of Lie derivative of Y with respect to X.
 
Last edited:
13
0
So it's basically a section in the product space of tangent- and cotangent space?
 

lavinia

Science Advisor
Gold Member
3,076
535
So it's basically a section in the product space of tangent- and cotangent space?
Sort of. You have the right idea.

It is a section of the tensor product of the vector bundle with the cotangent bundle.
 
13
0
That was the physicist in me saying "tensor product of the bundles" ;)
 

lavinia

Science Advisor
Gold Member
3,076
535
That was the physicist in me saying "tensor product of the bundles" ;)
!

BTW: This idea of covariant differentiation does not require a Riemannian metric on the vector bundle either. In Physics it is known as a gauge potential.
 
13
0
So what is needed for a connection to exist? Fanciest thing I've seen so far was the curvature form on Lie-valued bundles.
 

lavinia

Science Advisor
Gold Member
3,076
535
So what is needed for a connection to exist? Fanciest thing I've seen so far was the curvature form on Lie-valued bundles.
It is a theorem that there is always a connection on a vector bundle over a smooth manifold.

This theorem is not hard to prove.

Wierdly, one always gets curvature even in connections that are not compatible with a Riemannian metric.

I do not know about Lie valued bundles.
 
Last edited:
13
0
Oh wow.
Does the theorem/proof have a name or something?
 

lavinia

Science Advisor
Gold Member
3,076
535
Oh wow.
Does the theorem/proof have a name or something?
I don't know. The argument is typical rather than peculiar.
 

lavinia

Science Advisor
Gold Member
3,076
535
i think the argument is by induction on a finite open cover of the manifold. Choose frames on each open set in the cover.On an open ball choose any connection. assume inductively that the connection has extended to n-1 open sets. on the n'th you need to extend the connection that already exists on its intersection with the previous open sets. The connection on the n'th open set must obey a transformation rule with respect to the chosen frames on the intersection. Use this rule to extend across the entire open set.
 
13
0
Sounds reasonable. Let's see how long it takes me to stumble upon that in literature.
 

lavinia

Science Advisor
Gold Member
3,076
535
Sounds reasonable. Let's see how long it takes me to stumble upon that in literature.
Can you tell me what the role of gauge potentials is in Physics?
 
13
0
Can you tell me what the role of gauge potentials is in Physics?
I guess you're asking that on a level I'm not able to answer. I'm just getting into the topic, and all I've got is half-knowledge both on the mathematical and physical side. If you're still interested ...
 

lavinia

Science Advisor
Gold Member
3,076
535
I guess you're asking that on a level I'm not able to answer. I'm just getting into the topic, and all I've got is half-knowledge both on the mathematical and physical side. If you're still interested ...
Anything you have to say would interest me
 
13
0
Alright then, Electrodynamics is surely the easiest example.
Take R^4 as a manifold and assign an additional degree of freedom to each point*. In the EM case, that is an element of U(1), think of some ring everywhere in spacetime. There's now a connection on this bundle, and the connection coefficients turn out to be 1-forms that describe the electromagnetic potential. The associated curvature is F=dA, the electromagnetic field strength.
*: I guess that's the fiber, but I haven't found it stated like that explicitly
 

Related Threads for: Covariant derivative vs. Lie derivative

Replies
3
Views
4K
Replies
3
Views
3K
Replies
19
Views
6K
  • Posted
Replies
1
Views
2K
  • Posted
Replies
1
Views
2K
  • Posted
Replies
13
Views
5K

Physics Forums Values

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving
Top