# Covariant derivative vs. Lie derivative

1. Sep 8, 2010

### Quchen

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

2. Sep 9, 2010

Lie derivative depends on the derivatives of the vector vector field along which you take it, the covariant derivative does not:

$$\nabla_{fX} = f\nabla_X$$

for any function $$f$$

Lie derivative does not have this nice property.

3. Sep 9, 2010

### Quchen

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?

4. Sep 9, 2010

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).

5. Sep 13, 2010

### lavinia

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.

6. Sep 13, 2010

### Quchen

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.

7. Sep 13, 2010

### lavinia

The unit circle bundle is just the vectors of length 1 in the vector bundle. So really you have already encountered it.

Last edited: Sep 13, 2010
8. Sep 13, 2010

### Quchen

>So really you have already encountered it.
Ah, it's been called the "unit tangent bundle", so yes.

9. Sep 13, 2010

### lavinia

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.

10. Sep 14, 2010

### lavinia

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.

11. Sep 16, 2010

### lavinia

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: Sep 16, 2010
12. Sep 16, 2010

### Quchen

So it's basically a section in the product space of tangent- and cotangent space?

13. Sep 16, 2010

### lavinia

Sort of. You have the right idea.

It is a section of the tensor product of the vector bundle with the cotangent bundle.

14. Sep 16, 2010

### Quchen

That was the physicist in me saying "tensor product of the bundles" ;)

15. Sep 16, 2010

### lavinia

!

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.

16. Sep 16, 2010

### Quchen

So what is needed for a connection to exist? Fanciest thing I've seen so far was the curvature form on Lie-valued bundles.

17. Sep 16, 2010

### lavinia

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: Sep 16, 2010
18. Sep 16, 2010

### Quchen

Oh wow.
Does the theorem/proof have a name or something?

19. Sep 16, 2010

### lavinia

I don't know. The argument is typical rather than peculiar.

20. Sep 16, 2010

### lavinia

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.