# Riemannian Connections: Summary

1. Nov 21, 2008

### WWGD

Hi:
I have been aching over figuring out Riemannian Connections Conceptually
( It's a Heartache! :) )

We use connections to make sense of the difference quotient in differentiation:

Lim _h->0: ( F(x+h)-F(x)/h )

In IR^n, there is no problem, because the tangent spaces at any two points

are naturally isomorphic, i.e., the fact that F(x+h) and F(x) live in different tangent

spaces ( T_x and T_(x+h) , resp. ) is not a serious problem: we translate F(x+h)

to the tangent space based at x ; translation gives us the isomorphism.

Now: If we are working on a non-flat manifold , then vector spaces are not

naturally isomorphic anymore. If the manifold M is emnbedded in IR^n , then

we make sense of the difference quotient by seeing it as the directional derivative

of one vector field in the direction of another V.Field. ( seeing T_x, the tangent

space of x in M , as a point in IR^n , so that T_xM ~ T_x IR^n ).

If the directional derivative does not live in the tangent space of M , we decompose

IR^n as T_p (+) (T_p )^Perp. , i.e., the orthogonal decomposition of the space.

Then the covariant derivative of X in the direction of Y is the projection of the

derivative into T_p.

Main Case:
If M is stand-alone , i.e., not embedded in any ambient space , and M not flat.

Then we need to specify how we will be tranlation the vector F(x+h) based at

T_(x+h)M , ( tangent space of the point x+h in M , into the tangent space

T_x M, of the point x in M ) . We do this , by specifying an isomorphism between

the two tangent spaces . The choice of the isomorphism between any two points

is the Christoffel symbol, and it determines the connection.

Thanks For Any Comment/Suggestion/ Correction.

P.S: I am kind of used to using ASCII. If others are not, let me know, and I will

use Latex.

2. Nov 21, 2008

### Hurkyl

Staff Emeritus
(Hopefully I have the following correct)

Actually, what you are describing here is the Lie derivative, which is an intrinsic property of any differentiable manifold at all, and (a priori) doesn't have anything to do with connections.

There are two problems with this.

The first is that the "choice of isomorphism" is parallel transport. In some sense, the Christoffel symbol is the derivative of the gadget that describes parallel transport.

The second is that the choice of isomorphism depends on the path chosen between the two points! A connection (in one perspective) is nothing more than a gadget that takes as input a curve on your manifold, and gives as output an isomorphism between the tangent spaces at the endpoints of the curve. (called parallel transport along the curve) It has to satisfy some obvious properties, of course, such as if you concatenate two curves, you compose the corresponding isomorphisms.

Now recall that one of the ways to name a tangent vector at a point is by specifying a differentiable curve passing through that point. Now that we have both a curve and a connection, we can compute the difference quotient. If the result doesn't depend on which curve we used to represent our tangent vector, then our connection is 'differentiable', and we can define the Christoffel symbols.

3. Nov 23, 2008

### WWGD

Do we make a distinction re connections (defining them, laying them out)

, between embedded manifolds and stand-alone manifolds?.

Also: can we define the Lie Derivative as above , i.e., of one V.Field in the

dir. of another V.Field, in a stand-alone manifold?. We would obviously not be able to

do a decomposition using T_p(+)T_p^(Perp.) of the tangent space , (together with

the fact that for a stand-alone, the tangent space is an abstract space of derivations)

Can we still define this derivative?

Thanks.

case T_p i) so how w

4. Nov 24, 2008

### Hurkyl

Staff Emeritus
The Lie derivative follows from the differential operator interpretation of tangent vectors.

Specifically, if X and Y are the differential operators defined by two tangent vector fields, then their lie bracket is the vector field [X, Y] = XY - YX.

More explicitly, when applied to a scalar functon, you get
[X, Y](f) = X(Y(f)) - Y(X(f))​

(Other methods of definition are seen on Wikpiedia)

And one can define a lie derivative for scalar and vector fields by
$$\mathcal{L}_X Y = [X, Y]$$
$$\mathcal{L}_X f = X(f)$$
and from there, I believe you can extend it to all tensors.

5. Nov 24, 2008

### WWGD

Thanks for everything, Hurkyl.