# Connection on a principal bundle Intuition

1. Mar 21, 2014

### center o bass

I am reading up on principal bundles and currently I'm trying to get to grips with the definition of a connection on such a space. The definition is as follows:

A connection on P is a unique separation of the tangent space $T_uP$ into the vertical subspace $V_u P$ and the horizontal subspace $H_u P$ such that
$$(i) T_u P = H_u P \oplus V_u P.\\ (ii) \text{A smooth vector field X on P is separated into smooth vector fields} \ \ X_H \in H_u P \ \ \text{and} \ \ X_V ∈ V_u P \ \ \text{as} \ \ X = X_H + X_V .\\ (iii) H_{ug} P = R_{g∗} H_u P \ \ \text{for arbitrary} \ \ u \in P \ \ \text{and} \ \ g \in G.$$

First of all; is this a generalization of the usual affine connection? Do the affine connection also determine a direct sum composition of the tangent space on a manifold? Are there other similarities?

Any intuitive and enlightening thoughts on why this is a good definition of a connection?

2. Mar 21, 2014

### homeomorphic

I'm not feeling up to a very detailed response right now. Connections on manifolds are about parallel translating vectors. With a principal bundle, you want to have some notion of parallel translating group elements. The horizontal tangent planes give you subspace of directions that are to be considered "parallel".

It helps to think about the example of an orthogonal frame bundle of a Riemannian manifold.

3. Mar 22, 2014

### center o bass

If you feel up to a more detailed response later, perhaps you could elaborate on the example of the orthogonal frame bundle?
Every fibre above each point consists of all the possible orthonormal frames for the tangent space at that point. I would suppose that parallel transport along a curve within the bundle would mean something equivalent with parallel transporting each of the orthonormal basis vectors individually along a corresponding curve on the original manifold in accordance with the Levi-Civita connection. I.e. such that the basis vectors remain orthogonal and thus remain a basis.

4. Mar 22, 2014

### homeomorphic

You just have to think about what parallel translation actually is. You have a path downstairs in the manifold. You want to lift that to a path upstairs. Downstairs, you have a well-defined tangent vector along the path. You have to think about what the corresponding tangent vector is upstairs. It's not going to have any component in the direction of the fiber because you don't want to move within the fiber. That's where the horizontal subspaces come in. The horizontal subspaces project isomorphically to the tangent spaces downstairs. So, each tangent vector downstairs lifts to a unique tangent vector upstairs. Just follow the lifts of the tangent vectors along and that will give you a lift of the path. So, that's how the horizontal subspaces give you a way to parallel transport group elements from one fiber to another.

You want these lifts to be smooth, and you want it to commute with the group action, which are the other two parts of the definition (I'm not sure if I can completely convince myself WHY you want it to commute with the group action, but it's a fair thing to ask for, and if you think about that frame bundle example, that's the way it works).

Yes, that's pretty much how it works. But note that you could start with the frames and specify a way to parallel transport them. You could then apply that to vectors to get back the Levi-Civita connection.

5. Mar 22, 2014

### Ben Niehoff

My intuition is that commuting with the group action is what makes the transport "parallel". I.e. if we act with group element $g$ along the entire path, the new path is also horizontal.

6. Mar 22, 2014

### homeomorphic

I probably should have been better about what I meant with regard to the lift commuting with the group action. You could interpret it in two ways. The 3rd condition says horizontal planes map to horizontal planes under the group action, and it follows that the lift starting at one point in the fiber will map to the lift starting at the other point, under the action (which is almost what you are saying). The other way you could interpret commuting with the group action is if you think of parallel transport as a map from one fiber to another, that map commutes with the group action. This seems analogous to the "compatibility with the metric" condition that defines the Levi-Civita connection (which is saying you don't want to change the lengths of tangent vectors or angles between different ones when you parallel transport).

7. Mar 28, 2014

### center o bass

Perhaps I can come with a related followup question without starting a new thread: With a connection on a principal bundle one can also define curvature forms. Are these to be interpreted similarly to the Riemann curvature tensor of Riemannian geometry?

In Yang-Mill theory such curvature give rise to the Lagrangian - does this then express the "curvature of the principal bundle" on which the relevant connection lives?

8. Mar 28, 2014

### Ben Niehoff

Yes. The Riemann tensor can be viewed as the curvature of an SO(n) bundle (the tangent bundle).

Not sure what you mean here.

9. Mar 28, 2014

### homeomorphic

The Riemannian curvature tensor tells you something about parallel transport around an infinitesimal parallelogram (components of the difference vector between the parallel-transported vector and the original). You can interpret that roughly as a Lie-algebra-valued 2-form, in the sense that it is giving you an infinitesimal rotation, an element of the Lie algebra of SO(n), for each pair of vectors in a bilinear fashion. Technically, you should say it's an ad-P-valued 2-form, rather than a Lie-algebra-valued 2-form. On a general principal G-bundle, you will just be replacing SO(n) by G, and then you'll get, roughly, a Lie algebra-valued 2-form, the Lie algebra being the Lie algebra of G this time. This 2-form is telling you something about what happens when you parallel-transport a group element around an infinitesimal loop. It gets hit by an infinitesimal element of the group, rather than an element of G.

To define bundle curvature, you can try to work with the horizontal subspace definition, but usually you would define it in terms of the Lie-algebra-valued 1-form definition of a connection. It will be a modified (covariant) version of the exterior derivative of the connection form.

To be more concrete, I find it helpful to go back to the simplest case of the differential geometry of surfaces and compare how connection forms and curvature work in that case, or even the geometry of curves in R^3 and the Frenet formulas.

Electromagnetism can be view in terms of connections and curvature of U(1)-bundles over space-time. Yang-Mills theory generalizes that to different (non-Abelian) gauge groups, like the SU(3) X SU(2) X U(1) Standard Model gauge group. You can use the Lagrangian analogous to that of electromagnetism or you can just try to generalize Maxwell's equations directly.

10. Oct 1, 2014

### lavinia

An affine connection is a connection on a principal bundle with structure group the affine group.

Another kind of connection is a linear connection. Its structure group is the general linear group.
If the bundle is the bundle of orthonormal frames its structure group is the orthogonal group.
The concept of connection on a principal bundle includes all of these and so may be viewed as generalization.

- It is a theorem that parallel translation commutes with the action of the structure group.
This follows because horizontal spaces are preserved by the action. This implies that parallel translation is an isomorphism of fibers.

In a general principal bundle parallel translation does not mean translation of vectors or frames. Rather it means translating the fibers of the bundle. However, given a representation of the structure group in the general linear group one can define parallel translation on the induced vector bundle from parallel translation of the fibers.

- I know little about Physics but here is the idea of a Yang Mills action according to a paper that I looked at.

Given a local section of the principal bundle,one may pullback the connection 1 form and the curvature 2 form to the base manifold. The pullback of the connection form is called a gauge field. The pullback of the curvature form is called the field strength.

Since the field strength transforms by the adjoint representation of the coordinate transformation,one can define its norm with an ad invariant inner product on the Lie algebra of the structure group. The integral of the square of this norm with respect to a volume element on the manifold is called a Yang Mills action. Note that this action is defined for each connection on the principal bundle and so may be viewed as a function on the space of connections. Further since the space of connections is convex one can take the directional derivative of the action. I think the Yang Mills equations can be derived from critical points of the action in the space of connections.

- The Riemann curvature tensor is not the same as the curvature 2 form on a principal bundle. First of all it is defined on the tangent bundle of the manifold not on the tangent bundle to a principal bundle and it requires a metric on the tangent bundle while a general connection on a principal bundle does not. It is defined in terms of a covariant derivative on vector fieds while the curvature 2 form is defined as a covariant exterior derivative of a Lie algebra valued 1 form.

The Riemann tensor is defined wrsp to a Levi-Civita connection on the tangent bundle. A general curvature form is defined for an arbitrary connection on a principal bundle.

The principal bundle corresponding to the tangent bundle is the bundle of tangential frames. For each Levi-Civita connection there corresponds a compatible connection on the bundle of frames. In the presence of a Riemann metric, the principal bundle can be taken to be the bundle of orthonomal frames with structure group the orthogonal group. A local pullback of the cuvature 2 form can be retrieved from the Riemann curvature tensor. This is the relationship between the two.

BTW: a couple of unrelated comments

- A seemingly important induced vector bundle which seems to be used in the mathematical side of Yang Mills theory is the adjoint bundle. This is the vector bundle induced by the adjoint representation of the structure group. While the field strength is only defined locally, since it transforms by the adjoint representation of the transition functions, it becomes a global 2 form on the base manifold with values in this vector bundle. Note that this is an even more general type of differential form, a form with values in a vector bundle.

- Instead of an ad invariant inner product on the Lie algebra of the structure group,one can take ad invariant symmetric polynomials. Plugging the curvature 2 form into these polynomials produces global closed differential forms on the manifold whose de Rham cohomology classes are the Chern classes, Pontryagin classes and the Euler class. Pontryagin classes and the Euler class are defined for any oriented vector bundle, Chern classes for any complex vector bundle.

Last edited: Oct 10, 2014
11. Oct 1, 2014

### WWGD

Note that this is an Ehresmann connection, and there are other types of connections :
http://en.wikipedia.org/wiki/Ehresmann_connection

12. Oct 2, 2014

### Ben Niehoff

You want it to commute with the group action in order to preserve the group structure of the bundle. This is what makes it a principal bundle rather than a generic fiber bundle. For example, you can have a $S^3$ fiber over some base and define parallel transport however you like; however, this will only be an $SU(2)$ bundle if the parallel transport agrees with the $SU(2)$ structure.

As you point out, in the tangent bundle (or was it the frame bundle?), a metric-compatible connection is one that preserves an $SO(n)$ structure*. The Levi-Civita connection is then the unique metric-compatible connection with vanishing torsion.

* Locally, at least. On a non-orientable manifold, this will be an $O(n)$ structure.

Edit: Wow, I didn't remember I had already replied to this thread, addressing exactly the same question! I think it's the change to the new format; I didn't recognize it was a thread I had seen before. :P

Last edited: Oct 2, 2014