# I Curvature forms and Riemannian curvatures of connections

1. Mar 6, 2017

### RockyMarciano

I'm trying to think of the curvature form of a connection on a tangent frame pricipal bundle as an alternative description of the Riemannian curvature of the connection(see i.e. https://en.wikipedia.org/wiki/Curvature_form)

One thing I want to confirm is does a non-vanishing curvature-form imply loss of the information about the vector direction when parallel transporting it like in the Riemannian curvature case?

2. Mar 7, 2017

### Ben Niehoff

The curvature 2-form contains exactly the same information as the Riemann tensor; it's just a change of basis:

$$\Omega^a{}_b = \frac12 e_\mu{}^a e^\nu{}_b \, R^\mu{}_{\nu\rho\sigma} \, dx^\rho \wedge dx^\sigma$$
I'm not sure what you mean by "loss of information about vector direction when parallel transporting", though.

3. Mar 7, 2017

### RockyMarciano

I mean the usual path dependence of tangent vectors in the presence of curvature.

4. Mar 7, 2017

### Ben Niehoff

I wouldn't call that "loss of information". And yes, of course it's still there.

5. Mar 7, 2017

### RockyMarciano

Ok, thanks.
To give you more context about what I have in mind. I'm thinking about curvature forms of connections in principal G-bundles. In order to keep this correspondence with Riemann curvature in the general case, if we are modelling a theory with gauge invariance and we need path independence so that we can define a gauge covariant derivative, we must impose a trivial principal bundle and a base manifold with topology compatible with this triviality, right?

And in this special context of a trivial bundle the curvature form and path independence is just a special case of the Riemannian path dependent more general case in the same way that a (pseudo)Riemannian manifold with flat curvature preserves path independence trivially as a special case of general Riemannian curvature.

So for example in the specific physical case of Yang-Mills fields we have a connection with curvature due to the structure group being non-abelian, and also path independence as it is a gauge theory, the pricipal bundle is trivial and the base manifold is Minkowskian with contractible topology and all works fine.
But in a more general mathematical case where we didn't demand bundle triviality(therefore no gauge invariance) and a specific base manifold compatible with it in general we wouldnt' have both path independence and non-vanishing curvature (just from the information of the formula in #2), correct?

6. Mar 7, 2017

### Ben Niehoff

But gauge connections are path-dependent! Unless your gauge group is Abelian.

There is no need for your gauge bundle to be trivial; in fact the instanton number (i.e. second Chern class) computes its non-triviality.

Except that they're really the same thing. The Riemann tensor is the field strength of an $SO(p,q)$-connection on the tangent bundle.

Again, I'm not sure why you think gauge invariance has to do with path-dependence.

I should point out that the "charge" carried by a particle charged under a non-Abelian gauge group is in fact not gauge-invariant, but merely gauge-covariant.

7. Mar 7, 2017

### RockyMarciano

I think we are basically on the same pagen as far as I can see from your replies. It's just terminology.

True, the connection in Yang-Mills is non-trivial in this sense because the structure group G is not-commutative, I'm using the path independent notion in the sense that you refer to covariance of the gauge above. What I was trying to convey by path independence is the covariance of the gauge covariant derivative, that is able to parallel transport not only information about the change in the field due to translation of the vector from one point to another, but is also able to keep track of the rotation of the local coordinate axes on each point of the base manifold(oriented differently on each point in a non-commutative way), and it can do these because the actual vector is pointing actually at the same point if we abstract it of the particular local coordinates at each point, due to the actual flatness of Minkowski space, and the triviality of the principal bundle that has it as base manifold. See below for more on what I meant by path independence/dependence in relation with gauge covariance.
Well, yes with qualifiers that go beyond the level I'm keeping this so far. This requires Euclidean spacetime, the non-perturbative regime, restriction to a point in spacetime.... I know they use this in QCD, but this discussion is more in the perturbative side(although I'd be interested in maybe starting another thread to talk about this if I have the chance)

Yes, I know what you mean but let me explain what I meant: This "field strength" measures the deviation from flatness in the orientation of the vector being transported(that would have kept pointing in the same direction if the manifold had been flat).

In the Yang-Mills case we are keeping track not only of this but of the change in the field as it is trasported from point to point. And we have the same number of degrees of freedom since the gauge covariance is about redundance of degrees of freedom, so we achieve it by having a flat manifold and a trivial bundle.

We would achieve the equivalent thing in the simple Riemannian manifold setting by choosing a flat manifold that allows as to preserve both lengths and angles(path independence), not just lengths or just angles like in the general case in the presence of curvature.

It is just that the more flexible structure of the principal bundle lets us switch between connections and gauge groups G, like abelian U(1) that has vanishing curvature, or SU(n) that is non-abelian and therefore has non-vanishing curvature form and keep gauge covariance thanks to the flatness of the base manifold and the triviality of the bundle.

8. Mar 8, 2017

9. Mar 8, 2017

### Ben Niehoff

Sorry, I still have no idea what you're trying to get at with your use of "path dependence". I would recommend you stick to standard terminology.

Gauge connections parallel-propagate vectors in an "internal" vector space in exactly the same way that the Levi-Civita connection parallel-propagates vectors in the tangent space. There is no fundamental difference between the two concepts.

I don't believe we are on the same page at all, because you're continuing to say things that are obviously false:

Gauge bundles do not have to be trivial, and gauge invariance does not require that the base manifold be flat Minkowski!

Achieve what? You're being very unclear here.

OK, you need to drop this terminology "path independence" because you're using it in a totally nonsensical way.

Also, ALL metric-compatible connections (of which the Levi-Civita connection is one) preserve both lengths AND angles. That's what metric-compatible means.

Of course a U(1) gauge bundle can have curvature! That's what electromagnetism is!

Also gauge covariance has absolutely nothing to do with flatness of the base or triviality of the gauge bundle.

10. Mar 8, 2017

### RockyMarciano

Fair enough, I actually agree with what you explain here so I'm obviously doing a bad job trying to get across my point. Let me try and clarify once more. Also please be a bit indulgent with me here:I have received certain physics instruction long ago but I'm basically trying to express complex things from a layman point of view, and it is not easy at all. I'm here to correct my misunderstandings relying on people that is professionally dedicated to this like may be your case so the last thing I'd want is to appear as argumentative.

You are of course right, actually, for instance, global(spacetime) anomalies in QFT are examples of nontrivializations. But let me give you the context in wich I refer to trivial bundles in physics(wich I also tried earlier with the reference to the stack.exchange page) and please let me know if I'm misunderstanding it. I'm referring only to QFT, and the standard model of particle physics, and I understand this gauge QFT does use flat minkowski space as base manifold. If this were so(please correct it if this is not so), there is a theorem that says: "Given a fibre bundle with base space X and structure group G, if either X or G is contractible then the bundle is trivial." Since Minkowski space is contractible I understand that the bundle is trivial, but I'm open to be corrected if I'm wrong following this reasoning.

EDITED: I originally wrote something confusing and that deviates from the main discussion.
Yes, the inner product is preserved by the connectio along the specific curve, and depending on the specific path. But only line lengths are preserved independent of curve, position or direction(path).
The distinction I was doing was because in flat space the inner product is preserved independently of path.I was referring to what can be preserved independent of path.
I think an important distinction here between the Riemannian covariant derivative and the gauge covariant derivative is that the connection in the first case is not a tensorial object on the manifold, the Christoffel symbols are not tensors, the curvature is tensorial; while the pricipal connection used in gauge theory is an Ehresmann connection and is constructed as a tensorial object in the bundle(in the total space instead of in the manifold), and in this sense it is frame independent already from the parallel transport moment, this was the difference I was trying to stress when referring to path dependence/independence, is this any clearer?

.
Yes, what I meant was that in the U(1) case being abelian there is no contribution to the curvature from non-commutativity, as it happens for Yang-Mills fileds. So the curvature is computed only with the term corresponding to the exterior derivative of the connection $dA$.
Right. I was again referring to the special case that doesn't enter into nontrivialities of the bundle.

But I'd rather wait till I get some feed-back before I attempt to explain again the link I see and how it relates to path dependence/independence in order to avoid saying wrong things if it turns out I'm actually misunderstanding the basics.

Thanks for the assistance.

Last edited: Mar 8, 2017
11. Mar 9, 2017

### RockyMarciano

Ok, I think I see what you mean now. Gauge principal bundles are by definition local(local gauge), and they can of course be nontrivial like for instantons.
When I was thinking of trivial bundle I was going unconsciously to the global case, whis is the global symmetry of spacetime rather than a trivial bundle.
Ok, it's just that I switched from the local gauge to the global gauge(global symmetries) unconsciously, of course "global nontrivializations"(the ones I was mistakenly thinking of when speaking about nontrivial bundles) are not allowed in physics as long as we are only considering Minkowski spacetime global symetry.

Do you agree with this?

12. Mar 10, 2017

### RockyMarciano

And yet, let's consider the EM gauge connection U(1), unlike the Yang-Mills case, local changes in phase (infinitesimal rotations in the "internal" space) won't affect the "curvature" you referred to above. Compare with the LC connection in the tangent space of a Riemannian manifold, an infinitesimal loop of the vector will not leave it pointing in the original direction in general and this will be measured as curvature. So i would say the quoted claim is not exact in general for any gauge connection.

13. Mar 10, 2017

### Ben Niehoff

The Aharonov-Bohm effect is precisely a loop around which a U(1) phase fails to end up where it started.

14. Mar 11, 2017

### RockyMarciano

I think we have different concepts about what "exactly the same way" and "no fundamental difference" mean.
I was counting that particular type of holonomy as a further difference, whereas the connections in internal space can make compatible curvatures (as field strengths that you referred to earlier) of the connection and holonomy as a type of monodromy(A-B effect, looping around singularities) because all those curvatures are in the context of the flat connection in flat Minkowski base space, the LC connection on tangent spaces can't: a curved connection's holonomy in a curved manifold is never a monodromy as being a curved connection is not compatible with being a flat connection in the tangent space for a given metric, and only the latter have associated monodromy as their holonomy.

15. Mar 11, 2017

### Ben Niehoff

No, this is still not a distinction. If I parallel-transport a vector around the tip of a cone, even though my connection is locally flat, I still get a holonomy.

You don't even need the cone to be singular. Just cut off the tip and attach a spherical cap, and everything is smooth.

Furthermore, the Aharonov-Bohm effect is just a special case. The Maxwell field strength F is telling you about infinitesimal phase differences obtained by going around infinitesimal loops, just as the curvature 2-form is telling you about tangent-space vectors.

16. Mar 12, 2017

### RockyMarciano

You are missing the distinction I was drawing. This holonomy is a global holonomy, a monodromy, that is the holonomy associated to flat connections. My point was that a Riemann manifold with a curved metric(so the flat cone example is irrelevant here) has a curved LC connection, and doesn't have any notion of monodromy on its tangent space, while the structure of a principal bundle admits all these different concepts of holonomy even if its base manifold is the same flat Minkowski space.
The Maxwell field strength, or curvature of the U(1) principal connection is telling you about the horizontal subspace, how the fiber U(1) changes around loops, not how the phase at a particular point(vertically) changes, phase changes at a point are gauge transformation that leave the physics unaltered.
For the manifold at the tangent space the curvature of the LC connection does tell you at that particular point and vertical subspace how the vector fails to recover its original orientation and this brings it closer to the Yang-Mills field curvature, that has an additional term coming from the fiber's noncommutativity already as a nontrivial property.

17. Mar 12, 2017

### Ben Niehoff

I am telling you that both LC connections and gauge connections can have both types of features! I even gave examples.

Now you're deliberately missing my point. The Maxwell field strength measures the holonomy of infinitesimal Wilson loops. These are gauge-invariant.

The Riemann curvature is given by

$$R = d \omega + \omega \wedge \omega$$
The gauge curvature is given by

$$F = dA + A \wedge A$$
They are the same, just change the letters! Literally the only difference is that $F$ lives in a bundle other than the tangent bundle. ALL of the other types of properties you've mentioned (i.e., whether they can have holonomies, monodromies, etc.) are shared by both concepts.

And if you are going to reply to me again with further hangups about U(1) connections being Abelian, then look at LC connections on 2-dimensional manifolds!

Please take some time to work with these things on paper a bit.

18. Mar 12, 2017

### RockyMarciano

At this point it just gets absurd to go on, I don't even know anymore what the discussion is about and it is far from the original question.

19. Mar 13, 2017

### lavinia

Maybe this will help with your original question about the relation of Riemannian curvature to connections on principal bundles.

Via a local section $s$ of the principal bundle (a global section may not exist) over a domain $U$ over which the bundle is trivial, one can pull back the connection 1 form $ω$ to $U$ to get a Lie algebra valued 1 form $s^{*}ω$ on the tangent bundle. In case the principal bundle is the bundle of orthonormal frames (a principal $O(n)$ bundle) $s^{*}ω$ is a skew symmetric matrix of ordinary 1 forms defined on tangent vectors. If $s_{i}$ are the vector fields in this local frame then one can define a connection on the tangent bundle by $∇s_{i} = Σ_{j}\hat{ω_{ij}}⊗s_{j}$. This is the usual formula for a connection on a vector bundle with a Riemannian metric. The covariant derivative with respect to a tangent vector $X$ is then $∇_{X}s_{i} = Σ_{j}\hat{ω_{ij}}(X)s_{j}$.

The pull back of the Lie algebra valued curvature 2 form is the curvature of the connection on $U$. This is a skew symmetric matrix of 2 forms defined on the tangent space to $U$. In terms of the Riemann curvature tensor, this is the matrix of the linear map $R(x,y)$ and the usual skew symmetry $R(x,y)= - R(y,x)$ is just the skew symmetry of the curvature 2 form.

More generally if the principal bundle is a bundle of frames of a vector bundle (possibly a complex vector bundle and not necessarily with a metric) then for any local frame $s_{i}$ the pull back of the connection 1 form gives a connection by the same rule $∇s_{i} = Σ_{j}\hat{ω_{ij}}⊗s_{j}$ where now the $s_{i}$ are local sections of an arbitrary smooth vector bundle. In this general case the matrix $\hat{ω_{ij}}$ may not be skew symmetric since it does not necessarily take values in the Lie algebra of the orthogonal group.

As Ben suggested a good example to work through is a connection on a oriented two dimensional surface with a Riemannian metric. In this case the unit tangent circle bundle is a principal $SO(2)$ bundle. The action of $SO(2)$ on the unit circle bundle is by rotation of fiber circles. Since there is a Riemannian metric the pull back of the curvature 2 form can be written as $Ω= -K$vol where $K$ is the usual Gauss curvature and vol is the volume element of the surface.

Last edited: Mar 20, 2017
20. Mar 13, 2017