# A LLT, GCT and gauge transformations

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1. Aug 1, 2016

### Ravi Mohan

It has been sometime since I have been thinking about this question and I have been quite successful in confusing myself.
In Einstein's General Relativity, we say that the general coordinate transformations (or diffeomorphisms) on a manifold are the gauge transformations of the theory. The local Lorentz transformations are the orthogonal rotations in the tensor bundle of the manifold. Thus the structure group of the fibre bundle is essentially the Lorentz group (for manifold with a Lorentzian metric).

Now the structure group is a set of transformations which essentially performs rotations in the bundle which, means it changes the basis in a specific way. And in cases where we use natural basis (coordinate basis), it should mean just changing the coordinates. But that is essentially a diffeomorphism (which is gauge transformation). But Lorentz symmetry cannot be a gauge symmetry. So I was wondering where I am going wrong.

On another note, I wonder if GR is a gauge theory in typical sense. Generally, the structure group in the fibre bundle of a manifold, in a gauge theory, forms a gauge group. But for GR, the structure group of the fibre bundle is giving the actuall symmetries of the theory (Lorentz transformations).

2. Aug 2, 2016

### haushofer

Hi Ravi,

I´m not that familiar with bundle-jargon (I've never seen what it contributes to my understanding), but I consider in GR both the gct's and the LLT's (!) as gauge transformations. One way to see this is to consider the gauging of the Poincaré algebra. This procedure is reviewed in e.g. http://arxiv.org/abs/1011.1145. One gauges the algebra and imposes a co-called conventional constraint; this effectively removes the local translations from your theory, and as such the remaining gauge transformations are gct's and LLT's.

3. Aug 2, 2016

### Markus Hanke

How does the Lanczos tensor fit into all of this ? That's something I have always wondered about.

4. Aug 2, 2016

### Ravi Mohan

Interesting! I will study the reference. But if I can find a conserved Noether charge corresponding to LLTs, that certainly means that they are not the gauge transformations.

Last edited: Aug 2, 2016
5. Aug 12, 2016

### Ravi Mohan

I had a discussion with Prof. Distler and he also confirmed that LLTs are actually the gauge transformations. In fact, when we introduce the vielbeins
$$\hat{e}_{(\mu)}=e_{\mu}^{a}\hat{e}_{(a)}$$
we basically introduce $d^2$ variables out of which $\frac{d(d-1)}{2}$ are redundant. It can be seen from the equation
$$g_{\mu\nu}=e^a_{\mu}e^b_{\nu}\eta_{ab}.$$

Next, it is clear from the first equation (or its inverse), the coordinate transformations and fibre basis transformations are separate beasts. Hence LLTs and GCTs are not related to each other in any sense but both of them are essentially the gauge symmetries of the theory.

6. Aug 12, 2016

### haushofer

Yes, they are different beasts, hence the different indices (flat,a and curved,mu). They can be relatedthough by taking field-dependent parameters, i.e. in soft algebras. This fact is crucial in gauging the (super) Poincare algebra: it allows one to remove the local translations via a conventional constraint.