Graduate LLT, GCT and gauge transformations

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SUMMARY

The discussion centers on the relationship between general coordinate transformations (GCTs) and local Lorentz transformations (LLTs) in Einstein's General Relativity (GR). Participants assert that both GCTs and LLTs function as gauge transformations, particularly when gauging the Poincaré algebra, as detailed in the reference provided (http://arxiv.org/abs/1011.1145). The conversation highlights that while LLTs and GCTs are distinct, they both serve as gauge symmetries within the theory, with implications for the structure group of the fibre bundle in GR. The introduction of vielbeins complicates the relationship, indicating a redundancy in the variables involved.

PREREQUISITES
  • Understanding of Einstein's General Relativity (GR)
  • Familiarity with gauge theories and gauge transformations
  • Knowledge of Poincaré algebra and its gauging
  • Concept of fibre bundles and their structure groups
NEXT STEPS
  • Study the gauging of the Poincaré algebra as outlined in the paper at http://arxiv.org/abs/1011.1145
  • Explore the role of vielbeins in General Relativity and their implications for gauge symmetries
  • Investigate the relationship between Noether charges and gauge transformations in GR
  • Examine the distinctions between local Lorentz transformations and general coordinate transformations in detail
USEFUL FOR

The discussion is beneficial for theoretical physicists, particularly those specializing in General Relativity, gauge theories, and the mathematical foundations of physics. It is also relevant for researchers exploring the implications of gauge symmetries in modern theoretical frameworks.

Ravi Mohan
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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).
 
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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.
 
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How does the Lanczos tensor fit into all of this ? That's something I have always wondered about.
 
haushofer said:
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.

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:
haushofer said:
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.

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
<br /> \hat{e}_{(\mu)}=e_{\mu}^{a}\hat{e}_{(a)}<br />
we basically introduce ##d^2## variables out of which ##\frac{d(d-1)}{2}## are redundant. It can be seen from the equation
<br /> g_{\mu\nu}=e^a_{\mu}e^b_{\nu}\eta_{ab}.<br />

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.
 
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.
 
In an inertial frame of reference (IFR), there are two fixed points, A and B, which share an entangled state $$ \frac{1}{\sqrt{2}}(|0>_A|1>_B+|1>_A|0>_B) $$ At point A, a measurement is made. The state then collapses to $$ |a>_A|b>_B, \{a,b\}=\{0,1\} $$ We assume that A has the state ##|a>_A## and B has ##|b>_B## simultaneously, i.e., when their synchronized clocks both read time T However, in other inertial frames, due to the relativity of simultaneity, the moment when B has ##|b>_B##...

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