Timeline for Gravity as a gauge theory
Current License: CC BY-SA 4.0
12 events
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| Nov 15, 2023 at 19:28 | history | edited | hft | CC BY-SA 4.0 |
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| Jul 26, 2023 at 13:05 | vote | accept | riemannium | ||
| Feb 16, 2016 at 18:25 | history | edited | Diego Mazón | CC BY-SA 3.0 |
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| Jul 22, 2013 at 19:45 | history | edited | Diego Mazón | CC BY-SA 3.0 |
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| Jul 22, 2013 at 19:40 | history | edited | Diego Mazón | CC BY-SA 3.0 |
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| Jul 22, 2013 at 18:01 | comment | added | Diego Mazón | Thanks, @John I don't get your point. Can you write EH action only in terms of a connection (as in Yang-Mills) or you additionally have to write the tetrad field explicitly (which doesn't transform as a connection)? What is the analog of torsion in Yang-Mills? It seems to me you are saying the same thing as me in other variables. | |
| Jul 19, 2013 at 21:33 | comment | added | John | You can look in Blagojevic, "gravitation and gauge symmetries", which is freely downloadable. The matter is quite old. Take connection of the Poincare group $A=\frac12\omega^{a,b}L_{ab}+e^a P_a$, $L_{ab}$ and $P_a$ are Lorentz and translation generators. Then spin-connection and tetrad are the gauge fields corresponding to these generators. Compute the Yang-Mills field strength $F=dA+A^2=R^{a,b}L_{ab}+T^a P_a$. Then $T^a$ is a torsion two-form and $R^{a,b}$ is a Riemann two-form. Einstein-Hilbert action reads $\int R^{a,b} e^c...e^u \epsilon_{abc...u}$ | |
| Jul 19, 2013 at 20:06 | comment | added | Diego Mazón | Hello @John Your comment seems very interesting to me. Can you expand it? Some link or reference? Name of the formulation or key word to search? | |
| Jul 19, 2013 at 19:59 | comment | added | John | Drake, the tetrad field can be viewed as a part of the Yang-Mills connection for Poincare or (anti)-de Sitter group. In this case, tetrad and spin-connection are different parts of a single connection. Then gravity looks almost like YM-type gauge theory. There are important differences however. | |
| Jul 19, 2013 at 2:34 | comment | added | Diego Mazón | Hello @AlexNelson . Yes, in the first case, the metric tensor transform as you write. In the second — linearized diff — $h$ transforms with the covariant derivatives replaced by partial derivatives. The metric $g$ is invariant only in the case of a isometry. | |
| Jul 19, 2013 at 0:37 | comment | added | Alex Nelson | Wait, with the term "diffeomorphism invariance", doesn't it refer to the symmetry $g_{\mu\nu}\to g_{\mu\nu}-\nabla_{\mu}\xi_{\nu}-\nabla_{\nu}\xi_{\mu}$, i.e., the Lie Derivative along $\xi$ of the metric vanishes? (See section 3.3 of these MIT notes). | |
| Jul 18, 2013 at 22:40 | history | answered | Diego Mazón | CC BY-SA 3.0 |