Why is the Torsion Tensor Overlooked in General Relativity?

bchui
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I ahve always wondered why only the curvature term R_{\mu,\nu} been considered in GR. From differential Geoetry, how about the torsion tensor?
 
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The simple answer is that in standard formulation of general relativity, we require a unique connection on a manifold which is torsion-free and metric compatible, which lead to the properties of the covariant derivative that we need.
 
It always bothered me as well that the torsion term was simply ignored. Many people have examined what happens when you keep the term, which has led to the Einstein-Cartan theory as a way to try and incorporate spin. Although this theory doesn't appear to be fundamental, it does seem to embed itself into supergravity which is in turn one of the limits of M-Theory.


Here is a link to Hehl's 1976 article from Rev. Mod Phs...
http://prola.aps.org/abstract/RMP/v48/i3/p393_1


And here is a much more recent article
http://arxiv.org/abs/0711.1535
 
I asked a question here, probably over 15 years ago on entanglement and I appreciated the thoughtful answers I received back then. The intervening years haven't made me any more knowledgeable in physics, so forgive my naïveté ! If a have a piece of paper in an area of high gravity, lets say near a black hole, and I draw a triangle on this paper and 'measure' the angles of the triangle, will they add to 180 degrees? How about if I'm looking at this paper outside of the (reasonable)...
From $$0 = \delta(g^{\alpha\mu}g_{\mu\nu}) = g^{\alpha\mu} \delta g_{\mu\nu} + g_{\mu\nu} \delta g^{\alpha\mu}$$ we have $$g^{\alpha\mu} \delta g_{\mu\nu} = -g_{\mu\nu} \delta g^{\alpha\mu} \,\, . $$ Multiply both sides by ##g_{\alpha\beta}## to get $$\delta g_{\beta\nu} = -g_{\alpha\beta} g_{\mu\nu} \delta g^{\alpha\mu} \qquad(*)$$ (This is Dirac's eq. (26.9) in "GTR".) On the other hand, the variation ##\delta g^{\alpha\mu} = \bar{g}^{\alpha\mu} - g^{\alpha\mu}## should be a tensor...
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