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Nov 21, 2017 at 20:33 comment added Gareth Meredith (by the way, I am not a mathematician, just someone who is very keen) about physics too.
Nov 21, 2017 at 20:31 comment added Gareth Meredith I wrote them in commutator language, I picked it up from a tutorial. Never mind, it really isn't important. The last term (would be) completely analogous to a torsion, the complete symmetry of the Berry curvature and then the final commutator matches the full Riemann tensor with non-zero torsion. The similarities are striking.
Nov 18, 2017 at 17:30 comment added ACuriousMind But the object in the definition of the curvature is not a commutator! The curvature of a connection form $A$ is $F = \mathrm{d}A + A \wedge A$ in component-free notation, which yields $F_{ij} = \partial_i A_j - \partial_j A_i + [A_i,A_j]$ - the last term is a genuine commutator, but the first two are not. Where did you get the impression they are commutators, and why do you think the last term is "torsion"?
Nov 17, 2017 at 19:14 comment added Gareth Meredith The Berry geometric phase is related to the Berry curvature - I use $$[\partial_i,A_j]$$ because its standard notation for me to write things as commutators - I am asking about the last commutator $$[A_i,A_j]$$ which appears identical to $$[\Gamma_i,\Gamma_j]$$ - where the latter here is the torsion related to the gravitational field equations. If there is a deep connection (like suggested to me in my previous thread) between the Berry curvature and the curvature tensor, then I would like to know why the last object would not have similar meaning, hence a ''Berry Torsion.''
Nov 17, 2017 at 18:58 comment added Qmechanic Related post by OP: physics.stackexchange.com/q/369233/2451
Nov 17, 2017 at 18:56 history edited Qmechanic CC BY-SA 3.0
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Nov 17, 2017 at 18:53 comment added ACuriousMind 1. Your notation is idiosyncratic - why do you write $[\partial_i, A_j]$ instead of just $\partial_i A_j$? 2. Are you familiar with the more general concept of (Yang-Mills) gauge theory? 3. The notion of torsion needs soldered bundles, since the Berry curvature is not on a tangent bundle, it has no solder form. What do you mean by "geometric Berry torsion"?
Nov 17, 2017 at 18:49 history asked Gareth Meredith CC BY-SA 3.0