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The Berry Curvature is defined as (for invariant gauge transformations)

$$F_{ij} = [\partial_i, A_j] - [\partial_j,A_i] + [A_i,A_j]$$

The gauge covariance satisfies the transformation

$$A_i \rightarrow g^{-1}A_ig + g^{-1}\partial_ig$$

Where $$g \in U(N)$$ , in which $$N$$ is the degeneracy satisfying the transformation of the curvature

$$F_{ij} \rightarrow g^{-1}F_{ij}g$$

But this is not really pertinent to my question, but it is interesting to note that the initial part of the first equation describing the curvature will be invariant with respect to Abelian gauge transformations and not the whole $$U(N)$$ group (see reference).

The extra commutator in the first equation $$[A_i,A_j]$$ arises as a property of the gauge transformations - and in an earlier question posted at physics.stack, I asked a question concerning the similarities between the Berry curvature and the curvature tensor

Berry Curvature and Curvature Tensor

So I am almost at my question, I was informed there was actually a deep physical meaning why both the Berry curvature and the curvature share common dynamics, because they are both rooted in the same ''beautiful'' theory. This (is an oversimplification) at this moment in time to make things short. If there is an underlying deep reason they are similar, if we take a look at the Einstein equations, with a non-zero torsion, we have

$$[\partial_i \Gamma_j] - [\partial _j\Gamma_i] + [\Gamma_i,\Gamma_j]$$

Where the final commutator $$[\Gamma_i,\Gamma_j]$$ is just $$T_{ij} = \Gamma_i,\Gamma_j - \Gamma_j,\Gamma_i$$ which is a simplification of the torsion tensor in general relativity.

If the relationships are as deep as I have been led to believe rooted from the same common physics, why is the last commutator not part of (let's call it some kind of) geometric Berry torsion? The objects are so similar, again the only difference here is the name of the connections.

Calculating the Berry curvature in case of degenerate levels (Non abelian Berry curvature): issue

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    $\begingroup$ 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"? $\endgroup$
    – ACuriousMind ♦
    Commented Nov 17, 2017 at 18:53
  • $\begingroup$ Related post by OP: physics.stackexchange.com/q/369233/2451 $\endgroup$
    – Qmechanic ♦
    Commented Nov 17, 2017 at 18:58
  • $\begingroup$ 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.'' $\endgroup$
    – Gareth Meredith
    Commented Nov 17, 2017 at 19:14
  • $\begingroup$ 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"? $\endgroup$
    – ACuriousMind ♦
    Commented Nov 18, 2017 at 17:30
  • $\begingroup$ 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. $\endgroup$
    – Gareth Meredith
    Commented Nov 21, 2017 at 20:31

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