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When the curvature tensor (from Einstein's theory) has a non-zero torsion, it is said to be an antisymmetric tensor in the last two indices composed of the connections of the field. Alternatively, the Berry curvature used in the Berry phase however, is considered an antisymmetric tensor. The Berry curvature is given as

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

Where [A] is the Berry connection. In comparison, take a look at the curvature tensor with zero torsion

$$R_{ij} = -[\partial_i, \Gamma_j] + [\partial_j, \Gamma_i]$$

Now take a look at the gauge invariant Berry curvature

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

It has a structure identical to the non-zero torsion formulation of the field equations

$$R_{ij} = -[\partial_i, \Gamma_j] + [\partial_j, \Gamma_i] + [\Gamma_i, \Gamma_j]$$

The only formal difference between the Berry curvature and the Einstein curvature appears to be the definition of the connection. But I wish it was that simple, because the curvature tensor is a symmetric tensor in absence of the torsion yet the Berry curvature is written in literature as an antisymmetric tensor.

What are the intrinsic differences (even though they look formally similar) that distinguishes the Berry curvature as antisymmetric, when the curvature tensor not? A second question is, does the curvature tensor, even in absence of the torsion, need to be symmetric? Sorry if these seem like naive questions, but when investigating the Berry curvature, and writing it out, I couldn't help but notice the strong similarity to the Riemann curvature tensor.

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  • $\begingroup$ Which "curvature tensor" are you talking about? The Riemann tensor is antisymmetric, not symmetric, both with torsion and without. You may be thinking of the Einstein tensor, which is symmetric, but your equations don't describe it. Can you rewrite your question using more standard notation? $\endgroup$ – knzhou Nov 16 '17 at 18:42
  • $\begingroup$ Riemann geometry predicts symmetric Ricci tensor and symmetric Riemann connection. In presence of torsion situation changes.? $\endgroup$ – Gareth Meredith Nov 16 '17 at 18:46
  • $\begingroup$ en.wikipedia.org/wiki/Ricci_curvature - clearly we are working in a two indice case, but it doesn't matter. I was just interested in the symmetry of Berry curvature with that of the curvature tensor. $\endgroup$ – Gareth Meredith Nov 16 '17 at 18:48
  • $\begingroup$ So again I need to ask, why is the curvature tensor symmetric when the Berry curvature is considered not? What is it that makes them different, because the structure of the equations are not giving anything away for me. $\endgroup$ – Gareth Meredith Nov 16 '17 at 18:52
  • $\begingroup$ It seems, that one property of the curvature tensor is that it is skew symmetric in the last two indices, which helps give rise the symmetric cyclic Bianchi identities. So I have an incline that it is related to the properties of the indices - that is even with the torsion free affine connection. $\endgroup$ – Gareth Meredith Nov 16 '17 at 19:13
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Indeed the structure is similar for a very good reason: Both of these are particular examples of a beautiful concept in mathematics.

You have some underlying manifold $M$. In addition to the manifold you have a family of vector spaces parametrized by $M$. In other words over each point of $p \in M$ you have a vector space $V_p$ and as you move around in $M$ the vector spaces also change in a smooth way. Together with some consistency conditions the structure you get out of this is whats called a vector bundle.

Now once we have a vector bundle, we can consider a "vector field". This is simply an assignment of a particular element $v \in V_p$ to each point $p \in M$ where again the assignment is done in a smooth way. In the math language this is called a section of the vector bundle. Now since we want to do calculus, it's natural to ask how to differentiate sections. There is a clear problem with that however: differentiation instructs us to subtract values of functions at two nearby points of your domain, but in this situation the images of two nearby points are valued in different vector spaces, therefore subtraction makes no sense! This is where the connection comes in. It's a tool that allows one to connect nearby vector spaces and thus define an appropriate notion of a derivative (covariant derivative). That's really all a connection is: a rule that allows one to connect nearby vector spaces ("parallel transport"). If you think about it, just this alone will lead you to the right index structure $A_{i\,\,b}^{\,\,a}$ (locally it's an "$\text{End}(V)$-valued one form") along with the right transformation property ("gauge transformation"). Here $i$ is an index of your underlying manifold and $a,b$ are indices of your vector space $V$ .

Now given a vector bundle $V$ over $M$ with some specified connection $A$, I can ask what will happen if I use this connection to take some initial vector and move it around some small loop in the underlying manifold. That's essentially what the curvature of the connection $A$ is describing. In index notation it's written as $F_{ij\,\,\,\,b}^{\,\,\,a}$. One should also note here that curvature defined as such is always anti-symmetric in $i$ and $j$ ("$\text{End}(V)$-valued two form").

Now the specifics. In the case of Riemann tensor, your underlying manifold is some smooth Riemannian (or Lorentzian in the case of GR) manifold $M$ and the vector bundle is simply the tangent bundle (i.e the possible arrows/directions one can draw at a given point on $M$) and sections are simply vector fields on $M$ (think of a sphere with some arrows tangent to it at each point). Given this structure, there is a unique connection that makes sure that the smooth and Riemannian structure get along well, which is the well-known Levi-Civita connection (Christoffel symbols) and the curvature of this connection is the Riemann tensor.

In the Berry case, your underlying manifold is the space of control parameters that appear in the Hamiltonian and the vector bundle over the space of control parameters is simply the Hilbert space (or some subspace of "low energy states")/space of states at those values of parameters. The connection is given by Berry's connection and the Berry curvature is the curvature of this connection.

The reason you get some extra structure in the Riemannian case on the Riemann tensor is that the connection you define comes from (is uniquely defined by, in fact) an assumption known as "compatibility": the covariant derivative of the metric vanishes. A general connection on some vector bundle need not have such an extra structure, so you don't see this in the Berry case.

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  • $\begingroup$ I have a question, just to tackle this antisymmetry thing. The Reimann tensor is antisymmetric in the last two indices. How does this differ with the Berry curvature? $\endgroup$ – Gareth Meredith Nov 17 '17 at 7:53
  • $\begingroup$ I know the Berry curvature is antisymmetric so I am guessing at the core, there is no difference. $\endgroup$ – Gareth Meredith Nov 17 '17 at 8:06
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    $\begingroup$ The right way to say this is that curvature is a matrix valued 2-form. What this means is that the curvature for any connection on any bundle always have two manifold indices under which it's antisymmetric and two vector space indices under which there is no specific property in general. The Riemannian case is special in that the two vector space indices are also manifold indices, since the vector bundle is simply the tangent bundle. $\endgroup$ – childofsaturn Nov 18 '17 at 6:29
  • $\begingroup$ Are you sure that the Berry connection is not compatible with the metric? The answer here: physics.stackexchange.com/questions/116985/… shows that it is. $\endgroup$ – TheQuantumMan Jan 2 '18 at 12:47
  • $\begingroup$ Hello, I wonder if you are still about? Is this a generic feature of many aspects of mathematics applied to various definitions of systems? It keeps popping up in physics, now that I am fully aware of it. I will post an example. $\endgroup$ – Gareth Meredith Mar 20 '19 at 17:01

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