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Can anyone explain to me what is actually "curved" when we speak of a Berry Curvature?

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The surface or skin of the berry, to form usually a spherical like shape. The actual curvature depends on the type of berry and the individual berry picked. – Mew Mar 11 '13 at 5:47
@Mew That's the funniest thing I've read for a while: the context (Physics SE) makes it subtler than it otherwise would be and I think I was nearly to the end before I realised you were having a lend of us. Spike Milligan would be proud of you: mostly puns are very lame, but his and yours are priceless! – WetSavannaAnimal aka Rod Vance Jul 30 '14 at 10:01
up vote 7 down vote accepted

It's the curvature of a connection on a principal U(1) bundle over parameter space.

In describing the quantum Hall effect, we have a Hamiltonian, which depends on a number of parameters $H(R_1,R_2,..R_N)$. Suppose we have the system in its ground state. We now vary the parameters adiabatically (slowly!). As we vary the parameters, we can think of tracing out a curve $(R_1(\lambda), (R_2(\lambda)...(R_N(\lambda))$ in parameter space. As we twiddle the parameters we actually evolve the state using the Schroedinger equation. If we transport it round a closed curve in parameter space, i.e. we return to our starting parameters, we find that the state picks up a phase factor relative to the starting state. (Phase factors live in $U(1)$ the group of unit modulus complex numbers).

The mechanism that allows us to go from a state at one set of parameters to a state at another set is called a connection. In this case the connection is provided by the Schroedinger equation. Saying that the connection has curvature just means that transport of a state round a closed curve using this connection doesn't quite get you back to the state you started with (in fact mathematically the object which defines the curvature is obtained by transport round a little closed parallelogram).

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Let me say it in more heuristic words. When we make a round trip, the wave function may return to itself up to a transformation, a phase in this case. This is similar to the case of parallel transport around curved loops that also rotates the original vector. That rotation is the result of the usual Riemann curvature, $\delta V^a \sim R^a_{bcd} V^b dS^{cd}$, and similarly curvature (field strength) of gauge fields and similar fields results in rotations in other spaces induced by the round trips around infinitesimal loops. – Luboš Motl Mar 11 '13 at 8:34

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