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The Berry connection and the Berry phase should be related. Now for a topological insulator (TI) (or to be more precise, for a quantum spin hall state, but I think the Chern parities are calculated in the same fashion for a 3D TI). I can follow the argument up to defining the Chern parity $\nu$, where the Berry connection $\textbf{A}$ is explicitly within this equation where it gets integrated over half the Brillouin zone as it was described in Liang Fu's thesis from 2009 on page 31 to calculate the Chern parity or the $Z_2$ invariant(

$\nu = \frac{1}{2\pi} \left( \oint\limits_{\partial(A + B)} d\textbf{k} \cdot \textbf{A} - \iint\limits_{A + B} dk_x dk_y \nabla \times \textbf{A} \right) \text{mod } 2 $

It is known that the Berry phase for the surface state is $\pi$ and it is also known how the Berry connection $\textbf{A}$ leads to a Berry phase ( But I can't see how this is equal to $\pi$:

$\left( \oint\limits_{\partial(A + B)} d\textbf{k} \cdot \textbf{A} - \iint\limits_{A + B} dk_x dk_y \nabla \times \textbf{A} \right) \text{mod } 2 = \pi$ ?

Is there a simple way to show that the surface states of a topological insulator must have a $\pi$ Berry phase? Maybe by invoking time-reversal symmetry on this equation?

For the Integer Quantum Hall Effect the conductance was used to show how the Berry phase is connected to the Berry connection. But I would like to avoid using charge polarization, I just want to see the direct relation between the Berry connection on the BZ and the resulting Berry phase of the wavefunctions that is exactly equal to $\pi$, e.g. a fermion must undergo two complete rotations to acquire a phase of $2\pi$. Is there a simple argument to relate this equation with the time reversal contraint to the Berry phase?

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@NanoPhys answered a very similar question few days ago. See the question for the $\pi$-phase-shift. –  FraSchelle Jul 18 '13 at 5:05
Thanks for pointing this out! Unfortunately he does not explain it in terms of the Berry connection, but he already assumes a correct Hamiltonian that immediately leads to a Dirac cone. –  Mike Jul 18 '13 at 13:06
I think you need the wave-function to compute the Berry connection, isn't it ? If you just impose some symmetries, you will just be able to prove that the Chern-number is an odd/even integer, or zero (trivial case), or only $\pm 1$, but you should not end up with a specific value. The Berry phase of a fermions around a Dirac cone is $\pi$, as NanoPhys shown. At a topological insulator edge, you have a Dirac cone. –  FraSchelle Jul 19 '13 at 10:25
The Berry connection is always defined as $\mathbf{A}=\mathbf{i}\left\langle n\left(\mathbf{k}\right)\right|\nabla_{\mathbf{k}}\left|n\left(\mathbf{k}\right)‌​\right\rangle $ whereas the Berry phase is *always* defined as $\gamma=\int_{\text{BZ}}\mathbf{A}\left(\mathbf{k}\right)\cdot d\mathbf{k}$, with BZ the Brillouin zone. So they are always connected. The state $\left|n\left(\mathbf{k}\right)\right\rangle$ is the Block wave function of the band $n$, depending on the wave vector $\mathbf{k}$. –  FraSchelle Jul 19 '13 at 10:29
Thanks for continuing this discussion. Yes, I agree that they are always connected. But the Berry phase is explicitly in this equation I posted, when you apply Stokes' theorem you can always bring it in this form: $\gamma = \oint\limits_{\partial(A + B)} d\textbf{k} \cdot \textbf{A} - \iint\limits_{A + B} dk_x dk_y \nabla \times \textbf{A}$. –  Mike Jul 19 '13 at 14:58

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