How is the topological $Z_2$ invariant related to the Chern number? (e.g. for a topological insulator)

This question relates to the $Z_2$ invariant defined e.g. for topological insulators:

Is it correct to relate $Z_2$ = 1 to an odd Chern number and $Z_2$ = 0 to an even Chern number?

If yes, is it also correct to think of an even or odd Chern number in terms of an even or odd number of band crossings across the Fermi energy? (If it's odd, there must be a band connecting the valence to the conduction band and therefore provide a topological protected surface state.)

Edit: These lecture notes* (under Point H) state: "The formula (49) was not the first definition of the two-dimensional Z2 invariant, as the original Kane-Mele paper gave a definition based on counting of zeros of the “Pfaffian bundle” of wavefunctions. However, (49) is both easier to connect to the IQHE and easier to implement numerically."

and furthermore:

"...and the Chern numbers of the two spheres are equal so that the total Chern number is zero. The above argument establishes that the two values of the Z2invariant are related to even or odd Chern number of a band pair on half the Brillouin zone."

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The answer of David Aasen is correct, but let me add some comments which connect to your question of the relation of between the $\mathbb Z_2$ invariant $\nu$ and the first Chern-Number $C_1$.

Such a relation does not exist unless you require some extra symmetry than the generic symmetries usually required in the classification of topological insulators (such as time-reversal invariance in this case). Say the Hamiltonian is invariant under spin rotations along the $z$-axis (so a $U(1)$ subgroup of $SU(2)$ in left invariant), then the Hamiltonian can be block-diagonalized as

$H = \begin{pmatrix} H_\uparrow & \\ & H_\downarrow \end{pmatrix},$

where the indices refer to spin-up and down degrees of freedom. Due to time reversal symmetry we have that $H_\downarrow(k) = H^*_\uparrow(-k)$. The system now consist of two copies of Quantum Hall effects with counter propagating edge states of opposite spin. As Davis Aasen says, the chern number is zero $C_1 = C_1^\uparrow + C_1^\downarrow = 0$. The difference however, the "spin Chern number", $C_1^\uparrow - C_1^\downarrow = 2C_{spin}$ can be non-zero and can be calculated by the Chern-numbers of the spin up/down sectors. As long as $S_z$ is preserved the spin Chern-number can be any integer $C_{spin}\in\mathbb Z$.

But if we add off-diagonal elements, and thus break the rotation symmetry along $z$, the invariant breaks down to $\nu = C_{spin}\,\text{mod}\,2\in\mathbb Z_2$ (as was shown by Kane and Mele). So topological trivial/non-trivial phases are characterized by even and odd spin-Chern numbers $C_{spin}$, not the original Chern number $C_1$. This however only makes sense when you have this extra symmetry.

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Thank you, this explanation did help! But I'm still trying to make an analogy to the Quantum Hall state where the Chern number was exactly equal to the number of one-way edge states (i.e. = winding number). How would you describe the relation of the "spin Chern number" to the number of edge states or the # of band crossings for a TI? I'm focusing on this picture because it would help me understand the topology (e.g. compare the "Seven bridges of Koenigsberg" from Euler".) –  Matthias Dec 16 '12 at 2:51

For a time reversal invariant bloch hamiltonian (such as in a $\mathbb{Z}_2$ topological insulator) the Chern number is always zero.

The topological invariant $\nu = 0,1$ classifies the insulator as trivial or topological. This can be found by counting the number of times the surface energy bands intersect the Fermi energy mod 2 as you mentioned above.

For a reference see the RMP by Hasan and Kane, http://rmp.aps.org/pdf/RMP/v82/i4/p3045_1 Sections II.B.1 and II.C.

I hope this was helpful. I am trying to learn about these topics as well.

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Hello David, thanks for your answer. That's what I thought too until I discovered these lecture notes* where it says: "as the original Kane-Mele paper gave a definition based on counting of zeros of the “Pfaffian bundle" and "The above argument establishes that the two values of the Z2 invariant are related to even or odd Chern number of a band pair on half the Brillouin zone." (under Point H): *socrates.berkeley.edu/~jemoore/Moore_group,_UC_Berkeley/… I hope somebody could comment on this. –  Matthias Dec 4 '12 at 2:18
I see. Good point. Thanks for pointing out these notes, I will read them when I have more time. –  Stackexchange_user23 Dec 4 '12 at 2:28
Note that in those same notes, Joel writes at the beginning of section F that "nonzero Chern numbers cannot be realized with time-reversal invariance". I think you might have a misunderstanding with the arguments he is presenting to justify the Z2 invariant, but this is the correct answer. –  wsc Dec 4 '12 at 2:54
Thanks for pointing this out. This is exactly the point I don't understand. He shows that the Chern number can be even or odd of a band pair on half the Brillouin zone and I think he explicitly includes time-reversal. Does he then sum then up to get back to the full Brillouin zone and they somehow always cancel and produce zero? –  Matthias Dec 4 '12 at 3:57
@Matthias The usual Chern number is defined on a vector bundle over the BZ torus and is zero when you have time-reversal symmetry. I think that the odd/even comment you are talking about is related to the invariant $D$ given in eq. (49). I haven't read the argument but I think what he does is the following. Time-reversal symmetry is a non-local symmetry in the BZ, so he defines the effective BZ by identifying points on the BZ related by a time-reversal transformation. We can now define a Chern number on the vector bundle over the EBZ, but this is NOT the same as the original chern number. –  Heidar Dec 4 '12 at 11:21