condensed matter - 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.)

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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."

Answer

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.