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According to Fu & Kane (2006), systems with simultaneous time-reversal invariance and inversion symmetry have their $\mathbb{Z}_2$ topological invariant given by the product of the parity eigenvalue at the four TRIMs. Thus, there are clearly topologically nontrivial such systems.

According to Moore & Balents (2006) (page 2, the paragraph right after equation (4), "As a quick example..."), systems with inversion symmetry and time-reversal invariance are always topologically trivial because their "effective brillouin zone" is basically $S^2$ and thus all maps $S^2 \to S^4$ (where $S^4$ is the space of all $4\times4$ Dirac-Hamiltonians with no other degeneracies other than the Kramer's degeneracies) are topologically in the same class because $\pi_2(S^4) = 0$.

Why do these two articles disagree? Or am I misunderstanding something basic? Furthermore, later on, Moore et al then transform the EBZ into a sphere by contraction and claim that the maps now from the sphere to the space of Hamiltonians are not trivial. Why is suddenly the sphere domain produce non-trivial Chern numbers?

EDIT: The EBZ, as defined by Moore, is the minimal part of the torus necessary in order to specify $H(k)$ fully. The point is that due to time-reversal symmetry, we have $H(-k) = \Theta H(k) \Theta^{-1}$ and so we may "discard" half to the torus and still keep all the information. There is a subtlety, however, at the boundaries of this half. If we take the half torus we want as $k_x \geq 0$ (many other possibilities exist) then we basically stitch together the bottom and top parts of the half-torus $k_y = \pi$ with $k_y = -\pi$ to get a cylinder (this is part of the original topology of the torus) but now at the lines (which for the cylinder become circles) $k_x = 0$ and $k_x = \pi$ we need to keep the special time reversal condition $H(-k) = \Theta H(k) \Theta^{-1}$, so that same-distance-from-zero points on these two circles are related by $H(-k) = \Theta H(k) \Theta^{-1}$. This is the EBZ.

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    $\begingroup$ What's a "TRIM"? $\endgroup$
    – hft
    Commented Mar 2, 2015 at 22:38
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    $\begingroup$ Time-Reversal-Invariant-Momenta = The four points on the torus for which $k = -k$, which are basically given by $\{(0,0),(0,\pi),(\pi,0),(\pi,\pi)\}$. $\endgroup$
    – PPR
    Commented Mar 2, 2015 at 22:41
  • $\begingroup$ Also EBZ (effective Birllouin zone ?) should be define (please do that in the post, not as comment). I feel part of this question might already been answered in the following posts: physics.stackexchange.com/q/70057/16689 , physics.stackexchange.com/q/104258/16689 or physics.stackexchange.com/q/111440/16689 (still not answered) $\endgroup$
    – FraSchelle
    Commented Mar 4, 2015 at 10:58
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    $\begingroup$ @FraSchelle, thanks for the links. I feel they are slightly different though: I can see homotopy arguments as to why at certain times you can classify maps $T^2\to M$ using $\pi_2(M)$ instead (and thus you take the torus to be a sphere). But in this case, Moore seems to be doing something else. First he passes through a cylinder, and only in special circumstances is this cylinder a sphere, but otherwise, he treats it as a special cylinder, basically if I understand correctly, as in here: math.stackexchange.com/questions/1174342/… $\endgroup$
    – PPR
    Commented Mar 4, 2015 at 11:56

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