Hourglass fermions + going from glide symmetry eigenvalues to energy eigenvalues In this paper, the authors describe how you'd get an hourglass fermion.
The gist (page 2, second column) is that you have the operator $ \bar M_x$ consisting of a translation along z, $ t( c  \hat z /2) $ paired with a reflection along x. Fine, you can then argue that because $ \bar M_x ^2 = t( c \hat z ) \bar E $-- where $ \bar E $ is a rotation by 2 pi in the spin-- and because this is equivalent to $ - exp(-i k_z c/2 ) $, it means that the eigenvalues of $ \bar M_x $ are $ \pm i exp (-i k_z c/2 ) $.
Ok,and then you can argue that as you interpolate along $ k_z\in [0, \pi/c] $, you get different possible topologies where either you get a pair goes to a pair, or a crossing is created, generating the typical hourglass shape in the $ \bar M_x$ eigenvalue vs. $\textbf{k}$ diagram.
What I'm less sure about is why this implies a crossing in the energy vs. k space. If there's degeneracy in the  $ \bar M_x $ eigenvalues, why is there a crossing in the energies? I can understand that the eigenvectors of $ \bar M_x $ should match that of the energies if the matrices commute, unless the eigenvectors are degenerate (ex. at the crossing point), but why do they have to follow the same function and still cross? Is it because you expect that you must have another degeneracy somewhere is E vs. k?
 A: ok, the issues in my question are (relatively silly) misconceptions.
The relevant diagrams in the linked paper, for instance FIG. 1D, are clearly energy versus momentum. In some diagram that would plot $ \bar M_x $, you wouldn't get the hourglass at all, since the corners of the hourglass would be separated ($\bar M_x$'s eigenvalues would be $ \pm i $ ) ; there's no degeneracy at the corners.
For anyone else's benefit: a reason I was confused was why a certain set of $ \bar M_x $ eigenvals corresponded to one of H's, for ex. $ {1,1} $ with $ 1 $ (or the positive H eigenvalue at $ k_z = \pi $, whatever you want to call it).
for the eigenvalue branch, ex. $ i exp (- i k_z  /2 ) $, at $ k_z = 0 $, this is just $ i $. this particular branch becomes $ 1 $ when $ k_z = \pi $. the other branch goes from $ -i $ to $ -1 $.
now, we know that at each (Ham) energy at $ k_z = 0 $ is degenerate cuz its a TRIM*, $ | 1\rangle $ and $ | 2 \rangle $. These will split as you interpolate along $ k_z $.
Fine, we also know that because of time reversal symmetry at the TRIMs, we should have that any degenerate points within the final hamiltonian eigenvalue should be TRS invariant (Hamiltonian commutes with TR at the high symmetry points, and in theory the $ \bar M_x$ commutes with the hamiltonian). This is why we need $ i, -i $ in a set in a specific eigenvalue, and same for $ -1, -1$.
So, the reason why there's a crossing is because, I guess, you can associate each branch for $ \bar M_x $ to a branch for the Hamiltonian since they commute, and since you need to switch the members of each set as described above, you end up needing to cross the Hamiltonian energy bands as well.
*in this case, I'm guessing we wouldn't use the argument involving Kramer's thm. for TRIMs since its mirror glide which isn't exactly time reversal, but close enough (seems to be the same argument as the paper)
