Currently, I am puzzled by the formula 2.8 in the paper arXiv;1907.03363, where the author states that $Z_R(\beta)=0$ for $N$ being odd. I am having difficulty understanding the discussion in the paragraph on page 13.
"Now let us consider the Ramond sector. For odd $N $, the path integral of the SYK model in the Ramond sector (with no operator insertions) is actually identically zero. To see this, first note that if $H = 0$, so that the action (2.4) consists only of the kinetic energy, then each of the $ψ_k$ has a zero-mode in the Ramond sector, so in all there are an odd number of zero-modes. Now including the Hamiltonian, a term that is proportional to $H^r$ for some $r$ has an insertion of $qr$ fermions, which is an even number since $q$ is even. An even number of these fermions can be paired up by propagators, and the remaining ones – also an even number – can be used to soak up zero-modes. Since we started with an odd number of fermion zero-modes, there is always an odd number left over. In particular, we can never soak up all of the zero-modes, and therefore the Ramond sector path integral of the odd $N$ SYK model is identically zero."
Is there a more detailed discussion regarding this formula? From the perspective of the path integral, there are also zero modes present when $N$ is even. I am wondering why these zero modes, when $N$ is even, do not cause the path integral to vanish. Specifically, when considering the case $H=0$, the zero mode always appears in the determinant $\prod_i\omega_i$, right?