Time reversal symmetry implies that fermions are massless? In TASI Lectures on Emergence of
Supersymmetry, Gauge Theory and String in
Condensed Matter Systems some continuous limit of lattice model with fermions considered. And on page 6 there is a statement:

The fermions are massless without any fine tuning, which is protected by
the time reversal symmetry and the inversion symmetry.

Why time-reversal symmetry imply that fermions are massless?
 A: For relativistic fermions in 2+1 and 6+1 (mod 8) dimensions a Dirac mass changes sign under time reversal.  This is because we do usually do time reversal using the matrix ${\mathcal T}$ that obeys
$$
{\mathcal T}\gamma^\mu {\mathcal T}^{-1} = (\gamma^\mu)^T
$$
and setting
$$
{\mathfrak I^{-1}}  \psi(x,t) {\mathfrak I}= \eta_T {\mathcal T}\psi(x,-t).
$$
Here ${\mathfrak I}$ is the antiunitary time reversal operator.
The ${\mathcal T}$   matrix does not exists in 3 and 7 (mod 8) spacetime dimensions but we can still  implement a form of time reversal by using the matrix ${\mathcal C}$ that obeys
$$
{\mathcal C}\gamma^\mu {\mathcal C}^{-1} = -(\gamma^\mu)^T
$$
and setting
$$
{\mathfrak I^{-1}}  \psi(x,t) {\mathfrak I}= \eta_T {\mathcal C}\psi(x,-t).
$$
This does all the usual things that we want from time reversal, but it  flips the sign of $\bar\psi\psi$.  The net reults is that while a mass violates parity ${\mathsf P}$ in any odd space-time dimension, it only violates ${\mathsf T}$ in 3 and 7 (mod 8) dimensions. In 5=1+4 (mod 8) a mass does not violate   ${\mathsf T}$.
