Fredholm) equipped with an action by the $n$th Clifford algebra $\operatorname{Cliff}(n)$ (which should (super)commute with $G$). The presence of a time-reversal symmetry equips ${\mathcal H}$ with a real structure, acted on by the real Clifford algebra $\operatorname{Cliff}(n,{{\mathds}R})$. The supersymmetric ground states are then a finite-dimensional $\operatorname{Cliff}(n,{{\mathds}R})$-module $V$.[^18]

The usual [supersymmetric index]{} of the SQM model ignores the time-reversal symmetry: it depends just on $V\otimes {{\mathds}C}$ as a $\operatorname{Cliff}(n,{{\mathds}C})$-module. When $n$ is even, $\operatorname{Cliff}(n,{{\mathds}C})$ has two irreducible modules, differing by parity. Choose one of them arbitrarily to be “the” irrep $I$; then $V \otimes {{\mathds}C}\cong I^{a|b} = I \otimes_{{\mathds}C}{{\mathds}C}^{a|b}$, where ${{\mathds}C}^{a|b}$ means the (complex) supervector space with graded dimension $(a,b)$. The ordinary [*index*]{} of $V$ is simply $a-b$.[^19] But in the presence of a time-reversal symmetry, we don’t just have the $\operatorname{Cliff}(n,{{\mathds}C})$-module $V \otimes {{\mathds}C}$ — we have the $\operatorname{Cliff}(n,{{\mathds}R})$-module $V$. It turns out that when $n = 2 \pmod 4$, there is only one irreducible $\operatorname{Cliff}(n,{{\mathds}R})$-module $J$, with complexification $J \otimes {{\mathds}C}\cong I^{1|1}$. Thus the index vanishes when $n = 2 \pmod 4$. And when $n = 4 \pmod 8$, there are two irreducible $\operatorname{Cliff}(n,{{\mathds}R})$-modules, $J^{1|0}$ and $J^{0|1}$, but $J^{1|0} \otimes {{\mathds}C}= I^{2|0}$ splits as two copies of the irreducible 