ground states $V$ are isomorphic, as a $\operatorname{Cliff}(2,{{\mathds}R})$-module, to ${{\mathds}C}^{a|a}$ for some integer $a$, and the [*mod-2 index*]{} is $a \pmod 2$. A deformation will involve adding or subtracting from $V$ the underlying $\operatorname{Cliff}(2,{{\mathds}R})$-module of some $\operatorname{Cliff}(3,{{\mathds}R})$-module. The even subalgebra of $\operatorname{Cliff}(3,{{\mathds}R})$ is isomorphic to the quaternion algebra ${{\mathds}H}$. Because ${{\mathds}H}$ is a skew field, $\operatorname{Cliff}(3,{{\mathds}R})$ is irreducible as a module over itself. Call this irreducible module ${{\mathds}H}^{1|1}$. Since $\dim_{{\mathds}C}{{\mathds}H}$ is even, if you add or subtract some multiple of ${{\mathds}H}^{1|1}$ to ${{\mathds}C}^{a|a}$, you do not change the value of $a \pmod 2$.

The story repeats when $n = 1$ or $2 \pmod 8$, because the category of $\operatorname{Cliff}(n,{{\mathds}R})$-supermodules depends only on the value of $n \pmod 8$. What about when $n = 3$? The ground states $V$ for an SQM model are then isomorphic to ${{\mathds}H}^{a|a}$ for some nonnegative integer $a$, and so we may still contemplate a [*mod-2 index*]{} defined to be the value of $a \pmod 2$. But now this mod-2 index is not a deformation invariant because $\operatorname{Cliff}(4,{{\mathds}R})$ is not irreducible over itself. In fact, both irreps of $\operatorname{Cliff}(4,{{\mathds}R})$ restrict over $\operatorname{Cliff}(3,{{\mathds}R})$ to copies of ${{\mathds}H}^{1|1}$, and so only the dataless “$a \pmod 1$” is a 