${\mathcal X}$ wasn’t really “noncompact” at all, because it flows to a compact theory, and the factor of $2$ is the one coming from (\[eqn.mofn\]) when $n=4$.

If, on the other hand, supersymmetry is not spontaneously broken in the boundary theories ${\mathcal Y}_\pm$, then the index of ${\mathcal X}$ receives a fractional contribution from each eigenvalue that lands on $0$ in the limits $p \to \pm \infty$. After multiplying by the factor of $2$ coming from (\[eqn.mofn\]), we find that the index of ${\mathcal X}$ is odd depending on the number of supersymmetric ground states in the boundary theories ${\mathcal Y}_\pm$. And this number is exactly the non-deformation-invariant mod-2 index of ${\mathcal Y}_\pm$!

The same argument holds whenever ${\mathcal X}$ has degree $n = 4 \pmod 8$. When $n = 0 \pmod 8$, a similar argument holds without a factor of $2$. To give a unified formula, we complexify and strip off the spectator fermions. Then, after complexifying, the supersymmetric ground states in ${\mathcal Y}_\pm$ form a vector space isomorphic to ${{\mathds}C}^{a|a}$, with $a \in m{{\mathds}Z}$, and the mod-2 index is $\frac a m \pmod 2$. We will call this number $a$ the “bosonic index” of ${\mathcal Y}_\pm$. The end 