of ${\mathcal B}$ can be converted into a noncompact SQFT ${\mathcal F}$ by reversing the steps above, i.e. promoting the parameter of the deformation family to a dynamical $(0,1)$ chiral multiplet.

What is the elliptic genus of ${\mathcal F}$? The question is subtle because, being noncompact, ${\mathcal F}$ has continuous spectrum in its Hamiltonian. Because of this, the usual supersymmetric cancelation argument verifying that the elliptic genus is holomorphic breaks down. Let us work in the IR limit. In this limit, the Witten genus $\eta(\tau)^n Z_{RR}({\mathcal F})(\tau,\bar\tau)$ will automatically transform as a weight $(\frac n2,0)$ modular form under the action of $\mathrm{SL}(2,{{\mathds}Z})$ acting simultaneously on both $(\tau,\bar\tau)$, since modularity is manifest from the path-integral description of the index.

Let us try to prove that $Z_{RR}({\mathcal F})$ is holomorphic. We will fail, and by failing we will instead compute the [*holomorphic anomaly*]{} of ${\mathcal F}$. We will try to apply the usual proof of holomorphicity of the index. For any QFT,[^13] $$\frac{\partial}{\partial\bar{\tau}}Z_{RR}({\mathcal F}) = - 2\pi i (\text{torus one-point function of } \bar{T}  \mbox{ in ${\mathcal F}$}).$$ Use the supersymmetry: $\bar{T} = \frac12 [\bar{G}_0,\bar{G}]$, where the commutator $[,]$ means the supercommutator, $\bar{G}_0$ is the generator of ${\mathcal N}=(0,1)$ supersymmetry and $\bar{G}$ 