Quantum determinant–permanent reciprocity conjecture

Let $\operatorname{Sing}_{n,q}^{\mathrm{per}}$ and $\operatorname{Sing}_{n,q}$ be the singular sets for the quantum permanent and quantum determinant, respectively. For $\lambda\in{\widetilde{\mathcal{L}}_n}^{\mathrm{dom}}$, let $\lambda'$ denote the transposed diagram, and let $m_q^\lambda(\alpha)_{\mathrm{per}}$ and $m_q^\lambda(\alpha)$ be the corresponding multiplicities. Quantum determinant–permanent reciprocity conjecture. (1) If $\alpha(q)\in\operatorname{Sing}_{n,q}^{\mathrm{per}}$, then $\alpha(q^{-1})\in\operatorname{Sing}_{n,q}$. (2) The map

$$\operatorname{Sing}_{n,q}^{\mathrm{per}}\ni\alpha(q)\longmapsto\alpha(q^{-1})\in\operatorname{Sing}_{n,q}$$

is bijective. (3) For every $\alpha(q)\in\operatorname{Sing}_{n,q}^{\mathrm{per}}$ and every $\lambda\in{\widetilde{\mathcal{L}}_n}^{\mathrm{dom}}$,

$$m_q^\lambda(\alpha(q))_{\mathrm{per}}=m_q^{\lambda'}(\alpha(q^{-1})).$$

This expresses a mirror symmetry under $q\mapsto q^{-1}$ and transposition of diagrams. The source notes that it is true when $q=1$, but leaves the general case conjectural.