On homomorphisms from finite subgroups of $SU(2)$ to Langlands dual pairs of groups

TL;DR

This paper investigates a conjecture from physics relating homomorphisms from finite subgroups of SU(2) to Langlands dual groups, providing proofs for specific cases and encouraging a more general proof approach.

Contribution

The authors prove the conjectured equality of homomorphism counts for certain non-Abelian finite subgroups and specific Lie groups, extending known results beyond Abelian cases.

Findings

Proved the conjecture for (SU(n),PU(n)) and (Sp(n),SO(2n+1)) for arbitrary finite subgroups.

Established the conjecture for (PSp(n),Spin(2n+1)) with exceptional finite subgroups.

Presented a refined conjecture and proofs for some cases, inviting a unified proof approach.

Abstract

Let $N(\Gamma,G)$ be the number of homomorphisms from $\Gamma$ to $G$ up to conjugation by $G$. Physics of four-dimensional $\mathcal{N}=4$ supersymmetric gauge theories predicts that $N(\Gamma,G)=N(\Gamma , \tilde G)$ when $\Gamma$ is a finite subgroup of $SU(2)$, $G$ is a connected compact simple Lie group and $\tilde G$ is its Langlands dual. This statement is known to be true when $\Gamma=\mathbb{Z}_n$, but the statement for non-Abelian $\Gamma$ is new, to the knowledge of the authors. To lend credence to this conjecture, we prove this equality in a couple of examples, namely $(G,\tilde G)=(SU(n),PU(n))$ and $(Sp(n),SO(2n+1))$ for arbitrary $\Gamma$, and $(PSp(n),Spin(2n+1))$ for exceptional $\Gamma$. A more refined version of the conjecture, together with proofs of some concrete cases, will also be presented. The authors would like to ask mathematicians to provide a more uniform proof applicable to all cases.