Home > representation theory > A generalized Stone-von Neumann theorem

## A generalized Stone-von Neumann theorem

Continuing our terminology from the last post, let $G$ be an $\ell$-group, $U \subset G$ a closed normal abelian $\ell_c$-subgroup, and $\chi : U \to \mathbb{T}$ a smooth character. Now $G$ acts on $U$ by conjugation, which induces an action on the Pontryagin dual $\widehat{U}$.

If $M$ is a smooth representation of $G$, we can restrict to $U$ and consider the corresponding sheaf $\widetilde{M}$ on $\widehat{U}$, which comes from the equivalence proved last time. The stalk $\widetilde{M}(\chi)$ is identified with the space of $\chi$-coinvariants of $M$, i.e. the quotient of $M$ by the subspace spanned by vectors of the form $\chi(u) \cdot m - m$ for $u \in U$ and $m \in M$. Notice that for any $g \in G$ the action of $G$ on $M$ induces an isomorphism of stalks $\widetilde{M}(\chi) \to \widetilde{M}(\chi^g)$, and in particular $\widetilde{M}(\chi)$ has a natural action by the stabilizer $Z_G(\chi)$.

It is easy to see that the support $\text{Supp } \widetilde{M} \subset \widehat{U}$ is a $G$-invariant subspace: in particular, if $\text{Supp } \widetilde{M}$ is contained in the $G$-orbit $\chi^G$ of $\chi$, then either $\text{Supp } \widetilde{M} = \chi^G$ or $\text{Supp } \widetilde{M} = \varnothing$.

Theorem (Rodier) Suppose the orbit $\chi^G$ is locally closed in $\widehat{G}$ and that the natural map $Z_G(\chi) \backslash G \to \chi^G$ is a homeomorphism. Then the functor $\widetilde{M} \to \widetilde{M}(\chi)$ is an equivalence from the category of smooth representations of $G$ whose restriction to $U$ is supported on $\chi^G$ and the category of smooth representations of $Z_G(\chi)$ whose restriction to $U$ is supported at $\chi$.

Let us motivate this rather general theorem by deducing the Stone-von Neumann theorem.

Corollary (Stone-von Neumann) Let $G$ be the Heisenberg group over a non-Archimedean local field $F$ of characteristic different from $2$ and choose a nontrivial smooth character $\psi : F \to \mathbb{T}$. Then there is a unique irreducible smooth representation of $G$ with central character $\psi$.

Proof. Recall that as a space $G = V \times F$, where $V$ is a finite-dimensional symplectic vector space over $F$. Choose a lagrangian subspace $L \subset V$ and let $U = L \times F$, which is a closed normal abelian $\ell_c$-subgroup of $G$. We can inflate $\psi$ to a smooth character $\chi : U \to \mathbb{T}$ along the natural projection $U \to F$.

Now $U$ is a finite-dimensional vector space over $F$, as is $\widehat{U}$, which is noncanonically isomorphic to $U$. There is a natural projection $U \to L$ with kernel $F$, which corresponds to a linear subspace $\widehat{L} \subset \widehat{U}$ of codimension one, and the orbit $\chi^G$ is just the affine subspace $\chi \cdot \widehat{L}$. So $\chi^G$ is actually closed in $\widehat{U}$. Note that in our situation $Z_G(\chi) = U$, and the map $L \backslash V = U \backslash G \to \chi \cdot \widehat{L}$ is the composition of the isomorphism $L \backslash V \to \widehat{L}$, which comes from $\psi$ and the symplectic form, with the obvious isomorphism $\widehat{L} \to \chi \cdot \widehat{L}$.

Since $Z_G(\chi) = U$, the theorem says that there is an equivalence between the category of smooth representations of $G$ whose restriction to $U$ is supported on $\chi^G$ and the category of smooth representations of $U$ supported at $\chi$. Obviously the latter category has a unique simple object, namely the character $\chi$ itself, so there is a corresponding unique irreducible smooth representation of $G$ whose restriction to $U$ is supported on $\chi^G$.

So it remains to show that a smooth irreducible representation of $G$ has central character $\psi$ if and only if its restriction to $U$ is supported on $\chi^G$. But this is just a test of eyesight: the characters of $U$ in $\chi^G$ are precisely those which restrict to $\psi$ on $F$.

$\Box$

I think I’ll save the proof of the theorem for next time, which will make the next post a bit boring but will prevent this one from being extremely long.

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