Home > representation theory > Representations of totally disconnected abelian groups

## Representations of totally disconnected abelian groups

So last time I promised to start talking about the Weil representation, but I decided to backtrack a bit and give a rather interesting proof of the Stone-von Neumann theorem which goes through, of all things, sheaf theory. As far as I know this is due to François Rodier, and pretty much everything I am going to say can be found in his 1977 paper Decomposition Spectrale des Représentations Lisses. As you can probably tell from the title, the article was written in French, so I hope it will be useful to some people to have these results in English. Also, some of my proofs will be slightly different (hopefully a bit cleaner) than Rodier’s.

Let us introduce some convenient terminology. I will call a Hausdorff, locally compact, and totally disconnected topological space an $\ell$-space. A topological group whose underlying space is an $\ell$-space is called an $\ell$-group. For example, if $\mathbb{G}$ is an algebraic group over a non-Archimedean local field $F$ and $G = \mathbb{G}(F)$ is the group of $F$-points, then $G$ is an $\ell$-group. Recall that a Hausdorff topological group is an $\ell$-group if and only if it has a basis at the identity consisting of compact open subgroups. We also remind the reader that a complex representation of an $\ell$-group $G$ is called smooth provided that every vector in the representation space has open stabilizer in $G$.

I call an $\ell$-group an $\ell_c$-group provided that it is a filtered union of its compact open subgroups. If $G$ is an abelian $\ell$-group, then $G$ is an $\ell_c$-group if and only if the Pontryagin dual group $\widehat{G}$ is totally disconnected. Any finite-dimensional vector space over $F$ is an $\ell_c$-group under addition, but the multiplicative group $F^{\times}$ is not, since the units in the ring of integers are the maximal compact subgroup.  A more interesting (in particular, nonabelian) example of an $\ell_c$-group is a unipotent group over $F$, e.g. the Heisenberg group.

Of course we hope to eventually prove something about nonabelian groups, but the purpose of this post will be to establish a connection between sheaves and representations of abelian $\ell_c$-groups via the following theorem.

Theorem (Rodier) Let $G$ be an abelian $\ell_c$-group. Then there is an equivalence between the category $\mathcal{R}(G)$ of smooth representations of $G$ and the category $\text{Sh}(\widehat{G})$ of sheaves of complex vector spaces on the Pontryagin dual $\widehat{G}$.

If $X$ is any topological space, let us write $C^{\infty}_c(X)$ for the $\mathbb{C}$-algebra of locally constant, compactly supported complex-valued functions under pointwise multiplication. This algebra is unital only when $X$ is compact, so it will be necessary to work with the following definition. If $A$ is a ring (not necessarily commutative or unital), we say that an $A$-module $M$ is nondegenerate provided that $A \cdot M = M$.

Lemma Let $X$ be an $\ell$-space. The category $\text{Sh}(X)$ of sheaves of complex vector spaces on $X$ is equivalent to the the category of nondegenerate $C^{\infty}_c(X)$-modules.

Proof. We will write $\mathfrak{m}_x$ for the maximal ideal of $C^{\infty}_c (X)$ consisting of functions which vanish at a fixed $x \in X$, and similarly $\mathfrak{m}_K$ denotes the ideal of functions which vanish on some compact open subset $K \subset X$. Given a nondegenerate $C^{\infty}_c (X)$-module $M$, it is natural to consider the vector space $M/\mathfrak{m}_KM$: this will be the space of sections over $K$. It is not hard to see that the associated presheaf whose sections locally have this form (recall that an $\ell$-space has a basis of compact open subsets) is a sheaf, which we call $\widetilde{M}$. Indeed, any compact open cover of a compact open set has a finite refinement consisting of pairwise disjoint compact open sets, so locally patching is easy. The stalk of $\widetilde{M}$ at a point $x \in X$ is just the space $M/\mathfrak{m}_xM$. It is not hard to check that the mapping $M \mapsto \widetilde{M}$ extends to a functor from the category of nondegenerate $C^{\infty}_c (X)$-modules to $\text{Sh}(X)$, and we claim that this functor is an equivalence.
First, observe that $M \mapsto \widetilde{M}$ has a right adjoint which sends a sheaf $\mathcal{F} \in \text{Sh}(X)$ to its space $\Gamma_c(X,\mathcal{F})$ of compactly supported global sections, with the obvious $C^{\infty}_c(X)$-action: the level sets of any $f \in C^{\infty}_c(X)$ are open. The unit morphisms $M \to \Gamma_c(X,\widetilde{M})$ are defined as follows: given $m \in M$, choose a compact open subset $K \subset X$ which contains $\text{Supp } m = \{ x \in X \ | \ m \notin \mathfrak{m}_xM \}$, or equivalently $e_K \cdot m = m$, where $e_K$ denotes the indicator function for $K$. Then pass to the quotient $M/ \mathfrak{m}_K M = \widetilde{M}(K)$, from which we obtain $\widetilde{m} \in \Gamma_c(X,\widetilde{M})$ by extending by zero outside of $K$. This construction is easily seen to be independent of the choice of $K$. Now pick $\mathcal{F} \in \text{Sh}(X)$: the counit is a morphism of sheaves, so we define it locally (over compact open sets). For any compact open subset $K \subset X$, consider the restriction map $\Gamma_c(X,\mathcal{F}) \to \mathcal{F}(K)$. This vanishes on $\mathfrak{m}_K \Gamma_c(X,\mathcal{F})$, and is clearly compatible with the restriction maps of $\mathcal{F}$ and $\widetilde{\Gamma_c(X,\mathcal{F})}$. We leave it to the reader to check that the unit-counit relations are satisfied.
But in fact the unit and counit are isomorphisms. It is clear enough that $M \to \Gamma_c(X,\widetilde{M})$ is surjective: if $s \in \Gamma_c(X,\widetilde{M})$ choose compact open $K \subset X$ containing $\text{Supp} s$ and pick a representative $m \in M$ for $s|_K \in \widetilde{M}(K) = M/ \mathfrak{m}_K M$, then note that $e_K \cdot m$ is sent to $s$ by checking locally. Injectivity of the unit is much harder: observe that it suffices to prove that for $m \in M$, we have $\text{Supp } m = \varnothing$ if and only if $m = 0$. The nontrivial part is showing that if $m \in \mathfrak{m}_xM$ for all $x \in X$, then $m = 0$. First pick a compact open subset $K \subset X$ such that $e_K \cdot m = m$, which we know is possible because $M$ is nondegenerate. Then we can identify $e_K \cdot C^{\infty}_c(X)$ with the (unital) algebra $C^{\infty}_c(K)$, and the maximal ideals of $C^{\infty}_c(K)$ all have the form $e_K \cdot \mathfrak{m}_x$ for $x \in K$. Moreover, $e_K \cdot M$ is a module for $C^{\infty}_c(K)$, and we just need to show $m = e_K \cdot m = 0$, so by all of this we are reduced to the case that $X$ is compact. Thus we are dealing with an ordinary unital, commutative $\mathbb{C}$-algebra $A = C^{\infty}_c(X)$, and it is not hard to see that for each $x \in X$ the localization map $A \to A_{\mathfrak{m}_x}$ is surjective with kernel $\mathfrak{m}_x$, so we have an isomorphism $A/\mathfrak{m}_xA \to A_{\mathfrak{m}_x}$ (actually both are canonically identified with $\mathbb{C}$). But it is a well-known fact of commutative algebra that if $m$ is zero in every localization $M_{\mathfrak{m}_x}$ then $m = 0$.

As for the counit, notice first that that the canonical maps $\Gamma_c(X,\mathcal{F}) \to \mathcal{F}_x$ for $x \in X$ are all surjective since $X$ has a basis of compact open subsets. This shows that the counit is an epimorphism, and it is just as clear that if $s \in \Gamma_c(X,\mathcal{F})$ goes to zero in $\mathcal{F}_x$ then $s \in \mathfrak{m}_x\Gamma_c(X,\mathcal{F})$. Thus the counit induces an isomorphism on stalks, so it is an isomorphism.

$\Box$

Back to groups: for any $\ell$-group $G$ we write $\mathcal{H}(G)$ for the Hecke algebra of $G$, i.e. the algebra of locally constant, compactly supported complex-valued functions under convolution. It is well-known (and not so hard to see) that the category of nondegenerate $\mathcal{H}(G)$-modules is isomorphic to the category of smooth representations of $G$.

Proof of Rodier’s theorem. Let $G$ be an abelian $\ell_c$-group. As we just remarked, $\mathcal{R}(G)$ is isomorphic to the category of nondegenerate $\mathcal{H}(G)$-modules. The Fourier transform provides an isomorphism of algebras $\mathcal{H}(G) \to C^{\infty}_c(\widehat{G})$, so $\mathcal{R}(G)$ is in fact isomorphic to the category of nondegenerate $C^{\infty}_c(\widehat{G})$-modules. Now the $\ell_c$ assumption on $G$ implies that $\widehat{G}$ is totally disconnected and hence an $\ell$-space, so we can apply the lemma to see that the category of nondegenerate $C^{\infty}_c(\widehat{G})$-modules is equivalent to $\text{Sh}(\widehat{G})$ as desired.

$\Box$