# The discrete spectrum is discrete

No, this post is not an exercise in tautological reasoning: the point is that the word “discrete” is relatively overloaded. In the theory of automorphic forms, “discrete spectrum” (or “spectre discret”) is the same as “cuspidal spectrum”, and refers to those automorphic representations (of a given group $G$ over a given global field $F$) which are realized as closed subspaces of the relevant representation space on $L^2(G(\mathbf{Q})\backslash G(\mathbf{A}_F))$, by opposition with the “continuous spectrum”, whose components may fail to be actual subrepresentations.

However, we can also think of automorphic representations as points in the unitary dual of the relevant adélic group $G(\mathbf{A}_F)$, and this has a natural topology (the Fell topology), for which it then makes sense to ask whether the set of all cuspidal automorphic representations is discrete or not.

The two notions might well be unrelated: for instance, if we take the group $\mathrm{SL}_2(\mathbf{R})$ and the direct sum over $t\in\mathbf{Q}$ of the principal series with parameter $t$, we obtain a representation containing only “discrete” spectrum, but parameterized by a “non-discrete” subset of the unitary dual.

It is nevertheless true that the discrete spectrum is discrete in the unitary dual, in the automorphic case. I am sure this is well-known. (Unless the topology on the unitary dual is much weirder than I expect; my expertise in this respect is quite limited, but I remember asking A. Venkatesh who told me that the pathologies of the unitary dual topology are basically irrelevant as far as the automorphic case is concerned). I think this must be well-known, but I don’t remember seeing it mentioned anywhere (hence this post…)

Here is the argument, at least over number fields, and for $\mathrm{GL}_n$. Let $\pi$ be a given cuspidal representation. The unitary dual topology on the adélic group is, if I understand right, the restricted direct product topology with respect to the unramified spectrum. So a neighborhood of $\pi$ is determined by a finite set $S$ of places and corresponding neighborhoods $U_s$ of $\pi_s$ for $s\in S$. We want to find such a neighborhood in which $\pi$ is the unique automorphic cuspidal representation.

First, we can fix a neighborhood $U_{\infty}$ of the archimedean (Langlands) parameters that is relatively compact. Next, we note (or claim…) that for a finite place $v$, the exponent of the conductor is locally constant, so we get neighborhoods $U_v$ of $\pi_v$, with $U_v$ the unramified spectrum when $\pi$ is unramified at $v$, such that all automorphic representations in $U=U_{\infty}\times \prod_{v} U_v$
have arithmetic conductor equal to that of $\pi$. With the archimedean condition, it follows that the Iwaniec-Sarnak analytic conductor of cuspidal representations in $U$ is bounded.

However, a basic “height-like” property is that there are only finitely many cuspidal representations with bounded analytic conductor (this is proved by Michel-Venkatesh in Section 2.6.5 of this paper, and a different proof based on spherical codes, as suggested by Venkatesh, is due to Brumley). Thus $U$ almost isolates $\pi$, in the sense that only finitely many other cuspidal representations can lie in $U$. Denote by $X$ this set of representations.

Now, the unitary dual is (or should be…) at least minimally separated so that, for any two cuspidal representations $\pi_1$ and $\pi_2$, there is an open set which contains $\pi_1$ and not $\pi_2$, say $V_{\pi_1,\pi_2}$. Then $U'=U\cap \bigcap_{\rho\in X-\{\pi\}} V_{\pi,\rho}$
is an open neighborhood of $\pi$ which only contains the cuspidal representation $\pi$