5
$\begingroup$

Let $k$ be a global field, and let $G = \mathbf G(\mathbb A_k)$ for a connected, reductive group $\mathbf G$ over $k$. In these notes by Jayce Getz and Heekyoung Hahn, a unitary representation of $G$ is a Hilbert space $V$ together with a continuous homomorphism $\pi: G \rightarrow \operatorname{GL}(V)$ whose image is contained in the group $U(V)$ of unitary operators on $V$.

What is the topology on $\operatorname{GL}(V)$ (which I assume is the group of bounded linear operators on $V$) being considered here? Is it the induced topology coming from the norm topology?

I am trying to compare this definition with one given by Gerald Folland in A Course in Abstract Harmonic Analysis, which requires that for each $v \in V$ the map $g \mapsto \pi(g)v$ be continuous $G \rightarrow V$, where $V$ is taken in the norm topology. Are these two definitions of unitary representations different?

This matters because one later defines the Fell topology on the unitary dual $\hat{G}$ of $G$, and I want to know which representations are actually in $\hat{G}$.

$\endgroup$
  • 7
    $\begingroup$ The usual definition is the one by Folland and it means that you equip GL(V) with the strong operator topology SOT. $\endgroup$ – user1688 Oct 14 '18 at 20:32
7
$\begingroup$

It is absolutely essential that the space of (bounded/continuous) operators be given the "strong" operator topology (strictly weaker than the norm topology), and the map $G\times V\to V$ to be jointly continuous.

This is not a pathology: even in very simple cases, such as $G=\mathbb R$ acting on $V=L^2(\mathbb R)$, that joint continuity fails when operators are given the uniform norm topology, since there is no sufficient uniform bound on change in $g\in G$ so that $g\cdot v$ is close to $v$ for all $v\in V$. E.g., tent functions with ever-narrowing support illustrate this.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ And the continuity of $G \times V \rightarrow V$ is equivalent to the continuity of $g \mapsto \pi(g)v$ for each $v \in V$, right? I think this comes from the uniform boundedness principle $\endgroup$ – D_S Oct 15 '18 at 5:32
  • 1
    $\begingroup$ @D_S, I'd have to think about that... true, we have some "automatic" joint continuity of separately continuous maps, but I'm not sure that this falls into that. Anyway, in practice, one would have more than just the bare assertion that $g\to \pi(g)v$ is continuous for each $v$. $\endgroup$ – paul garrett Oct 15 '18 at 13:00
4
$\begingroup$

Considering a topological group $G$, a Hilbert space $V$ and a corresponding unitary representation, that is a homomorphism $\pi:G\to U(V)$, the following are equivalent:

  1. $\pi$ is continuous when $U(V)$ is taken with the weak operator topology.

  2. $\pi$ is continuous when $U(V)$ is taken with the strong operator topology.

  3. For every $v\in V$, the orbit map $G\to V$ given by $g\mapsto gv$ is continuous.

  4. The action map $G\times V\to V$ given by $(g,v)\mapsto \pi(g)(v)$ is continuous.

In fact, the implications $4 \Rightarrow 3\Rightarrow 2\Rightarrow 1$ are trivially true, while $1\Rightarrow 4$ follows from the uniform convexity of $V$ (thus an analogue is valid for any isometric representation on a uniformly convex space).

The standard terminology is to refer to $\pi$ as a continuous unitary representation if it satisfies the properties above.


It should be mentioned that for non-discrete locally compact groups (eg for your $\mathbf{G}(\mathbb{A}_k)$) unitary representations are almost never continuous when $U(V)$ is endowed with the norm topology, so a careful writer is unlikely to make such an assumption without mentioning it explicitly.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.