5
$\begingroup$

Consider the left-regular representation $\lambda : G \to B(L^2(G))$, $\lambda_g f(h) = f(g^{-1}h)$, for a locally compact group.

It is well-known that this is a unitary faithful and strongly-continuous representation, but is it also a homeomorphism onto its image $\lambda(G)$ (equipped with the strong-operator topology?

It would suffice to show that for any net $g_\alpha$, convergence in the strong operator topology of $\lambda_{g_\alpha}$ to the identity $I_{L^2(G)}$ implies convergence of $g_\alpha$ to the neutral element $e$ in $G$.

Improve this question
$\endgroup$
5
  • 2
    $\begingroup$ I suspect the answer is yes, by taking the contrapositive. (Minor nitpick: in general you should use nets rather than sequences.) That is, suppose $g_\alpha$ is a net which does not converge to $e$; then take a basic WOT-open-nhd ${\mathcal V}\ni I$, use it to define a suitable open neighbourhood $U\ni e$; we know that $g_\alpha\notin U$ infinitely often, and then by building bump functions supported in $U$ using vectors in $L^2(G)$, we should get $\lambda_{g_n} \notin {\mathcal V}$ infinitely often. (We can use WOT, since WOT and SOT agree on the group of unitaries of a Hilbert space.) $\endgroup$
    – Yemon Choi
    Commented Oct 30, 2022 at 15:59
  • $\begingroup$ I understand the reduction to such sequences only assuming that $G$ is second countable. In this case, using local compactness, we reduce to proving that $g_n\to\infty$ implies that $\lambda_{g_n}$ does not converge to the identity. But indeed for all $f$ it holds that $\langle\lambda_{g_n}f,f\rangle$ tends to zero (the regular representation is $C^0$. So $\lambda_{g_n}f$ can tend to $f$. The same argument works for arbitrary $G$, using nets instead of sequences, and works for every faithful $C^0$ representation. $\endgroup$
    – YCor
    Commented Oct 30, 2022 at 17:12
  • $\begingroup$ Thanks! Yes, you are right. I'll edit it. $\endgroup$
    – Lau
    Commented Oct 31, 2022 at 0:35
  • $\begingroup$ @YemonChoi Your idea is basically what Nick suggested in his answer, right? $\endgroup$
    – Lau
    Commented Nov 3, 2022 at 12:42
  • $\begingroup$ Yes, this is what I had in mind, I was writing in a hurry between meetings. So Nick's answer supplies the details. $\endgroup$
    – Yemon Choi
    Commented Nov 3, 2022 at 23:27

2 Answers 2

Reset to default
8
$\begingroup$

Yes. It's more generally true for every faithful $C^0$ unitary representation $\pi$ of $G$. (Recall that a unitary representation $\pi$ is $C^0$ if for all $v,w$ in the Hilbert space, one has $\langle \pi(g)v,w\rangle\to 0$ when $g$ leaves compact subsets of $G$.)

Indeed if this is not a homeomorphism onto its image, there exists an ultrafilter $\eta$ on $G$, not converging to $1$, such that $\lim_{g\to\eta}\lambda_g=\mathrm{id}$ (for the strong topology, i.e., $\lim_{g\to\eta}\pi(g)v=v$ for every $v$ in the Hilbert space.

If $\eta$ is unbounded (i.e. no compact subset of $G$ is in $\eta$), then we get a contradiction, since the $C^0$ property implies $\lim_{g\to\eta}\langle\pi(g)v,v\rangle=0$ for every $v$ in the Hilbert space.

If $\eta$ is bounded, then $\eta$ has a limit $g_0$ (which by assumption is not $1_G$), and we deduce $\lim_{g\to\eta}\pi(g)=\pi(g_0)$. So $\pi(g_0)=\mathrm{id}$, contradicting the faithfulness assumption.

Improve this answer
$\endgroup$
4
  • $\begingroup$ Thank you! could you explain what you mean by $C^0$? $\endgroup$
    – Lau
    Commented Oct 31, 2022 at 0:32
  • $\begingroup$ Thanks for adding in the definition. The left regular rep of a compact group does not have this property, right? $\endgroup$
    – Lau
    Commented Oct 31, 2022 at 7:49
  • $\begingroup$ @Lauritz yes, it does. Every unitary representation of a compact group is $C^0$ (this is a tautological $\forall x\in\emptyset$ condition). $\endgroup$
    – YCor
    Commented Oct 31, 2022 at 9:07
  • $\begingroup$ Is not the left representation a map $\lambda: L^1(G) to B(L^2(G)$ via convlution? As another question is te left and right representation gives the same reduce C^* structure? $\endgroup$
    – Ali Taghavi
    Commented 21 hours ago
3
$\begingroup$

In an attempt of an alternative answer specifically for the left-regular representation (in addition to the very nice and general solution by @YCor):

Firstly, if $V$ is a neighbourhood of $1$ let $W$ be a symmetric neighbourhood of $1$ such that $W\cdot W\subset V$. If $g\not\in V$ then $gW\cap W=\varnothing$. Because if not, then there is $h\in W$ such that $gh\in W$, but since $W$ is symmetric we then have $g = (gh)h^{-1} \in W\cdot W\subset V$.

Now, we prove the contraposition of the question: Let $g_\alpha$ be a net in $G$ that does not converge to $1$. Hence, there exists a neighbourhood $V$ of $1$ and a subnet of $g_\alpha$ that never enters $V$, so we assume, without loss of generality, that $g_\alpha\not\in V$ for all $\alpha$. Choosing $W$ as before and setting $\psi = \chi_W$ to be the characteristic function gives $\|\lambda_{g_\alpha}\psi - \psi\|_2^2 = 2\mu(W)>0$.

Improve this answer
$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .