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Nov 3, 2022 at 23:27 comment added Yemon Choi Yes, this is what I had in mind, I was writing in a hurry between meetings. So Nick's answer supplies the details.
Nov 3, 2022 at 12:42 comment added Lau @YemonChoi Your idea is basically what Nick suggested in his answer, right?
Nov 1, 2022 at 0:02 answer added Nick timeline score: 3
Oct 31, 2022 at 13:06 vote accept Lau
Oct 31, 2022 at 0:36 history edited Lau CC BY-SA 4.0
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Oct 31, 2022 at 0:35 comment added Lau Thanks! Yes, you are right. I'll edit it.
Oct 30, 2022 at 23:13 answer added YCor timeline score: 8
Oct 30, 2022 at 17:12 comment added YCor 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.
Oct 30, 2022 at 15:59 comment added Yemon Choi 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.)
Oct 30, 2022 at 14:49 history edited Lau
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Oct 30, 2022 at 14:31 history asked Lau CC BY-SA 4.0