Timeline for Does $\mathrm{SO}(3)$ act faithfully on a countable set?
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18 events
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| Sep 8, 2021 at 22:36 | comment | added | YCor | @ErikWalsberg you mean "faithfully", not "freely". The groups acting freely on a countable groups. The question which groups act faithfully on a countable set has been quite studied from the mid 1950s on (de Bruijn, Scott, McKenzie...). | |
| Sep 8, 2021 at 21:59 | comment | added | Erik Walsberg | @YCor Oh yeah, and the small index property gives you automatic continuity for morphisms to $S_\infty$. (Apparently I have forgotten everything I used to know about Polish groups.) So that rules out a lot of interesting examples of Polish groups. This makes the question "which groups act freely on a countable set" more interesting, cool. | |
| Sep 8, 2021 at 7:05 | comment | added | YCor | @ErikWalsberg for Polish groups there are many known counterexamples, e.g., $\mathrm{Homeo}(S^1)$ (keyword: "small index property"). The question about locally compact Polish groups would be worth asking separately. | |
| Sep 8, 2021 at 0:58 | comment | added | Erik Walsberg | oh, of course, thanks. So I guess what I really wonder is if every Polish group is a subgroup of $S_\infty$, or at least if every locally compact Polish group is. | |
| Sep 7, 2021 at 22:25 | comment | added | YCor | @ErikWalsberg instead of separable you should ask "second countable" or equivalently "metrizable separable". Indeed, there are separable compact groups of cardinality $2^c$, while $|S_\omega|=c$. | |
| Sep 7, 2021 at 22:19 | comment | added | Erik Walsberg | by Hilbert's fifth problem any locally compact connected topological group is pro-Lie. Now, there certainly will be locally compact groups that are not subgroups of $S_\infty$, for example for cardinality reasons. I wonder if any separable locally compact topological group is a subgroup of $S_\infty$. | |
| Sep 7, 2021 at 22:11 | comment | added | Erik Walsberg | @Ycor I accepted your answer as the answer. I am still curious to know about general Lie groups. I would guess that if the universal cover of $\mathrm{SL}(2,\mathbb{R})$ works out then any connected Lie group works. | |
| Sep 6, 2021 at 8:53 | comment | added | YCor | I've already thought about this question for non-linear Lie groups, say for the two-fold cover of $\mathrm{SL}(2,\mathbf{R})$ (or the universal cover). It sounds to me delicate and should involve some K-theory. | |
| Sep 6, 2021 at 8:41 | comment | added | YCor | This very argument was, to my knowledge, first used by Simon Thomas about 20 years ago (he told me later he wasn't aware of Ulam's question). The argument actually extends to the case when $\mathrm{SO}(3,\mathbf{R})$ is replaced with an arbitrary subgroup $G$ of $\mathrm{GL}_n(K)$ whenever $K$ is a field of cardinal $\le 2^{\aleph_0}$. [Oops, I see this latter fact was already pointed out in the comments. Anyway it's known.] | |
| Sep 6, 2021 at 3:42 | comment | added | Erik Walsberg | @MarkSapir Yeah, if I remember correctly every connected lie group can be built up by taking semidirect products of subgroups of $\mathrm{Gl}_n(\mathbb{C})$, so it might be enough to show that a semidirect product of groups which act freely on countable sets acts freely on a countable set, which seems plausible. | |
| Sep 6, 2021 at 3:39 | comment | added | markvs | If it works for $GL_n$ then it should work for its subgroups? | |
| Sep 6, 2021 at 3:34 | comment | added | Will Sawin | @MarkSapir One group that the argument doesn't obviously work for is the universal cover of $SL_2(\mathbb R)$. I'm not sure if it can be modified to handle that case. | |
| Sep 6, 2021 at 3:30 | comment | added | markvs | @ErikWalsberg: Hence for every connected Lie group? | |
| Sep 6, 2021 at 3:25 | comment | added | Erik Walsberg | this feels like it should work for $\mathrm{Gl}_n(\mathbb{C})$ as well. that and the structure theory of lie groups might get one up to connected lie groups. | |
| Sep 6, 2021 at 3:03 | history | edited | Will Sawin | CC BY-SA 4.0 |
added 16 characters in body
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| Sep 6, 2021 at 3:02 | comment | added | Will Sawin | @GHfromMO yes, exactly. | |
| Sep 6, 2021 at 3:02 | comment | added | GH from MO | Very nice. In the fourth paragraph, did you mean that an element acts trivially if and only if all its conjugates lie in $\mathrm{SO}_3( \overline{\mathbb Z_p})$? | |
| Sep 6, 2021 at 2:50 | history | answered | Will Sawin | CC BY-SA 4.0 |