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I read the claim in a preprint that "A simple topological group is either discrete or connected". However, the explanation given was "a connected component of a topological group that contains the identity is a normal subgroup", which only seems to show that one such group is either connected or totally disconnected. In these notes I found this last assertion made (i.e. connected or totally disconnected). I am looking for confirmation that indeed the first assertion is not true.

Question $(1)$: Are there simple topological groups that are not connected and not discrete?

$\pi$-Base did not seem to help with this. The questions is negative for locally connected spaces, since in this case a totally disconnected group must be discrete. I have reasons to believe that the question might also have a negative answer for compact groups:

Question $(2)$: Is every compact simple topological group either discrete (i.e. finite, by compactness) or connected?

There is another assertion, inspired by proofs in the aforementioned preprint, that I have reasons to believe is true but have not been able to find in the internet.

Question $(3)$: Let $G$ be a compact topological group which acts continuously on a topological space $X$, if the quotient space $X/G$ is compact then $X$ is compact.

A simpler version of Question $3$ would be the case where $X$ happens to be a topological group and $G$ is a compact subgroup.

Anguepa
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    There are many examples of (even locally compact) totally disconnected nondiscrete groups, e.g. $PSL(n, \mathbb Q_p)$. – Moishe Kohan Jul 01 '24 at 14:35
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    Most likely the authors meant Lie groups, for which the claim is indeed true. – Moishe Kohan Jul 01 '24 at 14:45
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    The claim is also true for compact groups (since simple compact groups are totally disconnected, hence, profinite). – Moishe Kohan Jul 01 '24 at 14:53
  • @MoisheKohan Is the group you mentioned simple? – Anguepa Jul 01 '24 at 16:53
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    @Anguepa: yes, $PSL_n(F)$ is a simple group for nearly all $n$ and all fields $F$ except for a couple of small finite fields and small $n$. – Qiaochu Yuan Jul 01 '24 at 16:54
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    Yes, this was discussed numerous times on MSE. – Moishe Kohan Jul 01 '24 at 16:55
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    https://math.stackexchange.com/questions/3364118/which-groups-can-be-proven-to-be-simple-using-iwasawa-criterion, https://math.stackexchange.com/questions/4727019/when-is-the-special-linear-group-sl-n-bbb-f-q-over-a-finite-field-bbb-f-q/4727086#4727086, https://math.stackexchange.com/questions/683296/is-mathrmpsl-2-mathbbq-a-simple-group/683330#683330 – Moishe Kohan Jul 01 '24 at 17:17

1 Answers1

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In order to convert my comments to an answer.

Q1. It is well-known and discussed numerous times on this site (see for instance here, here, here) that for every infinite field $F$ and $n\ge 2$, the group $PSL(n, F)$ is simple (as an abstract group). Taking $F=\mathbb Q_p$, will give you a totally disconnected locally compact nondiscrete simple group. (If you do not care about local compactness, you can take, for instance, $F=\mathbb Q$.)

Q2. Suppose $G$ is a simple topological group. For a general topological group $G$, the identity component of $G$ is a normal subgroup. Hence, assuming $G$ is simple, either $G$ is connected or its identity component is trivial, i.e. $G$ is totally disconnected. Thus, the question reduces to the case of totally disconnected compact groups. A totally disconnected compact group $G$ is known to be profinite, i.e. the inverse limit of a family of finite discrete groups; see the references in the Wikipedia article and also, for instance, Theorem 1.34 in

Karl H. Hofmann, Sidney A. Morris, The Structure of Compact Groups. A Primer for the Student – A Handbook for the Expert, de Gruyter Studies in Mathematics, Volume 25, 1998.

In particular, $G$ cannot be simple unless it is a finite simple group.

Q3. The general rule on stackexchange is one question per post (could be several closely related questions). Your last question is quite different from the first two and (assuming $X$ is Hausdorff) is a duplicate of several earlier MSE questions. The actual result is that if $X$ is a Hausdorff space, $G$ is a compact (Hausdorff) group and $G\times X\to X$ is a continuous action, then the quotient map $X\to X/G$ is proper. (For instance combine the answers here and here.)

Lastly: I did not read the papers that you linked but, most likely, the authors assume that groups in question are Lie groups. For a manifold, "totally disconnected" is equivalent to "discrete" and, thus, a simple disconnected Lie group is necessarily discrete.

Moishe Kohan
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