Definition. A group $G$ is called complete (resp. non-topologizable) if each Hausdorff group topology on $G$ is Raikov-complete (resp. discrete). It is clear that each non-topologizable group is complete.

Question 1. Does there exist a complete topologizable group?

In particular:

Question 2. Is the group $SO(3,\mathbb R)$ complete?

Question 3. Is the group $Sym(\mathbb N)$ complete?

A simple Baire category argument shows that each complete topologizable group is uncountable.

Remark. There are many examples of Polish groups admitting a unique $\omega$-narrow Hausdorff group topology (so, each $\omega$-narrow Hausdorff group topology on such a group is complete), see http://www.math.uiuc.edu/~ssolecki/papers/AutomaticContinuity13.pdf.
In particular, $Sym(\mathbb N)$ is such a group.

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    $\begingroup$ There is a discontinuous homomorphism from $SO(3,\mathbb R)$ to ${\rm Sym}(\mathbb N)$. This will likely give a non-complete topology. $\endgroup$ – Andreas Thom May 5 '17 at 5:30
  • $\begingroup$ @Andreas Thom Thank you. This indeed gives a non-complete (even second countable) topology on $SO(3,\mathbb R)$! So, unlike to the group $Sym(\mathbb N)$ the group $SO(3)$ does not possess a unique $\omega$-bounded group topology? Only the unique totally bounded group topology! This is very interesting. Well, maybe the group $Sym(\mathbb N)$ will be an example of a complete topologizable group? $\endgroup$ – Taras Banakh May 5 '17 at 5:55
  • $\begingroup$ @Andreas Thom Could you give me a reference to the existence of a discontinuous homomorphism $SO(3)\to Sym(\mathbb N)$. Thanks. $\endgroup$ – Taras Banakh May 5 '17 at 6:01
  • $\begingroup$ It is Example 1.5 in [C. Rosendal. Automatic Continuity of Group Homomorphisms; Bulletin of Symbolic Logic 15, no.2 (2009), 184-214.] with references to the work of Kallman and Thomas. $\endgroup$ – Andreas Thom May 5 '17 at 7:34
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    $\begingroup$ I think one can come up with a non-topolizable group along the following line: find a non-toplogizable ring $R$ and consider $\text{SL}_3(R)$. Note that group operation and suitable commutators operation retrive the ring structure on the subgroup having 1's on the diagonal and 0's elsewhere but the upper-right corner. I suppose the algebraic closure of a finite field is an example of an infinite non-topolozable ring. So I suggest $\text{SL}_3(\bar{\mathbb{F}}_p)$. The above sketch is not a proof, though. $\endgroup$ – Uri Bader May 5 '17 at 8:17

A simple way to come up with non-complete topologies on $\text{SO}(3,\mathbb{R})$ is by embedding $$ \text{SO}(3,\mathbb{R}) \to \text{SL}(3,\mathbb{R})\to \text{SL}(3,\mathbb{C})$$ and then using a discontinuous automorphism of $\mathbb{C}$ to produce a discontinuous automorphism of $\text{SL}(3,\mathbb{C})$. Knowing all the (not so many) closed subgroups of $\text{SL}(3,\mathbb{C})$, it is easy to deduce that the image of $\text{SO}(3,\mathbb{R})$ under such an automorphism is not closed, hence the induced topology not complete.

Another nice way to finish the argument is by making an identification of $\mathbb{C}$ with another algebraically closed field of the same cardinality, carrying a different topological structure. For this one can use the "$p$-adic complex field", $\mathbb{C}_p$. One gets the embedding $$ \text{SO}(3,\mathbb{R}) \to \text{SL}(3,\mathbb{R})\to \text{SL}(3,\mathbb{C}) \to \text{SL}(3,\mathbb{C}_p).$$ Finally, to cope with Andreas' remark, note that $\text{SL}(3,\mathbb{C}_p)$ has a countable permutation action. Indeed, this Polish group has an open subgroup, namely $\text{SL}(3,\mathcal{O})$, where $\mathcal{O}<\mathbb{C}_p$ is the ring of integers.

  • $\begingroup$ I think that this is also the original proof of the statement I made (due to Kallman). $\endgroup$ – Andreas Thom May 5 '17 at 7:36
  • $\begingroup$ @UriBader Inspired by your answer: is it true that each infinite compact metrizable group does admit a non-compact second-countable Hausdorff metrizable topology? $\endgroup$ – Taras Banakh May 5 '17 at 16:18
  • $\begingroup$ @TarasBanakh this is true for infinite compact Lie groups, and more generally for compact groups with non-trivial identity component (as the latters are inverse limits of the formers). The remaining class is the class of profinite groups. I am not sure about this interesting case. Maybe one can (in some cases) use Nikolov-Segal theorem to show that the algebraic structure forces the topology. $\endgroup$ – Uri Bader May 6 '17 at 6:54
  • $\begingroup$ @UriBader Thank you for the comment. I looked at the survey paper of Klopsch (arxiv.org/pdf/1601.00343.pdf). It indeed can be helpful in resolving the profinite case. I wrote the problem on the existence of discontinuous homomorphism on compact topological groups here (mathoverflow.net/questions/269124). $\endgroup$ – Taras Banakh May 6 '17 at 7:23

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