Timeline for Is there a topologizable group admitting only Raikov-complete group topologies?
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| May 31, 2017 at 16:42 | history | edited | Taras Banakh | CC BY-SA 3.0 |
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| May 6, 2017 at 18:07 | history | edited | Taras Banakh |
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| May 6, 2017 at 7:04 | comment | added | Uri Bader | @TarasBanakh no, I am not sure at all. After making the comment I thought about it a bit and it indeed seems to me that all infinite fields are topolizable. While writing the comment I had in mind that, by valuation theory, all fields of char 0 and all fields of poitive char which have at least one transendental element are topolizable. | |
| May 5, 2017 at 16:16 | comment | added | Taras Banakh | @UriBader Are you sure that the algebraic closure of a finite field is non-topologizable? It can happen that it is topologizable using the technique of $T$-sequences of Protasov and Zelenyuk: just take a sequence of algebraic numbers whose algebraicity degrees tend to infinity very quickly and take the strongest ring topology in which this sequence tends to zero. | |
| May 5, 2017 at 11:19 | comment | added | Taras Banakh | @Andeas Thom You are right, using discontinuous homomorphism into a second-countable group one can produce a stronger a second-countable (but not complete) group topology on $G$. | |
| May 5, 2017 at 8:17 | comment | added | Uri Bader | 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. | |
| May 5, 2017 at 7:58 | comment | added | Andreas Thom | If $h$ is not continuous, then I do not see how one can use the separability of $G$ to get $H$ separable. Moreover, even if $H$ is separable, then $(1 \times h)(G)$ is not closed in $G \times H$. So the induced topology from the product is not polish (as it is not completely metrizable). | |
| May 5, 2017 at 7:45 | comment | added | Taras Banakh | This argument implies that the compact topology of $SO(3)$ is not a unique Polish topology on this group. Or I am missing something? | |
| May 5, 2017 at 7:45 | comment | added | Taras Banakh | @Andreas Thom If a group $G$ has a unique Polish topology, then each homomorphism $h:G\to H$ to an $\omega$-narrow topological group is continuous: to prove this fact one should first reduce the problem to metrizable (and hence separable and eventually Polish) group $H$, then take the diagonal product of $G$ into $G\times H$ and use the uniqueness of the Polish topology to conclude that the topology on $G$ inherited form product coincides with the original Polish topology, which yields the continuity of the homomorphism $h$. | |
| May 5, 2017 at 7:39 | comment | added | Andreas Thom | You are right. What I know is that it has a unique polish topology and every homomorphism to a polish SIN group is continuous. | |
| May 5, 2017 at 7:37 | comment | added | Taras Banakh | @Andreas Thom: SO(3) has at least two SIN-topologies: compact and discrete. Maybe you had in mind a unique $\omega$-narrow SIN-topology? Also thanks for the reference to the paper of Rosendal. | |
| May 5, 2017 at 7:35 | comment | added | Andreas Thom | $SO(3)$ has also a unique SIN topology (not only totally bounded). | |
| May 5, 2017 at 7:34 | comment | added | Andreas Thom | 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. | |
| May 5, 2017 at 6:59 | answer | added | Uri Bader | timeline score: 5 | |
| May 5, 2017 at 6:15 | history | edited | Taras Banakh | CC BY-SA 3.0 |
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| May 5, 2017 at 6:01 | comment | added | Taras Banakh | @Andreas Thom Could you give me a reference to the existence of a discontinuous homomorphism $SO(3)\to Sym(\mathbb N)$. Thanks. | |
| May 5, 2017 at 5:55 | comment | added | Taras Banakh | @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? | |
| May 5, 2017 at 5:30 | comment | added | Andreas Thom | There is a discontinuous homomorphism from $SO(3,\mathbb R)$ to ${\rm Sym}(\mathbb N)$. This will likely give a non-complete topology. | |
| May 4, 2017 at 18:19 | history | edited | Taras Banakh | CC BY-SA 3.0 |
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| May 4, 2017 at 17:55 | history | asked | Taras Banakh | CC BY-SA 3.0 |