Timeline for Find a "natural" group that contains the quotient of the infinite symmetric group by the alternating subgroup
Current License: CC BY-SA 4.0
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| Oct 23, 2019 at 15:54 | history | edited | YCor | CC BY-SA 4.0 |
added link to Vitali's paper
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| Feb 26, 2019 at 11:15 | comment | added | YCor | In addition, using automatic continuity for $S_\infty$, one has, for every infinite set $X$ that there is no nontrivial (abstract) group homomorphism of $S_X/A_X$ to any Polish group. Here $S_X=S$ and $A_X=A$ (I just emphasize the dependency on $X$.) | |
| Feb 26, 2019 at 8:29 | comment | added | YCor | @AndreasBlass yes indeed, but with this too restricted interpretation of "nice". I mentioned the embedding of $S/F$ into $\mathrm{Homeo}(\beta X-X)$ as possibly nice, and the latter is not a Polish group, even if $X$ is countable. Furthermore it's too restrictive to consider Borel functions: the induced homomorphism $S\to\mathrm{Homeo}(\beta X-X)$ (vanishing on $F$) is not Borel when $S$ it endowed with its standard topology (induced by action on $X$). | |
| Feb 26, 2019 at 2:05 | comment | added | Andreas Blass | There has been a lot of work in set theory on the classification of Borel equivalence relations on Polish spaces (= separable, completely metrizable spaces). On the basis of my admittedly incomplete knowledge of this work, I expect that there is no Borel function from $S$ into a Polish space that is constant on exactly the cosets of $F$ (or of $A$). That is, one cannot "nicely" embed $S/F$ (or $S/A$) into a "nice" space. | |
| S Feb 25, 2019 at 22:42 | history | answered | YCor | CC BY-SA 4.0 | |
| S Feb 25, 2019 at 22:42 | history | made wiki | Post Made Community Wiki by YCor |