Timeline for Does every mapping class group embed into some $\mathrm{Out}(F_n)$?
Current License: CC BY-SA 4.0
7 events
| when toggle format | what | by | license | comment | |
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| Nov 15 at 5:29 | answer | added | Ian Agol | timeline score: 7 | |
| Nov 13 at 18:10 | comment | added | Matt Zaremsky | @HJRW Oh yes, that's true, "since the surface has free $\pi_1$,'' is not the whole story. But right, up to embedding just beef it up enough. | |
| Nov 13 at 16:06 | comment | added | HJRW | I'm not so sure the answer is obviously "yes" if the surface has boundary components (as opposed to punctures). For instance, the mapping class group of a pair of pants is $\mathbb{Z}^3$, which does not embed in $SL_2(\mathbb{Z})$, the outer automorphism group of its fundamental group. ADDED: I guess it's OK, since you can always glue some other stuff onto the boundary components to make a punctured surface. But there is a little something to do! | |
| Nov 13 at 1:12 | history | edited | Matt Zaremsky | CC BY-SA 4.0 |
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| Nov 13 at 1:12 | comment | added | Matt Zaremsky | @YCor Ah good point, I originally wrote "sufficient", and then absentmindedly changed it to "equivalent", assuming that direction was easy. But, right, this is unclear. I'll change it to "sufficient". | |
| Nov 13 at 0:51 | comment | added | YCor | Aut$(F_n)$ embeds into Out$F_{n+1}$, but is it true that Out($F_n$) embeds into Aut$(F_m)$ for some $m$? | |
| Nov 13 at 0:21 | history | asked | Matt Zaremsky | CC BY-SA 4.0 |