Timeline for Morphism from a surface group to a symmetric group, lifted to the braid group
Current License: CC BY-SA 3.0
11 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jul 18, 2019 at 6:48 | vote | accept | Gael Meigniez | ||
| Jul 18, 2019 at 6:48 | vote | accept | Gael Meigniez | ||
| Jul 18, 2019 at 6:48 | |||||
| Jul 18, 2019 at 6:48 | vote | accept | Gael Meigniez | ||
| Jul 18, 2019 at 6:48 | |||||
| Jun 6, 2017 at 19:57 | comment | added | Ian Agol | @WillSawin: for $H_2(\tilde{H})$, every class can be represented by a map of a surface (this is true for $H_2$ in general). Then one can stabilize (by "tubing" along generators) to obtain a homologous surface group which maps onto $H$. This is why $g$ needs to be large, as well as for the Dunfield-Thurston result. | |
| Jun 6, 2017 at 19:45 | comment | added | Will Sawin | To answer the problem for $g$ large with respect to $n$, doesn't one also need to know that every class in $H^2( \tilde{H},\mathbb Z)$ actually arises from a homomorphism from the surface group? | |
| Jun 6, 2017 at 19:43 | comment | added | Will Sawin | Yeah that was a mistake. I first thought it was the free group on ordered pairs where reversing the order is negation, but now believe it is the free group on unordered pairs. The sign is clockwise/counterclockwise and is not reversed by negation. I edited, but forgot to delete the second part. | |
| Jun 6, 2017 at 19:40 | comment | added | Ian Agol | @WillSawin: Okay, this seems plausible, although I'm a bit confused by your statement " the free group on unordered pairs of elements of {1,…,n} where reversing the order is negation. " - I think you mean ordered pairs? Anyway, I looked up a presentation of the pure braid group (because I'm lazy), and the abelianization is free abelian. So I think I see what you mean. So this should imply with the above argument that for fixed $n$ and large enough $g$, one may lift the surface group. | |
| Jun 6, 2017 at 19:08 | comment | added | Will Sawin | So I think the differential vanishes and hence this map is always onto and there is no obstruction of this type. | |
| Jun 6, 2017 at 19:08 | comment | added | Will Sawin | I think there is a Leray-Serre spectral sequence where $H_p(H, H_q(P_n))$ converges to $H_{p+q} (\tilde{H})$, where $P_n$ is the pure braid group. The obstruction to this map being onto should be the differential $H_2(H, H_0(P_n)) \to H^0(H, H_1(P_n))$. Now $H_1(P_n)$ is the abelianization of $P_n$, which is freely generated by braids where two strands wrap around each other i.e. the free group on unordered pairs of elements of $\{1,\dots,n\}$ where reversing the order is negation. The $H$-coinvariants of that is the free group on orbits of unordered pairs, which is torsion-free. | |
| Jun 6, 2017 at 14:49 | history | edited | Ian Agol | CC BY-SA 3.0 |
added 203 characters in body; deleted 278 characters in body
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| Jun 6, 2017 at 9:04 | history | answered | Ian Agol | CC BY-SA 3.0 |