Timeline for Morphism from a surface group to a symmetric group, lifted to the braid group
Current License: CC BY-SA 3.0
21 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 9:04 | answer | added | Ian Agol | timeline score: 6 | |
| Jun 6, 2017 at 5:54 | answer | added | Gael Meigniez | timeline score: 5 | |
| Jun 5, 2017 at 22:01 | comment | added | Gael Meigniez | @BS Many thanks, these are obviously the right references. | |
| Jun 5, 2017 at 21:56 | comment | added | Gael Meigniez | Well, as an immersed 1-manifold, it is orientable; but I cannot think of any natural orientation. | |
| Jun 5, 2017 at 21:55 | comment | added | BS. | The problem is considered in Petersen, Peter "Fatness of covers". J. Reine Angew. Math. 403 (1990), 154–165, MR1030413. The answer is positive when the monodromy (image of $p$) is solvable, in particular when $n\leq 4$. According to Melikhov, S.A."Transverse fundamental group and projected embeddings" Proc. Steklov Inst. Math. (2015) MR3488789 the case $n=5$ seems open in 2015. See also Hansen, V.L. Math. Ann. 236 (1978) doi:10.1007/BF01351369. Hope this helps. | |
| Jun 5, 2017 at 17:15 | comment | added | Will Sawin | @GaelMeigniez How do you intend to orient the curve downstairs? | |
| Jun 5, 2017 at 16:57 | comment | added | Gael Meigniez | @Will I mean, think to the orientations of the curve! the image downstairs is a curve parametrized twice, in two different directions; so it is trivially null-homologous but what about the underlying once-parametrized curve $C_f$? | |
| Jun 5, 2017 at 16:51 | comment | added | Gael Meigniez | @Will Of course, the curve in $S_1\times_{S_2} S_1$ is null-homologous; but why is its image in $S_1\times_{S_2} S_1/Sym_2$ null-homologous there? | |
| Jun 5, 2017 at 16:48 | comment | added | Gael Meigniez | Clearly, if $n\ge 2$, $C_f$ can never be empty, since $S_1$ is connected. | |
| Jun 5, 2017 at 16:46 | comment | added | Will Sawin | @GaelMeigniez The argument is that the vanishing locus of any real-valued function on any compact surface is null-homologous (because pulling back the cycle is the same as pullback on cohomology, but $H^1(\mathbb R)$ vanihses.) | |
| Jun 5, 2017 at 16:44 | comment | added | Gael Meigniez | @Jason I agree with the pertinency of $C_f$, a wonderful idea, and that a necessary condition is $C_f\cdot C_f=0$. I don't see why $C_f$ should be null-homologous in $S_1\times_{S_2} S_1/Sym_2$ (the involution reverses the orientation of the curve upstairs). | |
| Jun 5, 2017 at 15:57 | comment | added | Will Sawin | @JasonStarr You're right except I didn't think much about the connected components of $C_f$. Even if the connected components are homologically trivial, there is no guarantee such a $g$ exists. | |
| Jun 5, 2017 at 15:47 | comment | added | Jason Starr | @WillSawin. Are you saying that you can find a choice of $f$ such that $C_f$ is empty? I agree that the curve $C_f$ may be disconnected and the union of the connected components may be homologically trivial (after all, we have the involution of $S_1\times_{S_2} S_1 \setminus\Delta$ permuting components). However, if the connected components of $C_f$ are nontrivial, there is still an issue. (I bet that you already know everything that I said and you are only pointing out that it is insufficient to consider the homology class of $C_f$). | |
| Jun 5, 2017 at 15:30 | comment | added | Will Sawin | @JasonStarr This curve is always homologically trivial as it is the vanishing locus of $f(x)-f(y)$ and we can deform $f(x)-f(y)$ to a nowhere vanishing smooth function. | |
| Jun 4, 2017 at 21:28 | comment | added | Gael Meigniez | @Will Sawin Exactly! That is the topological translation of the question. I'm not sure if it is easier (for $n=3$ we had no proof from this topological viewpoint, we got it by brute computation), but at least it gives some moremotivation. | |
| Jun 4, 2017 at 20:59 | comment | added | Jason Starr | For a single real analytic function $f:S_1\to \mathbb{R}$, I would expect a (real analytic) curve $C_f\subset (S_1\times_{S_2} S_1)/\mathfrak{S}_2$ parameterizing pairs of points in fibers that have equal image under $f$. If the self-intersection of this curve in the closed orientable surface is nonzero, then for two functions $f$ and $g$, the curves $C_f$ and $C_g$ intersect. | |
| Jun 4, 2017 at 19:37 | comment | added | Will Sawin | An equivalent form of this question for $S_1 \to S_2$ a degree $n$ unramified cover of closed orientable Riemann surfaces, can we embed $S_1$ into $\mathbb R^2 \times S_2$ such that the the covering map is the second projection? I can prove this for $\mathbb R^3$ by a standard projection argument, but not $\mathbb R^2$ yet. | |
| Jun 4, 2017 at 18:09 | history | asked | Gael Meigniez | CC BY-SA 3.0 |