21
$\begingroup$

Let $\Sigma_g$ be the fundamental group of the closed orientable surface of genus $g\ge 2$; let $B_n$ be the braid group on $n\ge 3$ braids; let $S_n$ be the symmetric group on $n$ letters; let $p:B_n\to S_n$ be the canonical epimorphism.

Does every homomorphism $f:\Sigma_g\to S_n$ lift to $B_n$? That is, is there a homomorphism $\bar f:\Sigma_g\to B_n$ such that $f=p\circ\bar f$?

This is known for $n=3$, the proof is by ad hoc elementary computations (Hector, Meigniez, Matsumoto, "Ends of leaves of Lie foliations", J. Math. Soc. Japan 57 (2005), no. 3, 753--779.) Is it true for every $n$? The question is crucial for the construction of some Lie foliations.

$\endgroup$
  • 5
    $\begingroup$ 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. $\endgroup$ – Will Sawin Jun 4 '17 at 19:37
  • 2
    $\begingroup$ 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. $\endgroup$ – Jason Starr Jun 4 '17 at 20:59
  • $\begingroup$ @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. $\endgroup$ – Gael Meigniez Jun 4 '17 at 21:28
  • $\begingroup$ @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. $\endgroup$ – Will Sawin Jun 5 '17 at 15:30
  • 4
    $\begingroup$ 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. $\endgroup$ – BS. Jun 5 '17 at 21:55
6
$\begingroup$

In certain very special cases, I think this can be answered. In particular, if $f:\Sigma_g \twoheadrightarrow S_n$ is onto, $n\geq 4$ and $g \gg 0$, then Theorem 6.20 of Dunfield-Thurston implies that the map $f$ is determined up to the action of the mapping class group by the image $f_\ast:H_2(\Sigma_g) \to H_2(S_n) \cong \mathbb{Z}/2$ (when $n\geq 4$). In the case that the image is zero, then $f$ can be chosen to factor through a handlebody of genus $g$, and since the map factors through a free group, this will lift to $B_n$ since there is no obstruction.

If $f_\ast$ is non-trivial, then there will be a lift iff $\mathbb{Z}\oplus \mathbb{Z}/2 = H_2(B_n)\twoheadrightarrow H_2(S_n)=\mathbb{Z}/2$ is surjective. I'm pretty sure that this is true, and it should be represented by a torus whose fundamental group is generated by $\sigma_1, \sigma_3$ in the standard braid group generators. One need only check that this torus maps homologically non-trivially into $H_2(S_n)$, which I think follows from the presentation of the double cover of $S_n$.

If $f$ is not onto, then one could still attempt to apply Theorem 6.20 to its image. Let $f(\Sigma_g)= H < S_n$. Then Theorem 6.20 implies that for $g$ large enough, $f$ is classified up to the mapping class group by the image of $f_*: H_2(\Sigma_g)\to H_2(H)$, up to the action of $Out(H)$. Let $\tilde{H} = p^{-1}(H)$ be the preimage of $H$ in $B_n$. If $p_{|\ast}: H_2(\tilde{H}) \to H_2(H)$ is not onto (again, up to the action of $Out(H)$), then one could find a counterexample. I think there's a good chance of such a subgroup existing.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – Will Sawin Jun 6 '17 at 19:08
  • $\begingroup$ So I think the differential vanishes and hence this map is always onto and there is no obstruction of this type. $\endgroup$ – Will Sawin Jun 6 '17 at 19:08
  • $\begingroup$ @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. $\endgroup$ – Ian Agol Jun 6 '17 at 19:40
  • $\begingroup$ 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. $\endgroup$ – Will Sawin Jun 6 '17 at 19:43
  • 1
    $\begingroup$ @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. $\endgroup$ – Ian Agol Jun 6 '17 at 19:57
5
$\begingroup$

As BS pointed out, the question runs since the 70's. For $n=3, 4$, after Petersen the answer is positive, since $S_n$ is a solvable group (his example 5.8). For $n\ge 5$, the question is open (see Melikhov, problem 1.1); Melikhov even asks if every generic smooth map $\Sigma_1\to\Sigma_2$ between two orientable surfaces lifts to an embedding $\Sigma_1\to\Sigma_2\times R^2$.

Hansen, V.L. "Embedding finite covering spaces into trivial bundles" Math. Ann. 236, 3 (1978), 239-243, doi:10.1007/BF01351369

Petersen, Peter "Fatness of covers". J. Reine Angew. Math. 403 (1990), 154–165, MR1030413.

Melikhov, S.A."Transverse fundamental group and projected embeddings" Proc. Steklov Inst. Math. (2015) MR3488789

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.