## Can the n-string sphere braid group embed in to the (n+1)-string sphere braid group?

This question was originally posted on math.SE by myself nearly a year ago. I've been thinking again about the problem after it recently received a little attention, but little progress was made in finding a solution. So, I feel it's sufficiently difficult to post on mathoverflow.

It's an easy exercise to show that if $B_n$ is Artin's classical braid group on $n$ strings, then $B_n$ can be embedded in $B_{n+1}$ (and in a canonical way). A similar statement can be proved for the pure braid group $P_n$. This property is useful for proving various properties of the classical braid groups. One therefore asks if a similar property holds for braid groups on other surfaces such as the sphere.

Let $P\mathcal{S}_n$ be the pure $n$-string braid group on the sphere $S^2$. Fox's definition of this group is the fundamental group of the configuration space $F_{n}S^2=\prod_n S^2\setminus\{(x_1,\ldots,x_n)|\exists i\neq j, x_i=x_j\}$ and then the full braid group $\mathcal{S}_n$ is defined to be the fundamental group of the configuration space $B_nS^2=F_nS^2/\Sigma_n$ where $\Sigma_n$ is the action of the symmetric group by permuting coordinates of the elements of $F_nS^2$.

It was proven by Fadell and Van Buskirk that the braid group $\mathcal{S}_n$ has presentation given by the braid generators $\sigma_1,\ldots,\sigma_{n-1}$ and relations $$\begin{eqnarray}\sigma_i\sigma_j\sigma_i&=&\sigma_j\sigma_i\sigma_j&(\mbox{ for }|i-j|=1)\\ \sigma_i\sigma_j&=&\sigma_j\sigma_i&(\mbox{ for }|i-j|>1)\\ \gamma&=&1&\end{eqnarray}$$ where $\gamma=(\sigma_1\sigma_1\ldots\sigma_{n-1})(\sigma_{n-1}\ldots\sigma_2\sigma_1)$.

With that framework now built up, my question is, can $\mathcal{S}_n$ be embedded in to $\mathcal{S}_{n+1}$ for $n\geq 3$ (and similarly for their pure counter parts)? The naive 'add a string on the end' map will not work because, for instance, the braid $\gamma$ becomes non-trivial when a string is added on the end.

I would think that the answer is no because of the dependence of the relation $\gamma=1$ on $n$, but a proof eludes me.

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Similar to Greg's response, there's an embedding $\mathcal S_3 \to \mathcal S_5$. $\mathcal S_3$ is the symmetric group on three letters / dihedral group of the triangle. Each motion of a triangle induces a corresponding motion of the double cone over the triangle, and the double cone has five vertices. – Ryan Budney Jan 21 at 20:47
If instead of looking at the configuration space of points in the sphere you enhance it so that each point has a tangent vector attached, then you get very nice maps $F_n' S^2 \to F_{n+1}' S^2$ that come from "doubling" points in the direction of the tangent vector. This gives you the corresponding maps of the associated "braid groups" (fundamental groups of these spaces). – Ryan Budney Jan 22 at 1:07
@Ryan: In fact, the sections $F_n S^2 \to F_{n+1} S^2$ obtained in the answers below can be nicely described in terms of your modified configuration spaces --- although for that we actually only need a tangent vector associated to the first point (instead of all points) in the configuration, say. For each $n\geq 3$, there is a section $s$ of the projection $F'_n S^2\to F_n S^2$ (constructed using similar arguments to my answer below), and post-composing $s$ with your doubling map gives rise to a section $F_n S^2 \to F_{n+1} S^2$ which is essentially the same as the ones in the answers below. – Ricardo Andrade Jan 22 at 2:15
Any such sequence of embeddings would certainly have to be very funny-looking, and in particular would lack some of the properties that makes the braid group story so appealing. For instance: the H_1 of the spherical braid group on n strands is (Z/2nZ) (or something like that) and so your maps, maybe apart from some order-2 business, are going to have to send each spherical braid group to the commutator of the next, just because (Z/2nz) doesn't map so well into (Z/2(n+1)Z). – JSE Jan 23 at 4:04

Revision: For $n>6$, there is no embedding of $\mathcal{S}_n \hookrightarrow \mathcal{S}_{n+1}$.

First, recall that there is an extension $\mathbb{Z}/2\mathbb{Z} \to \mathcal{S}_n \to Mod(S_{0,n})$, where $Mod(S_{0,n})$ is the (orientation preserving) mapping class group of the $n$-punctured sphere.

Inside $Mod(S_{0,n})$, there is a subgroup isomorphic to $\mathbb{Z}/(n-2)\mathbb{Z}$, which is a rotation of order $n-2$ of $S^2$, and fixes the north and south poles. The $n$ punctures include the north and south poles and one orbit of size $n-2$. The preimage of this group in $\mathcal{S}_n$ is isomorphic to $\mathbb{Z}/(n-2)\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z}$ or to $\mathbb{Z}/2(n-2)\mathbb{Z}$ (of course if $n$ is odd, these are isomorphic). Then we get a corresponding subgroup of $\mathcal{S}_{n+1}$ from the injection. Project this group to $Mod(S_{0,n+1})$ via the map $\mathcal{S}_{n+1}\to Mod(S_{0,n+1})$. The kernel of this projection will have size at most $2$. By the Nielsen Realization Theorem, any finite subgroup of $Mod(S_{0,n+1})$ must preserve a complete hyperbolic metric of finite area on $S_{0,n+1}$, and in particular by uniformization extends to a finite group of conformal automorphisms of $S^2$ which permutes $n+1$ marked points. The finite subgroups of $PSL_2(\mathbb{C})$ lie inside a conjugate of $SO(3)$, so the image is an abelian subgroup of $SO(3)$. The only abelian subgroups of $SO(3)$ are cyclic (or $\mathbb{Z}/2\mathbb{Z}^2$), so the image is either $\mathbb{Z}/(n-2)\mathbb{Z}$ or $\mathbb{Z}/2(n-2)\mathbb{Z}$ (in which case we may take an index 2 subgroup isomorphic to $\mathbb{Z}/(n-2)\mathbb{Z}$; here we need $n>6$ to conclude that the image is not $\mathbb{Z}/2^2$). However, for $n>5$, there is no subgroup of $SO(3)$ isomorphic to $\mathbb{Z}/(n-2)\mathbb{Z}$ which permutes $n+1$ points, and therefore there is no such subgroup of $Mod(S_{0,n+1})$, a contradiction. To see this, note that a cyclic group of rotations of $S^2$ isomorphic to $\mathbb{Z}/(n-2)\mathbb{Z}$ has two fixed points, and every other orbit of size $n-2$. Thus, there must be some $k$ and $e$ such that there are $k$ orbits of size $n-2$, and $e$ orbits of size $1$, where $e\leq 2$. If $k\leq 1$, then we get $n+1=k(n-2)+e \leq n$, a contradiction. If $k\geq 2$, then $n+1=k(n-2)+e \geq 2(n-2)$, so $n\leq 5$, a contradiction.

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 I think most would agree that this completely answers the question as only a few low $n$ cases remain. I don't fully understand the proof yet but I'll work on it. Thank you @Agol. – Daniel Rust Jan 24 at 12:44 I'm pretty sure the cases $3\leq n\leq 6$ can be settled by considering finite subgroups more carefully. The argument is a bit simpler for $Mod(S_{0,n})$ vs. $\matchcal{S}_n$. – Agol Jan 24 at 21:10

The answer of Gregory Arone can be generalized to give a section for all $n\geq 3$ using a remark I learned from the writings of Fred Cohen. For $n\geq 3$ there is a homotopy equivalence $$SO(3)\times F_{n−3}(S^2 \text{ minus 3 points}) \simeq F_n S^2$$ Note that $S^2$ minus three points is homeomorphic to ${\mathbb R}^2$ minus two points.

Here are some details. For $n=3$ the map $SO(3) \to F_3(S^2)$ takes a linear map L to the configuration (L(1,0,0),−L(1,0,0),L(0,1,0)) (other variations on this map are obviously possible). The case for general $n$ follows from the case $n=3$ using the action of $SO(3)$ on $S^2$. In particular, the map giving the homotopy equivalence $$T_n:SO(3)\times F_{n−3}(S^2 \setminus \{ (1,0,0),(-1,0,0),(0,1,0) \} ) \longrightarrow F_n S^2$$ is given by $$T_n(L,(x_1,x_2,\ldots,x_{n-3})) = (L(1,0,0),-L(1,0,0),L(0,1,0),L(x_1),L(x_2),\ldots,L(x_{n-3}))$$

Further, the projections $\pi:F_n S^2 \to F_m S^2$ for $n\geq m$ correspond under the above homotopy equivalences to the map $$\textrm{id}_{SO(3)}\times\textrm{proj} : SO(3)\times F_{n−3}(S^2 \text{ minus 3 points}) \longrightarrow SO(3)\times F_{m−3}(S^2 \text{ minus 3 points})$$ which is the product of the identity map on $SO(3)$ with the projection on configuration spaces of $S^2$ minus three points. More precisely, we have a commutative square $$\pi\circ T_n = T_m \circ (\textrm{id}_{SO(3)}\times\textrm{proj})$$

Thus the projection $\pi:F_n S^2\to F_3 S^2$ has a section. We simply need to observe that $\pi$ is a fibration, and that the following homotopy equivalent fibration $$SO(3)\times F_{n−3}(S^2 \text{ minus 3 points}) \longrightarrow SO(3)$$ is trivial, and therefore has a section. This recovers Gregory Arone's answer, except for the neat expression he gave.

More generally, we can find a section of $\pi:F_n S^2 \to F_m S^2$ for $n\geq m$. We use the fact that $S^2$ minus three points is homeomorphic to ${\mathbb R}^2$ minus two points, together with a section of the projection $$\textrm{proj}:F_{n−3}({\mathbb R}^2 \text{ minus 2 points}) \longrightarrow F_{m−3}({\mathbb R}^2 \text{ minus 2 points})$$ constructed by adding points "near infinity" (i.e. far away from the points already in the configuration) on the positive $x$-axis. Together with the above homotopy equivalences and commutative square, we thus obtain a section of $\pi:F_n S^2 \to F_m S^2$.

[Edit: I corrected the following paragraph to account for Gregory Arone's first comment below. In any case, the geometric description given in my first comment below is simpler.]

Intuitively, what is the above section of $\pi$ doing? Given a configuration $(x_1,\ldots,x_m)$ of $m$ points in $S^2$, it is simply taking the first three points $x_1$, $x_2$, $x_3$, and using them to add points to the configuration --- thus obtaining a configuration $(x_1,\ldots,x_m,\ldots,x_n)$ --- in the manner now described. View $S^2$ as a Riemann surface and take the unique Moebius transformation (i.e. complex automorphism of the Riemann sphere) $f:S^2\to S^2$ with $$f(1,0,0)=x_1 \qquad f(-1,0,0)=x_2 \qquad f(0,1,0)=x_3$$ and then place the new points $x_{m+1}, \ldots, x_n$ along the image by $f$ of the shortest geodesic from $(-1,0,0)$ to $(0,1,0)$, but very, very near to $x_2$. The neat expression given by Gregory Arone almost works, except that (it introduces an extra rotation by 90 degrees and) you have to further deform it in some way which depends on the rest of the points $x_4,\ldots,x_m$, to make sure the point you add is sufficiently close to $x_2$.

For completeness, I would like to add a couple of remarks for the cases $n=1$ and $n=2$, which are not interesting from the question's perspective as $F_1 S^2$ and $F_2 S^2$ are both equivalent to $S^2$ and thus are simply connected. The projection $F_2 S^2\to F_1 S^2=S^2$ is a fibration and a homotopy equivalence, and hence has a section. On the other hand, the projection $\pi:F_3 S^2\to F_2 S^2$ does not admit a section: composing it with the homotopy equivalences $T_n$ and $\pi:F_2 S^2\to F_1 S^2 =S^2$, we get the composite map $$f : SO(3) \overset{T_3}{\longrightarrow} F_3 S^2 \overset{\pi}{\longrightarrow} F_2 S^2 \overset{\pi}{\longrightarrow} S^2$$ defined by $f(L)=L(1,0,0)$. This map $f$ is the projection onto $S^2$ of the principal $SO(2)$-bundle associated with the tangent bundle of $S^2$; since $S^2$ is not parallelizable, that bundle does not admit a section.

Unfortunately, the above answer only works for the spherical pure braid groups. In any case, perhaps the above decompositions can still give some practical geometric insight into the spherical braid groups which will help to answer the original question.

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Are you sure that the generalization works? The homeomorphism of $S^2\setminus x_1$ with ${\mathbb R}^2$ has to depend continuously on $x_1$. It seems that constructing such a homeomorphism is equivalent to trivializing the tangent bundle of $S^2$. – Gregory Arone Jan 22 at 0:03
@Greg: Indeed, that is correct. The choice of homeomorphism is certainly not continuous; that is why I only said it was an intuitive description. In any case, I don't actually need the homeomorphism. Perhaps I could have phrased it better in terms of the plane passing through $x_1$, $x_2$, $x_3$. The intersection of that plane with $S^2$ is a circle. Place the new point very close to $x_2$ within the half-circle delimited by $x_1$ and $x_2$ which contains $x_3$. – Ricardo Andrade Jan 22 at 0:31
Greg, in this one case the bundle $F_n S^2 \to F_3 S^2$ is trivial, since $F_3 S^3 \simeq SO_3$, i.e. the base space is a group that acts on the total space. – Ryan Budney Jan 22 at 0:32
That last $F_3 S^3$ should have been $F_3 S^2$. – Ryan Budney Jan 22 at 0:33
@Ricardo - OK, I think I got it, thanks. – Gregory Arone Jan 22 at 0:41
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Just in the case $n=3$, it seems that the natural fibration $F_4S^2\to F_3S^2$ (forget the last point) has a section. The section $F_3S^2 \to F_4 S^2$ is defined by the formula $$(v_1, v_2, v_3)\mapsto (v_1, v_2, v_3, \frac{(v_2-v_1)\times (v_3 - v_1)}{||(v_2-v_1)\times (v_3 - v_1)||}).$$ Here $v_1, v_2, v_3$ are three distinct unit vectors in ${\mathbb R}^3$, representing a point in $F_3S^2$. The idea is that you consider the line through the origin perpendicular to the plane containing the points $v_1, v_2, v_3$, and take the intersection of that line with $S^2$ to be your fourth point. Use the ordering of the points to decide which of the two points to take.

Such a section defines a split injection from the third pure spherical braid group to the fourth.

Edit: Ricardo's answer shows a simpler way to add a fourth point, and also generalizes the argument to produce a section of the fibration $F_mS^2 \to F_nS^2$ for all $m>n \ge 3$.

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