(Alternative) Presentation for the pure braid group of the sphere

First I need some notation (it's all standard I think). For a manifold $M$, let $F_nM = F_{0,n}M$ be the space of $n$-tuples of distinct points on $M$ ; let $B_nM = B_{0,n}M = F_nM / \Sigma_n$. When $M= \mathbb{R}^2$ the fundamental group of $B_nM$ is the braid group $B_n$, and that of $F_nM$ is the pure braid group $P_n$.

Further, $F_{m,n} M$ is defined to be $F_n N$ where $N$ is $M$ with $m$ points removed. Likewise for $B_{m,n} M$.

Now this question is about $\pi_1 F_n S^2$. The fundamental group of $B_n S^2$(the "braid group of the sphere") is usually presented as a quotient of $B_n$, adding just one relation to Artin's usual ones ; so the fundamental group of $F_n S^2$ is a quotient of $P_n$.

However, a classical result asserts that there is a fibration

$$F_{m+r, n-r}M \to F_{m,n}M \to F_{m,r}M$$

for all $m,n$ and $r\le n$. Taking $M=S^2$, $m=0$ and $r=1$ this becomes

$$F_{n-1} \mathbb{R}^2 \to F_n S^2 \to S^2$$

And the long exact sequence of homotopy groups gives in particular

$$\mathbb{Z} \to P_{n-1} \to \pi_1 F_n S^2 \to 0$$

using that $\pi_1(S^2) = 0$ and $\pi_2(S^2) = \mathbb{Z}$.

This gives a presentation of $\pi_1 F_n S^2$ as a quotient of $P_{n-1}$ rather than $P_n$, adding just one relator (with a very simple proof indeed). The group $P_n$ can be generated by $n(n-1)/2$ generators and no fewer, and so $P_{n-1}$ can be generated by $(n-1)(n-2)/2$ generators, giving a much smaller set of generators for $\pi_1 F_n S^2$.

So my questions are: (EDITED)

(1) Did I get something wrong in the above argument?

(2) Does someone know what the image of $\mathbb{Z}$ in $P_{n-1}$ is?

(3) Has this presentation been studied algebraically? Is it easier to work with the group $\pi_1 F_n S^2$ presented as a quotient of $P_{n-1}$ than with the presentation as a quotient of $P_n$ ? Any reference to a work in this direction?

The answer by Ryan Budney below covers (1) and (2), I think. Any help with (3) appreciated.

Thanks !

Pierre

I think this is in Birman's book. Certainly all this appears in many of Fred Cohen's papers.

The bundle $F_n S^2 \to S^2$ can be understood perhaps best by bringing in the principal bundle for the tangent bundle of $S^2$. This is $SO_3$. There's the bundle:

$$SO_2 \to SO_3 \to S^2$$

where the map on the right is the one that takes a single column vector from your $3 \times 3$ matrix in $SO_3$.

Let $C_n \mathbb R^2$ be the configuration space of $n$ ordered points in the plane. $SO_2$ acts on this space by rotation, so you can form the product $C_n \mathbb R^2 \times SO_3$ and mod out by the diagonal action of $SO_2$, i.e.

$$C_n \mathbb R^2 \times_{SO_2} SO_3$$

This space has a natural identification with $C_{n+1} S^2$.

So the copy of $\mathbb Z$ corresponding to the connecting map in your homotopy long exact sequence consists of an $n$-component trivial braid, after you twist it an even number of times (in the plane). And this basically just boils down to the tangent bundle of $S^2$ having Euler class $2$.

edit: Regarding your 3rd question, there is a sense in which this extension is very natural. This copy of $\mathbb Z$ in the pure braid group is the index-2 subgroup of the centre of the group. The centre of the pure braid group, and the braid group is precisely the "one full twist" braid and its powers. So you can think of the pure braid group as a central extension over the spherical braid group. Given that the extension is so canonical, it appears in many places. For example, Bigelow and I used it to get a faithful linear representation of mapping class groups of punctured spheres and the genus two mapping class group.

• Thanks. I have Birman's book right here and I haven't found this presentation in it (yet). If you have references for Cohen's papers, or any general references, it would be great. Let me think about what you wrote. – Pierre Nov 20 '14 at 22:01
• Hey it took me less time than I though to understand the first part. I see how you build a fibration with the same spaces as in the fibration I mentioned, and I'm ready to believe that the two can be identified. Unfortunately I have no idea how you draw your conclusion in the last paragraph. – Pierre Nov 20 '14 at 22:13
• This is general bundle nonsense. $SO_3$ as a bundle over $S^2$ has fibre a circle. If you decompose $SO_3$ as two copies of $D^2 \times S^1$, glued along a common $\partial D^2 \times S^1$, there is a bundle gluing map $\partial D^2 \to Aut(S^1)$ and the fact that the Euler class of this bundle is $2$ can be stated that this map is twice the generator of $\pi_1 Aut(S^1) \simeq \mathbb Z$. You can do a similar decomposition to $C_{n+1} S^2 \to S^2$ and you get a similar map $\partial D^2 \to Aut(C_n \mathbb R^2)$ but in this case the map is induced from $\partial D^2 \to Aut(S^1)$ via the – Ryan Budney Nov 20 '14 at 23:06
• action of $SO_2$ on $C_n \mathbb R^2$. – Ryan Budney Nov 20 '14 at 23:06
• OK thanks again, I think I'll be able to understand that (after I brush up on my general bundle nonsense). There still an (algebraic) aspect of my question that has not been answered, I'll edit to clarify. Mind you I will accept your answer of noone else joins the conversation. – Pierre Nov 22 '14 at 22:22