9
$\begingroup$

Setup: Consider the braid group $B_n$. One way to define this is as the fundamental group of the unordered configuration space $UC_n(\mathbb{C}) = \{\{z_1,\dotsc,z_n\}\subset \mathbb{C} \mid z_i \not= z_j\}$. By translating, we can assume that any set of $n$ points sums to $0$. By scaling, we can assume that $\sum |z_i|^2 = 1$. This gives a deformation retraction $UC_n \simeq (S^{2n-3} - X)$ where $X$ is some real codimension-$2$ submanifold of the sphere which encodes the fact that we forbid repetitions in our set of points. There is a circle action on both spaces given by simply multiplying each of our $n$ points by $\exp(i\theta)$. I'm wondering if anyone knows how this story goes for $n=4$. Here's what we know for the case of $n=3$:

Topology and algebra: we have $UC_3 \simeq (S^3 - T_{2,3})$ where $T_{2,3}$ is the trefoil knot. The circle action gives $S^3-T_{2,3}$ its familiar structure as a Seifert fiber space. The base orbifold of this space is the $(2,3,\infty)$ turnover orbifold, i.e. a once-punctured sphere with one cone point each of order $2$ and $3$. The orbifold fundamental group of this space has a presentation as $$\langle a,b \mid a^2=b^3=1 \rangle.$$ This is also known as a presentation for the central quotient $B_3/Z(B_3)$, which makes sense. The center of the braid group is cyclic as generated by the full twist braid, and the circle action we have described is exactly a full twist on a set of points in the plane. So quotienting by the center and taking the base orbifold of the Seifert fibration are really the same thing.

Geometry: Now we can fill in the puncture of this orbifold with an order $p$ cone point, giving the $(2,3,p)$ turnover orbifold. This orbifold has a spherical, Euclidean, or hyperbolic metric (away from its cone points) when $1/2+1/3+1/p$ is greater than, equal to, or less than $1$, respectively. In particular, the universal cover of $(2,3,p)$ is compact if and only if $p<6$. This shows that the group $$B_3/\langle Z(B_3),\sigma_1^p \rangle = \langle \sigma_1,\sigma_2 \mid \sigma_1^p=1, (\sigma_1\sigma_2)^3=(\sigma_1\sigma_2\sigma_1)^2=1 \rangle$$ is finite if and only if $p<6$.

My question: Does anyone know what happens with $n=4$? We should have the complement of some 3-submanifold in $S^5$ being fibered by circles, and the leaf space of this fibration would be a $4$-orbifold. Do people know what this 4-orbifold is? Its orbifold fundamental group would be the central quotient $B_4/Z(B_4)$, which has an analogous presentation as $$\langle a,b \mid a^4=b^3=1, a^2\cdot bab=bab \cdot a^2 \rangle$$ in case that jogs some recognition in someone's mind. There should similarly be some codimension-2 thing that we can fill in with an order $p$ singular locus. Would there be a neat geometric classification as with the 2-orbifold above that would explain when $B_4/\langle Z(B_4), \sigma_1^p \rangle$ is finite? As a check, this quotient group would be finite only for $p=2,3$ as explored in Coxeter's paper Factor Groups of the Braid Group.

$\endgroup$

0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.