# Configuration space of 4 points as an orbifold

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.