Are any braid groups linear algebraic group?

Question

$\DeclareMathOperator\PSU{PSU}\DeclareMathOperator\U{U}\DeclareMathOperator\SU{SU}\DeclareMathOperator\O{O}\DeclareMathOperator\SO{SO}\DeclareMathOperator\GL{GL}\DeclareMathOperator\ASL{ASL}\DeclareMathOperator\SL{SL}\DeclareMathOperator\Sp{Sp}\DeclareMathOperator\Spin{Spin}\DeclareMathOperator\PSL{PSL}\DeclareMathOperator\PGL{PGL}\DeclareMathOperator\G{G}\DeclareMathOperator\F{F}\DeclareMathOperator\E{E}\DeclareMathOperator\Tr{Tr}$Recall that the braid group $ B_3 $ is closely related to the group of integer points of the linear algebraic group $ \PSL_2 $. Indeed $ B_3 $ fits into the short exact sequence $$ 1 \to \mathbb{Z} \to B_3 \to \PSL_2(\mathbb{Z}) \to 1 $$ and is the unique torsion free central extension of the modular group $\PSL_2(\mathbb{Z})$ by $ \mathbb{Z} $ (see the comment from Ian Agol here). Since this map is just the restriction of the universal covering map $$ 1 \to \mathbb{R} \to \widetilde{\SL_2(\mathbb{R})} \to \PSL_2(\mathbb{R}) \to 1 $$ and $ \widetilde{\SL_2(\mathbb{R})} $ is famously the unique example of a simple Lie group which is not a linear algebraic group I am a bit dubious that $ B_3 $ is isomorphic to the group of integer points of a linear algebraic group. But it does at least seem closely related, so you never know.

At least $ B_2 \cong \mathbb{Z} $ is the group of integer points of a linear algebraic group.

My question: Are any $ B_n, n \geq 3 $ isomorphic to the group of integer points of some linear algebraic group?

My thoughts so far: Every braid group is torsion free. I think a torsion free linear algebraic group is always solvable (not sure how to prove that or if it is even true). And I think the only solvable braid group is $ B_2 $. Thus $ B_2 \cong \mathbb{Z} $ should be the only braid group which is a linear algebraic group.

Answer 1

The following is just a sketch of a proof. You can find a somewhat different argument using the notion of rank due to Ballmann and Eberlein, in

Ivanov, N. V. The rank of Teichmüller modular groups. Math. Notes 44 (1988), no. 5-6, 829–832

where a similar result is proven for a more general class of groups, namely, mapping class groups of surfaces.

For the Braid group $B_n, n\ge 4$ let $\bar{B}_n$ denote the quotient of $B_n$ by its center. It is a finite index subgroup in the mapping class group of $n+1$-times punctured sphere. Then $\bar{B}_n$ contains a subgroup isomorphic to $\mathbb Z\times F_2$. It follows that $\bar{B}_n$ cannot act faithfully and discretely on any rank 1 symmetric space. Suppose now that $B_n$ is commensurable to a lattice $\Gamma$ in a connected Lie group $G$ with solvable radical $S$. For simplicity, I will assume that $G$ is linear (but it's not essential).
The quotient $H=G/S$ is a linear semisimple Lie group. Any lattice $\Gamma< G$ will project to a lattice in $H$ (and also intersect $S$ in a uniform lattice in $H$), this should be in Raghunathan's book on discrete subgroups of Lie groups. A lattice in a semisimple Lie group has to be Zariski dense, hence, must have finite center. In our case, because $B_n$ contains central subgroup $\mathbb Z$, this subgroup (up to passing to a finite index sublattice again) has to be in the kernel of the homomorphism $G\to H$. Thus, we obtain an embedding (up to passing to a finite index subgroup) $\bar B_n\to H$. (You have to invoke here the fact that $\bar B_n$ has no nontrivial solvable normal subgroups.) Thus, up to commensuration, $\bar B_n$ is isomorphic to a lattice in $H$. As I noted earlier, $H$ cannot have rank 1. Assume for a moment that $H$ is a simple Lie group. Then every lattice in $H$ has Property T, which $\bar B_n$ does not have since its pure subgroup (of finite index) admits an epimorphism to $\mathbb Z$. Thus, $H$ cannot be simple. In general, lattices in a semisimple Lie group $H$ virtually split as direct products of irreducible lattices in products of direct factors of $H$. But $\bar B_n$ does not (virtually) split as a direct product. Hence, $\bar B_n$ is isomorphic to an irreducible lattice in $H$ and $rank(H)\ge 2$. But, by a theorem of Margulis, such lattices have the property that every normal subgroup is either finite or has finite index. Groups $\bar B_n$ violate this property.

To eliminate the linearity assumption, take $H$ which is the quotient of $G$ by its solvable radical followed by the quotient by the (discrete) center. Then $H$ is linear, projection of a lattice in $G$ to $H$ is again a lattice and the same argument as above goes through.