Is the braid group with $n$ strings $\mathcal{B}_n$ a lattice in a connected semi-simple Lie group?

Question

Is the braid group with $n$ strings $\mathcal{B}_n$ known to be a lattice in a connected semi-simple Lie group ? (for $n$, say, bigger than $3$)

Or is it known that it cannot be such a lattice ?

Answer 1

Proposition: no finite index subgroup of $B_n$, for any $n\ge 4$, is isomorphic to a lattice in any virtually connected Lie group.

Assume that some finite index subgroup $H$ of $B_n$, $n\ge 4$, is a lattice in a virtually connected Lie group $G$. Modding out if necessary, we can assume that the compact radical (=largest compact normal subgroup) of $G$ is trivial. Let $M$ be the largest amenable closed normal subgroup in $G$ (it can be obtained by first taking the largest solvable closed normal [not necessarily connected] subgroup $R$ and defining $M$ as the inverse image of the compact radical of $G/R$). Then classical arguments (see Raghunathan's book) show that $M\cap H$ is a lattice in $G$. Since the amenable radical of $H$ is cyclic, it follows that $M\cap H$ is cyclic. So $M$ is a Lie group, with trivial compact radical, with a cyclic lattice; hence $M$ is either trivial or isomorphic to either $\mathbf{Z}$ the infinite (discrete) dihedral group, or $\mathbf{R}$, or the group of isometries of $\mathbf{R}$.

Define $H'=H/(M\cap H)$ and $S=G/M$. Then $H'$ is a lattice in $S$. Since $S$ has a trivial amenable radical, it is semisimple with no compact factors and trivial center. Actually $M\cap H$ is necessarily infinite, since a lattice in a semisimple group with trivial center cannot have a nontrivial cyclic normal subgroup.

There exists a unique decomposition $S=S_1\times\dots\times S_k$ with each $S_i$ semisimple, such that for every $i$, $H'\cap S_i$ is an irreducible lattice in $S_i$. For every $i$ such that $S_i$ has real rank $\ge 2$, the Kazhdan-Margulis theorem implies that $H'\cap S_i$ has no finite index subgroup with a surjective homomorphism onto $\mathbf{Z}$. On the other hand, $H'$ is virtually residually (torsion-free nilpotent), which implies that it has a finite index subgroup all of whose infinite subgroups have a surjective homomorphism onto $\mathbf{Z}$. We deduce that for every $i$, $S_i$ has real rank 1.

It is known that the quotient $B_n/Z$ of $B_n$ by its center and its finite index subgroups satisfy: for any two normal subgroups centralizing each other, at least one of the two is trivial. This follows, for instance, from the existence of a faithful linear representation of $B_n/Z$ such that the Zariski closure of the image is simple with trivial center. We deduce that $k=1$ (because otherwise $H'\cap S_1$ and $H'\cap S_2$ centralize each other and are non-abelian).

We deduce that $S$ is a simple group of real rank 1. (It follows that the unit component of $G$ is isomorphic to either $\mathbf{R}\times S$ or, in case $S$ has an infinite fundamental group, to the universal covering of $S$.)

Now let me get a contradiction by proving the following result (using only a properness assumption with no finite covolume requirement):

Proposition: if $\Gamma$ is the group obtained by modding out any finite index subgroup $\Lambda$ of $B_4$ by its center, then $\Gamma$ has no proper homomorphism $f$ into any $S$ where $S$ is any connected simple Lie group of real rank 1.

Fix $n$ such that $t^n\in\Lambda$ for all $t\in B_4$. Let $x_1\dots,x_3$ be the canonical generators of $B_4$, so that $x_1^n,\dots,x_3^n$ belong to $\Lambda$. Define $w=(x_1x_2x_1)^{2n}$; it is a nontrivial central element of $B_2=\langle x_1,x_2\rangle$. Then $\langle x_1^n,w\rangle$ is free abelian of rank 2, and so is its image in $\Gamma$. If $f(x_1^n)$ is loxodromic, then its axis is preserved by $f(w)$ and we obtain a proper action of $\mathbf{Z}^2$ on this axis, a contradiction, so $f(x_1^n)$ is a horocyclic isometry (I mean "parabolic" but the latter is in conflict with the meaning of "parabolic" for algebraic subgroups of $S$); the same argument shows that $f(w)$ is horocyclic as well, with the same fixed point at infinity. A symmetric argument shows that $f(x_3^n)$ is horocyclic as well, with a priori another fixed point at infinity, but since it commutes with $f(x_1^n)$ this has to be the same fixed point. On the other hand, it is easy to show that $\langle (x_1x_2x_1)^2,x_3^2\rangle$ generate a free group of rank 2, and hence the subgroup of $\Gamma$ generated by $w$ and $x_3^n$ is also free of rank 2. But the horocyclic subgroups of $S$ are nilpotent and we obtain a contradiction.