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

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 ?

• At least they are linear (Krammer, Bigelow): ams.org/journals/jams/2001-14-02/S0894-0347-00-00361-1 . – few_reps May 23 '15 at 15:55
• There's a positive case : the Burau rep. at $t=-1$ furnishes an epi. $B_3\to SL_2(\mathbf Z)$. This lifts as an iso. $B_3\to \tilde{SL}_2(\mathbf Z)$, the latter group being the preimage of $SL_2(\mathbf Z)$ in the universal covering $\tilde{SL}_2(\mathbf R)$ of $SL_2(\mathbf R)$. The natural bijection $\tilde{SL}_2(\mathbf R)/\tilde{SL}_2(\mathbf Z)\to \tilde{SL}_2(\mathbf R)/SL_2(\mathbf Z)$ shows that this subgroup is a lattice, and $\tilde{SL}_2(\mathbf R)$ is semi-simple since its Lie algebra is $\mathfrak{sl}_2$. – few_reps May 23 '15 at 20:18
• sorry for the typo, the bijection is $\tilde{SL}_2(\mathbf R)/\tilde{SL}_2(\mathbf Z)\to {SL}_2(\mathbf R)/{SL}_2(\mathbf Z)$. – few_reps May 23 '15 at 21:48
• For $n\ge 4$ $B_n$ (as well as any of its finite index subgroups) is certainly not a lattice in any connected Lie group (semisimple or not). Due to usual rigidity results, the only possible candidates would be $\mathbf{R}\times S$ for $S$ simple of rank 1, and $\tilde{S}$ when $S$ is simple of rank 1 with infinite $\pi_1$. This can probably be discarded (I have a simple algebraic argument for $n\ge 9$ but probably there's a general one). – YCor May 24 '15 at 2:55
• @YCor : Thank you for the answer, although I don't understand the details. Which rigidity results are you thinking of ? Could you be a bit more specific ? – Selim G May 24 '15 at 6:22

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.

• Very nice, indeed. – few_reps May 30 '15 at 2:21