Perfect quotients of braid groups

Question

$\DeclareMathOperator\SL{SL}\DeclareMathOperator\PSL{PSL}\DeclareMathOperator\Sp{Sp}\DeclareMathOperator\PSp{PSp}$The braid groups $ B_1=1 $ and $ B_2\cong \mathbb{Z} $ have no perfect quotients.

$ B_3 $ has quotients $ \SL(2,\mathbb{Z}) $ and $ \PSL(2,\mathbb{Z})\cong B_3/Z(B_3) $. As a result, $ B_3 $ has infinitely many perfect quotients $ \SL(2,p) $ and $ \PSL(2,p) $ (of course $ p \neq 2,3 $ for it to be perfect). Indeed since $ B_3/Z(B_3) \cong \PSL(2,\mathbb{Z}) \cong C_2 * C_3 $ then every $ (2,3) $ generated group is a quotient of $ B_3 $. This includes all except for a few of the infinite families of finite simple groups see Is every finite simple group a quotient of a braid group? and On $(2,3)$-generation of finite simple classical groups

Does this hold for all $ n\geq 3 $? In other words, do all braid groups $ B_n, n \geq 3 $ have infinitely many non-isomorphic (finite) perfect quotients?

Update: Here is an update based on the accepted answer from Ian Agol. Considered the reduced Burau representation https://en.wikipedia.org/wiki/Burau_representation evaluated at $ t=-1 $ $$ \rho: B_n \to U(n-1) $$ Note that this representation is not continuous i.e. the image is not a closed subgroup of $ U(n-1) $. The image of $ \rho $ turns out to be a finite index subgroup of $ \Sp(2m,\mathbb{Z}) $ by some topological/geometric argument I don't quite understand about preserving a symplectic form on the homology. Also the way that $ m $ is related to $ n $ is somehow topological. I guess $ m $ is the rank of the first homology $ H_1 $ of some hyperelliptic curve. Again not exactly sure what's going on with the topology/geometry/algebraic curve stuff.

An example of this phenomenon where there is a quotient of $ B_n $ which is a (finite index subgroup of) $ \Sp(2m,\mathbb{Z}) $ can already be seen for the first non-trivial case $ n=3 $. For $ B_3 $, there is a central quotient of $ B_3 $ isomorphic to $ \SL(2,\mathbb{Z}) \cong \Sp(2,\mathbb{Z}) $.

So for $ n=3, B_3 $ we have $ m=1 $ and the index of $ \rho(B_n) $ in $ \Sp(2,\mathbb{Z}) $ is just $ 1 $.

For higher values of $ n $ the values of $ m $ increase strictly monotonically with $ n $ (at least I think so, again there is some topology I don't quite understand here). And for higher values of $ n $ the index $ [\Sp(2m,\mathbb{Z}): \rho(B_n)] $ also form a strictly increasing sequence of integers, an explicit form for the sequence is given in a corollary to theorem 1 in the paper "Tresses, monodromie et le groupe symplectique."

Now that you have a quotient of $ B_n $ which is (finite index) subgroup of $ \Sp(2m,\mathbb{Z}) $ you can take congruence subgroups modulo $ p $ prime. Since $ p $ is prime then the image of $ \rho(B_n) $ in the simple group $ \Sp(2m,p) $ must be either everything or trivial. Direct inspection of the matrices for the reduced Burau representation https://en.wikipedia.org/wiki/Burau_representation reveals that they are not congruent to the identity mod any $ p $ (there is a $ 1 $ off the diagonal). Thus the image is all of $ \Sp(2m,p) $.

Answer 1

A’Campo showed that the finite symplectic group $Sp(2m,p)$, $p>2$ prime, is a quotient of the braid group $B_n$ for some $m$ depending on $n$. Hence the finite groups $PSp(2m,p)$ are quotients of $B_n$. These groups are simple non-abelian hence perfect for most $m, p$.

A’Campo, Norbert, Tresses, monodromie et le groupe symplectique, Comment. Math. Helv. 54, 318-327 (1979). ZBL0441.32004 MR0535062

Answer 2

A useful general strategy to tackle such questions is to use small cancellation theory. For instance, in Small cancellation in acylindrically hyperbolic groups, Michael Hull proved the following statement (known before for hyperbolic and relatively hyperbolic groups):

Theorem. Two finitely generated acylindrically hyperbolic groups admit a common acylindrically hyperbolic quotient.

As a consequence, because there exist many perfect finitely generated acylindrically hyperbolic groups and that quotients of perfect groups are perfect, every finitely generated acylindrically hyperbolic group admits "many" perfect quotients. This applies in particular to braid groups: even though $B_n$ is not acylindrically hyperbolic, its quotient $B_n/Z(B_n)$ is acylindrically hyperbolic for every $n \geq 3$.

The basic idea is the following. You start with two finitely generated acylindrically hyperbolic groups, say $A$ and $B$. Fix two finite generating sets $\{a_1, \dots, a_n\} \subset A$ and $\{b_1, \ldots, b_m \} \subset B$. Given some words $u_1, \ldots, u_n$ and $v_1, \ldots, v_m$ written respectively over $\{1, \ldots, b_m\}$ and $\{a_1, \ldots, a_n\}$, the group $$Q:= \left( A \ast B \right) / \langle \langle a_1=u_1, \ldots, a_n=u_n, b_1=v_1, \ldots, b_m=v_m \rangle \rangle$$ is clearly a common quotient of $A$ and $B$. The difficulty is to prove something about $Q$, and it's where small cancellation is useful. If the $u_i$ and $v_j$ are chosen "sufficiently complicated", then some information can be obtained about $Q$. Of course, there is a lot a freedom in the construction (i.e. in the choice of the $u_i$ and $v_j$), which explains why "many" quotients can be constructed using this strategy.

For instance, $Q$ can be constructed so that all its finite subgroups are subgroups of $A$ or $B$. So given your favorite acylindrically hyperbolic group $G$ (say $B_n/Z(B_n)$ for some $n \geq 3$), a finite group $F$, and a torsion-free perfect acylindrically hyperbolic group $H$ (say Higman's group), then $G$, $F \ast F \ast F$, and $H$ have a common quotient, which is necessarily perfect, and all of whose finite subgroups are isomorphic to subgroups of $G$ or of $F$. Thus, if $G$ contains only finitely many isomorphism classes of finite subgroups (like $B_n/Z(B_n)$), then you obtain infinitely many perfect quotients up to isomorphism.

In fact, the theorem above has been generalised to infinite countable collections of acylindrically hyperbolic groups by Minasyan and Osin. The construction relies on the same idea: you start with the free product of your groups, $G_1 \ast G_2 \ast \dots$, and you add complicated relations (i.e. satisfying some good small cancellation condition) in order to merge each $G_i$ into each $G_j$. Of course, you have to add infinitely many relations, which makes the quotient infinitely presented, but this also means that you have much more freedom in the construction. In particular, it is possible to create uncountably many perfect quotients up to isomorphism.

However, in such constructions, the quotients are always infinite. If you are looking for finite quotients, then the strategy is not relevant.