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$\DeclareMathOperator\SL{SL}\DeclareMathOperator\PSL{PSL}\DeclareMathOperator\Sp{Sp}\DeclareMathOperator\PSp{PSp}$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).

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?

At the other extreme, are there any $ B_n $ other than $ n=1,2 $ that have no 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) $.

$\DeclareMathOperator\SL{SL}\DeclareMathOperator\PSL{PSL}\DeclareMathOperator\Sp{Sp}\DeclareMathOperator\PSp{PSp}$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) $.

$\DeclareMathOperator\SL{SL}\DeclareMathOperator\PSL{PSL}\DeclareMathOperator\Sp{Sp}\DeclareMathOperator\PSp{PSp}$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).

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?

At the other extreme, are there any $ B_n $ other than $ n=1,2 $ that have no perfect quotients?

Update: Here is an update based on the accepted answer from Ian Agol. Considered the reduced Burau representation $$ \rho: B_n \to U(n-1) $$ 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) $.

$\DeclareMathOperator\SL{SL}\DeclareMathOperator\PSL{PSL}\DeclareMathOperator\Sp{Sp}\DeclareMathOperator\PSp{PSp}$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).

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?

At the other extreme, are there any $ B_n $ other than $ n=1,2 $ that have no 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) $.

$\DeclareMathOperator\SL{SL}\DeclareMathOperator\PSL{PSL}$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).

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?

At the other extreme, are there any $ B_n $ other than $ n=1,2 $ that have no perfect quotients?

Update: Here is an update based on the accepted answer from Ian Agol. Considered the reduced Burau representation $$ \rho: B_n \to U(n-1) $$ 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 the the image of $ \rho(B_n) $ in the simple group $ Sp(2m,p) $ must be either everything or trivial. There is probably some standard argument for why the image of $ \rho(B_n) $ mod $ p $ cannot be trivial. Maybe some argument about divisibility of the index $ [Sp(2m,\mathbb{Z}): \rho(B_n)] $ since the paper gives an explicit form for that? Or maybe just argue directly by looking at the generators of the (symplectic image of the) braid group mod $ p $ and showing that they are nontrivial.

$\DeclareMathOperator\SL{SL}\DeclareMathOperator\PSL{PSL}\DeclareMathOperator\Sp{Sp}\DeclareMathOperator\PSp{PSp}$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).

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?

At the other extreme, are there any $ B_n $ other than $ n=1,2 $ that have no perfect quotients?

Update: Here is an update based on the accepted answer from Ian Agol. Considered the reduced Burau representation $$ \rho: B_n \to U(n-1) $$ 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) $.