Is $PSL(n, \mathbb Z)$ isomorphic to a subgroup of $GL(n,\mathbb C)$ or even $GL(n+1,\mathbb C)$?
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4$\begingroup$ When $n$ is odd, $PSL(n,\mathbb Z)$ is the same as $SL(n,\mathbb Z)$. So yes to the first question in that case. $\endgroup$– Tom GoodwillieCommented Apr 23, 2011 at 23:04
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$\begingroup$ It's an intriguing question all by itself, but it might be equally interesting to know what kind of consequences the answer would have. $\endgroup$– Jim HumphreysCommented Apr 25, 2011 at 15:29
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2 Answers
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You can look for a finite subgroup of $PSL(n,\mathbb Z)$ such that every faithful representation of it has dimension $>n+1$. The following works for even $n\ge 6$.
In $GL(n,\mathbb Z)$ take the group of monomial matrices, i.e. products of permutation matrices and diagonal matrices. Intersect with $SL(n,\mathbb Z)$ and let $G$ be the image in $PSL(n,\mathbb Z)$.
$G$ maps onto the symmetric group $S_n$ with kernel $D\cong (\mathbb Z/2)^{n-2}$. In any faithful action of $G$ on $\mathbb C^d$ there is a one-dimensional subspace $L$ acted on nontrivially by $D$. In the action of $S_n$ on the set of nontrivial $1$-dimensional characters of $D$ every orbit has at least $n(n-1)/2$ elements. It follows that there at least that many independent choices of $L$ and therefore $d\ge n(n-1)/2>n$.
One step breaks down when $n=4$, so you'd need a different subgroup in that case.
answered Apr 24, 2011 at 13:03
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$\begingroup$ Thanks Tom. I was, of course, only interested in the n even case. Your answer definitively answers the question for my purposes. $\endgroup$ Commented Apr 24, 2011 at 15:39
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$\begingroup$ Rather than add this as an answer it would be preferable just to accept Tom's answer. $\endgroup$ Commented Apr 25, 2011 at 15:21
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$\begingroup$ This is definitely a creative approach, though I still wonder whether something more conceptual (in a more general context of arithmetic groups) is possible. Also, I wonder what an efficient subgroup choice would be for
$n=4$? $\endgroup$ Commented Apr 25, 2011 at 15:25 -
$\begingroup$ I think you can use a subgroup of $PSL(4,\mathbb Z)$ that is conjugate in $PSL(4,\mathbb R)$ to a subgroup of $PSO(4)\cong SO(3)\times SO(3)$ that is isomorphic to $A_4\times A_4$. $\endgroup$ Commented Apr 26, 2011 at 1:25
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If $n \ge 3$, perhaps one can use Margulis superrigidity to deduce that any homomorphism from $\mathrm{SL}(n,\mathbb{Z})$ to $\mathrm{GL}(m,\mathbb{C})$ extends to a representation of $\mathrm{SL}(n,\mathbb{C})$. The standard representation of $\mathrm{SL}(n,\mathbb{Z})$ factors through $\mathrm{PSL}(n,\mathbb{Z})$ if and only if $n$ is odd. The next smallest representation of $\mathrm{SL}(n,\mathbb{C})$ has dimension $> n+1$, (assuming that $n > 2$). If $n = 2$, then $\mathrm{PSL}(2,\mathbb{Z})$ is certainly a subgroup of $\mathrm{GL}(3,\mathbb{Z})$. So the answer seems to be: if and only if $n = 2$ or $n$ is odd.
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2$\begingroup$ What Margulis superrigidity tells you is that a homomorphism $SL_n(\mathbb{Z})\rightarrow GL_m(\mathbb{C})$ is a direct sum of a homomorphism extending to $SL_n(\mathbb{R})$ and of a homomorphism factoring through some congruence subgroup. $\endgroup$ Commented Apr 24, 2011 at 9:05