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Consider the following two symplectic matrices $$ A \ = \ \left(\begin{array}{rrrr}% 1&0&0&0\\% 0&1&0&0\\% 0&0&-1&1\\% 0&0&-1&0\\% \end{array}\right), \ \ \ B \ = \ \left(\begin{array}{rrrr}% -1&0&0&-1\\% 0&0&-1&0\\% 0&1&-1&0\\% 1&0&0&0\\% \end{array}\right). $$ Is it true that the (Zariski-dense) group $\langle A,B \rangle$ generated by $A$ and $B$ has infinite index in ${\rm Sp}(4,\mathbb{Z})$ (i.e., $\langle A, B \rangle$ is thin in ${\rm Sp}(4,\mathbb{Z}))$?

This question emerged from some calculations performed by Vincent Delecroix and myself with the monodromy of certain square-tiled surfaces (and, in their turn, these calculations were motivated by a question posed by Peter Sarnak to Alex Eskin and Alex Wright).

More precisely, our considerations led to a representation $p: G \to {\rm Sp}(4,\mathbb{Z})$ of the level $4$ congruence group $G = \langle a,b \rangle$ generated by the order three matrices $$ a \ = \ \left(\begin{array}{rr}% 0&-1\\% 1&-1\\% \end{array}\right), \ \ \ b \ = \ \left(\begin{array}{rr}% 1&-3\\% 1&-2\\% \end{array}\right) $$ in ${\rm SL}(2,\mathbb{Z})$ such that $p(a) = A$ and $p(b) = B$.

As it turns out, $G = \langle a \rangle * \langle b \rangle$ is the free product of two copies of $\mathbb{Z}/3\mathbb{Z}$ (since $\{a,a^2\}$ and $\{b,b^2\}$ play ping-pong with some cones in $\mathbb{R}^2$). Moreover, Vincent and I believe that the group $\langle A, B \rangle$ is thin because some numerical experiments with non-trivial words on $A, A^2$ and $B, B^2$ of length $< 25$ seem to indicate that the representation $p$ might be faithful (and, thus, $<A,B>$ would be thin as ${\rm Sp}(4,\mathbb{Z})$ doesn't contain finite-index subgroups isomorphic to free groups).

Nevertheless, after trying a couple of standard tricks (e.g., testing the injectivity of $p$ on finite-index free subgroups of $G$ or playing ping-pong in $\mathbb{R}^4$, its exterior powers [and $p$-adic variants], etc.), Vincent and I are still unable to establish the thinness of $\langle A, B \rangle$ and/or the faithfulness of $p$, so that we would be thankful to any help with these problems!

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  • $\begingroup$ Would not a torsion free subgroup of finite index be free, by Kurosh? (math.stackexchange.com/questions/29264/…) If that's true, then your group is thin, since $Sp(4, \mathbb{Z})$ has no free subgroup of finite index, by property-T reasons. $\endgroup$
    – Igor Rivin
    Commented Oct 30, 2015 at 22:52
  • $\begingroup$ A torsion free subgroup of $\langle a, b \rangle$ would be free as $\langle a,b \rangle$ is an amalgam of finite groups. But if the representation $p$ is not faithful, I am not sure how you would conclude that a torsion free subgroup of $\langle A,B \rangle$ is free. $\endgroup$
    – Geoff Robinson
    Commented Oct 31, 2015 at 2:20
  • $\begingroup$ @IgorRivin: In fact, ${\rm Sp}(4,\mathbb{Z})$ itself does have torsion free subgroups of finite index. The kernel of reduction (mod $p$) for any prime $p > 5$ is a torsion free subgroup of finite index. $\endgroup$
    – Geoff Robinson
    Commented Oct 31, 2015 at 2:24
  • $\begingroup$ @GeoffRobinson Any matrix group has torsion free subgroups of finite index, but $Sp(4, \mathbb{Z})$ cannot have FREE subgroups of finite index. Of course, my comment missed the point, since it is the faithfulness which is the key issue (otherwise, any f.g. group is quotient of a free group...) $\endgroup$
    – Igor Rivin
    Commented Oct 31, 2015 at 9:38
  • $\begingroup$ Yes, I understood the distinction between free and torsion free, and I know you did not say that Sp(4,Z) has no torsion free subgroup of finite index. As for your first statement immediately above, that's true (for integral matrix groups) if you interpret the trivial group as torsion free, which I usually don't. $\endgroup$
    – Geoff Robinson
    Commented Oct 31, 2015 at 9:43

3 Answers 3

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Your representation $p$ is not faithful, since we have $$ (ABA^{-1}BA^{-1}BAB^{-1})^3 \ = \ 1. $$ In particular, this means that $$ (aba^{-1}ba^{-1}bab^{-1})^3 \ = \ \left(\begin{array}{rr}% -24587&42408\\% 15048&-25955\\% \end{array}\right) $$ lies in the kernel of $p$.

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    $\begingroup$ This seems to be a non-trivial word of length 24. This doesn't actually contradict what the OP says, but I would like to understand what is going on. Presumably you produced this word with some computer algebra system? And did the OP check all words of length < 25 in $A,B,A^{2},B^{2}$? $\endgroup$
    – Geoff Robinson
    Commented Oct 31, 2015 at 0:55
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    $\begingroup$ @GeoffRobinson: It says that the representation $p$ is not faithful; apparently Matheus did not check all words of length $< 25$ in $A$, $B$, $A^2 = A^{-1}$ and $B^2 = B^{-1}$, unless he counted $A^2$ and $B^2$ as having length $2$. I found the word with GAP; actually I computed spheres of radii $1, 2 \dots$ about $1$, and noticed that sphere sizes dropped below $2^{r+1}$ at $r=12$. This then led to a nontrivial relation of length $2 \cdot 12 = 24$. $\endgroup$
    – Stefan Kohl ♦
    Commented Oct 31, 2015 at 10:11
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    $\begingroup$ @StefanKohl Thank you! In fact, you're right: while I tested all words of length $\leq 23$, I did only a partial test for words of length 24... $\endgroup$
    – Matheus
    Commented Oct 31, 2015 at 10:17
  • $\begingroup$ @StefanKohl The fact that $p$ is not faithful rules out the strategy pointed out in my post and in Igor Rivin's comments to show the thinness of $<A,B>$, but, if I understand it correctly, $<A,B>$ might be thin anyway. Thus, if you don't mind, I propose to wait for a couple of days for an eventual answer to the original thinness question in my post before I give you the corresponding credit by accepting your answer, what do you think? $\endgroup$
    – Matheus
    Commented Oct 31, 2015 at 10:42
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    $\begingroup$ @Stefan: Thanks, yes I saw that it meant that $p$ is not faithful, I was just trying to reconcile your conclusion with what Matheus had said about words of length < 25. $\endgroup$
    – Geoff Robinson
    Commented Oct 31, 2015 at 10:52
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After talking to Gabriela Weitze-Schmithuesen, I think that we can show the arithmeticity of $\langle A, B\rangle$ using the argument in Section 2 of this paper of Singh and Venkataramana here (http://www.ams.org/mathscinet-getitem?mr=3165424).

Indeed, let us consider the permutation matrix $P=\left(\begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 \end{array}\right)$ exchanging the second and fourth basis vectors and let us show that the conjugate $P\cdot \langle A, B\rangle\cdot P$ of $\langle A, B \rangle$ is arithmetic, i.e., it has finite-index in $Sp(4,\mathbb{Z})$.

For this sake, we asked Sage to look words on $A$, $B$, $A^2$ and $B^2$ of size $\leq 10$ fixing the first basis vector, and we found that the matrices $x=P(A^2 B)^2(AB^2)^2P$, $y=PABA^2BA(AB^2)^2P$ and $z=PA^2BA^2(B^2A)^2BP$ are interesting because $$[y,x]=yxy^{-1}x^{-1} = \left(\begin{array}{cccc} 1 & 0 & 0 & 18 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array}\right), \quad x^6[y,x] = \left(\begin{array}{cccc} 1 & 0 & 18 & 0 \\ 0 & 1 & 0 & 18 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array}\right)$$ $$y^6[y,x]^{-1} = \left(\begin{array}{cccc} 1 & 18 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & -18 \\ 0 & 0 & 0 & 1 \end{array}\right), \quad z^6 \beta^{-1} = z^6 (x^6 [y,x])^{-1} = \left(\begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 1 & -18 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array}\right)$$ generate the positive root groups of $Sp(4,\mathbb{R})$ and, thus, $P\cdot\langle A, B \rangle\cdot P$ intersects the subgroup $U(\mathbb{Z})$ of unipotent upper triangular matrices of $Sp(4,\mathbb{Z})$ in a finite-index subgroup.

Since we know that $\langle A, B\rangle$ is Zariski-dense (see my comment above to a question of Venkataramana), we can apply a result of Tits (http://www.ams.org/mathscinet-getitem?mr=424966) saying that Zariski dense subgroups of $Sp(4,\mathbb{Z})$ containing a finite-index subgroup of $U(\mathbb{Z})$ are arithmetic to get the desired conclusion.

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I don't know if it helps, but you can compose the representation of $\langle a, b \rangle \to \langle A,B \rangle$ with reduction (mod $q$) for any prime $q$, and you get a free kernel $K_{q}$ of $\langle a,b \rangle$ whose rank you can calculate with the Euler characteristics of CTC Wall. But you still have to figure out what $p(K_{q})$ looks like inside $\langle A,B \rangle$.

Note that the kernel of $p$ is the intersection of all the $K_{q}$, and is a free normal subgroup of infinite index of $\langle a,b \rangle$.

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  • $\begingroup$ You're right that this would help if one can figure out what is $p(K_q)$ (or something like it). Indeed, a simple-minded way to prove $H=<A,B>$ has infinite index is to find $g\in Sp(4,\mathbb{Z})$ of infinite order so that $\{g^n: n\in\mathbb{N}\}$ is disjoint from $H$, and, for this sake, I tried to play with some elements $g_0$ whose reduction mod 2 doesn't belong to the reduction mod 2 of $H$ (which is isomorphic to the alternate group inside $Sp(4,\mathbb{F}_2)\sim S_6$ if I'm not mistaken), but I got stuck since I could not figure out how to prove that $g_0^{2n}\notin H$ for all $n>0$. $\endgroup$
    – Matheus
    Commented Nov 3, 2015 at 14:05

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