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Let $G$ be the subgroup of integer matrices in $\mathrm{SO}(3,2)$.

(The invertible linear maps from a $5$ dimensional real vector space to itself which leave invariant a nondegenerate symmetric billinear form of signature $(3,2)$, and have determinant $1$.)

$G$ is a group of real rank $2$, so conjecturally, it should be boundedly generated.

(There exists an $m \in \mathbb{N}$, and cyclic subgroups $C_1, \dots, C_m \leq G$ such that $G = C_1 \cdots C_m$.)

I would like to know whether the following much weaker statement holds:

There exists a positive integer $m$, and finitely generated subgroups $H_1, \dots, H_m \leq G$ of infinite index, such that $G = H_1 H_2 \cdots H_m$.

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    $\begingroup$ If you want to define the integral points, you have to specify more than the real type of the quadratic form, but its integral type. I guess you want something such as $x_1^2+x_2^2+x_3^2-x_4^2-x_5^2$ or $x_1x_2+x_3x_4+x_5^2$, or, better any form that is of isotropic rank 2 over the rationals (which provide infinitely many, all commensurable, possible lattices). Or possibly you also allow rational rank 1? (rational rank 0 means cocompact, in which case you don't expect bounded generation) $\endgroup$
    – YCor
    Commented Nov 11, 2015 at 12:46
  • $\begingroup$ @YCor - Thanks! I have missed this point - I was thinking about $x_1^2 + x_2^2 +x_3^2 - x_4^2 - x_5 ^2$. If someone can prove the result for a slightly different bilinear form I will probably still be happy. I think that I am just looking for an example of a lattice with real rank 2, or just an example of an arithmetic group where bounded generation is conjectured to hold but has not been established yet. $\endgroup$
    – Pablo
    Commented Nov 11, 2015 at 13:06

1 Answer 1

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$\mathrm{SO}(3,2)_{\mathbb{Z}}$ is an orthogonal group of $\mathbb{Q}$-rank 2, so it is boundedly generated. This was proved by Tavgen [Math. USSR. Izvestiya 36 (1991) 101-127; MR1044049] for the quasi-split case, which includes $\mathrm{SO}(3,2)_{\mathbb{Z}}$, but the general case is due to Erovenko and Rapinchuk [J. Number Theory 119 (2006) 28-48; MR2228948]. For most arithmetic groups of real rank $\ge 2$, it is not known whether they are boundedly generated.

The Erovenko-Rapinchuk proof is by induction, and the induction step writes the group as a product of finitely many subgroups $H_i$, as in the question's weaker statement. (Each $H_i$ is the stabilizer of a vector.) Although it does not seem to be spelled out in their paper, I have heard from the authors (in talks and elsewhere) that their methods to write the group as a product also work for unitary groups (and maybe others?), not just orthogonal groups. So the methods may suffice to establish the weaker statement for isotropic arithmetic groups of classical type. Or perhaps the weaker statement is fairly easy for isotropic groups?

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