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The hyperbolic orthogonal group $O_{g,g}(\mathbb{Z})$ often appears in the study of high-dimensional manifolds, see e.g. work of Kreck or Galatius and Randal-Williams. Let $H$ denote the lattice $\mathbb{Z}\{e,f\}$ with symmetric bilinear form $\lambda$ determined by $\lambda(e,e) = \lambda(f,f) = 0$ and $\lambda(e,f) = 1$. Then $O_{g,g}(\mathbb{Z})$ is defined to be the group of automorphisms of the orthogonal direct sum $H^{\oplus g}$.

This is one of the "classical groups" studied in Hahn-O'Meara. In particular, they give a presentation of $O_{g,g}(\mathbb{Z})$ for $g \geq 3$, or rather the subgroup given by elements with determinant and spinor norm equal to $1$. It is also easy to see that $O_{1,1}(\mathbb{Z}) \cong (\mathbb{Z}/2)^2$. However, I do not know about the case $g=2$:

Does there exist an explicit finite presentation of (a finite index subgroup of) $O_{2,2}(\mathbb{Z})$ in the literature?

I think this may be contained in the literature about integral Chevalley groups from the 60's and 70's, but am not very familiar with the notation. If this is not known, maybe at least $H_1$ is:

Is the abelianization of (a finite index subgroup of) $O_{2,2}(\mathbb{Z})$ known?

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    $\begingroup$ $O_{2,2}$ over $\mathbb Q$ is isomorphic to $SL_2\times SL_2$ modulo the diagonals in the centre of $SL_2\times SL_2$ (this is isomorphic to $\pm 1$) . Thus every finite index subgroup of $2O_{2,2}(\mathbb Z)$ contains one isomorphic to $F\times F$ where $F$ is a free group of finite index in $SL_2({\mathbb Z})$.Hence there are many f.i. subgroups with infinite abelianization. $\endgroup$
    – Venkataramana
    Jul 16, 2019 at 0:35
  • $\begingroup$ When you define $O_{g,g}$ over $\mathbf{Q}$ or $\mathbf{R}$, it does not matter if you choose the quadratic form, say, $\sum_{i=1}^gx_ix_{2g-i}$ or $\sum_{i=1}^gx_i^2-\sum_{i=g+1}^{2g}x_i^2$. However, it matters when you consider the $\mathbf{Z}$-points. (Still they will be commensurate subgroups.) So "they give a presentation of $O_{g,g}(\mathbf{Z})$" is sensitive to the choice of quadratic form of signature $(g,g)$. $\endgroup$
    – YCor
    Jul 16, 2019 at 7:42
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    $\begingroup$ Well, the quadratic form has been explicitly chosen by the OP $\endgroup$
    – Venkataramana
    Jul 16, 2019 at 9:03
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    $\begingroup$ Actually for this choice of the form ("split") the same is true over $\mathbb Z$: the group is isomorphic to $SL_2\times SL_2$ up to some discrete subgroup. A presentation of $SL_2({\mathbb Z})$ is well known, it is an amalgamated product of $C_4$ and $C_6$ over $C_2$. So it is not difficult to find a presentation for $O_{2,2}({\mathbb Z})$; however I don't see how it can help. $\endgroup$
    – Victor Petrov
    Jul 16, 2019 at 21:08

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