$\DeclareMathOperator\SL{SL}\DeclareMathOperator\SO{SO}\DeclareMathOperator\O{O}$I’m studying the group $\O(5,5,\mathbb{Z})$, the indefinite orthogonal matrices with integer entries. In particular, I want to know about its homology/cohomology. It doesn’t seem like much is known about this, so I’m starting with the first integral homology, which is its abelianization. To do this, I'm trying to determine the commutator subgroup of $\O(5,5,\mathbb{Z})$. From the Cartan-Dieudonné theorem it is pretty easy to show that
$$[\O(5),\O(5)] = \SO(5)$$
I'm pretty sure we can use this same theorem to show that
$$[\O(5,5),\O(5,5)] = \SO(5,5)$$
Now, I'm interested in $[\O(5,5,\mathbb{Z}),\O(5,5,\mathbb{Z})]$. Given the above, my guess is that it is equal to $\SO(5,5,\mathbb{Z})$ (we certainly have that $[\O(5,5,\mathbb{Z}),\O(5,5,\mathbb{Z})]\subseteq \SO(5,5,\mathbb{Z})$). However, we cannot use the Cartan-Dieudonné theorem in this case since we are not working over a field. Furthermore, the only calculation of a commutator subgroup of integral matrices that I have seen is that for $\SL(n,\mathbb{Z})$, but this relied on knowing generators for $\SL(5,\mathbb{Z})$. I don't know any generators for $\O(5,5,\mathbb{Z})$.


Update: I misspoke in the comments, the quadratic form I am considering is $\sum\_{n=1}^5 x\_ix\_{i+5}$. This is the one that is relevant to string theory. That being said, the generators for $O(5,5,\mathbb{Z})$ are given in Becker, Becker, Schwarz “String Theory Aand M-Theory”, equation (7.76) and (7.77):
$$\begin{pmatrix} 
0 & I\_5 \\
I\_5 & 0
\end{pmatrix} \text{ and } \begin{pmatrix} I\_5 & 0\\
N\_{IJ} & I\_5 
\end{pmatrix}
 $$
where $N\_{IJ}$ is an anti-symmetric matrix of integers.



Computing a Commutator Subgroup
With the latest update, indicating that the quadratic form in question is, up to reordering, $\sum x\_i x\_{-i}$ (here we number the basis elements as $e\_1,\ldots,e\_5,e\_{-5},\ldots,e\_{-1}$), the answer is affirmative.


This can be seen as follows:


1. Note that the group in question is the group of $\mathbb{Z}$-points of a Chevalley—Demazure group scheme of type $\mathsf{D}\_5$ (for an intermediate weight lattice). For this see, for example, "Structure of Chevalley groups over commutative rings" by N. Vavilov (<https://mahalex.net/misc/Structure.djvu>, p. 255). Below I denote the split group of this type by $\operatorname{SO}(10,\mathbb{Z})$.
2. Introduce the elementary subgroup $\operatorname{EO}(10,R)\leqslant \operatorname{SO}(10,R)$, generated by elementary orthogonal transvections (here $R$ is a commutative ring). The detail can be found in the same paper.
3. Use Corollary 4.4 of ["Generators, relations and coverings of Chevalley groups over commutative rings"](https://doi.org/10.2307/2373742) by M. Stein to see that $\operatorname{EO}(10,R)$ is perfect.
4. Use the stability results of Stein from ["Stability theorems for K1, K2 and related functors modeled on Chevalley groups"](https://doi.org/10.4099/math1924.4.77) or the fact that $\mathbb{Z}$ is Euclidean to show that $\operatorname{EO}(10,\mathbb{Z})=\operatorname{SO}(10,\mathbb{Z})$.
5. Conclude that $[\operatorname{O}(10,\mathbb{Z}),\operatorname{O}(10,\mathbb{Z})] = [\operatorname{SO}(10,\mathbb{Z}), \operatorname{SO}(10,\mathbb{Z})] = \operatorname{SO}(10,\mathbb{Z})$.


