Computing a Commutator Subgroup

Question

$\DeclareMathOperator\SL{SL}\DeclareMathOperator\SO{SO}\DeclareMathOperator\O{O}$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.

Answer 1

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" 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" or the fact that $\mathbb{Z}$ is Euclidean to show that $\operatorname{EO}(10,\mathbb{Z})=\operatorname{SO}(10,\mathbb{Z})$.
5. Conclude that $[\operatorname{O}(10,\mathbb{Z}),\operatorname{O}(10,\mathbb{Z})] = [\operatorname{SO}(10,\mathbb{Z}), \operatorname{SO}(10,\mathbb{Z})] = \operatorname{SO}(10,\mathbb{Z})$.