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

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    $\begingroup$ I guess your question (i.e., the structure of indefinite orthogonal group over $\mathbf{Z}$) might be sensitive on choices, notably on whether you pick the quadratic form $\sum_{i=1}^5x_ix_{i+5}$ or $\sum_{i=1}^5x_i^2-x_{i+5}^2$. Could you clarify this? $\endgroup$ Commented May 1, 2023 at 18:23
  • $\begingroup$ Hahn-O'Meara might be relevant: link.springer.com/book/10.1007/978-3-662-13152-7. $\endgroup$ Commented May 1, 2023 at 18:29
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    $\begingroup$ @YCor I’m using the quadratic form $\sum_{i=1}^5 x_i^2 - x_{i+5}^2$. $\endgroup$ Commented May 1, 2023 at 18:31
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    $\begingroup$ @skupers I checked out Hahn-O’Meara. As far as I know, this is not addressed in that text. $\endgroup$ Commented May 1, 2023 at 18:33
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    $\begingroup$ The m.se post is math.stackexchange.com/questions/4687187/… $\endgroup$ Commented May 2, 2023 at 2:03

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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 $[\operatorname{O}(10,\mathbb{Z}),\operatorname{O}(10,\mathbb{Z})] = [\operatorname{SO}(10,\mathbb{Z}), \operatorname{SO}(10,\mathbb{Z})] = \operatorname{SO}(10,\mathbb{Z})$.
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  • $\begingroup$ Great answer. Thank you so much! Also, assuming you have a stack exchange account, you should post your answer over there as well. The link is in the other comment thread. $\endgroup$ Commented May 3, 2023 at 22:21
  • $\begingroup$ @slowspider I think it is perfect as well, but there are some difficulties in identifying the group from the invariant forms concerning whether 2 is invertible in the base ring. $\endgroup$ Commented May 4, 2023 at 6:29
  • $\begingroup$ I took a closer look at step 4 in your proof and have a question. Both the paper by stein and the fact about the Euclidean domain are seemingly only relevant for simply connected chevalley-demazure group schemes. From step 1, it doesn’t seem that we have that. $\endgroup$ Commented May 11, 2023 at 16:01
  • $\begingroup$ @slowspider Yes, but the actual calculation takes place inside the orthogonal group (which is the image of the minimal representation of Spin group), which is reflected by the use of the corresponding weight diagram. $\endgroup$ Commented May 11, 2023 at 21:41
  • $\begingroup$ My apologies, but the reason that $EO(2n,\mathbb{Z}) = SO(2n,\mathbb{Z})$ is still not clear to me. For example, section 2.5 of "Commutator width in Chevalley groups" says that $EO(n,K)$ does not necessarily equal $SO(n,K)$ when $K$ is a field. $\endgroup$ Commented May 24, 2023 at 2:38

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