I have been looking at the orthogonal matrix group

$$ \mathrm{O}_n(\mathbb{R}) := \{ M \in \mathbb{R}^{n \times n} : M^T M = I_n \} $$

This group for $n \geq 2$ is infinite because e.g. $\begin{pmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{pmatrix} \in \mathrm{O}_2(\mathbb{R})$ can be embedded into higher dimensions. Since there are infinitely many Pythagorean triples $(x, y, z)$, we also know $\frac{1}{z}\begin{pmatrix} x & -y \\ y & x \end{pmatrix} \in \mathrm{O}_2(\mathbb{Q})$, so $\mathrm{O}_n(\mathbb{Q})$ is infinite.

On the other hand, I know that the group over integers is finite: each column/row must be of norm $1$, so they must contain a single nonzero element from $\pm 1$. In fact, $\mathrm{O}_n(\mathbb{Z}) \cong S_n \wr \{-1, 1\}$.

My question is the generalisation of this $\mathrm{O}_n(\mathbb{Q})$ vs $\mathrm{O}_n(\mathbb{Z})$ comparison: For a number field $K = \mathbb{Q}(X)$ with ring of integers $\mathcal{O}_K$, is the matrix group $\mathrm{O}_n(\mathcal{O}_K)$ finite? In either case, can we describe the group structure or even provide generators, probably depending on $K$ itself?

I looked at $\mathrm{O}_2(\mathcal{O}_K)$, which has the condition

$$ \begin{pmatrix} a & b \\ c & d \end{pmatrix} \in \mathrm{O}_2(\mathcal{O}_K) \iff a^2 + c^2 = b^2 + d^2 = 1 \land ab + cd = 0 $$

However, I can't seem to get anything beyond a few special cases. This also doesn't seem to give any insight for $n \geq 2$.

Looking at the "opposite" direction, we can show that $\mathrm{O}_n(\mathcal{O}_K) \cong S_n \times \mathcal{O}_K^{\times}$ doesn't hold in general: Fix an element $a \in \mathcal{O}_K$ and consider $L = K(\sqrt{1 - a^2})$. Then, $\begin{pmatrix} a & -\sqrt{1 - a^2} \\ \sqrt{1 - a^2} & a \end{pmatrix} \in \mathrm{O}_2(L)$.

(This problem came from implementing .is_finite() in Sage)

Update: I found a slightly nontrivial result, which is that if $\sqrt{-k} \in R$ for any positive non-square integer $k$, then $\mathrm{O}_n(\mathcal{O}_K)$ is infinite. The reason is that

$$ \begin{pmatrix} a & b \\ -b & a \end{pmatrix} \in \mathrm{O}_n(\mathcal{O}_K) \iff a^2 + b^2 = 1 $$

And the Pell's equation $x^2 - ky^2 = 1$ has infinitely many solution, so $x^2 + (\sqrt{-k}y)^2 = 1$!

In particular, for $m \geq 3$ and $K = \mathbb{Q}(\zeta_{2^m})$, we have $\frac{1}{\sqrt{2}}\left(1 + i\right) \in \mathcal{O}_K \implies \sqrt{-2} \in \mathcal{O}_K$.

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    $\begingroup$ The cross product in $S_n\times \{-1,1\}$ is actually the wreath product, I guess? $\endgroup$ Sep 20, 2023 at 16:01
  • $\begingroup$ @colt_browning I haven't heard of that before and I need more time to read on it, it takes me a while to digest the wiki definition... I have edited it $\endgroup$
    – Gareth Ma
    Sep 20, 2023 at 22:52

1 Answer 1


$\def\cO{\mathcal{O}}\def\RR{\mathbb{R}}\def\QQ{\mathbb{Q}}\def\ZZ{\mathbb{Z}}$This answer resolves the question of when $O_n(\cO_K)$ is finite. I will show:

(1) If $K$ is a totally real field, then $O_n(\cO_K)$ is finite for all $n$.

(2) If $K$ is not totally real and is not $\QQ(\sqrt{-1})$, then $O_2(\cO_K)$ is infinite.

(3) $O_2(\ZZ[\sqrt{-1}])$ is finite, but $O_3(\ZZ[\sqrt{-1}])$ is infinite.

Proof of (1): We will show that there are only finitely many solutions to $z_1^2+z_2^2+\cdots+z_n^2=1$ in $\cO_K$. Thus, $O_n(\cO_K)$ is finite in this case.

Let $[K:\QQ] = d$ and let $\sigma_1$, $\sigma_2$, ..., $\sigma_d$ be the embeddings $K \to \RR$. We recall that the trace map $K \to \QQ$ is defined by $$T(z) = \sum_{i=1}^d \sigma_i(z).$$ Here are the basic facts that we will need about trace:

Lemma 1: For $z \in \cO_K$, we have $T(z) \in \ZZ$. Proof: $T(z)$ is a sum of algebraic integers, so it is an algebraic integer, and it is rational. $\square$

Lemma 2: For $z \in K$, $z \neq 0$, we have $T(z^2) > 0$. Proof: We have $T(z^2) = \sum \sigma_i(z^2) = \sum \sigma_i(z)^2$, so it is a sum of real squares. That shows that $T(z^2) \geq 0$, and we have $T(z^2)=0$ if and only if $\sigma_1(z) = \sigma_2(z) = \cdots = \sigma_d(z) = 0$, in which case $z=0$. $\square$

Now, suppose that $z_1^2+z_2^2 + \cdots + z_n^2 = 1$ with the $z_i$ in $\cO$. Taking $T$ of both sides, $\sum T(z_i^2) = d$. By the lemmas above, each of the $T(z_i^2)$ is a nonnegative integer; there are only finitely many options for ways to write $d$ as a sum of $n$ nonnegative integers.

For each nonnegative integer $r$, the equation $T(z^2) = r$ is a sphere is $\cO_K \otimes_{\ZZ} \RR$ (since Lemma 2 shows that the quadratic form $T(z^2)$ is positive definite) and $\cO_K$ is a discrete lattice in $\cO_K \otimes_{\ZZ} \RR$. So there are only finitely many solutions to $T(z^2) = r$ for fixed $r$. $\square$

Proof of (2): We need to break into two cases.

Case 1: Suppose that $\sqrt{-1} \in K$. Since $K \neq \QQ(\sqrt{-1})$, we must have $[K:\QQ] > 2$ and thus the unit group of $\cO_K$ is infinite. The finite index subgroup of units $u$ which are $\equiv 1 \bmod 2 \cO_K$ is thus also infinite.

Let $u$ be a unit which is $\equiv 1 \bmod 2 \cO_K$. Put $$a = \frac{u+u^{-1}}{2},\ b = \frac{u-u^{-1}}{2\sqrt{-1}}.$$ Then $a$ and $b$ are both in $\cO_K$, and we have $$a^2+b^2 = (a+b\sqrt{-1})(a-b\sqrt{-1}) = u \cdot u^{-1} = 1.$$ so the matrix $\begin{bmatrix} a&-b \\ b&a \end{bmatrix}$ is in $SO_2(\cO_K)$. $\square$

Case 2: $\sqrt{-1} \not\in K$. Put $L = K(\sqrt{-1})$, so $[L:K] = 2$. Let $K$ have $r$ real embeddings and $2s > 0$ complex emebeddings; then $L$ has $2(r+2s)$ complex embeddings (and no real embeddings). By Dirichlet's unit theorem, the ranks of $\cO_K^{\times}$ and $\cO_L^{\times}$ are $r+s-1$ and $r+2s-1$ respectively; in particular, since $s>0$, the rank of $\cO_L^{\times}$ is greater than the rank of $\cO_K^{\times}$.

The norm map $N_{L/K}$ is a group homomorphism $\cO_L^{\times} \to \cO_K^{\times}$. Since these groups have different ranks, the kernel of $N_{L/K}$ is infinite. If we intersect that kernel with the finite index subgroup of units that are in $\cO_K + \cO_K \sqrt{-1}$, it will still be infinite. Thus, we can find infinitely many $(a,b) \in \cO_K^2$ such that $N_{L/K}(a+b\sqrt{-1}) = 1$. We have $N_{L/K}(a+b\sqrt{-1}) = (a+b \sqrt{-1})(a-b \sqrt{-1}) = a^2+b^2$. So, again, we have infinitely many matrices of the form $\begin{bmatrix} a&-b \\ b&a \end{bmatrix}$ in $SO_2(\cO_K)$.

Proof of (3): If $O_2(\ZZ[\sqrt{-1}])$ were infinite then $SO_2(\ZZ[\sqrt{-1}])$ would also be infinite, so we would have infinitely many solutions to $a^2+b^2=1$ with $a$, $b \in \ZZ[\sqrt{-1}]$. Rewrite this as $(a+b\sqrt{-1})(a-b \sqrt{-1}) = 1$, so $a+b \sqrt{-1}$ is a unit of $\ZZ[\sqrt{-1}]$. The unit group of $\ZZ[\sqrt{-1}]$ is finite.

Now to handle $SO_3$. Recall that, if $R$ is a commutative ring, and $(a,b,c,d) \in R$ with $a^2+b^2+c^2+d^2=1$, then $$\begin{bmatrix} a^2+b^2-c^2-d^2&2bc-2ad &2bd+2ac \\ 2bc+2ad &a^2-b^2+c^2-d^2&2cd-2ab \\ 2bd-2ac &2cd+2ab &a^2-b^2-c^2+d^2\\ \end{bmatrix} \in SO_3(R).$$ (This is the quaternion parametrization of $SO_3$.) This map is $2 \to 1$; $(a,b,c,d)$ and $(-a, -b, -c, -d)$ have the same image.

Thus, it is enough to find infinitely many solutions to $a^2+b^2+c^2+d^2=1$ in $\ZZ[\sqrt{-1}]$. There are surely more principled ways to do this, but just taking $(a,b,c,d) = (n, n \sqrt{-1}, 1, 0)$ is enough to show that there are infinitely many solutions. $\square$.

Switching to a much more sophisticated perspective: Borel and Harish-Chandra showed that, if $G$ is an affine algebraic group of finite type over $\ZZ$ then $G(\ZZ)$ is finitely generated. If $R$ is a ring which is finite free over $\ZZ$ and $H$ is an affine algebraic group of finite type over $R$, then we can make an algebraic group $G$ with $G(\ZZ) = H(R)$. (Let $R = \ZZ \alpha_1 \oplus \ZZ \alpha_2 \oplus \cdots \oplus \ZZ \alpha_k$. Take each $R$-valued variable in the definition of $H$ and replace it with $k$ many $\ZZ$-valued variables, then use the multiplication table for $R$ to replace each polynomial expression in the $R$-valued variables with $k$ polynomial expressions in the $R$-valued variables.) So $O_n(R)$ is always finitely generated. I have no idea what the actual generators are, though.

  • $\begingroup$ Thanks a lot for the great answer, it's easy to understand (even for me :)). I will leave the question open for a couple more days but I have given you the bounty :D $\endgroup$
    – Gareth Ma
    Sep 20, 2023 at 22:49

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