# Group structure of $\mathrm{O}_n(R)$ over a **ring** $R$

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 $$\frac{1}{\sqrt{2}}\left(1 + i\right) \in \mathcal{O}_K \implies \sqrt{-2} \in \mathcal{O}_K$$.

• The cross product in $S_n\times \{-1,1\}$ is actually the wreath product, I guess? Sep 20, 2023 at 16:01
• @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 Sep 20, 2023 at 22:52

## 1 Answer

$$\def\cO{\mathcal{O}}\def\RR{\mathbb{R}}\def\QQ{\mathbb{Q}}\def\ZZ{\mathbb{Z}}$$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.

• 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 Sep 20, 2023 at 22:49