Is $ SU_2(\Bbb Z[1/\sqrt{2}]) $ finite? Let $ G=SU_2(\mathbb{Z}[1/\sqrt{2}]) $ be the group of $ 2 \times 2 $ unitary determinant one matrices with entries whose real and imaginary parts are from $\mathbb{Z}[i,1/\sqrt{2}]$.
Is $ G $ finite?
For $ SU_2(\mathbb{Z}[1/\sqrt{p}]) $ for $ p=3,5 $ etc I have already found some matrices of infinite order. But for $ p=2 $ I am having difficulty finding any matrices of infinite order or any other indication that $ G $ is infinite.
 A: EDIT: My earlier "answer" was toooooo glib, and not accurate! EDIT-EDIT: some precise details filled in...
In general, we should tend to expect that a semi-simple algebraic group's points over a localization of the defining ring (here $\mathbb Z$, actually), is infinite, for general reasons related to Strong Approximation.
In the present example, there is a potentially confusing point, that the definition of the group seems to involve $i=\sqrt{-1}$... yes, but the points of the algebraic group using $i$ in their coordinates are defined over $\mathbb Q$, not $\mathbb Q(i)$. Namely, if $G$ is the standard $SU(2)$, and $A$ is a $\mathbb Z$-algebra, what is $G(A)$? It is $\{g\in GL_2(\mathbb Z[i]\otimes_{\mathbb Z} A):g^*g=1_2, \;\det g=1\}$. That is, the appearance of complex numbers is an artifact of the construction.
Partly to avoid that pitfall, and partly to have a more efficient approach with this small group, we use the isomorphism of $SU(2)$ with Hamiltonian quaternions $\mathbb H^1$ of norm $1$. Specifically, letting $\mathbb H_{\mathbb Q}=\{a+bi+cj+dk: a,b,c,d\in\mathbb Q\}$, the isomorphism $H^1_{\mathbb Q}\to SU(2)(\mathbb Q)$ is by
$$
a+bi+cj+dk\;\to\;\pmatrix{a+bi & c+di \cr c-di & a-bi}
$$
For a $\mathbb Z$-algebra $A$, the $A$-valued points of the group are identified with the elements of $\mathbb H\otimes_{\mathbb Z} A$ with (extended) norm $1$.
Without worrying about adjoining $\sqrt{2}$, $\sqrt{3}$, and such things, and/or their inverses, we can address a more direct question: for prime $p$, does $G(\mathbb Z[1/p])$ have any infinite-order elements? The discussion of impossibility is a little simpler than the case of possibility, and we are perhaps most immediately interested in $A=\mathbb Z[1/2]$, as a preliminary to $A=\mathbb Z[1/\sqrt{2}]$.
If an element $\alpha$ of infinite order were in $G(\mathbb Z[1/2])$, then $\mathbb Q_2(\alpha)$ is an extension $k$ of $\mathbb Q$ of either degree one or two, since it is inside the quaternion algebra $\mathbb H_{\mathbb Q}$. Since $G(\mathbb Z)$ is finite, the extension must be of degree $2$. By Hilbert's Theorem 90, since it has norm $1$, $\alpha=\beta/\beta^\sigma$ for some $\beta$ in $k$. If $\beta=\beta^\sigma$, this is just the identity, not of infinite order. Meanwhile, $\sigma$ is of order two. Thus, it cannot be that $\beta^\sigma=u\cdot \beta$ with a unit $u$ other than $u=\pm 1$. Thus, the only way to make such $alpha$ is as $\beta/\beta^\sigma$ with $\beta$ divisible by at least one prime $\pi$ that has split in the extension. That is, for some rational $p$ splitting as $p=\pi_1\cdot \pi_2$, take $\alpha=\pi_1/\pi_2$.
Of course, the ring of integers $\mathfrak o_k$ in $k$ may not be a PID, but its class number $h$ is finite, so even if $p$ splits into non-principal ideals $\pi_1$, $\pi_2$, the power $\pi_1^h$ and $\pi_2^h$ are principal, and we can manufacture $\alpha$.
Indeed, this trick succeeds for any quadratic extension $k$ of $\mathbb Q$ that admits an imbedding to the rational Hamiltonian quaternions, since all primes $p\ge 3$ split these quaternions, there is no constraint at any primes other than $2$ and $\infty$. At $\infty$, the quadratic (algebra) extension $k$ must be $\mathbb C$ (as opposed to $\mathbb R\oplus \mathbb R$).
So, with a little work, $G(\mathbb Z[1/p])$ has torsion elements for any $p\ge 3$. We don't even need $1/\sqrt{p}$.
In contrast, since $\mathbb H_{\mathbb Q}$ is ramified at $2$, there is no quadratic extension $k$ of $\mathbb Q$ splitting at $2$ that imbeds into that quaternion algebra. So there are no elements of infinite order in $G(\mathbb Z[1/2])$.
But, still, what about $1/\sqrt{2}$? If this extension splits the quaternion algebra, then the construction above will succeed in producing an infinite-order element... :)
A: It's not finite. This is closely related to the Golden Gates in quantum computation. Roughly, not only the group is not finite, its elements can actually be used to approximate any element in $PU(2)$. Details can be found in the notes by Peter Sarnak.
To illustrate the key ideas, with some effort, we can show that any element in $G$ can be written as $$g=\frac{1}{2^{h/2}}\begin{pmatrix} x_1+x_2i && x_3+x_4i \\ -x_3+x_4i && x_1 - x_2i\end{pmatrix}$$ where $x_1, x_2, x_3, x_4\in \mathbb Z[\sqrt 2]$ and $x_1^2+x_2^2+x_3^2+x_4^2=2^h$. (It's slightly harder to show the decomposion exsits, but it's straightforward to show such a matrix indeed lies in $G$, which is all we need.)
Now let $x_1=\sqrt 2 + 1, x_2 = \sqrt 2 - 1, x_3 = \sqrt 2, x_4 = 2\sqrt 2, h=4$. And to show $g$ has inifnite order, we only need to show its eigenvalues are not roots of unity. Since $\det(g)=1, \text{Tr}(g)=\frac{2}{2^{h/2}} x_1=\frac{x_1}{2}$, the characteristic polynomial is $$\lambda^2-\frac{x_1}{2}\lambda + 1 = 0$$
Let $\cos(\theta)=\frac{x_1}{4}=\frac{\sqrt 2 + 1}{4}$, then we have $\lambda = \cos(\theta) \pm i \sin\theta=e^{i\theta}$. Now it suffices to show $\theta$ is not a rational multiple of $\pi$. OK... I don't how to do this, but if $e^{i\theta}$ is a root of unity, so is $e^{-i\theta}$, hence $2\cos(\theta)=e^{i\theta} + e^{-i\theta}$ is integral but $\frac{x_1}{2}=\frac{\sqrt 2 + 1}{2}$ is not.
