Diophantine approximation on spheres

Question

I am trying to read this letter by Sarnak, where, among other things, he discusses the problem of intrinsic diophantine approximation on $S^3 = \left\{\mathbf{x} \in \mathbb{R}^4 : x_1^2+x_2^2+x_3^2+x_4^2 = 1\right\}$. In particular, on page 22, he studies diophantine approximation of $S^3$ by points on $S^3(\mathbb{Z}[\tfrac{1}{5}])$. I am aware that the circle method can be used to obtain these approximations.

My question is as follows: Is there a more direct way to prove that $S^3(\mathbf{Z}[\tfrac{1}{5}])$ is dense in $S^3$? Also, is the statement true if we replace $5$ with any other prime $p \neq 2$?

Edit: Following the comments of Dodd and Wojowu, the condition $p \neq 2$ has been added, since this is clearly necessary.

Answer 1

Here is a proof that $S(\mathbb{Z}[\frac{1}{p}])$ lies dense in $S^3$ for all primes $p \equiv 1 \pmod{4}$.

Since this is a wholly algebraic/arithmetical question, it is easier to switch to algebro-geometric terminology. We consider the map $$ \phi \, \colon S^3 \rightarrow \mathbb{A}^1 $$ where $\mathbb{A}^1$ is the affine line, which sends $(x_1,x_2,x_3,x_4)$ to $x_1^2+x_2^2=1-x_3^2-x_4^2$.

The fibres of $\phi$ consist of the sets of points $(x_1,x_2,x_3,x_4)$ with both $x_1^2+x_2^2$ and $x_3^2+x_4^2$ constant. These fibres have an action of $S^1 \times S^1$ on them, where $S^1$ is the algebraic variety defined by $u^2+v^2=1$ (except I guess for degenerate fibres, which can be safely ignored): namely by interpreting the points of $S^1$ as complex numbers of unit $1$, and interpreting also the tuples $(x_1,x_2)$ and $(x_3,x_4)$ as complex numbers, and considering complex multiplication, we arrive at the action $$ ((u_1,v_1),(u_2,v_2)) \circ (x_1,x_2,x_3,x_4) = (u_1x_1-v_1x_2,u_1x_2+v_1x_1,u_2x_3-v_2x_4,u_2x_4+v_2x_3) $$ whose virtue is that the entire orbit of a point in $S^3(\mathbb{Z}[\frac{1}{p}])$ under $(S^1 \times S^1)(\mathbb{Z}[\frac{1}{p}])$ lies wholly in $S^3(\mathbb{Z}[\frac{1}{p}])$, as is clear from the formula.

I first claim that the set $S^1(\mathbb{Z}[\frac{1}{p}])$ lies dense in the set of real points of $S^1$. This is a pretty easy exercise, so I will just leave this claim without a proof for the time being (but this is where the congruence condition on $p$ comes in). Hence also the points $(S^1 \times S^1)(\mathbb{Z}[\frac{1}{p}])$ lie dense in the real points of $S^1\times S^1$. It then also follows that if any fibre of $\phi$ contains a point defined over $\mathbb{Z}[\frac{1}{p}]$, the points lie dense within the real points of that fibre.

I next claim that the image $\phi(S(\mathbb{Z}[\frac{1}{p}]))$ is dense in $\phi(S^3(\mathbb{R})) = [0,1]$. But this follows directly from the density of $S^1(\mathbb{Z}[\frac{1}{p}])$ in $S^1(\mathbb{R})$: already the image under $\phi$ of the subvariety of $S^3$ where $x_1=x_4=0$ lies dense in $[0,1]$.

So for any real point $P$ of $S^3$ I can find a point $P' \in S(\mathbb{Z}[\frac{1}{p}])$ which lies on a fibre of $\phi$ arbitrarily close to the fibre that $P$ is on. But then we are done, because the points on this fibre that are defined over $\mathbb{Z}[\frac{1}{p}]$ are dense on it.

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The easy exercise. I claim: let $p$ be a prime congruent to $1\pmod{4}$, then $S^1(\mathbb{Z}[\frac{1}{p}])$ lies dense in the set of real points of $S^1$. Proof: it is a classical result that there exist integers $a,b$ such that $a^2+b^2=p$, in other words there exists a complex number $a+bi$ with absolute value $\sqrt{p}$. Then $u+vi=\frac{1}{p} \cdot (a+bi)^2$ is a complex number with absolute value $1$ and $u,v \in \mathbb{Z}[\frac{1}{p}]$. Then, writing $(u+vi)^k = u_k + iv_k$, we get an infinite number of pairs $(u_k,v_k)$ satisfying $u_k^2 + v_k^2=1$ and $u_k,v_k \in \mathbb{Z}[\frac{1}{p}]$. Since $\arg(u_k + iv_k) = k \arg(u+iv)$, these points indeed lie dense on the unit circle.

Answer 2

dodd is right. Every point over $\mathbb{Z}[1/2]$ must have coordinates in $\frac{1}{2} \mathbb{Z}$, since by clearing denominators we get four squares of integers, not all even, summing to a power of two. Since the square of an odd number is 1 mod 8, this power of two can only be 1 or 4. Hence the number of points over this ring is finite.

I wonder what happens for p = 3 though.