# Is the sequence equidistributed on the sphere?

Let $\Gamma=\{\vec{h}\in \mathbb{Z}^3: \vec{h}\neq \vec{0}\}$ and let $\{ \vec{h}_n\}_{n\geq 1}$ be an ordering of the elements of $\Gamma$ such that $\Gamma=\{\vec{h}_n : n\geq 1\}$ and for all $n, m$ $$n\geq m\Rightarrow \vert \vec{h}_n\vert \geq \vert \vec{h}_m\vert$$ I'm wondering if the following sequence is equidistributed on the sphere $S^2 = \{\vec{x}\in\mathbb{R}^3: \vert\vec{x}\vert=1\}$ : $$\bigg\{ \frac{\vec{h}_n}{\vert \vec{h}_n\vert} : n\geq 1\bigg\}$$ By equidistributed I mean that, for any $f:S^2\to\mathbb{R}$ continuous, it holds $$\lim_{N\to\infty} \frac{1}{N} \sum_{n=1}^N f\bigg(\frac{\vec{h}_n}{\vert \vec{h}_n\vert} \bigg) = \frac{1}{4\pi}\int_{S^2} f(x)dx \quad (\ast)$$ The only criterion I know in order to establish equidistribution is Weyl's criterion for tori $\mathbb{T}^n=[0,1]^n$; I think that by a change in polar coordinates I can write the integral in $(\ast)$ as an integral on the two-dimensional torus, so maybe using the same change of coordinates on the sequence, I can reduce myself to the case of establishing equidistribution of a sequence on $\mathbb{T}^2$, where Weyl criterion can be applied. But I don't know if it's correct to do that (polar coordinates are not defined on the whole $S^2$, maybe it could create problems) and it feels like lots of calculations. So is such reasoning correct? Are there any other criteria that can be applied?

Weyl's criterion is based on the fact that any continuous function can be uniformly approached by trigonometric polynomials. In fact, any family of test functions that span a dense subset will do; it's best to choose one adapted to the problem, so I wouldn't advise to try to change coordinates to have a local homeomorphism with the torus.

What I would suggest are the following steps to prove your claim (which is indeed true):

1) Consider the set $\mathcal{F}$ of functions $F$ on the sphere which are supported on small geodesic triangles of the sphere, which are constant equal to $1$ in the interior of the triangle, zero outside, and the value on the boundary of the triangle is irrelevant (arbitrary, but bounded).

2) Check the result is true for $F\in \mathcal{F}$. This amount to estimate the number of integer points in $\mathbb{Z}^3$ in a conical sector (defined by the triangle), in a ball of radius $R\simeq N^{1/3}$ (using your parametrisation): it grows like $R^3$ times the (sphere) measure of the triangle. The fact that the $1$-one neighborhood of boundary of intersection of the conical sector with the ball of radius $R$ grows like $R^2=o(R^3)$ is crucial, so you don't miss too much points by approximating the number of integer points by computing a volume.

3) Take a continuous function $f$ on the sphere, and $\epsilon>0$. Prove that by taking a sufficiently refined triangulation of the sphere, one can approximate uniformly $f$ by a finite linear combination of functions in $\mathcal{F}$, i.e. $$|f-\sum_{i=1}^k \lambda_i F_i|<\epsilon,$$ where $F_i \in \mathcal{F}$.

4) Deduce the result for $f$ from the two points above.

• Tha's a very nice answer! Points 1), 3) and 4) are clear, i'll need a bit of time to think about point 2). I have one question, arising from the structure of the answer: I'm only interested in the case of $\mathbb{R}^3$ with $S^2$, but it feels like the reasoning works for any $\mathbb{R}^n$ with $S^{n-1}$ ($\vec{h}_n$ properly redefined). Is it true? – Lucio Feb 17 '18 at 8:34
• yes, it's true in any dimension. – user120527 Feb 17 '18 at 9:49