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?


1 Answer 1


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.

  • $\begingroup$ 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? $\endgroup$
    – Lucio
    Commented Feb 17, 2018 at 8:34
  • 1
    $\begingroup$ yes, it's true in any dimension. $\endgroup$
    – user120527
    Commented Feb 17, 2018 at 9:49

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .