# Are lattice points in thin spherical shells uniformly distributed?

Consider the spherical shell (annulus) $$A(R,r) = \{ x \in \mathbb{R}^3 : R \leq | x|\leq R+r \}.$$ Think of the limit $R \to \infty$. Assume that $r$ depends on $R$ as $r(R) = R^{-\delta}$. We are interested in the distribution of lattice points in $A(R,r)$.

From results on the Gauss circle problem in three dimensions (see e.g. https://arxiv.org/abs/math/0410522) I know that the number of lattice points in the ball $B(R)$ is given by the volume of the ball, up to an error (the lattice point discrepancy) which is bounded by $\mathcal{O}(R^{\frac{42}{32}+\epsilon})$, for all $\epsilon > 0$. So by taking the difference we can obtain the number of lattice points in the annulus for $\delta >0$ not too large. We find that the number of lattice points is of order $R^{2}r$ (surface of the sphere times width of the shell).

In https://arxiv.org/abs/1204.0134 I found the reference to Duke & Golubeva-Fomenko, showing that the the lattice points exactly on the sphere are uniformly distributed, if $R^2 \neq 0,4,7 \mod 8$. (However there are only order $R^{1-\epsilon}$ lattice points exactly on the sphere.)

My question is: Is anything known about the distribution of lattice points in the thin spherical shell? For example, consider a spherical cap on $B(R)$ and 'fatten' it up to a radius $R+r$. We now have a segment of the annulus (if you want, the intersection of a cone with the annulus) and ask whether the number of lattice points in this segment is to leading order given by the area of the spherical cap times $r$. If yes, does this stay true if the solid angle defining the cap goes to zero (not too fast) as $R \to \infty$? If it is not true for all such segments, could I at least say that it is true for most segments, or for a specific sequence of radii R?

Another possible way of asking could be: consider summing a continuous function over $x/ |x|$, $x \in A(R,r)$ (and normalize by the number of points). Does the sum converge to the integral w.r.t. the uniform measure on the unit sphere?

Is anyone aware of results in this direction?

• The surface of the sphere is proportional to $R^2$, not $R^{2/3}$. – GH from MO Jun 9 '18 at 1:37
• There is a relevant picture in an answer here on MO – მამუკა ჯიბლაძე Jun 9 '18 at 4:21
• Sorry, the $R^{2/3}$ was a typo, I've fixed it. Concerning the picture, that is an interesting phenomenon. I'll think that through tomorrow. Thank you already for your anwers, I need a few days to find time to go through it. – NBat Jun 10 '18 at 21:42
• If you like my answer, please accept it officially (so that it turns green). Thanks in advance! – GH from MO Jun 23 '18 at 23:13

Yes, they are equidistributed as long as $\delta<11/16$ and $r=R^{-\delta}$ and $R\to\infty$. Without loss of generality, we shall assume that $\delta>-1$ (i.e. $r<R$).
To see this, let $\mathcal{F}\subset S^2$ be a fixed convex region with piecewise smooth boundary on the unit sphere. Let $\mu(\mathcal{F})$ be the normalized area of $\mathcal{F}$, where normalization is such that $\mu(S^2)=1$. Let $r(n)$ be the number of representations $n=|\mathbf{x}|^2$ with $\mathbf{x}\in\mathbb{Z}^3$, and let $r(n,\mathcal{F})$ be the number of representations with the additional constraint that $\mathbf{x}/|\mathbf{x}|\in\mathcal{F}$.
Let $k\in\mathbb{N}$ be fixed. By Theorem 1 in the 1990 Inventiones paper of Duke and Schulze-Pillot, \begin{align*}\sum_{\substack{R^2\leq n\leq (R+r)^2\\4^k\nmid n}}r(n,\mathcal{F})&=\bigl(\mu(\mathcal{F})+o(1)\bigr)\sum_{\substack{R^2\leq n\leq (R+r)^2\\4^k\nmid n}}r(n)\\&=\bigl(\mu(\mathcal{F})+o(1)\bigr)\sum_{R^2\leq n\leq (R+r)^2}r(n)+O\Bigl(\sum_{\substack{R^2\leq n\leq (R+r)^2\\4^k\mid n}}r(n)\Bigr).\end{align*} On the right hand side, by a result of Heath-Brown (1999), $$\sum_{R^2\leq n\leq (R+r)^2}r(n)=\bigl(4\pi+o(1)\bigr)R^2r.$$ For $4^k\mid n$, we have that $r(n)=r(n/4^k)$, hence Heath-Brown's result also yields $$\sum_{\substack{R^2\leq n\leq (R+r)^2\\4^k\mid n}}r(n)=\sum_{4^{-k}R^2\leq m\leq 4^{-k}(R+r)^2}r(m)=\bigl(4\pi\cdot 8^{-k}+o(1)\bigr)R^2r.$$ To summarize so far, $$\sum_{\substack{R^2\leq n\leq (R+r)^2\\4^k\nmid n}}r(n,\mathcal{F})=\bigl(\mu(\mathcal{F})+o(1)\bigr)\sum_{R^2\leq n\leq (R+r)^2}r(n)+\bigl(O(8^{-k})+o(1)\bigr)R^2r.$$ Combining this with the earlier two displays (consequences of Heath-Brown's result), we infer that \begin{align*}\sum_{R^2\leq n\leq (R+r)^2}r(n,\mathcal{F})&=\bigl(\mu(\mathcal{F})+o(1)\bigr)\sum_{R^2\leq n\leq (R+r)^2}r(n)+\bigl(O(8^{-k})+o(1)\bigr)R^2r\\[6pt]&=\bigl(\mu(\mathcal{F})+O(8^{-k})+o(1)\bigr)\sum_{R^2\leq n\leq (R+r)^2}r(n). \end{align*} In other words, $$\limsup_{R\to\infty}\left|\frac{\sum_{R^2\leq n\leq (R+r)^2}r(n,\mathcal{F})}{\sum_{R^2\leq n\leq (R+r)^2}r(n)}-\mu(\mathcal{F})\right|=O(8^{-k}).$$ As $k\in\mathbb{N}$ is arbitrary, and the left hand side is independent of $k$, the left hand side is zero. That is, $$\lim_{R\to\infty}\frac{\sum_{R^2\leq n\leq (R+r)^2}r(n,\mathcal{F})}{\sum_{R^2\leq n\leq (R+r)^2}r(n)}=\mu(\mathcal{F}).$$