# 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. Ivic, Krätzel, Kühleitner, and Nowak - Lattice points in large regions and related arithmetic functions: Recent developments in a very classic topic) 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 Bourgain, Sarnak, and Rudnick - Local statistics of lattice points on the sphere, I found the reference to Duke - Hyperbolic distribution problems and half-integral weight Maass forms & Golubeva–Fomenko - Asymptotic distribution of lattice points on the three-dimensional sphere, showing that the the lattice points exactly on the sphere are uniformly distributed, if $$R^2 \neq 0,4,7 \bmod 8$$. (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 $R^{2/3}$. Jun 9 '18 at 1:37
• There is a relevant picture in an answer here on MO 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
• I edited in the names of the papers you referenced. The Bourgain–Sarnak–Rudnik paper you reference cites two papers by Duke, but only one with Duke as sole author, so I assumed you meant that one. What is up with all these papers whose authors are not listed in alphabetical order? Feb 1 at 16:03

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. $$r).
To see this, let $$\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}).$$