0

This is a question from Hoffstein cryptography book.

I'm trying to show that

$\lim_{R \to \infty}$$\frac{{\#(\mathbb{B}_R(\mathbf{0})}\cap L)}{{Vol(\mathbb{B}_R(\mathbf{0}))}}$$=\frac{1}{Vol(\mathcal{F})}$ for a lattice L and its fundamental domain F

Following the questions, I've already proved that

$\bigcup_{\mathbf{v}\in L ,{\mathcal{F}+\mathbf{v}}\subset{\mathbb{B}_R(\mathbf{0})}}$$(\mathcal{F}+\mathbf{v})\subset{\mathbb{B}_R(\mathbf{0})}\subset$ $\bigcup_{\mathbf{v}\in L ,{\mathcal{F}+\mathbf{v}}\cap{\mathbb{B}_R(\mathbf{0})}\neq\emptyset}{({\mathcal{F}+\mathbf{v}})}$

Taking volumes, I've also proved

$\#\{\mathbf{v}\in L:\mathcal{F}+\mathbf{v}\subset{\mathbb{B}_R(\mathbf{0})}\}\cdot Vol(\mathcal{F}) \leq Vol ({\mathbb{B}_R(\mathbf{0})})\leq \#\{\mathbf{v}\in L:(\mathcal{F}+\mathbf{v})\cap{\mathbb{B}_R(\mathbf{0})}\neq\emptyset\}\cdot Vol(\mathcal{F}).$

Now I'm struggling with this part: Prove that the number of translates $\mathcal{F}+\mathbf{v}$ that intersect ${\mathbb{B}_R(\mathbf{0})}$ without being entirely contained within ${\mathbb{B}_R(\mathbf{0})}$ is comparatively small compared to the number of translates that are entirely contained within ${\mathbb{B}_R(\mathbf{0})}$.

I understand this is true by instinct in geometric way, but I don't know how to prove this mathematically. Can someone help me with this?

Thank you a lot in advance.

Measurezero
  • 23
  • Putting $\mathcal{F}$ around each lattice point in $B_R(0)$ you get a volume in between $B_{R-c}(0)$ and $B_{R+c}(0)$ (where $c$ is large enough so that $\mathcal{F} \subset B_c(0)$) so it suffices to show $\lim_{R \to \infty} \frac{B_{R+c}(0)}{B_{R-c}(0)}=1$ – reuns Apr 13 '19 at 03:07
  • You mean that the volume of translates around $B_R (0)$ which is not entirely contained within $B_R (0)$ is between 0 and $B_{R+c} (0) - B_{R-c} (0)$? – Measurezero Apr 13 '19 at 04:35

1 Answers1

1

Let $$L_R = \bigcup_{v \in B_R(0) \cap L} (v+\mathcal{F})$$ then $$B_{R-c}(0) \subset L_R \subset B_{R+c}(0)$$ where $c$ is large enough such that $\mathcal{F} \subset B_c(0)$.

Then $Vol(L_R) = Vol(\mathcal{F})|B_R(0) \cap L|$ and $\lim_{R \to \infty} \frac{Vol(B_{R-c}(0))}{Vol(B_{R}(0))} = 1=\lim_{R \to \infty} \frac{Vol(B_{R+c}(0))}{Vol(B_{R}(0))}$ so $$1=\lim_{R \to \infty} \frac{Vol(L_R)}{Vol(B_{R+c}(0))}=\lim_{R \to \infty} \frac{Vol(\mathcal{F})|B_R(0) \cap L|}{Vol(B_{R}(0))}$$

reuns
  • 77,999