samples/texts/1897687/page_7.md

of all integer vectors having integer inner products with all vectors in $\mathcal{L}$. It is easy to check that $(\mathcal{L}^*)^* = \mathcal{L}$ and that, if $\mathbf{B}$ is a basis for $\mathcal{L}$, then $\mathbf{B}^* = \mathbf{B}(\mathbf{B}^T\mathbf{B})^{-1}$ is a basis for $\mathcal{L}^*$. The fundamental parallelepiped of a lattice $\mathcal{L}$ with respect to basis $\mathbf{B}$ is the set

It is easy to see that $\mathcal{P}(\mathbf{B})$ is a fundamental domain of $\mathcal{L}$. I.e., it tiles $\mathbb{R}^n$ with respect to $\mathcal{L}$. For any lattice $\mathcal{L}(\mathbf{B})$ and point $\mathbf{x} \in \mathbb{R}^n$, there exists a unique point $\mathbf{y} \in \mathcal{P}(\mathbf{B})$ such that $\mathbf{y} - \mathbf{x} \in \mathcal{L}(\mathbf{B})$. We denote this vector by $\mathbf{y} = \mathbf{x} \bmod \mathbf{B}$. Notice that $\mathbf{y}$ can be computed in polynomial time given $\mathbf{B}$ and $\mathbf{x}$. We sometimes write $\mathbf{x} \bmod \mathcal{L}$ when the specific fundamental domain is not important, and we write $\mathbb{R}^n/\mathcal{L}$ for an arbitrary fundamental domain.

The determinant of a lattice $\mathcal{L}$, is defined to be $\det(\mathcal{L}) = \sqrt{\det(\mathbf{B}^T\mathbf{B})}$. It is easy to verify that the determinant does not depend on the choice of basis and that $\det(\mathcal{L})$ is the volume of any fundamental domain of $\mathcal{L}$.

The minimum distance of a lattice $\mathcal{L}$, is the length of the shortest non-zero lattice vector,

$${\lambda }_1(\mathcal{L}):=\underset{\mathbf{y}\in \mathcal{L}\setminus \{\mathbf{0}\}}{\min }\mathrm{\parallel }\mathbf{y}\mathrm{\parallel }\mathrm{.}\backslash lambda\_1(\backslash mathcal\{L\})\; :=\; \backslash min\_\{\backslash mathbf\{y\}\; \backslash in\; \backslash mathcal\{L\}\; \backslash setminus\; \backslash \{\backslash mathbf\{0\}\backslash \}\}\; \backslash |\backslash mathbf\{y\}\backslash |.$$

Similarly, we define

$${\lambda }_n(\mathcal{L}):=\underset{i}{\min }\underset{j}{\max }\mathrm{\parallel }{\mathbf{y}}_i\mathrm{\parallel },\backslash lambda\_n(\backslash mathcal\{L\})\; :=\; \backslash min\_i\; \backslash max\_j\; \backslash |\backslash mathbf\{y\}\_i\backslash |,$$

where the minimum is taken over linearly independent lattice vectors $\mathbf{y}_1, \dots, \mathbf{y}_n \in \mathcal{L}$. The covering radius of a lattice $\mathcal{L}$ is

$$\mu (\mathcal{L}):=\underset{\mathbf{t}\in {\mathbb{R}}^n}{\max }\mathrm{d}\mathrm{i}\mathrm{s}\mathrm{t}(\mathbf{t},\mathcal{L})\mathrm{.}\backslash mu(\backslash mathcal\{L\})\; :=\; \backslash max\_\{\backslash mathbf\{t\}\; \backslash in\; \backslash mathbb\{R\}^n\}\; \backslash mathrm\{dist\}(\backslash mathbf\{t\},\; \backslash mathcal\{L\}).$$

The Voronoi cell of a lattice $\mathcal{L}$ is the set

$$\nu (\mathcal{L}):=\{\mathbf{x}\in {\mathbb{R}}^n:\mathrm{\parallel }\mathbf{t}\mathrm{\parallel }\le \mathrm{\parallel }\mathbf{y}-\mathbf{t}\mathrm{\parallel },\mathrm{\forall }\mathbf{y}\in \mathcal{L}\setminus \{\mathbf{0}\}\}\backslash nu(\backslash mathcal\{L\})\; :=\; \backslash \{\backslash mathbf\{x\}\; \backslash in\; \backslash mathbb\{R\}^n\; :\; \backslash |\; \backslash mathbf\{t\}\; \backslash |\; \backslash le\; \backslash |\; \backslash mathbf\{y\}\; -\; \backslash mathbf\{t\}\; \backslash |,\; \backslash forall\; \backslash mathbf\{y\}\; \backslash in\; \backslash mathcal\{L\}\; \backslash setminus\; \backslash \{\backslash mathbf\{0\}\backslash \}\backslash \}$$

of vectors in $\mathbb{R}^n$ that are closer to 0 than any other point of $\mathcal{L}$. It is easy to check that $V(\mathcal{L})$ is a symmetric polytope and that it tiles $\mathbb{R}^n$ with respect to $\mathcal{L}$. The following claim is an immediate consequence of the fact that an $n$-dimensional unit ball has volume at most $(2\pi e/n)^{n/2}$.

Claim 2.1. For any lattice $\mathcal{L} \subset \mathbb{R}^n$,

$$\mu (\mathcal{L})\ge \sqrt{n\mathrm{/}(2\pi e)}\cdot \det (\mathcal{L}{)}^{1\mathrm{/}n}\mathrm{.}\backslash mu(\backslash mathcal\{L\})\; \backslash geq\; \backslash sqrt\{n/(2\backslash pi\; e)\}\; \backslash cdot\; \backslash det(\backslash mathcal\{L\})^\{1/n\}.$$

Lemma 2.2. For any lattice $\mathcal{L} \subset \mathbb{R}^n$ and $r \ge 0$,

$$\mathrm{\mid }\mathcal{L}\cap r{B}_2^n\mathrm{\mid }\le (5\mathrm{/}\sqrt{n}{)}^n\cdot \frac{(r+\mu (\mathcal{L}){)}^n}{\det (\mathcal{L})}\mathrm{.}|\backslash mathcal\{L\}\; \backslash cap\; r\; B\_2^n|\; \backslash le\; (5/\backslash sqrt\{n\})^n\; \backslash cdot\; \backslash frac\{(r\; +\; \backslash mu(\backslash mathcal\{L\}))^n\}\{\backslash det(\backslash mathcal\{L\})\}.$$

Proof. For each vector $\mathbf{y} \in \mathcal{L} \cap rB_2^n$, notice that $\nu(\mathcal{L}) + \mathbf{y} \subseteq (r + \mu(\mathcal{L}))B_2^n$. And, for distinct vectors $\mathbf{y}, \mathbf{y}' \in \mathcal{L}, \nu(\mathcal{L}) + \mathbf{y}$ and $\nu(\mathcal{L}) + \mathbf{y}'$ are disjoint (up to a set of measure zero). Therefore,

$$\mathrm{v}\mathrm{o}\mathrm{l}((r+\mu (\mathcal{L}))B_2^n)\ge \mathrm{v}\mathrm{o}\mathrm{l}(\bigcup _{\mathbf{y}\in \mathcal{L}\cap r{B}_2^n}(\nu (\mathcal{L})+\mathbf{y}))=\mathrm{\mid }\mathcal{L}\cap r{B}_2^n\mathrm{\mid }\mathrm{v}\mathrm{o}\mathrm{l}(\nu (\mathcal{L}))=\mathrm{\mid }\mathcal{L}\cap r{B}_2^n\mathrm{\mid }\det (\mathcal{L})\mathrm{.}\backslash mathrm\{vol\}((r\; +\; \backslash mu(\backslash mathcal\{L\}))B\_2^n)\; \backslash geq\; \backslash mathrm\{vol\}\backslash left(\backslash bigcup\_\{\backslash mathbf\{y\}\; \backslash in\; \backslash mathcal\{L\}\; \backslash cap\; rB\_2^n\}\; (\backslash nu(\backslash mathcal\{L\})\; +\; \backslash mathbf\{y\})\backslash right)\; =\; |\backslash mathcal\{L\}\; \backslash cap\; rB\_2^n|\; \backslash mathrm\{vol\}(\backslash nu(\backslash mathcal\{L\}))\; =\; |\backslash mathcal\{L\}\; \backslash cap\; rB\_2^n|\; \backslash det(\backslash mathcal\{L\}).$$

The result follows by recalling that for any $r' > 0$, $\mathrm{vol}(r'B_2^n) \le (5r'/\sqrt{n})^n$. □