《格》第一章习题选做

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Chapter 1 Introduction to Lattices

1.1 Let \(\Lambda=\Lambda(\bm{B})\) with \(\bm{B}\in \bbR^{n\times n}\) be a lattice. Show that for any \(\varepsilon>0\) there is a radius \(R := R(\varepsilon,n,\bm{B})\) so that

\[ (1-\varepsilon)\cdot \frac{\Vol_n(R\cdot B_2^n)}{\det(\Lambda)} \leq |R\cdot B_2^n \cap \Lambda| \leq (1 + \varepsilon) \frac{\Vol_n(R\cdot B_2^n)}{\det(\Lambda)}. \]

Proof. Consider the fundamental parallelepiped \(P = \calP(\bm{B})\), clearly we have \(\Vol_n(P) = \det(\bm{B}) = \det (\Lambda)\). Let \(d\) be the diameter of \(P\), which is a finite number, then for all points in \(\bm{x}\in (R-d)\cdot B_2^n \cap \Lambda\), the fundamental parallelepiped \(P + \bm{x}\) lies in \(R\cdot B_2^n\) and are pairwise disjoint. By the volume argument, we have

\[|R\cdot B_2^n \cap \Lambda| \geq \frac{\Vol_n((R-d)\cdot B_2^n)}{\det(\Lambda)}. \]

Similarly, for all points in \(\bm{x}\in (R+d)\cdot B_2^n \cap \Lambda\), the fundamental parallelepiped \(P + \bm{x}\) contains \(R\cdot B_2^n\) and are pairwise disjoint. Thus we have

\[|R\cdot B_2^n \cap \Lambda| \leq \frac{\Vol_n((R+d)\cdot B_2^n)}{\det(\Lambda)}. \]

In conclusion, we have

\[ \left(1 - \frac{d}{R}\right)^n \frac{\Vol_n(R\cdot B_2^n)}{\det(\Lambda)} \leq |R\cdot B_2^n \cap \Lambda| \leq \left(1 + \frac{d}{R}\right)^n \frac{\Vol_n(R\cdot B_2^n)}{\det(\Lambda)}. \]

When \(R\) is large enough, we have \((1-d/R)^n \geq 1-\varepsilon\) and \((1+d/R)^n \leq \varepsilon\). \(\square\)

1.2 Solve the following:
(a) Let \(K \subseteq \bbR^n\) be a symmetric convex set with \(\Vol_n(K)>k\cdot 2^n\) for some \(k\in \bbN\). Show that \(|K\cap \bbZ^n | \geq k + 1\).

Proof. Firstly we have \(\Vol_2(\frac 12 \cdot K) > k\). By pigeonhole principle, there exists \(\bm{x}\in \bbR^n\) such that there are at least \(k+1\) lattice points \(\bm{y}_0,\dots,\bm{y}_{k} \in \bbZ^n\) such that \(\bm{x} + \bm{y} \in \frac 12 \cdot K\), this shows that each \(\bm{y}_i - \bm{y}_0 \in \frac 12 \cdot K - \frac 12 \cdot K = K\). Thus we have \(k+1\) lattice points in \(K\). \(\square\)

(b) Is the following claim true: For any \(k \in {1,\dots,n}\) there is a value \(f (k)\) so that for any symmetric convex body \(K\) with \(\Vol_n(K)>f(k)\cdot 2^n\), \(K\cap \bbZ^n\) contains \(k\) linearly independent vectors.

Proof. No. Consider \(K = [-1/2, 1/2] \times [-n, n]\) in \(\bbR^2\), the volume of \(K\) can be arbitrarily large, but \(K\cap \bbZ^2\) contains at most one linearly independent vector. \(\square\)

1.3 This is an application of Dirichlet's Theorem: Let \(a \in (0, 1]^n\) be a real vector and consider the hyperplane \(H := \{\bm{x} \in \bbR^n \mid \ang{\bm{a},\bm{x}} = 0\}\). Then there is a rational vector \(\tilde{\bm{a}} \in \frac{\bbZ^n}{q}\) with
\(q\leq (2nR)^n\) so that \(\tilde H := \{\bm{x} \in \bbR^n \mid \ang{\tilde{\bm{a}}, \bm{x}} = 0\}\)
satisfies the following:

\[\forall \bm{x} \in \{-R,\dots,R\}^n : (\bm{x}\in H \implies \bm{x} \in \tilde{H}). \]

Proof. By Dirichlet's theorem, there exists \(p_1,\dots,p_n\in \bbZ_{\geq 0}\) and \(q\leq (2nR)^n\) such that

\[\left| \frac{p_i}{q} - a_i \right| \leq \frac {1}{2nR q}, \]

we take \(\tilde{\bm{a}} = (p_1/q,\dots,p_n/q)\), then for each \(\bm{x} \in \{-R,\dots,R\}^n\) and \(\bm{x}\in H\), we have

\[\begin{align*} \left|\ang{\tilde{\bm{a}}, \bm{x}}\right| &= \left|\ang{\tilde{\bm{a}}-\bm{a}, \bm{x}} +\ang{\bm{a}, \bm{x}}\right|\\ &= \left|\ang{\tilde{\bm{a}}-\bm{a}, \bm{x}}\right|\\ &\leq \norm{\tilde{\bm{a}} - \bm{a}}_1 \cdot \norm{\bm{x}}_\infty\\ &\leq n \cdot \frac 1{2nRq} \cdot R \\ &\leq \frac 1{2q}. \end{align*}\]

Note that \(\tilde{\bm{a}} \in \frac {\bbZ^n}{q}\) and \(\bm{x}\in \bbZ^n\), we have \(\ang{\tilde{\bm{a}}, \bm{x}} \in \frac 1{q} \cdot \bbZ\), thus \(\ang{\tilde{\bm{a}}, \bm{x}} = 0\), we have \(\bm{x}\in \tilde{H}\). \(\square\)

1.5 \item Let \(\Lambda\subseteq \bbR^n\) be a full rank lattice. Show that \(\lambda_1(\Lambda)\cdot \lambda_n(\Lambda^*) \geq 1\), where \(\Lambda^*\) is the dual lattice.

Proof. Let \(\bm{x}\) be the shortest non-zero vector in \(\Lambda\). Consider the lattices points of \(\Lambda^*\) in radius \(\lambda_n(\Lambda^*)\), since they span \(\bbR^n\), there must have one lattice point \(\bm{y}\) such that \(\ang{\bm{x}, \bm{y}} \in \bbZ \smallsetminus \{0\}\). Thus

\[ \lambda_1(\Lambda) \cdot \lambda_n(\Lambda^*) \geq \norm{\bm x}_2 \cdot \norm{\bm y}_2 \geq |\ang{\bm{x}, \bm{y}}| \geq 1. \square \]

1.6 Prove that for any lattice \(\Lambda \subseteq \bbR^n\), one has \(\lambda_1(\Lambda)\cdot \lambda_1 (\Lambda^*)\leq n\).

Proof. By Minkowski's first theorem, we have

\[ \lambda_1(\Lambda)\cdot \lambda_1 (\Lambda^*) \leq \sqrt n \cdot \det(\Lambda)^{1/n} \cdot \sqrt n \cdot \det(\Lambda^*)^{1/n} = n. \square \]

1.8 Let \(a=(a_1,\dots,a_n)\in \bbZ^n\) be a vector of integer numbers. The original Euclidean algorithm does the following:

• REPEAT
  • Select the index \(i\) with \(|a_i|\) minimal.
  • Forall \(j\neq i\) replace \(a_j\) by \(\min \{|a_j +z\cdot a_i| \mid z\in \bbZ\}\).

Prove that the algorithm terminates after at most \(O(\log\norm{\bm{a}}_\infty)\) many iterations.

Proof. Note that \(\min\{ |a_j + z\cdot a_i| \mid z\in \bbZ \} < |a_i| / 2\), we have the minimal value of \(|a_i|\) is halved in each iteration. Thus the algorithm terminates after at most \(1 + \log_2 \norm{a}_\infty\) iterations. \(\square\)

1.9 Let \(\bm{A}\in \bbZ^{m\times n}\) and \(\bm b\in \bbZ^m\) with \(m\leq n\) where \(\bm A\) has full row rank. Show that in polynomial time one can compute a vector \(\bm x\in \bbZ^n\) with \(\bm{A x} = \bm b\) (or decide that no such vector exists).

Proof. We know that one can compute a unimodular matrix \(\bm U \in \bbZ^{n\times n}\) such that \(\bm{AU} = \bm B\) in Hermite normal form. Then we have \(\bm B \bm{U}^{-1} \bm x = \bm b\), let \(\bm y = \bm{U}^{-1} \bm x\), then \(\bm y\) should also be integral. Since \(\bm A\) is full rank, each time we determine \(y_i\) by

\[\sum_{j=1}^i B_{ij} y_j = b_i, \]

in the increasing order of \(i\). \(\square\)

1.10 Let \(\Lambda\subseteq \bbR^n\) be a full-rank lattice. Assume that \(\bm b_1,\dots,\bm b_n \in \Lambda\) are linearly-independent and minimize \(|\det(\bm b_1,...,\bm b_n)|\). Prove that \(\bm b_1,\dots,\bm b_n\) are indeed a basis of \(\Lambda\).

Proof. Let \(\bm A\) be a basis of \(\Lambda\), then we have \(\bm B = \bm {AZ}\) for some \(\bbZ\in \bbZ^{n\times n}\), and since \(\bm B\) is linearly independent, we have \(\det(\bm B) = \det(\bm A)\det(Z) \neq 0\). Since \(\bm B\) minimizes the determinant, we have \(\bm Z\) is unimodular, thus \(\bm B\) is a basis.

1.11 We want to consider a relaxed version of a KZ-reduced basis. We say that a basis \(\bm B = (\bm b_1,\dots, \bm b_n)\in \bbR^{n\times n}\) for a lattice \(\Lambda\) is \(\alpha\)-KZ-reduced for \(\alpha \geq 1\) if \(\bm B\) is coefficient reduced and \(\norm{\bm{b}^*_i} \leq \alpha\cdot \lambda_1 (\pi_{U_i} (\Lambda))\) for all \(i = 1,\dots,n\). Here \(\pi_{U_i}\) is again the projection into \(U_i := \Span\{\bm b_1,\dots,\bm b_{i-1}\}^\perp\). Show that the orthogonality defect of such a basis is \(\gamma(B) \leq (\alpha n)^n\).

Proof. For each \(i\), we first show that \(\lambda_1(\pi_{U_i}(\Lambda)) \leq \lambda_i(\Lambda)\). By definition we have \(i\) linearly independent lattice vectors \(\bm x_1,\dots,\bm x_i\) with length \(\leq \lambda_i(\Lambda)\). Then at least one of them satisfies \(\pi_{U_i}(\bm x_j) \neq \bm 0\), thus \(\lambda_1(\pi_{U_i}(\Lambda)) \leq \norm{\pi_{U_i}(\bm x_j)}_2 \leq \norm{\bm x_j}_2 \leq \lambda_i(\Lambda)\). Thus we have

\[\begin{align*} \norm{\bm b_i}_2^2 &= \norm{\bm b_i^*}_2^2 + \sum_{j=1}^{i-1} \mu_{ji}^2 \norm{\bm b_j^*}_2^2\\ &\leq \norm{\bm b_i^*}_2^2 + \sum_{j=1}^{i-1} \norm{\bm b_j^*}_2^2 & (|\mu_{ji}|\leq 1/2)\\ &\leq n \cdot (\alpha \lambda_i(\Lambda))^2, \end{align*}\]

thus we have

\[ \prod_{i=1}^n \norm{\bm b_i}_2 \leq n^{n/2} \alpha^n \lambda_1(\Lambda) \cdots \lambda_n(\Lambda) \stackrel{\text{Mink.~2nd Thm}}{\leq} n^{n/2} \alpha^n \cdot n^{n/2} \det(\Lambda), \]

thus we have \(\gamma(\bm B) \leq (\alpha n)^n\). \(\square\)

1.12 Let \(\Lambda \subseteq \bbR^n\) be a full rank lattice and let \(\bm B \in \bbR^{n\times n}\) be an LLL-reduced basis for \(\Lambda\). Prove that for all \(i \in \{1,\dots,n\}\) one has \(\norm{\bm b_i}_2 \leq 2^{(n+1)/2}\lambda_i(\Lambda)\).

Proof. Firstly, we have

\[\begin{align*} \norm{\bm b_i}_2^2 &= \Norm{\bm b_i^* + \sum_{j=1}^{i-1} \mu_{ji} \bm b_j^*}_2^2\\ &= \norm{\bm b_i^*}_2^2 + \sum_{j=1}^{i-1} \mu_{ji}^2 \norm{\bm b_j^*}_2^2\\ &\leq \norm{\bm b_i^*}_2^2 + \frac 14 \sum_{j=1}^{i-1} \norm{\bm b_j^*}_2^2 & (|\mu_{ji}|\leq 1/2)\\ &\leq \norm{\bm b_i^*}_2^2 + \frac 14 \sum_{j=1}^{i-1} 2^{i-j}\norm{\bm b_i^*}_2^2 & (\text{Lovász condition})\\ &\leq 2^{i-1} \norm{\bm b_i^*}_2^2. \end{align*}\]

Note that

\[ \lambda_1(\pi_{U_i}(\Lambda))^2 \geq \min\{\norm{\bm b_i^*}_2^2,\dots, \norm{\bm b_n^*}_2^2\} \geq 2^{i-n} \norm{\bm b_i^*}_2^2, \]

we have

\[\norm{\bm b_i}_2^2 \leq 2^{n-1} \lambda_1(\pi_{U_i}(\Lambda))^2\leq 2^{n-1}\lambda_i(\Lambda), \]

thus \(\norm{\bm b_i}_2 \leq 2^{(n-1)/2}\lambda_i(\Lambda)\). \(\square\)

1.13 Let \(\Lambda\subseteq \bbR^n\) be a full rank lattice. Then for any \(k\)-dimensional sublattice \(\tilde \Lambda \subseteq \Lambda\) one has \(\det(\tilde \Lambda) \geq \left(\frac{\lambda_1(\Lambda)}{\sqrt k}\right)^k\).

Proof. Note that a sublattice can't have shortest vector shorter, we have

\[ \lambda_1(\Lambda) \leq \lambda_1(\tilde \Lambda) \stackrel{\text{Mink.~1st Thm}}{\leq} \sqrt k \cdot \det(\Lambda)^{1/k}, \]

then the inequality follows from rearrangement. \(\square\)