《图论与加性组合》第七章习题选做
Chapter 7 Structure of Set Addition
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7.1 Sets of Small Doubling: Freiman’s Theorem
Exercise 7.1.2 (Sumsets in abelian groups). Show that if \(A\) is a finite subset of an abelian group, then \(|A+A|\geq |A|\), with equality if and only if \(A\) is the coset of some subgroup.
Proof. Let \(A = \{a_1,\dots,a_n\}\), then \(a_1 + a_i \in A+A\) are distinct, thus \(|A+A|\geq |A|\). Note that \((A+A)-2a_1 = (A-a_1)+(A-a_1)\), we only need to consider the case \(a_1 = 0\), and show when the equality is attained, \(A\) should be a subgroup.
When the equality is attained, we have \(|A+A| = |A|\), and since \(a_1=0\), we have \(A+A\supset A\), thus \(A+A=A\). Since \(A\) is closed under addition, it is a subgroup. \(\square\)
7.2 Sumset Calculus I: Ruzsa Triangle Inequality
Exercise 7.2.4 (Iterated sumsets). Let \(A\) be a finite subset of an abelian group satisfying
Prove that
$|mA-mA| \leq K^{m-1}|A|\quad$ for every integer $m\geq 2$.
Proof. Suppose the inequality holds for \(m\). Let \(B = 2A-A\), \(C = mA - (m-1)A\), then we have \(A-B = 2A-2A\), \(A-C = mA-mA\) and \(B-C = (m+1)A - (m+1)A\). By Ruzsa's triangle inequality, we have
\[ \begin{align*} |A||B-C| &\leq |A-B||B-C|,\\ |A||(m+1)A-(m+1)A| &\leq |2A-2A| |mA-mA|\\ &\leq K|A| \cdot K^{m-1}|A|,\\ |(m+1)A-(m+1)A| &\leq K^m |A|. \end{align*} \]By induction, the inequality holds for all \(m\geq 2\). \(\square\)
7.3 Sumset Calculus II: Plünnecke’s Inequality
Exercise 7.3.9 (Sumset vs. difference set). Let \(A\subseteq \mathbb Z\). Prove that
Proof. Plug in \(B=C=-A\) in Corollary 7.3.6, we have
\[|A| |A+A| \leq |A-A|^2. \]Since \(|A-A|\leq |A|^2\), we have \(|A|^{-1} \leq |A-A|^{-1/2}\) we have
\[|A+A| \leq \frac{|A-A|^2}{|A|} \leq |A-A|^{3/2}, \]this proves the right hand side.
Plug in \(B=C=-A\) in Ruzsa's triangle inequality, we have\[|A| |A-A| \leq |A+A|^2. \]Similarly, since \(|A+A|\leq |A|^2\), we have
\[|A-A| \leq \frac{|A+A|^2}{|A|} \leq |A+A|^{3/2}, \]this proves the left hand side. \(\square\)
7.4 Covering Lemma
Exercise 7.4.3 (Chang’s covering lemma). Let \(A\) and \(B\) be finite sets in an abelian group satisfying
$|A+A|\leq K|A| \quad$ and $\quad |A+B| \leq K'|B|$.
Show that there exists some set $X$ in the abelian group so that$A\subseteq \Sigma X + B - B \quad$ and $\quad |X| = O(K\log(KK'))$.
where $\Sigma X$ denotes the set of all elements that can be written as the sum of a subset of elements of $X$ (including zero as the sum of the empty set).Proof. Let \(B_0 = B\) and \(B_t\) determined by the following: Try to find \(2K\) elements in \(A\), such that all \(a+B_{t-1}\) are disjoint, if cannot, terminate the process, otherwise let such \(2K\) elements be \(A_t\), and set \(B_t = B_{t-1} + A_t\).
Since each \(a+B_{t-1}\) are disjoint, we have \(|B_t| = |A_t| |B_{t-1}| = 2K |B_{t-1}|\), by induction we have \(|B_t| = (2K)^t |B|\). On another hand, we have \(B_t = B + A_1 + A_2 + \cdots + A_t \subseteq B+tA\). By Corollary 7.3.6, we have\[|A| |B+tA| \leq |A + B| |(t+1)A| \leq K'|B| |(t+1)A|, \]since \(|A+A|\leq K|A|\), by Plünnecke's Inequality we have \(|(t+1)A|\leq K^{t+1}|A|\), thus
\[|B+tA| \leq K' K^{t+1}|B|. \]Combine with \(|B+tA| \geq (2K)^t |B|\), we have \(2^t \leq KK'\), which is violated at \(t > \log_2 (KK')\), so the process must terminate at \(t \leq \log_2 (KK')\).
When the process terminates, we let \(A_{t+1}\) be a maximal set \(\subseteq A\) such that \(a+B_t\) are disjoint, we have \(|A_{t+1}| < 2K\). Now for any \(a\in A\), we have \(a+B_{t}\) intersects with \(A_{t+1}+B_t\), thus\[A \subseteq A_{t+1} + B_t - B_t = A_{t+1} + \sum_{i=1}^t (A_i-A_i) + B-B. \]Thus we can take
\[X = A_{t+1} \cup \bigcup_{i=1}^t (A_i \cup -A_i), \]we have \(A \subseteq \Sigma X + B - B\), and \(|X| \leq (2t+1) \cdot 2K \leq 2K(1 + 2\log(KK'))\). \(\square\)
7.5 Freiman’s Theorem in Groups with Bounded Exponent
Exercise 7.5.7. Show that for every real \(K\geq 1\) there is some \(C_K\) such that for every finite set \(A\) of an abelian group with \(|A+A|\leq K|A|\), one has \(|nA|\leq n^{C_K}|A|\) for every positive integer \(n\).
Proof. By Plünnecke's inequality, we have \(|A+(2A-A)| = |3A-A| \leq K^4 |A|\). By the Ruzsa covering lemma, we have a subset \(T\subseteq 2A-A\), such that \(|T|\leq |A+(2A-A)|/|A| \leq K^4\), and
\[2A-A\subseteq T+A-A. \]Adding \(A\) to both sides, we have
\[3A-A\subseteq T+2A-A \subseteq 2T+A-A. \]Then by induction, we have
\[(n+1)A-A \subseteq T+nA-A \subseteq nT + A-A, \]for all \(n\geq 1\). Then for all \(n\geq 1\), we have
\[nA \subseteq nA-A \subseteq (n-1)T+A-A. \]Therefore we have
\[|nA| \leq |(n-1)T| |A-A| \leq |(n-1)T| \cdot K^2|A|. \]Note that \(|nT| \leq \binom{n+|T|-1}{|T|-1} = O(n^{|T|-1})\), we have
\[|nA| \leq O\left(n^{\lfloor K^4 \rfloor -1}\right) \cdot |A|, \]where such \(O(\cdot)\) only depends on \(K\). \(\square\)
7.7 Modeling Lemma
Exercise 7.7.4 (Modeling arbitrary sets of integers). Let \(A\subseteq \bbZ\) with \(|A| = n\).
(a) Let \(p\) be a prime. Show that there is some integer \(t\) relatively prime to \(p\) such that \(\norm{at/p}_{\bbR/\bbZ}\leq p^{-1/n}\) for all \(a\in A\).
Proof. Let \(A = \{a_1,\dots,a_n\}\). For each \(x \in [p]\), let \(P_x\in (\bbR/\bbZ)^n\) defined by
\[P_x = (a_1x/p, a_2x/p, \dots, a_n x/p). \]For any \(r = p^{-1/n} + \epsilon\), consider the cube \(P_x + [0,r)^{n}\), each of the cube has volume \(r^n\), since \(pr^n > 1\), by pigeonhole principle, we must have some pair \(x, y\) such that their cube intersects. Thus take \(t=x-y\), we have \(\norm{a_i t/p}_{\bbR/\bbZ} < r\), since we only have finite choices of \(t\), take \(\epsilon\to 0\), we have a \(p \nmid t\) such that \(\norm{at/p}_{\bbR/\bbZ}\leq p^{-1/n}\) for all \(a\in A\). \(\square\)
(b) Show that \(A\) is Freiman \(2\)-isomorphic to a subset of \([N]\) for some \(N = (4 + o(1))^n\).
Proof. Let \(M = \max\{2A-2A\}\), we show that if \(M\) is sufficiently large, we can always find a Freiman \(2\)-isomorphic set \(A' \subseteq \bbZ\) such that \(M' = \max\{2A'-2A'\} < M\).
Choose the smallest prime \(p > M\), consider the composition of functions as follows\[\phi \colon \bbZ \xrightarrow{\bmod p} \bbZ/p\bbZ \xrightarrow{\cdot t} \bbZ/p\bbZ \xrightarrow{{\pmod p}^{-1}} \left\{ -\frac{p-1}2 , \dots, \frac{p-1}2 \right\}, \]where \(t\) is chosen to let each \(\norm{at/p}_{\bbR/\bbZ} \leq p^{-1/n}\). Since \(p > \max\{2A-2A\}\), the first composition is Freiman \(2\)-isomorphism. Since \(p\nmid t\), the dilation gives an isomorphism, thus a Freiman \(2\)-isomorphism.
By our choice of \(t\), the image \(A'=\phi(A)\) lies in \((-p^{(n-1)/n}, p^{(n-1)/n})\). When \(M>4^n\), we have \(\max\{2A'-2A'\}<4p^{(n-1)/n} < p\), it's not hard to verify that the whole \(\phi\) gives a Freiman \(2\)-isomorphism.
Since \(p = (1+o(1))M\), we have \(4p^{(n-1)/n}<M\) always holds for some bound \(M \geq (4+o(1))^n\). At last, we can tranlate the set to a subset of \([M]\). \(\square\)
Exercise 7.7.5 (Sumset with 3-AP-free set). Let \(A\) and \(B\) be \(n\)-element subsets of the integers. Suppose \(A\) is \(3\)-AP free. Prove that \(|A + B| \geq n(\log \log n)^{1/100}\) provided that \(n\) is sufficiently large.
Proof. Let \(\delta\) be a constant to be determined, suppose we have \(|A+A|\leq (\log \log n)^{\delta} |A|\), by Plünnecke's inequality, we have
\[|2A-2A| \leq (\log\log n)^{2\delta} |A|, \]by Ruzsa modeling lemma, we have a subset \(A'\subseteq A\) with \(|A'|\geq n/2\), such that \(A'\) is Freiman \(2\)-isomorphic to a subset of \(\bbZ/N\bbZ\), where \(N = n(\log\log n)^{2\delta}\). Note that \(A\) is \(3\)-AP free, i.e., the solutions of \(y+y=x+z\) can only be \(x=y=z\), such property is preserved by Freiman \(2\)-isomorphism. Thus by Roth's theorem, \(n/2 \leq O(N/\log \log N) =O(n(\log\log n)^{2\delta-1})\), which is only possible when \(\delta \geq 1/2\), and we can further conclude that \(|A+A| = \Omega(n(\log\log n)^{1/2})\).
By Corollary 7.3.6, we have\[|A+B| \geq \sqrt{n |A+A|} = \Omega(n(\log\log n)^{1/4}). \square \]
Exercise 7.7.6 (3-AP-free subsets of arbitrary sets of integers). Prove that there is some constant \(C>0\) so that every set of \(n\) integers has a 3-AP-free subset of size \(\geq n e^{-C\sqrt{\log n}}\).
Proof. First choose a prime \(q > \max\{A-A\}\), consider the composition of functions
\[\phi \colon \bbZ \xrightarrow{\bmod q} \bbZ/q\bbZ \xrightarrow{\cdot \lambda}\bbZ/q\bbZ \xrightarrow{{\pmod q}^{-1}} \{0,\dots,q-1\}, \]where \(\lambda \in \{1,\dots,q-1\}\) is randomly chosen.
For each pair \(\{x, y\} \subseteq A\), we have \(\lambda(x-y) \bmod q\) is uniform over \(\{1,\dots,q-1\}\), thus, take \(N=2n\), consider the probability that \(\phi(x) \equiv \phi(y) \pmod N\), this is not greater than \(2/N=1/n\), the expected number of collisions of \(\phi(A) \bmod N\) is \(\leq n(n-1)/2 \cdot 1/n \leq n/2\), for each collision we remove one of the number, thus obtain a subset \(A'\subseteq A\), with \(\bbE[|A'|] \geq n/2\).
Therefore, by pigeonhole principle, we have a subset \(A''\subseteq A'\) such that \(\phi(A'')\) has diameter \(< q/2\), with \(|A''|\geq |A'|/2\). In conclusion, there exists a \(\lambda\) such that \(|A''|\geq n/4\), and one can verify that \(x\mapsto \phi(x)\bmod N\) is an injective Freiman \(2\)-homomorphism over \(A''\).
By Behrend's construction, there is a \(3\)-AP free subset \(S \subseteq [N]\) with size \(N/e^{C\sqrt{\log N}}\), then consider a random translation \((\phi(A'')\bmod N) \cap (S + r)\), it has expected size \(|A''| / e^{C\sqrt{\log N}}\), so we have a subset \(A^{(3)} \subseteq A''\) such that \(\phi(A^{(3)})\bmod N\) is \(3\)-AP free, with \(|A^{(3)}| \geq n/4e^{C\sqrt{\log N}}\). Therefore, we have \(A^{(3)}\) itself is \(3\)-AP free, and\[|A^{(3)}| \geq \Omega\left(n / e^{C\sqrt{\log n}}\right). \square \]
7.8 Iterated Sumsets: Боголю́бов's Lemma
Exercise 7.8.7 (Боголю́бов with 3-fold sums). Let \(A\subseteq \bbF_p^n\) with \(|A|=\alpha p^n\). Prove that \(A+A+A\) contains a translate of a subspace of codimension \(O(\alpha^{-3})\).
Proof. First let
\[R = \left\{ r \in \widehat{\bbF_p^n} \smallsetminus \{0\} : \left|\widehat{1_A}(r) \right| > \alpha^2 \right\}, \]we have
\[|R| \alpha^4 \leq \sum_{r \in R} \left|\widehat{1_A}(r) \right|^2 < \alpha, \]thus we have \(|R| < 1/\alpha^3\). Then consider a uniform random \(x\in \bbF_p^n\) and the summation
\[\sum_r \widehat{1_A}(r)^3 \omega^{r(x)}, \]we have
\[ \bbE_x\left[\sum_r \widehat{1_A}(r)^3 \omega^{r(x)}\right] = \sum_r \widehat{1_A}(r)^3 \bbE_x[\omega^{r(x)}] = 0. \]Therefore, we have a specific \(x_0\) such that
\[ \Re \left[\sum_r \widehat{1_A}(r)^3 \omega^{r(x_0)} \right] \geq 0. \]Then we consider the rest part, we have
\[ \begin{align*} \sum_{r\notin R\cup \{0\}} \Re \left[\widehat{1_A}(r)^3 \omega^{r(x)}\right] &\leq \sum_{r\notin R\cup \{0\}} \left|\widehat{1_A}(r)\right|^2 \\ &\leq \left(\max_{r\notin R\cup\{0\}} \left|\widehat{1_A}(r)\right| \right) \sum_{r\notin R\cup\{0\}} \left|\widehat{1_A}(r)\right|^2\\ &< \alpha^2\cdot \alpha = \alpha^3. \end{align*}\]In conclusion, for \(x\) such that \(r(x) = 0\) for all \(r\in R\), we have
\[\begin{align*} (1_A*1_A*1_A)(x+x_0) &=\sum_r \Re \left[ \widehat{1_A}(r)^3 \omega^{r(x+x_0)} \right]\\ &= \alpha^3 + \sum_{r\in R} \Re \left[ \widehat{1_A}(r)^3 \omega^{r(x_0)} \right] - \sum_{r\notin R \cup \{0\}} \Re \left[ \widehat{1_A}(r)^3 \omega^{r(x+x_0)} \right]\\ &> \alpha^3 + 0 - \alpha^3 = 0, \end{align*}\]thus \(R^\perp + x_0 \subseteq A+A+A\), with codimension \(<1/\alpha^3\). \(\square\)
Exercise 7.8.8 (Боголю́бов with better bounds). Let \(A\subseteq \bbF_p^n\) with \(|A| = \alpha p^n\).
(a) Show that if \(|A+A| < 0.99\cdot p^n\), then there is some \(r\in \bbF_p^n \smallsetminus \{0\}\) such that \(|\widehat{1_A}(r)| > c \alpha^{3/2}\) for some absolute constant \(c > 0\).
Proof. Suppose \(|\widehat{1_A}(r)| \leq c \alpha^{3/2}\) for all \(r\neq 0\). Consider the random variable \(X = 1_A*1_A(x)\), where \(x\in \bbF_p^n\) is uniformly random, we have
\[\bbE[X] = \widehat{1_A}^2(0) = \alpha^2, \]furthermore, we have
\[\begin{align*} \bbE[X^2] &= \bbE [X \overline{X}]\\ &= \bbE \left[ \sum_{r\in \widehat{\bbF_p^n}} \widehat{1_A}(r)^2 \omega^{r(x)} \overline{\sum_{r'\in \widehat{\bbF_p^n}} \widehat{1_A}(r')^2 \omega^{r'(x)}} \right]\\ &= \sum_{r,r'\in \widehat{\bbF_p^n}} \widehat{1_A}(r)^2 \overline{\widehat{1_A}(r')^2} \bbE[\omega^{(r-r')(x)}], \end{align*}\]by orthonormality, we have
\[\begin{align*} &= \sum_{r\in \widehat{\bbF_p^n}} \left|\widehat{1_A}(r)\right|^4\\ &\leq \alpha^4 + \max_{r\neq 0} \left|\widehat{1_A}(r)\right|^2 \cdot \sum_{r\neq 0} \left|\widehat{1_A}(r)\right|^2\\ &\leq (1+c^2)\alpha^4. \end{align*}\]Therefore, we have \(\Var[X] \leq c^2 \alpha^4\), Чебышёв's inequality gives
\[\Pr[X = 0] \leq \frac{\Var[X]}{\bbE[X]^2} \leq c^2, \]therefore we have \(|A+A| \geq (1-c^2) p^n\). Setting \(c = 0.1\), we have \(|A+A| \leq 0.99 \cdot p^n\). \(\square\)
(b) By iterating (a), show that \(A+A\) contains \(99\%\) of a subspace of codimension \(O(\alpha^{-1/2})\).
Proof. Here we mimic the density increment argument of Lemma 6.2.7. Suppose \(|\widehat{1_A}(r)|\geq \delta\) for some \(r\neq 0\), consider
\[\widehat{1_A}(r) = \bbE [1_A(x)\omega^{-r(x)}] = \frac 1 p \sum_{i \in \bbF_p} \alpha_i \omega^{i}, \]where \(\alpha_i\) are the density on the cosets of \(r^\perp\). By triangle inequality,
\[\begin{align*} p \delta &\leq \left| \sum_{i} \alpha_i \omega^i \right|\\ &= \left| \sum_{i} (\alpha_i - \alpha) \omega^i \right|\\ &\leq \sum_{i} |\alpha_i - \alpha|\\ &= \sum_{i} (\alpha_i - \alpha + |\alpha_i - \alpha|), \end{align*}\]thus we have some \(\alpha_i - \alpha + |\alpha_i - \alpha| \geq \delta\), thus \(\alpha_i \geq \alpha + \delta/2\).
If \(|A+A| < 0.99 \cdot p^n\), we can have \(\delta = \alpha^{3/2}/10\), thus we have a hyperplane \(r^\perp\) and a translate such that \(|(A-x)\cap r^\perp| \geq \alpha + \alpha^{3/2}/20\).
Let \(A_1 = (A-x)\cap r^\perp\), if \(|A_1 + A_1| < 0.99\cdot p^{n-1}\) still holds, we can repeat the density increament. But after \(40/\alpha^{1/2}\) times, the subset has density \(\geq 4\alpha\), so after\[\sum_{k\geq 0} \frac{40}{(4^k\alpha)^{1/2}} \leq 80 \alpha^{-1/2} \]times, the density exceeds \(1\), which is not possible. Then we must have an \(|A_k+A_k| > 0.99\cdot p^{n-k}\) for some \(k\leq 80 \alpha^{-1/2}\), thus \(A+A\) covers \(99\%\) of a subspace of codimension \(\leq 80 \alpha^{-1/2}\). \(\square\)
(c) Deduce that \(2A-2A\) contains a subspace of codimension \(O(\alpha^{-1/2})\).
We already proved that \(A+A\) covers \(99\%\) of a subspace with codimension \(\leq 80\alpha^{-1/2}\), say, \(A' = W\cap (A-x)\) satisfies \(|2A'| > 0.99 |W|\), then for any \(y\in W\), since \(2A'+y\) and \(2A'\) cannot be disjoint, we have \(2A'-2A' = W\), thus \(W \subseteq 2A-2A\), where \(W\) has codimension \(\leq 80\alpha^{-1/2}\). \(\square\)
7.11 Proof of Freiman's Theorem
Exercise 7.11.2 (Improved bounds on Freiman’s theorem). Using a more efficient covering lemma from Exercise 7.4.3, prove Freiman's theorem with \(d(K) = K^{O(1)}\)
and \(f (K) = \exp(K^{O(1)})\).
Proof. Let \(A\subseteq \bbZ\) be a finite set and \(|A+A| \leq K|A|\).
By Plünnecke's theorem, we have \(|8A-8A|\leq K^{16}|A|\). By Bertrand's postulate, there is a prime number \(K^{16}|A| \leq N \leq 2K^{16}|A|\). By Ruzsa modeling lemma, we have some \(A'\subseteq A\) with \(|A'|\geq |A|/8\) is Freiman \(8\)-isomorphic to a subset \(B \subseteq \bbZ/N\bbZ\).
Apply Боголю́бов's lemma on \(B\subseteq \bbZ/N\bbZ\), with\[\alpha = \frac{|B|}{N} \geq \frac 1{16 K^{16}}, \]we deduce that \(2B-2B\) contains a Bohr set with dimension \(<256K^{32}\) and width \(1/4\). By the geometry of numbers argument (Theorem 7.10.1), \(2B-2B\) contains a proper GAP with dimension \(d<256K^{32}\) and volume \(\geq (4d)^{-d}N\).
Since \(B\) is Freiman \(8\)-isomorphic to \(A'\), \(2B-2B\) is Freiman \(2\)-isomorphic to \(2A'-2A'\). Hence the proper GAP in \(2B-2B\) is mapped to a proper GAP \(Q\subseteq 2A'-2A'\) with same dimension and volume. We have\[|A|\leq 8|A'| \leq 8N \leq 8(4d)^d |Q|. \]Since \(Q\subseteq 2A'-2A'\subseteq 2A-2A\), we have \(Q+A\subseteq 3A-2A\). By Plünnecke's inequality,
\[|Q+A| \leq |3A-2A| \leq K^5 |A| \leq 8K^5(4d)^d |Q|. \]Apply Chang's covering lemma with \(K'=8K^5(4d)^d\), we have a set \(X \subseteq \bbZ\) such that \(A\subseteq \Sigma X + Q - Q\) and \(|X| = O(K\log(KK')) = O(K^{33}\log K)\).
Now \(\Sigma X\) is contained in a GAP of dimension \(|X|\) and volume \(\leq 2^{|X|}\). Since \(Q\) is a proper GAP with dimension \(d = O(K^{32})\) and volume \(\leq K^4|A|\), \(Q-Q\) is a GAP with dimension \(d\) and volume \(2^d K^4|A|\). It follows that \(A\subseteq X+Q-Q\) is contained in a GAP with\[\mathrm{dimension} \leq |X| + d = O(K^{33}\log K), \]and
\[\mathrm{volume} \leq 2^{|X|+d} K^4|A| = \exp (O(K^{33}\log K)) |A|. \square \]
