Past exam of the mathematics course of the University of Cambridge / 2007 / iii / Paper 88 (Asymptotic Structure and Quasirandomness)

Assume $A$ is finite and nonempty; the empty case is immediate. Here $2A=A+A$. Choose a nonempty subset $X\subseteq A$ minimizingTaking $X=A$ as a competitor gives $K\le C$, and every subset $Y\subseteq X$ satisfies $|Y+A|\ge K|Y|$, including $Y=\varnothing$.

We first prove the Petridis minimal-growth lemma in the particular form needed: for every finite set $S$,List $S=\{s_1,\dots ,s_t\}$ and let $X_j$ consist of those $x\in X$ for which $x+s_j$ does not belong to an earlier translate $X+s_h$, $h<j$. The sets $X_j+s_j$ partition $X+S$, so $|X+S|={\sum }_j|X_j|$.

If $x\in X\setminus X_j$, then $x+s_j=x^{\prime }+s_h$ for some earlier $h$ and $x^{\prime }\in X$. Hence every point of $x+A+s_j$ already belongs to $X+A+s_h$. Therefore the new points contributed by $(X+A)+s_j$ are contained inTheir number is at mostSumming over $j$ proves the lemma. With $S=A$, it gives

We also prove the Ruzsa triangle inequality directly. For each $d\in U-V$, choose one representation $d=u_d-v_d$. The mapfrom $(U-V)\times X$ into $(X+U)\times (X+V)$ is injective: subtracting the output coordinates recovers $d$, after which either coordinate recovers $x$. Thus $|U-V||X|\le |X+U||X+V|$. Taking $U=V=2A$ now yieldsThis proves the required growth and difference set estimates directly.

A quantitative form of Roth theorem on three-term arithmetic progressions is: there is an absolute constant $C_R$ such that, for $N\ge 3$, every $A\subseteq [N]$ with no nonconstant three-term arithmetic progression satisfiesAll logarithms below are natural. We prove the classical Roth bound for three-term progressions with an unoptimized $C_R=100$. The empty set is immediate; write $\delta =|A|/N>0$.

First establish a quantitative density increment. Suppose $N\ge 4\cdot 10^6{\delta }^{-4}$. Put $M=2N+1$, embed $I=[N]$ in $G={\mathbb{Z}}_M$, and extend all interval functions by zero. Let $a=1_A$, $u=\delta 1_I$ and $f=a-u$, the balanced subset indicator. DefineThe relation in the sum is modulo $M$. For interval elements its integer discrepancy has absolute value less than $M$, so it is precisely the integer relation. Thus $\Lambda (a,a,a)=|A|=\delta N$. The full interval count is the number of endpoint pairs of the same parity:Consequently, since $N\ge 4{\delta }^{-2}$,

Orthogonality of the additive characters givesBoth identities follow by expanding the sums and using ${\sum }_r{e}^{2\pi irt/M}=M$ for $t=0$ in $G$, and zero otherwise; the latter follows from a finite geometric series. Since $M$ is odd, $r\mapsto -2r$ permutes the frequencies. Telescope the difference asThe Cauchy-Schwarz inequality and the displayed Parseval identity on a finite group bound each term by $\delta N{\max }_r|\hat{f}(r)|$. Indeed, $\Vert a{\Vert }_2^2=\delta N$ and $\Vert u{\Vert }_2^2={\delta }^{2}N$, so every product of the two remaining L2 norms is at most $\delta N$. HenceThe zero coefficient vanishes because $\sum f=0$, so choose a nonzero frequency $\theta =r/M$ giving this bound.

Let $Q=\lfloor \sqrt{N}\rfloor$. The pigeonhole principle applied to the $Q+1$ fractional parts $0,\theta ,\dots ,Q\theta$ in $Q$ intervals gives an integer $1\le q\le Q$ with $\Vert q\theta {\Vert }_{\mathbb{R}/\mathbb{Z}}\le 1/Q$. Partition each residue class modulo $q$ in $[N]$ into consecutive arithmetic progressions of exactlyelements, leaving fewer than $L$ remainder points per residue. The total remainder $R_0$ has size at most $qL\le {\delta }^{2}N/1000$. The size hypothesis gives $L\ge {\delta }^2\sqrt{N}/2000$. On a full arithmetic progression $P$, the phase $e^{-2\pi i\theta x}$ varies from its value at the first point by at most $2\pi L/Q\le 4\pi {\delta }^2/1000$.

Let $F_P={\sum }_{x\in P}f(x)$. Replacing the phase by its first value on each full arithmetic progression and estimating the remainder using $|f|\le 1$ givesAlso $|{\sum }_P{F}_P|=|{\sum }_{R_0}f|\le {\delta }^{2}N/1000$. Therefore the sum of the positive $F_P$ isSince the total length of these arithmetic progressions is at most $N$, at least one has $F_P/|P|\ge {\delta }^2/64$. Thus the step just proved isAn affine map taking this arithmetic progression to $[|P|]$ preserves the absence of nonconstant three-term arithmetic progressions.

Now iterate. Write $N_j,{\delta }_j$ for successive lengths and subset densities, with $N_0=N$ and ${\delta }_0=\delta$. Since every subset density is at most one,Consequently $T=\lceil 65/\delta \rceil$ successful steps are impossible. On the other hand, subset densities never decrease, so with $B=\log (2000/{\delta }^2)$,Suppose $\delta \log \log N>100$. Since $T\le 66/\delta$,Let $D=\log (4\cdot 10^6{\delta }^{-4})$. We have $2B+D<31+8\log (1/\delta )\le 39/\delta <e^{54/\delta }$. Therefore every length through step $T$ satisfies $\log N_j>D$, and hence $N_j\ge 4\cdot 10^6{\delta }^{-4}\ge 4\cdot 10^6{\delta }_j^{-4}$. The density increment lemma remains applicable for all $T$ steps, contradicting the reciprocal-density bound.

Thus $\delta \log \log N\le 100$, proving the boxed theorem. The proof includes the Fourier analysis on a finite abelian group detection, conversion to an integer arithmetic progression, and the length bookkeeping that produces the double logarithm.

A Freiman 2-isomorphism is an injective map onto its image preserving pair-sum equality in both directions, with repeated summands allowed. Put $D=2A-2A$, so $D=-D$, $0\in D$, and $|D|<N$. We prove the half-size cyclic Freiman model with a sharp difference-set bound. Assume $A\ne \varnothing$; for an empty set the conclusion is immediate.

Choose an odd auxiliary prime $P>N$ large enough that reduction modulo $P$ is injective on $A$ and no nonzero element of $D$ reduces to zero. Choose a multiplier $t$ uniformly in ${\mathbb{Z}}_P^{\times }$. The forbidden residues areThey form a symmetric set, exclude zero since $P>N$, and have cardinality at most $2\lfloor (P-1)/N\rfloor$. For each nonzero $d\in D$, the product $td$ is uniform on the $P-1$ nonzero residues. Moreover, $td\in \mathcal{F}$ is the same event as $t(-d)\in \mathcal{F}$, so one need only sum over the $(|D|-1)/2$ opposite pairs. The union bound givesFix a multiplier avoiding all forbidden events. The use of opposite pairs is what achieves the threshold $N>|D|$ without an extra factor of two.

Represent each $ta(\mathrm{mod}P)$ by an integer $r(a)\in \{0,\dots ,P-1\}$. Partition these residues into the two half-intervalsAt least half of $A$ has representatives in one of them; call this subset $A^{\prime }$. Both intervals have diameter less than $P/2$, so for $a_1,a_2,a_3,a_4\in A^{\prime }$ the integersatisfies $|R|<P$.

Define $\varphi (a)=r(a)(\mathrm{mod}N)$. If $a_1+a_2=a_3+a_4$, then $R\equiv 0(\mathrm{mod}P)$, hence $R=0$ and the image relation holds. Conversely, if the image relation holds, then $R=jN$. If the original difference $d=a_1+a_2-a_3-a_4$ were nonzero, $R\equiv td(\mathrm{mod}P)$ would be nonzero; since $|R|<P$, this would be a forbidden residue in $\mathcal{F}$. Thus $d=0$.

Finally $\varphi$ is injective. If $\varphi (a)=\varphi (b)$, choose any $c\in A^{\prime }$ and apply the converse to $\varphi (a)+\varphi (c)=\varphi (b)+\varphi (c)$, obtaining $a+c=b+c$. ThereforeThe construction actually works for any integer $N>|D|$; the prime hypothesis in the question is sufficient.

Fix $C$ and write $m=|A|$. Choose the permitted prime $N$ with $Cm<N<2Cm$. Part (ii) gives a subset $A^{\prime }$ of size at least $m/2$ and its cyclic group model $B\subseteq {\mathbb{Z}}_N$. ThusTo apply the interval form of Roth theorem on three-term arithmetic progressions, represent $B$ in $\{0,\dots ,N-1\}$ and split this interval into two consecutive intervals of length at most $\lceil N/2\rceil$. One has at least $|B|/2\ge m/4$ points. After translation, this is an integer subset $B_0$ of an interval of length at most $Cm+1$ and subset density bounded below in terms of $C$ only. Pad the interval to length $L=\lceil N/2\rceil$ if necessary. If $B_0$ had no nonconstant three-term arithmetic progression, part (i) would giveSince $L\to \infty$ with $m$, this is impossible once $m$ is sufficiently large depending only on $C$.

Thus $B$ contains distinct elements $b_1,b_2,b_3$ with $b_1+b_3=2b_2$ modulo $N$. Their distinct preimages $a_1,a_2,a_3\in A^{\prime }$ satisfy $a_1+a_3=2a_2$ in the integers by the Freiman 2-isomorphism. Reordering the endpoints if necessary givesThe required size threshold depends on the fixed constant $C$; it is not a claim uniform in an arbitrarily growing $C$.

For disjoint nonempty vertex sets $U,V$, write $d(U,V)=e(U,V)/(|U||V|)$. A regular pair of vertex sets is $\epsilon$-regular if every $U^{\prime }\subseteq U,V^{\prime }\subseteq V$ with $|U^{\prime }|\ge \epsilon |U|$, $|V^{\prime }|\ge \epsilon |V|$ satisfies $|d(U^{\prime },V^{\prime })-d(U,V)|\le \epsilon$.

The Szemerédi regularity lemma states that, for every $\epsilon >0$ and integer $m_0$, there are $M,n_0$ such that every finite simple graph on $n\ge n_0$ vertices has a set partitionwith $m_0\le m\le M$, $|V_0|\le \epsilon n$, $|V_1|=\cdots =|V_m|$, and at most $\epsilon m^2$ irregular unordered pairs among the nonexceptional classes. Prove it for $0<\epsilon \le 1/2$; larger parameters follow by using a smaller one.

For the energy-increment proof of Szemerédi regularity, keep every exceptional vertex as a singleton cell in the set partition used to measure regularity energy. Define the equitable regularity energyThe sum is ordered, and on diagonal cells subset density means the average of the adjacency indicator, with zero on loops. Thus $0\le q\le 1$. If $U,V$ are refined into $U_a,V_b$, expanding squares givesThe cross term vanishes because the weighted arithmetic mean of the refined subset densities is the original subset density. Consequently every refinement of a set partition increases or preserves regularity energy.

Suppose the current nonexceptional classes have common size $t$ and there are $k$ of them. For each irregular pair $V_i,V_j$, choose witnesses $A_{ij},B_{ij}$ of sizes at least $\epsilon t$ and discrepancy greater than $\epsilon$. Refine each class by all its incident witness subsets. There are at most $k-1$ cuts in each class and at most $2^k$ resulting atoms. The witness rectangle is a union of refined rectangles. By the Cauchy-Schwarz inequality, its contribution to the variance identity is at leastIndeed, the sum of squared deviations on the witness rectangle is at least its total weight times the square of their weighted arithmetic mean. If more than $\epsilon k^2$ unordered pairs are irregular, summing even just one orientation of each givesprovided the exceptional set has at most $n/2$ vertices. This is the strictly positive regularity energy increment.

Restore equal class sizes without losing that increment. Divide every witness atom into blocks of size $\ell =\lfloor t/4^k\rfloor$ and make every leftover vertex an exceptional singleton. This is a further refinement of a set partition, so it cannot decrease $q$. There are at most $k{2}^k$ atoms, hence fewer than $k{2}^k\ell \le n{2}^{-k}$ newly exceptional vertices. If $t\ge 2\cdot 4^k$, then $\ell \ge t/(2\cdot 4^k)$, so the number of full blocks is at most $2k{4}^k$. Each original class produces at least one full block: its leftovers occupy less than $2^k\ell$, whereas $t\ge 4^k\ell >2^k\ell$. Thus the new class count is at least $k$.

Choose $R_{\ast }=\lceil 4{\epsilon }^{-5}\rceil +1$, and choose an initial integer $k_0\ge m_0$ large enough that $R_{\ast }{2}^{-k_0}\le \epsilon /2$. Initially partition into $k_0$ equal classes and fewer than $k_0$ exceptional vertices. Take $n_0$ large enough that this initial exception is at most $\epsilon n/2$. During at most $R_{\ast }$ rounds, the additional exception is at most $R_{\ast }n{2}^{-k_0}\le \epsilon n/2$, so it never exceeds $\epsilon n\le n/2$.

A bound on class counts is obtained by iterating $k\mapsto 2k{4}^k$ at most $R_{\ast }$ times starting at $k_0$; call the resulting bound $M$. Increasing $n_0$ further to satisfy $n_0\ge 4M{4}^M$ guarantees $t\ge n/(2k)\ge 2\cdot 4^k$ at every round, so all full-block constructions are valid. If the desired regularity had not been reached after $R_{\ast }$ refinements, the regularity energy increments would total more than one, impossible. Therefore the process terminates with the required partition and bounds. The singleton treatment of discarded vertices is what preserves regularity energy monotonicity during equalization.

First derive the triangle removal lemma from the Szemerédi regularity lemma. Given an allowed deletion fraction $0<\eta <1$, take $d=\eta /10$, $m_0=\lceil 10/\eta \rceil$, and $0<\epsilon \le \min (d/4,\eta /100,1/8)$. Apply part (i), with upper class bound $M$. Delete edges incident with $V_0$, inside each nonexceptional class, between irregular pairs, and between regular pairs of subset density less than $d$. Their total number is at mostThese four terms are bounded respectively by $\epsilon n^2$, $n^2/(2m_0)$, $\epsilon n^2$, and $d{n}^2/2$.

If a triangle in a graph remains, its three vertices lie in three distinct equal classes, all three regular pairs with subset density at least $d$. The permitted regular triangle counting lemma gives at least $(d^3/4)t^3$ triangles in a graph in the original graph, where $t$ is the class size. This lower bound can also be checked directly: all but $2\epsilon t$ vertices of the first class have at least $(d-\epsilon )t$ neighbors in each of the other two. Those neighborhoods are large enough to use the regular pair condition on the remaining pair, giving at least $(d-\epsilon {)}^3{t}^2$ edges between them. Hence the total is at least $(1-2\epsilon )(d-\epsilon {)}^3{t}^3\ge (d^3/4)t^3$.

Since $t\ge n/(2M)$, a surviving triangle in a graph forces at least $d^3{n}^3/(32M^3)$ original triangles in a graph. Thus a graph with fewer than $\xi n^3$ triangles in a graph, where $\xi =d^3/(64M^3)>0$, can be made triangle-free by deleting fewer than $\eta n^2$ edges, for all sufficiently large $n$. This proves the removal lemma needed here.

Now use the triangle-removal proof of Roth theorem. Fix a subset density lower bound $\delta >0$ and suppose $A\subseteq [N]$, $|A|\ge \delta N$, has no nonconstant three-term arithmetic progression. Set $m=2N+1$ and $G={\mathbb{Z}}_m$. Use three disjoint copies $X,Y,Z$ of $G$ and put edges according toDivision by two is valid because $m$ is odd. A triangle in a graph corresponds to $a=y-x$, $b=z-y$, $c=(z-x)/2$ with $a,b,c\in A$ and $a+b=2c$. Since $A\subseteq [N]$ and $|a+b-2c|<m$, this congruence is an integer equality. Progression-freeness therefore forces $a=b=c$.

Every triangle in a graph is consequently of the form $(x,x+a,x+2a)$ with $x\in G,a\in A$, so there are exactly $m|A|$ triangles in a graph. These triangles in a graph are edge-disjoint: an $XY$ edge determines $(x,a)$; a $YZ$ edge determines $a$ and then $x$; and an $XZ$ edge determines $a=(z-x)/2$ and $x$. Eliminating them all requires at least $m|A|$ edge deletions.

The graph has $3m$ vertices but only $m|A|\le mN=O(m^2)$ triangles in a graph. Take $\eta =\delta /100$. For large enough $N$ its triangle in a graph count is below $\xi (3m{)}^3$, so removal requires fewer than $\eta (3m{)}^2=9\delta m^2/100$ deletions. Yeta contradiction. Therefore every fixed positive-density subset of a sufficiently long interval contains a nonconstant three-term arithmetic progression.

This proves the qualitative Roth theorem on three-term arithmetic progressions. The quantitative bound from part 2(i) came from the explicit Fourier iteration; the regularity-removal argument gives a density-dependent threshold without that double-logarithmic estimate.

Use octahedral quasirandomness of a three-uniform hypergraph, with its eighth-power normalization. Let $h:X\times Y\times Z\to \{0,1\}$ be the hypergraph edge indicator function, define $p={\mathbb{E}}_{x,y,z}h(x,y,z)$, and put $g=h-p$. All expectations are uniform and independent on the indicated finite sets. The hypergraph is $\alpha$-quasirandom whenThe eight triples form the faces of the tripartite octahedral configuration. Repeated choices within a part are included in this normalized average. For real $g$, the expression is nonnegative, since it equalsThus the condition says that the balanced hypergraph edge indicator function has three-dimensional box norm at most ${\alpha }^{1/8}$. If the norm itself is used as the quasirandomness parameter instead, the eighth-power parameter here is the eighth power of that parameter; stating this convention fixes the quantitative meaning of the counting error.

First prove the pair-factor correlation bound for the three-dimensional box norm. For real $g$ and three functions $a(y,z),b(x,z),c(x,y)$ bounded in absolute value by one, letApply the Cauchy-Schwarz inequality in $(y,z)$, placing the $x$ average inside and removing $a$. This givesExpand this square, with independent $x_0,x_1$. Apply Cauchy-Schwarz in $(x_0,x_1,z)$, removing the product $b(x_0,z)b(x_1,z)$. The result isExpand with independent $y_0,y_1$ and apply Cauchy-Schwarz in $(x_0,x_1,y_0,y_1)$, removing the four $c$ factors. ThenTherefore $|T|\le \Vert g{\Vert }_{{\square }^3}$, with no independence assumption on the three pair functions.

In this multipartite hypergraph, a simplex is a three-uniform tetrahedron: one vertex of a hypergraph from each of $X,Y,Z,W$, with all four possible triples present. Write the corresponding hypergraph edge indicator functions as $h_1(x,y,z),h_2(x,y,w),h_3(x,z,w),h_4(y,z,w)$, of subset densities $p,q,r,s$. Its normalized count isTelescope the product exactly:For the first term fix $w$. The remaining three factors are bounded pair functions of $(x,y)$, $(x,z)$ and $(y,z)$, so the inequality just proved bounds its $(x,y,z)$ average by $\Vert h_1-p{\Vert }_{{\square }^3}\le {\alpha }^{1/8}$. Averaging over $w$ preserves that bound. For each other term fix the vertex of a hypergraph outside its balanced triple and apply the same argument, allowing an unused pair factor to be identically one. The subset density prefactors are at most one. HenceMultiplying by $|X||Y||Z||W|$ givesThus one may take the absolute constant $C=4$. This proves the complete-pair-support case of the tetrahedron counting lemma directly; no lower bound on $p,q,r,s$ is needed.

Yes. The same telescoping argument proves the counting lemma for octahedrally quasirandom three-uniform hypergraphs for every fixed simple three-uniform hypergraph $F$.

Label its vertices of a hypergraph $1,\dots ,v$ and give vertex of a hypergraph $i$ a nonempty host part $V_i$. For each hypergraph edge $e=\{i,j,k\}$, let $h_e$ be the corresponding host hypergraph edge indicator function of subset density $p_e$, with $\Vert h_e-p_e{\Vert }_{{\square }^3}\le {\alpha }^{1/8}$. The normalized labelled hypergraph copy count isIf the host parts are disjoint, every such vertex of a hypergraph choice is automatically injective. Enumerate the $m$ hypergraph edges as $e_1,\dots ,e_m$ and telescopeIn the term for $e_j=\{a,b,c\}$, fix all variables outside this triple. Since $F$ is simple and three-uniform, every other hypergraph edge intersects $\{a,b,c\}$ in at most two vertices of a hypergraph. Its remaining factor therefore depends on at most two of $x_a,x_b,x_c$. Group all two-variable factors according to the pairs $(a,b),(a,c),(b,c)$. Absorb each one-variable factor into any pair containing its variable, and constants into any group. The three resulting pair functions remain bounded by one.

The balanced factor is now multiplied by exactly the sort of pair functions handled in part (ii). Its average has absolute value at most ${\alpha }^{1/8}$, independently of the fixed outside variables. Averaging those variables and summing the $m$ errors givesEquivalently,This counts labelled copies preserving the specified vertex of a hypergraph classes, for arbitrary intersection patterns among the hypergraph edges, and includes the three-uniform tetrahedron as a special case.

For a common host vertex of a hypergraph set of size $n$, the same product average counts homomorphisms. At most $\left(\genfrac{}{}{0ex}{}{v}{2}\right)n^{v-1}$ assignments identify a pair of distinct labelled vertices of a hypergraph, by the union bound. Thus restricting to injective copies changes the estimate by at most this amount. With a common normalized subset density $p$ the injective count is therefore $p^m{n}^v+O(m{\alpha }^{1/8}{n}^v+v^2{n}^{v-1})$. For induced copies, include the nonhypergraph edge indicator functions as additional factors; their balanced functions are negatives of the corresponding edge-balanced functions and have the same box norm. The same proof applies provided all the relevant triple parts satisfy the stated quasirandomness condition.