Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 147 (Introduction to Discrete Analysis)

For $x_1,x_2\in X$, let $d(x_1,x_2)$ be the number of vertices of $Y$ adjacent to both. The required number $Q$ of ordered quadruples isThe Cauchy-Schwarz inequality first givesReverse the order of counting. A vertex $y\in Y$ contributes $\mathrm{deg}(y{)}^2$ ordered pairs, so a second application of Cauchy-Schwarz and the assumed edge density of a bipartite graph giveTherefore the bipartite four-cycle count is

Write $1_A$ for the indicator function and use the normalized Fourier analysis on a finite abelian groupThere are $N^5$ sextuples satisfying the equation, since any five coordinates determine the sixth. The desired probability is consequentlyBy orthogonality of complex exponentials, the indicator of the equation isSubstitution makes all six sums independent and gives the sixth Fourier moment as a three-sum collision count:Because $1_A$ is real, $\hat{A}(-r)=\overline{\hat{A}(r)}$. Hence the probability is

We prove the polynomial nonvanishing below the field size by mathematical induction on $n$. The case $n=1$ is the Lagrange root bound over a field: a nonzero polynomial of degree below $p$ cannot have all $p$ elements of ${\mathbb{F}}_p$ as roots.

For the induction step, suppose that $P$ vanishes on all of ${\mathbb{F}}_p^n$ and writeThe upper limit is valid because the total degree of a polynomial is less than $p$. Fixing the first $n-1$ variables produces a univariate polynomial of degree below $p$ which vanishes at every $x_n\in {\mathbb{F}}_p$. It is therefore the zero polynomial, so every $P_j$ vanishes on all of ${\mathbb{F}}_p^{n-1}$. Each $P_j$ also has total degree below $p$, and the induction hypothesis gives $P_j=0$ for every $j$. Thus $P=0$.

Taking the contrapositive, every nonzero such $P$ has some $a\in {\mathbb{F}}_p^n$ for which

Remove the quadratic phase by settingThe phase around the parallelogram isso the hypothesis says exactly that $\Vert g{\Vert }_{U^2}^4\ge c$, with harmless negative choices of the increments. The Fourier identity for the Gowers U2 norm and the Parseval identity now givewhere the last inequality uses $\Vert f{\Vert }_{\infty }\le 1$. Thus the Quadratic phase detection by the Gowers U2 norm supplies an $r\in {\mathbb{Z}}_N$ such that

Let $I=[N]$ and let $f=1_A-\delta 1_I$ be the balanced indicator function, extended by zero to the integers. For finitely supported functions define the linear configuration countSince $A$ has no nonconstant three-term arithmetic progression, $\Lambda (1_A,1_A,1_A)=|A|=\delta N$. On the other hand, $\Lambda (1_I,1_I,1_I)\ge N^2/4$. Hence, once $N$ is sufficiently large in terms of $\delta$,for an absolute $c_0>0$.

Use the unnormalized Fourier transform $\hat{h}(\theta )={\sum }_{x}h(x)e(-\theta x)$ on $\mathbb{R}/\mathbb{Z}$. Telescoping $1_A=\delta 1_I+f$ writes the preceding difference asLet $M=\Vert \hat{f}{\Vert }_{\infty }$. The Fourier-integral formula for $\Lambda$, followed by the Cauchy-Schwarz inequality and the Parseval identity, bounds these three terms respectively byThey are all at most $M\delta N$, sofor an absolute $c_1>0$. Choose $\theta$ with $|\hat{f}(\theta )|=M$.

By the Dirichlet approximation theorem, some integer $1\le d\le \sqrt{N}$ satisfies $\Vert d\theta {\Vert }_{\mathbb{R}/\mathbb{Z}}\le N^{-1/2}$. Choose a sufficiently small absolute multiple $\eta$ of ${\delta }^2$. Partition each residue-class progression of common difference $d$ into progressions $P_j$ whose lengths lie between $\eta \sqrt{N}$ and $2\eta \sqrt{N}$; for sufficiently large $N$, short final pieces can be joined to the preceding piece. On each $P_j$, the linear phase $e(-\theta x)$ varies by at most $4\pi \eta$. Choosing $\eta$ small enough compared with $c_1{\delta }^2$ therefore yieldsThe sums over all cells add to ${\sum }_{I}f=0$, so their positive parts total half their absolute values. At least one $P_j$ consequently satisfiesThis is the Roth density-increment step, and it giveswith $c=c_1/4$ and $\eta >0$ depending only on $\delta$.

For nonempty $A\subseteq X$ and $B\subseteq Y$, their edge density of a bipartite graph isThe pair $(A,B)$ is $\epsilon$-regular when every $A^{\prime }\subseteq A$ and $B^{\prime }\subseteq B$ with $|A^{\prime }|\ge \epsilon |A|$ and $|B^{\prime }|\ge \epsilon |B|$ satisfyFor set partitions $X=X_1\bigsqcup \cdots \bigsqcup X_r$ and $Y=Y_1\bigsqcup \cdots \bigsqcup Y_s$, the regular pair of bipartite partitions condition is

Give $X\times Y$ the uniform probability measure and let $f$ be the indicator function of the edge set. For partitions $\mathcal{P}=\{X_i\}$ and $\mathcal{Q}=\{Y_j\}$, define the energyBecause conditional expectation is an orthogonal projection in L2 space, $0\le \mathcal{E}\le \Vert f{\Vert }_2^2\le 1$.

Suppose the pair of partitions is not $\epsilon$-regular. For each irregular $(X_i,Y_j)$ choose witnesses $A_{ij}\subseteq X_i$ and $B_{ij}\subseteq Y_j$ withRefine each $X_i$ by all sets $A_{ij}$ and each $Y_j$ by all sets $B_{ij}$. If the old partitions have $r,s$ cells, the refined ones have at most $r{2}^s,s{2}^r$ cells.

The Pythagorean theorem for the two nested conditional-expectation projections givesInside an irregular $X_i\times Y_j$, the witness rectangle occupies at least an ${\epsilon }^2$ proportion and the mean of the displayed difference over it has magnitude greater than $\epsilon$. The Cauchy-Schwarz inequality therefore gives an energy gain greater thanfrom that pair. Since irregular pairs have total weight greater than $\epsilon$, the complete refinement raises the energy by more than ${\epsilon }^5$.

Energy is at most one, so after at most $\lceil {\epsilon }^{-5}\rceil$ refinements the process stops at an $\epsilon$-regular pair of partitions. Iterating the cell-count bounds $r\mapsto r{2}^s$ and $s\mapsto s{2}^r$ from $r=s=1$ a bounded number of times produces a finite $K(\epsilon )$ independent of $G$. This proves the Bipartite Szemerédi regularity lemma.

Assume first that $A\subseteq {\mathbb{F}}_3^n$ is a cap set. Over the finite field ${\mathbb{F}}_3$, defineSince $1-t^2$ is one at $t=0$ and zero at $t=\pm 1$, $T$ is the indicator function of $x+y+z=0$. If $x,y,z\in A$ satisfy this equation, then either they are all equal or they are three distinct points; the latter is excluded. Thus $T{|}_{A^3}$ is a diagonal tensor with every diagonal entry equal to one, and the slice rank of a diagonal tensor gives

Expand $T$ as a polynomial. Every variable has exponent at most two, and every monomial has total degree at most $2n$. Splitting a monomial's degree among its $x$-, $y$-, and $z$-blocks, at least one block has degree at most $2n/3$. Assign each monomial to one such block and group together terms with the same low-degree block monomial. Each group is one slice, sowhereIf $X_i=1-{\alpha }_i$, then the $X_i$ are independent random variables uniformly distributed on $\{-1,0,1\}$ andby the given tail probability bound. Consequently every cap set satisfies

Put $C_0=3e^{-1/24}<3$. For $n\ge 27$, the preceding bound is at most $C_0^n$. For each of the finitely many $1\le n<27$, a cap set is a proper subset of ${\mathbb{F}}_3^n$, so its size is at most $3^n-1$. We may therefore chooseEvery cap set then has size strictly below $C^n$. Equivalently,