Past exam of the mathematics course of the University of Cambridge / 2010 / iii / Paper 11 / 2 / i / Solution

For disjoint nonempty vertex sets $U,W$, define their edge density of a bipartite graph by $d(U,W)=e(U,W)/(|U||W|)$. They form an $\epsilon$-regular pair of vertex sets if every $X\subseteq U$, $Y\subseteq W$ with $|X|\ge \epsilon |U|$, $|Y|\ge \epsilon |W|$ satisfies $|d(X,Y)-d(U,W)|\le \epsilon$.

The Szemerédi regularity lemma asserts that, given $\epsilon >0$ and a positive integer $m_0$, there are $M,n_0$ such that every graph on $n\ge n_0$ vertices has a set partition $V_0,V_1,\dots ,V_k$ withand at most $\epsilon k^2$ irregular ordered pairs $(V_i,V_j)$ with $1\le i\ne j\le k$.

Assume $0<\epsilon \le 1/2$, which suffices since a smaller regularity parameter gives the claim for a larger one. Treat the exceptional vertices as singleton cells in the full partition $\mathcal{P}$, and use equitable regularity energyHere the adjacency indicator function defines density even for diagonal cells. Clearly $0\le q\le 1$. If ${\mathcal{P}}^{\prime }$ refines $\mathcal{P}$, weighted averaging of densities gives the refinement variance identity for regularity energyTo verify it, expand each square and use ${\sum }_{A^{\prime },B^{\prime }}|A^{\prime }||B^{\prime }|d(A^{\prime },B^{\prime })=|A||B|d(A,B)$.

For every irregular ordered pair of nonexceptional cells, choose witnesses $X_{ij},Y_{ij}$ of relative sizes at least $\epsilon$ and density discrepancy greater than $\epsilon$. Split each $V_i$ along all witnesses involving it; symmetry allows at most one witness for each other cell, hence at most $2^k$ atoms in each $V_i$. In the witness rectangle, the weighted Cauchy-Schwarz inequality in the variance identity contributes more than ${\epsilon }^4|V_i||V_j|/n^2$.

If more than $\epsilon k^2$ pairs are irregular and the equal cell size is $L$, the energy increase is greater thanRestore equal cell sizes by cutting each witness atom into pieces of size $\ell =\lfloor L/4^k\rfloor$, with leftover vertices added to $V_0$ as singleton cells. This operation is a further refinement, so the energy does not decrease. There are at most $k{2}^k$ atoms, and fewer than $\ell$ leftover vertices in each, giving at most $k{2}^k\ell \le n{2}^{-k}$ new exceptional vertices.

Here is a uniform bound ensuring that the iteration is legitimate. Put $R=\lceil 4{\epsilon }^{-5}\rceil$ and chooseStart with $k_0$ equal classes and fewer than $k_0$ exceptional vertices. Taking $n\ge 2k_0/\epsilon$ makes the initial exception at most $\epsilon n/2$. At every refinement the number of nonexceptional cells does not decrease: each old cell loses at most $L{2}^{-k}\le L/2$ vertices and is cut into pieces of size at most $L/4^k$. Thus each of the at most $R$ rounds adds at most $\epsilon n/(2R)$ exceptional vertices.

When $L\ge 2\cdot 4^k$, the new cell count is at mostusing $|V_0|\le \epsilon n\le n/2$. Define $K_0=k_0$ and $K_{j+1}=4K_j{4}^{K_j}$, and take $M=K_R$. Choosing $n_0\ge \max \{2k_0/\epsilon ,4M{4}^M\}$ ensures $L\ge n/(2M)\ge 2\cdot 4^M$ at every stage. All rounds are therefore defined, with cell count between $m_0$ and $M$ and exceptional size at most $\epsilon n$.

If the lemma still failed after $R$ refinements, the energy would have increased by more than $R{\epsilon }^5/4\ge 1$, impossible because it lies in $[0,1]$. Hence some stage has the required regular pairs. This completes the energy-increment proof of Szemerédi regularity, including the equal-size and exceptional-set conditions.