Daniel Kane and I have just uploaded to the arXiv our paper “A bound on partitioning clusters“, submitted to the Electronic Journal of Combinatorics. In this short and elementary paper, we consider a question that arose from biomathematical applications: given a finite family ${X}$ of sets (or “clusters”), how many ways can there be of partitioning a set ${A \in X}$ in this family as the disjoint union ${A = A_1 \uplus A_2}$ of two other sets ${A_1, A_2}$ in this family? That is to say, what is the best upper bound one can place on the quantity

$\displaystyle | \{ (A,A_1,A_2) \in X^3: A = A_1 \uplus A_2 \}|$

in terms of the cardinality ${|X|}$ of ${X}$? A trivial upper bound would be ${|X|^2}$, since this is the number of possible pairs ${(A_1,A_2)}$, and ${A_1,A_2}$ clearly determine ${A}$. In our paper, we establish the improved bound

$\displaystyle | \{ (A,A_1,A_2) \in X^3: A = A_1 \uplus A_2 \}| \leq |X|^{3/p}$

where ${p}$ is the somewhat strange exponent

$\displaystyle p := \log_3 \frac{27}{4} = 1.73814\dots, \ \ \ \ \ (1)$

so that ${3/p = 1.72598\dots}$. Furthermore, this exponent is best possible!

Actually, the latter claim is quite easy to show: one takes ${X}$ to be all the subsets of ${\{1,\dots,n\}}$ of cardinality either ${n/3}$ or ${2n/3}$, for ${n}$ a multiple of ${3}$, and the claim follows readily from Stirling’s formula. So it is perhaps the former claim that is more interesting (since many combinatorial proof techniques, such as those based on inequalities such as the Cauchy-Schwarz inequality, tend to produce exponents that are rational or at least algebraic). We follow the common, though unintuitive, trick of generalising a problem to make it simpler. Firstly, one generalises the bound to the “trilinear” bound

$\displaystyle | \{ (A_1,A_2,A_3) \in X_1 \times X_2 \times X_3: A_3 = A_1 \uplus A_2 \}|$

$\displaystyle \leq |X_1|^{1/p} |X_2|^{1/p} |X_3|^{1/p}$

for arbitrary finite collections ${X_1,X_2,X_3}$ of sets. One can place all the sets in ${X_1,X_2,X_3}$ inside a single finite set such as ${\{1,\dots,n\}}$, and then by replacing every set ${A_3}$ in ${X_3}$ by its complement in ${\{1,\dots,n\}}$, one can phrase the inequality in the equivalent form

$\displaystyle | \{ (A_1,A_2,A_3) \in X_1 \times X_2 \times X_3: \{1,\dots,n\} =A_1 \uplus A_2 \uplus A_3 \}|$

$\displaystyle \leq |X_1|^{1/p} |X_2|^{1/p} |X_3|^{1/p}$

for arbitrary collections ${X_1,X_2,X_3}$ of subsets of ${\{1,\dots,n\}}$. We generalise further by turning sets into functions, replacing the estimate with the slightly stronger convolution estimate

$\displaystyle f_1 * f_2 * f_3 (1,\dots,1) \leq \|f_1\|_{\ell^p(\{0,1\}^n)} \|f_2\|_{\ell^p(\{0,1\}^n)} \|f_3\|_{\ell^p(\{0,1\}^n)}$

for arbitrary functions ${f_1,f_2,f_3}$ on the Hamming cube ${\{0,1\}^n}$, where the convolution is on the integer lattice ${\bf Z}^n$ rather than on the finite field vector space ${\bf F}_2^n$. The advantage of working in this general setting is that it becomes very easy to apply induction on the dimension ${n}$; indeed, to prove this estimate for arbitrary ${n}$ it suffices to do so for ${n=1}$. This reduces matters to establishing the elementary inequality

$\displaystyle (ab(1-c))^{1/p} + (bc(1-a))^{1/p} + (ca(1-b))^{1/p} \leq 1$

for all ${0 \leq a,b,c \leq 1}$, which can be done by a combination of undergraduate multivariable calculus and a little bit of numerical computation. (The left-hand side turns out to have local maxima at ${(1,1,0), (1,0,1), (0,1,1), (2/3,2/3,2/3)}$, with the latter being the cause of the numerology (1).)

The same sort of argument also gives an energy bound

$\displaystyle E(A,A) \leq |A|^{\log_2 6}$

for any subset ${A \subset \{0,1\}^n}$ of the Hamming cube, where

$\displaystyle E(A,A) := |\{(a_1,a_2,a_3,a_4) \in A^4: a_1+a_2 = a_3 + a_4 \}|$

is the additive energy of ${A}$. The example ${A = \{0,1\}^n}$ shows that the exponent ${\log_2 6}$ cannot be improved.