Additive energy ($E(A)$)

The additive energy of a finite set $S\subseteq G$ in an abelian group $G$ counts additive quadruples, with order and repeated members allowed:At a fixed sum $s$, independently choosing two of its $r_{S+S}(s)$ ordered pairs gives $r_{S+S}(s{)}^2$ additive quadruples. Thus additive energy weights heavily represented sums more strongly than the cardinality of the sumset does.

For $A=\{0,1,2,3\}$ and $B=\{0,1,4,10\}$, the sumset representation functions giveThe arithmetic progression has additional coincidences such as $0+3=1+2$. The set $B$ still has coincidences forced by commutativity, including $0+1=1+0$, so its additive energy is not merely the sixteen identical-pair comparisons.

The same additive energy can be counted using the difference set. Define $r_{S-S}(t)=|\{(a,b)\in S\times S:a-b=t\}|$. The identity $a+b=c+d$ is equivalent to $a-c=d-b$, and the coordinate rearrangement $(a,b,c,d)\mapsto (a,c,d,b)$ is a bijection. ConsequentlyFor $A$, the difference counts from $-3$ through $3$ are again $1,2,3,4,3,2,1$. For $B$, the zero difference has four representations and each of the twelve nonzero differences has one, giving $4^2+12\cdot 1^2=28$. The sumset representation functions for sums and differences can look different while their squared totals agree.

For every nonempty finite set $S\subseteq G$ in an abelian group $G$,The lower bound follows from the Cauchy-Schwarz inequality and the total sumset representation function count:For the upper bound, any three members $a,b,c$ determine at most one fourth member $d=a+b-c$ of an additive quadruple. See Yufei Zhao's MIT lecture 25, pages 1-3, for representation counts and the energy bounds.

In the two integer examples, the lower bounds are $256/7$ and $256/10$, and the upper bound is $64$. For the subgroup $H=\{0,2,4\}$ of $\mathbb{Z}/6\mathbb{Z}$, each element of $H$ has three sum representations, so $E(H)=3\cdot 3^2=27$ and both bounds are equalities.

For nonempty $S$, dividing additive energy by $|S{|}^3$ gives a relative measure between $1/K(S)$ and one. Small doubling constant therefore forces relatively large additive energy. A converse requires care: the Balog-Szemerédi-Gowers theorem extracts a large subset with small doubling constant, rather than asserting that all of $S$ has small doubling constant.

If a subset $A$ of a cyclic group has normalized additive energy at most ${\alpha }^4+ϵ$, every nonconstant Fourier coefficient on a finite abelian group of its indicator has modulus at most ${ϵ}^{1/4}$. Expanding the count of $a+c=2b$ gives the frequency product $\widehat{1_A}(r)\widehat{1_C}(r)\widehat{1_B}(-2r)$. The Cauchy-Schwarz inequality and Parseval identity on a finite group bound its nonconstant part by the displayed quantity. Character doubling has multiplicity $\gcd (2,N)$, so no assumption that the modulus is odd is needed. The error is a normalized density error multiplied by $N^2$.

For a set of shifts $H$ and a map $\theta$ into a circle group, this additive energy counts quadruples whose shift sums and frequency sums both agree. It is the additive energy of the graph of $\theta$ in the product abelian group.

Suppose $|f|\le 1$ on a finite cyclic group, $H$ has density $\beta$, and for each $h\in H$ the multiplicative derivative has a Fourier coefficient on a finite abelian group of magnitude at least $\eta$ at $\theta (h)$. Align the correlation phases, apply Cauchy-Schwarz, group pairs by their shift and frequency differences, and apply Parseval identity. This proves the displayed lower bound on additive energy of a frequency graph without replacing exact frequency equalities by approximate ones.

Use the unnormalized Fourier transform and $|f|\le 1$. Normalize each derivative correlation by $N$ and align its phase. Their absolute sum is at least $\alpha N$. One Cauchy-Schwarz inequality, expansion by shift/frequency differences, and a second Cauchy-Schwarz inequality bound its fourth power by $N$ times the additive energy of a frequency graph. Parseval identity on a finite group supplies the factor $N$; exact equality of both shift sums and frequency sums is retained.

For every $\theta >0$, some $\gamma (\theta )>0$ has this property: any nonempty finite $A\subseteq \mathbb{Z}$ with $E(A)\ge \theta |A{|}^3$ has an arithmetic progression of length at least $|A{|}^{\gamma (\theta )}$ in $2A-2A$. The small-difference-set form of the Balog-Szemerédi-Gowers theorem, Ruzsa modelling lemma, cyclic Bogolyubov lemma and nonwrapping progression in a cyclic Bohr set prove it. An order-eight Freiman s-isomorphism suffices to lift the progression because its consecutive second-difference equations expand into equalities of eight-term sums.

A popular sum of a finite set $A$ is an element having at least a specified number $u$ of ordered representations as $a+b$, with $a,b\in A$. Popular sums define a dense bipartite graph when additive energy is large. Since the total representation count is $|A{|}^2$, there are at most $|A{|}^2/u$ popular sums.

An additive quadruple is a tuple $(a,b,c,d)$ satisfying $a+b=c+d$.

If a finite set $A$ has at least $\eta |A{|}^3$ additive quadruples, then it contains $A^{\prime }$ with $|A^{\prime }|\ge {\eta }^C|A|$ and $|A^{\prime }+A^{\prime }|\le {\eta }^{-C}|A^{\prime }|$ for an absolute constant $C$.

There are absolute $c,C>0$ such that finite sets $A,B$ of size $n$, a bipartite graph of density at least $\delta$, and a restricted sumset of size at most $Kn$ yield subsets $A^{\prime },B^{\prime }$ with $|A^{\prime }|,|B^{\prime }|\ge c{\delta }^{C}n$ and $|A^{\prime }+B^{\prime }|\le C{\delta }^{-C}{K}^{C}n$. A graph formed by popular representations of sums turns this into the usual large-additive energy form of the Balog-Szemerédi-Gowers theorem.

Let a bipartite graph have $n$ vertices in each class and at least $\rho n^2$ edges. There is a subset $B$ of one class with $|B|\ge \rho n/4$ such that every pair of its vertices has at least ${\rho }^5{n}^3/4096$ four-edge walks between them. Declare a pair of left vertices bad if it has fewer than $\tau n$ common neighbours, with $\tau ={\rho }^2/32$. For a random right vertex $y$, put $U=N(y)$ and let $b(U)$ count ordered bad pairs inside $U$. Then $\mathbb{E}|U|\ge \rho n$ and $\mathbb{E}b(U)\le \tau n^2$. Some $y$ has $|U|-16b(U)/(\rho n)\ge \rho n/2$, so $|U|\ge \rho n/2$ and $b(U)\le |U{|}^2/8$. Remove vertices with more than $|U|/4$ bad partners. At least half remain. For any two remaining vertices, at least $|U|/2$ middle vertices are good partners of both. Each supplies at least $(\tau n{)}^2$ walks, giving the bound.

If $E(A)\ge \theta |A{|}^3$, a subset $B\subseteq A$ satisfies $|B|\ge c_{\theta }|A|$ and $|B-B|\le K_{\theta }|B|$. A dependent random choice argument in the popular sum graph finds many four-edge paths representing each member of the difference set. The Petridis minimal-growth lemma then bounds all higher iterated sumsets and difference sets.