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Additive combinatorics studies how the cardinalities and sumset representation functions of sumsets reveal structure in finite sets contained in abelian groups. Write the group operation as addition: the ambient abelian group may be the integers $\mathbb{Z}$, a finite cyclic group, or another abelian group. The same definitions apply, but the ambient group changes which bounds are possible.

The basic experiment is to add every ordered pair of members of a finite set. The sumset records the distinct answers; the sumset representation function records how often each answer occurs; the additive energy counts coincidences between those answers. An arithmetic progression produces many repeated sums, so its sumset grows slowly and its additive energy is relatively large. The doubling constant measures that growth relative to the original cardinality.

The examples below compare the arithmetic progression $A=\{0,1,2,3\}$ with the more spread-out integer set $B=\{0,1,4,10\}$. Their four members give sixteen ordered pairs in each case, but different numbers of distinct sums. This small comparison illustrates the definitions; it does not establish a claim about arbitrarily large sets. For a systematic introduction, see Tao and Vu's author page for Additive Combinatorics, with free sample chapters on sum set estimates and additive geometry.

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.

The doubling constant of a nonempty finite set $S\subseteq G$ in an abelian group $G$ isIt measures the cardinality of the sumset relative to the original set. For the integer examples, $K(A)=7/4$ and $K(B)=10/4=5/2$. A nonconstant arithmetic progression with $n$ members has doubling constant $(2n-1)/n$, which stays below two as $n$ grows. This is the sense in which its sumset has small growth.

In any abelian group, $K(S)\ge 1$: fixing $s_0\in S$ puts a translated copy $s_0+S$ inside $S+S$. The stronger integer bound gives $K(S)\ge 2-1/|S|$ only in that setting. A finite subgroup $H$ has $K(H)=1$, as does every coset $x+H$, since its sumset is $2x+H$. In particular $H=\{0,2,4\}\subseteq \mathbb{Z}/6\mathbb{Z}$ has doubling constant one.

Small doubling constant is useful because it constrains repeated growth, as in the Plünnecke-Ruzsa inequality. It signals structure such as arithmetic progressions and subgroups; precise descriptions depend on the ambient abelian group and are the subject of the Freiman-Ruzsa theorem. The additive energy bound below gives a simpler immediate consequence: $E(S)\ge |S{|}^3/K(S)$.

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.

For fixed $\theta >0$, a sufficiently large finite set of integers with additive energy at least $\theta |A{|}^3$ has a nonconstant three-term arithmetic progression. The Balog-Szemerédi-Gowers theorem and Ruzsa modelling lemma reduce the problem to the Roth theorem on three-term arithmetic progressions in a dense cyclic model.

Let $T_3(A)$ count ordered solutions $a+c=2b$ in $A^3$, including the trivial ones. The hypothesis gives $T_3(A)\ge c{n}^2$. Its relation to additive energy is the promised easy Fourier step. Translate $A$ into $[0,M]$ and embed it in an odd prime cyclic group of order $P>4M$. No relevant integer equality wraps modulo $P$. With the unnormalized Fourier transform $\hat{f}(r)={\sum }_x{1}_A(x)e(-rx/P)$, character orthogonality givesSince multiplication by $2$ permutes the characters, Cauchy-Schwarz and Parseval identity on a finite group implyThus $E(A)\ge c^2{n}^3$. Equivalently, one can apply Cauchy-Schwarz directly to $T_3(A)={\sum }_{b\in A}{r}_{A+A}(2b)$.

By part (i), some $B\subseteq A$ has $|B|\ge c_1(c)n$ and $|B+B|\le K(c)|B|$. Part 3(ii) then bounds $|2B-2B|\le K(c{)}^4|B|$. Apply part (ii) with $k=2$: there is $B^{\prime }\subseteq B$, $|B^{\prime }|\ge |B|/3$, Freiman 2-isomorphic to a subset $C\subseteq {\mathbb{Z}}_N$ with $N\le 4K(c{)}^4|B|+1$. Consequently $C$ has a positive subset density depending only on $c$, and $N$ tends to infinity with $n$ because $N\ge |B^{\prime }|$.

View the representatives of $C$ as a subset of $\{0,\dots ,N-1\}$. The assumed four-term case of Szemerédi's theorem gives four distinct integer representatives in arithmetic progression once $n$ is sufficiently large in terms of $c$. If their preimages are $b_0,b_1,b_2,b_3$, the two model equalitieslift through the Freiman 2-isomorphism to the same equalities in $\mathbb{Z}$. Hence their consecutive differences agree; injectivity makes the common difference nonzero. Therefore $A$ contains a nonconstant four-term arithmetic progression. This proves dense three-term progressions force four-term progressions without an assumption that $A$ initially lies in a short interval.

One quantitative form of the Balog-Szemerédi-Gowers theorem is: if a nonempty finite set $A$ in an abelian group has $|A|=n$ and additive energy $E(A)\ge \eta n^3$, where $0<\eta \le 1$, then there is $B\subseteq A$ with $|B|\ge c_{\eta }n$ and both $|B-B|$ and $|B+B|$ bounded by $K_{\eta }|B|$. The constants depend only on $\eta$. We prove explicit, nonoptimal polynomial bounds, thereby proving the required Balog–Szemerédi statement.

Let $r(s)=|\{(a,a^{\prime })\in A^2:a+a^{\prime }=s\}|$. Then $E(A)={\sum }_{s}r(s{)}^2$. Declare $S=\{s:r(s)\ge \eta n/2\}$ to be the set of popular sums. The remaining sums contribute at most $(\eta n/2){\sum }_{s}r(s)=\eta n^3/2$. Since $r(s)\le n$, it follows that ${\sum }_{s\in S}r(s)\ge \eta n^2/2$. Also $|S|\le 2n/\eta$. Form a bipartite graph on two copies of $A$, joining $a$ to $a^{\prime }$ when $a+a^{\prime }\in S$. Its density is at least $\rho =\eta /2$.

We prove the needed four-step path lemma for a dense bipartite graph. Call an ordered pair of left vertices bad if its number of common right neighbors is less than $\tau n$, where $\tau ={\rho }^2/32$. For a uniformly chosen right vertex $y$, let $U=N(y)$ and let $b(U)$ count bad ordered pairs in $U^2$. ThenConsequently some $y$ satisfiesFor this $U$, $|U|\ge \rho n/2$ and $b(U)\le \rho n|U|/16\le |U{|}^2/8$. Delete vertices having more than $|U|/4$ bad partners in $U$. At most half are deleted; the retained set $B$ has $|B|\ge |U|/2\ge \rho n/4$. For any $a,b\in B$, at least $|U|/2$ middle vertices $u\in U$ are good partners of both. Each such $u$ supplies at least $(\tau n{)}^2$ walks $a,v,u,w,b$ of four edges. Hence every pair in $B$ has at leastsuch walks. Repetition of vertices is allowed; it does not affect the counting argument.

Each walk has edge labelsall in $S$, and $a-b=s_1-s_2+s_3-s_4$. For a fixed pair $a,b$, these four labels determine $v,u,w$, so distinct walks have distinct label quadruples. For each element of $B-B$, fix one representing pair $a,b$. Label quadruples for distinct differences cannot overlap. ThereforeUsing $\rho =\eta /2$ and $|S|\le 2n/\eta$ yieldsThis proves the small-difference-set form of the Balog-Szemerédi-Gowers theorem. To obtain a small sumset too, write $K=2^{24}{\eta }^{-10}$ and apply the already proved Plünnecke inequality to starting set $-B$ and summand set $B$. Some nonempty $X\subseteq -B$ has $|X+2B|\le K^2|X|$. For any fixed $x\in X$, $x+2B\subseteq X+2B$, whence $|B+B|\le K^2|B|$. Thus the conventional small-sumset conclusion also follows with constants depending only on $\eta$.

There is a normalization omission in the printed conclusion: the discrepancy of a count of pairs must be bounded by a small function of the energy excess times $N^2$. We prove this normalized result, then give a counterexample to the literal bound without $N^2$. Write $ϵ$ for the energy excess parameter to avoid confusing it with an element of the third set.

For Fourier analysis on a finite abelian group, use $e_N(t)=e^{2\pi it/N}$ andCharacter orthogonality, ${\mathbb{E}}_x{e}_N(rx)=1$ for $r=0$ and zero otherwise, gives Fourier inversion and the Parseval identity on a finite group:Put $u=1_A$, $v=1_B$, $w=1_C$. Expanding the constraint $a_1+a_2=a_3+a_4$ by the same character orthogonality gives the additive energy identityThus ${\sum }_{r\ne 0}|\hat{u}(r){|}^4\le ϵ$, and every nonconstant Fourier coefficient on a finite abelian group of $u$ has modulus at most ${ϵ}^{1/4}$. Expanding the constraint in the triple count $T$ givesThe Cauchy-Schwarz inequality yieldsThe first squared norm is $\gamma$ by the Parseval identity on a finite group. For the second, multiplication by $2$ on ${\mathbb{Z}}_N$ has kernel size $h=\gcd (2,N)$ and every element of its image has exactly $h$ preimages. HenceThis proves that additive energy controls three-term progression mixing, including for even moduli:Thus the intended density-error function can be $F(ϵ)=\sqrt{2}{ϵ}^{1/4}$, which tends to zero. For odd moduli the factor $\sqrt{2}$ can be replaced by one.

To see why the literal unnormalized conclusion is false, take $N=q^2$ and let all three sets be the subgroup of multiples of $q$. Their densities are $1/q$. Every choice of three entries of that subgroup determines the fourth entry of an additive quadruple, so its energy is $q^3$. The excess parameter isEvery pair $(a,b)$ in the subgroup determines $c=2b-a$ there, giving $T=q^2$, whereas $\alpha \beta \gamma N^2=q$. Their difference is $q^2-q\to \infty$, which cannot be bounded by any function of $ϵ$ tending to zero. The missing $N^2$ is therefore a genuine source defect, not an alternative Fourier transform normalization.

Assume $n>0$; the empty case is immediate. Write $r(s)=\#\{(a,b)\in A^2:a-b=s\}$. The additive energy in the hypothesis is ${\sum }_{s}r(s{)}^2$: rearranging $x+y=z+w$ gives $x-z=w-y$. Since $r(s)\le n$, the energy is at most $n^3$, so the substantive range is $0<\alpha \le 1$.

Let $S=\{s:r(s)\ge \alpha n/2\}$. The contribution of the other differences to the energy is at most $(\alpha n/2){\sum }_{s}r(s)=\alpha n^3/2$. ConsequentlyMake a bipartite graph with two copies of $A$, joining a left vertex $a$ to a right vertex $b$ when $a-b\in S$. Put $\rho =\alpha /2$. This graph has at least $\rho n^2$ edges, and $|S|\le n/\rho$.

We prove the needed dense-graph selection rather than quote the Balog-Szemerédi-Gowers theorem. A pair of left vertices is bad if it has fewer than $\tau n$ common right neighbours, where $\tau ={\rho }^2/32$. Choose a right vertex $y$ uniformly, let $U=N(y)$, and let $b(U)$ be the number of ordered bad pairs in $U$. Averaging the degrees gives $\mathbb{E}|U|\ge \rho n$. Every bad pair lies in fewer than $\tau n$ of these neighbourhoods, so $\mathbb{E}b(U)\le \tau n^2$. Therefore some $y$ satisfiesFor this neighbourhood, $|U|\ge \rho n/2$ and $b(U)\le \rho n|U|/16\le |U{|}^2/8$. Delete vertices having more than $|U|/4$ bad partners in $U$. At most half are deleted, so the remaining set $B$ has $|B|\ge |U|/2\ge \rho n/4$.

For any $b,b^{\prime }\in B$, at least $|U|/2$ vertices $v\in U$ are good partners of both. Each such $v$ has at least $\tau n$ common neighbours with $b$ and at least $\tau n$ with $b^{\prime }$. Independently choosing those neighbours gives at leastfour-edge walks $b,u,v,w,b^{\prime }$. Repeated vertices cause no difficulty. This proves the four-step path lemma for a dense bipartite graph in the required form.

The labels of such a walk are $s_1=b-u$, $s_2=v-u$, $s_3=v-w$, $s_4=b^{\prime }-w$, all in $S$, andFor a fixed pair $(b,b^{\prime })$, its label tuple uniquely recovers $u=b-s_1$, $v=u+s_2$, and $w=v-s_3$, so distinct walks have distinct label tuples. Choose one pair representing each element of the difference set $B-B$. Tuples belonging to different differences are disjoint because their alternating sum determines that difference. HenceWe have obtained explicit constants for the small-difference-set form of the Balog-Szemerédi-Gowers theorem:

For a finite nonempty set $A$, its additive energy iswhere $r_{A+A}(s)$ counts ordered representations. Since ${\sum }_s{r}_{A+A}(s)=|A{|}^2$, Cauchy-Schwarz givesThus $|A+A|\le K|A|$ implies $E(A,A)\ge |A{|}^3/K$: a small doubling constant forces many additive quadruples.

A polynomial form of the Balog-Szemerédi-Gowers theorem gives the converse after passing to a subset. There are absolute constants $c,C>0$ such that if $E(A,A)\ge \eta |A{|}^3$, $0<\eta \le 1$, then some $A^{\prime }\subseteq A$ satisfiesThe exponent and constants here need not be optimal.

Precisely, a graph form of the Balog-Szemerédi-Gowers theorem states the following. If $A,B$ are finite sets of size $n$ in an abelian group, $\Gamma \subseteq A\times B$ has at least $\delta n^2$ edges, and its restricted sumsethas size at most $Kn$, then there are $A^{\prime }\subseteq A$, $B^{\prime }\subseteq B$ with $|A^{\prime }|,|B^{\prime }|\ge c{\delta }^{C}n$ andThese are uniform absolute constants, valid for $0<\delta \le 1$ and $K\ge 1$.

To apply this theorem to additive energy, let $n=|A|$ and retain the sums with $r_{A+A}(s)\ge \eta n/2$. The discarded sums contribute at most $(\eta n/2){\sum }_{s}r(s)=\eta n^3/2$ to the additive energy. Since $r(s)\le n$, the retained pairs number at least $\eta n^2/2$. There are at most $2n/\eta$ retained sums. They therefore define a bipartite graph of density at least $\eta /2$ with restricted sumset size at most $2n/\eta$. The graph theorem supplies large subsets $A^{\prime },B^{\prime }$ with polynomially bounded $|A^{\prime }+B^{\prime }|$.

For completeness, turn this cross-sumset bound into a doubling constant bound. The Ruzsa triangle inequality gives $|A^{\prime }-A^{\prime }|\le |A^{\prime }+B^{\prime }{|}^2/|B^{\prime }|$. If $D=|A^{\prime }-A^{\prime }|/|A^{\prime }|$, the Plünnecke inequality, applied to the pair $A^{\prime },-A^{\prime }$, gives $|A^{\prime }+A^{\prime }|\le D^2|A^{\prime }|$. The minimal-growth proof below in Question 5 applies to arbitrary finite pairs and proves this particular expansion estimate as well. All losses are fixed powers of ${\eta }^{-1}$, proving the displayed converse after increasing $C$.

The requested dense bipartite common-neighbour lemma is a dependent random choice assertion. Call an ordered pair in $V^2$ bad when it has fewer than $ϵn$ common neighbours. Choose $w\in W$ uniformly, and put $S=N(w)\subseteq V$. ThenFor each bad pair, the probability that both endpoints lie in $S$ is its number of common neighbours divided by $n$, which is less than $ϵ$. Thus the expected number $b(S)$ of bad ordered pairs in $S^2$ is at most $ϵm^2$. ConsequentlyChoose a neighbourhood $V^{\prime }=S$ attaining at least this expectation. It has $|V^{\prime }|\ge \delta m/\sqrt{2}\ge \delta m/2$, andHence at least the proportion $1-2ϵ/{\delta }^2$ of its ordered pairs have at least $ϵn$ two-edge connections. As usual these connections are counted by common neighbours, including a return walk for a diagonal ordered pair; assume $\delta >0$, as the formula requires.

The relevance to the graph theorem is that many two-edge connections turn a sparse collection of permitted sums into additive control on a large vertex subset. On a connection $a\to b\to a^{\prime }$, $a-a^{\prime }=(a+b)-(a^{\prime }+b)$ expresses a difference using two elements of the restricted sumset. The abundance of connections, followed by standard cleaning and further path counting, bounds the number of differences without throwing away most vertices. This is the role of dependent random choice in obtaining polynomial losses in the graph form of the Balog-Szemerédi-Gowers theorem.