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We take $\alpha ,\beta >0$, as is necessary for the assertion. The nonvacuous cases have $0<\alpha \le 1/2$ and $0<\beta \le 1$. Put $n=|A|$ andWe prove a lower bound for $K$ by transferring small sumset growth through algebraic operations on the scalars. This gives a self-contained polynomial skew-sumset expansion over a prime field proof using the permitted Ruzsa triangle inequality, Ruzsa covering lemma and Plünnecke-Ruzsa inequality.

Since $|B|>1$, choose $b_0\in B\setminus \{0\}$. The bound $|b_{0}A+A|\le K|b_{0}A|$ and the mixed-set Plünnecke-Ruzsa inequality giveIn particular $|A+A|\le K^{2}n$ and $|3A-2A|\le K^{5}n$. Here repeated-set sums such as $3A$ allow independent choices from each copy.

For a scalar $t$, write $D(t)=|A+tA|/n$, and put $Q=A-A$ for the difference set. The Ruzsa covering lemma gives $tA\subseteq T_t+Q$ with $|T_t|\le D(t)$. Indeed choose a maximal family of disjoint translates $u+A$, with $u\in tA$; counting their union bounds the family size, and maximality forces every remaining translate to meet one of them.

For any $r,s$, the containments $A+(r+s)A\subseteq A+rA+sA$ and $A+rsA\subseteq r{T}_s+A+rQ$ give, after covering each copy of $rA,sA$,For the second bound, specifically $rQ=rA-rA\subseteq T_r-T_r+2Q$, and $A+2Q=3A-2A$. The Ruzsa triangle inequality also gives $|A-rA|\le |A+A||A+rA|/n$, so $D(-r)\le K^{2}D(r)$. These are the closure bounds for small skew-sumset scalars.

We next prove a growth lemma for the scalar sets. For $S\subseteq {\mathbb{F}}_p$ with $m=|S|\ge 2$, letIf $R\ne {\mathbb{F}}_p$, then $R+1$ is not contained in $R$: otherwise equal cardinality would make $R$ invariant under addition of one, which generates the entire prime field. Choose $r=1+(a-b)/(c-d)\notin R$. The map $(x,y)\mapsto x+ry$ on $S^2$ is injective. Indeed a collision with different second coordinates would express $r$ as a ratio of two differences from $S$, while equal second coordinates force equal first coordinates. Clearing the nonzero denominator puts its image insideso the latter set has at least $m^2$ elements. Here $SS$ is the product set, and $3SS-3SS$ denotes three independent product terms minus three others, not multiplication by the scalar three.

If $R={\mathbb{F}}_p$, let $E_t$ count the collisions $x_1+t{y}_1=x_2+t{y}_2$ with all four variables in $S$. Summing over $t$ givesThe first term comes from equal second coordinates, which force equal first coordinates. For unequal second coordinates, each quadruple determines exactly one scalar. Some $t$ consequently satisfies $E_t\le m^2+m^4/p$. The Cauchy-Schwarz inequality on the representation counts givesRepresenting $t$ as a ratio from $R$ and clearing its denominator embeds a dilate of this sumset in $2SS-2SS\subseteq 3SS-3SS$, padding with the same product in the positive and negative sums. We have proved the quadratic growth of a sixfold product difference set estimate

Starting from $S_0=B$, define $S_{j+1}=3S_j{S}_j-3S_j{S}_j$. ChooseFor $p\ge \max (4,2^{2/\beta })$, the growth lemma implies $|S_k|\ge p/2$. To see the constants explicitly, put $v_j=|S_j|/2$. Then $v_0\ge p^{\beta /2}$ and $v_{j+1}\ge \min (p/4,v_j^2)$. Induction gives $v_j\ge \min (p/4,p^{\beta 2^{j-1}})$, and the chosen $k$ makes the exponent at least one.

Every $t\in S_0$ has $D(t)\le K$. If $D(t)\le K^{L_j}$ on $S_j$, the closure bounds show that each product has exponent at most $3L_j+5$, its negative at most $3L_j+7$, and a sum of six such terms has exponent at most $18L_j+67$. Thus, withwe have $|A+tA|\le K^{L_k}n$ for every $t\in S_k$. The integer $L_k$ depends only on $\beta$.

Now apply the collision count to $A$ instead of $S$. Summing over all scalars gives at most $p{n}^2+n^4$. Since $|S_k|\ge p/2$, some $t\in S_k$ has collision count at most $2n^2+2n^4/p$. The Cauchy-Schwarz inequality therefore givesIt follows thatFor $p\ge 4^{2/\alpha }$ this gives $K\ge p^{\alpha /(2L_k)}\ge n^{\alpha /(2L_k)}$, proving the desired positive exponent for all sufficiently large primes.

For completeness, the same conclusion with a possibly smaller exponent holds for the remaining primes. Choose an integerIf $p<P_0$, choose any nonzero $b\in B$. We have $1<n<p$. If $|A+bA|=n$, all its translates $A+u$ for $u\in bA$ would coincide. Two distinct such $u$ would make $A$ invariant under a nonzero translation, forcing $A={\mathbb{F}}_p$, a contradiction. Thus $|A+bA|\ge n+1\ge n^{1+c_{\alpha ,\beta }}$ by the definition of the second exponent. Together with the large-prime case this provesPositivity of the parameters cannot be dropped: at $\alpha =0$, taking $A={\mathbb{F}}_p$ prevents any expansion; at $\beta =0$, the allowed set $B=\{0\}$ prevents expansion of any proper $A$ with more than one element. No external sum-product theorem was invoked: the scalar growth and transfer arguments were proved above.

The printed statement needs $A\ne \varnothing$: for the empty set every linear subspace has cardinality at least one, so the requested bound is impossible. We prove the intended nonempty version of the Freiman-Ruzsa theorem over a finite field. Necessarily $C\ge 1$, because any one translate of $A$ inside $A+A$ has size $|A|$.

We first prove all sumset estimates needed below. For finite nonempty $X$ and finite $U,V$ in an abelian group, choose one representation $z=u_z-v_z$ for each $z\in U-V$. The mapis injective: subtracting the outputs recovers $z$, hence its chosen representation and then $x$. This proves the Ruzsa triangle inequality

Choose nonempty $X\subseteq B$ minimizing $c^{\prime }=|X+B|/|X|\le C$, where $B$ is any set with $|B+B|\le C|B|$. We prove the Petridis minimal-growth lemma $|X+B+D|\le c^{\prime }|X+D|$ for every finite $D$. List $D=\{d_1,\dots ,d_s\}$, and let $X_i$ consist of those $x\in X$ for which $x+d_i$ was not already in $X+\{d_1,\dots ,d_{i-1}\}$. The disjoint pieces $X_i+d_i$ partition $X+D$.

Every $(X\setminus X_i)+B+d_i$ was already covered in $X+B+\{d_1,\dots ,d_{i-1}\}$. The new part at step $i$ therefore has size at mostHere minimality of $X$ applies to every nonempty subset $X\setminus X_i$, and the empty case has size zero. Summing proves the lemma. Iteration gives $|X+kB|\le C^k|X|$. Apply the Ruzsa triangle inequality with $U=kB$, $V=\ell B$ to obtainThis derives the required Plünnecke-Ruzsa inequality. In characteristic two, subtraction is addition, so $|rB|\le C^r|B|$ for every positive integer $r$.

Fix $a\in A$ and translate to $B=A+a$, which contains zero, has size $|A|$, and has the same doubling constant. Choose $m=\lceil {\log }_2(2|12B|)\rceil$. A uniformly random linear map $L:{\mathbb{F}}_2^N\to G^{\prime }={\mathbb{F}}_2^m$ kills each specified nonzero vector with probability $2^{-m}$. The union bound shows that some $L$ satisfiesBecause $0\in B$, all lower iterated sumsets lie in $12B$. In particular $L$ is injective on $B$, since the difference of two elements belongs to $2B$. For $B^{\prime }=L(B)$, the density $\alpha =|B^{\prime }|/|G^{\prime }|$ is at least $1/(4C^{12})$. This proves injective linear modelling of a small sumset.

We next prove the Finite-field Bogolyubov lemma in this model. Write $f=1_{B^{\prime }}$ and use normalized real Fourier coefficients on a finite abelian group $\hat{f}(\xi )={\mathbb{E}}_{x}f(x)(-1{)}^{\xi \cdot x}$. LetBy Parseval identity, $|S|\le 2{\alpha }^{-2}$. Thus $W$ is a vector subspace of codimension at most $D=\lceil 32C^{24}\rceil$.

For $w\in W$, Fourier inversion of the fourfold normalized convolution givesAll terms with $\xi \in S$ are nonnegative, and the zero-frequency term is ${\alpha }^4$. The absolute contribution from the remaining frequencies is at mostHence the convolution is positive and $W\subseteq 4B^{\prime }$. Also $|W|\ge 2^{-D}|G^{\prime }|\ge 2^{-D}|B|$.

For each $w\in W$, choose its lift $\sigma (w)\in 4B$. It is unique because two lifts differ by an element of $8B\cap \ker L$, which is zero. For $w_1,w_2\in W$, the discrepancybelongs to $12B\cap \ker L$, so it vanishes. Also $\sigma (0)=0$. Therefore $\sigma$ is a linear map and its image $\tilde{W}\subseteq 4B$ is a vector subspace with $|\tilde{W}|=|W|\ge 2^{-D}|B|$. This establishes lifting a subspace through a Freiman model without assuming that arbitrary choices of preimages preserve addition.

Finally, $B+\tilde{W}\subseteq 5B$ is a union of $\tilde{W}$-cosets. The number of these cosets is at mostChoose representatives $b_1,\dots ,b_t$ from $B$. Then $B\subseteq \tilde{W}+\{b_1,\dots ,b_t\}$, so adjoining these representatives and $a$ to $\tilde{W}$ gives a vector subspace $V$ containing $A$. Over ${\mathbb{F}}_2$, adjoining each vector multiplies the size by at most two. Since $|\tilde{W}|\le |4B|\le C^4|B|$, the Freiman-Ruzsa bound from Bogolyubov and linear modelling yieldsThis bound depends only on $C$, as required, and every sumset estimate used in its derivation has been proved.

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.

We first prove the needed Petridis minimal-growth lemma, including the expansion estimate rather than assuming it. For finite nonempty sets $A,B$, choose a nonempty $X\subseteq A$ minimizing $r=|X+B|/|X|$. For any finite set $C=\{c_1,\dots ,c_s\}$, defineThen $|X+C|={\sum }_i|X_i|$. An element from $(X\setminus X_i)+B+c_i$ has already appeared in an earlier $X+B+c_j$, so the new contribution of $X+B+c_i$ has size at mostThe inequality follows from minimality of $r$ on subsets of $A$, with an empty subset contributing zero. Summing proves $|X+B+C|\le r|X+C|$. Iteration gives $|X+jB|\le r^j|X|$, for every $j\ge 0$; in particular, $|jB|\le r^j|X|$ by translating $jB$ into $X+jB$.

Apply this with $B=A$, so $r\le K$. It follows thatThis is the required Plünnecke inequality.

We will also use the mixed-sumset bound $|mA-nA|\le K^{m+n}N$. To justify it, the Ruzsa triangle inequality saysChoose one representation $z=p_z-q_z$ for each $z\in P-Q$. The map $(z,t)\mapsto (p_z-t,t-q_z)$ is injective: adding the two coordinates recovers $z$, then the chosen representation recovers $t$. This proves the inequality. Apply it with $P=mA$, $Q=nA$, $T=-X$ and the minimal-growth set already chosen. The bounds on $X+mA$ and $X+nA$ give the mixed-sumset estimate, including $0A=\{0\}$.

The Ruzsa covering lemma states: if $S,T$ are finite, $T\ne \varnothing$, and $|S+T|\le L|T|$, then there is $D\subseteq S$, $|D|\le L$, such that $S\subseteq D+T-T$. Choose $D$ maximal with the translates $d+T$, $d\in D$, pairwise disjoint. They lie in $S+T$, so $|D||T|\le |S+T|$. Maximality says each $s+T$ meets some $d+T$, which writes $s=d+t_1-t_2$. This proves both assertions.

Let $T=A-A$. The mixed-sumset bounds give $|T|\le K^{2}N$ and $|2T+A|=|3A-2A|\le K^{5}N$. Apply the Ruzsa covering lemma to $S=2T$ and the set $A$. There is $D\subseteq 2T$ of size $q\le K^5$ withInduction then gives $\ell T\subseteq (\ell -1)D+T$. The number of possible sums of $\ell -1$ elements from the $q$-element set $D$ is at most the number of multiplicity vectors, namely $\left(\genfrac{}{}{0ex}{}{\ell +q-2}{q-1}\right)$. Since $A-a_0\subseteq T$ for any $a_0\in A$, we obtain the explicit polynomial growth of iterated sumsetsFor fixed $K>1$, this polynomial in $\ell$ is eventually smaller than $K^{ϵ\ell }$. For example, take $q_{\ast }=\lceil K^5\rceil$ and choose ${\ell }_0(K,ϵ)$ so that $2\log K+q_{\ast }\log (\ell +q_{\ast })\le ϵ\ell \log K$ for every $\ell \ge {\ell }_0$; such a threshold exists because $\log \ell /\ell \to 0$. ThereforeFor $K=1$, a finite integer set has $|A+A|\ge 2|A|-1$ (list the increasing sums $a_1+a_1,\dots ,a_1+a_N,a_2+a_N,\dots ,a_N+a_N$). Thus $N=1$, and the asserted bound holds for every $\ell$. Nonemptiness is implicit in the doubling constant hypothesis.

An additive form of the Plünnecke inequality is the following. If $A,B$ are nonempty finite subsets of an abelian group and $|A+B|\le K|A|$, then there is one nonempty $X\subseteq A$ such that, simultaneously for every integer $m\ge 0$,Here $mB$ is an iterated sumset and $0B=\{0\}$. In particular, its Plünnecke-Ruzsa inequality consequence isWe prove both statements, so that the sumset and difference set formulations are covered.

Choose a nonempty $X\subseteq A$ minimizing the ratio $|X+B|/|X|$, and denote this minimum by $\kappa$. Such a minimizer exists because $A$ is finite, and $\kappa \le K$. Every $Z\subseteq X$ satisfies $|Z+B|\ge \kappa |Z|$, with the empty case also valid. We first prove the Petridis minimal-growth lemmafor every finite nonempty $C$.

List $C=\{c_1,\dots ,c_t\}$. Let $U_i=X+\{c_1,\dots ,c_i\}$ and $U_0=\varnothing$. DefineThe new points of $U_i$ are exactly $X_i+c_i$, so $|X+C|={\sum }_i|X_i|$. Since $Z_i+c_i\subseteq U_{i-1}$, the sumset $Z_i+B+c_i$ is already contained in $U_{i-1}+B$. It follows that the new points introduced into $U_i+B$ are contained inAs $Z_i+B\subseteq X+B$, their number is at mostSumming these increments proves the Petridis minimal-growth lemma. Taking $C=(m-1)B$ for $m\ge 1$ and iterating now givesThis proves the Plünnecke inequality with the same minimizing set $X$ for every $m$.

To obtain the difference set bound, we also prove the required Ruzsa triangle inequality. For finite $U,V,Z$ with $Z\ne \varnothing$, choose one representation $d=u_d-v_d$ of each $d\in U-V$. The mapis an injection: the sum of the output coordinates recovers $d$, and the first coordinate then recovers $z$ because $u_d$ was fixed. ThusUse $U=kB$, $V=\ell B$, and $Z=-X$. The already proved Plünnecke inequality yieldsThis finishes the proof of the stated Plünnecke-Ruzsa inequality. The use of an abelian group is essential in commuting the translates and sumsets in the minimal-growth argument.

For each element $x$ of the product set $A\cdot B$, let $s_x$ count its representations as $ab$ with $(a,b)\in A\times B$. The ratio equality $a/b=c/d$ is equivalent to $ad=cb$. Swapping the two $B$ coordinates is a bijection between the ratio-equality quadruples and the equal-product quadruples. Therefore the multiplicative energy satisfiesThe Cauchy-Schwarz inequality gives the energy lower bound:For the upper bound, place the Cartesian product $P=A\times B$ in the strictly positive quadrant. For every occupied Euclidean ray from the origin let $P_{\lambda }$ be its points and $r_{\lambda }=|P_{\lambda }|$. The slope is $\lambda =b/a$, soTake $\log$ to be the natural logarithm and put $L=\lceil \log |B|\rceil \ge 1$, using the non-triviality assumption $|B|\ge 2$. Partition the possible occupancies into $L$ classes: $[e^j,e^{j+1})$ for $0\le j<L-1$, and $[e^{L-1},|B|]$ for the last class. This last closed endpoint ensures that the partition works even when the top endpoint is attained. Within each class the ratio of any two occupancies is at most $e$.

Fix one class and order its occupied Euclidean rays by increasing slope, with occupancies $r_1,\dots ,r_t$. Write $S=|A+A||B+B|$, the cardinality of $(A+A)\times (B+B)$. If $t=1$, then $r_1^2\le |A||B|\le S$, since adding any fixed element gives an injection of $A$ into its sumset $A+A$, and similarly for $B$.

If $t\ge 2$, the sums $P_i+P_{i+1}$ contain exactly $r_i{r}_{i+1}$ different points, by injectivity of sums on two distinct rays. Indeed, the two direction vectors are linearly independent, so the coefficients of a sum uniquely recover its two summands. Positivity puts every sum strictly inside the open planar sector between those two Euclidean rays. The sectors between successive selected Euclidean rays are disjoint, even if there are additional unselected rays between them. All these sums belong to $(A+A)\times (B+B)$, and henceFor neighbouring occupancies their ratio lies between $e^{-1}$ and $e$, soSumming covers every $r_i^2$ at least once, and gives ${\sum }_i{r}_i^2\le 4S$. The same bound holds for empty or singleton classes by the preceding observations. Adding over the $L$ classes proves the multiplicative energy sumset bound $E(A,B)\le 4LS$. Combining both bounds yields the sum-product conclusion:If instead $\log$ is interpreted as base two, use the same $L=\lceil {\log }_2|B|\rceil$ occupancy classes with $2$ in place of $e$; $2+2^{-1}<4$ proves that convention as well. Positivity is essential to the sector argument.

Let $e_1,\dots ,e_m$ be the standard basis of ${\mathbb{F}}_2^m$ and takeIts sumset consists of zero, the $m$ basis vectors, and the $\left(\genfrac{}{}{0ex}{}{m}{2}\right)$ sums of two distinct basis vectors. ThusIf a coset $x+H$ contains $A_m$, then every difference of two elements of $A_m$ lies in $H$. In particular every $e_i$ lies in $H$, so $H={\mathbb{F}}_2^m$ andSince $m=2K_m+O(1)$, this ratio is eventually much larger than $K_m^{-1}{2}^{K_m}$. Hence no bound valid for all $K$ can replace the factor in part c by a function smaller than $K^{-1}{2}^K$.

Apply the preceding Fourier argument to $f=1_A$, retaining the frequencies needed to make the oscillation of $1_A\ast 1_A\ast 1_A$ strictly smaller than its mean ${\alpha }^3$. The standard optimized cutoff gives a set $\Gamma \subseteq \hat{G}$ withsuch that the nonnegative function $F=1_A\ast 1_A\ast 1_A$ cannot fall from a maximal value to zero under any shift in $B(\Gamma ,\alpha )$. If $x$ maximizes $F$, thenThe support of a convolution of indicator functions is the corresponding sumset, so

For fixed positive density $\delta$, every sufficiently large $A\subseteq [1,N]$ of size at least $\delta N$ has a long arithmetic progression in its triple sumset. Embed it in $\mathbb{Z}/(8N)\mathbb{Z}$, where its density is at least $\delta /8$. A translated Bohr neighborhood in a triple sumset has bounded rank and fixed positive radius depending only on $\delta$. A torus pigeonhole principle argument supplies a nonzero step whose small multiples stay in the Bohr set, with polynomially many multiples. All resulting residues have representatives in $[3,3N]$, an interval shorter than half the modulus, so their successive ordinary integer differences agree and give an actual integer arithmetic progression. The positive constant prefactor is absorbed by reducing the exponent for sufficiently large $N$.

The product set of two sets of real numbers is $A\cdot B=\{ab:a\in A,b\in B\}$. It is a multiplicative counterpart of a sumset, not the Cartesian product $A\times B$.

For $S\subseteq {\mathbb{F}}_p$ with $|S|\ge 2$, put $R=(S-S)/(S-S)$ using nonzero denominators. If $R$ is proper, some $r\in R+1$ lies outside $R$, since invariance under addition of one would give all of the prime field. The map $(x,y)\mapsto x+ry$ on $S^2$ is injective. Clearing the denominator in a representation of $r-1$ puts a dilate of its image inside $3SS-3SS$. If $R$ is the full field, averaging the collision count of $S+rS$ gives some $r$ with $|S+rS|\ge \frac{1}{2}\min (p,|S{|}^2)$. Clearing its denominator puts a dilate inside $2SS-2SS\subseteq 3SS-3SS$. Here $SS$ is a product set, and the prefactors denote repeated sumsets, not scalar multiplication.

For finite sets in an abelian group and a bipartite graph $\Gamma \subseteq A\times B$, the restricted sumset is $\{a+b:(a,b)\in \Gamma \}$. It retains only sums corresponding to graph edges.

Given numbers $s_1,\dots ,s_n$, the sum associated with a subset $A$ of indices is ${\sum }_{i\in A}{s}_i$. The empty subset gives zero. The set of all these values is the sumset $\{0,s_1\}+\cdots +\{0,s_n\}$.

The sum-product phenomenon says that a finite subset of a field cannot usually have both a small sumset and a small product set unless it has additional subfield-like structure.