The Balog–Szemerédi–Gowers theorem
Extract a substantial small-doubling subset from many additive collisions.
On this page
The question for today
What you’ll understand
- State a qualitative polynomial BSG theorem with all subset and constant scopes.
- Explain why large energy does not force whole-set small doubling.
- Identify the role of popular relations and paths in the extraction argument.
Small doubling immediately gives high additive energy, but reversing that statement requires discarding poorly behaved elements. Balog–Szemerédi–Gowers provides exactly that extraction. The theorem applies to finite nonempty sets in an abelian group and converts a global count of additive relations into a large subset with controlled growth. Its conclusion depends on an energy parameter, not on the ambient group size. Keeping the extracted subset visible is essential when using the result in an inverse argument.
The theorem controls a subset with polynomial losses
For a single finite nonempty set A, suppose its counting energy is at least the cube of its cardinality divided by K, with K at least one. There are absolute positive constants c and C such that one can find a subset A′ containing at least c times K to a negative absolute power times the size of A, and with doubling at most an absolute constant times a fixed power of K. Enlarging the same symbol C if necessary gives the concise formulation below. No specific optimal exponent is claimed.
The constants are uniform over the group and the set, but their existence does not give a numerical promise such as retaining half the elements when K equals two. The theorem does not say every subset of that size has small doubling, and it does not control the entire original set. A two-set version uses mixed energy with a geometric-mean normalization and gives two extracted subsets. That version needs its own careful bookkeeping; the single-set theorem already captures the essential extraction principle used here.
A structured core can coexist with broad growth
Take an integer interval I with n elements and add n widely separated positive integers R, chosen as successive powers of an integer M greater than twice n. The translates I+r for r in R are disjoint, so the mixed sums alone supply n squared outputs. The full set A has size twice n, and its doubling constant is therefore at least n divided by two. This growth becomes unbounded with n, despite the fixed-density interval core inside A.
The energy contributed by that interval alone is already at least two thirds of n cubed. Relative to the cube of the full size, this is at least one twelfth. Thus A has energy at least |A| cubed divided by twelve while its doubling grows without bound. This is a concrete family disproving a whole-set converse. The subset I supplies the small-doubling object that the extraction theorem predicts in qualitative terms, without telling us that every high-energy input must literally contain an interval.
Popular differences lead to robust connections
A proof first identifies differences with many representations. Their popularity captures a substantial part of the energy rather than just many arbitrary pairs. Regard the associated relations as edges of a bipartite graph. A graph lemma finds large vertex subsets with many short connecting paths between their members. Labels along a path express endpoint differences as combinations of popular differences, converting robust connectivity into a small collection of possible endpoint differences. Counting many paths is what controls multiplicity in this conversion.
Ruzsa inequalities then convert the resulting difference control into small doubling for an extracted set. This paragraph is a proof roadmap, not a replacement for the graph lemma or its quantitative proof. The reason to learn the roadmap is operational: energy is a second-moment statistic that may be concentrated, and the graph step removes that concentration before a structural theorem is invoked. In applications one should carry both the retained-size lower bound and the doubling upper bound into the next argument; retaining only the word structured loses the quantitative content.
Work it through
Locate the core in a small example
For , compute energy and compare its doubling with the interval core I={0,1,2}.
The outputs 0,1,2,3,4,100,101,102,200 have multiplicities 1,2,3,2,1,2,2,2,1. Their total is sixteen and their squared sum is thirty-two.
Since |A|³=64, this energy meets the hypothesis with K=2. The full sumset has nine outputs, giving doubling 9/4.
The three-point core has five sum outputs and energy nineteen, so its doubling is 5/3 and its retained fraction is 3/4. These are direct computations, not specific constants extracted from the general theorem.
The core visibly has lower doubling than the full set. BSG guarantees some controlled subset without prescribing this core or these numerical constants.
The living notebook
Explore the model
Interactive lab · modular
Recognize the relations of a structured core
Reset, choose modulus 12 and step 4, and inspect the orbit H={0,4,8}. Predict why choosing any three points in H determines a fourth still in H for the relation a+b=c+d. Count twenty-seven ordered relations manually, then compare H+H with H. The lab shows an exact subgroup orbit, not BSG subset extraction or a graph proof. Reset restores both controls and time; changes preserve the clock.
- Residue r0
- 0
- Orbit period · m/gcd(m,a)
- 12
- Distinct residues visited
- 1 of 12
Addition modulo m starts at zero: rₙ = na mod m. Playback advances two additions per second. The orbit has period m/gcd(m,a), including period one for a = 0. This is not a cryptographic security test or a primality algorithm. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.
Look forClosure creates cubic energy and doubling one; the extraction theorem finds related control inside more irregular inputs.
Pause & reason
Check your understanding
Your turn
Try it yourself
Let A={0,1,10} in the integers. Compute its sum multiplicities, energy, the smallest K for which E(A)≥|A|³/K, and the doubling of the subset {0,1}.
Take a moment to make a prediction or work it out on paper. Open a hint when you need one.
Give me a hint
List all nine ordered pairs, remembering that off-diagonal pairs have two orientations.
Compare with the solution
The outputs 0,1,2,10,11,20 have multiplicities 1,2,1,2,2,1, so energy is fifteen. The inequality requires 15≥27/K, giving K≥27/15=9/5. The two-element subset has self-sum {0,1,2}, so its doubling is three halves and it retains two thirds of A. These computations illustrate extraction but do not identify optimal universal BSG constants from one input.
Keep the idea
Take it with you
- High energy yields a large controlled subset, not necessarily small doubling of the whole set.
- Both the retained-size bound and the doubling bound depend polynomially on the energy parameter.
- Popular relations and robust short paths explain why a second moment can lead to structural control.
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