For a set of density in a finite cyclic group, select frequencies whose normalized Fourier coefficients on a finite abelian group have magnitude at least . Parseval identity bounds their number by . Fourier inversion of the fourfold convolution shows it is positive on the Bohr set in phase-distance convention of radius . This is an explicit form of the Bogolyubov lemma.
Past exam of the mathematics course of the University of Cambridge 2008 iii Paper 16 2 iii Solution Created 2026-10-03 Updated 2026-10-14
Let , , and use normalized averages on . Write and define the normalized convolution on a finite group by . With ,These identities follow by expanding the convolution and using character orthogonality. Also is a normalized nonnegative count of representations ; thus implies .
Take . The Parseval identity on a finite group givesFor in the Bohr set in phase-distance convention , all large-frequency terms have real part at least half their magnitude. Since is real,Thus . This proves the cyclic Bogolyubov lemma needed here, rather than citing it.
We now find an arithmetic progression in a cyclic Bohr set. Put . If , the whole group lies in and the result is immediate. Otherwise set . When , place the vectors , , into the boxes of side in the -dimensional unit cube. Two occupy the same box, giving with for every . The arithmetic progressionlies in . Its terms are distinct: the order of in is , greater than its number of steps. Its length is at least .
Let . Since , the exponent is at least . If , then and the singleton already exceeds the required lower bound. This also covers . Therefore one may takeThe arithmetic progression here is a set of distinct elements in the cyclic group; this part does not impose the additional integer nonwrapping condition used in Question 3.
Past exam of the mathematics course of the University of Cambridge 2009 iii Paper 13 1 i Solution Created 2026-10-03 Updated 2026-10-14
Assume is nonempty, so , and put . By the Parseval identity on a finite group, . Select the nonzero frequenciesThen , giving . Outside , the fourth Fourier moment has the boundThe fourfold normalized convolution on a finite group has transform , so Fourier inversion on a finite group gives . It is real and nonnegative, and implies because the convolution counts representations, with a positive normalization.
Use the Bohr set in phase-distance conventionOn this set all retained additive characters have positive real part. The zero-frequency contribution is , and the total absolute contribution of the discarded frequencies is at most . HenceThereforeThis is an explicit Bogolyubov lemma. In the chord-distance convention for a Bohr set, the phase-radius corresponds to chord radius ; alternatively the same positivity proof works directly at chord radius . Thus the requested lower bound on width holds in either convention. If is empty, the Bohr set is the whole group. Nonemptiness of is necessary for the density statement.
Past exam of the mathematics course of the University of Cambridge 2013 iii Paper 11 4 Solution Created 2026-10-03 Updated 2026-10-07
Use the phase-distance convention for a Bohr set:where the norm is distance to the nearest integer. This is the Bohr set in phase-distance convention; the alternative condition has a different radius and corresponding constants. Assume , as requires.
For each residue , form in the -dimensional torus. Around these points place translates of a box of side length in each coordinate. For , their volumes sum to . If , two boxes overlap, and their distinct indices differ by a nonzero with every phase distance less than . Thus the Bohr set contains a nonzero element at the stated threshold. For larger radii the entire group is already present.
Letting the box width decrease to and using the finite number of possible nonzero , obtain a nonzero with for every . By the triangle inequality, lies in whenever . Since is prime, these multiples are distinct up to length . Therefore, in the usual radius range , the arithmetic progressionhas the requested length. For literally unrestricted , the correct lower bound is : a progression of distinct residues cannot exceed . For example, , makes the uncapped printed bound impossible. If is empty, the Bohr set is the whole group and has a progression of length ; the displayed exponent is simply inapplicable.
Now transfer the dense integer set to a larger cyclic group. Choose a prime , using Bertrand's postulate, and regard as a subset of . Its density is at least . Normalize Fourier coefficients on a finite abelian group and convolution byThe Fourier inversion and Parseval identity formulas are and ; convolution turns into multiplication of Fourier coefficients on a finite abelian group.
For , set . Then Parseval identity boundsAlso, with , the nonnegative convolution is supported exactly on the modular set , andThe frequencies outside contribute in absolute value at most . For , the real part of each phase from is at least ; moreover and . Hence . We have proved the cyclic Bogolyubov lemma with explicit phase radius:The previous Bohr set argument gives a modular arithmetic progression of length at least . The remaining step is lifting a short modular progression to an integer progression. Every one of its elements has a unique integer lift belonging to . Successive lifted differences lie in and are congruent to the same residue. Since , at most one integer in this difference interval represents that residue, so every successive difference is the same. The lifts form a genuine integer arithmetic progression, not merely a modular one.
For sufficiently large , the constants can be absorbed into half the exponent. ThusNo attempt to optimize this absolute constant is needed.
A counterexample comes from dense integer sets with only logarithmic-length progressions. To show that itself need not have such a progression, choose a random subset by independent inclusion with probability . Its size has mean and variance at most , so Chebyshev's inequality shows . There are at most increasing arithmetic progressions of any fixed length , and each belongs to with probability . Taking makes the union bound at most . For large , some realization has at least elements and no length- arithmetic progression. Take a subset of exactly elements. It retains that avoidance. Such a set has no progression longer than , and therefore no progression of length for any fixed , eventually. Interpret the prescribed exact cardinality on values of where it is integral, or use its integer part.