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Statement

Definition (p. 3). A finite subset SS of an abelian group is independent if ∑s∈Sk(s)s≠0\sum_{s\in S}k(s)s\ne0 for every integer-valued function kk on SS unless all summands k(s)sk(s)s are 00; equivalently, the sum ⨁s∈S⟨s⟩\bigoplus_{s\in S}\langle s\rangle is direct. The edge boundary ∂S(A)=∣{(a,s)∈A×S:a+s∉A}∣\partial_S(A)=|\{(a,s)\in A\times S:a+s\notin A\}| is defined on p. 1.

Theorem 2 (p. 3). Let AA and SS be finite, non-empty subsets of an abelian group, with SS independent. Put n=∣S∣n=|S| and d=min⁡{ord⁡s:s∈S}d=\min\{\operatorname{ord}s:s\in S\}. If ∂S(A)≤(1−γ)n∣A∣\partial_S(A)\le(1-\gamma)n|A| with a real γ∈(0,1]\gamma\in(0,1], then

∣A∣≥4(1−1/d)γn.|A|\ge4^{(1-1/d)\gamma n}.

The paper says the statement is to be read in the expected way when some or all elements of SS have infinite order, and that when all of them do, the conclusion reads ∣A∣≥4γn|A|\ge4^{\gamma n} (p. 3).

Sharpness (pp. 1-3). The abstract calls the constant 44 best possible; the paper does not say which example shows this. In Example 3 on the page of Theorem 1, the case t=2t=2 is the box [0,1]n⊆Cmn[0,1]^n\subseteq C_m^n (m>2m>2) with the standard generating set, which is independent with d=md=m; there γ=1/2\gamma=1/2 and ∣A∣=4γn|A|=4^{\gamma n}, while the theorem there gives only ∣A∣≥4(1−1/m)γn|A|\ge4^{(1-1/m)\gamma n}. The paper says Example 2 shows that the coefficient 1−1/d1-1/d is best possible for d=2d=2 and cannot be replaced by a number larger than log⁡3/log⁡4≈0.792\log3/\log4\approx0.792 for d=3d=3 (p. 3).

Open questions (p. 10). The paper asks whether, for every generating subset SS of a finite abelian group GG, the hypothesis ∂S(A)≤(1−γ)n∣A∣\partial_S(A)\le(1-\gamma)n|A| with n=rk⁡Gn=\operatorname{rk}G and real γ∈(0,1]\gamma\in(0,1] implies ∣A∣≥4(1−1/d)γn|A|\ge4^{(1-1/d)\gamma n}, with dd the least order of an element of SS. It also asks whether the coefficient 1−1/d1-1/d there can be improved, or dropped, when GG is homocyclic with exp⁡(G)≥5\exp(G)\ge5; it calls the case exp⁡(G)≤4\exp(G)\le4 settled by Theorem 1 and Example 2.

Source. Vsevolod F. Lev, On Isoperimetric Stability, Discrete Analysis 2018:14, 11 pp., doi:10.19086/da.3699: the definition and Theorem 2 on p. 3, the proof in Section 3 on pp. 6-9, the questions on p. 10. The edition read is identified on the source card.

Read depth. Claims checked: the definition, the statement, the reading for infinite orders and the sharpness remarks were read clause by clause on the printed pages. The proof (pp. 6-9) was read but not checked step by step.

Proof pointer

Pages 6-9. One may assume that S={s1,…,sn}S=\{s_1,\ldots,s_n\} generates the group; the general case follows by the coset decomposition used for Corollary 1. Compressing AA along each sis_i (pushing the part of AA in each ⟨si⟩\langle s_i\rangle-coset to an initial segment of that coset) keeps ∣A∣|A| and, by Claims 1 and 2 (p. 7), produces a set compressed with respect to SS without increasing ∂S(A)\partial_S(A). For a compressed set the boundary in the direction sis_i is the number of ⟨si⟩\langle s_i\rangle-cosets that meet AA minus the number contained in AA, equations (6)-(8). With the hypothesis, equation (9) then gives an average weight (number of non-zero coordinates with respect to SS) of at least (1−1/d)γn(1-1/d)\gamma n. Corollary 2 (p. 8), which carries Theorem 3 over to compressed sets through the coordinate map into Zn\mathbb Z^n, bounds the same average by 12log⁡2∣A∣\frac12\log_2|A|.

Dependencies

Theorem 3, through Corollary 2 of the same paper. The theorem supplies the first estimate of Theorem 4.

Bears on

  • Problem 963: every non-zero real has infinite order, so for finite non-empty A⊂RA\subset\mathbb R and finite non-empty S⊂R∖{0}S\subset\mathbb R\setminus\{0\} with SS independent (no non-trivial integer relation) and ∂S(A)≤(1−γ)∣S∣∣A∣\partial_S(A)\le(1-\gamma)|S||A| for a real γ∈(0,1]\gamma\in(0,1], the theorem, in its infinite-order reading, gives ∣A∣≥4γ∣S∣|A|\ge4^{\gamma|S|}. Independence is stronger than the dissociativity the problem asks about, and the theorem gives no bound on the problem's f(n)f(n).