gpt 6 astra pro attempt on Math-199

Polynomial Freiman-Ruzsa Conjecture for General Abelian Groups

Definitions and mathematical statement

For a nonempty finite set $A$ in an abelian group, define $A+A=\{a+a':a,a'\in A\}$ and suppose $|A+A|\le K|A|$, $K\ge1$. A centered convex coset progression of dimension $d$ has the form $P=H+\phi(C\cap{\mathbb{Z}}^d)$, where $H$ is a finite subgroup, $\phi:{\mathbb{Z}}^d\to G$ is a homomorphism, and $C\subset{\mathbb{R}}^d$ is a centrally symmetric convex body centered at zero. The assertion is that an absolute $C_0$ allows \[ A\subseteq\bigcup_{t\in T}(t+P),\quad |T|\le K^{C_0}, \quad d\le C_0\log K,\quad |P|\le K^{C_0}|A|. \] No bounded exponent or torsion-free restriction is placed on the ambient group, and the selected progression is convex rather than necessarily rectangular.

Formulation and concept references: [S1].

Short English statement

Can every finite set with small additive doubling in any abelian group be covered efficiently by translates of a low-dimensional convex coset progression?

Research attempt

Current frontier.

The general-group statement concerns convex coset progressions with absolute polynomial bounds, as in Sanders' formulation [S1]. The bounded-torsion breakthrough proves polynomial covering for groups of a fixed exponent $m$, with a bound of the form $(2K)^{O(m^3)}$ [E1], Theorem 1.1. The dependence on $m$ prevents a direct reduction of arbitrary groups by a large-modulus quotient. We investigate a geometric large-subset criterion and prove its equivalence, up to absolute polynomial losses, to the requested covering statement.

Attack route.

An entropy argument might find a large structured piece without constructing a cover immediately. A covering lemma can then transfer it to all of $A$, provided enlargement of its convex progression has controlled size. Alternatively, one might reduce a finitely generated ambient group modulo a large exponent and invoke the bounded-torsion theorem. The first route reduces the target to a precise large-piece statement; the second fails to retain an absolute exponent. We check both quantitatively, including the endpoint $K=1$.

Attempt.

Assume first $K>1$. Suppose there are absolute nonnegative constants $a,b,c$ and, for every set $A$ in the question, a subset $B\subseteq A$ and a translate of a centered convex coset progression \[ B\subseteq x+P,\qquad P=H+\phi(C\cap{\mathbb{Z}}^d) \] such that \begin{equation}\label{wsref-0-65-71-31-39-35-3a-63-68-75-6e-6b} |B|\ge K^{-a}|A|,\qquad |P|\le K^b|A|, \qquad d\le c\log K. \end{equation} This is the missing large-piece assertion, not an assumption known for all groups.

Choose a maximal set $T\subseteq A$ for which the translates $t+B$, $t\in T$, are pairwise disjoint. They are contained in $A+B\subseteq A+A$, so \[ |T||B|\le K|A|,\qquad |T|\le K^{a+1}. \] Maximality gives $A\subseteq T+B-B$: otherwise $a+B$ could be added. Also \[ B-B\subseteq P-P\subseteq P_2:=H+\phi(2C\cap{\mathbb{Z}}^d). \] To finish, the size of $P_2$ must be controlled without assuming $\phi$ injective or the progression proper.

Convex doubling lemma. For a centered symmetric convex body $C\subset{\mathbb{R}}^d$, \begin{equation}\label{wsref-0-65-71-31-39-35-3a-63-6f-6e-76-65-78} 2C\cap{\mathbb{Z}}^d\subseteq Z+(C\cap{\mathbb{Z}}^d),\qquad |Z|\le5^d \end{equation} for a finite $Z\subseteq2C\cap{\mathbb{Z}}^d$. Choose $Z$ maximal with $z-z'\notin C$ for distinct selected points. The sets $z+\tfrac12\operatorname{int}C$ are disjoint: overlap would put the difference in $\operatorname{int}C$. They are all contained in $\tfrac52 C$. Comparing volumes gives $|Z|\le5^d$. Maximality gives the covering, and the differences are integral. The case $d=0$ is immediate. No statement about the number of lattice points by volume is used, so thin bodies cause no error.

Mapping (eq195:convex) by $\phi$ and adding $H$ gives \[ |P_2|\le5^d|P|\le K^{b+c\log5}|A|. \] Thus $P_2$ is still a centered convex coset progression of dimension $d$, and at most $K^{a+1}$ translates cover $A$. Enlarging a single absolute constant to dominate $a+1,b+c\log5,c$ gives exactly the polynomial and logarithmic form in the target.

Conversely, a cover by at most $K^C$ translates of a progression of size at most $K^C|A|$ has one translate meeting $A$ in at least $K^{-C}|A|$ points. Taking that intersection as $B$ proves (eq195:chunk). Hence large-piece control and covering control are equivalent up to explicit absolute exponent losses. This is an independently written proof of a standard covering mechanism, not a claim that the missing large piece has been constructed.

Endpoint check. If $|A+A|=|A|$, translate so $0\in B=A-a$. Then $B\subseteq B+B$ and equal cardinality forces $B+B=B$. A finite set containing zero and closed under addition is a subgroup: for any $b$, repetition among its nonnegative multiples gives a positive multiple zero and hence its inverse in $B$. Therefore $A$ is one coset of a finite subgroup. The target holds with dimension zero, one translate, and $|P|=|A|$, respecting the exact $K=1$ bounds rather than replacing $K$ by $2K$.

Why a large finite quotient loses uniformity. The subgroup generated by $A$ is finitely generated, so it has form ${\mathbb{Z}}^r\oplus T_0$. One can choose a modulus $q$ making reduction injective on the finite set $A+A$; the quotient has exponent dividing $\operatorname{lcm}(q,\exp T_0)$. Applying [E1] then incurs an exponent depending on this number, not only on $K$. For the interval $A=\{0,\ldots,N-1\}\subset{\mathbb{Z}}$, $K=(2N-1)/N<2$, while injection on $A+A$ requires $q\ge2N-1$. Thus the theorem's $(2K)^{O(q^3)}$ loss is not bounded by $K^{C_0}$ with an absolute $C_0$. A small modulus avoids that exponent growth only by losing information that must then be recovered through a separate lifting theorem.

Stress test.

The covering uses $A+B\subseteq A+A$, not an unjustified estimate for an unrelated set. The volume argument proves a lattice translate covering, so it remains valid after a noninjective homomorphism and in the presence of torsion. No conversion to rectangular progressions is made: such a conversion could introduce a dimension-dependent loss outside the stated target. A high affine rank of $A$ is not itself a counterexample, since the target permits polynomially many translates; sparse simplexes illustrate why a full-span argument would be too strong. The bounded-torsion result remains a genuine major special case but supplies no modulus-uniform version by the quotient trick above.

Outcome.

Outcome: Conditional reduction. It suffices to find a polynomially large subset of $A$ inside a polynomial-size convex coset progression of dimension $O(\log K)$. The passage from that subset to the exact requested cover is proved with explicit losses, including $K=1$. The unresolved input is the absolute large-piece theorem for arbitrary groups; the fixed-exponent theorem and large-quotient reduction do not establish it.

Sources

• S1 Tom Sanders, "The structure theory of set addition revisited," Bulletin (New Series) of the American Mathematical Society, Volume 50, Number 1, January 2013, Pages 93-127; S 0273-0979$2012$01392-7, published electronically October 2, 2012. https://www.ams.org/journals/bull/2013-50-01/S0273-0979-2012-01392-7/S0273-0979-2012-01392-7.pdf. Catalogue formulation source.
• E1 W. T. Gowers, Ben Green, Freddie Manners and Terence Tao, Marton's conjecture in abelian groups with bounded torsion, arXiv:2404.02244v2, Theorem 1.1; explicit dependence on the ambient exponent. https://arxiv.org/html/2404.02244v2. Research reference; consulted 27 September 2026.