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Source: Imre Leader, Paul A. Russell and Mark Walters, Transitive sets and cyclic quadrilaterals, Journal of Combinatorics 2 (2011), no. 3, 457--462: Theorem 1 and the remark after it on p. 458, the proof on pp. 460--461. The edition read is identified on the source card.

Statement

Theorem 1 (p. 458, quoted). "Let $x$, $y$, $z$ and $w$ be four distinct points lying on a circle such that

$$w=z+\alpha(x-z)+\beta(y-z),$$

where $\alpha\ne1$ and $\beta$ is transcendental over $\mathbb Q(\alpha)$. Then $xyzw$ does not embed into a transitive set."

Here a set is transitive when its isometry group acts transitively on it (p. 457), and the transitive sets meant are finite ones in a Euclidean space of any dimension: the abstract states the aim "in any dimension", and the proof works in the finite symmetry group of the transitive set. The numbers $\alpha,\beta$ are the quadrilateral's parameters in the sense of the parameter page; since three distinct points of a circle are not collinear, they are unique.

Remark after the theorem (p. 458, unlabeled). The paper states that the condition $\alpha\ne1$ is necessary: for every $\beta$ there is a trapezium with parameters $1$ and $\beta$ that embeds in a transitive set with symmetry group $D_8$. It also notes that the cyclic hypothesis is redundant, since every quadrilateral embedding in a transitive set is cyclic.

Read depth. Claims checked: the statement, its hypotheses, the remark, the label and the page were read clause by clause against the print. The proof sketch below was checked against the printed proof and the cited lemma.

Proof sketch

Pages 460--461. Suppose the quadrilateral sits in a finite transitive set $T$. Put the centroid of $T$ at the origin, so that the symmetries of $T$ act as a finite orthogonal group, and pick symmetries sending $w$ to $x$, $y$, $z$; let $H$ be the group they generate. The parameter relation becomes a linear relation among $w$ and its three images. The parameter $\alpha$ is not $0$, since $\alpha=0$ would put three distinct points of the circle on a line. Splitting the space into irreducible $H$-summands preserves the relation, and some summand carries a nontrivial projected quadrilateral; that summand's representation is nontrivial. On it, Lemma 4 gives a polynomial with algebraic coefficients that vanishes at $(\alpha,\beta)$ and is not identically zero in its second variable. So $\beta$ is algebraic over the field generated by $\alpha$ and the algebraic numbers, hence over $\mathbb Q(\alpha)$, against the hypothesis.

Two steps the print leaves brief are supplied here. The restricted symmetries form a finite group because a linear isometry fixing a spanning set pointwise is the identity, so they inject into the permutations of $T$. And algebraicity passes down to $\mathbb Q(\alpha)$: the finitely many algebraic coefficients of the polynomial generate a finite extension $K$ of $\mathbb Q$, so $\beta$ is algebraic over $K(\alpha)$, which is algebraic over $\mathbb Q(\alpha)$.

Dependencies

Parameters, projections and transitive spheres and Lemma 4 of the same paper.

Bears on

• Problem 174: the problem asks for a characterization of the Ramsey sets. Theorem 1 gives a sufficient condition for a cyclic quadrilateral, which is a spherical set, not to embed in any finite transitive set. It says nothing about whether such a quadrilateral is Ramsey. Together with Corollary 2, which exhibits quadrilaterals meeting the condition, it separates the spherical sets from the subtransitive ones, the two classes that the conjectures of Graham and of Leader, Russell and Walters propose as the Ramsey sets.