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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: Lemma 4 on p. 459, its proof on pp. 459--460. The edition read is identified on the source card.

Statement

Lemma 4 (p. 459), which the paper calls its key result. Let $G$ be a finite group generated by elements $g,h,k$, and let $A,B,C$ be the maps representing $g,h,k$ in a nontrivial irreducible real orthogonal representation of $G$. Call $(\alpha,\beta)$ attainable when, for some vector $w\ne0$, the quadrilateral $A(w),B(w),C(w),w$ has parameters $\alpha,\beta$, that is,

$$w=C(w)+\alpha\bigl(A(w)-C(w)\bigr)+\beta\bigl(B(w)-C(w)\bigr). \tag{1}$$

Then the attainable pairs are exactly the zero set of a polynomial $P(\alpha,\beta)$ with algebraic coefficients. Moreover, for every fixed $\alpha\ne0,1$, the polynomial $P$ viewed as a polynomial in $\beta$ is not identically zero.

Read depth. Claims checked: the statement, its hypotheses, the label and the page were read clause by clause against the print, and the proof sketch below was checked against the printed proof, including the slip recorded under Source correction.

Proof sketch

Pages 459--460. Equation (1) says that $w$ lies in the kernel of $L(\alpha,\beta)=\alpha A+\beta B+(1-\alpha-\beta)C-I$, so the attainable pairs are the zeros of $P=\det L$. The paper gets algebraic coefficients from the fact that a real representation of a finite group is equivalent to one with algebraic matrix entries, and the determinant does not change under equivalence. An alternative route needs no basis: the coefficients of $\operatorname{tr}(L^r)$ are integer combinations of traces of group elements, which are sums of roots of unity, and Newton's identities give $\det L$ as a rational polynomial in these traces.

For the second claim it suffices, for each $\alpha\ne0,1$, to find one $\beta$ at which no nonzero $w$ satisfies (1). A nonzero $w$ fixed by $A$, $B$ and $C$ is impossible, since its span would be an invariant line, forcing the irreducible representation to be one-dimensional and trivial. The choices are $\beta=(1-\alpha)/2$ when $0<\alpha<1$ or $\alpha>1$, and $\beta=\alpha$ when $\alpha<0$. In each case (1) can be rearranged so that one of the four vectors $w,A(w),B(w),C(w)$ is a convex combination, with positive weights, of the other three. All four have the same length, because the maps are orthogonal, so strict convexity of the Euclidean norm makes them all equal, which was just ruled out.

Source correction

After reducing the second claim to finding one $\beta$ at which $P$ is nonzero, the print (p. 460) says that this is equivalent to finding a $\beta$ "for which there exists non-zero $w$ such that $L(\alpha,\beta)(w)=0$". The equivalent condition is that no such nonzero $w$ exists, and the next sentence of the print, and its three cases, use that correct form. The slip is in all three versions read for the source card; this is a correction made here, not an erratum issued by the authors.

Dependencies

None beyond standard linear algebra and representation theory. Used by Theorem 1.

Bears on

• Problem 174: Lemma 4 is the algebraic step behind Theorem 1, which shows that certain cyclic quadrilaterals do not embed in finite transitive sets. It bears on the problem only through that separation of spherical from subtransitive sets and says nothing about which sets are Ramsey.