ARTICLE IN PRESS

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Journal of Mathematical Psychology 51 (2007) 319–324

www.elsevier.com/locate/jmp

On the existence of extremal cones and comparativeprobability orderings

Simon Marshall

Department of Mathematics, Princeton University, Fine Hall, Washington Road, Princeton, NJ 08544, USA

Received 1 May 2006; received in revised form 2 February 2007

Available online 10 July 2007

Abstract

We study the recently discovered phenomenon [Conder, M. D. E., & Slinko, A. M. (2004). A counterexample to Fishburn’s conjecture.

Journal of Mathematical Psychology, 48(6), 425–431] of existence of comparative probability orderings on finite sets that violate the

Fishburn hypothesis [Fishburn, P. C. (1996). Finite linear qualitative probability. Journal of Mathematical Psychology, 40, 64–77;

Fishburn, P. C. (1997). Failure of cancellation conditions for additive linear orders. Journal of Combinatorial Designs, 5, 353–365]—we

call such orderings and the discrete cones associated with them extremal. Conder and Slinko constructed an extremal discrete cone on a

set of n ¼ 7 elements and showed that no extremal cones exist on a set of np6 elements. In this paper we construct an extremal cone on a

finite set of prime cardinality p if p satisfies a certain number theoretical condition. This condition has been computationally checked to

hold for 1725 of the 1842 primes between 132 and 16,000, hence for all these primes extremal cones exist.

r 2007 Elsevier Inc. All rights reserved.

Keywords: Comparative probability ordering; Discrete cones; Quadratic residues

1. Introduction

A comparative probability ordering on a finite set X ofcardinality n is an order (any reflexive, complete andtransitive binary relation) on the power set 2X satisfyingthe following de Finetti’s axiom (deFinetti, 1931): for anyA;B;C � X

A � B () A [ C � B [ C whenever ðA [ BÞ \ C ¼ ;.

Without loss of generality, the set X can be assumed to bef1; 2; . . . ; ng with 1 � 2 � � � � � n.

Comparative probability orderings are normally studiedin terms of combinatorial objects associated with themcalled discrete cones (Conder & Slinko, 2004; Fishburn,1996, 1997). We may represent a subset A � X by ann-dimensional characteristic vector vA whose ith coordinateis 1 if i 2 A and 0 otherwise. Likewise the comparison A �

B can be represented by the vector vA � vB whosecoordinates will now lie in the set T ¼ f�1; 0; 1g. In thisway we may think of comparative probability orderings as

e front matter r 2007 Elsevier Inc. All rights reserved.

p.2007.05.003

ess: slm@math.princeton.edu

subsets of Tn by converting all comparisons to vectors,with de Finetti’s axiom ensuring that this correspondence iswell defined. The resulting objects, called discrete cones,proved to be a convenient tool for the study of comparativeprobability orderings.

Definition 1. A subset C of Tn is a discrete cone if thefollowing hold:

(i)

if x 2 C and y 2 C then xþ y 2 C providedxþ y 2 Tn;

(ii)

for x 2 Tn, either x 2 C or �x 2 C (not both forxa0);

(iii)

fe1 � e2; . . . ; en�1 � en; eng � C, where fe1; . . . ; eng is thestandard basis of Rn.

We define a discrete cone to be almost representable ifthere is a vector n with strictly positive, distinct entries suchthat x 2 C¼)x � nX0, and representable if, in addition,x � n ¼ 0()x ¼ 0. These two conditions correspond to theexistence of a probability measure on X which almost agreeor (respectively) agree with the comparative probabilityordering to which the cone corresponds. The concept of an

ARTICLE IN PRESSS. Marshall / Journal of Mathematical Psychology 51 (2007) 319–324320

almost representable ordering was introduced in Kraft,Pratt, and Seidenberg (1959).

A central problem in the study of comparative prob-ability orderings is deciding what conditions are requiredto ensure that such an ordering is representable. Condi-tions which are known to be necessary and sufficient arethe so-called cancellation conditions C1; . . . ;Ck; . . . ; devel-oped by Kraft et al. (1959). A discrete cone C (and thecorresponding comparative probability ordering) satisfiesthe mth cancellation condition Cm if there are no m non-zero vectors x1; . . . ;xm 2 C and positive integers a1; . . . ; am

such that

Xm

i¼1

aixi ¼ 0.

It is clear that any representable cone satisfies allcancellation conditions.

As conditions C1;C2;C3 are satisfied by any compara-tive probability ordering, Fishburn, (1996, 1997) defined afunction f ðnÞ as the smallest number k such that thecancellation conditions C4; . . . ;Ck ensure that a probabil-ity ordering on an n element set is representable. Conderand Slinko (2004) introduced a similar function gðnÞ whichis the smallest number k such that the cancellationconditions C4; . . . ;Ck ensure that an almost representableorder is representable. It is clear that gðnÞpf ðnÞ and it iseasy to show that f ðnÞpnþ 1 and gðnÞpn. Fishburnproved that f ð5Þ ¼ 4 and f ðnÞXn� 1. He conjectured thatf ðnÞ ¼ n� 1 for all nX5. Conder and Slinko (2004)confirmed for n ¼ 6 that f ð6Þ ¼ gð6Þ ¼ 5, but also theydisproved the hypothesis for n ¼ 7 by showing thatgð7Þ ¼ 7.

Definition 2. We will call a discrete cone C in Tn (and therespective comparative probability ordering) extremal if C

satisfies the cancellation conditions C1; . . . ;Cn�1 but is notrepresentable.

Thus, we may say that Conder and Slinko constructedthe first extremal almost representable cone in T7. The goalof this article is to prove gðnÞ ¼ n, when n is a primesatisfying a certain condition. More exactly, we prove

Theorem 1. Let p be a prime greater than 131. If

1þ

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi�1

p

� �p

s !p

� 1 ¼ aþ b

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi�1

p

� �p

s, (1)

where gcdða; bÞ ¼ p, then there exists an extremal almost

representable discrete cone in Tp and, in particular, gðpÞ ¼ p.

The odd primes satisfying (1) we will call optimus primes.The first few non-optimus primes are

3; 23; 31; 137; 191; 239; 277; 359; . . . .

Our calculations show that 1725 of the 1842 primesbetween 132 and 16,000 are optimus primes and Theorem 1is true for them. At this point, however, we know nothing

about the general distribution of such primes, or even ifthere are an infinite number of them.

2. The discrete cones and matrices

The idea behind all constructions of almost representablebut not representable cones is as follows. We choose aprobability measure p ¼ ðp1; . . . ; pnÞ on the set ½n� ¼f1; 2; . . . ; ng with all pi positive and distinct, such that thecorresponding ordering of the power set P½n� is not strict,and some subsets are tied, having equal probabilities. Thenwe break ties in a coordinated way, and with some luck wemay get a comparative probability ordering which is notrepresentable. In the language of cones, a tie means havinga pair of vectors �x in the cone, and breaking it meansthrowing one of them away.

Example 3. The non-representable comparative probabil-ity ordering � on P½5� constructed by Kraft et al. (1959)does not satisfy the condition C4, since it contains thefollowing comparisons:

f1; 3g � f2; 4; 5g; f2; 4g � f1; 5g; f2; 5g � f3; 4g; f4; 5g � f2g.

(2)

These are contradictory, which is reflected in the relationx1 þ x2 þ x3 þ x4 ¼ 0 for their respective vectors

x1 ¼ ð1;�1; 1;�1;�1ÞT ,

x2 ¼ ð�1; 1; 0; 1;�1ÞT ,

x3 ¼ ð0; 1;�1;�1; 1ÞT ,

x4 ¼ ð0;�1; 0; 1; 1ÞT .

This can be obtained from a representable but non-strictcomparative probability ordering with the measure p ¼124ð8; 7; 4; 3; 2Þ for which all pairs in (2) are tied.

Conder and Slinko (2004) clarified the conditions underwhich such a construction would work. Interchanging rowsand columns in their theorem, we have the following:

Theorem 2 (Conder and Slinko, 2004). Let X ¼ fx1; . . . ;xmg, mX4, be a system of non-zero vectors from Tn, such

thatPm

i¼1 aixi ¼ 0 for some positive integers a1; . . . ; am, and

such that no proper subsystem X 0 X is linearly dependent

with positive coefficients. Suppose further that the nm

matrix U having the vectors x1; . . . ;xm as its columns has the

property that pU ¼ 0 for some positive integer-valued

vector p ¼ ðp1; . . . ; pnÞ such that p14p24 � � �4pn40 andPni¼1 pi ¼ 1. Moreover,

spanfx1; . . . ; xmg \ Tn ¼ f�x1; . . . ;�xm; 0g. (3)

Let Cð�Þ be the discrete cone belonging to the weak

comparative probability ordering which arises from the

measure p, that is, Cð�Þ ¼ fx 2 Tn j p � xX0g. Then the

discrete cone

C0 ¼ Cð�Þnf�x1; . . . ;�xmg

ARTICLE IN PRESSS. Marshall / Journal of Mathematical Psychology 51 (2007) 319–324 321

corresponds to an almost representable comparative prob-

ability ordering which almost agrees with p, and satisfies Ci

for all iom, but does not satisfy Cm.

It should be noted that condition (3) is most demandingand very difficult to achieve.

For our construction of extremal cones we will use theabove theorem. We will construct a p p matrix U withcolumns u1; . . . ; up which has the following properties:

(i)

The only dependence between the columns ui isPpi¼1 ui ¼ 0 (and its scalar multiples).

(ii)

The only vectors in colðUÞ \ Tp are �ui and 0, wherecolðUÞ is the column space of U.

(iii)

None of the uk are of the form ei or �ei � ej, for iaj.

We claim that the conditions of the Theorem 2 will thenbe possible to satisfy. Indeed, the only thing to check is theexistence of a positive integer-valued vector p with theproperty pU ¼ 0. Since U has rank p� 1, such a vector psatisfying pU ¼ 0 is unique up to a scalar multiple and hasrational co-ordinates. From the conditions on the columnsof U we know that neither ei nor �ei � ej belongs toColðUÞ for and i and j. This implies that pia0 and jpijajpjj

for all iaj. If any of the co-ordinates of p are negative, wechange U multiplying the respective rows by �1. We thenknow that all co-ordinates of p are distinct, and so bypermuting the rows of U we may assume thatp14p24 � � �4pn40. Finally p can be normalised so thatPn

i¼1pi ¼ 1. Summarising we have the following:

Theorem 3. If a p p matrix U satisfying properties (i)–(iii)exists, then there exists an extremal discrete cone in Tp.

3. The construction of U

Our construction is based on the distribution ofquadratic residues modulo a prime p43. We use the

Legendre quadratic residue symbol ð ipÞ, for i ¼ 0; 1; 2; . . . ;

p� 1, where for convenience we take ð0pÞ to be 0. The idea is

to alter the vector of quadratic residue symbols

r ¼ 0;1

p

� �;

2

p

� �; . . . ;

p� 1

p

� �� �T

in the first two co-ordinates as follows:

q ¼ 1;1

p

� �� 1;

2

p

� �; . . . ;

p� 1

p

� �� �T

,

and then form a circulant matrix

Q ¼ ½q;Sq;S2q; . . . ;Sp�1q�

from q, where S is the standard matrix of the circular shiftoperator (which translates all coordinates one place down,with the last co-ordinate being placed at the top). Thequadratic residue nature of these vectors means that 1’sand �1’s are evenly distributed, making it hard for anynon-trivial linear combination to have co-ordinates all

from T, while at the same time the matrix still has enoughstructure that we may effectively prove that the desiredconditions hold.Further we will also need the matrix

R ¼ ½r;Sr;S2r; . . . ;Sp�1r�,

and we note that Q ¼ Rþ I � S, where I is the identitymatrix.Now finally we form U from Q ¼ ðqijÞ by subtracting 1

from q11 and adding 1 to q1p, and denote its columns byu1; . . . ; up. We note that ui 2 Tp for all i ¼ 1; 2; . . . ; p. Hereis an example of this construction for p ¼ 5:

U ¼

0 1 �1 �1 1

0 1 1 �1 �1

�1 0 1 1 �1

�1 �1 0 1 1

1 �1 �1 0 1

26666664

37777775.

Because Q and R are circulant, they lie in the one-generated subalgebra of the matrix algebra generated by S.Being in a one-generator subalgebra, any two circulantmatrices commute. If A is a circulant matrix with firstcolumn ða0; a1; . . . ; ap�1Þ

T then A ¼Pp�1

k¼0 akSk. Let the pthroots of unity over Q be 1;o;o2; . . . ;op�1. Then theseroots of unity are exactly the eigenvalues of S and, hence,the eigenvalues of A will be

li ¼Xp�1

k¼0akoki ði ¼ 1; . . . ; pÞ

(see also Ortega, 1987). We then know that the eigenvaluesof Q are given by

li ¼ 1� oi þXp�1k¼0

oki k

p

� �ði ¼ 1; . . . ; pÞ.

These eigenvalues can be rewritten as

li ¼ 1� oi þi�1

p

� �Xp�1k¼0

oki ki

p

� �

¼ 1� oi þi�1

p

� �Xp�1k¼0

ok k

p

� �.

Denote by t the Gauss sumPp�1

k¼1ðkpÞok. It is known

(Ribenboim, 2001) that t2 ¼ ð�1pÞp, so if iap (as lp ¼ 0) we

have

li ¼ 1� oi �

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi�1

p

� �p

s.

It should be noted that lia0 for iap, so Q has rank p� 1.Denote by W the space spanned by the eigenvectorscorresponding to non-zero eigenvalues of Q. Since theseeigenvectors are of the form

mi ¼ ð1;oi;o2i; . . . ;oðp�1ÞiÞT ; i ¼ 1; 2; . . . ; p� 1,

where o is a pth root of the unity, we have W ¼ n?, wheren ¼ ð1; . . . ; 1Þ is the eigenvector belonging to 0.

ARTICLE IN PRESSS. Marshall / Journal of Mathematical Psychology 51 (2007) 319–324322

Let us now define the integer span of the columns of U asfollows:

intspanfu1; . . . ; upg ¼Xp

i¼1

niui j ni 2 Z

( ).

We will now prove that U satisfies properties (i)–(iii) givenin Section 2. We wish to split the proof of (ii) into twosmaller statements, the first that the only integer vectors incolðUÞ \ Tp are those of the form

Ppi¼1niui with ni 2 Z, and

the second that there are no vectors of this form in Tp otherthan �u1; . . . ;�un and 0. They will be proved in thefollowing two lemmata.

Lemma 1. If the condition of Theorem 1 is satisfied,

colðUÞ \ Tp intspanfu1; . . . ; upg.

Proof. We will make use here of the natural homomorph-ism

f : Z�!Zq ¼ Z=qZ,

for a given prime q, and denote the image of a under f bya. This mapping can be extended in an obvious way to amapping of integer vectors, or integer matrices, and theimage of a vector u or matrix M under this mapping will besimilarly denoted u or M, respectively.

Assuming the contrary of the lemma, there must be someai; bi 2 Z, for i ¼ 1; 2; . . . ; p, such that

Xp

i¼1

ai

bi

ui 2 Tp and u ¼Xp

i¼1

ai

bi

uieintspanfu1; . . . ; upg.

(4)

Since u1 þ � � � þ up ¼ 0 we may always obtain a relation (4)with ai ¼ 0 for an arbitrary i. It is clear that ua0 and afterrepresenting u in the form

u ¼Xp

i¼1

ni

nui,

where gcdðn1; . . . ; npÞ is relatively prime to n, we mayassume that it is not true that n1 ¼ � � � ¼ np (otherwiseu ¼ 0). As n41, let q be any prime divisor of n. Then

Xp

i¼1

niui 2 qZp,

where at least one ni is not divisible by q, sincegcdðn1; . . . ; npÞ is relatively prime to n. Hence

Xp

i¼1

niui ¼ 0,

and such relations may be obtained with ni ¼ 0 forarbitrary i. Therefore, any p� 1 element subset offu1; . . . ; upg must have a linear dependency. As a result,the determinant of any principal minor of U is 0, orequivalently the determinant of any principal minor of U isdivisible by q.

Because Q has row sum and column sum both equal tothe zero vector, the formula describing how the determi-

nant changes under row and column operations impliesthat all principal minors of Q must have determinant �D

for some D. If q is a prime dividing the determinants of allprincipal minors of U, it must also divide D as Q and U

share some principal minors. The determinant of the ði; 1Þstprincipal minor of U will be the determinant of thecorresponding principal minor of Q plus the determinantof the matrix V i obtained by removing from Q rows 1 andi, and columns 1 and p. This implies that q divides detðV iÞ

for all i. Therefore, Q must have nullity at least 2, andbecause the sum of the rows of Q is 0 it will also have a setof p� 2 dependent columns. We will now show that thisleads to a contradiction.To show that Q has no p� 2 columns with a dependency

modulo any prime dividing D, let us write Q ¼ Rþ I � S,where I is the identity matrix and S is the standard matrixof the shift as before. We wish to show that W ¼ ðR�

I þ SÞðRþ I � SÞ has no p� 2 dependent columns, and,hence, Q ¼ Rþ I � S does not have them either. Becauseany two circulant matrices commute, RðI � SÞ ¼ ðI � SÞR

and W ¼ R2 � ðI � SÞ2.We may now calculate R2 by expanding the identity

t2 ¼ ð�1pÞp, where t is the Gauss sum mentioned previously.

We have

�1

p

� �p ¼ t2 ¼

Xp�1i¼1

i

p

� �oi

!2

¼Xp�1n¼0

onX

iþj�n

i

p

� �j

p

� �,

(5)

where the congruence in the subscript is modulo p. Since

Xi��j

i

p

� �j

p

� �¼Xp�1i¼1

�i2

p

� �¼�1

p

� �ðp� 1Þ, (6)

formula (5) implies

�1

p

� �¼Xp�1n¼1

onX

iþj�n

i

p

� �j

p

� �

or

��1

p

� �Xp�1n¼1

on ¼�1

p

� �¼Xp�1n¼1

onX

iþj�n

i

p

� �j

p

� �.

Because fongp�1n¼1 is linearly independent over Q, we must

haveXiþj�n

i

p

� �j

p

� �¼ �

�1

p

� �(7)

for all na0.The ðijÞth entry of R2 is equal to

Xp

k¼1

i � k

p

� �k � j

p

� �¼

Xmþn�i�j

m

p

� �n

p

� �.

Therefore, due to (6) and (7), R2 has entries �ð�1pÞ

everywhere except for ð�1pÞðp� 1Þ on the main diagonal.

Therefore, R2 ¼ �ð�1pÞJ þ ð�1

pÞpI , where I is the identity

matrix and J is the matrix whose entries are all 1.

ARTICLE IN PRESSS. Marshall / Journal of Mathematical Psychology 51 (2007) 319–324 323

Because W ¼ �ð�1pÞJ þ ð�1

pÞpI � I þ 2S � S2 is circu-

lant, we may calculate the eigenvalues of W in thealgebraic closure Zq of Zq in a similar way as for circulantmatrices in Q. We first assume that q is different from p.Then if we let 1;p; p2; . . . ;pp�1 denote the solutions to xp ¼

1 in Zq we find that the eigenvalues of the circulant matrixwith first column ða0; a1; . . . ; ap�1Þ

T will be

li ¼Xp�1k¼0

akpkiq ði ¼ 1; . . . ; pÞ.

The eigenvalues of W are therefore li ¼ �ð�1pÞPp�1

k¼0 pkiþ

ð�1pÞp� 1þ 2pi � p2i, which when iap can be rewritten as

li ¼ ð�1pÞp� 1þ 2pi � p2i (as we know the sum of all

solutions to xp � 1 ¼ 0 is 0). We know that the matrix W

has nullity at least 2 so li ¼ 0 for some ipp, i.e. there mustbe some y ¼ pj such that

y2 � 2y��1

p

� �pþ 1 ¼ 0. (8)

We first consider the case q ¼ 2. Because we assumed thatpaq we know that p is odd, so (8) reduces to

y2 ¼ 0,

which contradicts our assumption that y satisfied yp¼ 1.

We assume from here on that q is odd. Suppose that somep� 2 columns of W have a linear dependence. Write thedependency in the form

Ppi¼1 niwi ¼ 0 where ni 2 Zq and

two of the ni are 0. The ith row of W gives the followingequation between the n1; . . . ; np

��1

p

� �Xp

i¼1

ni þ�1

p

� �pni � ni þ 2niþ1 � niþ2 ¼ 0, (9)

and by taking the difference of the equations correspond-ing to two consecutive rows we have the equation

ni � 3ni�1 þ 3��1

p

� �p

� �ni�2 þ

�1

p

� �p� 1

� �ni�3 ¼ 0.

(10)

We may consider the pth row and 1st row as consecutiverows as well and in the recurrence relation (10) we canconsider indices mod p, if we define niþp ¼ ni. When weconsider this as a recurrence relation, the characteristicpolynomial of the relation factors in Zq as

ðx� 1Þðx2 � 2x��1

p

� �pþ 1Þ ¼ ðx� 1Þðx� yÞðx� aÞ,

where y is as in (8). We recap that yp¼ 1. Then

ni ¼ Aai þ Byiþ C, as it may be easily shown that 1, y

and a are all different. If we suppose Aa0, the requirementthat niþp ¼ ni forces ap ¼ 1. Assume that both a and y areof the form pi and pj, respectively. Completing the squarein (8) we have

pi ¼ 1þ b and pj ¼ 1� b,

where b satisfies b2 ¼ ð�1pÞp. Raising both of these

equations to the power of p we obtain

ð1þ bÞp � 1 ¼ 0 and ð1� bÞp � 1 ¼ 0. (11)

Expanding these numbers as c� db with c; d 2 Zq, the factb is non-zero implies c and d are both 0. It should be notedthat if b is not in Zq we have c ¼ d ¼ 0 from only one ofthese equations. In general, however, b may lie in Zq so theequation cþ db ¼ 0 does not imply c ¼ d ¼ 0, and weneed both equations to ensure this.We may now construct a homomorphism

j:Z

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi�1

p

� �p

s !�!ZqðbÞ

defined by

j: xþ y

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi�1

p

� �p

s7�!xþ yb.

This may be routinely verified to be a homomorphism.Due to (11) we must have

j 1þ

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi�1

p

� �p

s !p

� 1

!¼ aþ bb ¼ 0,

from which, as noted above, both a and b in (1) must bedivisible by q, which contradicts gcdða; bÞ ¼ p.In the case A ¼ 0 we have ni ¼ Byi

þ C. If ni ¼ nj ¼ 0for some distinct i and j we have Bþ Cyi

¼ Bþ Cyj¼ 0

which implies B ¼ C ¼ 0. However, this contradicts theassumption that the dependence was non-trivial.We now consider the case q ¼ p. The recurrence relation

(9) becomes

��1

p

� �Xp

k¼1

ni � ni þ 2niþ1 � niþ2 ¼ 0, (12)

and the recurrence relation (10) in this case has character-istic polynomial ðx� 1Þ3. The latter has solutionni ¼ Aþ Bi þ Ci2. Firstly, we note that

Xp

k¼1

ni ¼Xp

k¼1

ðAþ Bk þ Ck2Þ ¼ 0.

Indeed, since pa2, we havePp

k¼1ðAþ BkÞ ¼ pAþ12pðpþ 1ÞB ¼ 0. Also if b is any primitive element of Zp,then bp�1

¼ 1, and

Xp

k¼1

Ck2¼ C

Xp�2i¼0

b2i¼ C

b2p�2� 1

b2 � 1¼ 0,

where b2a1, because p43. Therefore, (12) becomes

ni � 2niþ1 þ niþ2 ¼ 0,

and, substituting here ni ¼ Aþ Bi þ Ci2, we have 2C ¼ 0.Therefore, ni ¼ Aþ Bi, and so if ni takes the value 0

twice we must have A ¼ B ¼ 0, and ni must be identically0, which implies the dependence is trivial. Therefore, there

ARTICLE IN PRESSS. Marshall / Journal of Mathematical Psychology 51 (2007) 319–324324

are no p� 2 dependent columns of Q modulo q for anyqjD, including q ¼ p. &

This gives us (ii) when combined with the following:

Lemma 2. For p4131 satisfying the conditions of Theorem 1,

intspanfu1; . . . ; upg \ Tp ¼ f�u1; . . . ;�up; 0g.

Proof. Recall that li are the eigenvalues of Q, and W ¼

ð1; . . . ; 1Þ? is the subspace spanned by eigenvectors of Q

corresponding to non-zero eigenvalues. Because jlijXffiffiffipp�

2 for iap, all v 2W satisfy kQvkXðffiffiffipp� 2Þkvk: In general

we may estimate kQvk from below as

kQvkXðffiffiffipp� 2Þkwk, (13)

where w ¼ projW ðvÞ.If some k ¼ ðk1; . . . ; kpÞ 2 Z

p satisfies Uk 2 Tp we havekUkkp ffiffiffi

pp

. Denote projW ðkÞ by s ¼ ðs1; . . . ; spÞ. BecausekUkk and kQkk differ by at most jk1 � kpj by the triangleinequality, we may combine this with 13 to obtainffiffiffi

ppþ jk1 � kpjXð

ffiffiffipp� 2Þksk. (14)

The first and pth entries of s differ by k1 � kp, so the sumof their absolute values is at least jk1 � kpj. By thearithmetic mean–quadratic mean inequality this impliess21 þ s2pX

12jk1 � kpj

2, and

kskX1ffiffiffi2p jk1 � kpj.

Combining this with 14 gives

ffiffiffippþ jk1 � kpjXð

ffiffiffipp� 2Þ

1ffiffiffi2p jk1 � kpj

or ffiffiffiffiffi2pp

ðffiffiffipp� 2�

ffiffiffi2pÞXjk1 � kpj.

For p4131 this implies jk1 � kpjp1. We may use asimilar argument to show that kprojNðkÞkX

1ffiffi2p jki � kjj for

any i and j, so

ffiffiffippþ 1X

1ffiffiffi2p ð

ffiffiffipp� 2Þjki � kjj

(the 1 here arising from the estimate jk1 � kpjp1). Forp4100 this implies jki � kjjp1 as before. We may there-fore assume that some of the ki are 0 and the rest 1. If thereare m and ‘ of each, respectively, s will have m entries equalto �l

pand ‘ equal to m

p, and, since mþ ‘ ¼ p,

ksk2 ¼m2‘ þ ‘2m

p2¼

m‘

p¼

mðp�mÞ

p.

Since for m ¼ 1 and p� 1 we obtain vectors of the form�ui, we may assume that 2pmpp� 2. The minimumvalue of ksk2 will be at m ¼ 2 and p� 2, where it is 2ðp�2Þ

p.

Therefore, if there is to be a non-trivial vector in Tp, which

is a linear combination of the ui’s, by (14) we must have

ðffiffiffipp� 2Þ

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi2ðp� 2Þ

p

sp

ffiffiffippþ 1,

which can be shown implies po100. Therefore, for allp4131, no non-trivial integral linear combinations of theui are contained in Tp. &

We may now check properties (i) and (iii) are satisfied.Because the bottom left principal minor of U is the same asthat of Q, and it is known that this minor has non-zerodeterminant, U has rank p� 1. Therefore,

Ppi¼1ui ¼ 0 is

the only dependence among the columns of U. As (iii) istrivial, U satisfies all the requirements for the existence of amaximal cone.

4. Conclusion

Conder and Slinko (2004) conjectured that gðnÞ ¼ n forall nX7 but this hypothesis remains open. However, webelieve that gðnÞ ¼ n for all sufficiently large n for a numberof reasons. First, the construction used here may be variedin a number of ways, so that even for non-optimus primesit is likely that we may find a matrix of the desired type.Second, we think that with some work our ideas could beextended to numbers with no small prime factors. Third,computational checking of the matrices used here in caseswhere the primes are non-optimus has verified that theconstruction works for all primes between 7 and 23, and sowe believe that this construction works for all primes.More general computational investigation has in factproven that gðnÞ ¼ n for all n between 7 and 12.We note that it is also not known whether or not f ðnÞ can

be greater than gðnÞ.

Acknowledgment

We would like to thank Sam McCall for conducting acomputational check of the condition for primes up to16,000, and Marston Conder for checking that the matriceswork for primes between 7 and 23, and that gðnÞ ¼ n for7pnp12, as detailed above.

References

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