Salem numbers and Pisot numbers from stars

Salem numbers and Pisot numbers from stars J.F. McKee, P. Rowlinson and C.J. Smyth October 17, 2005 Abstract We use trees to construct families of algebraic integers. These include certain families of Salem numbers and of Pisot numbers, constructed from stars. The family of Pisot numbers we construct forms a proper closed subset of the set of all Pisot numbers. We show that this subset contains Pisot numbers of arbitrary trace. In particular, we exhibit a Pisot number of trace 5 and degree 141
731
565
070
951  1991 Mathematics Subject Classification: 11R06 

To Professor Andrzej Schinzel on the occasion of his sixtieth birthday

1 Introduction.



A Pisot number is an algebraic integer θ 1  all of whose other conjugates have  modulus  1  A Salem number is an algebraic integer τ 1  all of whose other conjugates have modulus  1  with at least one conjugate of modulus 1  The set of all Pisot numbers, traditionally if somewhat confusingly denoted by S, was shown by Salem [Sa] to be closed in   The structure of the set of Salem numbers is not known; its derived set includes S, however, and it may be that its derived set is precisely S. There is an algorithm (due essentially to Dufresnoy and Pisot [DP] but developed by Boyd [Bo1], [Bo2], [Bo3]) for determining the structure of S in a finite interval. The algorithm applies to any such interval containing only a finite number of elements of S  It also treats intervals containing certain limit points of  S  In particular, for any δ 0  S  1  2 δ can be determined. We give here a method for constructing Salem numbers (Corollary 9) and Pisot numbers (Theorem 1). While the method is believed to be new in this form, 1

the Salem number construction seems to underlie the work of Cannon and Wagreich[CW] and of Parry [P]. In these papers, minimal polynomials of Salem numbers were constructed as denominators of growth series of Coxeter groups. The links between our different approaches are the associated Coxeter graphs. We thank David Boyd for pointing out these papers to us. We also show how our methods can be used to construct Pisot numbers of arbitrary trace (Theorem 2). Theorem 1 Let r 2 be an integer, and k be an integer with 0 a1  a2      ar k be integers all at least 2, with not both a1 a2 k 0  Then








(i) The only roots of the equation











k  r Also let 2 if r 2 and



 k 1 (A) z  are a certain Pisot number θ   a      a
 say, and its conjugates. Denote r k ai 1 z

∑

i 1

z ai

1 1

1

rk

r k

by U the set of all such Pisot numbers.

(ii) For 0  k



r the set

   θ   a      a
 a

Ur k :

1

rk

r k

2

i



belongs to the kth derived set of U 

 



(iii) For 0  k  r 1 the set Ur k is a subset of the interval  θmin r k  r but of no min smaller interval. Here θ2 0 1  3247.. is the real root of θ3 θ 1 0  while for r k 2 0

        1 θ  r 1  r 1  4k 2 Note that U    r  min rk

2

rr

(iv) The set U is a proper closed subset of S  We can also show 2



r 1





Theorem 2 The Pisot number θr k a1 

   ar


k

 has trace r 1 

1       

S

 



∑

S 
a1 ar gcd S 1

k

This trace can, with suitable choice of r and k  be made to take any integer value. In particular there are, for infinitely many D  examples of θr 0 of degree D and log D r and r  24000   for D trace  4  log log exp 2 3

2 D




 





 

Here, for a set A a1      ar of positive integers, and a subset S a k1      a ks of A  gcd S denotes gcd ak1      aks with gcd 0/ ∞  This shows that the plausible conjecture that there are no Pisot numbers of negative trace is false. This conjecture was only recently formally (and, as it turned out, rashly!) proposed by the third author [Sm1]. As a question it is, however, certainly older. An example of a Pisot number of trace 5 is given at the end of the paper.

 



 



2 Graphs and trees.




To any graph G on n vertices is associated the characteristic polynomial PG x of its adjacency matrix AG  Recall that AG ai j  where ai j 1 if there is an edge joining the ith and jth vertices, and 0 otherwise. In fact PG x xn txn 2     where t is the number of edges of G  The roots of PG are called the eigenvalues of G As A is symmetric, all zeros of PG are real. If G is bipartite, for instance when G is a tree, then PG x

1 nPG x and we can define

  



RG



 

      y : y P   y  1  y  y   n t  y
      n t  y  n 2

n

G

n 1

1

which we call the reciprocal polynomial of G  For a tree T with vertex set V and vertex v  V  we denote by T  the induced graph on V  v  Of course T  is in general a forest of trees Ti i 1      r say. A  star is a tree with at most one vertex of degree 2  We have the following results:



 

  T  T

Lemma 3 For a tree T with T 

1

 

PT x

r





∏ PTi x i 1

r

x

we have T   ∑ PTi  x  r



i 1

3

x

P

i





This is a special case of e.g. [CDS], p.78, Problem 9. As an immediate consequence Corollary 4 For a tree T

 

RT z



 

r

∏ RTi z i 1

In particular

 

r

1 z∑



z

i 1

  z 

RTi z RTi



Lemma 5 Let T n be the n-vertex path (so it has n 1 edges). Then zn 1 1 z 1






RT  n z





More generally, let a1  a2      ar all be integers 2 and T a1  with r arms of a1 1  a2 1      ar 1 edges. Then

   z  ∏ r

RT  a1 

ar

i 1



z ai 1 z 1





z



r




z ai 1 1 ai 1 z 1 

1 z∑ i



    ar 

be the star



This follows easily from Corollary 4. We also need the following beautiful Proposition 6 (Interlacing Theorem) [LT] Let G be an n-vertex graph with vertex set V and eigenvalues λ1 λ2    λn  and let G be the induced graph on V  v obtained from G by deleting the vertex v and its incident edges. Then the    λn 1 of G satisfy eigenvalues λ1 λ2



λ1

λ1




λ2

λ2



λn




1



λn 

Corollary 7 If T is a forest with k vertices of degree 2  then λk   PT x has at most k zeros 2  and RT z has at most k zeros 1 










1



2  so that

Proof. If k 0  T is a union of paths so that, from Lemma 5, RT is a product of cyclotomic polynomials, and any root ω of RT is on z 1  Hence all roots ω 1 ω of P 2 2 are in  

     G  Now assume k 0 and that the result is true for k 1  For a forest T having  k vertices of degree 2  remove one such vertex, giving a forest T  with at most k 1 such vertices. By the induction hypothesis, its kth eigenvalue λ k is  2  so,  by Prop.6, λk 1  2  Thus at most k eigenvalues of T are 2 and hence, using   z 1   z λ  at most k zeros of RT are 1 












4





Lemma 8 The star T a1      ar has at most one eigenvalue  exactly one eigenvalue 2 unless



r or r or r or r or r



3 and a1 3 and a1 3 and a1 3 and a1 4 and a1

   





2  In fact it has

a2 2  a 3 2 2  a2 3  3  a 3  6 2  a 2 a3 4 a2 a3 3 a2 a3 a4 2 

  

 









 



The first sentence follows from Corollary 7. For the second part, see [CR] or [BS].



Corollary 9 If T is star, not one of the exceptional stars of Lemma 8, then τ defined by  τ 1   τ λ1  is a Salem number.





1

Here λ1 is the largest eigenvalue of T 

3 A M¨obius sum. For the M¨obius function µ Lemma 10 We have M :



∑



 

µd

d ai for some i

Proof. Now by inclusion-exclusion

∑



d ai for some i

 

∑ ∑ µ  d  ∑

µd

  

as ∑ µ d



dn

0 if n



1



     1

1 if n



S



1  Then the identity ∑S

A

i j d gcd  ai a j

i d ai

S A gcd  S  1

 1 ∑



∑

∑

∑

µd



    

∑

i j k d gcd  ai a j ak

µd



S



S A gcd  S 1

0 gives the result.

5

1

S

  

1 1

A

 

Corollary 11 Suppose that any m of the ai have gcd have gcd 1  Then



M



r  r  2 1

1



1  but any m

    1  r  m

m



1 of the ai



In particular, for r even, m r even, m r odd, m r odd, m



r 2 r 2




r 1 2 r 3 2




M 1 M M M

  1
  1 
  1 
  1

r 2
r r 2 r 2 1
r r 2 r 1 2
r 1  r 1  2  r 3 2
r 1  r 1  2 

1 2 1 2









This follows from Lemma 10, along with some simple binomial coefficient identities (which are left as an exercise for the reader). The following is easily checked:




r Lemma 12 Given m  2r  put R m  and label the R m-element subsets of 1  2      r by j 1      R  Define, for i 1      r ai : ∏ p j where p j is the jth prime, and the product is taken over those j for which i belongs to the jth such subset. Then the ai satisfy the conditions of Corollary 11.















The construction given here is that of a 1- R  Rm  r m design, with blocks of size r ([C], p257), [vanLW]). Note that each ai is the product of Rm  r
mr 11  primes.





4 Proof of Theorem 1.

 

We first re-write RT  a1  ar as given in Lemma 5 to show the terms containing ar explicitly. We will then be able to see what happens as ar  ∞  We obtain




 
    z    z 1 ∏ z ai 1

r 1

R T  a1 where

i 1

ar

r



 1 σ
 z  z
1 σ  z : ∑   z 1 ar 1


z

r 1

s

s

i 1

6

ai 1 ai



1 σr




 1

1 z





(B)

Note that



  zσ  z  (C) We must now study the equation σ  z   1 : Lemma 13 Suppose that s 2 and a 2  i  1      s   Then all roots of σ  z   1 lie in z  1 except one. The exceptional root lies in  1  ∞   being equal to 1 only when s  a  a  2  Proof. Using (C) and z  1  z for z  1  we see that z  1 is the only possible root of σ  z   1 of modulus 1. Since σ  1   ∑
, we see that z  1 is in fact a root only for s  a  a  2   Now suppose that α is a possible root of σ  z   1 with α 1  but α not on  1  ∞  Then one can take a δ  0 small enough so that the disc z α  δ has no other roots of σ  z   1 in it, and intersects neither the unit circle nor  1  ∞   Let B  min
 1 σ  z  B  max
 1 σ  1 z  Then for n large enough so that  α δ B B  we have

z  1 σ  z   1 σ  1 z on z α  δ  Hence, applying Rouch´e’s Theorem to (B), we see that R      z  has the same number of roots in z α  δ as z  1 σ  z   for a  large enough. But, by Corollary 9, R     has at most one root of modulus 1  and that is  real and 1  Thus α is not a root of σ  z   1  and σ  z   1 has no roots in  z z  1  z    1  ∞  Finally, since σ  z decreases and tends to zero on [1  ∞  and σ  1  1 we see that σ  z   1 has exactly one root on [1,∞   We can now prove part (i) of Theorem 1. First note that (A) has no roots on

z  1 for k  0 except in the trivial case r  k  1  by the same argument as for the k  0 case in the proof of Lemma 13. We exclude this trivial case. Next, take   a δ 0 so that (A) has no zero in the annulus 1  z  1 δ  Then write  k 1  σ  z 1  z 1 ∑ 1 σ
 z (D) z z 
z 1 where a
     a are chosen large enough so that z 1 1  k 1  min σ
 z  max  ∑ z 
z 1 z  σs

1 z

s

s

i

1

s

2

ai 1 i ai

s

s

1

2

s

s

z α

s

δ

z α

s

δ

n

n

s

s

T a1

ar 1

ar




T a1

s

r

ar

s

s

s

s

s

r

r

r k

i r k 1


r k 1

k




r

z

1 δ


i r k 1


ai

z

7

1 δ


r k

ai



       


Then, using Rouch´e’s Theorem (with 1  z as the complex variable) and (D), we see  that (A) and σr z 1 have the same number of roots in z 1 δ  As σr z 1 has only one such root, so does (A). Next, we clear the denominator of (A) by multiplying through by the lcm, C z say, of the polynomials zai 1  z 1 i  1      r k , and also z if k 0  No proper factor of C z will suffice to clear the denominator of (A) since, as is easily checked, the residue of σ r k z at a primitive  dth root of unity ωd is (1 ωd ∑ a1i 0 for d 1  Hence, doing this, we obtain









i:d ai










a monic polynomial with one root θr k a1      ar k in z 1  and all others in z  1  If this polynomial were reducible, one factor would have all its roots in z  1  which is impossible. Hence the polynomial is the minimal polynomial of the Pisot number θr k  Part (ii) of Theorem 1 now follows easily, by another ‘Rouch´e’s Theorem’ argument, showing that θr k is the limit, as ar k 1  ∞  of the root θr 1 k 1 of k 1 σr k 1 z 1 z  To prove part (iii), first note that, for fixed z 1  za 1 1  za 1 1 z as a  ∞  This shows that the supremum of the elements of Ur k is taken when all ai  ∞  giving in the limit r z k kz 1  i.e. z r Conversely, the smallest element of Uk r is when all ai 2  i.e. zr 1k kz 1  z 12 r 1 r 1 2 4k  except for r 2  k 0  when, a1 a2 2 not being allowed, the smallest element occurs when a1 2  a2 3  z 1 1 z2z z1 1 1  z3 z 1 0 










 


























 
      






 











 






  

      a    
      tending to a limit, for each i either the sequence  a    remains bounded, or








n

n

To prove part (iv): first note that, for any sequence θr a1

r

n 12

n

n 12

i

becomes unbounded. If it remains bounded, we can take an infinite subsequence  n whose ai are all equal, say to ai  If the sequence becomes unbounded, we can  n take an infinite subsequence of the ai which tends to ∞   n To simplify the notation, suppose ai n 1 2  remains fixed and equal to ai for i 1      j and tends to ∞ for i j 1      r  Then

         θ    a      a
  θ
a  a   U
  This follows from the fact that, for fixed z 1   z
1    z 1  is an in∞  Now  k so that j  r  and creasing function of a , with limit 1  z as a r j  k  Hence this limit point belongs to U   Now consider any convergent sequence of Pisot numbers in U  Such a sequence eventually belongs to the interval  r 1  r  for some r so, by (iii), has an n 1

r

n

r

rr j

1

j

ai 1

i

i



rk

8

rr j

ai



infinite subsequence belonging to Ur k for some r and k  r Hence, by the above, the sequence contains an infinite increasing subsequence tending to an element of Ur k  This shows that U is closed, and also that it is well-ordered by   since U contains no monotonically decreasing convergent sequences. Since it is wellknown that this is not true for S – the sequence of Pisot numbers with minimal polynomials zn z2 z 1 1 being an example – we see that U is a proper subset of S  In a following paper it is planned to exploit a more general class of graphs than stars, which will give a wider class of Salem and Pisot numbers.







5 Proof of Theorem 2.






We have seen that the minimal polynomial of θr k a1      ar k is obtained by mul tiplying (A) by the lcm of the zai 1  z 1  and also by z if k 0  Now, this lcm is Cd z  where Cd is the dth cyclotomic polynomial. Now, as ∏





1 d ai for some i

 



 

is well-known ([HW] p239), Cd has trace µ d  from which we readily get

    r

trace θr k

∑

 

µd

1 d ai for some i



Hence, by Lemma 10, we get the trace claimed. Now, taking k 0  choosing the ai as in Lemma 12, and applying Corollary 11, we have, for r 2 mod 4  m r  2  trace θr 0 r 1 12
r r2  . This is  0 for r 0  and can be made arbitrarily large and negative. Then, for the same r and ai ’s,



 



trace θr

   r

k 1

kk




1 r 2 r 2



  

which, by appropriate choice of k 0  be chosen to take any value trace θ r 0  This shows that there are Pisot numbers of every trace. To obtain the examples of θr 0 with large negative trace as estimated in the theorem, we take r 2 mod 4 and the ai as above. Put D deg θr 0  R 2r  log R r log 2  2 log 1 log log R  3 log 3  Then 1 1 2 4

 



 

 



       

 1 r    T :  trace  θ    r 1 2 r 2 r0

9



3 2r 8 r



  

for r

14  as is easily checked using e.g. Maple. Also D









∑

 

φd

1 d ai for some i



1



∑ ai ∑ gcd  ai  a j 

∑ ai 

i



rpRR

i j



gcd ai  a j  ak

∑

i j k

where pR is the Rth prime. Now from [RS] p69, pR



R

  R       : f  r 

log D 

1

1

4



3




so that





say. Now, since  log xx 3 2 is an increasing function of x 5  and f r is an increasing function of r  log ff   rr 3 2 is also an increasing function of r and



log D log log D





3 2

   log f  r  

  1  R   logr logr log2 2  log  1r log 2  1 2r r log 2 1

f r



3 2



4 1




r log 2

  o  1     

1 R

3 2

4 1




 

2r

8 3  1 3  log 2 8  r 3  21T 1 o 1



D using (D). One can check that actually  loglog log D 3 2



4T for r



24000 

6 Example and remark.





Example of a Pisot number of negative trace. Take r 6 
r r2  20  and p1      p20 the first 20 primes, this time taken in any order. The ai are as in Lemma 12, so that each of the ai is a product of 10 of the p j ’s, with a1 the smallest of them. Then θr 0 is just smaller than 6 (in fact it is approximately 6 7 6 a1 ), has trace 5 and degree






 D 1

∑ ai i

 ∑ gcd  a  a  i

i j

j

∑

i j k

10

6  6 6 gcd  a  a  a   5 6 4 i

j

k



We chose the order of the p j ’s to nearly equalise the ai ’s and so get D almost as small as possible for this construction. We chose a1 a2 a3 a4 a5 a6















3  7  11  13  19  31  47  59  67  71 3  7  11  23  29  31  37  41  53  61 2  5  13  23  29  31  43  47  61  71 2  5  11  17  23  37  53  59  67  71 2  7  17  19  29  37  41  43  47  59 3  5  13  17  19  41  43  53  61  67













23  23  23  23  23  24 

332  425  528  671  720  053 

090  603  019  849  970  038 

268  762  817  950  960  881 

487 407 510 430 694 105

giving D 141  731  565  070  951  We do not claim that this is the smallest degree for which there is a Pisot number of negative trace! Salem numbers of large negative trace? In this paper we have constructed Pisot numbers as limits of Salem numbers. This contrasts with Salem’s classical construction of Salem numbers from Pisot numbers. He took Sn z : zn P z
P 1  z , where P z is the minimal polynomial of a Pisot number. Now if (after multiplying by an appropriate power of z) Sn were irreducible, then its trace would be the same as that of P Thus the difficulty in computing the trace of Salem numbers constructed in this way lies in factoring Sn . Such a factorization was carried out for the Salem polynomials RT  2 a2 a3 z in [Sm2], where examples of Salem numbers of trace 1 and all degrees 4 were found. Although it is easy to see that all factors but one must be cyclotomic, such factorizations seem to present considerable difficulties in general. This work was done while the first author was at the University of Edinburgh.

 

 





  

References [BS]

Bell, F.K. and Simi´c, S.K., A note on the second largest eigenvalue of star-like trees (to appear).

[Bo1]

D.W. Boyd, Pisot and Salem numbers in intervals of the real line, Math. Comp. 32 (1978), 1244-1260.

[Bo2]

D.W. Boyd, Pisot numbers in the neighbourhood of a limit point I, J.No. Theory 21 (1985), 17-43.

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[Bo3]

D.W. Boyd, Pisot numbers in the neighbourhood of a limit point II, Math. Comp. 43 (1984), 593-602.

[C]

P.J. Cameron, Combinatorics: topics, techniques, algorithms, C.U.P. 1994.

[CDS]

D. Cvetkovi´c, M. Doob and H. Sachs, Spectra of graphs – theory and application, 3rd edition Johann Ambrosius Barth Verlag 1995.

[CR]

D. Cvetkovi´c and P. Rowlinson, The largest eigenvalue of a graph – a survey, Linear and multilinear algebra 28 (1993), 45-66.

[CW]

J.W. Cannon and Ph. Wagreich, Growth functions of surface groups, Math. Ann. 293 (1992), 239-257.

[DP]

J. Dufresnoy and Ch. Pisot, Sur un ensemble ferm´e d’entiers ´ alg´ebriques, Ann. Sci. Ecole Norm. Sup. (3) 70 (1953), 105-133.

[HW]

G.H. Hardy and E.M. Wright, An introduction to the theory of numbers, 4th Edition, Oxford 1960.

[LT]

P. Lancaster and M. Tismenetsky, The theory of matrices, 2nd edition, Academic Press, Orlando, Florida, 1985.

[P]

W. Parry, Growth series of Coxeter groups and Salem numbers, J. Alg. 154 (1993), 406-415.

[RS]

J.B. Rosser and L. Schoenfeld, Approximate formulas for some functions of prime numbers, Ill. J. Math. 6 (1962), 64-94.

[Sa]

R. Salem, A remarkable class of algebraic integers. Proof of a conjecture of Vijayaragharan, Duke Math. J. 11 (1994), 103-107.

[Sm1]

C.J. Smyth, Problem 5.4, Proceedings of the 5th Conference of the Canadian Number Theory Association, Ottawa, August 1996.

[Sm2]

C.J. Smyth, Salem numbers of negative trace, University of Edinburgh preprint MS-011 (1997).

[vanLW] J.H. van Lint and R.M. Wilson, A Course in Combinatorics, C.U.P. 1992. 12

Mathematical Institute 24-29 St Giles, Oxford OX1 3LB, UK email: [email protected] Department of Computer Science and Mathematics, University of Stirling, Stirling FK9 4LA, UK email:[email protected] Department of Mathematics and Statistics, Edinburgh University, JCMB, King’s Buildings, Mayfield Road, Edinburgh EH9 3JZ, UK email: [email protected]

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