aa r X i v : . [ m a t h . N T ] D ec SALEM NUMBERS AND PISOT NUMBERS VIA INTERLACING
JAMES MCKEE AND CHRIS SMYTH
Abstract.
We present a general construction of Salem numbers via rationalfunctions whose zeros and poles mostly lie on the unit circle and satisfy aninterlacing condition. This extends and unifies earlier work. We then considerthe ‘obvious’ limit points of the set of Salem numbers produced by our theo-rems, and show that these are all Pisot numbers, in support of a conjectureof Boyd. We then show that all Pisot numbers arise in this way. Combiningthis with a theorem of Boyd, we produce all Salem numbers via an interlacingconstruction. Introduction A Pisot number is a real algebraic integer θ >
1, all of whose other (algebraic)conjugates have modulus strictly less than 1. A
Salem number is a real algebraicinteger τ >
1, whose other conjugates all have modulus at most 1, with at leastone having modulus exactly 1. It follows that the minimal polynomial P ( z ) of τ is reciprocal (i.e., z deg P P (1 /z ) = P ( z )), that τ − is a conjugate of τ , that allconjugates of τ other than τ and τ − have modulus exactly 1, and that P ( z ) haseven degree. The set of all Pisot numbers is traditionally denoted S , with T beingused for the set of all Salem numbers.In [17], we constructed Salem numbers via rational functions associated to cer-tain rooted trees (the quotients of rooted Salem trees ). In this paper we abstractthe essential properties of these rational functions, and give a much more generalconstruction of Salem numbers (Theorems 3.1, 5.1 and 5.2) via rational functionswhose zeros and poles mostly lie on the unit circle and satisfy an interlacing con-dition. In addition to extending the work of [17], this also extends the interlacingconstruction of [16]. We then consider the ‘obvious’ limit points of the set of Salemnumbers produced by our theorems, and show that these are all Pisot numbers(Theorems 4.2 and 5.3). This supports a conjecture of Boyd [4, p. 327]. We thenshow that all Pisot numbers arise in this way (Theorem 6.4). Combining this with atheorem of Boyd, we show that all Salem numbers can be produced via interlacing.We conclude the paper with some applications to the study of small Salem numbersand negative-trace elements of S or T .It is our hope that these ideas will lead to further improvements in our under-standing of the set of Salem numbers, and may give a way to attack some outstand-ing problems: (i) is there a least Salem number, and, if so, what is it? (ii) is theset of Salem numbers below (say) 1 . f ( z ) is a monic polynomial with integer coefficients having a simple real Mathematics Subject Classification. root θ >
1, such that all roots other than θ have modulus strictly less than 1, andthe constant term of f ( z ) is not 0, then f ( z ) is irreducible, and is therefore the min-imal polynomial of θ (if f ( z ) split into two nontrivial factors, then the factor thatdoes not have θ as a root would have as its constant term something that on the onehand is a nonzero integer, and on the other hand is a product of numbers all withmodulus strictly less than one, which is absurd). For Salem numbers, the analogousstatement is not as pleasant: if g ( z ) is a monic polynomial with integer coefficients,having a simple real root τ >
1, such that all the other roots of g ( z ) have modulusat most one, with at least one having modulus equal to 1, and if the constant termof g ( z ) is not zero, then g ( z ) = t ( z ) u ( z ), where t ( z ) is the minimal polynomial of τ ,and u ( z ) is a cyclotomic polynomial (for us, following [4], meaning simply that allits roots are roots of unity: it need not be irreducible). It is the possibility that u ( z )might not equal 1 that renders explicit constructions of the minimal polynomials ofSalem numbers more difficult. For Pisot numbers it is enough to find a polynomialthat has all its roots in the right place; for Salem numbers one also has to dealwith the possibility of cyclotomic factors. A further annoyance is that t ( z ) mighthave degree 2, in which case one has that τ is a reciprocal quadratic Pisot numberrather than a Salem number.With these thoughts in mind, it is convenient to define a Pisot polynomial to bea polynomial of the form z k f ( z ), where k ≥ f ( z ) is the minimal polynomialof a Pisot number. And we define a Salem polynomial to be a polynomial of theform t ( z ) u ( z ), where u ( z ) is a cyclotomic polynomial, and t ( z ) is either the minimalpolynomial of a Salem number or is the minimal polynomial of reciprocal quadraticPisot number.The plan for the remainder of the paper is as follows. In § § § § § § §
9. Other interlacing constructions have appeared in [6] (Proposition4.1), [11], and [16]. For an encyclopaedic account of real interlacing, see [8].We use T to denote the unit circle, T = (cid:8) z ∈ C (cid:12)(cid:12) | z | = 1 (cid:9) .2. Flavours of interlacing
Several variants of interlacing will be seen to arise naturally as we study Salemnumbers. We are concerned with interlacing on the unit circle, but the differentflavours of interlacing are perhaps most easily understood when one moves to thereal line via a Tchebyshev transformation. In § §§ § ALEM NUMBERS AND PISOT NUMBERS 3
Moving to the real world.
Our ultimate objective is to understand Salemnumbers and Pisot numbers, and these are firmly rooted in the world of complexnumbers. We shall give constructions that involve reciprocal polynomials. More-over most (perhaps all) of their roots will be in T , and other roots will be real andpositive. It will be extremely convenient for the proofs to transform such polyno-mials to totally real polynomials. The transformation that we shall use is(2.1) x = √ z + 1 / √ z . It is a matter of historical accident (growing out of [14], where this particulartransformation was essential) that this variant of the Tchebyshev transformationis used rather than the more familiar x = z + 1 /z , which would serve just as well,but with many small differences in detail. In applying (2.1), a fixed branch of thesquare-root is used throughout the right-hand side, but since there is a choice ofbranch we generally find two possible values of x . If z ∈ T , or if z is real, then thecorresponding one or two values of x are real.The transformation (2.1) is generally a 2-to-2 map, with a reciprocal pair z ,1 /z mapping to a pair x , − x . The exceptions are important for us: the singlepoint z = − x = 0, and the single point z = 1corresponds to the pair x = 2, x = −
2. The inverse correspondence involves solvinga quadratic equation, but we shall never have need for it explicitly.2.2.
CC-interlacing.
Suppose that P ( z ) and Q ( z ) are coprime polynomials withinteger coefficients, and with positive top coefficients. We say that Q and P satisfythe CC-interlacing condition , or that
Q/P is a
CC-interlacing quotient if: • P and Q have all their roots in T ; • all their roots are simple; • their roots interlace on the unit circle, in the sense that between every pairof roots of P ( z ) there is a root of Q ( z ), and between every pair of roots of Q ( z ) there is a root of P ( z ).Extending to real coefficients, one recovers the circular interlacing condition of [16].If P and Q satisfy the CC-interlacing condition, then they must have the same de-gree. Moreover, both 1 and − P and Q isa reciprocal polynomial; the other is antireciprocal ( z − T , but need not be cyclotomic since they need not be monic.For an example (derived from the quotient attached to ˜ E (8) in [17, p. 220]) towhich we shall return later, take(2.2) P ( z ) = ( z − z + 1)( z + z + 1)( z + z + z + z + 1) ,Q ( z ) = z + z − z − z − z + z + 1 . Thus Q ( z ) is the thirtieth cyclotomic polynomial, and P ( z ) is the product of thefirst, second, third and fifth cyclotomic polynomials. The roots of P and Q interlaceon the unit circle, as shown in Figure 1: Q/P is a CC-interlacing quotient.Our definition is symmetric in P and Q : if Q/P is a CC-interlacing quotient,then so is
P/Q .Note that the definition of the CC-interlacing condition does not require either P or Q to be monic. When both are monic, then by a theorem of Kronecker [10] JAMES MCKEE AND CHRIS SMYTH ✫✪✬✩rr rr rrrr ❜❜❜❜❜❜ ❜ ❜
Figure 1.
CC-interlacing. The roots of ( z − z + 1)( z + z +1)( z + z + z + z + 1) [ • ] interlace on T with those of z + z − z − z − z + z + 1 [ ◦ ]. ✫✪✬✩❜ ❜❜❜rr r r Figure 2.
CS-interlacing with simple roots. The roots of Q =( z − z − z + 1) [ ◦ ] interlace on T \{ } with those of P =( z + z + 1)( z − z + 1) [ • ].they are cyclotomic. In this case, all interlacing examples have essentially beenclassified by Beukers and Heckman [2].2.3. CS-interlacing.
Now we turn to another flavour of interlacing, where onepolynomial has all its roots in T , and the other has all but two roots in T , withthese two roots being θ and 1 /θ for some real θ >
1. Here “CS” is short for“cyclotomic-Salem”, with the same caveat as before that the polynomials need notbe monic. One will be reciprocal, and the other will be antireciprocal.Suppose that P ( z ) and Q ( z ) are coprime polynomials with integer coefficients,and with positive top coefficients. We say that P and Q satisfy the CS-interlacingcondition and that
Q/P is a
CS-interlacing quotient if: • P is reciprocal, and Q is antireciprocal; • P and Q have the same degree; • all the roots of P and Q are simple, except perhaps at z = 1; • z − | Q ; • Q has all its roots in T ; • P has all but two roots in T , with these two being real, positive and = 1; • on the punctured unit circle T \{ } , the roots of Q and P interlace.Notice the strange interlacing condition. On the unit circle, Q has two more rootsthan P , and necessarily Q (1) = 0. The interlacing condition implies that either Q has a triple root at 1, or it has a pair of simple roots that are closer to 1 on theunit circle than any of the roots of P .A couple of pictures should clarify this: Figures 2 and 3.There is no symmetry in the CS-interlacing conditions: if Q/P is a CS-interlacingquotient, then
P/Q is not.2.4.
SS-interlacing.
For our third flavour of interlacing, “SS” suggests “Salem-Salem” with the usual caveats.
ALEM NUMBERS AND PISOT NUMBERS 5 ✫✪✬✩❜ ❜❞❢rr r r
Figure 3.
CS-interlacing with a triple root at 1. The roots of Q = ( z + 1)( z − [ ◦ ] interlace on T \{ } with those of P =( z + z + 1)( z − z + 1) [ • ]. ✫✪✬✩❜ ❜❜❜ ❜❜r rr rrr Figure 4.
Type 1 SS-interlacing. The roots of Q = z − z − z − z + 1 [ ◦ ] interlace on T with those of P = z − z + 2 z − • ].For type 2 SS-interlacing, interchange P and Q .Suppose that P ( z ) and Q ( z ) are coprime polynomials with integer coefficients,and with positive top coefficients. We say that P and Q satisfy the SS-interlacingcondition and that
Q/P is an
SS-interlacing quotient if: • P and Q have the same degree; • all the roots of P and Q are simple; • one of P and Q is reciprocal, the other is antireciprocal; • each of P and Q has all but two of its roots in T , with these two being real,positive and = 1; • on the unit circle, the roots of Q ( z ) and P ( z ) interlace.The behaviour of the real roots of P and Q gives us two possible types of SS-interlacing. If Q/P is an SS-interlacing quotient then we say that it is a type interlacing quotient if the largest real root of P Q is a root of P , and it is a type interlacing quotient if the largest real root of P Q is a root of Q . There is symmetryin the conditions for SS-interlacing, but between the two types: Q/P is a type 1SS-interlacing quotient if and only if
P/Q is a type 2 SS-interlacing quotient. Againit is helpful to see a picture: Figure 4.Swapping the roles of P and Q in the example in Figure 4 gives an example oftype 2 SS-interlacing. Notice that we do not insist that the roots on the positivereal axis interlace (although in this particular example they do).2.5. Real interlacing quotients.
From any of the above flavours and types ofinterlacing pairs, we shall consider transforming the pair to a rational function withonly real zeros and poles. These zeros and poles will generally interlace (though theinterlacing is not always perfect), and for convenience we shall refer to the rationalfunction as a (real) interlacing quotient.If P and Q satisfy the CC-interlacing condition, or the CS-interlacing condi-tion, or either type of SS-interlacing condition, then we transform the function √ zQ ( z ) / ( z − P ( z ) via the map (2.1) to get a quotient q ( x ) /p ( x ), with q and p JAMES MCKEE AND CHRIS SMYTH coprime polynomials in Z [ x ], and xq ( x ) /p ( x ) a rational function in x . Suppose P and Q have degree d . If z − | Q ( z ) (which must be the case for CS-interlacing),then when considering Q ( z ) / ( z − P ( z ) we have pulled out a root of Q , and theremaining roots of P and Q transform in a 2-to-2 or 1-to-1 manner, so that q hasdegree d − p has degree d . If z − | P ( z ), then the factor ( z − in thedenominator of Q ( z ) / ( z − P ( z ) transforms to x −
4: we conclude that q hasdegree d and p has degree d + 1. We call q ( x ) /p ( x ) the (real) interlacing quotient corresponding to Q ( z ) /P ( z ). The conditions on the roots of P and Q are sufficientto ensure that the roots of p and q are all real.For CC-interlacing, CS-interlacing and type 1 SS-interlacing the roots of p and q interlace perfectly: the zeros and poles of the interlacing quotient interlace. Thequotient q ( x ) /p ( x ) is decreasing wherever it is defined, and has partial fractionexpansion(2.3) deg p X i =1 λ i x − α i where the α i are the roots of p and the λ i are all positive.For type 2 SS-interlacing, there is perfect interlacing of the zeros of q and p within the interval [ − , x = 2 (and to the leftof x = −
2) with the top (and bottom) zeros of p and q being in the wrong orderfor perfect interlacing. The derivative of the quotient q ( x ) /p ( x ) changes sign twice,and the partial fraction expansion (2.3) has two of the λ i negative.Note that a real interlacing quotient q ( x ) /p ( x ), as defined, is always an oddfunction: one of p and q is an even polynomial and the other is an odd polynomial.The degree of the denominator is one more than the degree of the numerator, andthe top coefficients are positive. As x → ∞ , q ( x ) /p ( x ) → Lemma 2.1. (a) If Q /P and Q /P are CC-interlacing quotients, then sois their sum. (b) Suppose that Q /P is either a CS-interlacing quotient or an SS-interlacingquotient and that Q /P is a CC-interlacing quotient. Then Q /P + Q /P is either a CS-interlacing quotient or an SS-interlacing quotient.Proof. Part (a) is just Proposition 3.3 of [16].For (b), we transform to the real world, where it easy to see that everythingis of the right shape. Let q /p and q /p be the corresponding real interlacingquotients. Then the partial fraction expansions of q /p and q /p as in (2.3) willhave all the λ i positive, except in the case of type 2 SS-interlacing, when the λ i corresponding to the largest and smallest α i are negative: these correspond to theroots of p outside [ − , q /p + q /p will be of the same form: eitherall the numerators in the partial fraction expansion will be positive, or there will beprecisely two negative numerators corresponding to the roots of p outside [ − , q ( x ) /p ( x ) → −∞ as x approaches the largest pole from above, so the sameis true for the sum. Also, both q ( x ) /p ( x ) and q ( x ) /p ( x ) are positive for allsufficiently large x , so the sum has a zero to the right of this pole. (cid:3) ALEM NUMBERS AND PISOT NUMBERS 7 Salem numbers via CC-interlacing
We now show how to produce Salem numbers from CC-interlacing quotients.The first construction, which is essentially that of [16], uses a single quotient; wethen consider a product construction combining two interlacing quotients in a mul-tiplicative manner, inspired by (but greatly generalising) a formula for the quotientsof certain Salem trees [17].3.1.
A single pair.
Our first interlacing construction is a translation of Proposi-tion 3.2(a) of [16]. This is also a special case of our second construction, Theorem3.2.
Theorem 3.1.
Let
Q/P be a CC-interlacing quotient, with the additional con-straint that P is monic. Let q/p be the corresponding real interlacing quotient.If (3.1) lim x → q ( x ) /p ( x ) > , then the only solutions to the equation (3.2) Q ( z )( z − P ( z ) = 1 + 1 z are a Salem number (or a reciprocal quadratic Pisot number), its conjugates, andpossibly one or more roots of unity. This is proved in [16] using the transformation x = z + 1 /z . It also follows fromTheorem 3.2 on taking P = P , Q = Q , P = z + 1, Q = z −
1. Neverthelesswe give a proof here, using the transformation x = √ z + 1 / √ z , as this provides amodel for later generalisations. Proof.
Suppose P and Q have degree d . Since the real interlacing quotient q/p is decreasing (except for jumps at poles), the equation q ( x ) /p ( x ) = x has exactlyone (simple) root between each pair of consecutive roots of p (these all lie in theinterval [ − , q ( x ) /p ( x ) = x in the interval (2 , ∞ ). We have now accounted for all the roots of xp ( x ) − q ( x ), which is a monic polynomial (given that P is monic) of degree d + 1or d + 2 (according as z − | Q or z − | P ). Transforming back to the complexworld, we see that all but two of the solutions to (3.2) lie in T , and these two are areciprocal pair { τ, /τ } with τ >
1. Clearing denominators in (3.2) gives a monicpolynomial with integer coefficients, and degree d + 1 or d + 2 as appropriate, sowe are done. (cid:3) The condition on q/p at x = 2 translates to lim z → Q ( z ) / ( z − P ( z ) > P (1) = 0 or ( Q (1) = 0 and) Q ′ (1) > P (1). Thus thiscondition can be checked readily without computing q and p .For an example, take P and Q as in (2.2). We have CC-interlacing, and also P (1) = 0, and P is monic. Solving (3.2) gives the famous Lehmer polynomial z + z − z − z − z − z − z + z + 1.To see that cyclotomic factors may appear, consider P ( z ) = z + z − z − Q ( z ) = 2 z + z + 2 z + z + 2 z + z + 2 z + z + 2. Again we have P monic and P (1) = 0. Now (3.2) gives the four primitive eighth roots of unity as solutions, aswell as the degree-8 Salem number with minimal polynomial z − z − z − z − z − z + 1. JAMES MCKEE AND CHRIS SMYTH
A product construction.
The following extension of Theorem 3.1 exploitstwo CC-interlacing pairs ( P , Q ) and ( P , Q ). Of course, after Lemma 2.1, onepossible way of combining two such pairs is to write P /Q + P /Q = P /Q ,giving a third pair ( P , Q ) that could be used in Theorem 3.1. Instead of the sum,we consider now the product: this will no longer give CC-interlacing, but we canstill squeeze out Salem numbers. Theorem 3.2.
Let Q /P and Q /P be two CC-interlacing quotients, with P and P both monic. Let q /p and q /p be the corresponding real interlacing quotients.(i) Suppose that lim x → (cid:18) q ( x ) p ( x ) − (cid:19) (cid:18) q ( x ) p ( x ) − (cid:19) < . Then the only solutions to the equation (cid:18) Q ( z )( z − P ( z ) − − z (cid:19) (cid:18) Q ( z )( z − P ( z ) − − z (cid:19) = 1 z are a Salem number (or a reciprocal quadratic Pisot number), its conjugates, andpossibly one or more roots of unity.(ii) Suppose that lim x → q ( x ) q ( x ) p ( x ) p ( x ) > . Then the only solutions to the equation Q ( z ) Q ( z )( z − P ( z ) P ( z ) = 1 z are a Salem number (or a reciprocal quadratic Pisot number), its conjugates, andpossibly one or more roots of unity. Part (i) extends an explicit formula arising from a certain family of Salem trees[17, Lemma 7.1(ii)]. The proof makes use of the following lemma.
Lemma 3.3.
Let ψ ( x ) and ψ ( x ) be rational functions in Z ( x ) , strictly decreasingon the real line (over intervals for which they are defined), with simple zeros andpoles. Write ψ ( x ) ψ ( x ) = f ( x ) /g ( x ) , where f ( x ) and g ( x ) are coprime polyno-mials with integer coefficients. Suppose that g ( x ) has real zeros at a and b (with a < b ). Then, counted with multiplicity, the number of solutions to the equation ψ ( x ) ψ ( x ) = c for x ∈ ( a, b ) is independent of real c ≥ . It will be evident from the proof that all relevant solutions to ψ ( x ) ψ ( x ) = c are simple, except perhaps when c = 0. It is possible that ψ ψ has one or moredouble zeros, but it cannot have zeros of higher order. The application of interestto us will use only that the number of solutions when c = 1 is the same as when c = 0. Proof.
In intervals where ψ ψ is positive, it is strictly monotonic: it is decreasingif both ψ and ψ are positive, and it is increasing if both are negative. As x passesthrough a zero x = α of ψ ψ , the function either decreases from ∞ to 0 as x approaches α from below, or ψ ψ increases from 0 to ∞ as x increases from α (orboth, in which case ψ ψ has a double zero at α : note that if both ψ and ψ vanishat α then necessarily both have the same sign in a punctured neighbourhood of α ). ALEM NUMBERS AND PISOT NUMBERS 9
For any c ≥ ψ ψ the number ofsolutions to ψ ( x ) ψ ( x ) = c is independent of c . The result follows. (cid:3) The proof of Theorem 3.2 now follows. We take for ψ and ψ the rationalfunctions q /p − ax and q /p − ax , where a = 1 for part (i) and a = 0 for part (ii).These are decreasing where defined, since the q i /p i are real interlacing quotientscorresponding to CC-interlacing quotients. Write ψ ( x ) ψ ( x ) = f ( x ) /g ( x ), aftercancelling any common factors, so that f and g are coprime polynomials withinteger coefficients. Note that, from the remarks in Section 2.5, f ( x ) /g ( x ) is aneven function. The number of zeros of ψ ψ between its extreme poles is equal tothe degree of f ( x ), since all roots are real. By Lemma 3.3, this equals the numberof solutions to f ( x ) /g ( x ) = 1: all of these lie in the interval [ − , f /g ∼ x → ∞ as x → ∞ ; for part (ii), f /g tends to a finite non-positive numberas x → ∞ . The condition at x = 2 ensures a solution to f ( x ) /g ( x ) = 1 in theinterval (2 , ∞ ), and by evenness also in ( −∞ , − g ( x ) − f ( x ) is monic, andwe have accounted for all its roots, we are done when we transform back to thecomplex world. As before, the condition at x = 2 transforms to an easily-checkedcondition at z = 1. 4. Pisot numbers via CC-interlacing
We now construct Pisot numbers by taking limits of convergent sequences ofSalem numbers. There is a conjecture of Boyd [4, p. 327] which, if true, wouldimply that this process will always yield either a Salem number or a Pisot number.Our results in this paper give a confirmation of this conjecture for all the casesconsidered. In this section we consider CC-interlacing; subsequently ( §
5) we shalltreat briefly the other flavours of interlacing.4.1.
CC-limit functions.
We define a
CC-limit function to be a rational function h ( z ) such that there is a sequence of CC-interlacing quotients ( h n ( z )) for which h n ( z ) / ( z −
1) converges to h ( z ) uniformly in any compact subset of the exterior ofthe unit disc. For example, 1 /z is a CC-limit function, as we could take h n ( z ) =( z n − z − / ( z n +1 − | z | ≥ ε , for any ε > Lemma 4.1.
Take any non-negative integers A , r , r , r , r , not all zero, andpositive integers A i , a i ( ≤ i ≤ r ), B i , b i ( ≤ i ≤ r ), C i , c i ( ≤ i ≤ r ), D i , d i ( ≤ i ≤ r ). Then the rational function (4.1) Az − r X i =1 A i ( z a i − z − z a i + r X i =1 B i z b i ( z − z b i − r X i =1 C i ( z c i + 1)( z − z c i + r X i =1 D i z d i ( z − z d i + 1) is a CC-limit function.Proof. Using the Beukers-Heckman classification [2] (and see also [17], where allthese terms (or their reciprocals) appear as quotients of graphs (multiplied by z − natural number n a CC-interlacing quotient Q n /P n by(4.2) Q n ( z ) P n ( z ) = A ( z n +1) z n − + P r i =1 A i ( z ai − z n − z n + ai − + P r i =1 B i ( z n + bi − z bi − z n − + P r i =1 C i ( z ci +1)( z n − z n + ci +1 + P r i =1 D i ( z n + di +1)( z di +1)( z n − . An easy estimate shows that for any ε > Q n ( z ) / (cid:0) ( z − P n ( z ) (cid:1) converges to the advertised limit function, uniformly in | z | ≥ ε . (cid:3) We shall call a rational function of the shape (4.1) a special CC-limit function .For these we can exploit their explicit form to prove that certain limit points of theset of Salem numbers are in fact Pisot numbers.4.2.
A single interlacing quotient.
Given a single CC-interlacing quotient, wecan take the limiting form of our Salem number construction, and attempt to provethat the limit is a Pisot number.
Theorem 4.2.
Let
Q/P be either a CC-interlacing quotient or zero ( Q = 0 , P =1 ), with P monic, and put g ( z ) = Q ( z ) / (cid:0) ( z − P ( z ) (cid:1) . Let h ( z ) be a special CC-limit function, as in (4.1). Let f ( z ) = g ( z ) + h ( z ) − − /z (if this has a removablesingularity at z = 0 , then remove it). If (4.3) lim z → (cid:0) g ( z ) + h ( z ) (cid:1) > , then the only non-zero solutions to f ( z ) = 0 are a Pisot number θ , the conjugatesof θ , and possibly some roots of unity. Before proving this, let us make some remarks. One possible choice for h ( z ) is1 /z , giving simply f ( z ) = g ( z ) −
1. The construction of Pisot numbers in [14] isessentially that of Theorem 4.2 with h ( z ) = k/z for some positive integer k ; theconstruction in [16] uses h ( z ) = 1 / ( z − z / ( z − z −
1) is given in § Proof.
For the special CC-limit function h ( z ), as in (4.1), let Q n ( z ) /P n ( z ) be asin (4.2), and define f n ( z ) = g ( z ) + Q n ( z ) / ( z − P n ( z ) − − /z . We have thatfor | z | > f ( z ) is the limit of the sequence ( f n ( z )), with convergenceuniform in compact subsets of that region. Moreover, from Theorem 3.1 each f n ( z )has a unique root τ n in the exterior of the unit disc, at least for all sufficiently large n , say n ≥ n (so that (3.1) holds).Note that f ( z ) has no poles outside the unit disc, and has finitely many zerosthere (it cannot be identically zero, as the condition (4.3) would then fail). Thecondition near z = 1 gives lim z → f ( z ) >
0, and we plainly have lim z → + ∞ f ( z ) = −
1. So there is at least one θ in the real interval (1 , ∞ ) such that f ( θ ) = 0.Take any circle, centred on θ , with radius sufficiently small that it lies outsidethe unit disc and such that no zeros of f other than θ lie in or on the circle. Forall sufficiently large n , the function f dominates f n − f on this circle; hence byRouch´e’s Theorem (assuming also that n ≥ n ) there is exactly one root of f n inthis circle (for all sufficiently large n ), and this root must be τ n . We conclude that τ n → θ as n → ∞ . Since this Rouch´e argument could be applied to any root of ALEM NUMBERS AND PISOT NUMBERS 11 f outside the unit disc, but the sequence ( τ n ) has at most one limit, we concludethat θ is the only root of f outside the unit disc.We deduce that θ is either a Pisot number or a Salem number, and the theoremwill follow if we show that θ is not a Salem number. Suppose for a contradictionthat θ is a Salem number, and let z be a conjugate of θ that lies in T . For arational function k ( z ), write ˜ k ( z ) = k (1 /z ) /z . Since f ( z ) has all coefficients real, z = 1 /z is also a zero of f ( z ). Thus z is a zero of both f ( z ) = g ( z )+ h ( z ) − − /z and ˜ f ( z ) = g ( z ) + ˜ h ( z ) − − /z (using here that ( z − g ( z ) is a CC-interlacingquotient, so that g ( z ) is a quotient of reciprocal polynomials). Thus z is a zero of h ( z ) − ˜ h ( z ), and by Galois conjugation so is θ .For the five special cases h ( z ) = 1 / ( z − z a − / ( z − z a , z b / ( z − z b − z c + 1) / ( z − z c , z d / ( z − z d + 1) one checks explicitly that h ( z ) − ˜ h ( z ) hasno roots outside the unit disc, giving the desired contradiction. For the generalcase, we appeal to Salem’s theorem [18] that the set of Pisot numbers is closed.Write h ( z ) = h ( z ) + h ( z ), where h ( z ) is a single term in (4.1), and h ( z ) is therest. Then take Q n /P n as in the proof of Lemma 4.1, for the limit function h (rather than h ). Now for each sufficiently large n , we can apply our special resultto conclude that the unique root θ n of g ( z ) + Q n ( z ) / ( z − P n ( z ) + h ( z ) − − /z outside the unit disc is a Pisot number. (Here we use Lemma 2.1(a) again, to showthat ( z − g ( z ) + Q n ( z ) /P n ( z ) is a CC-interlacing quotient.) Now another Rouch´eargument shows that θ is the limit of the θ n , so that Salem’s theorem gives that θ is a Pisot number. (cid:3) A product of two quotients.Theorem 4.3.
Let Q /P and Q /P each be either a CC-interlacing quotient orzero, with P and P both monic, and define (for i = 1 , ) g i ( z ) = Q i ( z ) / ( z − P i ( z ) . Let h be a special CC-limit function, and let h be either a special CC-limit function or zero.(i) Suppose that lim z → (cid:0) g ( z ) + h ( z ) − − /z (cid:1)(cid:0) g ( z ) + h ( z ) − − /z (cid:1) < . Then the only non-zero roots of the rational function f ( z ) = (cid:0) g ( z ) + h ( z ) − − /z (cid:1)(cid:0) g ( z ) + h ( z ) − − /z (cid:1) − /z are a certain Pisot number, θ , its conjugates, and perhaps some roots of unity.(ii) Suppose that lim z → (cid:0) g ( z ) + h ( z ) (cid:1)(cid:0) g ( z ) + h ( z )) (cid:1) > . Then the only non-zero roots of the rational function f ( z ) = (cid:0) g ( z ) + h ( z ) (cid:1)(cid:0) g ( z ) + h ( z ) (cid:1) − /z are a certain Pisot number, θ , its conjugates, and perhaps some roots of unity.Proof. The proof is very similar to that of Theorem 4.2, so we merely spell outthe differences. We again use closure of S to reduce to the special case where h ( z ) = 0 and h ( z ) is one of the five special functions 1 / ( z − z a − / ( z − z a , z b / ( z − z b − z c + 1) / ( z − z c , z d / ( z − z d + 1). Any Salem number that isa root of f ( z ) is also a root of f (1 /z ) /z , so is a common root of (cid:0) g ( z ) + h ( z ) − a (1 + 1 /z ) (cid:1)(cid:0) g ( z ) − a (1 + 1 /z ) (cid:1) − /z and (cid:0) g ( z ) + ˜ h ( z ) − a (1 + 1 /z ) (cid:1)(cid:0) g ( z ) − a (1 + 1 /z ) (cid:1) − /z , so is a root of (cid:0) h ( z ) − ˜ h ( z ) (cid:1)(cid:0) g ( z ) − a (1 + 1 /z ) (cid:1) . As before, h − ˜ h has no zeros outside the unit disc. Here g ( z ) − a (1 + 1 /z ) has a singlezero outside the unit disc, but we see from the definition of f that this cannot bea zero of f . (cid:3) Salem and Pisot numbers via CS/SS-interlacing
Several of the results of the previous section extend to obvious analogues for CS-and SS-interlacing quotients. We record these here briefly.5.1.
Salem numbers.
The analogue of Theorem 3.1 for CS-interlacing quotientsis obvious from a sketch of the graph of q ( x ) /p ( x ). Indeed necessarily one has q (2) ≥ p (2) <
0, so that q (2) /p (2) ≤
0, making a single root of q ( x ) /p ( x ) = x in (2 , ∞ ) automatic. Theorem 5.1.
Let
Q/P be a CS-interlacing quotient, with the additional constraintthat P is monic. Then the only solutions to the equation (3.2) are a Salem number(or a reciprocal quadratic Pisot number), its conjugates, and possibly one or moreroots of unity. For SS-interlacing quotients, a sufficient condition that there should be a uniquesolution to q ( x ) /p ( x ) = x in the interval (2 , ∞ ) is that q (2) /p (2) ≤ q (2) /p (2) < q (2) /p (2) = 2, depending on the derivativeof q/p at x = 2. But it is simpler to restrict to a strong inequality. Theorem 5.2.
Let
Q/P be an SS-interlacing quotient (of either type), with theadditional constraint that P is monic. Suppose further that (5.1) lim z → Q ( z )( z − P ( z ) < . Then the only solutions to the equation (3.2) are a Salem number (or a reciprocalquadratic Pisot number), its conjugates, and possibly one or more roots of unity.For type 1 SS-interlacing, a weak inequality in (5.1) would suffice.
There is no analogue of Theorem 3.2, as the construction would give two rootsoutside the unit disc.5.2.
Pisot numbers.
Certain limiting cases of Theorems 5.1 and 5.2 yield Pisotnumbers. Armed with Lemma 2.1(b), we can give the analogue of Theorem 4.2 inthis setting.
Theorem 5.3.
Let
Q/P be either a CS-interlacing quotient or an SS-interlacingquotient, with P monic, and put g ( z ) = Q ( z ) / (cid:0) ( z − P ( z ) (cid:1) . Let h ( z ) be a specialCC-limit function, as in (4.1). Let f ( z ) = g ( z ) + h ( z ) − − /z (if this has aremovable singularity at z = 0 , then remove it). If lim z → (cid:0) g ( z ) + h ( z ) (cid:1) < , then the only non-zero solutions to f ( z ) = 0 are a Pisot number θ , the conjugatesof θ , and possibly some roots of unity. ALEM NUMBERS AND PISOT NUMBERS 13
Proof.
This is much as before, but now f , and each f n in the sequence of functionsconverging to f , has a pole outside the unit disc (the same pole for each f n andfor f , corresponding to the Salem zero of P ). When considering circles centredon roots of f outside the unit disc, the radii must be sufficiently small to avoidenclosing this pole. (cid:3) All Pisot numbers via interlacing
In this section we shall show that all Pisot numbers are produced by a specialcase of Theorem 5.3 (Theorem 6.4 below). We proceed in three steps: in § P k ) k ≥ , following Salem; in § k the pair ( P k , P k +1 ) is an SS-interlacing quotient (that thesepolynomials are Salem polynomials is contained in Salem’s work—the novelty hereis in establishing the interlacing property); in § A ( z ) ∈ Z [ z ] of exact degree d , define A ∗ ( z ) = z d A (1 /z ).6.1. The polynomials P k .Lemma 6.1. Let A ( z ) be any polynomial of degree d with integer coefficients. For k ≥ , define P k ( z ) = (cid:0) z k A ( z ) − A ∗ ( z ) (cid:1) / ( z − . Then for k ≥ we have (6.1) z k A ( z ) = P k +1 ( z ) − P k ( z ) . If k ≥ , then the polynomial P k has degree d + k − . If k ≥ , then P k is a reciprocalpolynomial; P is a power of z times a reciprocal polynomial. The polynomials P k satisfy the recurrence (6.2) P k +2 − ( z + 1) P k +1 + zP k = 0 , for k ≥ . For each k ≥ , the pair of polynomials ( P, Q ) = ( P k +1 , P k ) is theunique pair of reciprocal polynomials such that the degrees of P and Q are d + k and d + k − and such that z k A ( z ) = P ( z ) − Q ( z ) .Proof. This is a collection of simple assertions, all of which follow directly from thedefinitions. For the recurrence, its characteristic polynomial is X − ( z + 1) X + z =( X − z )( X − (cid:3) Suppose that A ( z ) is monic. If A (0) = 1, then the degree of P is d −
1, but if A (0) = 1 then this degree is at most d −
2. We record as a lemma the observationthat no further cancellation in the degree of P can occur if A ( z ) is the minimalpolynomial of a Pisot number θ , unless θ is a reciprocal quadratic Pisot number. Lemma 6.2.
Let A ( z ) be the minimal polynomial of a Pisot number θ . If A ( z ) isnot a reciprocal (and hence quadratic) polynomial, then P ( z ) = (cid:0) A ( z ) − A ∗ ( z ) (cid:1) / ( z − has degree at least d − . Thus, writing A ( z ) = z d + a d − z d − + · · · + a , this tells us that if a = 1 then a d − = a . In this sense, Pisot polynomials are strongly non-palindromic. Proof. If a = 1 then P has degree d −
1. If a = 1 and a = a d − , then expanding A ( z ) /A ∗ ( z ) about z = 0 gives A ( z ) /A ∗ ( z ) = 1 + u z + u z + · · · . This contradicts [7, Th´eor`eme 1] (which asserts that the coefficient of z in such anexpansion must be strictly positive), unless A is a reciprocal quadratic polynomial. (cid:3) A winding argument.Theorem 6.3.
Suppose that A ( z ) is the minimal polynomial of a Pisot number.For each k ≥ , define P k ( z ) as in Lemma 6.1.For all large enough k , both P k +1 and P k have all but two roots on the unit circle,with the other roots real and positive, and the roots of P k +1 ( z ) and ( z − P k ( z ) that lie on the unit circle interlace.For all k ≥ , the rational function ( z − P k ( z ) /P k +1 ( z ) is an interlacing quo-tient (either CC, CS or SS).Proof. Suppose that A ( z ) has degree d . Note that P k (1) = kA (1)+ A ′ (1) − ( A ∗ ) ′ (1),and this is negative for all large enough k , since A (1) < A ( z ) is the minimalpolynomial of a Pisot number). Hence P k ( z ) has at least one real root greater than1, for all large enough k . Since P k +1 and P k are both reciprocal, each has at leasttwo positive real roots, for all large enough k . From now on, we assume that k islarge enough (say k ≥ k ≥
1) for this to hold.For z on the unit circle, ( z − P k ( z ) = 0 if and only if z k A ( z ) = A ∗ ( z ) = z d A ( z ) = z d A ( z ) , which is equivalent to z k − d A ( z ) = | A ( z ) | , which is equivalent to z k − d A ( z ) beingreal and positive.Now z k − d A ( z ) winds round the origin k + d − z winds round 0. Hence( z − P k ( z ) has at least k + d − P k ( z ) has at least k + d − P k ( z ) has exactly k + d − P k +1 ( z ) has exactly k + d − z windsround 0 in the positive sense (anticlockwise), on the unit circle, starting at z = 1.When z = 1, the argument of z k − d A ( z ) is 0. As z winds round the unit circle, theargument increases to ( k + d − π , not necessarily monotonically. The argument isan integer multiple of 2 π precisely when P k ( z ) = 0 (or when z = 1). The argumentequals that of 1 /z (modulo integer multiples of 2 π ) precisely when P k +1 ( z ) = 0. Itis clear from Figure 5 that this must happen at least once (and hence, by counting,exactly once) between each two consecutive zeros of ( z − P k ( z ), as claimed: theline running from bottom left to top right (which need not be a straight line!) mustcross one of the short diagonals at least once between each pair of horizontal lines.For the final assertion of the theorem, we need to consider what happens forsmaller values of k ≥
1. The winding argument still accounts for all but two ofthe zeros of each P k . We need to pin down the other two roots, and establish theclaimed interlacing property. Let θ > A ( z ). If A ∗ ( θ ) > P k ( θ ) <
0, in which case P k always has a real root greater than θ . In this casethere is nothing more to prove: we have SS-interlacing. ALEM NUMBERS AND PISOT NUMBERS 15 ❅❅❅❅❅❅❅❅❅❅❅❅❅❅❅❅❅❅❅❅❅❅❅❅❅❅❅❅❅❅❅❅❅❅❅❅✘✘✘✘✘✘✘✘✘✘✘✘✘✘✘✘✘✘✘✘✘✘✘✘✘✘✘✘✘✘✘✘✘ s s s s❝ ❝ ❝ ❝ π π π π ✲ arg (cid:0) z k − d A ( z ) (cid:1) π arg z ✻ Figure 5.
The case d + k − z − P k ( z ) [ ◦ ]and of P k +1 ( z ) [ • ] interlace.We are left with the case that A ∗ ( θ ) <
0. Then P k has no real root greater than θ , for any k . In particular, for k ≥ k the Salem root τ k of P k satisfies 1 < τ < θ .Then from z k A ( z ) = P k +1 ( z ) − P k ( z ) we have P k +1 ( τ k ) <
0, and hence τ k +1 > τ k .If follows that ( z − P k ( z ) /P k +1 ( z ) is a type 1 SS-interlacing quotient for k ≥ k ,and the roots and poles of the corresponding real interlacing quotient p k ( x ) /p k +1 ( x )interlace perfectly on the real line. The recurrence (6.2) translates to the real worldas p k +1 ( x ) = xp k ( x ) − p k − ( x ) . Since the zeros of p k and p k +1 interlace, one deduces that those of p k − and p k interlace, and then those of p k − and p k − , and so on. Thus P k +1 ( z ) and( z − P k ( z ) interlace for all k ≥ p k (2) = 0, then P k has a double zero at z = 1, and ( z − P k has a triple zero there: in this case we have CS-interlacing. If P k +1 has all roots on the unit circle, then we have CC-interlacing (if p k +1 (2) = 0,then P k +1 has a double zero at z = 1, and P k +1 ( z ) / ( z −
1) interlaces with P k ( z )).If P k +1 is Salem but P k is cyclotomic, we have CS-interlacing. (cid:3) Theorem 5.3 gives all Pisot numbers.
Now we are in a position to showthat the interlacing construction given in Theorem 5.3 produces all Pisot numbers,with h ( z ) = 1 /z . Theorem 6.4.
Given any Pisot number θ , there exists an SS-interlacing quotient Q ( z ) /P ( z ) satisfying the conditions of Theorem 5.3 (with h ( z ) = 1 /z ), such thatthe only solutions to Q ( z ) /P ( z ) = 1 are θ , its conjugates, and .Proof. Let A ( z ) be the minimal polynomial of θ . We consider P ( z ) = P k +1 ( z ) = (cid:0) z k +1 A ( z ) − A ∗ ( z ) (cid:1) / ( z − Q ( z ) = ( z − P k ( z ) = z k A ( z ) − A ∗ ( z ).We have seen (Theorem 6.3) that for all large enough k the quotient Q/P isan SS-interlacing quotient. To apply Theorem 5.3 with h ( z ) = 1 /z we need thecondition lim z → Q ( z ) / ( z − P ( z ) <
1. Butlim z → Q ( z )( z − P ( z ) = lim z → P k ( z ) P k +1 ( z ) = kA (1) + A ′ (1) − ( A ∗ ) ′ (1)( k + 1) A (1) + A ′ (1) − ( A ∗ ) ′ (1) , and this is less than 1 if k is large enough, since A (1) < Finally we note that Q ( z ) / ( z − P ( z ) = 1 is equivalent to z k A ( z ) = 0, whichhas as its roots θ , all the conjugates of θ , and 0 (assuming k > (cid:3) For smaller values of k the quotient Q/P in the proof of Theorem 6.4 might beCS-interlacing or CC-interlacing. The case k = 0 and P (0) = 0 is exceptional, asever.The proof of Theorem 6.4 uses Salem’s method to construct the P k from A , andthen shows, conversely, how A can be recovered from P k via (6.1) of Lemma 6.1,for k sufficiently large.7. All Salem numbers via interlacing
Boyd’s theorem.
We recall the following fundamental result of Boyd [4].
Theorem 7.1 ([4, Theorem 4.1]) . Let τ be a Salem number, with minimal polyno-mial R ( z ) . Define S ( z ) = z + 1 , S − ( z ) = z − . Then for each choice of ε = ± there exist infinitely many Pisot polynomials A ( z ) such that (with A ∗ ( z ) as before) (7.1) S ε ( z ) R ( z ) = zA ( z ) + εA ∗ ( z ) . All Salem numbers via interlacing.
Armed with Theorem 7.1, we showfirst (Lemma 7.2) that we can produce all Salem numbers via SS-interlacing quo-tients, but with a “right hand side” other than 1 + 1 /z , as used in Theorems 3.1,5.1 and 5.2. Lemma 7.2.
Let τ be any Salem number, and choose ε = ± . Then for allsufficiently large k there exists an SS-interlacing quotient Q ( z ) /P ( z ) such that theonly non-zero solutions to (7.2) Q ( z )( z − P ( z ) = z k − + εz k + ε are τ , its conjugates, and perhaps some roots of unity.Proof. Let R ( z ) be the minimal polynomial of τ , and let A ( z ) be a Pisot polynomialsuch that (7.1) holds with our choice of ε . As in the proof of Theorem 6.4, we put P = P k +1 ( z ) = (cid:0) z k +1 A ( z ) − A ∗ ( z ) (cid:1) / ( z −
1) and Q = ( z − P k ( z ) = z k A ( z ) − A ∗ ( z ),and repeat the observation that for all sufficiently large k the quotient Q ( z ) /P ( z )is an SS-interlacing quotient. With z k A ( z ) = P k +1 ( z ) − P k ( z ) we have A ∗ ( z ) = P k +1 ( z ) − zP k ( z ), and hence (with S ε ( z ) = z + 1 or z − ε = 1 or −
1) from (7.1) we have z k − S ε ( z ) R ( z ) = z k A ( z ) + εz k − A ∗ ( z )= ( P k +1 ( z ) − P k ( z )) + εz k − ( P k +1 ( z ) − zP k ( z ))= (1 + εz k − ) P k +1 ( z ) − (1 + εz k ) P k ( z ) , from which the result follows. (cid:3) Instead of taking large k in Lemma 7.2, we can consider choosing k of any size.The choice of k = 1 gives the following theorem. Theorem 7.3.
Consider the equation (7.3) Q ( z )( z − P ( z ) = 2 z + 1 . Define four types of Salem number I , II , III , IV as follows. A Salem number τ is of type I (respectively, II , III , IV ) if there exist monic polynomials P ( z ) , Q ( z ) ALEM NUMBERS AND PISOT NUMBERS 17 such that Q ( z ) /P ( z ) is a CC-interlacing quotient (respectively, CS-interlacing, type SS-interlacing, type SS-interlacing) and for which the only non-zero solutionsto (7.3) are τ , its conjugates, and pehaps some roots of unity. Then every Salemnumber is of at least one of these four types.Proof. We take k = 1 and ε = 1 in the proof of Lemma 7.2. For the interlacingproperties, we appeal to Theorem 6.3. (cid:3) Comparison with the Bertin-Boyd classification
Let τ be any Salem number, with minimal polynomial R ( z ). Bertin and Boyd[1] showed that there exist reciprocal polynomials K ( z ) and L ( z ) such that L ( z )interlaces with K ( z ) R ( z ) on the unit circle. Their Theorem B is most relevanthere, as it relates to expressing K ( z ) R ( z ) in the shape zA ( z ) + εA ∗ ( z ), where A ( z ) = z m A ( z ) is a Pisot polynomial, with A (0) = 0.In the case ε = 1, which they use only when A (0) <
0, their polynomial L ( z )is A ( z ) + A ∗ ( z ); in the case ε = −
1, which they use only when A (0) >
0, their L ( z ) is our P ( z ). Our proof of interlacing comes from a winding argument; theirsis via a characterisation of “entrances” and “exits” to/from the unit disc for theassociated algebraic curve zA ( z ) + εtA ∗ ( z ) = 0 ( t ≥ Final remarks
Pisot numbers of negative trace.
As one application of the constructionin Theorem 4.2, we produce an example of a Pisot number that has trace − h ( z ) = 1 / ( z − g ( z ) = ( z − z + z − z − z − z + z + 1) / ( z − z − z − h ( z ) = z / ( z − z − −
1; its minimal polynomial is z + z − z − z − z − z − z − z − z − z − z − z + z +2 z +2 z +1 . The choice of g ( z ) (see § − h ( z ) ismotivated by the desire to introduce a new, negative-trace, low-degree cyclotomicfactor into the denominator.We used the Dufresnoy-Pisot-Boyd algorithm [5] to search for small Pisot num-bers of small degree and negative trace. For Pisot numbers below 2, we found 10examples, of degrees between 22 and 48. The degree-16 example above is for aPisot number slightly larger than 2. Finding the smallest degree (perhaps 16?) fora Pisot number of negative trace remains a challenge.9.2. Salem numbers of large negative trace.
For Salem numbers of trace − − g ( z ) = ( z − z + z − z − z − z + z + 1) / ( z − z − z −
1) from § z − / ( z − z − z −
1) + ( z − / ( z − z − z − −
3, the smallest degree currently known for this trace: z + 3 z + 2 z − z − z − z − z + · · · . (One needs to check that this polynomial has no cyclotomic factors. This can bedone using the algorithm of Beukers and Smyth [3], or by checking irreducibility.)A real transform of this polynomial can be found in [13, § Small Salem numbers.Proposition 9.1.
Let τ be any Salem number below the real root of z − z − (sothat τ is smaller than any Pisot number). Let R ( z ) be the minimal polynomial of τ , and let A ( z ) be any Pisot polynomial such that (7.1) holds with ε = 1 . Then A ( z ) has at least three real roots, with at least one between /τ and .Proof. The conditions on τ imply that A ( τ ) <
0. Putting z = τ in (7.1) we deducethat A ∗ ( τ ) > A ∗ has a real root between 1 and τ . Thus A has at leasttwo real roots: the Pisot number, and another root between 1 /τ and 1. Since A has odd degree it must have at least three real roots. (cid:3) For example, taking τ to be Lehmer’s number, one possibility for A ( z ) is z − z − z − z − z − z + z + 3 z + 4 z + 3 z + 1. Sure enough, this has realroots approximately equal to − . . . Theorem 9.2.
If there is a smallest Salem number, τ , then it is of type IV (asdefined in Theorem 7.3), and not of any other type.Proof. Suppose that there is a smallest Salem number, τ . We take A ( z ) and thesequence P k ( z ) as in the proof of Lemma 7.2, and claim that Q ( z ) /P ( z ) = ( z − P ( z ) /P ( z ) must be a type 2 SS-interlacing quotient, showing that τ is of typeIV.If Q ( z ) /P ( z ) were either CS-interlacing or type 1 SS-interlacing, then the Salemroot of P ( z ) would be smaller than τ , giving a contradiction.Finally we eliminate the possibility that Q ( z ) /P ( z ) is CC-interlacing. In thiscase we increase k until P k +1 becomes Salem, with P k still cyclotomic; then we get(using (7.2) with Q/P = ( z − P k /P k +1 , as in the proof of Lemma 7.2) that theroot of P k +1 is smaller than our Salem number, again contradicting the minimalityof τ . (cid:3) Note that this is not saying that there are no examples of small Salem numbersthat come from CC or CS interlacing: merely that if we go via Boyd’s theorem (asin the definition of types) we will not see small Salem numbers arising other thanas type IV.9.4.
Further consequences of CC-interlacing.
We conclude with two amusingremarks concerning CC-interlacing quotients, which we record as a single proposi-tion.
Proposition 9.3.
Let
Q/P be a CC-interlacing quotient. Then (a) P + Q has all its roots in T ; ALEM NUMBERS AND PISOT NUMBERS 19 (b) P + Q has all its roots in the open unit disc | z | < .Proof. For (a) we simply apply Lemma 2.1(a): the sum
P/Q + Q/P is a CC-interlacing quotient, so its numerator has all roots in T . We can even say furtherthat these roots interlace with those of P Q .For (b) we use another winding argument. Let d be the common degree of P and Q , and suppose that z − | Q . We observe that it is enough to show that f ( z ) = P ( z ) + Q ( z ) (a polynomial of degree 2 d ) has all its roots in the open unitdisc.Write f ( z ) = z d g ( z ) = z d (cid:0) P ( z ) /z d + Q ( z ) /z d (cid:1) . Since P is reciprocal and Q is antireciprocal, P ( z ) /z d and Q ( z ) /z d give the real and imaginary parts of g ( z )when z is on the unit circle. As z goes round the unit circle, anticlockwise, theargument of z d increases by 2 dπ . The roots of P and Q interlace on the unit circle,so as z goes round the unit circle the pair (cid:0) ℜ g ( z ) , ℑ g ( z ) (cid:1) cycles d times throughone of the patterns (+ , +), (+ , − ), ( − , − ), ( − , +) or (+ , +), ( − , +), ( − , − ), (+ , − ).We do not (yet) know which of these patterns occurs, nor at what point in thepattern we start, but d complete cycles through one of these two patterns must bemade. In either case, g ( z ) winds d times round the origin: in the former case itwinds clockwise, and in the latter case anticlockwise. We conclude that as z goesanticlockwise around the unit circle, the argument of f ( z ) increases by either 0 or4 dπ . It follows that P ( z ) + Q ( z ) has either all of its roots in the open unit discor none of them, and the same holds for P ( z ) + Q ( z ). But since P is reciprocaland Q is antireciprocal, we have P (0) + Q (0) = 1 + ( −
1) = 0, so that P + Q has atleast one root, namely 0, that has modulus strictly less than 1. Hence all d rootsmust have modulus strictly less than 1. (cid:3) Acknowledgment
We are profoundly grateful to the referee who pointed out that in our firstsubmitted draft we had been cavalier with our leading coefficients, leading us toassert several falsehoods.
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