The geometry of bi-Perron numbers with real or unimodular Galois conjugates
aa r X i v : . [ m a t h . G T ] S e p THE GEOMETRY OF BI-PERRON NUMBERS WITHREAL OR UNIMODULAR GALOIS CONJUGATES
LIVIO LIECHTI AND JOSHUA PANKAU
Abstract.
Among all bi-Perron numbers, we characterise those all of whoseGalois conjugates are real or unimodular as the ones that admit a powerwhich is the stretch factor of a pseudo-Anosov homeomorphism arising fromThurston’s construction. This is in turn equivalent to admitting a powerwhich is the spectral radius of a bipartite Coxeter transformation. Introduction
Particular geometric situations often give rise to particular algebraic numbers,and it is a natural question to characterise these numbers by their geometry.In this note, we provide a description of those bi-Perron numbers all of whoseGalois conjugates are real or unimodular. We relate those numbers to pseudo-Anosov stretch factors arising via Thurston’s construction, and to the spectralradii of bipartite Coxeter transformations.A bi-Perron number λ is a real algebraic unit > λ − , λ ), except for λ itself and possi-bly one of ± λ − . Bi-Perron numbers feature prominently as stretch factors ofpseudo-Anosov homeomorphisms, for example in the following problem posed byFried [4]: does every bi-Perron number have a power that arises as the stretchfactor of a pseudo-Anosov homeomorphism? Recently, the second author gavea positive answer to Fried’s problem for the class of Salem numbers [10]. Ourmain result extends this positive answer to the class of bi-Perron numbers withreal or unimodular Galois conjugates. This condition on the Galois conjugatesturns out to precisely characterise the pseudo-Anosov stretch factors that arisefrom Thurston’s construction, and the spectral radii of bipartite Coxeter trans-formations. Theorem 1.
For a bi-Perron number λ , the following are equivalent. (a) All Galois conjugates of λ are contained in S ∪ R . (b) For some positive integer k , λ k is the stretch factor of a pseudo-Anosovhomeomorphism arising from Thurston’s construction. (c) For some positive integer k , λ k is the spectral radius of a bipartite Cox-eter transformation of a bipartite Coxeter diagram with simple edges. This result is of optimal quality. Indeed, the smallest stretch factor of a pseudo-Anosov homeomorphism arising from Thurston’s construction as well as thesmallest spectral radius > Fried’s problem is sometimes cited in a stronger form that does not allow powers, andindeed the powers might not be required. However, the version with powers suffices to ensurethat every bi-Perron number arises as the growth rate of a surface homeomorphism.
Lehmer’s number λ L ≈ . S ∪ R . This is thecontent of the following proposition. Proposition 2.
There exist bi-Perron numbers arbitrarily close to and all ofwhose Galois conjugates are contained in S ∪ R > . The proof is short and we choose to give it here.
Proof.
Choose any ε >
0. By Robinson’s work on Chebyshev polynomials [12],there exist infinitely many algebraic integers that lie, together with all theirGalois conjugates, in the interval [ − ε, ε ]. On the other hand, by aresult due to P´olya described in Schur [14], only finitely many algebraic integerslie, together with all their Galois conjugates, in the interval [ − ε, , ε ], there exist infinitely many Perron numbersall of whose Galois conjugates are contained in the interval [ − ε, ε ].Let p ( t ) be the minimal polynomial of such a Perron number and define thepolynomial f ( t ) = t deg( p ) p ( t + t − ). Then every root x of f ( t ) is related tosome root y of p ( t ) by x + x − = y and vice versa. In particular, all theroots of f ( t ) are contained in S ∪ R > . Furthermore, if 2 < y < ε ,then 1 < x < ε + √ ε + ε . Now, let x be the maximal real root of f ( t ).By construction, no other root of f ( t ) is as small as x − in modulus, so x isa bi-Perron number all of whose Galois conjugates are contained in S ∪ R > .Choosing ε arbitrarily small yields the desired result. (cid:3) Remark 3.
The largest root x of the polynomial t − t − t + 4 t − x is a Mahler measure and asks whether x is the stretch factor of apseudo-Anosov homeomorphism. While we do not answer the question for x ,Theorem 1 shows that some power of x is a pseudo-Anosov stretch factor.For the Galois conjugates of a bi-Perron number, we have the following result;the statement is different from the one of Theorem 1 in that we only have touse squares for the characterisation, and we only need Coxeter diagrams thatare trees. Theorem 4.
For a Galois conjugate λ of a bi-Perron number, the following areequivalent. (a) All Galois conjugates of λ are contained in S ∪ R . (b) The number λ is an eigenvalue of a Coxeter transformation associatedwith a tree. We note that the bi-Perron number in the statement might not be the spectralradius of the Coxeter transformation. Furthermore, we do not include a state-ment concerning stretch factors, since in the setting of Thurston’s constructionwe cannot assure that λ is actually a Galois conjugate of a stretch factor, butonly an eigenvalue of the action induced on the first homology of the surface bya pseudo-Anosov homeomorphism.Again, no result of the generality of Theorem 4 can be obtained without takingsquares: by a result of A’Campo [1], a Coxeter transformation associated witha tree has no negative real eigenvalue, except for possibly − I-PERRON NUMBERS WITH REAL OR UNIMODULAR CONJUGATES 3
Organisation . In the next section, we discuss some Galois-theoretic propertiesof Perron and bi-Perron numbers and prove a result related to trace fields ofPerron numbers. We also recall a result of the second author [10] that is key forour purposes. In Section 3, we describe the key input of Coxeter transformationsand Thurston’s construction of pseudo-Anosov homeomorphisms, and we proveTheorem 1 and Theorem 4.2.
Perron and bi-Perron numbers
In this section, we prove a result about the trace fields of Perron and bi-Perronnumbers. A Perron number λ is a real algebraic integer > , λ ), except for λ itself. Thefollowing statement is given in the proof of Lemma 8.2 of Strenner [15]. Lemma 5.
Let λ be a Perron number of degree l , and let λ , . . . , λ l be its Galoisconjugates. Then for all positive integers k , λ k is also of degree l and λ k , . . . , λ kl are its Galois conjugates. The property of Perron numbers highlighted by this lemma is a key ingredientto proving the following proposition.
Proposition 6.
Let λ be a Perron number. If [ Q ( λ ) : Q ] is odd, then for all k we have Q ( λ + λ − ) = Q ( λ k + λ − k ) . If [ Q ( λ ) : Q ] is even, then for all k we have Q ( λ + λ − ) = Q ( λ k +1 + λ − (2 k +1) ) and Q ( λ + λ − ) = Q ( λ k + λ − k ) with Q ( λ + λ − ) = Q ( λ + λ − ) if and only if − λ − is not a Galois conjugateof λ . Example 7.
The golden ratio φ = √ is a bi-Perron number with minimalpolynomial t − t −
1. By definition, − φ − = −√ is a Galois conjugate of φ ,and we have that Q ( φ ) = Q ( φ + φ − ) = Q ( √
5) but Q ( φ + φ − ) = Q . Whilethe golden ratio is not the stretch factor of an orientation-preserving Anosovmap of the torus, it is the stretch factor of an orientation-reversing one: indeed,the spectral radius of the matrix (cid:18) (cid:19) is the golden ratio.We now prove Proposition 6, which will be important in the proof of Theorem 1below. Proof of Proposition 6.
We start by noting that since λ is a Perron number,then Lemma 5 tells us that [ Q ( λ ) : Q ] = [ Q ( λ k ) : Q ] for all positive integers k .This immediately implies that Q ( λ ) = Q ( λ k ) for all k , since the former is afield extension of the latter.Now, we assume that [ Q ( λ ) : Q ] is odd. Since Q ( λ k ) is a field extensionof Q ( λ k + λ − k ), and since λ k is a root of t − ( λ k + λ − k ) t + 1, then we musthave that [ Q ( λ k ) : Q ( λ k + λ − k )] = 1 or 2 . LIVIO LIECHTI AND JOSHUA PANKAU
But clearly this cannot equal 2 since [ Q ( λ ) : Q ] is assumed to be odd. Hence, Q ( λ ) = Q ( λ k ) = Q ( λ k + λ − k ) for all positive integers k . In particular, we havethat Q ( λ + λ − ) = Q ( λ k + λ − k ) for all k .We now assume that [ Q ( λ ) : Q ] is even. We will first prove that equality Q ( λ + λ − ) = Q ( λ + λ − ) holds if and only if − λ − is not a Galois conjugateof λ . Note that − λ − can only be a Galois conjugate in the even degree casesince if it is a conjugate, then if σ ( λ ) is any other conjugate, then so is − σ ( λ ) − .Hence the minimal polynomial has an even number of roots.Now, because Q ( λ + λ − ) ⊆ Q ( λ + λ − ) ⊆ Q ( λ ) = Q ( λ ) then we have thefollowing tower, with the possible degree of each extension listed: Q ( λ ) = Q ( λ ) (cid:12)(cid:12)(cid:12) Q ( λ + λ − ) (cid:12)(cid:12)(cid:12) Q ( λ + λ − )We now assume that Q ( λ + λ − ) = Q ( λ + λ − ). This means that the topextension must be degree 1, which implies that Q ( λ + λ − ) = Q ( λ ) = Q ( λ ),and the tower collapses to Q ( λ + λ − ) = Q ( λ ) = Q ( λ ) (cid:12)(cid:12)(cid:12) Q ( λ + λ − ) . This immediately implies that t − ( λ + λ − ) t + 1 is the minimal polynomialfor λ over Q ( λ + λ − ). Thus, we see that the non-identity automorphism φ ∈ Gal( Q ( λ ) / Q ( λ + λ − )) maps λ to λ − . Hence, [ φ ( λ )] = λ − , and weget φ ( λ ) = ± λ − . It cannot be the case that φ ( λ ) = λ − because this wouldimply that Q ( λ + λ − ) = Q ( λ ) is fixed by Gal( Q ( λ ) / Q ( λ + λ − )), whichcontradicts the definition of the Galois group. Therefore, φ ( λ ) = − λ − , whichimplies that − λ − is a Galois conjugate of λ .Now, running the argument in reverse, if − λ − is a Galois conjugate of λ thenthere exists a Q -automorphism φ of Q ( λ ) such that φ ( λ ) = − λ − . This imme-diately implies that Q ( λ + λ − ) is fixed by φ but Q ( λ + λ − ) is not, therefore, Q ( λ + λ − ) = Q ( λ + λ − ).We now generalize the argument. Suppose that − λ − is a Galois conjugate of λ .Then, the automorphism φ fixes Q ( λ k + λ − k ) for all positive integers k . Thisimplies that [ Q ( λ k ) : Q ( λ k + λ − k )] = 2for all k . On the other hand Q ( λ k +1 + λ − (2 k +1) ) is not fixed for any k , hence Q ( λ k +1 + λ − (2 k +1) ) = Q ( λ + λ − )for all k . I-PERRON NUMBERS WITH REAL OR UNIMODULAR CONJUGATES 5
Finally, in the case when Q ( λ + λ − ) = Q ( λ + λ − ), then both of these fieldsmust equal Q ( λ k ) for all positive integers k , since otherwise − λ − would be aGalois conjugate of λ by above. The equality Q ( λ k ) = Q ( λ + λ − ) = Q ( λ + λ − )implies that these are all equal to both Q ( λ k + λ − k ) and Q ( λ k +1 + λ − (2 k +1) ),for all k . To see this, if there is a k for which equality fails, then there is a Q -automorphism that either exchanges λ k with λ − k , or exchanges λ k +1 with λ − (2 k +1) . Since − λ − is not a Galois conjugate of λ , then λ − must be, andthe automorphism exchanges λ with λ − . This implies that both Q ( λ + λ − )and Q ( λ + λ − ) are fixed by the automorphism, but this is impossible sincethey equal Q ( λ k ). Therefore, when Q ( λ + λ − ) = Q ( λ + λ − ) we must have Q ( λ ) = Q ( λ k ) = Q ( λ k + λ − k ) for all k . (cid:3) The following result is the key to the construction of a geometric situation thatcorresponds to the power of a bi-Perron number. It was used by the secondauthor to show that every Salem number has a power that is the stretch factorof a pseudo-Anosov homeomorphism arising from Thurston’s construction [10].We now show that the proof works almost identically for bi-Perron numbersall of whose Galois conjugates are contained in S ∪ R . This extension is alsopresented in the second author’s thesis [11]. Proposition 8.
Let λ be a bi-Perron number all of whose Galois conjugatesare contained in S ∪ R . Then there exists a positive integer k so that λ k + λ − k equals the spectral radius of a positive symmetric integer matrix.Proof. We very closely follow the second author’s proof in [10]. For the conve-nience of the reader, we summarise the key steps and mention where we have topay attention because our setting is slightly more general than in the originalargument.Let λ be a bi-Perron number all of whose Galois conjugates lie in S ∪ R .Then λ + λ − is a totally real Perron number. Let f ( t ) be the minimal poly-nomial of λ + λ − , and denote by n its degree. Without loss of generality, weassume that − λ − is not a Galois conjugate of λ . Indeed, if − λ − is a Galoisconjugate of λ , we can simply run the argument for λ . Step 1.
By a result of Estes [3], there exists a rational symmetric matrix Q of size ( n + e ) × ( n + e ) with characteristic polynomial f ( t )( x − e , where e equals 1 or 2. Step 2.
By conjugation with an element in O( n + e, Q ) and possibly a smallperturbation, we may assume that the eigenvector of the matrix Q for the eigen-value λ + λ − is positive, compare with the discussion starting with Proposi-tion 5.2 in [10]. Step 3.
Define the matrix M = (cid:18) Q − II (cid:19) . We now describe the characteristic polynomial of M . In the proof of Propo-sition 5.3 in [10], it is shown that µ is an eigenvalue for M with eigenvec-tor ( v , µ − v ) ⊤ exactly if µ + µ − is an eigenvalue for Q with eigenvector v .Hence, the characteristic polynomial of M equals t n f ( t + t − )( t − t + 1) e . We LIVIO LIECHTI AND JOSHUA PANKAU note the following discrepancy with Proposition 5.3 in [10]: if the character-istic polynomial g ( t ) of λ is not reciprocal, then the polynomial t n f ( t + t − )equals g ( t ) g ∗ ( t ), where g ∗ ( t ) = t n g ( t − ). On the other hand, if g ( t ) is recipro-cal, which is the case exactly if λ has a Galois conjugate on the unit circle (forexample if λ is a Salem number), then t n f ( t + t − ) equals g ( t ). In any case, thecharacteristic polynomial of M has integer coefficients and det( M ) = 1. Step 4.
By Proposition 5.4 in [10], for any positive integer k , M k + M − k is ablock diagonal matrix with two blocks Q k . Here, Q k is a rational symmetricmatrix with characteristic polynomial f k ( t )( t − a ) e , where f k ( t ) is the minimalpolynomial of λ k + λ − k and a is among the numbers − , − , ,
2. The proof doesnot depend on whether the characteristic polynomial g ( t ) of λ is reciprocal ornot. Also, by the discussion right above Proposition 5.5 in [10], the eigenspacesof Q and Q k agree. In particular, the eigenvector v for the eigenvalue λ k + λ − k of Q k is positive. Step 5.
Since the eigenvector v for λ k + λ − k is positive, the matrix Q k is positivefor k large enough. To see this, we write e i = c i v + w i for every basis vector e i ,where w i is a fixed vector (independent of k ) in the orthogonal complementof v , and c i >
0. In particular, we have Q k e i = c i ( λ k + λ − k ) v + Q k w i , which becomes positive for large k . Indeed, since w i lies in the orthogonalcomplement to v , it is a linear combination of eigenvectors of Q k other than v .In particular, the modulus of every coefficient of Q k w i is bounded from aboveby | λ k + λ − k | · || w i || ∞ , where λ + λ − is the second-largest root in modulusof f ( t ). However, since λ is a bi-Perron number and − λ − is not among itsGalois conjugates, the ratio between λ k + λ − k and λ k + λ − k becomes arbitrar-ily large when k tends to infinity. Hence, Q k e i eventually becomes positivesince c i > v is a positive vector. Step 6.
For large enough k , the matrix M k has integer coefficients by Propo-sition 5.5 in [10]. Hence, also M − k has integer coefficients for large enough k ,since det( M ) = 1. In particular, also Q k has integer coefficients for largeenough k . This finishes the proof that for k large enough, the number λ k + λ − k equals the spectral radius of a positive symmetric integer matrix Q k . (cid:3) The Coxeter transformations and Thurston’s construction
The Coxeter transformation.
Coxeter groups are abstract generalisa-tions of reflection groups. They admit a presentation encoded in a graph withweighted edges, the so-called Coxeter diagram, and they are linear by Tits’ rep-resentation. As we can single out the only input we need from the theory ofCoxeter groups in Lemma 9 below, we do not give the definitions and insteadrefer to Bourbaki’s classic [2].In case the underlying graph of a Coxeter diagram is bipartite, there is a well-defined conjugacy class of matrices obtained via Tits’ representation, the so-called bipartite Coxeter transformation, see, for example, McMullen [9]. By aresult of A’Campo, the spectrum of this matrix is contained in S ∪ R > , anddetermines, for example, whether the group is finite [1]. All we need for ourpurposes is the following formula relating the spectra of the Coxeter adjacency I-PERRON NUMBERS WITH REAL OR UNIMODULAR CONJUGATES 7 matrix and the bipartite Coxeter transformation. We do not give the definitionof the Coxeter adjacency matrix, but simply note that in our case of Coxeterdiagrams with simple edges (which, in the language of Coxeter groups, meansthat every edge is of weight 3), the Coxeter adjacency matrix equals the ordinaryadjacency matrix of the underlying abstract graph.
Lemma 9.
Let Ω be the adjacency matrix of a finite bipartite graph with simpleedges, understood as a Coxeter diagram Γ with edge weights equal to . Thenthe eigenvalues λ i of the bipartite Coxeter transformation associated with Γ arerelated to the eigenvalues α i of Ω by α i − λ i + λ − i . Proof.
This is exactly what is shown in the proof of Proposition 5.3 of Mc-Mullen’s article [9]. (cid:3)
Pseudo-Anosov stretch factors and Thurston’s construction.
Ahomeomorphism φ of a closed surface Σ is called pseudo-Anosov if there existsa pair of transverse, singular measured foliations of Σ so that φ stretches oneof them by a real number λ > /λ . Thenumber λ is called the stretch factor of the pseudo-Anosov homeomorphism,and it is known to always be a bi-Perron number by a result of Fried [4].Thurston gave a construction of pseudo-Anosov homeomorphisms in terms oftwists along multicurves [16]. We do not review the whole construction, butinstead summarise the main input we need from it in Lemma 10 below. Lemma 10.
For a bi-Perron number λ > , the following are equivalent. (1) The number λ is the stretch factor of a pseudo-Anosov homeomorphismarising from Thurston’s construction. (2) The number λ is the spectral radius of a product of matrices (cid:18) r (cid:19) , (cid:18) − r (cid:19) , where r is the spectral radius of an adjacency matrix of a finite bipartitegraph with simple edges. In order to give the proof, we first review some notions of surface topology. Amulticurve α in a surface Σ is a disjoint union of simple closed curves. Two mul-ticurves are said to intersect minimally if any pair of components has the mini-mal number of intersection points among all representatives in their respectiveisotopy classes. Two multicurves fill a surface Σ if the complement of their unionconsists of discs, once-puntured discs and boundary-parallel annuli. Further-more, given two multicurves α = α ∪· · ·∪ α n and α ′ = α n +1 ∪· · ·∪ α n + m , we de-fine their geometric intersection matrix to be the matrix of size ( n + m ) × ( n + m )whose ij -th entry equals the number of intersection points of α i and α j . Proof of Lemma 10.
Thurston’s construction [16] directly implies the statementfor the case where r is the spectral radius of a geometric intersection matrix ofmulticurves α and α ′ that intersect minimally and fill a surface Σ. It thereforesuffices to show that the set of numbers that appear as spectral radii of suchintersection matrices equals the set of numbers that appear as the spectral radiiof adjacency matrices of finite bipartite graphs with simple edges. LIVIO LIECHTI AND JOSHUA PANKAU
Given two multicurves α and α ′ that intersect minimally and fill a surface Σ,their geometric intersection matrix is, by definition, a symmetric nonnegativeinteger matrix. In the proof of Proposition 2.1 of Hoffman [5], it is shown thatany spectral radius of such a matrix is also the spectral radius of an adjacencymatrix of a finite bipartite graph with simple edges. This proves one direction.Conversely, given an adjacency matrix A of a finite bipartite graph with simpleedges, it is straightforward to abstractly construct a surface Σ filled by two mul-ticurves α and α ′ that have the matrix A as their geometric intersection matrix.These multicurves must intersect minimally, since simple closed curves with zeroor one point of intersection always minimise the number of intersections withintheir respective isotopy classes. (cid:3) Proof of Theorem 1.
We prove the following implications: (a) implies (c)implies (b) implies (a). (a) implies (c) : Let λ be a bi-Perron number all of whose Galois conjugates arecontained in S ∪ R . By Proposition 8, there exists a positive symmetric integermatrix M that has λ k + λ − k as its spectral radius, for some positive integer k .In the proof of Proposition 2.1 of Hoffman [5], it is shown that any numberthat is the spectral radius of a positive symmetric integer matrix is also thespectral radius of an adjacency matrix of a finite bipartite graph with simpleedges. In particular, λ k + λ − k is the spectral radius of an adjacency matrix Ωof a bipartite graph Γ with simple edges. By Lemma 9, the spectral radius x of the bipartite Coxeter transformation associated with Γ equals λ k . Indeed,we have ( λ k + λ − k ) − x + x − , which yields λ k + λ − k = x + x − andhence x = λ k , as x x + x − is a strictly monotonic function on [1 , ∞ ). (c) implies (b) : In the above implication, we have seen that λ k is the spectralradius of a bipartite Coxeter transformation associated with a bipartite Coxeterdiagram with simple edges if and only if λ k + λ − k is the spectral radius of anadjacency matrix Ω of a finite bipartite graph Γ with simple edges. We now useLemma 10 for the matrix product (cid:18) λ k + λ − k (cid:19) (cid:18) − ( λ k + λ − k ) 1 (cid:19) = (cid:18) − ( λ k + λ − k ) λ k + λ − k − ( λ k + λ − k ) 1 (cid:19) , the trace of which equals 2 − ( λ k + λ − k ) . In particular, the eigenvalues mustsatisfy the equation − t − t − = ( λ k + λ − k ) − λ k + λ − k . Hence, theeigenvalue with larger modulus is − λ k and so λ k is the spectral radius ofthe matrix product. By Lemma 10, the number λ k is the stretch factor of apseudo-Anososv homeomorphism arising from Thurston’s construction. (b) implies (a) : Assume that λ k is the stretch factor of a pseudo-Anosov home-omorphism arising from Thurston’s construction. By a result of Hubert andLanneau, the associated trace field Q ( λ k + λ − k ) is totally real [6].We first consider the case where − λ − is not a Galois conjugate of λ . FromProposition 6, we know that Q ( λ + λ − ) equals Q ( λ k + λ − k ) for all positiveintegers k . Hence, if the field Q ( λ k + λ − k ) is totally real, then obviously so mustbe Q ( λ + λ − ), and all Galois conjugates of λ + λ − must be real. We note thatall Galois conjugates of λ are roots of the polynomial t n p ( t + t − ), where p ( t ) isthe minimal polynomial of λ + λ − . In particular, all Galois conjugates λ i of λ must satisfy λ i + λ − i ∈ R and so λ i ∈ S ∪ R . I-PERRON NUMBERS WITH REAL OR UNIMODULAR CONJUGATES 9
In the case where − λ − is a Galois conjugate of λ , Proposition 6 shows thatone out of Q ( λ + λ − ) and Q ( λ + λ − ) equals Q ( λ k + λ − k ). In the formercase, we are done by the above argument. In the latter case, the same argumentgives that all Galois conjugates λ i of λ are contained in S ∪ R . Hence, allGalois conjugates λ i of λ are contained in S ∪ R ∪ i R . We are done by theobservation that no Galois conjugate of λ can be totally imaginary. Indeed,assume λ i is such a Galois conjugate. Then also λ i is a Galois conjugate of λ ,and we have λ i = λ i . As an irreducible integer polynomial has no multiplezeroes, this implies deg( λ ) < deg( λ ), a contradiction by Lemma 5.3.4. Proof of Theorem 4.
We prove (a) implies (b) implies (a). (a) implies (b) : Let λ be a Galois conjugate of a bi-Perron number all of whoseGalois conjugates are contained in S ∪ R . Then λ + λ − is a totally realalgebraic integer, so by a theorem of Salez [13], λ + λ − is an eigenvalue of anadjacency matrix Ω of a finite tree Γ. By Lemma 9, the eigenvalues ρ i of thebipartite Coxeter transformation associated with Γ seen as a bipartite Coxeterdiagram with simple edges are related to the eigenvalues α i of Ω by α i − ρ i + ρ − i . By plugging in λ + λ − for α i we see that λ + λ − = ρ i + ρ − i . Hence we have ρ i = λ , that is, λ is an eigenvalue of the Coxeter transformationassociated with Γ. (b) implies (a) : This follows from the result that all the eigenvalues of theCoxeter transformation of a tree are contained in S ∪ R > , due to A’Campo [1].In particular, we have that all Galois conjugates of λ are contained in S ∪ R > .Now, since λ is a Perron number, if λ , . . . , λ l are the Galois conjugates of λ ,then λ , . . . , λ l are the Galois conjugates of λ by Lemma 5. Hence, all Galoisconjugates of λ lie in S ∪ R . References [1] N. A’Campo:
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Every totally real algebraic integer is a tree eigenvalue , J. Comb. Theory,Ser. B (2015), 249–256.[14] I. Schur: ¨Uber die Verteilung der Wurzeln bei gewissen algebraischen Gleichungen mitganzzahligen Koeffizienten , Math. Z. (1918), no. 4, 377–402.[15] B. Strenner: Algebraic degrees of pseudo-Anosov stretch factors , Geom. Funct. Anal. (2017), no. 6, 1497–1539.[16] W. Thurston: On the geometry and dynamics of diffeomorphisms of surfaces , Bull. Am.Math. Soc. (1988), no. 2, 417–431. Department of Mathematics, University of Fribourg, Chemin du Mus´ee 23, 1700Fribourg, Switzerland
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