Linear Recurrences in the Degree Sequences of Monomial Mappings
aa r X i v : . [ m a t h . D S ] O c t Linear Recurrences in the DegreeSequences of Monomial Mappings
Eric Bedford and Kyounghee Kim §
1. Introduction.
Let A denote a k × k matrix of rank k which has integer entries.The monomial map f A : C k → C k defined by f A ( x ) = x A = Y j x a ,j j , . . . , Y j x a n,j j (1 . f nA = f A ◦ · · · ◦ f A = f A n . If A ∈ GL ( n, Z ), then f A is birational, and f − A = f A − . A rational map f on projectivespace P k induces a linear map f ∗ on H p,p ( P k ; Z ) ∼ = Z . We define the degree of f n incodimension p to be d ( n ) p := ( f n ) ∗ | H p,p ( P k ); or equivalently (see [RS]), if ω is a K¨ahlerform on P k normalized so that R P k ω k = 1, then d ( n ) p = Z P k f ∗ ( ω p ) ∧ ω n − p . If f : P k → P k is rational, if X is a compact K¨ahler manifold, and if π : X → P k isholomorphic and bimeromorphic, then we have a map ˜ f = π − ◦ f ◦ π : X → X . Therewill be an induced linear map ˜ f ∗ : H p,p ( X ) → H p,p ( X ) . Let χ ( x ) = x m + α m − x m − + · · · + α ∈ Z [ x ] be the characteristic polynomial of ˜ f ∗ | H p,p .If we have ( ˜ f n ) ∗ = ( ˜ f ∗ ) n on H p,p ( X ) (1 . { d ( n ) p } n ∈ N satisfies the linear recurrence d ( n + m ) p + α m − d ( n + m − p + · · · + α d ( n ) p = 0 (1 . n ∈ N .In dimension k = 2, there is only the case p = 1 to consider. Favre [F] has given neces-sary and sufficient conditions for a monomial map in dimension 2 to have a regularization π : X → P satisfying (1.2). Diller and Favre [DF] showed that for every bimeromorphicsurface map there is such a regularization. Favre and Jonsson [FJ] have shown that thedegree sequence of a polynomial map of C always satisfies (1.3).In dimension k = 3, Hasselblatt and Propp [HP] showed that there is a matrix A ∈ GL (3 , Z ) such that the degree sequence { d ( n )1 } = { deg( f nA ) } does not satisfy any linearrecurrence. 1n homogeneous coordinates we have f A [ x : x : . . . : x k ] = [1 : Y j ( x j /x ) a ,j : · · · : Y j ( x j /x ) a n,j ] . (1 . f A so that the coordinates are homogeneous polynomials, then theirdegree is d (1)1 = D ( A ), with D ( A ) := max(0 , k X j =1 a ,j , . . . , k X j =1 a n,j ) + k X j =1 max(0 , − a ,j , . . . , − a n,j ) . (1 . C consisting of ( k + 1) k +1 linear functionals L C : M k → R on the space of real k × k matrices such that D ( A ) = max { L C ( A ) : C ∈ C} (1 . L on the set of k × k real matrices, the sequence { L ( A n ) } n ∈ N satisfies the linear recurrence (1.3) (see Lemma 2.1), where χ ( x ) is the characteristic poly-nomial of A . It follows that the degree sequence { d ( n )1 } = { D ( A n ) } is “almost” the solutionto a linear recurrence: it is the finite maximum over C ∈ C of the sequences { L C ( A n ) } n ∈ N ,each of which satisfies (1.3) but which may have different initial conditions. Another wayof describing this phenomenon is that the space of matrices is divided into different cellsdefined by S C := { M ∈ M k : D ( M ) = L C ( M ) } . (1 . n , A n belongs to one of the cells S C ( n ) , so d ( n )1 = L C ( n ) ( A n ). Although thereare only finitely many cells, the cell C ( n ) might change in a sufficiently irregular way that { d ( n )1 } n ∈ N does not satisfy any linear recurrence at all. This is the approach taken in [HP]and also used here in § Theorem 1.1.
Let A be a k × k integer matrix with rank k . Suppose that for everyeigenvalue λ of A with | λ | ≥ , ¯ λ/λ is a root of unity. Then { d ( n )1 } n ∈ N satisfies a linearrecurrence. On the other hand, suppose that the eigenvalues of largest modulus consist ofa conjugate pair λ and ¯ λ of simple eigenvalues, and that ¯ λ/λ is not a root of unity. Then { d ( n )1 } n ∈ N does not satisfy any linear recurrence. By the duality between H , and H k − ,k − we see that f ∗ A | H k − ,k − is dual to( f A ) ∗ | H , = f ∗ A − | H , . Thus in dimension 3, { d ( f nA ) } = { d ( f nA − ) } . By this dual-ity, Theorem 1.1 gives a rather complete treatment of the cases in dimension 3: Theorem 1.2.
Suppose that the matrix A ∈ GL (3 , Z ) has no eigenvalues of modulus one.In case all three eigenvalues are real, then both { d ( n )1 } n ∈ N and { d ( n )2 } n ∈ N satisfy linearrecurrences. In case there is a non-real eigenvalue, we may write the eigenvalues as λ , ¯ λ ,and ±| λ | − . If | λ | > , then { d ( n )1 } n ∈ N does not satisfy a linear recurrence, but { d ( n )2 } n ∈ N does. And vice versa if | λ | < . In dimension 4 Theorem 1.1 and duality, applied to the map f A : P → P , give:2 heorem 1.3. Suppose that the eigenvalues of A ∈ GL (4 , Z ) are two conjugate pairs λ j , ¯ λ j , j = 1 , , such that neither ¯ λ /λ nor ¯ λ /λ is a root of unity, and | λ | < < | λ | .Then neither { d ( n )1 } n ∈ N nor { d ( n )3 } n ∈ N is given by a linear recurrence. In particular, thereis no map ˜ f A satisfying (1.2) for p = 1 or p = 3 . In § § §
2. Monomial mappings: linear recurrence for d ( n )1 . Recall the linear functions L C in (1.6) and the cells S C in (1.7). From the form of D ( A ) in (1.5) we see that theboundary ∂S C is contained in a union of hyperplanes in the set { A = ( a i,j ) } ⊂ M k of k × k matrices. These hyperplanes are defined, for fixed i , j , and m , by { a i,j = 0 } , { a i,j = a m,j } , { P σ a i,σ = 0 } , and { P σ a i,σ = P σ a m,σ } . We now make two observations aboutlinear recurrences. Lemma 2.1.
Let a k × k matrix A be given. For fixed ≤ i, j ≤ k , the sequence of i, j elements { ( A n ) i,j } n ∈ N satisfies (1.3), and thus for fixed C ∈ C the sequence { L C ( A n ) } n ∈ N satisfies (1.3) Proof.
By the Cayley Hamilton Theorem, the characteristic polynomial satisfies χ ( A ) = 0.Thus each entry ( A n ) i,j satisfies (1.3), and so the Lemma follows. Corollary 2.2.
Suppose A is a k × k integer matrix, and N < ∞ is such that A n ∈ S C for some fixed C ∈ C and for all n ≥ N . Then the degree sequence { d ( n )1 } for f A satisfiesa linear recurrence relation with constant coefficients. Let λ , . . . , λ t denote the distinct non-zero eigenvalues of A, and let µ j ≥ λ j . Then there existconstants α i,j ( s, ℓ ) which do not depend on n such that:( A n ) i,j = t X ℓ =1 µ ℓ − X s =0 α i,j ( s, ℓ ) (cid:18) ns (cid:19) λ n − sℓ ! (2 . Case 1: All eigenvalues are positive: λ > · · · > λ t > . For a linear functional L : M k → R , we define Q L ( n ) := L ( A n ). With reference to (2.1) we set Q L,ℓ ( n ) = µ ℓ − X s =0 L ( α i,j ( s, ℓ )) (cid:18) ns (cid:19) λ n − sℓ , and so we have Q L ( n ) = t X ℓ =1 Q L,ℓ ( n ) . (2 . Lemma 2.3.
In Case 1, there is an integer N such that either (i) Q L ( n ) ≥ for all n ≥ N or (ii) Q L ( n ) ≤ for all n ≥ N . Proof.
Let us look at the form of Q L,ℓ ( n ). It is the sum of terms (cid:0) ns (cid:1) which are polynomialin n and λ n − sℓ , which are exponential in n . Thus Q L,ℓ ( n ) may be written as a polynomial3 ℓ ( n ) in n multiplied by λ nℓ . Let ℓ denote the first value of ℓ for which P ℓ is not the zeropolynomial. It follows that | P ℓ ( n ) | grows like a power of n , and times λ nℓ . Summing verthe remaining ℓ , we see that Q L ( n ) has the same growth. Thus Q L ( n ) is ultimately ≥ ≤ Lemma 2.4.
In Case 1, there exists a positive integer N such that for all n ≥ N , A n belongs to one particular cell. Proof.
For C ∈ C , we consider a linear functional that defines one of the sides of a cell S C . By Lemma 2.3, we know that the sequence L ( A n ) is ultimately ≥ ≤
0. We applythis to all of the linear functionals defining S C , and we find that for all n sufficiently large,either A n ∈ S C or A n / ∈ S C . On the other hand, the set of all S C , C ∈ C , exhausts M k ,so A n must belong to one particular cell. Case 2: Eigenvalues whose modulus is ≥ are positive. Let us choose t such thatthe eigenvalues of A are given by λ > λ > · · · > λ t ≥ > | λ t +1 | ≥ · · · ≥ | λ t | >
0. Asbefore, we let L : M k → R be a linear functional, but now we suppose that it has integercoefficients. Then with Q L and Q L,ℓ as above, we define R L ( n ) by Q L ( n ) = t X ℓ =1 Q L,ℓ ( n ) + R L ( n ) . Lemma 2.5.
In Case 2, there is an integer N such that either (i) Q L ( n ) ≥ for all n ≥ N or (ii) Q L ( n ) ≤ for all n ≥ N . Proof.
As in the Proof of Lemma 2.3, each Q L,ℓ ( n ) is either identically zero or grows like apower of n times λ nℓ . The conclusion of the Lemma must hold, then, unless Q L,ℓ ( n ) = 0 for ℓ equal to 1 , . . . , t . This means that Q L ( n ) = R L ( n ). But now we recall that L has integercoefficients, so that Q L ( n ) ∈ Z . On the other hand, R L ( n ) is made with exponentials withmodulus <
1, so we have R L ( n ) → n → ∞ . We this can happen only if Q L ( n ) = 0for all n .We observe that the hyperplanes bounding S C are defined by functions with integercoefficients, so as in Lemma 2.4 we have: Lemma 2.6.
In Case 2, there exists a positive integer N such that for all n ≥ N , A n belongs to one particular cell. Theorem 2.7.
Let A be a k × k integer matrix with rank k . Suppose that for everyeigenvalue λ of A with | λ | ≥ , ¯ λ/λ is a root of unity. Then { d ( n )1 } n ∈ N satisfies a linearrecurrence. If the eigenvalues | λ | ≥ are all positive, then the degree sequence satisfiesthe linear recurrence relation given by the characteristic polynomial of A . Proof.
For each such eigenvalue λ with | λ | ≥ τ > λ/ | λ | ) τ = 1.Let ˜ τ be the least common multiple of all such τ . Now the eigenvalues of A τ are in Case2. By Lemma 2.6, we find that for fixed ν with 0 ≤ ν ≤ ˜ τ − { d ˜ τn + ν ) } satisfies a linearrecurrence whose coefficients are given by the characteristic polynomial of A τ . Thus thefull sequence also satisfies a linear recurrence.4
3. Monomial mappings: no linear recurrence for d ( n )1 . We will use a fact fromCombinatorics ( see [S, Chapter 4]): If { c k } and { d k } both satisfy linear recurrence rela-tions, then the indices k for which c k − d k = 0 is eventually periodic. Proposition 3.1.
Suppose that A = ( a i,j ) is an integer matrix of rank k , and suppose A has exactly two eigenvalues δ, ¯ δ of maximum modulus, and ¯ δ/δ is not a root of unity.Then the degree sequence d ( n )1 for f A does not satisfy a linear recurrence. Proof.
Let m denote the size of the largest Jordan block with eigenvalue δ or ¯ δ . Wechoose C ∈ C such that A n ∈ C for infinitely many n . Writing λ ℓ , 1 ≤ ℓ ≤ t for the othereigenvalues of A , we have c n := L C ( A n ) = m − X s =0 (cid:18) ns (cid:19) Re (cid:0) β ( s ) δ n − s (cid:1) + t X ℓ =1 µ ℓ − X s =0 α i,j ( s, ℓ ) (cid:18) ns (cid:19) λ n − sℓ ! For the values of n such that A n ∈ S C we have d ( n )1 = L C ( A n ). Since d ( n )1 grows like | δ | n ,and since | λ j | < | δ | , we see that not all the coefficients β ( s ) can be equal to zero.By Lemma 2.1, { c n } satisfies a linear recurrence. If { d ( n )1 } also satisfies a (possiblydifferent) linear recurrence, then by the combinatorial fact above, the indices for which d ( n )1 = c n are eventually periodic. This means that they agree for n belonging to anarithmetic progression An + B . Now we write η = ( δ/ | δ | ) A . Since η is not a root of unity,the numbers Re ( η n ) are dense in the unit circle. Restricting to this arithmetic sequence,we have d ( An + B )1 = c An + B = | δ | An m − X s =0 (cid:18) An + Bs (cid:19) Re (cid:0) δ B − sA β ( s ) η n (cid:1) + O ( | λ | An ) . We let s denote the largest value of s such that β ( s ) = 0, and for large n this will givethe dominant term in the summation. However, there are arbitrarily large values of n forwhich Re ( · · · ) ≤ − | δ B − s A β ( s ) | . So, for such a value of n which is sufficiently large, d ( n )1 will be negative, which is a contradiction. Proof of Theorem 1.1.
This follows directly from Theorems 2.7 and 3.1.
Proof of Theorem 1.2.
The characteristic polynomial of A is an irreducible cubic so ¯ λ/λ cannot be a root of unity, and Theorem 1.2 follows from Theorem 1.1. Example of Hasselblatt and Propp.
The example given in [HP] is f A ( x , x , x ) := ( x x , x x , x ) . The eigenvalues of A are a conjugate pair λ, ¯ λ with | λ | >
1, and ¯ λ/λ is not a root of unity.By [HP] (or by Proposition 3.1), the degree sequence d ( n )1 = deg( f n ) does not satisfy anylinear recurrence relation with constant coefficients. On the other hand the inverse map isa polynomial map g := f − A ( x , x , x ) = ( x , x x , x x ) .
5y Theorem 2.7, the degree sequence d − k := deg( f − kA ) satisfies d − k = d − k − + d − k − + d − k − .Finally, we show that in fact g can be made 1-regular in the sense of (1.2). We startwith the induced map on P g : [ x : x : x : x ] [ x : x x : x x : x x ] . The indeterminacy locus is I ( g ) = { x = x = 0 } ∪ { e := [0 : 0 : 0 : 1] } . The orbits of theexceptional hypersurfaces are: g : { x = 0 } 7→ { x = x = 0 } 7→ e ∈ I ( g ) { x = 0 } 7→ [1 : 0 : 0 : 0] [1 : 0 : 0 : 0] . We see that g does not satisfy (1.2) by the criterion of [FS]: an exceptional hypersurfaceis mapped, after two iterates, completely inside the indeterminacy locus. We consider thecomplex manifold π : X → P obtained by blowing up the point e and then the lineΣ = { x = x = 0 } . By E and S we denote the exceptional fibers over e and Σ respectively. Under the induced map g X , we have g X : Σ := { x = 0 } 7→ S ∩ Σ E ∩ Σ E ∩ Σ S ∩ Σ . It follows from [BK, Theorem 1.4] that the induced map g X is 1-regular. And by duality, g − X = ˜ f A is 2-regular. References [BK] E. Bedford and KH. Kim, Degree growth of matrix inversion: birational maps ofsymmetric, cyclic matrices, arXiv:math/0512507v1[BV] M.P. Bellon and C.-M. Viallet, Algebraic entropy, Comm. Math. Phys.. 204 (1999),425-437.[DF] J. Diller and C. Favre, Dynamics of bimeromorphic maps of surfaces, Amer. J. ofMath., 123 (2001), 1135–1169.[F] C. Favre, Les applications monomiales en deux dimensions, Mich. Math. J. 51 (2003),467-475.[FJ] C. Favre and M. Jonsson, Dynamical compactifications of C .[FS] J.-E. Fornæss and N. Sibony, Complex dynamics in higher dimension, II. ModernMethods in Complex Analysis (Princeton, NJ, 1992) Ann. of Math. Stud. Vol. 137,Princeton Univ. Press, Princeton, NJ, 1995, 135–182.[HP] B. Hasselblatt and J. Propp, Degree-growth of monomial maps, arxiv.org/math.DS/0604521[RS] A. Russakovskii and B. Shiffman, Value distribution for sequences of rational mappingsand complex dynamics, Indiana U. Math. J., 46 (1997), 897–932.[S] R.P. Stanley, Enumerative Combinatorics , Vol 1. Wadsworth and Brooks/Cole, Mon-terey, CA, 1986 Indiana UniversityBloomington, IN 47405 [email protected]
Florida State UniversityTallahassee, FL 32306 [email protected]@math.fsu.edu