An algebraic approach to certain cases of Thurston rigidity
aa r X i v : . [ m a t h . D S ] O c t An algebraic approach to certaincases of Thurston rigidity
JOSEPH H. SILVERMAN
Abstract.
In the moduli space of polynomials of degree 3 withmarked critical points c and c , let C ,n be the locus of maps forwhich c has period n and let C ,m be the locus of maps for which c has period m . A consequence of Thurston’s rigidity theorem is thatthe curves C ,n and C ,m intersect transversally. We give a purelyalgebraic proof that the intersection points are 3-adically integraland use this to prove transversality. We also prove an analogousresult when c or c or both are taken to be preperiodic with taillength exactly 1. Introduction
The moduli space P d of polynomials of degree d ≥ z αz + β .Working over C and choosing appropriate values for α and β , everypolynomial can be put into the form f ( z ) = z d + a z d − + · · · + a d , so P d ∼ = C d − . The polynomial f has d − P crit d for the moduli space of poly-nomials f with marked critical point ( c , . . . , c d − ). Imposing naturalrelations on these critical points gives subvarieties of P crit d , and an im-portant consequence of Thurston’s rigidity theorem [2] is that in manycases these subvarieties have transversal intersection. For example,transversality holds if we require c , . . . , c d − to be periodic with re-spective periods n , . . . , n d − , or more generally if they are preperiodicwith specified tail lengths and periods. Thurston’s theorem also givesanalogous results for rational functions.The proof of Thurston’s general theorem is quite difficult and re-quires deep tools; see [2]. Adam Epstein has asked if one might proveat least some cases of Thurston rigidity using p -adic and/or algebraic Date : November 5, 2018 .2010
Mathematics Subject Classification.
Primary: 37F10; Secondary: 37P0537P45.The author’s research is supported by NSF DMS-0650017 and DMS-0854755. methods. In this note we give an algebraic 3-adic proof of the followingspecial case of Thurston rigidity for cubic polynomials.
Theorem 1.
Let P crit3 be the moduli space of polynomials of degree with marked critical points, i.e., points in P crit3 are equivalence classesof triples ( f, c , c ) , where f ∈ C [ z ] is a polynomial of degree and c and c are the critical points of f .For integers n, m ≥ , let C ,n = (cid:8) ( f, c , c ) ∈ P crit3 : c is periodic with f n ( c ) = c (cid:9) ,C ,m = (cid:8) ( f, c , c ) ∈ P crit3 : c is periodic with f m ( c ) = c (cid:9) . Then C .n and C ,m intersect transversally at all of their points of in-tersection. Our proof of Theorem 1 may be compared with the analogous 2-adicproof for quadratic polynomials that is due independently to Adlerand Gleason; see [1, Lemma 19.1] and [4], and also [3, Appendix] fora generalization. Our proof of Theorem 1 may also be compared withthe recent, independently discovered, 3-adic proof by Epstein [3]. (Wenote that Epstein’s paper contains results stronger than our Theo-rem 1. The primary purpose of our paper is to provide a conceptuallydifferent proof.) Both our proof and Epstein’s proof deduce the finalconclusion, namely that a certain Jacobian determinant is non-zero,by showing that it does not vanish modulo 3. The most difficult partof the proof is to show that the critical points of suitably normalizedpost-critically finite cubic polynomials are 3-adically integral, and thisis where the two proofs differ. Epstein’s proof uses a detailed analysisof the sequence of 3-adic valuations ord (cid:0) f n ( c ) (cid:1) of the points in theforward orbit of a critical point. Thus it makes extensive use of a ( p -adic) metric and has a dynamical flavor. Our proof uses an estimate forthe degrees of the curves C ,n and C ,m , followed by a resultant calcu-lation, so is much more algebraic in nature. We mention in particularthe interesting explicit formula (Lemma 8) for the resultantRes( x p n − x − A, x p m − x − B ) ∈ F p [ A, B ] . This formula is used to show (Theorem 11) that a certain resultant hasmaximal degree by showing that it has maximal degree when reducedmodulo 3.Thurston’s theorem deals also with the case that the critical pointsare preperiodic, i.e., have finite orbits. Using an algebraic trick, we areable to give an algebraic proof of this result for cubic polynomials inthe case that the critical points have tail length at most 1. We give the n algebraic approach to Thurston rigidity 3 exact statement and proof in Section 3. It would be quite interestingto extend this result to allow arbitrary preperiodic behavior.2.
Proof of Thurston Rigidity for Cubic Polynomials
In this section we give the proof of Theorem 1. Making a change ofvariables, we may assume that our cubic polynomials have the form f x,y ( z ) = z − x z + y with marked critical points ± x . For the given integers n, m ≥
1, we let F ( n ) ( x, y ) = f nx,y ( x ) − x and G ( m ) ( x, y ) = f mx,y ( − x ) + x. (1)Then the solutions to F ( n ) ( x, y ) = G ( m ) ( x, y ) = 0 (2)are exactly the pairs ( α, β ) with the property that the critical pointsof f α,β ( z ) have period n and m , respectively.Let ( α, β ) ∈ C be a solution to (2). The curves F ( n ) = 0 and G ( m ) = 0are transversal at ( α, β ) if and only if the Jacobian determinant doesnot vanish, i.e., det F ( n ) x ( α, β ) G ( m ) x ( α, β ) F ( n ) y ( α, β ) G ( m ) y ( α, β ) ! = 0 . In general, the Jacobian determinant is the polynomial J ( x, y ) = det F ( n ) x ( x, y ) G ( m ) x ( x, y ) F ( n ) y ( x, y ) G ( m ) y ( x, y ) ! ∈ Z [ x, y ] . (3)Then the curves F ( n ) = 0 and G ( m ) = 0 intersect transversally at all oftheir intersection points if and only if the ideal (cid:0) F ( n ) ( x, y ) , G ( m ) ( x, y ) , J ( x, y ) (cid:1) ⊂ C [ x, y ]is the unit ideal.We will prove that ( F ( n ) , G ( m ) , J ) = (1) by proving the following twoassertions. • All solutions ( α, β ) to F ( n ) = G ( m ) = 0 are 3-adically integral. • J ( x, y ) ≡ Remark 2.
Our proof of Theorem 1, mutatis mutandis , can be usedto show the following more general result. Let p ≥ f x,y ( z ) = z p − px p − z − y. JOSEPH H. SILVERMAN
The critical points of f x,y are the points ζ x , where ζ ∈ µ p − . Let ζ and ζ be distinct ( p − st -roots of unity. Fix integers n, m ≥
1. Thenthe curves f nx,y ( ζ x ) = ζ x and f mx,y ( ζ x ) = ζ x intersect transversally.We begin with a lemma that describes the iterates of f x,y ( z ) evalu-ated at z = x . Lemma 3.
Let f x,y ( z ) = z − x z + y. Then f nx,y ( z ) = f n − x,y ( z ) , (4) The iterates of f x,y evaluated at x have the following properties :(a) As a polynomial in x , f nx,y ( x ) = n X k =0 a k ( y ) x n − k ∈ Z [ y ][ x ] (5) with deg a k ( y ) ≤ (cid:22) k (cid:23) − k (6) a ( y ) = ( − n − ≡ , (7) a n ( y ) = y n − + (lower order terms) . (8)( By convention, a polynomial with negative degree is the zero poly-nomial. )(b)
Reducing modulo , we have f nx,y ( x ) ≡ x n + y + y + y + · · · + y n − (mod 3) . (c) For n, m ≥ , define F ( n ) ( x, y ) = f nx,y ( x ) − x and G ( m ) ( x, y ) = f mx,y ( − x ) + x. Then F ( n ) ( x, y ) ∈ Z [ x, y ] and G ( m ) ( x, y ) ∈ Z [ x, y ] . Further G ( m ) ( x, y ) = F ( m ) ( − x, y ) , and F ( n ) ( x, y ) ≡ x n − x + y + y + y + · · · + y n − (mod 3) ,G ( m ) ( x, y ) ≡ − x m + x + y + y + y + · · · + y m − (mod 3) . n algebraic approach to Thurston rigidity 5 Remark 4.
The upper bound in the right-hand side of (6) has theform k . . . ⌊ k/ ⌋ − k − − − . . . Experimentally, it seems that the polynomials a k ( y ) appearing in theexpansion (5) satisfydeg a k ( y ) = 4 ⌊ k/ ⌋ − k for all k except k = 3 n − a n − ( y ) = 0. It would probably not be hard to prove this byinduction. Proof.
It is clear that we can write f nx,y ( x ) in the form (5) for somepolynomials a k ( y ) ∈ Z [ y ], so it remains to prove that these polynomialssatisfy (6), (7) and (8). We begin with the proof of (6), which is byinduction on n . To indicate the dependence on n , we write a ( n ) k ( y ). For n = 1 we have f x,y ( x ) = z − x z + y (cid:12)(cid:12) z → x = − x + y, so a (1)0 ( y ) = − , a (1)1 ( y ) = a (1)2 ( y ) = 0 , a (1)3 ( y ) = y. Next we assume that (6) is true for n and we compute f n +1 xy ( x ) = f xy (cid:0) f nxy ( x ) (cid:1) = f nxy ( x ) − x f nxy ( x ) + y = n X k =0 a ( n ) k ( y ) x n − k ! − x n X k =0 a ( n ) k ( y ) x n − k ! + y. (9)We consider first the cubed expression in (9). If it is multiplied out,we obtain a sum of terms of the form a ( n ) i ( y ) x n − i a ( n ) j ( y ) x n − j a ( n ) k ( y ) x n − k = a ( n ) i ( y ) a ( n ) j ( y ) a ( n ) k ( y ) x n +1 − i − j − k with 0 ≤ i, j, k ≤ n . Applying (6) to a ( n ) i ( y ), a ( n ) j ( y ), and a ( n ) k ( y ), wefind thatdeg (cid:0) a ( n ) i ( y ) a ( n ) j ( y ) a ( n ) k ( y ) (cid:1) ≤ (cid:22) i (cid:23) − i + 4 (cid:22) j (cid:23) − j + 4 (cid:22) k (cid:23) − k ≤ (cid:22) i + j + k (cid:23) − i − j − k, where the last line follows from the elementary inequality (see Sec-tion 4) ⌊ t ⌋ + ⌊ t ⌋ + ⌊ t ⌋ ≤ ⌊ t + t + t ⌋ for all t , t , t ∈ R . (10) JOSEPH H. SILVERMAN
Thus terms coming from the cubed expression in (9) satisfy (6) for n + 1. Since it is easy to see that the other terms in (9) satisfy (6) for n + 1, this completes the proof by induction that (6) holds for all n ≥ x and z and weight 0 to y , then the terms of weight 3 n in f nxy ( z ) areprecisely the ones that come from repeatedly cubing the degree 3 ex-pression z − x z , i.e., f nxy ( z ) = ( z − x z ) n − + (lower weight terms) . Hence a ( n )0 ( y ) = coefficient of x n in f nxy ( x )= coefficient of x n in ( − x ) n − = ( − n − . The proof of (8) is a trivial induction on n . More precisely, if welet y have weight 1 and x and z have weight 0, then f n +1 x,y ( x ) = f nx,y ( x ) − x f nx,y ( x ) + y = ( y n − + (lower weight terms)) − x ( y n − + (lower weight terms)) + y = y n + (lower weight terms) . This completes the proof of (a).For (b) we are working modulo 3, so f x,y ( z ) ≡ z + y (mod 3) . An easy induction gives the desired result, f n +1 x,y ( x ) ≡ f (cid:0) f nx,y ( x ) (cid:1) (mod 3) ≡ f nx,y ( x ) + y (mod 3) ≡ ( x n + y + y + y + · · · + y n − ) + y (mod 3) ≡ x n +1 + y + y + y + · · · + y n + y (mod 3) . To prove the first part of (c), we evaluate (4) at z = − x to obtain f nx,y ( − x ) = f n − x,y ( − x ). Substituting this into the definition of G ( n ) ( x, y )yields G ( n ) ( x, y ) = f nx,y ( − x ) + x = f n − x,y ( − x ) + x = F ( n ) ( − x, y ) . Finally, the values of F ( n ) and G ( n ) modulo 3 follow from the valueof f nx,y ( x ) modulo 3 computed in (b). (cid:3) n algebraic approach to Thurston rigidity 7 An immediate consequence of Lemma 3 is the mod 3 value of theJacobian.
Proposition 5.
The Jacobian determinant J ( x, y ) ∈ Z [ x, y ] definedby (3) satisfies J ( x, y ) ≡ . Proof.
Differentiating the formulas for F ( n ) ( x, y ) and G ( m ) ( x, y ) in Lemma 3(c)and reducing modulo 3 yields J ( x, y ) = det F ( n ) x ( x, y ) G ( m ) x ( x, y ) F ( n ) y ( x, y ) G ( m ) y ( x, y ) ! ≡ det (cid:18) − (cid:19) ≡ . (cid:3) Before tackling the 3-integrality of the common roots of F ( n ) ( x, y )and G ( m ) ( x, y ), we prove two elementary lemmas. With an eye towardsgeneralizations, we work over F p . Lemma 6.
Let p be a prime, let m, n ≥ be integers, let d = gcd( m, n ) ,and let τ denote the p -power Frobenius map. Then Y u ∈ F pn Y v ∈ F pm ( T − u − v ) = τ d ◦ ( τ n − ◦ ( τ m − τ d − T ) ∈ F p [ T ] . (11) Remark 7.
The meaning of the right-hand side of (11) is as follows.The rational expression τ d ◦ ( τ n − ◦ ( τ m − τ d − is actually a polynomial in τ ,since d divides m . In other words, it is an element of Z [ τ ]. We thenuse the natural action of Z [ τ ] on F p [ T ] defined by (cid:16)X a i τ i (cid:17) (cid:0) f ( T ) (cid:1) = X a i f ( T ) p i . Proof.
We first observe that if u , u ∈ F p n and v , v ∈ F p m satisfy u + v = u + v , then u − u = v − v ∈ F p n ∩ F p m = F p d . Hence Y u ∈ F pn Y v ∈ F pm ( T − u − v ) = (cid:18) Y w ∈ ( F pn + F pm ) / F pd ( T − w ) (cid:19) p d . Let ϕ ( T ) = Y w ∈ ( F pn + F pm ) / F pd ( T − w ) and ψ ( T ) = ( τ n − ◦ ( τ m − τ d − T ) . JOSEPH H. SILVERMAN
Our earlier observation shows that ϕ ( T ) has distinct roots, and it ismonic of degree p n + m − d .We next observe that for any u ∈ F p n and v ∈ F p m , we have ψ ( u + v ) = ( τ n − ◦ ( τ m − τ d − u + v )= (cid:18) τ m − τ d − (cid:19) ◦ ( τ n − u ) + (cid:18) τ n − τ d − (cid:19) ◦ ( τ m − v )= 0 , since τ n ( u ) = u and τ m ( v ) = v . Thus ψ ( T ) vanishes at each of theroots of ϕ ( T ), and ϕ ( T ) has simple roots, so ϕ ( T ) | ψ ( T ). But ψ ( T ) ismonic and has the same degree p n + m − d as ϕ ( T ). Hence ψ ( T ) = ϕ ( T ),and therefore Y u ∈ F pn Y v ∈ F pm ( T − u − v ) = ϕ ( T ) p d = ψ ( T ) p d = τ d (cid:0) ψ ( T ) (cid:1) = τ d ◦ ( τ n − ◦ ( τ m − τ d − T ) . This completes the proof of Lemma 6. (cid:3)
Lemma 8.
Let p be a prime, let m, n ≥ be integers, let d = gcd( m, n ) ,and let τ denote the p -power Frobenius map. Then working in F p [ A, B ] ,we have Res( x p n − x − A, x p m − x − B ) = τ d ◦ ( τ m − τ d − A ) − τ d ◦ ( τ n − τ d − B ) . Remark 9.
Lemma 8 uses Frobenius to give a compact expression forthe resultant, but we can also write it out explicitly asRes( x p n − x − A, x p m − x − B ) = m/d X i =1 A p id − n/d X i =1 B p id . Proof.
Let α, β ∈ F p ( A, B ) be roots, respectively, of x p n − x − A and x p m − x − B. The extensions ¯ F p ( α ) / ¯ F p ( A ) and ¯ F p ( β ) / ¯ F p ( B ) are Artin–Scheier exten-sions. The conjugates of α over ¯ F p ( A ) are (cid:8) α + u : u ∈ F p n (cid:9) , and similarly for β , so we have factorizations x p n − x − A = Y u ∈ F pn ( x − α − u ) and x p m − x − B = Y v ∈ F pm ( x − β − v ) . n algebraic approach to Thurston rigidity 9 We now computeRes( x p n − x − A, x p m − x − B )= Y u ∈ F pn Y v ∈ F pm ( α + u − β − v ) see [5, 2.13(b)],= Y u ∈ F pn Y v ∈ F pm (cid:0) ( α − β ) − u − v )= τ d ◦ ( τ n − ◦ ( τ m − τ d − α − β ) from Lemma 6,= τ d ◦ ( τ m − τ d − A ) − τ d ◦ ( τ n − τ d − B ) since ( τ n − α ) = A and ( τ m − β ) = B .This completes the proof of Lemma 8. (cid:3) Remark 10.
We observe that for m = n , Lemma 8 can be provendirectly from the Sylvester matrix. To ease notation, let N = p n . Thenthe Sylvester matrix for the resultant of x N − x − A and x N − x − B is the 2 N -by-2 N matrix S ( A, B ) = − A − A . . . . . .1 0 0 − A − B − B . . . . . .1 0 0 − B . (12)If we subtract each row in the top half from the corresponding row inthe bottom half, we obtain an upper-triangular matrix whose diagonalis (1 , , . . . , , B − A, B − A, . . . , B − A ). HenceRes( x p n − x − A, x p m − x − B ) = det S ( A, B ) = ( B − A ) N = B p n − A p n . Proposition 11.
Let F ( n ) ( x, y ) and G ( m ) ( x, y ) be as defined by (1) .Then Res x (cid:0) F ( n ) ( x, y ) , G ( m ) ( x, y ) (cid:1) ∈ Z [ y ] is a polynomial of degree n + m − with integer coefficients and leadingcoefficient relatively prime to .Proof. As in (5) of Lemma 3(a), we write f nx,y ( x ) = n X k =0 a k ( y ) x n − k ∈ Z [ y ][ x ] with polynomials a k ( y ) satisfying (6), (7), and (8). We similarly write f mx,y ( − x ) = n X k =0 b k ( y ) x m − k ∈ Z [ y ][ x ] . (We adopt this notation as being less clumsy for the present proofthan our earlier notation, which would have been a k ( y ) = a ( n ) k ( y ) and b k ( y ) = ( − i +1 a ( m ) k ( y ).) Then F ( n ) ( x, y ) = n X k =0 a k ( y ) x n − k − x and G ( m ) ( x, y ) = m X k =0 b k ( y ) x m − k + x. In order to surpress the extra ± x for the moment, we write F ( n ) ( x, y ) = n X k =0 A k ( y ) x n − k and G ( m ) ( x, y ) = m X k =0 B k ( y ) x m − k , where A k = a k except A n − = a n − −
1, and similarly for B k . Weobserve that the degree estimates for a k given by (6) are true for A k and B k , since the extra ± x is within the specified bound for the degree.To ease notation, we let N = 3 n and M = 3 m . Then the x -resultant of F ( n ) ( x, y ) and G ( m ) ( x, y ) is given by the deter-minant of the Sylvester matrix A A A A · · · A N − A N A A A A · · · A N − A N . . . . . . A A A A · · · A N − A N B B · · · B M − B M B B · · · B M − B M . . . . . . B B · · · B M − B M B B · · · B M − B M B B · · · B M − B M . The Sylvester matrix, which we denote by S , is a square matrix ofsize M + N . Its top M rows have A k coefficients and its bottom N rows have B k coefficients. When we entirely expand det S , it is a sumof terms of the form ( − sign( σ ) M + N Y i =1 S i,σ ( i ) , n algebraic approach to Thurston rigidity 11 where σ is a permutation of { , , . . . , M + N } . We are interested inbounding the degree of this term, so we assume that all of the S i,σ ( i ) are nonzero and computedeg (cid:18) M + N Y i =1 S i,σ ( i ) (cid:19) = M X i =1 deg( S i,σ ( i ) ) + M + N X i = M +1 deg( S i,σ ( i ) )= M X i =1 deg( A σ ( i ) − i ) + M + N X i = M +1 deg( B σ ( i ) − ( i − M ) )= M X i =1 deg( A σ ( i ) − i ) + N X i =1 deg( B σ ( i + M ) − i ) . We now apply the bound (6) from Lemma 3(a), which as we notedearlier applies to A k and B k . This yieldsdeg (cid:18) M + N Y i =1 S i,σ ( i ) (cid:19) ≤ M X i =1 (cid:18) (cid:22) σ ( i ) − i (cid:23) − ( σ ( i ) − i ) (cid:19) + N X i =1 (cid:18) (cid:22) σ ( i + M ) − i (cid:23) − ( σ ( i + M ) − i ) (cid:19) . We rewrite this last expression using fractional part notation, { t } = t − ⌊ t ⌋ , to obtaindeg (cid:18) M + N Y i =1 S i,σ ( i ) (cid:19) ≤ M X i =1 (cid:18) σ ( i ) − i − (cid:26) σ ( i ) − i (cid:27)(cid:19) + N X i =1 (cid:18) σ ( i + M ) − i − (cid:26) σ ( i + M ) − i (cid:27)(cid:19) = 13 (cid:18) M + N X j =1 j − M X i =1 i − N X i =1 i (cid:19) − M X i =1 (cid:26) σ ( i ) − i (cid:27) − N X i =1 (cid:26) σ ( i + M ) − i (cid:27) = M N − M X i =1 (cid:26) σ ( i ) − i (cid:27) − N X i =1 (cid:26) σ ( i + M ) − i (cid:27) ≤ M N m + n − . Since the determinant of the Sylvester matrix is a sum of terms of thisform, we have proven thatdeg(det S ) ≤ m + n − . We are next going to evaluate det S modulo 3. To ease notation, welet Y n = y + y + y + · · · + y n − and Y m = y + y + y + · · · + y m − . Then Lemma 3(c) says that F ( n ) ( x, y ) ≡ x N − x + Y n (mod 3) ,G ( m ) ( x, y ) ≡ − x M + x + Y m (mod 3) . Working modulo 3, this allows us to computeRes x (cid:0) F ( n ) ( x, y ) , G ( m ) ( x, y ) (cid:1) ≡ Res x ( x N − x + Y n , − x M + x + Y m ) (mod 3) ≡ − Res x ( x N − x + Y n , x M − x − Y m ) (mod 3) . We apply Lemma 8 with A = − Y n and B = Y m . Letting d = gcd( m, n )and τ denote 3-power Frobenius, this givesdet( S ) = Res x (cid:0) F ( n ) ( x, y ) , G ( m ) ( x, y ) (cid:1) ≡ − τ d ◦ ( τ m − τ d − − Y n ) + τ d ◦ ( τ n − τ d − Y m ) (mod 3) ≡ τ m ( Y n ) + τ m − d ( Y n ) + · · · + τ d ( Y n )+ τ n ( Y m ) + τ n − d ( Y m ) + · · · + τ d ( Y m ) (mod 3) ≡ Y m n + Y n m + (lower order terms) (mod 3) ≡ y m + n − + (lower order terms) (mod 3) . We have now proven thatdeg(det S ) ≤ m + n − and det S ≡ y m + n − + (l.o.t.) (mod 3) . It follows that det S has degree exactly equal to 3 m + n − and that itsleading coefficient is relatively prime to 3, which completes the proofof Proposition 11. (cid:3) We now have all of the tools needed to prove Theorem 1.
Proof of Theorem 1.
Let ( α, β ) be a solution to F ( n ) ( x, y ) = G ( m ) ( x, y ) = 0 . To ease notation, let R ( n,m ) ( y ) = Res x (cid:0) F ( n ) ( x, y ) , G ( m ) ( x, y ) (cid:1) . n algebraic approach to Thurston rigidity 13 A standard property of the resultant of two polynomials says that itis in the ideal generated by those polynomials [5, 2.13(c)]. Thus thereare polynomials U ( x, y ) , V ( x, y ) ∈ Z [ x, y ] such that U ( x, y ) F ( n ) ( x, y ) + V ( x, y ) G ( m ) ( x, y ) = R ( n,m ) ( y ) . Substituting ( x, y ) = ( α, β ), we find that R ( n,m ) ( β ) = 0. Proposition 11says that R ( n,m ) ( y ) ∈ Z [ y ] has leading coefficient prime to 3, whichproves that β is 3-adically integral. We next use Lemma 3(a) to write F ( n ) ( x, y ) = f nx,y ( x ) − x = ( − n − x n + n X k =1 a k ( y ) x n − k − x. Substituting y = β we see that α is a root of the polynomial F ( n ) ( x, β )whose coefficients are 3-adically integral and whose leading coefficientis a 3-adic unit. Hence α is also 3-adically integral.Now consider the value J ( α, β ) of the Jacobian determinant (3).Proposition 5 says that there is a polynomial K ( x, y ) ∈ Z [ x, y ] satisfy-ing J ( x, y ) = 1 + 3 K ( x, y ) . We know that α and β are 3-adically integral, so the same is trueof J ( α, β ) and K ( α, β ). Taking norms down to Q , we find that N Q ( α,β ) / Q J ( α, β ) = N Q ( α,β ) / Q (cid:0) K ( α, β ) (cid:1) ≡ . In particular, J ( α, β ) = 0. It follows that the ideal (cid:0) F ( n ) ( x, y ) , G ( m ) ( x, y ) , J ( x, y ) (cid:1) ⊂ C [ x, y ]is the unit ideal, since if it weren’t, then F ( n ) , G ( n ) , and J would have acommon root. This completes the proof that the curves F ( n ) ( x, y ) = 0and G ( m ) ( x, y ) = 0 intersect transversally. (cid:3) Preperiodic critical points — a modest beginning
Generalizing the notation from Theorem 1, for i, j ≥ C ,n,i = (cid:26) ( f, c , c ) ∈ P crit3 : f i + n ( c ) = f i ( c ) and f i − n ( c ) = f i − ( c ) (cid:27) ,C ,m,j = (cid:26) ( f, c , c ) ∈ P crit3 : f j + m ( c ) = f j ( c ) and f j − m ( c ) = f j − ( c ) (cid:27) . In words, ( f, c , c ) ∈ C ,n,i if c is purely preperiodic with tail length i and cycle length dividing n , and similarly for C ,m,j . For convenience,we let C ,n, = C ,n and C ,m, = C ,m . Thurston’s theorem implies that C ,n,i and C ,m,j intersect transver-sally. We sketch a 3-adic proof of a very special case. The key to theproof is the following elementary identity. Lemma 12.
We have f n +1 x,y ( x ) − f x,y ( x ) = (cid:0) f nx,y ( x ) − x (cid:1) (cid:0) f nx,y ( x ) + 2 x (cid:1) . (13) In particular, we have F ( n, ( x, y ) = f n +1 x,y ( x ) − f x,y ( x ) (cid:0) f nx,y ( x ) − x (cid:1) ∈ Z [ x, y ] , and F ( n, ( x, y ) satisfies F ( n, ( x, y ) ≡ F ( n ) ( x, y ) (mod 3) ,F ( n, x ( x, y ) ≡ ,F ( n, y ( x, y ) ≡ − . Proof.
The polynomial f x,y ( z ) has a critical point at z = x , so thedifference f x,y ( z ) − f x,y ( x ) should be divisible by ( z − x ) . Explicitly,we find that f x,y ( z ) − f x,y ( x ) = ( z − x ) ( z + 2 x ) . Substituting z = f nx,y ( x ) gives (13). Then the function we have called F ( n, ( x, y ) is given by F ( n, ( x, y ) = f nx,y ( x ) + 2 x. Reducing modulo 3 gives F ( n, ( x, y ) = f nx,y ( x ) + 2 x ≡ f nx,y ( x ) − x = F ( n ) ( x, y ) (mod 3) . The formulas for the partial derivatives of F ( n, ( x, y ) mod 3 then followby differentiating the formula for F ( n ) ( x, y ) given in Lemma 3(c). (cid:3) Theorem 13. (a) C ,n, and C ,m, intersect transversally. (a) C ,n, and C ,m, intersect transversally.Proof. As usual, let f x,y ( z ) = z − x z + y be a cubic polynomial normalized to have critical points ± x . Then thepoints ( f x,y , x, − x ) in C ,n, are the points satisfying f n +1 x,y ( x ) = f x,y ( x ) and f nx,y ( x ) = x. From Lemma 12, these points satisfy F ( n, ( x, y ) = f nx,y ( x ) + 2 x = 0 , n algebraic approach to Thurston rigidity 15 so the locus F ( n, ( x, y ) = 0 contains the curve C ,n, . We will showthat the curves F ( n, ( x, y ) = 0 and G ( m ) ( x, y ) = 0intersect transversally.The first part of the proof is to show that the intersection pointsare 3-adically integral. This can be proven using the resultant meth-ods, mutatis mutandis , of this paper. It is also proven in a more generalsetting by Epstein [3]. We then compute the Jacobian using the congru-ences for the derivatives F ( n, x ( x, y ) and F ( n, y ( x, y ) given in Lemma 12and differentiating the formula for G ( m ) ( x, y ) given in Lemma 3(c).Thus J ( x, y ) = det F ( n, x ( x, y ) G ( m ) x ( x, y ) F ( n, y ( x, y ) G ( m ) y ( x, y ) ! (mod 3) ≡ det (cid:18) − (cid:19) ≡ . The proof for C ,n, and C ,m, is almost identical, since replacing x by − x in Lemma 12 gives G ( m, x ( x, y ) ≡ G ( m, y ( x, y ) ≡ . (cid:3) Proof of (10)For the convenience of the reader, we prove the elementary inequal-ity (10) used in the proof of Lemma 3. For t ∈ R , write t = ⌊ t ⌋ + { t } ,where 0 ≤ { t } < t . Then (10) is equivalentto the inequality { t + t + t } ≤ { t } + { t } + { t } . This inequality is invariant under t i → t i + k for any k ∈ Z , so withoutloss of generality, we may assume that 0 ≤ t i < ≤ i ≤ { t + t + t } ≤ t + t + t , which is trivially true. (There is nothing special about a sum of threeterms. The same proof shows that P ⌊ t i ⌋ ≤ ⌊ P t i ⌋ .) Acknowledgements.
I would like to thank Adam Epstein for suggestinggeneralizing Gleason’s 2-adic proof to prove other cases of Thurston’stheorem, and Adam Epstein and Bjorn Poonen for ongoing discussions of related matters. I would also like to thank Xander Faber for orga-nizing and the CRM for funding the May, 2010 workshop on “ModuliSpaces and the Arithmetic of Dynamical Systems” at the Bellairs Re-search Institute in Barbados, where these discussions began.
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E-mail address : [email protected]@math.brown.edu