On commuting polynomial automorphisms of C2

Publ. Mat. 48 (2004), 227–239 ON COMMUTING POLYNOMIAL AUTOMORPHISMS OF C2 Cinzia Bisi∗ Abstract We characterize the commuting polynomial automorphisms of C2, using their meromorphic extension to P2and looking at their dynamics on the line at infinity. 1. Introduction The group of polynomial automorphisms of C2, Aut(C2), consists of bijective maps: f: (z, w)∈C2→(f1(z, w), f2(z, w)) ∈C2 where f1, f2∈C[z, w]. When fis polynomial and bijective, then the inverse f−1is polynomial. Following [4], we introduce two subgroups of Aut(C2), the group Eof elementary maps E={(z, w)→(αz +p(w), βw +γ) : α, β, γ ∈C, αβ 6= 0, p ∈C[w]} and the group Aof affine maps A={(z, w)→(a1z+b1w+c1, a2z+b2w+c2) : ai, bi, ci∈C, a1b2−a2b16=0}. An elementary map preserve the horizontal foliation dw = 0. We denote by AT =A ∩ E the group of the automorphisms affine and triangular, i.e.: AT ={(z, w)→(a1z+b1w+c1, b2w+c2) : a1, bi, ci∈C, a1b26= 0}. 2000 Mathematics Subject Classification. Primary: 32H50, 14R10; Secondary: 37F10, 58F23. Key words. H´enon maps, indeterminacy points, Green functions, filled Julia set. ∗Partially supported by Progetto MURST di Rilevante Interesse Nazionale Propriet`a geometriche delle variet`a reali e complesse. 228 C. Bisi We recall now a theorem on the structure of Aut(C2) which is known only in dimension 2. It is due to Jung, [5]; it was reproved in several different ways [9] and recently also in [8]. Jung’s Theorem asserts that the group Aut(C2) is the amalgamated product of its subgroups Eand Awith respect to their intersection AT . By this theorem, each automorphism ϕ∈Aut(C2)− AT can be written as a composition of elementary and affine automorphisms which can start or finish indifferently with an affine or an elementary map. A finite composition of maps of the form: hj(z, w) = (pj(z)−ajw, z) = (−ajz+pj(w), w)◦(w, z) = ej◦a (where aj∈C∗,pjis a polynomial of degree dj≥2, ej∈ E,a∈ A and it is the inversion of the coordinates) is called a H´enon map. The set of H´enon maps is a semigroup and it is denoted by H. Proposition 1.1. [4]A polynomial automorphism of C2is conjugate, in the group of polynomial automorphisms, to an elementary map or to a map in H. Let f= (f1, f2) be a polynomial automorphism of C2of algebraic degree d≥2. We will denote by fits meromorphic extension to P2. The graph Γ of fis the closure in P2of the graph of f. Let (z, w) be affine coordinates in C2and let [z:w:t] be corresponding homogeneous coordinates in P2, then the line at infinity L∞has equation {t= 0}. We will denote respectively I+and I−the indeterminacy subsets of f and of f−1. These are two analytic subsets of codimension at least 2 in P2, contained in L∞. It is known, [11, p. 106], that they are both composed by at most one point. If pis an indeterminacy point, we define f(p) as the analytic subset of Γ which projects on p, it coincides with ∩>0f(B(p, )−I); we call f(p) the blow-up at p. Definition 1.2. [11] A polynomial automorphism is regular if I+(f)6= I−(f). The H´enon maps are regular, whereas for elementary maps we have I+=I−. Observe that the notion depends on choice of coordinates. We study, in this paper, the equation f◦g=g◦ffor polynomial automorphisms of C2. The first result that we will prove is the following Main Lemma: On Comm. Polyn. Autom. of C2229 Lemma 1.3. Suppose that f,gare two commuting polynomial automorphisms of C2, not of affine type, then at least one of the two following cases occurs: (i) I+ f=I+ g(which implies also I− f=I− g); (ii) I+ f=I− g(which implies also I− f=I+ g). As a consequence of it, we have that a regular map cannot commute with a non affine elementary map. We get: Proposition 1.4. Let CA(f)be the group of affine automorphisms of C2 which commute with f. If fis regular, then CA(f)is a finite cyclic subgroup of A. Theorem 1.5. Let f,gbe two regular automorphisms of C2respectively of degree d1and d2. Suppose that f◦g=g◦f. Then there exist n0, m0∈ Nsuch that dn0 1=dm0 2and there exists an affine automorphism hsuch that fn0=gm0◦h. Proposition 1.4 and Theorem 1.5 were proved by Lamy, [6], [7] using the action of Aut(C2) on the tree whose vertices are the cosets of the subgroups Aand E; unfortunately this action can be defined only when the group is an amalgamated product, [10], hence Lamy’s approach depends on Jung’s structure theorem and it cannot be generalized to higher dimensions. Since the analogue of Jung’s Theorem is not available in higher dimension, we have introduced a new approach. We think that the approach we follow here will give the centralizer of a regular polynomial automorphism in higher dimension. About commuting elementary maps, first Wright, [14], proved that the group generated by two commuting elementary maps contains Z⊕Z, then Lamy, [7], mentioned that this group is not countable. Acknowledgements. The author is very grateful to N. Sibony and C. Favre: indeed the first one has suggested the interest of a new approach to the problem, the second one has suggested a new approach to Proposition 1.4 and Theorem 1.5. 2. Characterization of commuting polynomial automorphisms of C2 We start recalling the following preliminary result: 230 C. Bisi Proposition 2.1. [11]If fis a non-affine polynomial automorphism of C2, then f(L∞−I+) = I−; f−1(L∞−I−) = I+. This is an immediate consequence of the following elementary property: Suppose that Fand F−1are the lifts of fand f−1to C3, then F◦F−1(x, y, t) = F−1◦F(x, y, t) = td2−1(x, y, t) where d= deg(f) = deg(f−1). Proof of Lemma 1.3: We show first that if I+ f=I+ gthen I− f=I− g. Suppose by contraddiction, that I− f6=I− g. Then (i) either I− f6=I+ g; (ii) or I− g6=I+ f. In case (i), I− f6=I+ g=I+ fhence fis regular. In case (ii), I− g6=I+ f=I+ g hence gis regular. Hence up to change fand g, we can suppose that I− f6=I+ gand that fis regular. We know that the closure of the set K− f intersects the line at infinity only in one point I− fwhich is different from I− g. Therefore it exists at least one point z∈C2such that z∈K− f but g(z)/∈K− f. Hence the sequence {f−n(z)}is bounded, and also the sequence g◦f−n(z) is bounded; on the contrary the sequence f−n◦g(z) is not bounded: this contradicts that f−n◦g=g◦f−n. Assume now that I+ f6=I+ g. Then we have: ∀q∈L∞− {I+ f, I+ g, I− f, I− g}, f−1◦g−1(q) = f−1(I+ g) = I+ f (2.1) unless I+ g=I− f,(2.2) and g−1◦f−1(q) = g−1(I+ f) = I+ g (2.3) unless I+ f=I− g.(2.4) On Comm. Polyn. Autom. of C2231 By the commutation property of fwith g, we have: f−1◦g−1(q) = g−1◦f−1(q) hence, except for the two cases (2.2) and (2.4), we have I+ f=I+ g, which is in contradiction with the assumption. Therefore we have I+ g=I− for I+ f=I− g. But it turns out that one of the relations implies the other one by an argument similar to the starting one. Lemma 1.3 allows us to assume in the rest of the paper that we are in case (i). Corollary 2.2. A non affine elementary map cannot commute with a regular one. The proof follows immediately from Lemma 1.3. In Corollary 2.2 the system of coordinates is fixed, indeed one can give an example of a regular map that after conjugation it is no more regular, [11]. Example 2.3. If f(z, w) = (z2+aw, z) with a6={0}, then fis regular. Let h(z, w) = (w, z +w2), then g=h−1◦f◦his no more regular. Corollary 2.4. Suppose that fand gare two commuting polynomial automorphisms of C2, where fis not of affine type. If fis conjugate to a regular map, then the same holds for g, in the same coordinates, or g is affine. Proof: By hypothesis, there exists an automorphism ρsuch that ˜ f= ρ◦f◦ρ−1is regular. Since ˜g=ρ◦g◦ρ−1commutes with ˜ fthen ˜gis regular or affine. Proof of Proposition 1.4: Recall, [11, p. 132], that a regular biholomorphism fhas infinitely many distinct periodic orbits (this follows from Bezout Theorem), no subvariety of dimension greater or equal than 1 is periodic. First we want to prove that all the periodic points of fcannot lie on the same complex line. Suppose on the contrary that there exists a complex line Lsuch that Sn∈ZFix(fn)⊂L(indeed a periodic point for fis a periodic point also for f−1of the same period). Of course L6=L∞and Lis at the same time f-invariant and f−1-invariant. Let {p}=L∩L∞, then phas to be equal to I−, because f(I−) = I−, and it has also to be equal to I+because f−1(I+) = I+. Since I+6=I−, this is a contradiction. 232 C. Bisi If his affine and f◦h=h◦f, then, for all N∈N,hinduces a permutation on Fix(fN) = {periodic points of order Nfor f}. So we have a group homeomorphism ϕfrom CA(f) into the group ΣNof the permutations of the points of Fix(fN). ϕ:CA(f)→ΣN. If Nis large enough, the points of Fix(fN) do not lie on the same line and hence ϕis injective (an affine map cannot fix more than 5 points not on the same line). Hence CA(f) is a finite group of a suitable order p. To prove the cyclicity of CA(f), we prove that: (1) CA(f) is abelian. (2) The eigenvalues of the linear part of each affine automorphism h∈ CA(f) are roots of unity of the same order. (3) For all h1, h2∈CA(f) of the same order q, there exist n0, m0∈N such that hn0 1=h2and hm0 2=h1. In order to prove (1), we recall that if h◦f=f◦h, then h(I− f) = I− f and h(I+ f) = I+ f. Then,up to conjugation, we can assume that I− f= [1 : 0 : 0] and I+ f= [0 : 1 : 0]. In these coordinates (2.5) h([x:y:t]) = [αx +γt :βy +δt :t]. Consider now the commutator [h1, h2] of two maps h1, h2∈CA(f), then its linear part in C2is the identity 2×2 matrix, because the linear part of each of them is diagonal, see (2.5). But [h1, h2] cannot be a translation of C2because CA(f) is a finite group. Hence the only possibility is [h1, h2] = Id. Since CA(f) is abelian, it follows that all the elements in CA(f) have a common fixed point, hence, up to conjugation, we can suppose that they are all rotations fixing the origin, therefore they are of type (αz, βw). In order to prove (2), we recall that, since the order of the group CA(f) is p, then for all h∈CA(f) there exists k∈Nwhich divides psuch that hk= Id. This means that αk=βk= 1, and the eigenvalues of h are k-roots of unity. But suppose that they have different orders, then there exists a n∈Nwhich divides ksuch that hnis the identity in one component but not in the other one. Suppose that αn= 1 and βn6= 1. This means that all the points (z, 0) are fixed by hn. Since for all m∈Z,fmcommutes with hn, the line {w= 0}is invariant for all fm, with m∈Z. For the invariance of the line {w= 0}by fand by f−1, it follows that the unique point p={w= 0} ∩ L∞has to be equal to I+ f and at the same time to I− f, but this contradicts the regularity of f. On Comm. Polyn. Autom. of C2233 The assertion in (3) follows directly from (1) and (2): since the order q of the rotation is exactly the common order of its eigenvalues, there exist an0∈Nsuch that hn0 1◦h−1 2has an eigenvalue equal to 1. But hn0 1◦h−1 2 is still an element in CA(f) and hence its two eigenvalues have the same order; this implies also that the second eigenvalue has to be equal to 1 and hn0 1=h2. The cyclicity of the group CA(f) follows from (1), (2), (3). If h0is one of the elements of CA(f) of maximal order s≤p, then hh0i=CA(f). Indeed for each h∈CA(f), the order of hhas to be a divisor of the maximal order s; hence there exists an element in hh0i,hr 0, which has the same order of h, but, by (3), his a power of hr 0and so h∈ hh0i. In conclusion CA(f) is isomorphic to Zp. We recall two examples, see [7], to show that it is possible to construct either regular maps fsuch that some element in CA(f) has two equal eigenvalues, or regular maps fsuch that some element in CA(f) has two different eigenvalues, but of the same order. Example 2.5. 1) Consider f= (y, yn+1 +x). Let αbe equal to β and αn= 1, then h= (αx, βy) commutes with f. 2) Consider f= (y, yp+x) and g= (y, yq+x). Let αbe different from βbut αp=βand βq=α, then h= (αx, βy) commutes with f◦g. We now prove Theorem 1.5. We recall that, [11], if fis a regular polynomial automorphism of C2, we can associate to it the sets: K+={z∈C2:{fn(z)}n∈Nis bounded}, K−={z∈C2:{f−n(z)}n∈Nis bounded}, K=K+∩K−, U+=C2−K+, U−=C2−K−, and the Green functions: G+(z, w) = lim n→+∞ 1 dnlog+|fn(z, w)|, G−(z, w) = lim n→+∞ 1 dnlog+|f−n(z, w)|, where d= deg(f) = deg(f−1), GK(z, w) = sup G+(z, w), G−(z, w). 234 C. Bisi Proposition 2.6. [1],[2]If fis a regular polynomial automorphism of C2of algebraic degree d≥2, then •G+and G−are continuous functions on C2and K+={G+= 0}, K−={G−= 0}. •G+and G−are pluriharmonic (p.h.) respectively on U+and U−, and plurisubharmonic (p.s.h.) on C2. •G+◦f=d·G+and G−◦f−1=d·G−. •The closure K+and K−of K+and K−in P2verify: K+=K+∪I+, K−=K−∪I−. •I+is an attractive point for f−1and I−is an attractive point for f. •K=K+∩K−is a compact subset of C2. Proof of Theorem 1.5: First of all we want to prove that: (i) If dm 2≤dn 1with n, m ∈N, then dm 2divides dn 1. Then we will prove that: (ii) If for m, n ∈N,dm 2≤dn 1implies that dm 2divides dn 1, then there exist n0, m0∈Nsuch that dn0 1=dm0 2. A first way to prove (i) is to prove that the Green functions of the two commuting regular automorphisms are equal. The Green function’s approach extends to Ck,k≥3. Let G+ fand G+ gthe Green functions associated to fand g. Consider the function: H1=G+ f◦g d2 . H1is a solution of the following equation, because fcommutes with g: (2.6) H1◦f=d1◦H1. Hence H1and G+ fsatisfy the same functional equation. But from [11], G+ fis the largest solution of the equation (2.6) among the p.s.h. functions bounded by log+|z|+O(1) at infinity. Hence H1=G+ f◦g d2 ≤G+ f On Comm. Polyn. Autom. of C2235 and also, ∀n∈N, Hn=G+ f◦gn dn 2 ≤G+ f (2.7) because G+ f◦gn dn 2 also solves the equation (2.6). On the other hand, limn→+∞ G+ f(gn) dn 2=G+ g. Indeed, if (z, w)∈K+ g, then gn(z, w) is bounded when n→+∞; by the continuity of G+ fwe have that G+ f◦gn(z, w) is also bounded when n→+∞, and hence lim n→+∞ G+ f◦gn dn 2 = 0 on K+ g. Hence any limit function of G+ f(gn) dn 2 is equal to G+ gon K+ g={G+ g= 0}. On the other hand, if (z, w)∈U+ g=C2−K+ gand (z, w) is in a neighborhood of I+ g, log+|z|+c2≤G+ f(z, w)≤log+|(z, w)|+c1 we get log+|z◦gn(z, w)|+c2 dn 2 ≤G+ f◦gn dn 2 ≤log+|gn(z, w)|+c1 dn 2 . The first and the last member of the sequence of inequalities tend to G+ g(z, w). Therefore limn→+∞ G+ f(gn) dn 2=G+ geverywhere on C2. Hence we have that G+ g≤G+ fand interchanging fand gwe get that G+ g= G+ f=G+. Observe that it would be sufficient that fn◦gm=gm◦fn, for some n, m > 1, in order to have G+ f=G+ g. It follows that: (2.8) G+(fn◦g−m) = dn 1G+(g−m) = dn 1 dm 2 G+. Suppose that d2≤d1(if this is not the case, we have that d1< d2 and we can use the same argument exchanging gwith f) and consider h:= g−1◦f. The map his a polynomial automorphism of C2which commutes with a regular one (for example for g), hence, by Corollary 2.4, his affine or regular.