On Hilbert s 17th Problem for global analytic functions in dimension 3

Transcription

1 Comment. Math. Hev. 83 (008), Commentarii Mathematici Hevetici Swiss Mathematica Society On Hibert s 17th Probem for goba anaytic functions in dimension 3 José F. Fernando Abstract. Among the invariant factors g of a positive semidefinite anaytic function f on R 3, those g whose zero set Y is a curve are caed specia. We show that if each specia g is a sum of squares of goba meromorphic functions on a neighbourhood of Y, then f is a sum of squares of goba meromorphic functions. Here sums can be (convergent) infinite, but we aso find some sufficient conditions to get finite sums of squares. In addition, we construct severa exampes of positive semidefinite anaytic functions which are infinite sums of squares but maybe coud not be finite sums of squares. Mathematics Subject Cassification (000). 11E5, 3B10, 3S05. Keywords. Hibert s 17th Probem, sum of squares, irreducibe factors, specia factors. 1. Introduction The representation of positive semidefinite functions on a rea variety as a sum of squares has attracted a ot of attention from speciaists in number theory, quadratic forms, rea agebra and rea geometry; the probem goes back to the famous Hibert s 17th Probem for poynomia functions. The soution of that probem (see [Ar]) was the starting point for the deveopment of rea methods in agebra and geometry. Such deveopment ed to the theory of the rea spectrum due to Coste Roy (for more detais see [BCR]) which has been the suitabe technique for an agebraic approach to many probems in rea geometry. This too has been proved fruitfu to understand and sove Hibert s 17th Probem for poynomia functions, Nash functions, anaytic function germs at points and compact sets,, but it has faen short to dea with goba anaytic functions in dimension n 3 without compactness assumptions. Maybe this ack of a suitabe machinery is the main reason why the probem for genera goba anaytic functions has been apart from any substantia progress unti now. Author supported by Spanish GAAR BFM and GAAR Grupos UCM

2 68 J. F. Fernando CMH As it is we known, the probem is whether or not: H17. Every positive semidefinite anaytic function f : R n R is a finite sum of squares of meromorphic functions. Let us mention the best resut we can state today: A positive semidefinite goba anaytic function whose zero set is discrete off a compact set is a finite sum of squares of meromorphic functions ([BKS], [Rz], and [Jw]; see aso [ABR]). Such resut dates back to the 80s. On the other hand, note that in the anaytic setting infinite convergent sums have a meaning, which gives a new viewpoint on the probem (see [ABFR]). Nevertheess, the definition of an infinite sum of squares of anaytic functions, which wi be recaed ater, must be done carefuy to keep the anayticity of such sum. It is cear that an infinite sum of squares, whatever it means, is positive semidefinite and the cassica Hibert s 17th Probem can be weakened to ask whether: h17. Every positive semidefinite anaytic function f : R n R is an infinite sum of squares of meromorphic functions, that is, there exists a nonzero anaytic function g O(R n ) such that g f is an infinite sum of squares of anaytic functions. This is aso a quaitative question, and suggests the study of the finiteness question for the fied M(R n ) of meromorphic functions of R n : Is every infinite sum of squares in M(R n ) aso a finite sum of squares? Obviousy, H17 is equivaent to h17 pus finiteness. Quite remarkaby, finiteness is equivaent to the finiteness of the Pythagoras number p R n of the fied M(R n ) of meromorphic functions on R n (see [ABFR]). We reca that p R n is either the east integer p such that every sum of squares in M(R n ) can be written as a sum of p squares in M(R n ) or + if such integer does not exists. Now, et us fix some terminoogy. Given a cosed set Z R n and an anaytic function f : R n R we say that f is a sum of squares at Z if there exist an open neighbourhood R n of Z in R n such that f is a (possibe infinite) sum of squares of meromorphic functions on. One of the most reevant resuts in [ABFR] is the foowing: ( ) To represent a positive semidefinite anaytic function f as a sum of squares it suffices to represent it at X = f 1 (0). In this work we go further and we search the obstructions for a positive semidefinite anaytic function f to have the foowing property: ( ) To represent f as a sum of squares it suffices to represent its irreducibe factors at their respective zero sets. The most satisfactory resuts hod for dimension 3 and, in fact, we wi prove that ( ) hods for R 3. Furthermore, notice that to represent as sum of squares each irreducibe factor at its zero set is much ess than to represent f at its zero set.

3 Vo. 83 (008) On Hibert s 17th Probem for goba anaytic functions in dimension 3 69 Of course, first of a we have to define carefuy the irreducibe factors of a rea anaytic function on R n. For that, it is crucia to introduce the irreducibe factors of a hoomorphic function F : U C on an open set U C n and to reca some of their main properties. Throughout the paper hoomorphic functions wi refer to the compex case and anaytic functions to the rea case. In both cases the notion of irreducibiity is very simiar, however the behaviours are extremey different. Given an open set R n, we wi say that an anaytic function f O( ) is irreducibe if it cannot be written as the product of two anaytic functions with nonempty zero set. Anaogousy, a hoomorphic function F H(U) on an open set U C n of C n is irreducibe if it cannot be written as the product of two hoomorphic functions on U with nonempty zero set. We reca aso that an anaytic set X of U is irreducibe if it cannot be written as the union of two goba anaytic sets X 1,X X both different from X. First, we consider hoomorphic functions. Here irreducibiity behaves neaty. If every ocay principa sheaf of ideas on U is principa (which happens for instance if H (U, Z) = 0) then there exists a bijection between the irreducibe anaytic sets of U of codimension 1 and the principa prime ideas of the ring H(U) of hoomorphic functions on U. Next, we turn to the rea case. The situation for the irreducibe functions of O(R n ) is competey different and the behaviour of the zero set of an irreducibe function is unpredictabe. The zero set of an irreducibe function of O(R n ) can have any dimension; for instance, if k n the anaytic function f k (x) = f k (x 1,...,x n ) = x 1 + +x k is irreducibe in O(Rn ) but its zero set has dimension n k. Furthermore, there exist irreducibe anaytic functions of O(R n ) with the same zero set but which do not generate the same idea of O(R n ). Take, for instance, f 1 (x) = x 1 + x and f (x) = x 1 + 4x, whose common zero set is {x 1 = 0,x = 0}. Even more, as we wi see in Section, we can produce exampes of rea anaytic functions which are irreducibe but whose zero set is reducibe, and which can even have infinitey many irreducibe components. Thus, one is ed to define the irreducibe factors of a rea anaytic function f through the irreducibe factors of a hoomorphic extension F of f to a suitabe open neighbourhood U of R n in C n. As usua the uniqueness of the irreducibe factors wi be up to mutipication by units of the respective ring, O(R n ) or H(U), that is, never vanishing anaytic or hoomorphic functions. The process to construct the irreducibe factors of a rea anaytic function f wi be deveoped carefuy in Section, but we can describe roughy the main steps. (i) First, we consider a hoomorphic extension F : U C of f to a suitabe open neighbourhood U of R n in C n, invariant under conjugation, and decompose S = F 1 (0) as the union of its irreducibe components {S i } i I. We show that we can assume that S i R n = for a i I.

4 70 J. F. Fernando CMH (ii) Next, for each S i we construct a hoomorphic function H i which generates the idea sheaf J Si of S i. For those S i which are invariant under conjugation we prove that the hoomorphic function H i can be chosen to restrict on R n to a rea anaytic function. (iii) The hoomorphic function germs {H i,r n} i at R n, which are unique, wi be caed the irreducibe factors of f. Moreover, f O(R n ) is irreducibe if either (1) f has one irreducibe compex factor, whose zero set germ at R n is invariant under conjugation, or () two irreducibe compex factors whose respective zero set germs at R n are conjugated. In case (1) F is irreducibe, and in case () s reducibe. Given anaytic functions g, f O(R n ), we say that g divides f(in O(R n )) with mutipicity k 1ifg k divides f but g k+1 does not. As it can be checked, taking germs at any point of R n at which both f and g vanish, if g divides f, there exist an integer k 1 with the previous property. An irreducibe factor h O(R) of an anaytic function f O(R n ) is specia if the zero set germ at R n of a hoomorphic extension of h is invariant under conjugation, it divides f with odd mutipicity and 1 dim h 1 (0) n. In cose reation to the irreducibe factors of positive semidefinite anaytic functions we wi prove in Section the foowing decomposition resut that wi be crucia for our purposes. Lemma 1.1. Let f : R n R be a positive semidefinite anaytic function. Then there exist anaytic functions f 0,f 1,f,f 3 : R n R such that f 1, f, f 3 are positive semidefinite, f = f0 f 1f f 3 and (i) f1 1 (0) is a discrete set (hence, by [BKS], f 1 is a finite sum of squares of meromorphic functions on R n ), (ii) f is a sum of two squares of anaytic functions on R n, and (iii) the irreducibe factors of f 3 are a specia and divide f 3 with mutipicity one. In fact, we aso see that the irreducibe factors of f 3 are the specia irreducibe factors of f or just the specia factors of f. Moreover, if n we may take f 3 1 in Lemma 1.1. Hence, we get that f is a finite sum of squares of meromorphic functions (this is of course we known: [BKS] and [Jw1]). Thus, in what foows we may assume n 3. Next, we reca the suitabe definition of infinite sums of squares introduced in [ABFR]: Definition 1.. Let R n be an open set. An infinite sum of squares of anaytic functions on is a series k 1 f k where a f k O( ), such that:

5 Vo. 83 (008) On Hibert s 17th Probem for goba anaytic functions in dimension 3 71 (i) the f k s have hoomorphic extensions F k s, a defined on the same neighborhood U of in C n, and (ii) for every compact set L U, k 1 sup L F k < +. Accordingy, the infinite sum of squares k 1 f k defines we an anaytic function f on = U R n and we write f = k 1 f k O( ); of course, this triviay incudes finite sums. Hence, it makes sense to say that an eement of the ring O( ) is a sum of p squares in O( ), even for p =+. We reca that an anaytic function f : R is a sum of p + squares (of meromorphic functions on ) if there is g O( ) such that g f is a sum of p squares of anaytic functions on. The zero set {g = 0} is caed the bad set of that representation as a sum of squares. The choice of a suitabe sum of squares representation wi be a crucia matter and we wi need often to have a controed bad set, that is, a bad set contained in the zero set {f = 0}. Concerning the difference between arbitrary and controed bad sets, we reca this Proposition 1.3 ([ABFR, 4.1]). Let R n be an open set and et f : R be an anaytic function which is a sum of p + squares of meromorphic functions. Then f is a sum of q n p squares with controed bad set. Moreover, on a smaer neighborhood of {f = 0} we can assume q n 1 p. Our main resut here is the foowing: Theorem 1.4. Let f : R n R be a positive semidefinite anaytic function and et {h j } j J be the specia irreducibe factors of f. Assume that for each j J the positive semidefinite anaytic function h j is a (possiby infinite) sum of squares of meromorphic functions at X j = hj 1 (0). Let {Y i } i I be the famiy of the irreducibe components of the goba anaytic set X = j J X j. Suppose that one of the two foowing conditions hods true: (a) Y i Y k is a discrete set for i = k. (b) Y i is a compact set for a i I. Then f is a possiby infinite sum of squares of meromorphic functions on R n with controed bad set. The proof of the previous resut goes aong the same ines of the one of [ABFR, 1.5], but there are severa aspects that go far beyond a mere updating of [ABFR, 1.5]. Moreover, one of the main difficuties for the proof of Theorem 1.4 and the reason why the hypotheses (a) and/or (b) appear in its statement is that it cannot exist a genera formua to mutipy infinitey many sums of squares; even, if these sums of squares are finite. On the other hand, if n = 3, then the condition (a) in the statement of Theorem 1.4 is aways satisfied, since dim X = 1, and we get the foowing reevant consequence:

6 7 J. F. Fernando CMH ( ) To represent a positive semidefinite anaytic function f : R 3 R as a sum of squares it is enough to represent its specia factors at their zero sets. Thus, the probem h17 for R 3 is reduced to study if every specia positive semidefinite irreducibe anaytic function on R 3 is a sum of squares of meromorphic functions. Moreover, from Theorem 1.4 (b) and [ABR, VIII.5.8] one gets amost straightforwardy the foowing: Coroary 1.5. Let f : R n R be a positive semidefinite anaytic function and et {h j } j J be its specia factors. Suppose that for a j J, the set X j = hj 1 (0) is compact. Then f is a (possiby infinite) sum of squares of meromorphic functions in R n. The previous resut points out that the obstruction to be an infinite sum of squares concentrates on the specia irreducibe factors whose zero set is not compact. Concerning finite sums of squares the situation is quite more deicate and we ony have some partia resuts. Theorem 1.6. Let r 0 be an integer and et f : R n R be a positive semidefinite anaytic function. Let {h k } k K be the specia factors of f and et X k = h 1 k (0). Suppose that each proper intersection X k X is a discrete set. If h k is a sum of r squares at X k for a k K, then f is a sum of r+n squares. In fact, the previous resut can be improved if we find a suitabe distribution of the specia factors. Namey, Coroary 1.7. Let r 0 be an integer and et f : R n R be a positive semidefinite anaytic function. Let {h k } k K be the specia factors of f and et X k = h 1 k (0). Suppose that there exists a partition P ={A 1,...,A m } of K such that each proper intersection X k X, where k, beong to the same A j, is a discrete set. If h k is a sum of r squares at X k for a k K, then f is a sum of r+n squares. As we wi show in Section 3, this more technica statement aows us to represent as finite sum of squares certain positive semidefinite anaytic functions to which Theorem 1.6 does not appy. Athough the situation described in Coroary 1.7 is quite genera for positive semidefinite functions on R 3, we aso construct in Section 3 two exampes of positive semidefinite anaytic functions on R 3 to which we can appy Theorem 1.4 (hence, they are infinite sums of squares) but to which we cannot even appy our best resut Coroary 1.7 about finite sum of squares: The first function f has the foowing properties: (1) the zero sets of a its specia factors, which are infinitey many, have a infinitey many irreducibe components; () its specia factors are sums of four squares in O(R 3 ); and

7 Vo. 83 (008) On Hibert s 17th Probem for goba anaytic functions in dimension 3 73 (3) it has a hoomorphic extension F to an open neighbourhood U of R 3 in C 3 which is the uniform imit of a sequence {F k } k of sums of four squares of hoomorphic functions on U whose restrictions to R 3 are rea anaytic functions. 1 However, any two of the specia factors of this function share infinitey many irreducibe components, which makes Coroary 1.7 useess. The second function f has the foowing properties: (1) the zero sets of a its specia factors, which are infinitey many, have a finitey many irreducibe components; () its specia factors are sums of four squares in O(R 3 ); (3) each irreducibe component of the zero set of f is contained in no more than two of the zero sets of its specia factors; and (4) it is the uniform imit of sums of four squares in O(R 3 ) in the sense of the previous exampe. Nevertheess, there does not exist an integer s such that the number of irreducibe components of the zero set of each specia factor is bounded by s. Again, as we wi see ater, Coroary 1.7 is useess here. For the moment, we do not know whether or not the previous exampes are finite sums of squares of meromorphic functions on R 3. Both exampes have been constructed with the purpose of having a measure of the imitations of Coroary 1.7. This is done avoiding the hypothesis of Coroary 1.7 about the distribution of the zero sets of the specia factors, whie keeping a the other hypotheses that seem essentia to have a finite sum. In fact, the second exampe avoids the hypothesis about the distribution of the zero sets in a quite subte way. The reevance of such exampes arises from the way and the purpose for what they have been constructed. In fact, they seem to be at the border between finite and infinite sums of squares. Thus, they are suitabe candidates to be counterexampes to Hibert s 17th Probem for goba anaytic functions. On the other hand, if one is abe to prove that some or both of them are finite sums of squares, it seems pausibe to find reevant information to prove some version of Theorem 1.4 for finite sums of squares. The paper is organized as foows. In Section we prove some key resuts concerning the definition and computation of irreducibe factors of a rea anaytic function and the decomposition of positive semidefinite anaytic functions described in Lemma 1.1. Section 3 is devoted to the introduction and deveopment of the exampes previousy mentioned. Finay, Theorems 1.4 and 1.6, are proved in Section 4. The author woud ike to thank Prof. J. M. Ruiz for friendy hepfu discussions during the preparation of this work. 1 As it is we known, the best way to guarantee the anayticity of the uniform imit of a sequence of rea anaytic functions on R n is to consider ony sequences of such functions which have hoomorphic extensions to a common open neighbourhood of R n in C n.

8 74 J. F. Fernando CMH. Irreducibe factors We gather here some notations and technica emmas for ater purposes. Athough our probem concerns rea anaytic functions, we wi of course use some compex anaysis. For hoomorphic functions we refer the reader to the cassica [GuRo]..1 Genera terminoogy. Denote the coordinates in C n by z = (z 1,...,z n ), with z i = x i + 1y i, where x i = Re(z i ) and y i = Im(z i ) are respectivey the rea and the imaginary parts of z i. Consider the usua conjugation σ : C n C n, z z = (z 1,...,z n ), whose fixed points are R n. A subset A C n is (σ -) invariant if σ (A) = A; obviousy, A σ (A) is the biggest invariant subset of A. Thus, we see rea spaces as subsets of compex spaces. The notations Int and C stand for topoogica interiors and cosures, respectivey. Let U C n be an invariant open set and et F : U C be a hoomorphic function. We say that s (σ -) invariant if F(z) = F(z). This impies that F restricts to a rea anaytic function on U R n. In genera, we denote by and R(F ): U C, I(F ): U C, z z F(z)+ F(z), F(z) F(z), 1 the rea and the imaginary parts of F, which satisfy F =R(F ) + 1 I(F ). Note that both are invariant hoomorphic functions. Given a cosed set Z C n, germs (of sets or of hoomorphic functions) at Z are defined exacty as germs at a point, through neighborhoods of Z in C n ; we wi denote by F Z the germ at Z of a hoomorphic function F defined in some neighborhood of Z. For instance, if F : U C is an invariant hoomorphic function such that R n U and Z = R n, then the germ F Z is the same as the rea anaytic function f = F R n. In [ABFR,.3] we showed how to extend a convergent sum of squares of hoomorphic functions moduo another. Here such resut wi be again a powerfu too and we reca the precise statement for the sake of the reader. Proposition.. Let U be an invariant open Stein neighborhood of R n in C n and et : U C be an invariant hoomorphic function. Let V be an open invariant neighborhood of the connected components of 1 (0) that meet R n, and suppose that V does not meet the other connected components of 1 (0). Let C k : V C be a famiy of invariant hoomorphic functions such that k sup L C k < + for every compact set L V. Then, there exist invariant hoomorphic functions A k : U C, such that k sup K A k < + for every compact set K U and V divides a the differences A k V C k.

9 Vo. 83 (008) On Hibert s 17th Probem for goba anaytic functions in dimension Compex irreducibe factors of a rea anaytic function. Let f : R n R be an anaytic function. We define the irreducibe compex factors of f as foows. Let F : U 0 R be a hoomorphic extension of f to a sma enough open neighborhood U 0 of R n in C n. By [WBh, 6. Proposition 9], there exists a unique ocay finite famiy of irreducibe germs {S i } i 1 at R n such that S i S j if i = j and F 1 (0) R n = i I S i. By [WBh, 6. Prop. 8, Cor. ], for each i I there exists an open neighborhood U i of R n in C n and an irreducibe anaytic set T i in U i such that T i,r n = S i. Shrinking the open sets U i (if necessary), we may assume that the famiy {T i } i I is ocay finite in C n. In fact, we can take the U i s arbitrariy sma. Consider the open set in C n : U = ( C n \ ) C C n(t j ) j 1 j 1 ( U j \ i =j(c ) C n(t i ) \ U i ). A straightforward computation shows that R n U and that T i U is cosed in U for a i I. Hence, for each i I there exists an anaytic set T i T i in U, which for simpicity we denote again by T i, such that T i,r n = S i. Let U U be an open invariant Stein neighborhood of R n in C n such that R n is a deformation retract of U ([Ca]). Denote again by T i the intersection T i U. Taking the connected component of T i that intersects R n instead of T i we may aso assume that T i is irreducibe. Fix i I. Since the dim T i = n 1, for each z T i there exists a hoomorphic function germ h i,z O(C n z ) that generates the idea of the anaytic germ T i,z. Consider the subsheaf J (i) of the structure sheaf O U defined by { J z (i) h i,z O(C n z = ) if z T i, O(C n z ) if z T i. Note that J (i) is a ocay principa coherent idea sheaf. Hence, it defines a cocyce in H 1 (U, OC ). Since U is a Stein manifod, this group is isomorphic to H (U, Z), which is 0 since R n is a deformation retract of U. Hence, J (i) is in fact a principa idea sheaf, say generated by a hoomorphic function H i : U C. We aso have, by the definition of J (i), that Hi 1 (0) = T i and that h i,z O(C n ) z = H i,z O(C n ) z for a z T i. Thus, the germ S i = T i,r n is determined by the hoomorphic function H i. Moreover, since F 1 (0) = i I T i, each H i divides F. Furthermore, since the germs S i are uniquey determined by F and the function germ at R n of each H i is uniquey determined by S i, the hoomorphic function germs H i,r n are uniquey determined by f F R n. Thus, we wi say that {H i,r n} are the (compex) irreducibe factors of f. Next we caim: If S i is invariant, we may assume that H i is aso invariant. Indeed, since S i is invariant, the function H i σ has the same properties as H i. Hence, there exists a hoomorphic function A i : U C not vanishing on U such

10 76 J. F. Fernando CMH that H i σ = H i A i. Thus, we have that H i = H i σ A i σ = H i A i A i σ and therefore A i A i σ = 1. Now, et B i : U C be a hoomorphic function such that Bi = A i and B i B i σ = 1. Indeed, (B i B i σ) = A i A i σ = 1; hence, B i B i σ = ±1. Since the function B i B i σ restricts on R n to a sum of two squares in O(R n ),we deduce that B i B i σ = 1. Then H i = H ib i : U C is invariant: H i σ = H i σ B i σ = H i σ B i σ B i = H i σ A i σb i = H i B i = H i. Moreover, it is cear that H i satisfies the same properties as H i with respect to the germ S i. To simpify notations we denote again H i by H i. Reca that, H i being invariant, its restriction to R n is a rea anaytic function. Definitions.4. (a) We say that H i,r n is a specia irreducibe (compex) factor of f or just a specia factor of f if the germ of Hi 1 (0) at R n is invariant, the dimension d of the rea anaytic set Hi 1 (0) R n satisfies the inequaities 1 d n (which for n = 3givesd = 1) and H i divides F with odd mutipicity. Moreover, since Hi 1 (0) R n has dimension n, we may aso assume that the specia factors H i of f are invariant and that h i = H i,r n is a rea positive semidefinite anaytic function. (b) If f has ony one irreducibe (compex) factor and it is specia, we say that f is a specia anaytic function. We reca that a goba anaytic set (in an open set of R n )isirreducibe if it cannot be written as the union of two goba anaytic sets different from itsef. By [WBh, 8. Prop. 11] any goba anaytic set X in an open set of R n can be written as the union of a unique irredundant ocay finite famiy of irreducibe goba anaytic sets X i with X = i X i. Exampes.5. (a) f(x,y,z) = (x + y ) z + x 6 + y 6 defines a specia anaytic function whose rea zero set is {x = 0,y = 0}, which is irreducibe. (b) f(x,y,z) = x 4 y + y 4 z + z 4 x 3x y z (Motzkin s poynomia) defines a specia anaytic function whose rea zero set is {x 4 y = y 4 z = z 4 x }, that is, which is reducibe. {x = 0,y = 0} {x = 0,z= 0} {y = 0,z= 0} {y =±x,z =±x} {y =±x,z = x}, To introduce more exotic exampes we need the foowing resut:

11 Vo. 83 (008) On Hibert s 17th Probem for goba anaytic functions in dimension 3 77 Lemma.6. The homogeneous poynomia F(x,y,z) = (z + ax) (z + by) + z 4 + c x y R[x,y,z] is irreducibe in C[x,y,z] for a a,b,c R such that a,b,c = 0, c = a b. Proof. First, note that the zero set of n R 3 is the union of the two ines x = 0,z= 0 and y = 0,z = 0. Next, we write where F = Ax + azb x + z C, A = a z + a byz + (a b + c )y = (az + aby + 1 cy)(az + aby 1 cy), B = z + by, C = (z + by) + z = (z + by + 1 z)(z + by 1 z). Let us show now that if a, b, c, a b c = 0, then s irreducibe. Suppose that s reducibe. First, since gcd(a, zb, z C) = 1 (because gcd(b, C) = 1 and z does not divide A) we have that F cannot be written as F = G 1 G, where G 1 C[x,y,z] is a poynomia of degree 0 with respect to x. Next, we see that F cannot be written as the product of two inear rea factors with respect to x, namey, F = (α 1 x + β 1 )(α x + β ) where α i,β i R[y,z]. If this were the case, the set {α 1 x + β 1 = 0} R 3, which has dimension, woud be a subset of {F = 0} R 3, which has dimension 1, a contradiction. Thus, if s reducibe, it has two conjugated roots in C(y, z), namey, azb ± z a B 4 AC. A Hence, a B 4 AC 1R(y, z) and, in fact, since R[y,z] is a norma domain, we have that AC a B 4 R[y,z]. Therefore, AC a B 4 = H, where H R[y,z] is a quadratic form. Thus, AC = a B 4 + H and ooking at the factors of A and C we essentiay have the foowing possibiities: (i) C divides ab + ih. Since C R[y,z], we have that C divides B, a contradiction.

12 78 J. F. Fernando CMH (ii) E 1 = (az + aby + 1 cy)(z + by + 1 z) divides ab + ih, that is, ab + ih = ηe 1 for some η C. Then, there exists λ, μ R such that λ(az + (ab c)yz + b ay ) + μ(az + (ab + c)yz + cby ) = ab = a(z + by) = az + abyz + ab y. Thus, λ + μ = 1, λ(ab c) + μ(ab + c) = ab and λab + μcb = ab. Therefore, λ = 1 μ, μ(c ab) = c, bμ(c ab) = 0. Hence, b, c being nonzero, we deduce that c = ab, a contradiction. (iii) E = (az + aby + 1 cy)(z + by 1 z) divides ab + ih. Proceeding as in the previous case, we deduce that c = ab, a contradiction. Whence, we concude that s irreducibe in C[x,y,z]. Exampe.7. Let H : C C be an invariant hoomorphic function such that H 1 (0) = {k Z : k 0} and H has a zero of order one at each point of its zero set. Such a function exists by the Weierstrass Factorization Theorem. Let a k = H (k) = 0 for a integer k 0 and et M>0be a rea number such that M = 1 ak for a coupe of integers k, 0. Let a F(x,y,z) = (z + sin(πx)) (z + sin(πy)) + z 4 + M H(x) H(y), which is an invariant hoomorphic function on C 3. Let us show that f = F R 3 is a specia anaytic function whose rea zero set is the net S = {x = k, z = 0} =, z = 0} k 0 0{y which has infinitey many irreducibe components. Proof. Indeed, a straightforward computation shows that {F = 0} R n = S. To show that f is a specia anaytic function we have to check that the restrictions of F to sma enough invariant neighbourhoods U of R n in C n cannot be written as the product of two hoomorphic functions G 1,G : U C such that G 1 i (0) R n =. First, we show some crucia properties of F to prove the irreducibiity of f. (a) For each pair of non-negative integers k, 0 consider the point p k, = (k,, 0). Then the function germs F pk, are irreducibe in the ring O C 3,p k, for a k,. This is because their initia forms at the points p k, are irreducibe by Lemma.6.

13 Vo. 83 (008) On Hibert s 17th Probem for goba anaytic functions in dimension 3 79 (b) For each integer k 0 and each λ R \ Z consider the point p k,λ = (k, λ, 0). Then, for a k, λ as before the function germs F pk,λ are the product in O C 3,p k,λ of two irreducibe factors of order 1 that vanish at the ine germ x = k, z = 0. Indeed, after transating the point p k,λ to the origin, the initia form of F pk,λ is (z + ax) b + c x for some rea number a,b,c > 0. Thus, by cassification of singuarities, F pk,λ is anayticay equivaent either to x + z or to a poynomia of the type x + z + εy k where k 3 and ε =±1. Since the zero set of F pk,λ is the ine germ x = k, z = 0, we concude that F pk,λ is anayticay equivaent to x + z. Hence F pk,λ = F 1 + F = (F 1 + 1F )(F 1 1F ), where the factors F 1 + 1F,F 1 1F have order 1, are irreducibe and vanish at the ine germ x = k, z = 0. Suppose now that there exist an open invariant neighbourhood U of R n in C n and two hoomorphic functions G 1,G : U C such that G 1 i (0) R n = and F = G 1 G. Then, for each p S = F 1 (0) R n we have that F p = G 1,p G,p. Since the ine x = 0,z = 0 is irreducibe and it is contained in F = 0, we may assume that it is aso contained in G 1 = 0. As the germ F p is irreducibe for the points p = p 0, = (0,,0), we have that a the ines y =, z = 0 are contained in G 1 = 0. Furthermore, since the germ F p is irreducibe for the points p = p k,0 = (k, 0, 0),we have that a the ines x = k, z = 0 are contained in G 1 = 0. Hence, S is a subset of G 1 = 0. Again, since the germ F p is irreducibe for the points p = p k, = (k,, 0), we have that no ine of S can be contained in G = 0. Hence, the ines being irreducibe, we deduce that G 1 (0) Rn has to be a discrete set contained in S but which does not intersect the set {(k,, 0) : k, 0}. Next, we take p G 1 (0) Rn ; we may assume p = (k, λ, 0) for certain integer k 0 and certain λ R which is not a non negative integer. Since F p = G 1,p G,p is a product of two irreducibe factors of order 1 that vanish at the ine germ x = k, z = 0, we concude that G,p must vanish at the ine germ x = k, z = 0, a contradiction. Thus, f is a specia anaytic function..8 Decomposition of rea anaytic functions. Now we proceed to prove the decomposition resut Lemma 1.1 announced in the introduction. We first reca a we-known resut to get rid of the squares. Lemma.9. Let f : R n R be a positive semidefinite anaytic function. Then we can factorize f = f 0 f, where f 0, f are anaytic functions on R n such that f is squares free in O(R n ) and its zero set has codimension. Proof. Firsty, at each zero x of f, we write f x = ζ x η x O R n,x, η x without mutipe factors; this factorization is unique up to units. The germ {η x = 0} has codimen-

14 80 J. F. Fernando CMH sion, because otherwise some irreducibe factor ξ x of η x woud be rea, and f x woud change sign at x. Now, the ζ x s generate a ocay principa coherent sheaf J O R n. Since H 1 (R n, Z ) = 0, J is gobay generated by a goba anaytic function f 0 : R n R. An easy computation shows that f0 divides f, and we have f = f 0 f. Each germ f x coincides with η x up to a unit, hence its zero set has codimension, f x does not change sign and it is squares free. Now we are ready to prove Lemma 1.1: Proof of Lemma 1.1. By Lemma.9 there exist anaytic functions f 0,f : R n R such that f is squares free, dim{f = 0} n and f = f 0 f. Hence, a the specia factors of f divide it with mutipicity one. By.3, there exist: An open invariant Stein neighborhood U of R n in C n such that R n is a deformation retract of U, A hoomorphic extension F of f to U, and Hoomorphic functions H j : U C, j J, such that {S j = Hj 1 (0) R n} j J are the (compex) irreducibe components of the germ F 1 (0) R n and H j generates the idea of Hj 1 (0). Furthermore, if S j is invariant we may assume that H j is aso invariant, hence h j = H j R n defines an anaytic function on R n. Let J 1 ={j J : dim(s j R n ) = 0, S j = σ(s j )}, J ={j J \ J 1 : S j = σ(s j )} and J 3 = J \ (J 1 J ). Consider the bijection ˆσ : J J defined by S ˆσ(i) = σ(s i ). This bijection defines on J (together with the identity) an equivaence reation. For each equivaence cass α we choose a representative j α and consider the set J J of such representatives. We have J ˆσ(J ) = and J ˆσ(J ) = J. Next, et D 1 = j J 1 S j R n, which is a discrete set. Thus, we can define the foowing sheaf J x = { j J 1,x S j h j O R n,x if x D 1, O R n,x if x D 1. This sheaf J is a ocay principa coherent idea sheaf whose zero set is D 1. Since the group H 1 (R n, Z ) = 0, a ocay principa sheaves are principa, and J has a goba generator f 1. Since D 1 has dimension 0 we may assume that f 1 is positive semidefinite on R n. By the definition of J we have that f 1 divides f. Let Z = j J H j 1 (0) which is an anaytic subset of U. Consider the coherent sheaf of ideas defined on U by { j J x =,x S H j j O C n,x if x Z, O C n,x if x Z.

15 Vo. 83 (008) On Hibert s 17th Probem for goba anaytic functions in dimension 3 81 As before (in.3), the ocay principa coherent sheaf J is gobay principa, say generated by a hoomorphic function Ɣ : U C, whose zero set is Ɣ 1 (0) = Z. Moreover, by the definition of we have that Ɣ divides F, and since F is invariant, aso the hoomorphic function Ɣ σ divides F. By the properties of the set J we have that Ɣ and Ɣ σ do not have irreducibe common factors in the ring H(U) whose zero set intersect R n. Hence, their product F = Ɣ Ɣ σ divides F in a perhaps smaer neighbourhood of R n in C n. Moreover, F is an invariant hoomorphic function such that f = F R n is a sum of two squares of anaytic functions on R n. Next, since f 1,f are positive semidefinite anaytic functions that divide f and which do not have common irreducibe factors, we concude that f 3 = f /(f 1 f ) is a positive semidefinite anaytic function on R n. Moreover, a straightforward computation shows that the irreducibe compex factors of f 3 are H j, j J 3, which are a specia, as wanted. Remark.10 Since f 1 and f are finite sums of squares of meromorphic function on R n, to prove that f is a finite or convergent sum of squares of meromorphic functions on R n it is enough to check that for f 3. That is, we may aways assume that a the compex irreducibe factors of f are specia and divide f with mutipicity one. 3. Exampes In this section we construct two exampes of positive semidefinite anaytic functions on R 3 which are infinite sums of squares of meromorphic functions on R 3, but for which we have not been abe to decide whether or not they are finite sums of squares of meromorphic functions. We aso produce an exampe of a positive semidefinite anaytic function to which we cannot appy Theorem 1.6 but to which we can appy Coroary 1.7; hence, it is a finite sum of squares. Let us begin with such exampe. Exampe 3.1. Let f 0 : R 3 R be the anaytic function given by f 0 (x,y,z)= (z + x) (z + y) + z 4 + 4x y, which by Lemma.6 is a specia anaytic function. Note that f is a sum of three squares of anaytic functions. For each integer 1, et f (x,y,z) = f 0 (x q,y ( q ), z) where q = [ ]. The zero set of f is X ={x = q,z= 0} {y = q,z= 0}. Since the famiy {X } is ocay finite, the set X = X is cosed in R 3. Hence,

16 8 J. F. Fernando CMH the presheaf,x X f O(R 3 ) x if x X, J x = O(R 3 ) x if x X is a subsheaf of the structure sheaf O R 3. J is a ocay principa coherent sheaf of ideas whose zero set is X. Since H 1 (R 3, Z ) = 0, a ocay principa sheaves are principa, and J has a goba generator f. Note that since X has codimension we may assume that f is positive semidefinite on R 3. Ceary, the specia factors of f are the functions f, 1. Next, note that {x = q,z= 0} if is even, X X +1 = {y = q,z= 0} if is odd, which is not a discrete set for a 1. Thus, we cannot appy Theorem 1.6 to f. However, since X i X j is discrete if i j mod, we can appy Coroary 1.7 with the partition {A 1,A }, where A 1 is the set of the non negative odd numbers and A the set of the non negative even ones. Thus, we concude that f is a sum of 5 squares. Now et us construct the exampes we have announced in the introduction which are infinite sums of squares but to which we cannot appy Coroary 1.7. Exampes 3.. (a) Let f 0 : R 3 R be the specia anaytic function described in Exampe.7, which is a sum of three squares in O(R 3 ). For each integer 1 consider the anaytic function f (x,y,z)= f 0 (x, y, z), whose zero set is X = {x = k, z = 0} = k, z = 0} k k {y Since the famiy {X } is ocay finite, the set X = X is cosed in R 3, and as in Exampe 3.1 there exists a positive semidefinite anaytic function f : R n R whose specia factors are the functions f, 1 (and each one divides f with mutipicity one). Thus, by Theorem 1.4, f is a convergent sum of squares of anaytic functions on R 3. Moreover, we caim: There exist an open invariant neighbourhood U of R 3 in C 3, a hoomorphic extension F of f to U, and a sequence of invariant hoomorphic functions {G k } k on U which converges uniformy to n the compact sets of U. Indeed, et U 0 be an open invariant neighbourhood of R 3 in C 3 to which we can extend hoomorphicay f. We denote such extension by F. Let {K k } k 1 be an

17 Vo. 83 (008) On Hibert s 17th Probem for goba anaytic functions in dimension 3 83 exhaustion by compact sets of U 0, that is, U 0 = k 1 K k and K k Int(K k+1 ) for a k 1. We may assume that (1, 1, 0) K 1. For each k 1weset J k ={ 1 : F 1 (0) K k = }. We have that 1 J k and J k J k+1 for a k 1. We write H k = J k F k, which is a finite product of sums of 4 squares of anaytic functions; hence, H k is a sum of 4 squares itsef. For each k 1, the anaytic function k = H F k does not vanish on the compact set K k. Let ε k = 1 k 1 sup Kk H k +1 inf Kk k inf Kk k +1 > 0 and B k = k + ε k for each k 1. As one can check, k + ε k does not vanish on K k R n. Hence, B k is hoomorphic on an open set U k C n which contains K k R 3. Using that {K k } k is an exhaustion of U 0, one can verify that there exists an open neighbourhood U of R n in C n contained in k 1 U k. The functions G k = B k H k are sum of four squares of invariant hoomorphic functions on U. Moreover, a straightforward computation shows that the sequence {G k } k converges to F uniformy in the compact sets of U. However, we do not know whether or not f is a finite sum of squares of meromorphic functions on R 3. Note that the ines {x =, z = 0} and {y =, z = 0} beong exacty to the zero set of f 1,...,f for a 1. Hence, we cannot appy Coroary 1.7 to this exampe. (b) The description of the foowing exampe requires an initia preparation. Consider the foowing distribution of the natura numbers into an infinite array

18 84 J. F. Fernando CMH and the finite sets S k ={a k + 1,...,a k + k} where a k = 1 k(k 1), k 1. The set S k corresponds to the obique ine {α 1k,α,k,...,α k,1 } of the previous array. For each k 1 we aso construct (inductivey) a set T k = { b kj : k [ ] } k j k 1 such that b kj S j \ k 1 =1 T. We take T 1 = and T ={1}. By the definition of the sets T k, for any given j, wehaves j T k = if and ony if k [ k ] j k 1. Thus, j + 1 k j, and this means that S j intersects exacty j of the T k s, which are T j+1,...,t j. Since the set S j has j different eements then the T k s can be constructed with the desired conditions. We denote C k = S k T k. Next, for each k 1 we consider the hoomorphic function F k = (sin(πx) + z) (sin(πy) + z) + z 4 + M (x k) C k (y ), where M>0 is a positive rea number such that M C k, =j (j ) = 1 for a j C k. We have that the rea anaytic function f k = F k R n is a specia anaytic function whose rea zero set is X k ={x = k, z = 0} C k {y =, z = 0}. One can check, proceeding simiary to Exampe.7, that f is a specia factor. Once again, the famiy {X k } k is ocay finite; hence, the set X = k X k is cosed in R 3 and there exists a positive semidefinite anaytic function f : R n R whose specia factors are the functions f k, k 1. Hence, by Theorem 1.4, f is a convergent sum of squares of anaytic functions on R 3. Proceeding as in the previous exampe (a), one can produce a sequence {g } 1 of sums of four squares of anaytic functions on R 3 which converges uniformy to f (in the sense described in the introduction). However, we have do not know whether or not f is a finite sum of squares of meromorphic functions on R 3. Note that the ines {y =, z = 0} beong exacty to the zero set of two f k s, and the ines {x =, z = 0} beong exacty to the zero set of f. Moreover, for a k 1 the zero set X k has finitey many irreducibe components. Let us expain why we cannot appy Coroary 1.7 to this exampe. For, we have to check that there does not exist a finite partition P ={A 1,...,A r } of N such that for each j = 1,...,r and each pair k, A j the intersection X k X is a discrete set. Let r be an integer. By the definition of the sets X k, it foows that X shares an irreducibe component of dimension 1 with X +1,...,X. This means that the integers, + 1,..., shoud beong to different eements of the partition P. But this is impossibe because the partition has r<+ 1 eements.

19 Vo. 83 (008) On Hibert s 17th Probem for goba anaytic functions in dimension 3 85 A natura question is if it is possibe to go a itte bit further determining whether or not there is a positive semidefinite anaytic function f : R 3 R to which we cannot appy Coroary 1.7 but which satisfy the foowing conditions: (i) The function f has infinitey many specia factors whose zero sets have a finitey many irreducibe components. (ii) The specia factors of f are a sums of p squares in O(R 3 ) for certain integer p 1. (iii) f is the uniform imit of a convergent sequence of anaytic functions which are sum of q squares in O(R 3 ) for certain integer q 1. (iv) There exists and integer r 1 such that the number of irreducibe components of a specia factor of f is r. (v) Each irreducibe component of the zero set of f beongs to at most the zero sets of s of the specia factors of f for the same fixed integer s 1. Let f : R 3 R be a positive semidefinite anaytic function satisfying the conditions (i) to (v) above. Let {f k } k 1 be the specia factors of f, X k = fk 1 (0) and {Y } 1 the irreducibe components of f 1 (0) = k 1 X k. Consider the set S = S f ={(k, ) : Y X k } N and the projections π i : S N, (x 1,x ) x i for i = 1,. The set S has the foowing properties: (1) π i (S) = N for i = 1,. () The fibers π1 1 1 (k) and π () have respectivey ess than or equa to s and r points for a k, N. Conversey, for each set S N satisfying the properties (1) and () above there exists a positive semidefinite anaytic function f : R 3 R such that S f = S. To check that, it is enough to proceed as in the exampe 3. (b). Thus, the existence of an anaytic function f : R 3 R satisfying (i) to (v) above to which we cannot appy Coroary 1.7 is equivaent to the existence of a set S N satisfying the conditions (1) and () above and the foowing one: (3) There is no finite partition P ={A 1,...,A m } of N such that for a j = 1,...,m and a α, β A j, we have that π (π1 1 (α)) π (π1 1 (β)) =. As far as we known, after consuting severa speciaists in the matter, this is an open probem which seems to be difficut. 4. Proofs of the main resuts The purpose of this section is to prove Theorem 1.4 and Theorem 1.6 announced in the introduction. Before that we need some preiminary resuts.

20 86 J. F. Fernando CMH Proposition 4.1. Let U C n be an open set and et F k,g k : U C be invariant hoomorphic functions such that the series k 1 F k, k 1 G k converge in the sense of Definition 1. (ii), that is, k 1 sup K F k, k 1 sup K G k < + for a compact sets K U. Then the series j,k 1 F j G k converges on U to j 1 F j k 1 G k in the sense of Definition 1. (ii). The proof of the proposition foows straightforwardy from the foowing resut whose proof is a standard exercise of the theory of convergent series that we do not incude here. Lemma 4.. Let {a i } i N, {b i } i N be two sequences of compex numbers such that the series i 1 a i, i 1 b i converge to compex numbers a, b and the series i 1 a i, i 1 b i converge to non-negative rea numbers a, b. Then the series i,j 1 a ib j converges to ab, that is, for a ε>0there exists a finite subset I ε N such that if I N is a finite subset that contains I ε then (i,j) I a ib j ab <ε. Moreover, the series i,j 1 a ib j converges to a b. Lemma 4.3. Let f, f : R n R be two positive semidefinite anaytic functions such that f 1 (0) = f 1 (0) = S. Suppose that there exists a discrete set D S such that the meromorphic function f/f is anaytic on R n off the discrete set D. Then, there exist anaytic functions h 1,h : R n R such that h 1 i (0) D, h is a sum of n + n squares and h 1 f = h f. Proof. Indeed, consider the coherent sheaf (f :f)o R n. This sheaf is generated in a neighborhood of each y D by finitey many sections δ 1,...,δ ry O(R n ). By the standard sum of squares trick, f y /f y = η y/δ y for δ = k δ k and some η y O(R n y ). Furthermore, y is an isoated zero of δ. For that, suppose that there is x = y arbitrariy cose to y with δ(x) = 0. Then, a δ k s vanish at x, and since the idea (f x : f x) is generated by them, it contains no unit. This means that f x /f x is not anaytic, a contradiction. The ideas y = (δ y ), y D, gue to define a ocay principa coherent sheaf of ideas of O R n, whose zero set is D. Since H 1 (R n, Z ) = 0, a ocay principa sheaves are principa, and has a goba generator. This means that each germ y /δ y is a unit for a y D. This is a non-negative anaytic function on R n whose zero set is D, and f = f/f is anaytic. Moreover f is stricty positive on R n \ D. Thus, by [BKS] there exists an anaytic function : R n R such that 1 (0) f 1 (0) = D and f is a sum of n + n squares. Hence, ( ) f = ( f )f and taking h 1 = and h = f, we are done.

21 Vo. 83 (008) On Hibert s 17th Probem for goba anaytic functions in dimension 3 87 Lemma 4.4. Let U C n be an open set and et F k : U R, k 1, be invariant hoomorphic functions such that the sum k 1 F k converges in the sense of Definition 1. (ii). Let Y U be a cosed set and et G: U R be a rea anaytic function. If G x divides F k,x for a k 1 and a x Y, then G divides F = k 1 F k in O(W ), where W is a sma enough neighbourhood of Y in U. Proof. It is enough to check that for a x Y, G x divides F x. Fix x Y, and et G x = r i=1 G d i i,x be the decomposition of the germ G x into irreducibe compex factors. We take a compact neighborhood W x U of x such that: G 1,...,G r are hoomorphic on W x, and G 1 i (0) W x is an irreducibe compex anaytic set in W x for a i = 1,...,r. It is enough to see that G d i i,x divides F x for i = 1,...,r. By hypothesis, for each k 1 the germ G d i i,x divides F k,x, and an easy computation shows that G i,x divides a derivatives D α F k,x of degree α <d i. That means that D α F k vanishes on the intersection of G 1 i (0) with a sma neighborhood of x (depending on k). Thus, since G 1 i (0) W x is irreducibe, D α F k vanishes on G 1 i (0) W x. As this hods for each k 1 and D α F W x = k 1 Dα F k, we have that D α F W x vanishes on G 1 k (0) W x. Whence, G k,x divides a derivatives D α F x with α <d k. This concudes the proof up to the emma that foows. Lemma 4.5. Let G, F C{z} =C{z 1,...,z n } be anaytic germs such that G is irreducibe and et d be a positive integer. Suppose that G divides a derivatives D α F of degree α <d. Then, G d divides F. Proof. We proceed by induction on d. Ifd = 1 the resut is cear. Suppose the resut true for d and that G divides D α F = α F z α for α <d+ 1. By induction, G d divides F, z F 1,..., z F n. In particuar, there exists H C{z 1,...,z n } such that F = G d H. Hence, F d 1 G = dg H + G d H. z i z i z i Since G d divides a derivatives z F i, we see that G divides a products G z i H. But G is irreducibe and cannot divide a its derivatives G z i, hence G divides H. Thus, G d+1 divides F, as wanted. Next, we proceed to prove Theorem 1.4. Proof of Theorem 1.4. We wi spit the proof into severa steps. Step 1: Preparation. By the decomposition resut Lemma 1.1 we may assume that a the compex irreducibe factors {h j } j J of f are specia and divide f with mutipicity

22 88 J. F. Fernando CMH one. As we have seen in.3 we may assume that there exist hoomorphic extensions H j of h j to a Stein neighborhood U 0 of R n in C n. We write X j = H 1 j (0) R n for a j J. By hypothesis and by Proposition 1.3, for each j there are invariant hoomorphic functions G j,b jk : U j C, defined on an open neighborhood U j U 0 of X j in C n, such that Gj H j Uj = k B jk (the series converging in the strong sense of Definition 1. (ii)) and Gj 1 (0) R n X j. We write T j = Gj 1 (0) U j and shrinking U j if necessary, we may assume that the famiy {T j } j J is ocay finite in C n. Consider the open set in C n : ( U = C n \ ) C C n(t j ) ( U j \ ) C n(t k ) \ U k ). j 1 j 1 k =j(c A straightforward computation shows that T j U is cosed in U for a j J. Let U U U 0 be an open invariant Stein neighborhood of R n in C n such that R n is a deformation retract of U ([Ca]). Denote again by T j the intersections of T j with U which are anaytic (compex) subsets of U. We denote V j = U j U and keep F for the restriction of F to U, and G j,b jk for those of G j,b jk to V j. It hods: T j V j, and a T j s are cosed anaytic subsets of U, as we as their union T = j T j. Step : Extension of denominators. Fix j J and consider the coherent sheaf of ideas J defined on U by { G j O C J x = n,x if x T j, O C n,x if x T j. As it has been done before, the ocay principa coherent sheaf J is gobay principa, say generated by a hoomorphic function Ɣ j : U C, whose zero set is Ɣj 1 (0) = T j. In a sma enough neighborhood of T j we have that Ɣ j = G j v j where v j is a hoomorphic unit. Hence, in that neighborhood Ɣ j σ = G j v j σ, and therefore E j = Ɣ j σ/ɣ j is a unit in H(U). Moreover, one can check that E j E j σ = 1. Let j : U C be a hoomorphic function such that j = E j and j j σ = 1. A straightforward computation, aready done in.3, shows that G j = Ɣ j j is an invariant hoomorphic function that generates J. Consider aso the rea anaytic function g j = G j R n. The zero set of G j is T j and the zero set of g j is T j R n X j hj 1 (0). Now, since G j generates J, G j generates J Vj, and these functions are invariant, there exist an invariant hoomorphic function Q j : V j C such that G j V j = Q j G j. We deduce: G j Hj = Qj (G j H j ) = Qj Bjk = Cjk, k k