A Galoisian proof of Ritt theorem on the differential transcendence of Poincaré functions
aa r X i v : . [ m a t h . D S ] F e b A Galoisian proof of Ritt theorem on the differential transcendence ofPoincar´e functions
Lucia Di Vizio ∗ & Gwladys Fernandes † February 17, 2021
Abstract
Using Galois theory of functional equations, we give a new proof of the main result of the paper “Tran-scendental transcendency of certain functions of Poincar´e” by J.F. Ritt, on the differential transcendence ofthe solutions of the functional equation R ( y ( t )) = y ( qt ), where R ( t ) ∈ C ( t ) verifies R (0) = 0, R ′ (0) = q ∈ C ,with | q | >
1. We also give a partial result in the case of an algebraic function R .2010 Mathematics Subject Classification: 30D05, 34M15, 39B12. We fix a rational function R , with complex coefficients, such that R (0) = 0, R ′ (0) = q ∈ C , with | q | >
1. Tolinearize the rational map R at 0, one has to solve the functional equation(1.1) R ( σ ( t )) = σ ( qt ) . It means that, up to a conjugation by σ , the rational function R acts linearly in the neighborhood of 0.H. Poincar´e noticed that such an equation admits a formal solution σ ∈ t C [[ t ]], called a Poincar´e function,which is actually the expansion at zero of a uniform function, i.e. a meromorphic function over the whole C .See [Poi90, page 318]. The functional equation (1.1) plays a key role in rational dynamics and for this reasonhas been studied by many authors, in particular relaxing the assumption on the absolute value of q .In [Rit26], J.F. Ritt addresses the question of the differential algebraicity of σ over the field of rationalfunctions C ( t ), i.e., the fact that σ is solution of an algebraic differential equation with coefficients in C ( t ).This means that there exists a non-negative integer n and a non-zero polynomial P ∈ C [ t, X , X , . . . , X n ] suchthat P ( t, σ, σ ′ , . . . , σ ( n ) ) = 0. We say that σ is differentially transcendental over C ( t ) if it is not differentiallyalgebraic. Ritt’s paper is at the origin of a large literature on differential transcendence of solutions of functionalequations linked to dynamical systems, which is surveyed in [Fer21]. Theorem 1.1 ([Rit26]) . Let σ be a solution of a functional equation of the form (1.1) , where R ( t ) ∈ C ( t ) isnot a homography and R (0) = 0 , R ′ (0) = q ∈ C , with | q | > . If σ satisfies an algebraic differential equationthen it is the composition of a homography with a periodic function belonging to the following list: exp( αt p ) , cos( αt p + β ) , ℘ ( αt p + β ) , ℘ ( αt p + β ) , ℘ ( αt p + β ) or ℘ ′ ( αt p + β ) , where ℘ is the Weierstrass function and fora convenient choice of p ∈ Q , of α ∈ C and of a fraction β ∈ C of the period. The list above appears in [Rit22], where Ritt classifies periodic Poincar´e functions. For more comments andexplanations on such a list we refer to [Fer21, § σ is rational if and only if R ( t ) is a homography(see Lemma 2.1 below).The original proof of the theorem above is organised in the following way:1. Ritt supposes that σ is solution of a general algebraic differential equation. Replacing t by qt and using(1.1), he derives a new differential equation that allows him to perform an Euclidean division in order tolower what he calls the rank of the differential equation. He iterates this operation several times, choosingcarefully the terms to eliminate in the process: this part of the proof (namely from § §
11 in loc.cit. )is really a tour de force . He narrows the investigation to three possible differential equations satisfied by σ , that are called ( A ), ( B ) and ( C ) (see §
11 in loc.cit. ), and he notices that the solutions of ( A ) and ( B )are actually solutions of a differential equation of type ( C ), so that he is left with this last case. ∗ Lucia Di Vizio, CNRS, Universit´e Paris-Saclay, UVSQ, Laboratoire de math´ematiques de Versailles, 78000, Versailles, France. [email protected] , https://divizio.perso.math.cnrs.fr † Gwladys Fernandes, Universit´e Paris-Saclay, UVSQ, CNRS, Laboratoire de math´ematiques de Versailles, 78000, Versailles,France. [email protected] , https://fernandes.perso.math.cnrs.fr This project has received funding from the ANR project DeRerumNatura, ANR-19-CE40-0018.
1. He shows that the Poincar´e functions that are also solutions of ( C ), are obtained by composing a periodicPoincar´e function with a rational power of t .3. He uses his results in [Rit22] on the classification of periodic Poincar´e functions to make the list explicitand concludes a posterori that actually only the differential equation, that he calls ( A ), can occur:( A ) ( y ′ ) r = t j A ( y ) , where r, j ∈ Z and A ( t ) ∈ C ( t ).Ritt’s proof of Theorem 1.1 is extremely technical and it may be difficult to understand how the differentialequation ( A ) is singled out at the very last line of the paper. We think that the Galoisian proof presentedbelow, which is much shorter modulo the Galois theory, may give a different insight on the existence of thedifferential equation ( A ). The price to pay for a theoretical understanding of the existence of the equation ( A )is the algorithmic nature of Ritt’s proof. Another proof of Ritt theorem can be found in [Cas06, Cas15], wherethe same three differential equations as in Ritt’s paper appear.The purpose of this paper is to give a Galoisian proof of the following theorem, from whom one can deduceRitt’s theorem (see § Theorem 1.2.
Let σ ∈ C [[ t ]] be a formal solution of a functional equation of the form (1.1) , for some R ( t ) ∈ C ( t ) not a homography, such that R (0) = 0 , R ′ (0) = q = 0 not a root of unity. The function σ is differentiallyalgebraic over C ( t ) if and only if it is solution of a differential equation “of Ritt type ( A ) ”, with r = 0 . We close the paper by proving a generalisation of the last statement to the case of a series R ( t ) ∈ C [[ t ]]algebraic over C ( t ), such that R (0) = 0 and R ′ (0) = q , with q ∈ C , q = 0 not a root of unity.The paper is organized as follows. In § § § §
5, wegeneralize Theorem 1.2 to the case of an algebraic function R . Acknowledgement.
It is a pleasure to thank the participants of the
Groupe de travail sur les marches dansle quart de plan , where both authors have given several talks on Ritt’s theorem and on some subsequent worksin differential algebra. We would like to thank Guy Casale and Federico Pellarin for their interest for the presentwork and Alin Bostan for his attentive reading of the manuscript and his remarks that have allowed to improvea previous version of our result.
The proof below follows the main ideas of Ritt (see [Rit26, § R ( t ) ∈ C ( t ) be such that R (0) = 0 and R ′ (0) = q ∈ C , with | q | >
1. We consider the only solution σ ∈ t + t C [[ t ]] of the functional equation (1.1), namely: σ ( qt ) = R ( σ ( t )) . As we have already pointed out in the introduction, σ is the expansion at zero of a uniform meromorphicfunction over the whole C . Lemma 2.1 ([Rit26, § . The following assertions are equivalent:1. The Poincar´e function σ is rational.2. σ is a homography.3. R ( t ) is a homography.Proof. If R ( t ) is a homography, it must have the form R ( t ) = qtat +1 , for some a ∈ C . Then σ ( t ) = ( q − tat +( q − . Weconclude proving by contradiction that, if σ is rational, then R ( t ) is a homography. First, notice that Equation(1.1) implies that:(2.1) σ ( q t ) = R ( R ( σ ( t ))) . We suppose that σ ∈ C ( t ) and assume that the degree of the numerator of R , after eliminating any commonfactor with the denominator, is at least 2. In this case, R ( t ) has at least two finite zeros, namely 0, which is2imple zero by assumption, and a = 0. The rationality of R implies that there exists a such that R ( a ) = a .Note that a / ∈ { , a } and R ( R ( a )) = 0. The rationality of σ implies that there exists b i such that σ ( b i ) = a i ,for i ∈ { , } . Since b = b , b and b cannot be both equal to ∞ . If b is finite, we deduce from (1.1) that σ ( qb ) = R ( σ ( b )) = 0. Recursively we obtain that σ ( q n b ) = 0 for any positive integer n , and hence that σ has an infinite number of zeros, which contradicts its rationality. If b is finite, we use (2.1) to conclude that σ ( q n b ) = 0 for any positive integer n , and hence that σ has an infinite number of zeros, which contradictsits rationality. Therefore the numerator of R cannot have degree greater than 1, and it is equal to qt . If thedenominator of R ( t ) is a polynomial of degree at least 2, then the numerator of R ( R ( t )) has degree at least 2.Applying the previous reasoning to R ◦ R and Equation (2.1), we obtain a contradiction. So both the numeratorand the denominator of R have degree at most 1, which means that R is a homography. Corollary 2.2 ([Rit26, § § . If R ( t ) is not a homography, then:1. σ has an infinite set of zeros and an essential singularity at ∞ .2. σ is transcendental.Proof. Let us suppose that the numerator and the denominator of R ( t ) do not have common factors. If R isnot a homography, then either the numerator or the denominator of R has degree at least 2. As in the proofof the previous lemma, we conclude that σ has an infinite set of zeros, that accumulate at ∞ . Since σ is notidentically zero, it must have an essential singularity at ∞ . This also implies that σ is transcendental. Proof of Theorem 1.1.
By Theorem 1.2, we have ( σ ′ ) r = t j A ( σ ), for some r, j ∈ Z , with r = 0, and A ( t ) ∈ C ( t ),and hence: 0 = ord t =0 ( σ ′ ) r = j + ord t =0 A ( t ) ord t =0 σ ( t ) . We conclude that ord t =0 A ( t ) = − j . We know from the previous corollary that σ has a finite zero b = 0. Let p := ord t = b σ ( t ) ≥
1. We observe that: r ord t = b σ ′ ( t ) = ord t =0 A ( t ) ord t = b σ ( t ) = − jp, hence r ( p − jp = 0. We consider the change of variable t = u p and we set z ( u ) := σ ( u p ). A direct calculationshows that:(2.2) z ( q /p u ) = R ( z ( u )) and z ′ ( u ) r = p r A ( z ( u )) . Notice that for any non-zero constant e u , the function z ( u + e u ) is also a solution of the differential equationabove. Since σ has an essential singularity at ∞ , then σ takes all the values in C , with at most one exception.The same holds for z ( u ). Therefore the unique solution of the differential equation y ′ ( u ) r = p r A ( y ( u )) withinitial condition y ( u ) = c , with u , c ∈ C , is constructed in the following way: we find u c such that z ( u c ) = c and we chose the solution z ( u − u + u c ) of the differential equation above. Given the freedom in the choice ofboth u and c , one avoids the missing value of z ( u ) and concludes that all solutions of y ′ ( u ) r = p r A ( y ( u )) areobtained composing z ( u ) with a translation.By construction, the solution z ( u ) has a non-trivial monodromy. Let e z ( u ) be another branch of z ( u ). Weobserve that e z ( u ) is solution of the system (2.2), therefore e z ( u ) = z ( u + e u ), for some non-zero e u ∈ C . Theuniqueness of the analytic continuation implies that e z ( q /p u ) = R ( e z ( u )) and hence that: z ( q /p u + e u ) = e z ( q /p u ) = R ( e z ( u )) = R ( z ( u + e u )) = z ( q /p ( u + e u )) . We deduce that z is periodic of period ( q /p − e u . We conclude as Ritt does, using his result [Rit22] on theclassification of periodic Poincar´e functions. For an introduction to the Galois theory of difference equations, we refer the reader to [vdPS97], or to [OW15, § § § C is algebraically closed. For this reason we can naively identify the Galoisgroups with their C -points, which makes things slightly easier. We remind below the notions that are essentialto the understanding of the proof in the next section.Let F / K be a field extension, such that F comes equipped with an endomorphism Φ : F → F , which inducesa non-periodic endomorphism of K . It means that there exists x ∈ K such that Φ n ( x ) = x , for any non-zero3nteger n . We suppose that the field C of elements of F that are left invariant by Φ is algebraically closed andthat C ⊂ K . We consider the linear system of the form(3.1) Φ ~y = (cid:18) a b a (cid:19) ~y, where a , a , b ∈ K , with a a = 0, and we suppose that there exists an invertible matrix (cid:18) z w z (cid:19) ∈ GL ( F )that satisfies (3.1). Definition 3.1 (see [DV19, Def. 3.5]) . We call Picard-Vessiot ring of (3.1) over K the ring R = K [ z ± , z ± , w ] ⊂ F . We define the Galois group of (3.1) to be:(3.2) G := { ϕ : R → R , automorphism of K -algebras, such that ϕ ◦ Φ = Φ ◦ ϕ } . The elements of G extend to automorphisms of the field of fractions L of R .The system (3.1) boils down to the equations Φ( z i ) = a i z i , for i = 1 ,
2, and Φ( w ) = a w + bz . Any ϕ ∈ G ,being a ring automorphism, must leave globally invariant the space of solutions of (3.1). Therefore there existsnon-zero c i ∈ C such that ϕ ( z i ) = c i z i , for i = 1 ,
2. As far as ϕ ( w ) is concerned, it must be a solution ofΦ( y ) = a y + bϕ ( z ) = a y + bc z , hence there exists d ∈ C such that ϕ ( w ) = dz + c w . We conclude that ϕ (cid:18) z w z (cid:19) = (cid:18) z w z (cid:19) (cid:18) c d c (cid:19) = (cid:18) c z dz + c w c z (cid:19) , for some (cid:18) c d c (cid:19) ∈ GL ( C ).We now state the main properties of the Galois group of a functional equation: Theorem 3.2 ([DV19, Thm. 4.9 and 5.3]) .
1. The Galois group G is an algebraic subgroup of GL ( C ) , andits dimension as an algebraic variety over C is equal to the transcendence degree of L / K .2. K = { f ∈ L : ϕ ( f ) = f ∀ ϕ ∈ G } .Example . Let us consider a special case of the system (3.1), with b = 0 and, therefore, w = 0. In thiscase the Picard-Vessiot ring coincides with R := K [ z ± , z ± ] and its Galois group G is a subgroup of the groupof the invertible diagonal matrices of rank 2, that we can naively identify with ( C ∗ ) . It means that for anyautomorphism ϕ ∈ G there exist two non-zero constants c , c ∈ C such that ϕ ( z i ) = c i z i , for i = 1 , z , z are algebraically dependent over K if and only if G has dimensions 0 or 1, or equivalentlyif it is a proper algebraic subgroup of ( C ∗ ) . We know that the proper algebraic subgroups of ( C ∗ ) are definedby equations of the form X α X α = 1, for some α , α ∈ Z r { (0 , } . It means that c α c α = 1. We concludefrom the theorem above that z α z α ∈ F is invariant under any automorphism of the Galois group G and hencethat z α z α ∈ K . Example . If in (3.1) we set a = a = 1, then we can take z = z = 1. The Picard-Vessiot ring boils downto R = K [ w ] and the Galois group G is a subgroup of the group of matrices (cid:26)(cid:18) d (cid:19) , d ∈ C (cid:27) , that we canidentify to ( C, +). It means that for every ϕ ∈ G , there exists a constant d ∈ C such that ϕ ( w ) = w + d .The solution w is algebraically dependent over K if and only if G is a proper algebraic subgroup of ( C, +),but the only proper algebraic subgroup of ( C, +) is the trivial group 0. Hence, if w is algebraic over K , then d = 0 and w is tautologically left fixed by all morphisms of G . We conclude that w ∈ K . Let C be an algebraically closed field and F := C (( x )). We fix a formal power series R ∈ F such that R is nota homography, R (0) = 0, R ′ (0) = q = 0 is not a root of unity, so that we can define the morphism:Φ R : F → F ,f f ( R ( x )) . Notice that Φ R is an automorphism of F . Moreover, Φ R is not periodic. Lemma 4.1.
The field of constants F Φ R := { f ∈ F : Φ R ( f ) = f } of F with respect to Φ R coincides with C . roof. Let f ∈ F r C be such that f ( R ( x )) = f ( x ). Replacing f with f − f (0), we can suppose that f has noconstant term and, replacing f with its inverse for the Cauchy product, we can suppose that f = P n ≥ N f n x n ,for some positive integer N that we chose so that f N = 0. The coefficient of x N in f ( R ( x )) is f N q N . We deducefrom f ( R ( x )) = f ( x ) that f N q N = f N , with q N = 0 ,
1, and hence that f N = 0, against our assumption. Weconclude that, if f has no constant term, then f = 0. This means exactly that F Φ R = C .Let K := C ( x ) be the field of rational functions. We assume that R is (the expansion of) a rational function.In this case Φ R an endomorphism of K and we consider the functional equation:(4.1) Φ R ( y ( x )) = qy ( x ) . We use the notation y for the unknown function for reasons that will be clearer in a few lines. The equationabove has a formal solution τ ∈ F . It satisfies necessarily τ (0) = 0. Notice that τ is determined up to amultiplicative constant. We have: Theorem 4.2.
Let τ ∈ F be a formal solution of (4.1) . The following assertions are equivalent:1. τ is differentially algebraic over C ( x ) .2. τ satisfies a differential equation with coefficients in C ( x ) of order .3. τ is solution of the differential equation ( y ′ ) r = A ( x ) y j , for some ( r, j ) ∈ Z , with r = 0 , and A ( x ) ∈ C ( x ) .Proof. Notice that the implications 3 ⇒ ⇒ ⇒ ⇒ ⇒ ⇒
3. We consider the system of functional equations(4.2) ( Φ R ( y ) = qy , Φ R ( y ) = qR ′ y . We are in the case of Example 3.3. The fact that τ is solution of a differential equation of order 1 means thatthe system above has a basis of solutions, namely τ, τ ′ , which are algebraically dependent over K . Followingthe example, we conclude that τ α ( τ ′ ) α ∈ K , for some integers α , α , which cannot be simultaneously zero.If α = 0, it is enough to rephrase the latter statement in Ritt’s notation: There exist r, j ∈ Z , with r = 0, and A ( x ) ∈ K such that τ is solution of the differential equation ( y ′ ) r = A ( x ) y j , as claimed. On the other hand, if α = 0, then τ α ∈ K , and we can conclude by taking the logarithmic derivative.We now prove by contradiction that 1 ⇒
2. So let us suppose that τ is solution of an algebraic differentialequation of order n such that n > K n := K ( τ, τ ′ , . . . , τ ( n − ). Thedefinition of n implies that τ ( n ) is algebraic over K n , but τ ( n ) K n , otherwise we would obtain a differentialequation of order n − τ . Notice that deriving Φ R ( y ) = qy , calling y k the k -th derivative of y and usingthe Fa`a di Bruno’s formula we obtain:(Φ R ( y )) ( n ) = n X k =1 B n,k ( R ′ , . . . , R ( n − k +1) )Φ R ( y k ) = qy n , where B n,k ( x , . . . , x n − k +1 ) are the Bell polynomials defined by the multivariate identity: exp (cid:16)P ∞ j =1 x j t j j ! (cid:17) = P n ≥ k ≥ B n,k ( x , . . . , x n − k +1 ) t n n ! . As B n,n ( x ) = x n , replacing the Φ R ( y k )’s recursively up to y n − , we obtainan expression of the form: Φ R ( y n ) = q ( R ′ ) n y n + n − X k =1 A n,k ( x ) y k , where A n,k ( x ) ∈ C ( x ) is a rational function, which is actually a rational expression in the derivatives of R . Weset b := P n − k =1 A n,k ( x ) τ ( k ) ∈ K n and z = ( τ ′ ) n τ − n ∈ K ⊂ K n , to simplify the notation. We deduce from thefunctional equation above that ω n := τ ( n ) z verifies the functional equation(4.3) Φ R ( ω n ) = ω n + b ( R ′ ) n qz . We are in the situation of Example 3.4. Since ω n is algebraic over K n , we conclude that ω n ∈ K n . Since z ∈ K n ,we conclude that also τ ( n ) ∈ K n , which is in contradiction with our choice of n . This ends the proof.We deduce theorem Theorem 1.2 from the statement above. Proof of Theorem 1.2.
Let σ be the inverse of τ for the composition. First of all, let us notice that σ isdifferentially algebraic over C ( t ) if and only if τ is differentially algebraic over C ( x ) (see [BR86, page 344] or[Moo96, page 55, (n)]). We consider the local change of variable t = τ ( x ) in (1.1), or equivalently x = σ ( t ),which transforms (4.1) and ( τ ′ ) r = A ( x ) τ j into (1.1) and ( σ ′ ) − r = t j A ( σ ), respectively.5 The algebraic case
We change a little bit the notation with respect to the previous section. Let us consider the relative closure K of C ( x ) inside F := C (( x )) and R ( x ) ∈ K , such that R is not a homography, R (0) = 0, R ′ (0) = q = 0 is not aroot of unity. Then the functional equation τ ( R ( x )) = qτ ( x )has a formal solution at 0.We can define the automorphism Φ R of F as in the previous section. Then Φ R induces an automorphism of K and the field of constants is C , as in Lemma 4.1. Reasoning word by word as in the previous section one canprove: Theorem 5.1.
Let τ ( x ) be a formal solution of τ ( R ( x )) = qτ ( x ) in C [[ x ]] . If τ ( x ) is differentially algebraicover K , then it satisfies a differential equation of the form ( y ′ ) r = y j A ( x ) , where r, j are integers, with r = 0 ,and A ( x ) ∈ K .Remark . The fact that K /C ( x ) is algebraic implies that being differentially algebraic over K is equivalent tobeing differentially algebraic over C ( x ). Since A ( x ) ∈ K , there exists P ( x, T ) ∈ C [ x, T ] such that P ( x, A ( x )) = 0.Then P ( x, ( y ′ ) r y − j ) = 0 provides a differential equation for τ over C ( x ). References [BR86] Michael Boshernitzan and Lee A. Rubel,
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Difference Galois theory for the applied mathematician , In “Arithmetic and geometryover local fields”, LNM, Springer (2019), 20 pages, to appear.[Fer21] Gwladys Fernandes,
A survey on the hypertranscendence of the solutions of the Schr¨oder’s, B¨ottcher’sand Abel’s equations , Proceedings of the conference “Transient Transcendence in Transylvania”. Pro-ceedings in Mathematics & Statistics, Springer. (2021), 31 pages.[Moo96] Eliakim Hastings Moore,