L Oger

STUDY OF COMPOSITION OPERATORS ON THE UNIT BALL OF C N

Keywords: 2020 Mathematics Subject Classification. 47B33, 32A10, 47A10 composition operator, spectrum, holomorphic functions in several variables, Fréchet space

We study the composition operators C φ induced by holomorphic self-maps φ of the unit ball B N of C N , and acting on the Fréchet space Hol(B N ) of complex-valued holomorphic functions. In particular, we consider the general properties (continuity, invertibility, compactness) and the spectra of these operators. We give a complete description of the spectra in several cases: hyperbolic, positive-step parabolic, elliptic attractive with diagonal or nilpotent Jacobian, bijective periodic, and finally elliptic non attractive with « separable » components. We also provide inclusions otherwise.

The operator C φ : Hol(B N ) → Hol(B N ) is said to be a composition operator. It is clear that such operators are linear and multiplicative.

) considers these operators on Banach

spaces of analytic functions, such as the Hardy spaces H p . Indeed, they appear in numerous problems, such as the research of isometries (see [START_REF] Banach | Théorie des opérations linéaires[END_REF][START_REF] Chalendar | Weighted composition operators: isometries and asymptotic behaviour[END_REF][START_REF] Forelli | The isometries of H p[END_REF]).

In recent years, W. Arendt et al. ([3,[START_REF] Arendt | Spectral properties of weighted composition operators on Hol(D) induced by rotations[END_REF][START_REF] Arendt | In Koenigs' footsteps: diagonalization of composition operators[END_REF]) studied one-variable composition operators directly on the Fréchet space Hol(D), where D is the unit disc of C. Indeed, the spectral study of composition operators on Hol(D) is a first step in understanding the spectrum for bounded composition operators induced by a symbol φ on arbitrary Banach spaces X of holomorphic functions on D which embeds continuously in Hol(D). We aim to do the same for Hol(B N ), and Banach spaces of holomorphic functions in several variables.

The results are summarized in the following table. Denote by T the unit circle of C, by N 0 the set of all nonnegative integers, and N = N 0 \{0}. To prove these results, they mainly used Denjoy-Wolff's ( [START_REF] Bracci | Continuous semigroups of holomorphic self-maps of the unit disc[END_REF][START_REF] Cowen | Composition operators on spaces of analytic functions[END_REF]), Koenigs' ( [START_REF] Célariès | Opérateurs et semi-groupes d'opérateurs sur des espaces de fonctions holomorphes -Applications à la théorie de l'universalité[END_REF][START_REF] Koenigs | Recherches sur les intégrales de certaines équations fonctionnelles[END_REF]), Valiron's ( [START_REF] Valiron | Sur l'itération des fonctions holomorphes dans un demi-plan[END_REF]) and Baker-Pommerenke's theorems ( [START_REF] Baker | On the iteration of analytic functions in a half-plane II[END_REF][START_REF] Pommerenke | On the iteration of analytic functions in a halfplane[END_REF]).

Properties of φ σ p (C φ ) σ(C φ ) φ with a fixed point α ∈ D φ ≡ 0 {0, 1} φ ′ (α) = 0 {1} {0, 1} 0 < |φ ′ (α)| < 1 {φ ′ (α) n : n ∈ N 0 } σ p (C φ ) ∪ {0} φ(z) = e iθ z, θ/2π ∈ Q {e inθ : n ∈ N 0 } {λ ∈ C : λ p = 1}
, where p ∈ N is the smallest integer such that e ipθ = 1. φ(z) = e iθ z, θ/2π ̸ ∈ Q ⊂ T φ without any fixed point on D φ invertible C * φ not invertible C * C Table 1. Results in the one-dimensional case This article can be considered as a sequel to [START_REF] Arendt | Denjoy-Wolff theory and spectral properties of weighted composition operators on Hol(D)[END_REF][START_REF] Arendt | Spectral properties of weighted composition operators on Hol(D) induced by rotations[END_REF][START_REF] Arendt | In Koenigs' footsteps: diagonalization of composition operators[END_REF]. The aim is to fill as much as possible a similar table in the multidimensional case. Note that the behaviour of the point spectrum σ p (C φ ) and spectrum σ(C φ ) of composition operators strongly depends on the properties of the map φ.

In order to study this function, we need a generalization of Denjoy-Wolff's theorem, by Kubota and MacCluer ([25,[START_REF] Maccluer | Iterates of holomorphic self-maps of the unit ball in C n[END_REF]), presented here as a single result. Recall that S N -1 is the unit sphere of C N . Let φ [n] be the n-th iteration of φ.

Theorem 1.1. Let φ be a self-map of B N . Denote by Γ(φ) the set of all subsequential limits of (φ [n] ) n≥0 on Hol(B N ) equipped with the uniform convergence on all compact subsets of B N .

(i) If φ does not have any fixed point on B N , then there exists a point ζ φ ∈ S N -1 such that Γ(φ) = {f }, with f ≡ ζ φ . In this case, we say that ζ φ is the Denjoy-Wolff point of φ.

(ii) Assume that φ has a fixed point in B N .

(a) If there exists z 0 ∈ B N such that Γ(φ) contains f ≡ z 0 , then Γ(φ) = {f }. In this case, z 0 is the unique fixed point of φ, and its Denjoy-Wolff point. This coefficient, applied at the Denjoy-Wolff point ζ φ allow us to categorise the self-maps of B N , depending on whether φ has a fixed point on D or not. Definition 1.2. Let φ be a self-map of B N .

r Assume that φ has a fixed point on B N .

-In the case (a), we say that φ is elliptic attractive.

-In the case (b), we say that φ is elliptic non attractive.

r Assume that φ does not have any fixed point on B N .

-If δ(φ, ζ φ ) = 1, we say that φ is parabolic.

-If δ(φ, ζ φ ) < 1, we say that φ is hyperbolic.

The paper is organized as follows. First, in Section 2, we focus on the general properties of composition operators. We show that C φ is continuous (Proposition 2.1), and we characterize when the operator is invertible (Proposition 2.4) and compact (Proposition 2.8).

Then, in Section 3, we use generalizations of Valiron's and Baker-Pommerenke's theorems to obtain a description of the point spectrum and spectrum of C φ when φ is hyperbolic (Proposition 3.3) or positive-step parabolic (Proposition 3.7). L. Arosio's results ( [START_REF] Arosio | Canonical models for the forward and backward iteration of holomorphic maps[END_REF]) will be in the spotlight in this section.

Next, in Section 4, we restrict ourselves to elliptic attractive maps such that if α is the Denjoy-Wolff point of φ, then φ ′ (α) is invertible. To go into detail, we use a generalization of Koenigs' theorem (Theorem 4.3). This implies that the eigenvalues of C φ are exactly the products and powers of eigenvalues of φ ′ (α) (Proposition 4.5). In order to describe the spectrum (Proposition 4.10), we consider a sequence of projections (Theorem 4.7) to decompose the space Hol(B N ).

Section 5 contains results about elliptic attractive maps such that if α is the Denjoy-Wolff point of φ, then φ ′ (α) is not invertible. In this case, we give some partial results about the spectrum and point spectrum of C φ . If the matrix φ ′ (α) is nilpotent, we are able to give a complete description of the spectra, and we provide inclusions in general.

In Sections 4 and 5, the main difficulty compared to the one-variable case is that φ ′ (0) is not anymore a scalar, but instead a matrix, so we have to work with its eigenvalues.

Finally, in Section 6, we focus on elliptic non attractive symbols. We completely describe the point spectrum of composition operators associated with bijective φ (that are unitary matrices, see e.g. [START_REF] Itzykson | Unitary groups: Representations and decompositions[END_REF] for some work on the representation theory of the unitary group), and the spectrum in the periodic case. We prove that in general, the spectrum is contained in the unit circle. The case when the symbol φ has « separable » variables (Theorem 6.9) is quite well understood, combining ideas from the elliptic attractive case and and the bijective case.

General properties of composition operators

In this section, we study the basic properties of the composition operator C φ : continuity, invertibility and compactness. We work on the Fréchet space Hol(B N ), equipped with the semi-norms ∥•∥ ∞,j , j ∈ N, defined by ∥f ∥ ∞,j = sup

z∈K j |f (z)| , with K j = 1 - 1 j B N . Proposition 2.1 (Continuity). Let φ be a self-map of B N .
The operator C φ is continuous on Hol(B N ).

Proof. Note that φ(B N ) ⊂ B N , φ is continuous, and K k is a compact subset of B N . Therefore, φ(K k ) is also a compact subset of B N , and there exists j

∈ N 0 such that φ(K k ) ⊂ K j .
Hence, for all f ∈ Hol(B N ),

∥C φ (f )∥ ∞,k = ∥f • φ∥ ∞,k = sup z∈K k |(f • φ)(z)| ≤ sup w∈K j |f (w)| = ∥f ∥ ∞,j . □
Next, we describe when the operator C φ is invertible on L(Hol(B N )), the algebra of continuous and linear operators on Hol(B N ). To do this, we take inspiration from Chapter 1 of [START_REF] Cowen | Composition operators on spaces of analytic functions[END_REF]. Let us begin by recalling the concept of character. Definition 2.2. Let E be a unitary Fréchet algebra. A character on E is a continuous linear form k on E, which is also multiplicative, i.e. ∀x, y ∈ E, k(xy) = k(x)k(y).

The following lemma characterises the characters on Hol(B N ). Lemma 2.3. k is a character on Hol(B N ) if and only if there exists x ∈ B N such that

∀g ∈ Hol(B N ), k(g) = g(x) =: k x (g). Proof. Let x ∈ B N . It is immediate that k x is linear and multiplicative. Moreover, since {x} is a compact subset of B N , k x is continuous on Hol(B N ), so k x is a character.
Conversely, let k be a character on Hol(B N ). For p ∈ {1, • • • , N }, set e p (z) = z p , and

x = (k(e 1 ), • • • , k(e N )). For ⃗ ȷ = (j 1 , • • • , j N ) ∈ N N 0 , denote |⃗ ȷ| = j 1 + • • • + j N .
For all f ∈ Hol(B N ), if we consider the truncated Maclaurin series (p n ) n∈N 0 defined by

p n = ⃗ ȷ∈N N 0 , |⃗ ȷ|≤n a ⃗ ȷ N ℓ=1 e j ℓ ℓ ,
then (p n ) converges uniformly to f on all compact subsets of B N . Finally, using the properties of k, it follows that

k(f ) = k lim n→∞ p n = lim n→∞ k(p n ) = lim n→∞ k   |⃗ ȷ|≤n a ⃗ ȷ N ℓ=1 e j ℓ ℓ   = lim n→∞ |⃗ ȷ|≤n a ⃗ ȷ N ℓ=1 k (e ℓ ) j ℓ = lim n→∞ |⃗ ȷ|≤n a ⃗ ȷ N ℓ=1 x j ℓ ℓ = f (x). □
We can now prove the next proposition. Denote by Aut(B N ) the group of all automorphisms (i.e. bijective maps) of B N .

Proposition 2.4 (Invertibility).

The operator C φ is invertible if and only if φ ∈ Aut(B N ).

Proof. If φ ∈ Aut(B N ), we can consider its inverse φ -1 . Hence, for all f ∈ Hol(B N ),

(C φ • C φ -1 )(f ) = f • φ -1 • φ = f, and (C φ -1 • C φ )(f ) = f • φ • φ -1 = f. Therefore, C φ is invertible, and its inverse is C φ -1 .
Conversely, suppose that C φ is invertible. Then there exists an operator T on Hol(B N )

such that C φ • T = T • C φ = Id. Hence, for all n 1 , • • • , n N ∈ N 0 , if e j (z) = z j , (C φ • T ) N j=1 e n j j = N j=1 e n j j = N j=1 (C φ • T )(e j ) n j = C φ N j=1
T (e j ) n j .

We compose by C -1 φ to the left, and set ψ = (T e 1 , • • • , T e N ). Then

T N j=1 e n j j = N j=1 T (e j ) n j = N j=1 ψ n j j = N j=1 e n j j • ψ. Moreover, if z ∈ B N , ψ(φ(z)) = (T e 1 (φ(z)), • • • , T e N (φ(z))) = ([(C φ • T )(e 1 )](z), • • • , [(C φ • T )(e N )](z)) = (z 1 , • • • , z N ) = z.
We deduce that ψ • φ = Id, so φ is univalent.

To show that φ is onto, let x ∈ B N , and k be a character on Hol(B N ). Since C φ is linear, continuous (by Proposition 2.1), and multiplicative, k • C φ is also a character on Hol(B N ). Hence, we can choose k so that k • C φ = k x . Also, by Lemma 2.3, there exists y ∈ B N such that k = k y , so

k x = k • C φ = k y • C φ = k φ(y) .
Finally, x = φ(y), so φ is onto, and bijective. Moreover, ψ = φ -1 , and T = C ψ . □

To finish, let us describe when C φ is compact on L(Hol(B N )). The proof goes along the same lines as in [START_REF] Arendt | In Koenigs' footsteps: diagonalization of composition operators[END_REF] or [START_REF] Célariès | Opérateurs et semi-groupes d'opérateurs sur des espaces de fonctions holomorphes -Applications à la théorie de l'universalité[END_REF]. Recall the definition of a compact operator on a Fréchet space. We refer to J. Kelley and I. Namioka ( [START_REF] Kelley | Linear topological spaces[END_REF]) for this notion. Definition 2.5. An operator T on a Fréchet space E is compact if there exists a neighborhood V of 0 such that T (V ) is relatively compact.

Here, we focus on a particular family of neighborhoods of 0 on Hol(B N ). For all r ∈ [0, 1) and ε > 0, consider

V r,ε = f ∈ Hol(B N ) : sup |z|≤r |f (z)| < ε .
In the following, we will need a generalization of Runge's theorem, so that we can approximate holomorphic functions by polynomials. We start with a definition. Definition 2.6. Let K be a compact subset of C N . We say that K is polynomially convex if

K = z ∈ C N : ∀P ∈ C[z 1 , • • • , z N ], |P (z)| ≤ max x∈K |P (x)| =: K.
Theorem 2.7 (Oka-Weil, [START_REF] Alexander | Several complex variables and Banach algebras[END_REF], [START_REF] Oka | Sur les fonctions analytiques de plusieurs variables. I. Domaines convexes par rapport aux fonctions rationnelles[END_REF], [START_REF] Weil | L'intégrale de Cauchy et les fonctions de plusieurs variables[END_REF]). Let K be a compact, polynomially convex subset of C N . If Ω is an open set containing K, then for all f ∈ Hol(Ω), there exists a sequence of polynomials (p n ) converging to f uniformly on K.

We can now give a characterization of compactness for composition operators in the multidimensional case.

Proposition 2.8 (Compactness).

The operator C φ is compact if and only if

∥φ∥ ∞ := sup z∈B N |φ(z)| < 1. Proof. Suppose that ∥φ∥ ∞ = r 0 < 1. Then, for all f ∈ V r 0 ,1 , sup z∈B N |(f • φ)(z)| < 1.
The family C φ (V r 0 ,1 ) is uniformly bounded by 1 on B N . Therefore, by Montel's theorem ( [START_REF] Conway | Functions of one complex variable[END_REF]), C φ (V r 0 ,1 ) is relatively compact.

Conversely, if ∥φ∥ ∞ = 1, let ε > 0 and r ∈ (0, 1). We show that C φ (V r,ε ) is not relatively compact. Without loss of generality, set ε < 1. Since there exists z 0 ∈ B N such that |φ(z 0 )| > r, let w 0 = φ(z 0 ) and

ρ = min |w 0 | -r 3 , 1 -|w 0 | 2 .
Set K = B(0, r) ∪ {w 0 }, and Ω = B(0, r + ρ) ∪ B(w 0 , ρ).

Then K is a compact subset of B N , and Ω is an open set of B N containing K. Moreover, by definition of ρ, B(0, r + ρ) ∩ B(w 0 , ρ) = ∅. Also, we show that K is polynomially convex. To do so, remark that B(0, r) and {w 0 } are convex, compact and disjoint sets. Using the separation lemma of [START_REF] Kallin | Polynomial convexity: The three spheres problem[END_REF] with Hahn-Banach's theorem, we obtain

K = B(0, r) ∪ {w 0 } = B(0, r) ∪ {w 0 } = B(0, r) ∪ {w 0 } = K.
Finally, for all n ∈ N 0 , set

h n (z) = 0 if z ∈ B(0, r + ρ), n if z ∈ B(w 0 , ρ).
Then h n ∈ Hol(Ω). By Oka-Weil's theorem, there exists a sequence (p n ) of polynomials such that for all n ∈ N 0 , ∥h n -p n ∥ ∞,K < ε.

Therefore, p n ∈ Hol(B N ) ∩ V r,ε , so p n • φ ∈ C φ (V r,ε ), and
|(p n • φ)(z 0 )| = |p n (w 0 )| ≥ n -ε ---→ n→∞ +∞.
Finally, since no subsequence of (p n • φ) converges on Hol(B N ), we can conclude that C φ (V r,ε ) is not relatively compact. □

Hyperbolic and parabolic cases

In this section, we focus on the functions φ without fixed points on B N . We mainly use the results of L. Arosio ([6]). Recall that, for all N ≥ 2, the Siegel upper half-space is

H N = {z = (z 1 , z) ∈ C N = C × C N -1 : ℑ(z 1 ) > |z|}.
This space is homeomorphic to B N . In the one-dimensional case, we will denote it by

H 1 = {z ∈ C : ℑ(z) > 0}.
Theorem 3.1 ([6]). Let φ be a hyperbolic self-map of B N . Denote by δ ∈ (0, 1) the dilation coefficient of φ at its Denjoy-Wolff point, as defined in [START_REF] Abate | Iteration theory of holomorphic maps on taut manifolds[END_REF].

Then there exist k ∈ {1, • • • , N }, ψ ∈ Aut(H k ) of the form ψ(z 1 , z 2 , • • • , z k ) = 1 δ z 1 , e it 2 √ δ z 2 , • • • , e it k √ δ z k , with t 2 , • • • , t k ∈ R, and a holomorphic map f : B N → H k such that (2) f • φ = ψ • f.
We deduce the following corollary, which is a generalization of Valiron's theorem.

Corollary 3.2 ([6]

). Let φ be a hyperbolic self-map of B N . Denote by δ ∈ (0, 1) the dilation coefficient of φ at its Denjoy-Wolff point. Then there exists a holomorphic map η :

B N → H 1 such that (3) η • φ = δ -1 η.
The equation ( 3) is called Schröder's equation.

Proof. Consider ψ and f found in Theorem 3.1. Let π : H k → H 1 be the projection defined by

π(z 1 , • • • , z k ) = z 1 , and η = π • f . By composition, η : B N → H 1 is holomorphic. Moreover, by (2), η • φ = π • f • φ = π • ψ • f = δ -1 (π • f ) = δ -1 η. □
The next proposition gives the description of the point spectrum and spectrum of C φ .

L. OGER Proposition 3.3. Let φ be a hyperbolic self-map of B N . Then,

σ p (C φ ) = C * and σ(C φ ) = C * if φ ∈ Aut(B N ), C if φ ̸ ∈ Aut(B N ).
Proof. By Corollary 3.2, there exists a function η ∈ Hol(B N ) such that η

• φ = cη, with c ∈ R, c > 1. Let λ ∈ C * . Note that η λ ∈ Hol(B N ), and
C φ (η λ ) = (η • φ) λ = (cη) λ = c λ η λ . Hence, {c λ : λ ∈ C * } ⊂ σ(C φ ). Let α = |α| e iθ ∈ C * . For λ = ln |α| + iθ ln(c) , we have c λ = e λ ln(c) = e ln|α|+iθ = |α| e iθ = α. Hence, C * ⊂ σ p (C φ ). Also, since φ is non-constant, if f ∈ Hol(B N ) satisfies f • φ ≡ 0, then f ≡ 0. Thus, 0 ̸ ∈ σ p (C φ ), and σ p (C φ ) = C * . Finally, seeing that σ p (C φ ) ⊂ σ(C φ )
, we conclude using Proposition 2.4, depending on whether φ ∈ Aut(B N ) or not.

□

Similarly, we have results for some parabolic maps. Let us start with a definition.

Definition 3.4. Let φ be a parabolic self-map of B N , and k N be the Kobayashi distance on

B N . We say that φ is positive-step if ∀x ∈ B N , lim n-→∞ k N (φ [n] (x), φ [n+1] (x)) > 0.
On the other hand,

φ is zero-step if ∃x ∈ B N , lim n-→∞ k N (φ [n] (x), φ [n+1] (x)) = 0.
L. Arosio's work focus on the positive-step parabolic maps. We get the following theorem.

Theorem 3.5 ([6]

). Let φ be a positive-step parabolic self-map of

B N . Then there exist k ∈ {1, • • • , N }, ψ ∈ Aut(H k ) of the form ψ(z 1 , z 2 , • • • , z k ) = (z 1 ± 1, e it 2 z 2 , • • • , e it k z k ) (4) or (z 1 -2z 2 + i, z 2 -i, e it 3 z 3 , • • • , e it k z k ), (5)
with t 2 , • • • , t k ∈ R, and a holomorphic map f : B N → H k such that (6) f • φ = ψ • f.
We deduce a generalization of Baker-Pommerenke's theorem.

Corollary 3.6. Let φ be a positive-step parabolic self-map of B N .

Then there exists a function η ∈ Hol(B N ) such that

(7) η • φ = η ± 1.
The equation ( 7) is called Abel's equation.

Proof. Consider ψ and f found in Theorem 3.5. r If ψ is of the form (4), in a similar way as in Corollary 3.2, set π :

H k → H 1 defined by π(z 1 , • • • , z k ) = z 1 , and η = π • f .
Therefore, η is holomorphic on B N , and by [START_REF] Arosio | Canonical models for the forward and backward iteration of holomorphic maps[END_REF],

η • φ = π • f • φ = π • ψ • f = (π • f ) ± 1 = η ± 1.
r If ψ is of the form (5), set ρ :

H k → C the holomorphic map defined by ρ(z 1 , • • • , z k ) = iz 2 .
Once again using [START_REF] Arosio | Canonical models for the forward and backward iteration of holomorphic maps[END_REF], for η = ρ • f , η is holomorphic on B N , and we obtain

η • φ = ρ • f • φ = ρ • ψ • f = (ρ • f ) -i 2 = η + 1. □
Hence, we obtain a similar result as Proposition 3.3 for positive-step parabolic maps.

Proposition 3.7. Let φ be a positive-step parabolic self-map of B N . Then,

σ p (C φ ) = C * and σ(C φ ) = C * if φ ∈ Aut(B N ), C if φ ̸ ∈ Aut(B N ).
Proof. By Corollary 3.6, there exists a function η ∈ Hol(B N ) such that η • φ = η ± 1. Even if we take -η instead of η, we may assume that η

• φ = η + 1. Let λ = |λ| e iθ ∈ C * , and set, for z ∈ B N , g(z) = λ η(z) = |λ| η(z) e iθη(z) .
Then g ∈ Hol(B N ) by composing with the exponential map. Hence, for all z ∈ B N ,

(g • φ)(z) = |λ| η(φ(z)) e iθη(φ(z)) = |λ| η(z)+1 e iθ(η(z)+1) = λg(z). Thus, C * ⊂ σ p (C φ ). Also, since φ is non-constant, if f ∈ Hol(B N ) satisfies f • φ ≡ 0, then f ≡ 0, so 0 ̸ ∈ σ p (C φ ), and σ p (C φ ) = C * . Seeing that σ p (C φ ) ⊂ σ(C φ )
, we conclude using Proposition 2.4, depending on whether φ ∈ Aut(B N ) or not. □

« Invertible » elliptic attractive case

In this section, we consider φ elliptic attractive, such that φ(0) = 0, and φ ′ (0) is an invertible matrix. For a matrix A, we will denote

∥A∥ = sup{|Ax| : x ∈ S N -1 }.
Note that since automorphisms of B N are elliptic non attractive, we will consider φ non bijective. Moreover, consider the following lemma. Lemma 4.1. Let φ be an elliptic attractive self-map of B N , such that φ(0) = 0. Then, [START_REF] Abate | Iteration theory of holomorphic maps on taut manifolds[END_REF] implies that |φ(y)| = |y|, and Proposition 2.2.14 of [START_REF] Abate | Iteration theory of holomorphic maps on taut manifolds[END_REF] deduces that there exist a subspace V of C N containing y and a unitary matrix U such that φ = U on V ∩ B N . We conclude that on this subspace, the iterates of φ cannot converge to some point. Hence, φ is not elliptic attractive. □

∥φ ′ (0)∥ = sup{|φ ′ (0)x| : x ∈ S N -1 } < 1, where |•| denotes the euclidean norm on C N . Proof. By Schwarz's lemma ([1], [32]), if φ is a holomorphic self-map of B N such that φ(0) = 0, then ∥φ ′ (0)∥ ≤ 1. If ∥φ ′ (0)∥ = 1, there exists x ∈ S N -1 such that |φ ′ (0)x| = 1 = |x|. Let y = x/2 ∈ B N . Then |φ ′ (0)y| = |y|. Lemma 2.2.13 of

Study of the point spectrum.

In order to obtain the point spectrum, we need a generalization of Koenigs' theorem ( [START_REF] Célariès | Opérateurs et semi-groupes d'opérateurs sur des espaces de fonctions holomorphes -Applications à la théorie de l'universalité[END_REF][START_REF] Koenigs | Recherches sur les intégrales de certaines équations fonctionnelles[END_REF]) in several variables. Note that if U is a unitary matrix, without loss of generality, using the Schür decomposition ( [START_REF] Horn | Matrix analysis[END_REF]), we will assume that φ ′ (0) is a lower triangular matrix.

Denote by

λ = (λ 1 , • • • , λ N ) the diagonal of φ ′ (0).
Definition 4.2. We say that the eigenvalues are resonant if there exist

j ∈ {1, • • • , N } and k 1 , • • • , k N ∈ N such that k 1 + • • • + k N ≥ 2 and λ k 1 1 × • • • × λ k N N = λ j .
The following result, due to Poincaré ([29], [31, Theorems 1.3 and 1.1]), will be the wanted generalization. Following the definitions of [START_REF] Raissy | Linearization of holomorphic germs with quasi-Brjuno fixed points[END_REF], 0 is an attractive fixed point of φ. Hence, Theorem 4.3. Let φ be an elliptic attractive self-map of B N . Assume that φ(0) = 0, φ ′ (0) is invertible and the eigenvalues of φ ′ (0) are not resonant. Then there exists a holomorphic function κ :

B N → C N such that κ ′ (0) = Id and κ • φ = φ ′ (0)κ.
The second step is simplifying the expression of the derivatives of a composed map, under certain assumptions. We start by some notations.

Notation. For ⃗ α = (α 1 , • • • , α N ) ∈ N N 0 and z ∈ C N , we write |⃗ α| = N k=1 α k , ⃗ α! = N k=1 α k !, z ⃗ α = N k=1 z α k k , f (⃗ α) = ∂ |⃗ α| f ∂z ⃗ α .
We also define a strict order ≺ on

N N 0 . For ⃗ ı, ⃗ ȷ ∈ N N 0 , ⃗ ı ≺ ⃗ ȷ if |⃗ ı| < |⃗ ȷ| or |⃗ ı| = |⃗ ȷ| ∃k ∈ {1, • • • , N }, i 1 = j 1 , • • • , i k-1 = j k-1 , i k < j k . We write ⃗ ı ≼ ⃗ ȷ if ⃗ ı ≺ ⃗ ȷ or ⃗ ı = ⃗ ȷ.
The order ≼ is a well-order. Hence, we can define the predecessor and the successor of an element ⃗ ȷ ∈ N N 0 , denoted respectively by ⃗ ȷ -and ⃗ ȷ + . Finally, for ⃗ ȷ ∈ N N 0 and n ∈ N 0 , set

Hol ⃗ ȷ (B N ) = {f ∈ Hol(B N ) : ∀⃗ ı ≼ ⃗ ȷ, f (⃗ ı) (0) = 0}.
The next lemma gives the wanted simplification.

Lemma 4.4. Let ⃗ ȷ ∈ N N 0 , and n = |⃗ ȷ| ∈ N 0 . If f ∈ Hol ⃗ ȷ -(B N ), then (8) (f • φ) (⃗ ȷ) (0) = λ ⃗ ȷ f (⃗ ȷ) (0).
Proof. Faà di Bruno's formula ( [START_REF] Ma | Higher chain formula proved by combinatorics[END_REF]) gives the equation

(f • φ) (⃗ ȷ) (0) = ⃗ ȷ! (s,p,M,r)∈D ⃗ ȷ f (⃗ r) (0) s k=1 N ℓ=1 1 m kℓ ! 1 ⃗ p k ! φ (⃗ p k ) ℓ (0) m kℓ ,
where (s, p, M, r) ∈ D ⃗ ȷ if and only if

⃗ ȷ = s k=1 | ⃗ m k | ⃗ p k , r ℓ = s k=1 m kℓ , ⃗ r = (r 1 , • • • r N ).
However, if |⃗ r| < n, then f (⃗ r) (0) = 0. As a result, we focus on the terms with |⃗ r| = n.

If there exists 1 ≤ k 0 ≤ s such that |⃗ p k 0 | ≥ 2, then n = |⃗ ȷ| = s k=1 | ⃗ m k | |⃗ p k | ≥ s k=1 | ⃗ m k | + | ⃗ m k 0 | = |⃗ r| + | ⃗ m k 0 | > n.
Since it is impossible, we must have

|⃗ p 1 | = • • • = |⃗ p s | = 1.
Also, even if we write some ⃗ m k = ⃗ 0, we may assume that s = N , p ab = 1 {a=b} , and

| ⃗ m k | = j k . Hence, (f • φ) (⃗ ȷ) (0) = ⃗ ȷ! | ⃗ m 1 |=j 1 • • • | ⃗ m N |=j N f (⃗ r) (0) N k,ℓ=1 1 m kℓ ! ∂φ ℓ ∂z k (0) m kℓ =0 if k>ℓ .
If there exists one k > ℓ such that m kℓ > 0, then the product is zero. Thus, there remain only the terms with m kℓ = 0 for k > ℓ. Moreover, note that r 1 = m 11 < j 1 , unless m 11 = j 1 . In this case, r 2 = m 22 < j 2 , unless m 22 = j 2 . Iterating until the rank N , we get ⃗ r ≺ ⃗ ȷ, so f (⃗ r) (0) = 0, unless m kℓ = j k 1 {k=ℓ} , in which case ⃗ r = ⃗ ȷ. Finally, there remains only one term, which is

(f • φ) (⃗ ȷ) (0) = ⃗ ȷ!f (⃗ ȷ) (0) N k=1 1 j k ! λ j k k = λ ⃗ ȷ f (⃗ ȷ) (0). □
We finally reach the main result about the point spectrum of C φ .

Proposition 4.5. Let φ be an elliptic attractive self-map of B N . Assume that φ(0) = 0, and

φ ′ (0) is invertible with eigenvalues λ 1 , • • • , λ N not resonant. Then σ p (C φ ) = N k=1 λ j k k : j 1 , • • • , j N ∈ N 0 = {λ ⃗ ȷ : ⃗ ȷ ∈ N N 0 }.
Proof. For all k ∈ {1, • • • , N }, using Theorem 4.3 (in particular, the first coordinate of κ) and changes of basis, we can find maps

η 1 , • • • , η N such that η k • φ = λ k η k . Thus, λ k ∈ σ p (C φ ). Moreover, note that if µ, ν ∈ σ p (C φ ) and n ∈ N 0 , then µν, µ n ∈ σ p (C φ ) since C φ is multiplicative. This concludes the inclusion ⊃. Conversely, let µ ̸ ∈ {λ ⃗ ȷ : ⃗ ȷ ∈ N N 0 }. If µ ∈ σ p (C φ ), then there exists f ∈ Hol(B N )\{0} such that f • φ = µf . We show by induction that f (⃗ ȷ) (0) = 0 for all ⃗ ȷ ∈ N N 0 . r First, since f (φ(0)) = f (0) = µf (0) and µ ̸ = 1, we get f (0) = 0.
r Assume that for all ⃗ ı ≺ ⃗ ȷ, we have f (⃗ ı) (0) = 0. By Lemma 4.4, we get the equation ( 8):

(f • φ) (⃗ ȷ) (0) = λ ⃗ ȷ f (⃗ ȷ) (0). But f • φ = µf , so λ ⃗ ȷ f (⃗ ȷ) (0) = (f • φ) (⃗ ȷ) (0) = µf (⃗ ȷ) (0). Since µ ̸ = λ ⃗ ȷ , we have f (⃗ ȷ) (0) = 0. We conclude that f ≡ 0, which is impossible. Finally, µ ̸ ∈ σ p (C φ ). □ 4.2.
Study of the spectrum in the diagonal case.

We now study the spectrum of C φ , in the particular case where φ ′ (0) is unitarily diagonalizable. We take our inspiration from [START_REF] Bernard | Semi-groupes d'opérateurs de compositions pondérés[END_REF] and [START_REF] Célariès | Opérateurs et semi-groupes d'opérateurs sur des espaces de fonctions holomorphes -Applications à la théorie de l'universalité[END_REF].

Using Theorem 4.3, if φ ′ (0) is diagonal, we can find maps η 1 , • • • , η N such that for all k ∈ {1, • • • , N }, η k • φ = λ k η k , η k (0) = 0, ∂η k ∂z ℓ (0) = 0 if k < ℓ, 1 if k = ℓ.
Hence, we obtain the following result.

Lemma 4.6.

If ⃗ ı, ⃗ ȷ ∈ N N 0 satisfy ⃗ ı ≼ ⃗ ȷ, then (η ⃗ ȷ ) (⃗ ı) (0) = 0 if ⃗ ı ≺ ⃗ ȷ, ⃗ ȷ! if ⃗ ı = ⃗ ȷ. Proof. Note that for all k ∈ {1, • • • , N }, the Maclaurin series of η k is η k (z) = k ℓ=1 α kℓ z ℓ + o(|z|), with α kℓ ∈ C.
Hence, by the multinomial theorem,

η ⃗ ȷ (z) = N k=1 k ℓ=1 α kℓ z ℓ j k + o(|z| |⃗ ȷ| ) = N k=1   |⃗ p k |=j k j k ⃗ p k (α k z) ⃗ p k   + o(|z| |⃗ ȷ| ) = |⃗ p 1 |=j 1 • • • |⃗ p N |=j N β p z ⃗ r + o(|z| |⃗ ȷ| ), with β p ∈ C, p kℓ = 0 if k < ℓ, and
⃗ r = (p 11 + • • • + p N 1 , p 22 + • • • + p N 2 , • • • , p N N ).
However, for ⃗ p 1 , • • • , ⃗ p N satisfying these assumptions, |⃗ r| = |⃗ ȷ|. Also, note that p 11 = j 1 , so ⃗ r ≽ ⃗ ȷ. This gives the result for ⃗ ı ≺ ⃗ ȷ. If ⃗ ı = ⃗ ȷ, the only way to get ⃗ r = ⃗ ȷ is by taking p kℓ = j k 1 {k=ℓ} . Hence, we conclude since

β p = 1, because α kk = 1 for all k. □ Let η = (η 1 , • • • , η N ). For ⃗ ȷ = (j 1 , • • • , j N ) ∈ N N 0 and f ∈ Hol(B N ), set P ⃗ 0 (f ) ≡ f (0) and P ⃗ ȷ (f ) = 1 ⃗ ȷ! f -Q ⃗ ȷ -(f ) (⃗ ȷ) (0) × η ⃗ ȷ , with Q ⃗ ȷ (f ) = ⃗ ı≼⃗ ȷ P ⃗ ı (f ).
We get the following theorem.

Theorem 4.7. Let ⃗ ȷ ∈ N N 0 . (i) Set W ⃗ ȷ = Vect(η ⃗ ı : ⃗ ı ≼ ⃗ ȷ). Then, Hol(B N ) = W ⃗ ȷ ⊕ Hol ⃗ ȷ (B N ). (ii) Q ⃗ ȷ is the projection on W ⃗ ȷ in parallel to Hol ⃗ ȷ (B N ). (iii) P ⃗ ȷ • C φ = C φ • P ⃗ ȷ = λ ⃗ ȷ P ⃗ ȷ .
Proof. We prove most of the properties by induction.

(i) First, we show that W ⃗ ȷ ∩ Hol ⃗ ȷ (B N ) = {0}. Indeed, by considering

f = ⃗ ı≼⃗ ȷ c ⃗ ı η ⃗ ı ∈ Hol ⃗ ȷ (B N ), we obtain c ⃗ 0 = f (0) = 0. Let ⃗ k ≼ ⃗ ȷ such that for ⃗ ı ≺ ⃗ k, c ⃗ ı = 0. By Lemma 4.6, f ( ⃗ k) (0) = ⃗ k≼⃗ ı≼⃗ ȷ c ⃗ ı × (η ⃗ ı ) ( ⃗ k) (0) = ⃗ k!c ⃗ k . Since f ∈ Hol ⃗ ȷ (B N ), we get c ⃗ k = 0. Finally, f ≡ 0.
Now, we show that W ⃗ ȷ + Hol ⃗ ȷ (B N ) = Hol(B N ). The inclusion ⊂ is trivial, so we focus on the other one. Note that Im(Q ⃗ ȷ ) ⊂ W ⃗ ȷ . We use an induction argument.

r Let f ∈ Hol(B N ). Then,

(f -Q ⃗ 0 (f ))(0) = (f -P ⃗ 0 (f ))(0) = f (0) -f (0) = 0. Hence, f -Q ⃗ 0 (f ) ∈ Hol ⃗ 0 (B N ), so f ∈ W ⃗ 0 + Hol ⃗ 0 (B N ). r Suppose now that f -Q ⃗ ȷ -(f ) ∈ Hol ⃗ ȷ -(B N ). Then, Q ⃗ ȷ (f ) = Q ⃗ ȷ -(f ) + P ⃗ ȷ (f ) = Q ⃗ ȷ -(f ) + 1 ⃗ ȷ! f -Q ⃗ ȷ -(f ) (⃗ ȷ) (0) × η ⃗ ȷ . Let ⃗ ı ≼ ⃗ ȷ. By definition of Q ⃗ ȷ , (f -Q ⃗ ȷ (f )) (⃗ ı) (0) = (f -Q ⃗ ȷ -(f )) (⃗ ı) (0) - 1 ⃗ ȷ! f -Q ⃗ ȷ -(f ) (⃗ ȷ) (0) × (η ⃗ ȷ ) (⃗ ı) (0). If ⃗ ı ≺ ⃗ ȷ, we have (f -Q ⃗ ȷ -(f )) (⃗ ı) (0) = (η ⃗ ȷ ) (⃗ ı) (0) = 0
, by using Lemma 4.6 and the assumption. Hence, (f

-Q ⃗ ȷ (f )) (⃗ ı) (0) = 0. Otherwise, if ⃗ ı = ⃗ ȷ, (f -Q ⃗ ȷ (f )) (⃗ ȷ) (0) = (f -Q ⃗ ȷ -(f )) (⃗ ȷ) (0) - 1 ⃗ ȷ! [f -Q ⃗ ȷ -(f )] (⃗ ȷ) (0) × ⃗ ȷ! = 0. Finally, f -Q ⃗ ȷ (f ) ∈ Hol ⃗ ȷ (B N ), so f ∈ W ⃗ ȷ + Hol ⃗ ȷ (B N ). (ii) By (i), if f ∈ Hol(B N ), then f has the form f = g + h, g ∈ W ⃗ ȷ , h ∈ Hol ⃗ ȷ (B N ),
with g and h uniquely determined. Moreover, Q ⃗ ȷ (f ) = g, so Q ⃗ ȷ is indeed the projection on W ⃗ ȷ in parallel to Hol ⃗ ȷ (B N ).

(iii) First, we show that for all ⃗ ȷ ∈ N N 0 , C φ • P ⃗ ȷ = λ ⃗ ȷ P ⃗ ȷ . Indeed,

(C φ • P ⃗ ȷ )(f ) = 1 ⃗ ȷ! f -Q ⃗ ȷ -(f ) (⃗ ȷ) (0) × (η ⃗ ȷ • φ) = 1 ⃗ ȷ! f -Q ⃗ ȷ -(f ) (⃗ ȷ) (0) × (λ ⃗ ȷ η ⃗ ȷ ) = λ ⃗ ȷ P ⃗ ȷ (f ). Now, we prove that for all ⃗ ȷ ∈ N N 0 , Q ⃗ ȷ • C φ = C φ • Q ⃗ ȷ .
To do so, it suffices to show that W ⃗ ȷ and

Hol ⃗ ȷ (B N ) are invariant by C φ . Let ⃗ ı ≼ ⃗ ȷ. Since η ⃗ ı • φ = λ ⃗ ı η ⃗ ı , then η ⃗ ı • φ ∈ W ⃗ ȷ , so C φ (W ⃗ ȷ ) ⊂ W ⃗ ȷ . Let f ∈ Hol ⃗ ȷ (B N ). Then, (f • φ)(0) = f (0) = 0. In addition, for 0 ≺ ⃗ k ≼ ⃗ ȷ, by denoting ⃗ k -the predecessor of ⃗ k, f ∈ Hol ⃗ k -(B N ). Hence, by Lemma 4.4, (f • φ) ( ⃗ k) (0) = λ ⃗ k f ( ⃗ k) (0) = 0. We conclude that f • φ ∈ Hol ⃗ ȷ (B N ).
Thus, for all f ∈ Hol(B N ), there exist g, h

∈ Hol ⃗ ȷ (B N ) such that f = Q ⃗ ȷ (f ) + g and f • φ = Q ⃗ ȷ (f • φ) + h. Hence, (Q ⃗ ȷ (f ) • φ) + (g • φ) = Q ⃗ ȷ (f • φ) + h.
But f can only be written in a unique way on W ⃗ ȷ ⊕ Hol ⃗ ȷ (B N ), so

Q ⃗ ȷ (f ) • φ = Q ⃗ ȷ (f • φ).

Finally, we show that P

⃗ ȷ • C φ = C φ • P ⃗ ȷ . Since P ⃗ 0 = Q ⃗ 0 , we have P ⃗ 0 • C φ = C φ • P ⃗ 0 . Let ⃗ ȷ ∈ N N 0 . Using the equations above, P ⃗ ȷ • C φ = (Q ⃗ ȷ -Q ⃗ ȷ -) • C φ = C φ • (Q ⃗ ȷ -Q ⃗ ȷ -) = C φ • P ⃗ ȷ . □
In order to prove the main result of this section, we need the two following lemmas. They are proven in the one-dimensional case in [START_REF] Célariès | Opérateurs et semi-groupes d'opérateurs sur des espaces de fonctions holomorphes -Applications à la théorie de l'universalité[END_REF].

Lemma 4.8. Let φ be an elliptic attractive self-map of B N , such that φ(0) = 0. Let g ∈ Hol(B N ), and λ ̸ = 0. If there exist ε ∈ (0, 1) and f ∈ Hol(B(0, ε)) such that for all |z| < ε,

λf (z) -f (φ(z)) = g(z),
then there exists a function f ∈ Hol(B N ) such that f|B(0,ε) = f , and for all z ∈ B N ,

λ f (z) -f (φ(z)) = g(z).
Proof. Consider the set E = {r ∈ [ε, 1] : f has an analytic extension fr ∈ Hol(B(0, r))}.

Thus, E is non-empty, because by the hypothesis, ε ∈ E. Moreover, if f has an extension fr on B(0, r), then for ε ≤ ρ ≤ r, fr is also an extension of f on B(0, λ). Hence, E is an interval, of the form [ε, δ) or [ε, δ], with δ ≤ 1.

We show that δ ∈ E. Consider an increasing sequence (r n ) on E, converging to δ. Let z ∈ B(0, δ). By analytic continuation, if n, m ∈ N 0 satisfy r n , r m ≥ |z|, then frn (z) = frm (z). Therefore, for all z ∈ B(0, δ), we set f (z) := frn (z) if |z| < r n .

Hence, f is well-defined, holomorphic on B(0, δ), and extends f .

Next, we show that δ = 1. Assume δ < 1. Suppose that for all δ ′ ∈ (δ, 1], φ(B(0, δ ′ )) ̸ ⊂ B(0, δ). Then there exists

z δ ′ ∈ B(0, δ ′ ) such that |φ(z δ ′ )| > δ. By Schwarz's lemma ([1]), |z δ ′ | > |φ(z δ ′ )| > δ. Thus, we construct a sequence (z n ) ⊂ B N such that |z n | > δ, |z n | ---→ n→∞ δ, |φ(z n )| > δ.
Using once again Schwarz's lemma, we get

|φ(z n )| n→∞ ---→ δ. By considering a subsequence (z n k ) converging to a certain z, then |z| = |φ(z)| = δ, which is impossible since ∥φ ′ (0)∥ < 1. Hence, there exists δ ′ ∈ (δ, 1] such that φ(B(0, δ ′ )) ⊂ B(0, δ).
Let us define fδ ′ by

fδ ′ (z) = fδ (φ(z)) -g(z) λ , |z| < δ ′ .
Then fδ ′ ∈ Hol(B(0, δ ′ )), and if |z| ≤ ε,

fδ ′ (z) = fδ (φ(z)) -g(z) λ = f (φ(z)) -g(z) λ = f (z).
Finally, δ ′ ∈ E, which is impossible because δ ′ > δ = max E. Therefore, δ = 1, and if we set f = f1 , then f ∈ Hol(B N ). We conclude that for all |z| ≤ ε,

λ f (z) -f (φ(z)) = λf (z) -f (φ(z)) = g(z).

The equation holds on B N by analytic continuation. □

In the following, for p ∈ N 0 , set

Hol p (B N ) := Hol (p,0,••• ,0) (B N ) = {f ∈ Hol(B N ) : ∀ |⃗ ȷ| ≤ p, f (⃗ ȷ) (0) = 0}.
Lemma 4.9. Let φ be an elliptic attractive self-map of B N , such that φ(0) = 0. For all λ 0 > 0, there exist ε ∈ (0, 1) and p ∈ N 0 such that for all |λ| > λ 0 and g ∈ Hol p (B N ),

g • φ [n]
λ n converges uniformly on B(0, ε).

Proof. By Lemma 4.1, ∥φ ′ (0)∥ < 1. Let ∥φ ′ (0)∥ < ζ < 1. There exists p ∈ N 0 such that ζ p+1 < λ 0 . Note that lim sup z→0 |φ(z)| |z| ≤ ∥φ ′ (0)∥ .
Thus, there exists ε ∈ (0, 1) such that if |z| ≤ ε,

|φ(z)| ≤ ζ |z| < ε.
Using induction and Schwarz lemma, for all n ∈ N 0 , φ [n] (z) ≤ ζ n |z| < ε. In addition, the order of all Maclaurin coefficients of g is greater or equal to p + 1, so there exists C > 0 such that for all |z| ≤ ε, |g(z)| ≤ C |z| p+1 . Finally, for all n ∈ N 0 , |λ| > λ 0 and |z| ≤ ε,

(g • φ [n] )(z) λ n ≤ C |λ| n φ [n] (z) p+1 ≤ C |λ| n (ζ n |z|) p+1 ≤ ζ p+1 λ 0 n Cε p+1 . Since ζ p+1 < λ 0 , (g • φ [n] )/λ n converges normally on B(0, ε). □
Here is the main result of this section, about the spectrum of C φ . This is a generalisation of [5, Theorem 4.1]. Proposition 4.10. Let φ be an elliptic attractive self-map of B N , such that φ(0) = 0 and φ ′ (0) is invertible, unitarily diagonalizable, with eigenvalues not resonant. Then

σ(C φ ) = {0} ∪ σ p (C φ ). Proof. The inclusion ⊃ is trivial, since φ ̸ ∈ Aut(B N ).
We focus on the other inclusion. Let µ ̸ ∈ {0} ∪ σ p (C φ ). By Lemma 4.9, there exist 0 < ε < 1 and p ∈ N 0 such that for all g ∈ Hol p (B N ),

n≥0 g • φ [n]
µ n converges uniformly on B(0, ε). However, µ ̸ ∈ σ p (C φ ) = {λ ⃗ ȷ : |⃗ ȷ| ≤ p}, so S -µId is invertible.

Moreover, by

r Consider the function

h = - 1 µ n≥0 g • φ [n]
µ n ∈ Hol(B(0, ε)).

Then, on B(0, ε),

(h • φ) -µh = n≥0 g • φ [n] µ n - n≥0 g • φ [n+1] µ n+1 = g • φ [0] = g.
Hence, by Lemma 4.8, there exists a function h ∈ Hol(B N ) such that

( h • φ) -µ h = g. This shows that T -µId is onto. In addition, T -µId is univalent since µ ̸ ∈ σ p (C φ ) and C\σ p (C φ ) ⊂ C\σ p (T ). We conclude that T -µId is invertible on L(Hol p (B N )). Finally, C φ -µId is invertible, so µ ̸ ∈ σ(C φ ). □
Let us consider now a more general case. Let α ∈ B N . We define the holomorphic map [START_REF] Abate | Iteration theory of holomorphic maps on taut manifolds[END_REF] or [START_REF] Rudin | Function theory in the unit ball of C n[END_REF]), ψ(0) = α and ψ -1 α = ψ α . We get the following result. Corollary 4.11. Let α ∈ B N , and φ be an elliptic attractive self-map of B N such that φ(α) = α and φ ′ (α) is invertible. Denote by λ 1 , • • • , λ N the eigenvalues of φ ′ (α), and assume that they are not resonant. Then,

ψ α : B N → B N by ψ α (z) = α -P α (z) -s α Q α (z) 1 -⟨z, α⟩ , with P α (z) = ⟨z,α⟩ |α| 2 α, Q α (z) = z -P α (z) and s α = (1 -|α| 2 ) 1/2 . Then, ψ α ∈ Aut(B N ) (see e.g.
σ p (C φ ) = N k=1 λ n k k : n 1 , • • • , n N ∈ N 0 . Moreover, if φ ′ (α) is unitarily diagonalizable, σ(C φ ) = {0} ∪ σ p (C φ ). Proof. Consider the map γ = ψ α • φ • ψ α . Then, γ(0) = 0 and γ ′ (0) = ψ ′ α (α) × φ ′ (α) × ψ ′ α (0). Moreover, ψ ′ α (α) = [ψ ′ α (0)] -1
, so γ ′ (0) and φ ′ (α) are similar: they have the same eigenvalues. In addition, since the composition operators are multiplicative,

C γ = C -1 ψα • C φ • C ψα .
Thus, C γ and C φ are similar. By Proposition 4.5,

σ p (C φ ) = σ p (C γ ) = N k=1 λ n k k : n 1 , • • • , n N ∈ N 0 .
Similarly, when φ ′ (α) is unitarily diagonalizable, by Proposition 4.10,

σ(C φ ) = σ(C γ ) = {0} ∪ σ p (C φ ). □

« Non invertible » elliptic attractive case

Now, we consider φ elliptic attractive, such that φ(0) = 0 and φ ′ (0) is not invertible. Similarly to Section 4, in all the following results, we can replace 0 by some α ∈ B N , using the automorphism ψ α . 5.1. Symbols with nilpotent Jacobian matrix at 0.

The first case that we consider is when 0 is the unique eigenvalue of φ ′ (0), that is φ ′ (0) is nilpotent. We obtain the following results. Proposition 5.1. Let φ be an elliptic attractive self-map of B N . Assume that φ ̸ ≡ 0, φ(0) = 0, and φ ′ (0) is nilpotent. Then

{1} ⊂ σ p (C φ ) ⊂ {0, 1}.
Proof. The fact that 1 ∈ σ p (C φ ) was proved in Proposition 5.4.

Moreover, we may have 0 ∈ σ p (C φ ), for instance if the dimension of the range of φ is not N , by considering f such that f = 0 on ran(φ) and f ̸ = 0 (such a function exists by simple algebra).

Conversely, let µ ̸ ∈ {0, 1}. If µ ∈ σ p (C φ ), there exists f ∈ Hol(B N )\{0} such that f • φ = µf . We show that f (⃗ ȷ) (0) = 0 for all ⃗ ȷ ∈ N N 0 . r Note that f (φ(0)) = f (0) = µf (0). Since µ ̸ = 1, then f (0) = 0. r Let ⃗ ȷ ∈ N N 0 \{ ⃗ 0}. Assume that for ⃗ ı ≺ ⃗ ȷ, f (⃗ ı) (0) = 0. Lemma 4.4
gives us once again the formula [START_REF] Banach | Théorie des opérations linéaires[END_REF], with λ ⃗ ȷ = 0. Therefore,

(f • φ) (⃗ ȷ) (0) = λ ⃗ ȷ f (⃗ ȷ) (0) = 0. Finally, µf (⃗ ȷ) (0) = (f • φ) (⃗ ȷ) (0) = 0, so f (⃗ ȷ) (0) = 0 because µ ̸ = 0.
We conclude that f ≡ 0, which is impossible. Hence, µ ̸ ∈ σ p (C φ ). □ Proposition 5.2. Let φ be an elliptic attractive self-map of B N . Assume that φ ̸ ≡ 0, φ(0) = 0, and φ ′ (0) is nilpotent. Then

σ(C φ ) = {0, 1}.
Proof. Note that since the diagonal of φ ′ (0) only has zeroes, the matrix φ ′ (0) is nilpotent. Hence, there exists n 0 ∈ N such that

(φ [n 0 ] ) ′ (0) = (φ ′ (0)) n 0 = 0. Let µ ∈ C\{0, 1}. We consider two cases. r Let c ∈ C * . Setting f c (z) = c/(1 -µ), for all z ∈ B N , f c (φ(z)) -µf c (z) = c 1 -µ -µ c 1 -µ = c.
r Let g ∈ Hol(B N ) such that g(0) = 0. We know that there exists a constant d > 1 such that for all z ∈ B N ,

|g(z)| ≤ d |z| , |φ(z)| ≤ d |z| , φ [n 0 ] (z) ≤ d |z| 2 .
Using Schwarz's lemma, for all k ∈ N, k ≥ 3 and z ∈ B N ,

φ [kn 0 ] (z) ≤ d k |z| 2 k ≤ d k |z| 2k+2 .
For n ∈ N, if n ≥ 3n 0 , we can write n = kn 0 + p, with k ≥ 3 and 0 ≤ p < n 0 . Therefore, k = (n -p)/n 0 ≥ n/n 0 -1, and

g(φ [n] (z)) µ n ≤ d φ [n] (z) |µ| n ≤ d p+1 φ [kn 0 ] (z) |µ| n ≤ d k+p+1 |z| 2k+2 |µ| n ≤ d d n |z| 2n/n 0 |µ| n = d d |z| 2/n 0 |µ| n .
Now, we choose ε ∈ (0, 1) such that dε 2/n 0 < |µ|, so that

h = - 1 µ n≥0 g(φ [n] (z)) µ n ∈ Hol(B(0, ε)).
Moreover, the same calculations as in Proposition 4.10 show that

h • φ -µh = g.
Using Lemma 4.8, there exists h ∈ Hol(B N ) such that

h • φ -µ h = g.
Thus, if η ∈ Hol(B N ), then g = η -η(0) satisfies g(0) = 0, and

(f η(0) + h) • φ -µ(f η(0) + h) = η(0) + g = η.
Finally, C φ -µId is invertible, and µ ̸ ∈ σ(C φ ), so σ(C φ ) ⊂ {0, 1}.

To conclude, note that since φ ̸ ∈ Aut(B N ), using Proposition 2.4, 0 ∈ σ(C φ ). Moreover,

1 ∈ σ p (C φ ) ⊂ σ(C φ ). □ Remark 5.3. If φ ≡ 0, then for all f ∈ Hol(B N ), C φ (f ) ≡ f (0). In this case, σ p (C φ ) = σ(C φ ) = {0, 1}.
Indeed, let µ ∈ σ p (C φ ) and f ∈ Hol(B N )\{0} such that f (0) = µf . Evaluating the function at 0, we obtain µ ∈ {0, 1}. Also, 1 and z 1 are eigenvectors associated with 1 and 0.

To prove that σ(C φ ) = {0, 1}, let µ ̸ ∈ {0, 1}, and f ∈ Hol(B N ). Set

g = - f (0) µ(µ -1) - f µ .
Then g(0) -µg = f , so C 0 -µId is invertible.

Symbols with general Jacobian at 0.

In general, we can only get inclusions for the spectra of C φ .

Proposition 5.4. Let φ be an elliptic attractive self-map of B N . Assume that φ ̸ ≡ 0, φ(0) = 0 and φ ′ (0) is not invertible. Denote by λ 1 , • • • , λ p the non-zero eigenvalues of φ ′ (0). Then

{1} ⊂ σ p (C φ ) ⊂ p k=1 λ j k k : j 1 , • • • , j p ∈ N 0 ∪ {0, 1}.
Spectrum: Let g ∈ Hol(B N ), and µ ̸ ∈ σ p (C φ ). We want to find f ∈ Hol(B N ) such that

f • φ -µf = g. Writing f (z) = ⃗ ȷ∈N N 0 b ⃗ ȷ z ⃗ ȷ and g(z) = ⃗ ȷ∈N N 0 a ⃗ ȷ z ⃗ ȷ , we obtain ⃗ ȷ∈N N 0 b ⃗ ȷ p k=1 (λ k z k ) j k -µ N k=1 z j k k = ⃗ ȷ∈N N 0 b ⃗ ȷ z ⃗ ȷ λ ⃗ ȷ -µ = ⃗ ȷ∈N N 0 a ⃗ ȷ z ⃗ ȷ , with λ = (λ 1 , • • • , λ p , 0, • • • , 0). Hence, for all ⃗ ȷ ∈ N N 0 , b ⃗ ȷ = a ⃗ ȷ λ ⃗ ȷ -µ . Note that because |λ 1 | , • • • , |λ N | < 1, there exists n 0 ∈ N 0 such that if |⃗ ȷ| ≥ n 0 , then λ ⃗ ȷ < |µ| /2. Thus, |b ⃗ ȷ | ≤ C |a ⃗ ȷ | , with C = max max |⃗ ȷ|<n 0 1 |λ ⃗ ȷ -µ| , 2 |µ| .
Hence, we just proved that f ∈ Hol(B N ), so C φ -µId is invertible. Therefore, thanks to Proposition 5.5, σ(C φ ) = σ p (C φ ).

Elliptic non attractive case

We now consider symbols φ such that φ(0) = 0, and the iterates of φ do not converge to 0. We begin this section by looking at bijective symbols.

Automorphic symbols.

As in Section 4, even if we conjugate by ψ α , we will assume in the following that φ(0) = 0, so that φ = U is a unitary matrix. Since U is diagonalizable, and all its eigenvalues are of modulus 1, we can even consider

φ = D, where D = diag(e iθ 1 , • • • , e iθ N ).
We mainly take our inspiration from [START_REF] Arendt | Spectral properties of weighted composition operators on Hol(D) induced by rotations[END_REF]. Proposition 6.1. Let φ be an elliptic automorphism, such that φ(0) = 0. Denote by e iθ 1 , • • • , e iθ N the eigenvalues of the matrix asscoiated with φ. Then

σ p (C φ ) = N k=1 e ij k θ k : j 1 , • • • , j N ∈ N 0 . Proof. For all k ∈ {1, • • • , N }, consider f k (z) = z k . Then (f k • φ)(z) = f k (Dz) = [Dz] k = e iθ k z k = e iθ k f k (z).
Hence, e iθ k ∈ σ p (C φ ). Moreover, if λ, µ ∈ σ p (C φ ), then λµ ∈ σ p (C φ ) and for all n ∈ N 0 , λ n ∈ σ p (C φ ). This concludes for the inclusion ⊃.

For the converse, set e = (e iθ 1 , • • • , e iθ N ). Let µ ̸ ∈ {e ⃗ ȷ : ⃗ ȷ ∈ N N 0 }. If µ ∈ σ p (C φ ), there exists f ∈ Hol(B N )\{0} such that f • φ = µf . We show that f (⃗ ȷ) (0) = 0 for all ⃗ ȷ ∈ N N 0 . r Since f (φ(0)) = f (0) = µf (0) and µ ̸ = 1, we have f (0) = 0. r Let ⃗ ȷ ∈ N N 0 .
Assume that for all ⃗ ı ≺ ⃗ ȷ, f (⃗ ı) (0) = 0. Lemma 4.4 gives us once more the formula ( 8):

(f • φ) (⃗ ȷ) (0) = e ⃗ ȷ f (⃗ ȷ) (0). Therefore, µf (⃗ ȷ) (0) = e ⃗ ȷ f (⃗ ȷ) (0). Since µ ̸ = e ⃗ ȷ , we get f (⃗ ȷ) (0) = 0.
We conclude that f ≡ 0, which is impossible. Thus, µ ̸ ∈ σ p (C φ ). □ Proposition 6.2. Let φ be an elliptic automorphism, such that φ(0) = 0. Denote by e iθ 1 , • • • , e iθ N the eigenvalues of the matrix associated with φ. Recall that T is the unit circle of C.

Then N k=1 e ij k θ k : j 1 , • • • , j N ∈ N 0 ⊂ σ(C φ ) ⊂ T.
Proof. The first inclusion is immediate by Proposition 6.1.

To prove the other one, let µ ∈ C\T and g ∈ Hol(B N ). We write

g = ⃗ ȷ∈N N 0 a ⃗ ȷ z ⃗ ȷ . If f = b ⃗ ȷ z ⃗ ȷ satisfies f • φ -µf = g, then for all ⃗ ȷ ∈ N N 0 , b ⃗ ȷ (e ⃗ ȷ -µ) = a ⃗ ȷ . Hence, b ⃗ ȷ = a ⃗ ȷ e ⃗ ȷ -µ =⇒ 1 1 + |µ| ≤ b ⃗ ȷ a ⃗ ȷ = 1 |e ⃗ ȷ -µ| ≤ 1 |1 -|µ|| . Finally, f = ⃗ ȷ∈N N 0 b ⃗ ȷ z ⃗ ȷ ∈ Hol(B N ), so C φ -µId is invertible. □
We consider now a particular class of elliptic automorphisms. Let us start with a definition. Definition 6.3. Let φ = U be a unitary matrix, with eigenvalues e iθ 1 , • • • , e iθ N . We say that

φ is periodic if for all k ∈ {1, • • • , N }, θ k ∈ 2πQ.
Note the following lemma.

Lemma 6.4. Let φ be a periodic unitary matrix, with eigenvalues e iθ 1 , • • • , e iθ N . For all 1 ≤ k ≤ N , denote by p k the smallest positive integer such that p k θ k ∈ 2πZ. Then,

p = lcm(p 1 , • • • , p N )
is the smallest positive integer such that φ p is similar to Id.

Proof. Let k ∈ {1, • • • , N }. Since p is a multiple of p k and p k θ k ∈ 2πZ, then pθ k ∈ 2πZ.
Hence, (e iθ k ) p = 1. Finally, the eigenvalues of φ p are

(e iθ k ) p = 1, k = 1, • • • , N,
so 1 is the unique eigenvalue of φ p , which is unitary. Therefore, φ p is similar to the identity. Now, we show the minimality of p. To do so, let q ∈ N such that φ q is similar to Id. Then, 1 is the unique eigenvalue of φ q , so for all k ∈ {1, • • • , N },

e iqθ k = 1, that is qθ k ∈ 2πZ.
Hence, for all k, q is a multiple of p k , so q ≥ lcm(p 1 , • • • , p N ) = p. □

We reach the main result of this section. Proposition 6.5. If φ is a periodic unitary matrix, with eigenvalues e iθ 1 , • • • e iθ N , then

σ(C φ ) = {λ ∈ C : λ p = 1},
where p is defined in Lemma 6.4.

Proof. Note that since φ p is similar to the identity matrix, by polynomial functional calculus ( [START_REF] Bourbaki | Éléments de mathématique. Théories spectrales. Cha-pitres 1 et 2. Cham[END_REF], [START_REF] Dunford | Linear operators. I. General theory[END_REF]) and multiplicativity of C φ , we write

σ(C φ ) p = σ(C p φ ) = σ(C φ p ) = σ(C Id ) = {1}. Hence, if λ ∈ σ(C φ ), then λ p = 1.
Conversely, let λ ∈ C such that λ p = 1. Set e = (e iθ 1 , • • • , e iθ N ).

Step 1: We show that there exists ⃗ ȷ ∈ N N 0 such that p is the smallest positive integer verifying (e ⃗ ȷ ) p = 1. First, note that for all k ∈ {1, • • • , N }, we may write

θ k = 2πq k /p k , with p k ∧ q k = 1 and p = lcm(p 1 , • • • , p N ). We show that exp 2iπq k j p k : j ∈ N 0 = exp 2iπj p k : j ∈ N 0
Indeed, we only have to show the inclusion ⊃. Let j ∈ N 0 . By Bézout's identity, there exist ℓ ∈ Z and m ∈ N such that ℓp k + mq k = 1. Hence,

exp i 2πj p k = exp i 2π(jℓp k + jmq k ) p k = exp i 2π(jm)q k p k . Therefore, for all k ∈ {1, • • • , N }, there exists m k ∈ N such that exp 2iπm k q k p k = exp 2iπ p k .
We show now that by setting r k = p/p k , the r k are setwise coprime. Indeed, if all the r k were divisible by a certain r ≥ 2, then r divides p, and p/r is a common multiple of p 1 , • • • , p N , which is impossible.

Finally, there exists

ℓ 1 , • • • , ℓ N ∈ Z such that N k=1 ℓ k r k = 1.
Thus,

exp 2iπ N k=1 q k ℓ k m k p k = exp 2iπ N k=1 ℓ k p k = exp 2iπ p N k=1 r k ℓ k = exp 2iπ p .
By setting j k = ℓ k m k (even if we add a multiple of p k before, so that j k ≥ 0), e ⃗ ȷ = exp(2iπ/p), so p is the smallest positive integer verifying (e ⃗ ȷ ) p = 1.

Step 2: Since λ p = 1, λ p ∈ σ(Id) = σ(C p φ ) = σ(C φ ) p . Then, there exists γ ∈ σ(C φ ) such that (λ/γ) p = 1. If λ/γ = e iα , then

e iαp = 1 = (e ⃗ ȷ ) p .
Hence, there exist ℓ, q ∈ {0,

• • • , p -1} such that α = N k=1 j k θ k + 2qπ p and N k=1 j k θ k = 2ℓπ p ,
with ℓ ∧ p = 1. By Bézout's theorem, there exist s ∈ N and t ∈ Z such that ℓs + tp = 1. Thus,

λ γ = e iα = exp i N k=1 j k θ k + 2qπ p = exp i N k=1 j k θ k + 2ℓsqπ p + 2tqπ = exp i(sq + 1) N k=1 j k θ k = (e ⃗ ȷ ) sq+1 .
Step 3: Consider the map Γ ⃗ ȷ : Hol(B N ) → z ⃗ ȷ Hol(B N ) defined by Finally, since T is similar to e ⃗ ȷ C φ , we get σ(T ) = e ⃗ ȷ σ(C φ ). Hence, e ⃗ ȷ σ(C φ ) ⊂ σ(C φ ). Since λ = (e ⃗ ȷ ) qs+1 γ, with γ ∈ σ(C φ ), we conclude that λ ∈ σ(C φ ). □ Remark 6.6. If φ is not a periodic matrix, then by Propositions 6.1 and 6.2, the sets σ p (C φ ), T\σ p (C φ ), σ(C φ ) and T\σ(C φ ) are all dense in T.

[Γ ⃗ ȷ (f )](z) = z ⃗ ȷ f (z). Hence, Γ ⃗ ȷ is bijective. Set T = (C φ ) |z ⃗ ȷ Hol(B N ) . Thus, for f ∈ Hol(B N ), [(Γ -1 ⃗ ȷ • T • Γ ⃗ ȷ )(f )](z) = (z ⃗ ȷ f • φ)(z) z ⃗ ȷ = φ(z) ⃗ ȷ f (φ(z)) z ⃗ ȷ = e ⃗ ȷ f (φ(z)).
6.2. Non bijective symbols. Now, consider φ non automorphic. The following result gives an idea of the behaviour of the point spectrum. Proposition 6.7. Let φ be an elliptic non attractive self-map of B N such that φ(0) = 0 and φ ̸ ∈ Aut(B N ). Then, {1} ⊂ σ p (C φ ) ⊂ D.

Proof. Let λ ∈ C such that there exists f ∈ Hol(B N )\{0} satisfying

f • φ = λf.
Iterating this equality, we obtain, for all n ∈ N, f • φ [n] = λ n f . If |λ| > 1, then the right hand side diverges to +∞ at least at one point of B N . This is impossible.

Moreover, as in the previous proofs, since 1 • φ = 1, we get 1 ∈ σ p (C φ ). □ Remark 6.8. In the preceding proof, if |λ| < 1, then the right hand side goes to 0. Hence, any eventual eigenfunction must satisfy f • h = 0, with h the sublimit of (φ [n] ) found in Theorem 1.1.

If the symbol φ has « separable » variables (see Theorem 6.9), we can be more precise. Note that if φ is elliptic non attractive and not bijective, then by Theorem 1.1, there exists a subsequence of (φ [n] ) that converges to

h : z -→ (z 1 , • • • , z p , 0, • • • , 0).
Moreover, using one-variable Denjoy-Wolff's theory, the components that converge to the identity must be rotations, and the other ones must be elliptic non invertible maps (in one variable) fixing 0. Hence, the hypothesis of the following theorem are as general as possible, when φ has « separable » variables. Theorem 6.9. Let φ be an elliptic non attractive self-map of B N , such that φ is not invertible, φ(0) = 0, and

φ(z 1 , • • • , z N ) = (φ 1 (z 1 ), • • • , φ p (z p ), β p+1 z p+1 , • • • , β N z N ), with φ 1 , • • • , φ p non bijective, |φ ′ 1 (0)| , • • • , φ ′ p (0) < 1, and
|β p+1 | = • • • = |β N | = 1. Then, {0} ∪ p k=1 φ ′ k (0) n k × N k=p+1 β n k k : n 1 , • • • , n N ∈ N 0 ⊂ σ(C φ ) ⊂   n 1 ,••• ,np≥0 p k=1 φ ′ k (0) n k T   ∪ {0}.
Remark that if there exists j ∈ {1, • • • , n} such that φ ′ j (0) = 0, then all the terms

p k=1 φ ′ k (0) n k × N k=p+1 β n k k
with n k ̸ = 0 vanish. Then, we will consider in the proof one more hypothesis: we will assume that 0

< |φ ′ 1 (0)| , • • • , φ ′ p (0) < 1.
To prove this theorem, we will consider an other decomposition of Hol(B N ), using the Koenigs' functions κ

1 , • • • , κ p of φ 1 , • • • , φ p . Let us begin with a result in the two-dimensional case, with p = 1, that is φ(z 1 , z 2 ) = (φ 1 (z 1 ), β 2 z 2 ).
Proposition 6.10. For all m ∈ N, consider

W m = m-1 ℓ=0 κ ℓ 1 (z 1 )f ℓ (z 2 ) : f ℓ ∈ Hol(D) , X m = z m 1 Hol(B 2 ). Then (i) Hol(B 2 ) = W m ⊕ X m , (ii) C φ (W m ) ⊂ W m and C φ (X m ) ⊂ X m .
Proof. We prove (i) by induction. m = 1: We have to show that Hol(B 2 ) = W 1 ⊕ X 1 . Let f ∈ Hol(B 2 ). Using the Maclaurin coefficients of f , we may write

f (z) = j≥0 k≥0 a jk z j 1 z k 2 = k≥0 a 0k z k 2 ∈W 1 + z 1 j≥1 k≥0 a jk z j-1 1 z k 2 ∈X 1 . Hence, Hol(B 2 ) = W 1 + X 1 . Now, assume that f ∈ W 1 ∩ X 1 . We can write f (z) = f 0 (z 2 ) = z 1 f (z), with f 0 ∈ Hol(D) and f ∈ Hol(B 2 ). Considering z = (0, z 2 ), we get f 0 (z 2 ) = 0 for all z 2 ∈ D. Thus, f ≡ 0, so Hol(B 2 ) = W 1 ⊕ X 1 . m → m + 1: Assume that Hol(B 2 ) = W m ⊕ X m . Let f ∈ Hol(B 2 ). By induction hypothesis, f (z) = m-1 ℓ=0 κ ℓ 1 (z 1 )f ℓ (z 2 ) + z m 1 f (z), with f 0 , • • • , f m-1 ∈ Hol(D) and f ∈ Hol(B 2 ). Denote f (z) = j≥0 k≥0 b jk z j 1 z k 2 .
Hence, there exists g ∈ Hol(B 2 ) such that

z m 1 f (z) = z m 1 f m (z 2 ) + z m+1 1 g(z), f m (z) = k≥0 b 0k z k 2 .
Moreover, by the properties of κ 1 , there exists h ∈ Hol(D) such that

κ m 1 (z 1 ) = z m 1 + z m+1 1 h(z 1 ). Therefore, z m 1 f (z) = κ m 1 (z 1 )f m (z 2 ) + z m+1 1 (g(z) -h(z 1 )f m (z 2 )) .
Finally, we have

f (z) = m-1 ℓ=0 κ ℓ 1 (z 1 )f ℓ (z 2 ) + κ m 1 (z 1 )f m (z 2 ) ∈W m+1 + z m+1 1 (g(z) -h(z 1 )f m (z 2 )) ∈X m+1 , so Hol(B 2 ) = W m+1 + X m+1 . Now, assume that f ∈ W m+1 ∩ X m+1 . Then, f (z) = m ℓ=0 κ ℓ 1 (z 1 )f ℓ (z 2 ) = z m+1 1 f (z), with f 0 , • • • , f m ∈ Hol(D) and f ∈ Hol(B 2 ). Hence, m-1 ℓ=0 κ ℓ 1 (z 1 )f ℓ (z 2 ) = z m+1 1 f (z) -κ m 1 (z 1 )f m (z 2 ) = z m 1 (z 1 f (z) -f m (z 2 ) -z 1 h(z 1 )f m (z 2 )) ∈ W m ∩ X m = {0}. Thus, f (z) = κ m 1 (z 1 )f m (z 2 ) = z m+1 1 f (z), so z m 1 f m (z 2 ) = z m+1 1 ( f (z) -h(z 1 )f m (z 2 )).
Denote by (c jk ) the Maclaurin coefficients of z → z m 1 f m (z 2 ), and

(d jk ) those of z → z m+1 1 ( f (z)- h(z 1 )f m (z 2 )). Then, j ̸ = m =⇒ c jk = 0 and j = m =⇒ d jk = 0.
Finally, since c jk = d jk for all j, k ∈ N 0 , we obtain

z m 1 f m (z 2 ) = z m+1 1 ( f (z) -h(z 1 )f m (z 2 )) = 0.
Hence, f m ≡ 0, so f ≡ 0. We proved that Hol(

B 2 ) = W m ⊕ X m . Now, let us prove (ii). If f ∈ W m , then f = m-1 ℓ=0 κ ℓ 1 (z 1 )f ℓ (z 2 ), with f 0 , • • • , f m-1 ∈ Hol(D). Thus, (f • φ)(z) = m-1 ℓ=0 φ ′ 1 (0) ℓ κ ℓ 1 (z 1 )f ℓ (β 2 z 2 ) = m-1 ℓ=0 κ ℓ 1 (z 1 ) fℓ (z 2 ), where fℓ (z 2 ) = φ ′ 1 (0) ℓ f ℓ (β 2 z 2 ). Thus, f • φ ∈ W m . If f ∈ X m , then f (z) = z m 1 g(z), with g ∈ Hol(B 2 ). Hence, (f • φ)(z) = φ 1 (z 1 ) m g(φ 1 (z 1 ), β 2 z 2 ).
However, since φ 1 (0) = 0, we may write φ 1 (z 1 ) = z 1 ψ(z 1 ), with ψ ∈ Hol(D). Therefore, (f • φ)(z) = z m 1 ψ(z 1 )g(φ 1 (z 1 ), β 2 z 2 ) ∈ X m . □

We are now able to prove Theorem 6.9.

Proof of Theorem 6.9. Let g ∈ Hol(B N ). Denote z = (z 1 , z). Then, using multi-index for z instead of index for z 2 , in a same way as in Proposition 6.10, we may write for all m ∈ N g(z) = Using Schwarz's lemma, we obtain for all n ∈ N 0 |φ

[n] k (z k )| ≤ ζ n k |z k |. Hence, ( Gk • φ [n] )(z) λ n ≤ C k |φ [n] k (z k )| m k |λ| n ≤ C k |z k | m k ζ m k k |λ| n .
Therefore, the series ( Gk • φ [n] )(z)/λ n converges uniformly for |z k | < ε k .

Consider Fk the map defined by

Fk (z) = - 1 λ n≥0 Gk • φ [n] (z) λ n .
Then, Fk is holomorphic on B(0, ε k ), and we obtain Finally, we may find f ∈ Hol(B N ) such that f • φ -λf = g. It is defined as

f (z) = m 1 -1 ℓ 1 =0 • • • mp-1 ℓp=0 fℓ 1 ,••• ,ℓp (z) + p k=1 F k (z).
Hence, the operator C φ -λId is bijective, so λ ̸ ∈ σ(C φ ).

Conversely, let f 1 , • • • , f N : D → C holomorphic maps such that for all 1 ≤ k ≤ N ,

f k • φ k = φ ′ k (0)f k (1 ≤ k ≤ p), f k • φ k = β k f k (p + 1 ≤ k ≤ N ).
Such maps exist using the results in one variable ( [START_REF] Arendt | In Koenigs' footsteps: diagonalization of composition operators[END_REF][START_REF] Arendt | Spectral properties of weighted composition operators on Hol(D) induced by rotations[END_REF]), since φ has separable variables. For n 1 , • • • , n N ∈ N 0 , consider f : B N → C defined by f (z) = N k=1 f k (z k ) n k . Then,

(f • φ)(z) = p k=1 f k (φ k (z k )) n k × N k=p+1 f k (β k z k ) n k = p k=1 φ ′ k (0) n k f k (z k ) n k × N k=p+1 β n k k f k (z k ) n k = p k=1 φ ′ k (0) n k × N k=p+1 β n k k f (z).
Therefore, p k=1 φ ′ k (0) n k × N k=p+1 β n k k ∈ σ p (C φ ) ⊂ σ(C φ ). □ Remark 6.11. The proof of Theorem 6.9 says that if λ/ p k=1 φ ′ k (0) ℓ k ̸ ∈ σ(C ϕ ), with ϕ(z p+1 , • • • , z N ) = (β p+1 z N , • • • , β N z N ), then ( 9) is satisfied for some map f ℓ 1 ,••• ,ℓp . Hence, we may separate periodic and aperiodic ϕ. We obtain the following pattern: Example 6.12. Assume φ(z 1 , z 2 ) = (z 1 /2, iz 2 ) and ψ(z 1 , z 2 ) = (3z 1 /4, e i z 2 ).

Then, the spectra of C φ and C ψ are described below (see Figure 1).

( b )

 b Otherwise, there exists a function h ∈ Γ(φ) of the formh(z) = (z 1 , • • • , z r , 0, • • • , 0), with r ∈ {1, • • • N }.When φ does not have any fixed point on B N , the dilation coefficient of φ at ζ ∈ S N -1 is (1) δ(φ, ζ) := lim inf z-→ζ 1 -|φ(z)| 1 -|z| .

Theorem 4. 7 ,

 7 Hol(B N ) = W p ⊕ Hol p (B N ), with W p and Hol p (B N ) invariant by C φ . Hence, we write C φ = S 0 0 T , with S ∈ L(W p ) and T ∈ L(Hol p (B N )). r First, with respect to the basis (η ⃗ ȷ : |⃗ ȷ| ≤ p) of W ⃗ ȷ , S = diag(λ ⃗ ȷ : |⃗ ȷ| ≤ p).

Therefore, T

  is similar to e ⃗ ȷ C φ . Moreover, if C φ -βId is bijective, then β ̸ ∈ {e ⃗ ı : ⃗ ı ∈ N N 0 }, and for all g ∈ z ⃗ ȷ Hol(B N ), there exists f ∈ Hol(B N ) such that f • φ -βf = g. Denote respectively by (a ⃗ ı ) and (b ⃗ ı ) the Maclaurin coefficients of f and g. Hence, for all ⃗ ı ∈ N N 0 , a ⃗ ı = b ⃗ ı /(e ⃗ ı -β), so a ⃗ ı = 0 if and only if b ⃗ ı = 0. Therefore, f ∈ z ⃗ ȷ Hol(B N ), and T -βId is bijective. So σ(T ) ⊂ σ(C φ ).

1 (z 1 1 - 1 ℓ 1 =0m 2 - 1 ℓ 2 =0 κ ℓ 1 1

 111112121 )g ℓ (z) + z m 1 G 1 (z), with g 0 , • • • , g m-1 ∈ Hol(B N -1 ) and G 1 ∈ Hol(B N ). If we do the same decomposition for each g ℓ , with z = (z 2 , w), we obtain for all(m 1 , m 2 ) ∈ N 2 g(z) = m (z 1 )κ ℓ 2 2 (z 2 )g ℓ 1 ,ℓ 2 (z 3 , • • • , z N ) + z m 2 2 G 2 (z) + z m 1 1 G 1 (z), with g ℓ 1 ,ℓ 2 ∈ Hol(B N -2 ) and G 1 , G 2 ∈ Hol(B N ).Iterating this p times, and we get for all(m 1 , • • • , m p ) ∈ N p k (z k ) g ℓ 1 ,••• ,ℓp (z p+1 , • • • , z N ) + p k=1 z m k k G k (z), with g ℓ 1 ,••• ,ℓp ∈ Hol(B N -p ) and G k ∈ Hol(B N ). In the following, denote Gk (z) = z m k k G k (z). Now, let λ ∈ C * and ζ 1 , • • • , ζ p < 1 such that |λ| ̸ ∈ p k=1 |φ ′ k (0)| n k : n 1 , • • • , n p ∈ N 0 and |φ ′ k (0)| < ζ k , k = 1, • • • , p. Consider m 1 , • • • , m p ∈ N such that for all k = 1, • • • , p, ζ m k k < |λ|. r For all k ∈ {1, • • • , p}, there exists ε k > 0 such that if |z k | < ε k , then |φ k (z k )| ≤ ζ k |z k |.

(FkF k • φ -λ p k=1 F k = p k=1 z m k k G k , thanks to Lemma 4. 8 .

 k=18 Fk • φ)(z) -λ Fk (z) = n≥0 Gk • φ [n] (z) λ n -n≥0 Gk • φ [n+1] (z) λ n+1 = Gk • φ [0] = Gk . Hence, for |z| < ε = min(ε k : 1 ≤ k ≤ p), we get p k=1 k G k .The last thing to do is extending the map Fk to a mapF k ∈ Hol(B N ), satisfying p k=1 r Let ℓ 1 ∈ {0, • • • , m 1 -1}, • • • , ℓ p ∈ {0, • • • , m p -1}.We now look at the mapgℓ 1 ,••• ,ℓp = p k=1 κ ℓ k k (z k ) g ℓ 1 ,••• ,ℓp (z p+1 , • • • , z N ). Consider fℓ 1 ,••• ,ℓp ∈ Hol(D) of the form fℓ 1 ,••• ,ℓp = p k=1 κ ℓ k k (z k ) f ℓ 1 ,••• ,ℓp (z p+1 , • • • , z N ).Note that in this case, if we denote ϕ :(z p+1 , • • • , z N ) → (β p+1 z p+1 , • • • , β N z N ), ( fℓ 1 ,••• ,ℓp • φ) -λ fℓ 1 ,••• ,ℓp = ℓ k (f ℓ 1 ,••• ,ℓp • ϕ) -λf ℓ 1 ,••• ,ℓp . Remark that ϕ is bijective. Since 0 < |φ ′ k (0)| < 1 for all k and |λ| ̸ = p k=1 |φ ′ k (0)| ℓ k , by Proposition 6.2, there exists f ℓ 1 ,••• ,ℓp ∈ Hol(D) ℓ k (f ℓ 1 ,••• ,ℓp • ϕ) -λf ℓ 1 ,••• ,ℓp = g ℓ 1 ,••• ,ℓp .Hence, we obtain ( fℓ 1 ,••• ,ℓp • φ) -λ fℓ 1 ,••• ,ℓp = gℓ 1 ,••• ,ℓp .

Figure 1 .

 1 Figure 1. on the left: σ(C φ ), on the right: σ(C ψ )

Acknowledgments: This research is partly supported by the Bézout Labex, funded by ANR, reference ANR-10-LABX-58. The author is grateful to Isabelle Chalendar who brought this beautiful subject.

Proof. First, 1 ∈ σ p (C φ ) because 1•φ = 1, where 1 is the constant map equal to 1. Moreover, we may have 0 ∈ σ p (C φ ), for instance if there exists k ∈ {1, • • • , N } such that φ k is identically zero on B N . In this case, for f

Since µ ̸ = 1, we have f (0) = 0. r Suppose that for all ⃗ ı ≺ ⃗ ȷ, f (⃗ ı) (0) = 0. Then, by Lemma 4.4, we get the formula ( 8):

Then, similarly to Lemma 4.9, there exists ε > 0 such that for all h ∈ Hol 0 (B N ), the series (h • φ [n] )/µ n converges uniformly on B(0, ε). For g ∈ Hol(B N ), we define the functions h and f by

Hence, f ∈ Hol(B(0, ε)), h ∈ Hol 0 (B N ), and on B(0, ε),

Finally, by Lemma 4.8, there exists a function

□ We now give an example of φ for which we can find the spectrum and point spectrum.

Example 5.6. Let φ be defined by

Thus, 0 ∈ σ p (C φ ), and for 1 ≤ k ≤ p, λ k ∈ σ p (C φ ). Since the operator C φ is multiplicative, for all j 1 , • • • , j p ∈ N 0 , p k=1 λ j k k ∈ σ p (C φ ). Therefore, using Proposition 5.4,