F R Éd

Éric Bayart

COMPOSITION OPERATORS ON THE HARDY SPACE OF THE TRIDISC

Keywords: . 2020 Mathematics Subject Classification. 47B37 composition operators, polydisc, Hardy spaces, Carleson measures

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Introduction

Let U be a domain in C d , let X be a Banach space of holomorphic functions on U and let ϕ : U → U be holomorphic. The composition operator with symbol ϕ is defined by

C ϕ (f ) = f • ϕ, f ∈ X.
The first question to solve when studying composition operators is that of continuity: for which symbols ϕ do C ϕ induce a bounded composition operator on X? When X is the Hardy space or a weighted Bergman space of the unit disc D, the answer is easy: by the Littlewood subordination principle, this is always the case. In several variables, this statement does not hold, and there are simple examples of selfmaps of the euclidean ball (resp. of the polydisc) which do not induce a bounded composition operator on the corresponding Hardy space (see for instance [START_REF] Cima | On boundedness of composition operators on H 2 (B2)[END_REF][START_REF] Cowen | Composition operators on spaces of analytic functions[END_REF]). Therefore, it is interesting to find a characterization of the symbols ϕ inducing a bounded operator C ϕ , adding if necessary some regularity conditions on the symbol. For the ball, a very satisfying answer has been provided by Wogen in [START_REF] Wogen | The smooth mappings which preserve the Hardy space H 2 (Bn), Contributions to operator theory and its applications[END_REF]: he characterized the continuity of C ϕ acting on H 2 (B d ) or A 2 α (B d ) provided ϕ extends to a C 3 -function up to the boundary of B d . His characterization involves the first and second order derivatives of ϕ at any ξ ∈ ∂B d such that ϕ(ξ) ∈ ∂B d . As a consequence, he got that continuity on H 2 (B d ) is equivalent to continuity on A 2 α (B d ). In this paper we investigate the case of the polydisc, more precisely of the tridisc D 3 . It appears that characterizing the continuity of composition operators on H 2 (D d ) is a more difficult task than on H 2 (B d ). After a first attempt in [START_REF] Koo | Composition operators on the polydisc induced by smooth symbols[END_REF], where a necessary condition is given, the continuity problem was solved in [START_REF] Bayart | Composition operators on the polydisk induced by affine maps[END_REF] on A 2 (D 2 ) for symbols regular up to the boundary, and on H 2 (D d ) or on any A 2 α (D d ), α > -1, provided ϕ is an affine symbol. The case of the Hardy space requires a careful and specific argument since the method of [START_REF] Bayart | Composition operators on the polydisk induced by affine maps[END_REF] goes through Carleson measures and it is well known that Carleson measures on the Hardy space of the polydisc are difficult objects.

The recent paper [START_REF] Kosiński | Composition operators on the polydisc[END_REF] solves the continuity problem for H 2 (D 2 ) and for A 2 (D 3 ). Our main theorem (see Theorem 2.1 below) gives a characterization of the symbols ϕ : D 3 → D 3 , regular up to the boundary, such that C ϕ ∈ L(H 2 (D 3 )). As in Wogen's paper, this characterization involves conditions on the first and the second order derivatives of ϕ, whereas the characterization on H 2 (D 2 ) or A 2 (D 3 ) just involve conditions on the first derivative of the symbol. Theorem 2.1 needs some terminology that will be introduced in Section 2 to be stated but let us give a flavour of it. One of the main difficulty of working in the polydisc is that we have to play with its two boundaries: its topological boundary and its distinguished boundary. For ξ ∈ T 3 such that ϕ(ξ) / ∈ D 3 , we will discuss the properties of ϕ I at ξ, where I ⊂ {1, 2, 3} is maximal with ϕ I (ξ) ∈ T |I| , and ϕ I = (ϕ i 1 , . . . , ϕ ip ) if I = {i 1 , . . . , i p }. In particular, the case |I| = 2 will require a very careful analysis. Although our characterization is not very easy to state, we will provide several examples showing that it is effective: given a concrete ϕ ∈ O(D 3 , D 3 ) ∩ C 3 (D 3 ), where O(U, V) denotes the set of holomorphic functions between the two open sets U ⊂ C d and V ⊂ C d ′ , we will be able to decide whether C ϕ ∈ L(H 2 (D 3 )). Beyond continuity, the next question to handle is that of compactness. It seems that it was not considered at all for H 2 (D d ). We will first give a sufficient condition on ϕ ∈ O(D d , D d ) so that C ϕ induces a compact operator on H 2 (D d ), for any d ≥ 1. We will then use this condition to study in depth the cases d = 2 and d = 3.

Organization of the paper. Section 2 contains the terminology needed to state our main theorem, that is Theorem 2.1. Section 3 is devoted to some useful results which will be used throughout the paper. The proof of Theorem 2.1 is divided into Sections 4 to 6: Sections 4 and 5 are devoted to volume estimates of preimages of Carleson windows by ϕ at boundary points with order of contact equal to 2. These estimates are used to finish the proof of Theorem 2.1 in Section 6. Several examples are given in Section 7 to illustrate the different cases which might happen whereas Section 8 deals with compactness issues.

Notations. Throughout this paper, we shall use the following notations. For f, g : A → R, we shall write f ≲ g provided there exists C > 0 such that, for all x ∈ A, f (x) ≤ Cg(x). For ϕ : T d → A and θ ∈ R d , we shall write ϕ(θ) for ϕ(e iθ 1 , . . . , e iθ d ) and also sometimes e iθ = (e iθ 1 , . . . , e iθ d ). The Lebesgue measure on R d will be denoted by λ d whereas the normalized Lebesgue measure on T d will be denoted by σ d . The unit vector (1, . . . , 1) will be denoted by e. For a vector x ∈ X d and I = {i 1 , . . . , i p } ⊂ {1, . . . , d}, x I will mean (x i 1 , . . . , x ip ).

Terminology and statement of the main result

We start with a function φ ∈ O(D 3 , D) which admits a C 3 -smooth extension to D 3 . Assume that φ(e) = 1 and let us write

φ(z) = 1 + ∂φ ∂z 1 (e)(z 1 -1) + ∂φ ∂z 2 (z 2 -1) + ∂φ ∂z 3 (e)(z 3 -1) + j+k+l=2 a j,k,l (z 1 -1) j (z 2 -1) k (z 3 -1) l + O(|z 1 -1| 3 + |z 2 -1| 3 + |z 3 -1| 3 ).
By the Julia-Caratheodory theorem (see Lemma 3.8 below), ∂φ ∂z k (e) ≥ 0 for k = 1, 2, 3 and these derivatives cannot simultaneously be equal to 0. Hence, if we look at φ(θ), we may write

Imφ(θ) = L(θ) + o(|θ 1 | + |θ 2 | + |θ 3 |) Reφ(θ) = 1 -Q(θ) + O |θ 1 | 3 + |θ 2 | 3 + |θ 3 | 3
with L a nonzero linear form and Q a quadratic form. We now turn to a map ϕ ∈ O(D 3 , D 3 ) such that ϕ extends to a C 3 -smooth function on D 3 .

In order to prove or disprove continuity of C ϕ on H 2 (D 3 ), it will be important to deeply study the behaviour of ϕ near the points ξ ∈ T 3 such that ϕ(ξ) / ∈ D 3 . This means that there exists I ⊂ {1, 2, 3}, I ̸ = ∅ such that ϕ I (ξ) ∈ T |I| . For i ∈ I, the map ϕ i can be expanded as above, giving rise to a linear map L i and a quadratic form Q i . The mutual properties (and dependencies) of the maps L i , Q i , will be the key of our work. The most difficult case will be when there exists ξ ∈ T 3 , I ⊂ {1, 2} with |I| = 2, ϕ I (ξ) ∈ T 2 , ∇ϕ i 1 and ∇ϕ i 2 are linearly dependent. Without loss of generality, we may assume ξ = e, I = {1, 2} and ϕ I (ξ) = (1, 1). We may write for j = 1, 2,

Imϕ j (θ) = κ j L(θ) + o(|θ 1 | + |θ 2 | + |θ 3 |) (1) Reϕ j (θ) = 1 -Q j (θ) + O |θ 1 | 3 + |θ 2 | 3 + |θ 3 | 3 (2)
where L is a nonzero linear form, κ j are nonzero real numbers and Q j are quadratic forms. Since Reϕ j ≤ 1, the signature of each Q j should be (n j , 0) with n j ≥ 0. Therefore there exists m ∈ {0, . . . , 3} such that the signature of Q 1 + Q 2 is equal to (m, 0). We define s(ϕ, I, ξ) = m.

Each Q j may be written n j k=1 L 2 j,k . This decomposition is not unique. However, it is easy to check that span(L j,k :

j = 1, 2, k = 1, . . . , n j ) ⊥ = ker(Q 1 + Q 2 )
so that dim(span(L j,k ) ⊥ ) = 3 -m. Hence we may pick (L 1 , . . . , L m ) a basis of span(L j,k ). It should be observed that L belongs to span(L 1 , . . . , L m ). Otherwise, we could find θ = (θ 1 , θ 2 , θ 3 ) such that L(θ) ̸ = 0 and L j,k (θ) = 0 for j = 1, 2 and k = 1, . . . , n j . But this would contradict that

|Imϕ j (εθ)| 2 + |Reϕ j (εθ)| 2 ≤ 1
for small values of ε. In particular we get s(ϕ, I, ξ) = m ≥ 1. Now, we complete the family (L 1 , . . . , L m ) into (L 1 , . . . , L 3 ) so that this last family is a basis of (R 3 ) * . We then get

Imϕ j (θ) = κ j     L(θ) + 1≤k≤m k≤l≤3 a j,k,l L k L l + m+1≤k≤l≤3 a j,k,l L k L l     +O |θ 1 | 3 + |θ 2 | 3 + |θ 3 | 3 .
We then consider the quadratic form

R = κ 2   m+1≤k≤l≤3 a 1,k,l L k L l   -κ 1   m+1≤k≤l≤3 a 2,k,l L k L l  
and we define r(ϕ, I, ξ) = signature(R).

This last quantity does not depend on the way we define the linear forms L i . Indeed, let us assume that m = 1 which is the most difficult case. If we replace L i by L ′ i , we may write for k = 1, 2, 3

L k = β 1,k L ′ 1 + β 2,k L ′ 2 + β 3,k L ′ 3
where the matrix

B = β 2,2 β 2,3 β 3,2 β 33 is invertible since span(L 1 ) = span(L ′ 1 ). Writing R = j,k c j,k L j L k with a symmetric matrix C = (c j,k ) and the corresponding R ′ = j,k c ′ j,k L ′ j L ′ k with a symmetric matrix C ′ = (c ′ j,k
), then C ′ = BCB T , which proves the claim. The integers s(ϕ, I, ξ) and r(ϕ, I, ξ) we have just introduced will be the keys to understand the behaviour of ϕ at a boundary point of order 2, when the first derivatives are linearly dependent.

Theorem 2.1. Let ϕ : D 3 → D 3 be holomorphic such that ϕ extends to a C 3 -smooth function on D 3 . Then C ϕ is bounded on H 2 (D 3 ) if and only if ϕ satisfies the following properties:

(1) for any ξ ∈ T 3 such that ϕ(ξ) ∈ T 3 , dϕ(ξ) is invertible;

(2) for any I ⊂ {1, 2, 3}, I = {i 1 , i 2 }, such that ϕ I (ξ) ∈ T 2 , either (a) ∇ϕ i 1 (ξ) and ∇ϕ i 2 (ξ) are linearly independent (b) or ∇ϕ i 1 (ξ) and ∇ϕ i 2 (ξ) are linearly dependent, s(ϕ, I, ξ) = 3 (c) or ∇ϕ i 1 (ξ) and ∇ϕ i 2 (ξ) are linearly dependent, s(ϕ, I, ξ) = 2 and r(ϕ, I, ξ) = (1, 0) or (0, 1) (d) or ∇ϕ i 1 (ξ) and ∇ϕ i 2 (ξ) are linearly dependent, s(ϕ, I, ξ) = 1 and r(ϕ, I, ξ) =

(2, 0) or (0, 2).

Observe that conditions (b), (c) or (d) imply that, for any j ∈ {1, 2, 3} and any k ∈ {1, 2}, ∂ϕ i k ∂z j (ξ) ̸ = 0. Otherwise ϕ i 1 and φ 2 would not depend on the same variable. But setting m = s(ϕ, I, ξ) and (a, b) = r(ϕ, I, ξ), we would have m + a + b ≤ 2, which is not the case here. Therefore, as a corollary of Theorem 2.1 and of [Kos23, Theorem 10], we get:

Corollary 2.2. Let ϕ ∈ O(D 3 , D 3 ) ∩ C 3 (D 3 ). If C ϕ is continuous on H 2 (D 3 ), then it is continuous on A 2 (D 3 ).
The converse is false, a counterexample being given by ϕ(z) = (z 1 z 2 , z 3 , z 1 z 2 z 3 , 0) (see [START_REF] Kosiński | Composition operators on the polydisc[END_REF]).

Preliminaries

) ∈ D d : |z k -η k | ≤ δ k , 1 ≤ k ≤ d .
We shall use sometimes Carleson windows instead of Carleson boxes. For η = e iθ ∈ T d and δ ∈ (0, 2] d , the Carlson window at η with radius δ is

W (η, δ) = z = (r 1 e it 1 , . . . , r d e it d ) ∈ D d : 1 -δ k ≤ r k ≤ 1 and |θ k -t k | ≤ δ k , 1 ≤ k ≤ d .
Since there exists c > 0 such that, for all η ∈ T d and all δ ∈ (0, 2] d ,

S(η, δ) ⊂ W (η, δ) ⊂ S(η, cδ),
in all results that we will stay, we may replace S(η, δ) by W (η, δ), and conversely. We use dA to denote the normalized area measure on the unit disk D and for β > -1, we write

dA β (z) = (β + 1)(1 -|z| 2 ) β dA(z).
We also let

dV β (z) = dA β (z 1 ) . . . dA β (z d ),
the product measure on D d .

Lemma 3.1. (see [START_REF] Bayart | Composition operators on the polydisk induced by affine maps[END_REF]) Let ϕ : D d → D d be holomorphic. Assume that there exist C > 0 and ε > 0 such that, for all δ ∈ (0, ε] d , for all η ∈ T d , for all β ∈ (-1, 0],

V β ϕ -1 (S(η, δ)) ≤ Cδ 2+β 1 • • • δ 2+β d .
Then C ϕ is continuous on H 2 (D d ).

Regarding necessary conditions, we can use the following result, which is the easy part of [START_REF] Chang | Carleson measure on the bi-disc[END_REF].

Lemma 3.2. Let ϕ : D d → D d be holomorphic and belonging to C(D d ) Assume that C ϕ is continuous on H 2 (D d ).
Then there exists C > 0 such that, for all η ∈ T d , for all δ ∈]0, 2] d ,

(3)

σ d e iθ ∈ T d : ϕ(θ) ∈ S(η, δ) ≤ Cδ 1 • • • δ d .

Volume estimates.

As pointed out by Lemma 3.1 and 3.2, we shall need volume estimates. The forthcoming lemmas summarize what we need. We start by minorizing the volume of some subsets of R 2 .

Lemma 3.3. There exists C > 0 such that, for all δ ∈ (0, 1),

λ 2 (x, y) ∈ R 2 : |x| ≤ δ 1/3 , |y| ≤ δ 1/2 , |xy| ≤ δ ≥ -Cδ log(δ).
Proof. Let us consider y ∈ [δ 2/3 , δ 1/2 ]. Then any x ∈ 0, δ y satisfies |xy| ≤ δ and |x| ≤ δ • δ -2/3 = δ 1/3 . Therefore, the measure of the set we are looking for is greater than

δ 1/2 δ 2/3 δ y dy = - 1 6 δ log(δ). □ Lemma 3.4. Let a, b > 0.
There exists C > 0 such that, for all δ ∈ (0, 1),

λ 2 (x, y) ∈ -δ 1/3 , δ 1/3 2 : |ax 2 -by 2 | ≤ δ ≥ -Cδ log(δ).
Proof. We may assume a = b = 1. Setting x ′ = x -y, y ′ = x + y, we get

x 2 -y 2 ≤ δ ⇐⇒ |x ′ y ′ | ≤ δ.
Hence the result comes immediately from Lemma 3.3. □

We now majorize the measure of some subsets of R 2 .

Lemma 3.5. For all δ > 0, for all a ∈ R,

λ 2 (x, y) ∈ R 2 : |x 2 + y 2 -a| < δ ≤ 2πδ.
Proof. If a < -δ, the set is empty. If a ∈ [-δ, δ], the set is the disc of center 0 and radius

√ δ + a, its surface is π(δ + a) ≤ 2πδ. If a ≥ δ, the set is the corona a -δ ≤ x 2 + y 2 ≤ a + δ; its surface is π(a + δ) -π(a -δ) = 2πδ. □
Fubini's lemma will be useful to control the V β volume of some subsets of D d .

Lemma 3.6. Let E be a measurable subset of R d , let δ ∈ (0, 1] 3 . There exists C > 0 such that, for all β ∈ (-1, 0],

V β z = (1 -ρ k )e iθ k d k=1 ∈ D d : θ ∈ E, 0 ≤ ρ k ≤ δ k ≤ Cδ 1+β 1 • • • δ 1+β d λ d (E).
Proof. This follows from Fubini's theorem and from the inequality

1 1-δ (β + 1)r(1 -r 2 ) β dr ≤ Cδ 1+β
for some constant C > 0 which does not depend on δ ∈ [0, 1] and on β ∈ (-1, 0]. □ 3.3. Geometric lemmas. We will also need two geometric lemmas. The first one is the parametrized Morse lemma (see for instance [START_REF] Bruce | Curves and Singularities: a geometrical introduction to singularity theory[END_REF]Sec. 4.44]).

Lemma 3.7. Let f : R d × R n-d → R be a C 3 -function such that f (0) = 0, d y f (0) = 0 and d 2 y f (0) is nonsingular. There exists a neighbourhood U of 0 in R n , a diffeomorphism Γ : U → R n , (x, y) → (x, γ(x, y)) and a C 1 -map h : R d → R such that, for any (x, y) ∈ U, f (x, y) = n-d j=1 ±γ j (x, y) 2 + h(x).
The second one is a kind of Julia-Caratheodory theorem in several variables (see in particular [Bay11, Lemma 2.7]).

Lemma 3.8. Let φ : D 3 → D which is holomorphic in D 3 and C 1 on D 3 . Assume that φ(e) = 1.
Then for all k = 1, 2, 3, ∂φ ∂z k (e) ≥ 0. Moreover, if ∂φ ∂z k (e) = 0, then φ does not depend on z k .

3.4.

Positive definite quadratic forms. We will also need a simple lemma on quadratic forms.

Lemma 3.9. Let Q be a positive definite quadratic form on R p , let R be a quadratic form on R p × R n such that its restriction to {0} × R n is positive definite. Then there exists κ > 0 such that

(x, y) → κQ(x) + R(x, y) is positive definite.
Proof. We may assume

Q(x) = p k=1 x 2 k R(x, y) = n l=1 y 2 l + 1≤k,k ′ ≤p a k,k ′ x k x k ′ + 1≤k≤p 1≤l≤n b k,l x k y l . Let M ≥ sup{|a k,k ′ |, |b k,l |}. Let also δ > 0. Then one has k,l |b k,l x k y l | ≤ k,l M |x k | δ (δ|y l |) ≤ M 2 k,l x 2 k δ 2 + δ 2 y 2 l ≤ M n 2δ 2 k x 2 k + M pδ 2 2 l y 2 l .
A similar argument yields

k,k ′ |a k,k ′ x k x k ′ | ≤ M p k x 2 k .
We then adjust δ > 0 so that M pδ 2 /2 < 1 and we choose

κ > M n 2δ 2 + M p.
For any (x, y) ∈ R p × R n \{(0, 0)}, κQ(x) + R(x, y) > 0 so that the quadratic form is positive definite. □ 3.5. Self-maps of the unit disc. Our next lemmas will help us to provide examples of polynomials which are self maps of the unit disc.

Lemma 3.10. There exists ε 0 > 0 such that, for all ε ∈ [-ε 0 , ε 0 ], for all θ ∈ R,

1 + e iθ 2 + iε(e iθ -1) 2 ≤ 1 with equality if and only if θ ≡ 0 [2π]. Proof. Let g ε (θ) = 1 + e iθ 2 + iε(e iθ -1) 2 so that g ε (θ) = 1 + iθ 2 - θ 2 4 + iθ 3 g 1 (θ) + θ 4 g 2 (θ) + iεθ 2 g 3 (θ) + εθ 3 g 4 (θ)
where g 1 , . . . , g 4 are bounded real-valued functions. Hence,

|g ε (θ)| 2 = 1 - θ 2 4 + θ 3 ε 2 g 5 (θ) + εg 6 (θ) + g 7 (θ )
with g 5 , g 6 , g 7 bounded real-valued functions. Therefore there exists θ 0 > 0 such that, for all |θ| ≤ θ 0 , for all |ε| < 1,

|g ε (θ)| ≤ 1 with equality on [-θ 0 , θ 0 ] if and only if θ = 0. Let us now consider c ∈ (0, 1) such that ∀θ ∈ [-π, π]\[-θ 0 , θ 0 ], 1 + e iθ 2 ≤ c.
Then the choice ε 0 = 1-c 8 fulfills the requirements.

□ Lemma 3.11. For all z ∈ D, |3 + 6z -z 2 | ≤ 8 with equality if and only if z = 1. Proof. Let z = e iθ ∈ T. Then |3 + 6z -z 2 | 2 ≤ 64 ⇐⇒ (3 + 6 cos(θ) -cos(2θ)) 2 + (6 sin(θ) -sin(2θ)) 2 ≤ 64.
Developing, reducing and using cos(2θ

) = 2 cos 2 (θ) -1, sin(2θ) = 2 sin(θ) cos(θ), sin 2 (θ) + cos 2 (θ) = 1, we easily get that this is equivalent to cos(θ)(2 -cos(θ)) ≤ 1
and this last inequality is verified for all θ ∈ R with equality if and only if θ ≡ 0 [2π]. □ Lemma 3.12. There exists

ε 1 > 0 such that, for all ε ∈ [-ε 1 , ε 1 ], for all z ∈ D, 3 + 6z -z 2 8 + 2iε(z -1) 2 -iε(z -1) 3 ≤ 1
with equality if and only if z = 1.

Proof. We set

G(θ) = 3 + 6e iθ -e 2iθ 8 + 2iε(e iθ -1) 2 -iε(e iθ -1) 3 2
and we expand it around 0. We find

G(θ) = 1 + 4ε 2 - 3 64 θ 4 + θ 5 (G 1 (θ) + εG 2 (θ) + ε 2 G 3 (θ))
with G 1 , G 2 , G 3 bounded real-valued functions. Therefore there exists θ 0 > 0 such that, for all |θ| ≤ θ 0 , for all

ε 2 < 3/512, |G(θ)| ≤ 1 with equality on [-θ 0 , θ 0 ] if and only if θ = 0. Let us consider c ∈ (0, 1) such that ∀θ ∈ [-π, π]\[-θ 0 , θ 0 ], 3 + 6e iθ -e 2iθ 8 ≤ c.
Then the choice ε 1 = 1-c 32 < 3/512 fulfills the requirements. □

Discontinuity when the symbol has a boundary point of order of contact 2

The key point to prove Theorem 2.1 is the study of the case of holomorphic maps admitting points with order of contact equal to 2. We begin by showing that the conditions given in that theorem are necessary.

Theorem 4.1. Let ϕ : D 3 → D 3 be holomorphic on D 3 and C 3 -smooth on D 3 . Assume that there exist

I ⊂ {1, 2, 3} with |I| = 2, I = {i 1 , i 2 } and ξ ∈ T 3 such that ϕ I (ξ) ∈ T 2 .
Assume that ∇ϕ i 1 (ξ) and ∇ϕ i 2 (ξ) are linearly dependent, and that

• either s(ϕ, I, ξ) = 1 and r(ϕ, I, ξ) ̸ = (2, 0) or (0, 2)

• or s(ϕ, I, ξ) = 2 and r(ϕ, I, ξ) ̸ = (1, 0) or (0, 1).

Then C ϕ is not continuous on H 2 (D 3 ).
Proof. We may always assume that ξ = e, that I = {1, 2} and that ϕ I (e) = (1, 1). We shall prove that condition (3) breaks down. Because it only depends on what happens on the unit polycircle, we are allowed to make a linear change of variables on the coordinates (θ 1 , θ 2 , θ 3 ); this will not changed the volume estimates. Therefore, in what follows, we will always assume that the linear map appearing in (1) is L(θ) = θ 1 . We shall split the proof into several cases and we will intensively use the notations of Section 2.

Case 1. s(ϕ, I, ξ) = 1 and r(ϕ, I, ξ) = (0, 0). Since L(θ) = θ 1 and s(ϕ, I, ξ) = 1, we may choose L 2 = θ 2 and L 3 = θ 3 and write, for j = 1, 2, keeping only the terms of order 2 which can be expressed with θ 2 and θ 3 ,

Imϕ j (θ) = κ j (θ 1 + a j θ 2 2 + b j θ 2 θ 3 + c j θ 2 3 ) + O(θ 2 1 + |θ 1 θ 2 | + |θ 1 θ 3 | + |θ 2 | 3 + |θ 3 | 3 ). Since r(ϕ, I, ξ) = (0, 0), we get a 1 = a 2 =: a, b 1 = b 2 =: b, c 1 = c 2 =: c. Since s(ϕ, I, ξ) = 1, we may also write Reϕ j (θ) = 1 -d j θ 2 1 + O(|θ 1 | 3 + |θ 2 | 3 + |θ 3 | 3
) with d j > 0, for j = 1, 2. Let δ > 0 and set δ = (δ, δ, 2). We choose any (θ 2 , θ 3 ) ∈ [-δ 1/3 , δ 1/3 ] and then, any

θ 1 ∈ [-aθ 2 2 -bθ 2 θ 3 -cθ 2 3 -δ, -aθ 2 2 -bθ 2 θ 3 -cθ 2 3 + δ]. The volume of these θ = (θ 1 , θ 2 , θ 3 ) is greater than δ 5/3 ≫ δ 2 . Moreover, for these θ, observing that |θ 1 | ≲ θ 2/3 , it is easy to check that |Imϕ j (θ)| ≲ |θ 1 + aθ 2 2 + bθ 2 θ 3 + cθ 2 3 | + O(δ 4/3 + δ) ≲ δ and |Reϕ j (θ) -1| ≲ δ.
Therefore,

σ 3 e iθ ∈ T 3 : ϕ(θ) ∈ S(η, δ) ≫ δ 2
which prevents C ϕ to be continuous.

Case 2. s(ϕ, I, ξ) = 1 and r(ϕ, I, ξ) = (1, 0) or (0, 1). Keeping the notations of Section 2 and doing a linear change of variables, we may assume that

R(θ 2 , θ 3 ) = * θ 2 3 .
This means that we may write, for j = 1, 2,

Imϕ j (θ) = κ j (θ 1 + aθ 2 2 + bθ 2 θ 3 + c j θ 2 3 ) + O θ 2 1 + |θ 1 θ 2 | + |θ 1 θ 3 | + |θ 2 | 3 + |θ 3 | 3 and we always have Reϕ j (θ) = 1 -d j θ 2 1 + O |θ 1 | 3 + |θ 2 | 3 + |θ 3 | 3 .
We first assume that a ̸ = 0 and, without loss of generality, that a = 1. Writing

θ 2 2 + bθ 2 θ 3 = θ 2 + b 2 θ 3 2 - b 2 4 θ 2 3 and setting θ ′ 2 = θ 2 + b 2 θ 3 , θ ′ 3 = θ 3
, we may also assume that b = 0. Let δ > 0 and consider any (θ

1 , θ 3 ) ∈ [-2δ 2/3 , -δ 2/3 ] × [-δ 1/2 , δ 1/2 ].
For these couples (θ 1 , θ 3 ), consider any θ 2 in the interval

I θ 1 = -θ 1 -δ, -θ 1 + δ .
The length of I θ 1 satisfies

|I θ 1 | ≳ δ δ 1/3 = δ 2/3 so that σ 3 e iθ ∈ T 3 : (θ 1 , θ 3 ) ∈ [-2δ 2/3 , δ 2/3 ] × [-δ 1/2 , δ 1/2 ], θ 2 ∈ I θ 1 ≳ δ 4 3 + 1 2 ≫ δ 2 .
Moreover, for these θ, observing that |θ 2 | ≲ δ 1/3 , it is easy to check that, for j = 1, 2,

|Imϕ j (θ)| ≲ θ 1 + θ 2 2 + O |θ 1 | 2 + |θ 1 θ 2 | + |θ 1 θ 3 | + |θ 2 | 3 + |θ 3 | 2 ≲ δ
and |Reϕ j (θ) -1| ≲ δ. Therefore C ϕ cannot be continuous. We now handle the case a = 0. This means that, for j = 1, 2,

Imϕ j (θ) = κ j θ 1 + bθ 2 θ 3 + c j θ 2 3 + O θ 2 1 + |θ 1 θ 2 | + |θ 1 θ 3 | + |θ 2 | 3 + |θ 3 | 3 Reϕ j (θ) = 1 -d j θ 2 1 + O |θ 1 | 3 + |θ 2 | 3 + |θ 3 | 3 .
For δ > 0, we set

K δ = (θ 2 , θ 3 ) : |θ 2 | ≤ δ 1/3 , |θ 3 | ≤ δ 1/2 , |θ 2 θ 3 | ≤ δ which satisfies by Lemma 3.3 λ 2 (K δ ) ≳ -δ log(δ). Then any θ ∈ [-δ, δ] × K δ satisfies |Imϕ j (θ)| ≲ δ and |Reϕ j (θ) -1| ≲ δ. Since σ 3 e iθ : θ ∈ [-δ, δ] × K δ ≳ -δ 2 log(δ), C ϕ cannot be continuous.
Case 3. s(ϕ, I, ξ) = 1 and r(ϕ, I, ξ) = (1, 1). Doing a linear change of variables, we may and shall assume that

R(θ 2 , θ 3 ) = κ 1 κ 2 (θ 2 2 -θ 2 3
). This means that we may write, for j = 1, 2,

Imϕ j (θ) = κ j θ 1 + a j θ 2 2 + bθ 2 θ 3 + c j θ 2 3 + O θ 2 1 + |θ 1 θ 2 | + |θ 1 θ 3 | + |θ 2 | 3 + |θ 3 | 3 with a 1 = a 2 + 1, c 1 = c 2 -1. For δ ∈ (0, 1), we set E(δ) = (θ 2 , θ 3 ) ∈ -δ 1/3 , δ 1/3 , θ 2 2 -θ 2 3 ≤ δ .
For (θ 2 , θ 3 ) ∈ E(δ), let

I θ 2 ,θ 3 (δ) = θ 1 : θ 1 + a 1 θ 2 2 + bθ 2 θ 3 + c 1 θ 3 2 ≤ δ .
Then Lemma 3.4 implies that

λ 2 ({θ : (θ 2 , θ 3 ) ∈ E(δ), θ 1 ∈ I θ 2 ,θ 3 (δ)}) ≳ -δ 2 log(δ).
On the other hand, provided (θ 2 , θ 3 ) ∈ E(δ) and θ 1 ∈ I θ 2 ,θ 3 (δ), we easily verify that |θ 1 | ≲ δ 2/3 which in turn implies

|Imϕ 1 (θ)| ≲ δ + O |θ 1 θ 2 | + |θ 1 θ 3 | + |θ 1 | 2 + |θ 2 | 3 + |θ 3 | 3 ≲ δ, |Imϕ 2 (θ)| ≲ δ + θ 2 2 -θ 2 3 + O |θ 1 θ 2 | + |θ 1 θ 3 | + |θ 1 | 2 + |θ 2 | 3 + |θ 3 | 3 ≲ δ, |Reϕ j (θ) -1| ≲ δ, j = 1, 2.
Therefore, σ 3 e iθ ∈ T 3 : ϕ(θ) ∈ S(η, δ) ≫ δ 2 for δ = (δ, δ, δ), which prevents C ϕ to be continuous. Case 4. s(ϕ, I, ξ) = 2 and r(ϕ, I, ξ) = (0, 0). Using the notations of Section 2 and doing a linear change of variables, we may assume that L(θ) = θ 1 and that the linear forms L j,k , 1 ≤ j, k ≤ 2, are linear combinations of θ → θ 1 and θ → θ 2 . Therefore, we may write

Imϕ j (θ) = κ j θ 1 + a j θ 1 θ 3 + b j θ 2 θ 3 + cθ 2 3 + O |θ 1 | 2 + |θ 2 | 2 + |θ 3 | 3 Reϕ j (θ) = 1 + O |θ 1 | 2 + |θ 2 | 2 + |θ 3 | 3 .
Let δ > 0 and

E(δ) = (θ 2 , θ 3 ) : |θ 2 | ≤ δ 1/2 , |θ 3 | ≤ δ 1/3 , |θ 2 θ 3 | ≤ δ .
By Lemma 3.3, λ 2 (E(δ)) ≳ -δ log(δ). For θ 3 ∈ R, we set

I θ 3 (δ) = θ 1 : θ 1 + cθ 2 3 ≤ δ so that σ 3 e iθ : (θ 2 , θ 3 ) ∈ E(δ), θ 1 ∈ I θ 3 (δ) ≳ -δ 2 log(δ).
Moreover, for θ such that (θ 2 , θ 3 ) ∈ E(δ) and

θ 1 ∈ I θ 3 (δ), one has |θ 1 | ≲ δ 2/3 , |θ 2 | ≤ δ 1/2
and |δ 3 | ≤ δ 1/3 so that for j = 1, 2, |Reϕ j (θ) -1| ≲ δ and |Imϕ j (θ)| ≲ δ. Again, C ϕ cannot be continuous. □

Volume estimates when the symbol has a boundary point of order 2

The aim of this section is to prove the following result, which is the main step towards the proof of the positive part of Theorem 2.1.

Theorem 5.1. Let ϕ ∈ O(D 3 , D 3 ) ∩ C 3 (D 3 ). Let ξ ∈ T 3 , I ⊂ {1, 2} with |I| = 2, I = {i 1 , i 2 }, ϕ I (ξ) ∈ T 2
and ∇ϕ i 1 (ξ), ∇ϕ i 2 (ξ) are linearly dependent. There exist a constant C > 0 and a neighbourhood U of ξ such that, for all β ∈ (-1, 0], for all η ∈ T 2 , for all δ ∈ (0, +∞) 2 ,

V β (z 1 , z 2 , z 3 ) ∈ U ∩ D 3 : |ϕ i j (z) -η j | ≤ δ j , j = 1, 2 ≤ Cδ 2+β 1 δ 2+β 2
in the following cases:

• s(ϕ, I, ξ) = 3;

• s(ϕ, I, ξ) = 2 and r(ϕ, I, ξ) ∈ {(1, 0); (0, 1)};

• s(ϕ, I, ξ) = 1 and r(ϕ, I, ξ) ∈ {(2, 0); (0, 2)}.

Proof. To fix the notations, we assume that I = {1, 2}, that ξ = e = (1, 1, 1) and that ϕ I (e) = (1, 1). Without loss of generality we assume δ 1 ≤ δ 2 . We start dealing with the case s(ϕ, I, ξ) = 3 and we first prove an intermediate result:

Fact. There exist C > 0 and a neighbourhood O of 0 in R 3 such that, for all δ ∈ (0, +∞) 2 , for all η ∈ T 2 , (4)

λ 3 ({θ ∈ O : |ϕ j (θ) -η j | ≤ δ j , j = 1, 2}) ≤ Cδ 1 δ 2 .
We fix η ∈ T 2 and δ ∈ (0, +∞) 2 . We know that, for j = 1, 2,

Imϕ j (θ) = κ j (L(θ) + G j (θ))
with G j (0) = 0, ∇G j (0) = 0. Doing a linear change of variables, which does not affect neither the desired inequality (4) nor the signature of Q, we may assume that L(θ) = θ 1 . We set U (θ) = (θ 1 + G 1 (θ), θ 2 , θ 3 ). There exist two neighbourhoods O 1 and O 2 of 0 and

C 1 > 0 such that U is a diffeomorphism from O 1 onto O 2 and | det(dU )| ≤ C 1 on O 1 | det(dU -1 )| ≤ C 1 on O 2 .

We then write

Reϕ

j (θ) = 1 -Q j (θ) + H j (θ) with H j (θ) = O |θ 1 | 3 + |θ 2 | 3 + |θ 3 | 3 . Because dU (0) = I 3 , for u ∈ O 2 , Reϕ j • U -1 (u) = 1 -Q j (u) + K j (u) with K j (u) = O |u 1 | 3 + |u 2 | 3 + |u 3 | 3 . Let us now define E(δ) = θ ∈ R 3 : |ϕ j (θ) -η j | ≤ δ j , j = 1, 2
and let us write

η j = x j + iy j . Provided u ∈ U (E(δ) ∩ O 1 ), | -Q 1 (u) -Q 2 (u) + K 1 (u) + K 2 (u) + 2 -x 1 -x 2 | ≤ |Reϕ 1 • U -1 (u) -x 1 | + |Reϕ 2 • U -1 (u) -x 2 | ≤ 2δ 2 . (5) We set F (u) = -Q 1 (u) -Q 2 (u) + K 1 (u) + K 2 (u). Since the signature of Q 1 + Q 2 is (3, 0), the matrix ∂ 2 F ∂u i ∂u k (0) 2≤i,j≤3
is negative definite. Hence we may apply the parametrized Morse lemma (Lemma 3.7): there exist two neighbourhoods O ′ 2 ⊂ O 2 and O 3 of 0, a constant C 2 > 0 and a diffeomorphism

V : O ′ 2 → O 3 , V (u 1 , u 2 , u 3 ) = (u 1 , V 2 (u 1 , u 2 , u 3 ), V 3 (u 1 , u 2 , u 3 )) and a C 1 -map h : R → R such that, for all v ∈ O 3 , F • V -1 (v) = -v 2 2 -v 2 3 + h(v 1 ) and | det(dV )| ≤ C 2 on O ′ 2 | det(dV -1 )| ≤ C 2 on O 3 . We finally set O = U -1 (O ′ 2 ) ⊂ O 1 . Therefore, provided v ∈ V • U (E(δ) ∩ O), we infer from |Imϕ 1 -η 1 | ≤ δ 1 on E(δ) and from (5) that |v 1 -y 1 | ≤ δ 1 /κ 1 | -v 2 2 -v 2 3 + h(v 1 ) + 2 -x 1 -x 2 | ≤ 2δ 2
It then follows from Fubini's theorem and Lemma 3.5 that

λ 3 V • U (E(δ) ∩ O) ≲ δ 1 δ 2
where the involved constant does not depend on η 1 , η 2 and δ. We then get the fact by the change of variables formula.

We now prove the statement of the theorem (still in the case s(ϕ, I, ξ) = 3). We write each z k = (1 -ρ k )e iθ k . A Taylor expansion of ϕ j near e reveals that

|ϕ j (z)| 2 = 1 + 3 k=1 ρ k F j,k (ρ, θ) + G j (θ)
where F j,k (0, 0) = -2 ∂ϕ j ∂z k (e). Observe that ∂ϕ j ∂z k (e) ̸ = 0, otherwise we would get ∂ϕ 1 ∂z k (e) = ∂ϕ 2 ∂z j (e) = 0 (recall that ∇ϕ 1 (e) and ∇ϕ 2 (e) are linearly dependent). This would imply that neither ϕ 1 nor ϕ 2 would depend on z k (see Lemma 3.8), which contradicts s(ϕ, I, ξ) = 3. Choosing ρ k = 0, we get G j (θ) ≤ 0 for all θ. On the other hand, provided

|ϕ 1 (z)-η 1 | < δ 1 , which implies |ϕ 1 (z)| 2 ≥ 1 -2δ 1 , we necessarily get 3 k=1 ρ k F 1,k (ρ, θ) ≥ -2δ 1 .
Since F 1,k (0, 0) < 0, we get the existence of a neighbourhood U 1 of e in D 3 such that, for

all z ∈ U 1 ∩ ϕ -1 I (S(η, δ)), ρ k ≲ δ 1 , k = 1, 2, 3
where the implied constant does not depend on δ. Shrinking U 1 if necessary, we may assume that it is convex. Now, observe that for

z ∈ U 1 ∩ ϕ -1 I (S(η, δ)), writing always z k = (1 -ρ k )e iθ k , for j = 1, 2, |ϕ j (z) -ϕ j (θ)| ≤ sup U 1 ∥∇ϕ j ∥ • max k ∥z k -e iθ k ∥ ≲ δ 1 ≤ δ j .
Hence, there exists D > 0 such that, for all z ∈ U 1 ∩ ϕ -1 I (S(η, δ)), we have e iθ ∈ ϕ -1 I (S(η, Dδ)). Let O be given by the fact (for Dδ) and let us set

U = U 1 ∩ (1 -ρ k )e ikθ 3 k=1 : θ ∈ O, 0 ≤ ρ k ≤ δ 1 .
We get from Lemma 3.6 the existence of C > 0 such that, for all β ∈ (-1, 0], for all η ∈ T 2 , for all δ ∈ (0, 2] 2 ,

V β U ∩ ϕ -1 I (S(η, δ)) ≤ Cδ 3(1+β) 1 δ 1 δ 2 ≤ Cδ 2+β 1 δ 2+β 2 .
We turn to the case s(ϕ, I, ξ) = 2 and r(ϕ, I, ξ) = (1, 0) or (0, 1). Arguing as in the case s(ϕ, I, ξ) = 3, we just need to prove (4). Indeed, the last part of the proof only depends on the property that neither ϕ 1 nor ϕ 2 do not depend on some coordinate z k and the assumptions s(ϕ, I, ξ) = 2 and r(ϕ, I, ξ) = (1, 0) or (0, 1) also prevents this. Using the notations of Section 2 and doing a linear change of variables, we may always assume that L(θ) = θ 1 and that each L j,k is a linear combination of θ → θ 1 and θ → θ 2 . In particular, we may write

Imϕ j (θ) = κ j   θ 1 + 1≤k≤l≤3 a j,k,l θ k θ l + Gj (θ)   with Gj (θ) = O |θ 1 | 3 + |θ 2 | 3 + |θ 3 | 3 .
The condition r(ϕ, I, ξ) = (1, 0) or (0, 1) just means that a 2,3,3 -a 1,3,3 ̸ = 0. We first do the same change of variables as in the case s(ϕ, I, ξ) = 3:

U : O 1 → O 2 is defined by U (θ 1 , θ 2 , θ 3 ) =   θ 1 + 1≤k≤l≤3 a 1,k,l θ k θ l + G1 (θ), θ 2 , θ 3   and satisfies | det(dU )| ≤ C on O 1 | det(dU -1 )| ≤ C on O 2 . Therefore, Imϕ 1 • U -1 (u) = κ 1 u 1 Imϕ 2 • U -1 (u) = κ 2   u 1 + 1≤k≤l≤3 b k,l u k u l + H(u)   with b k,l = a 2,k,l -a 1,k,l and H(u) = O |u 1 | 3 + |u 2 | 3 + |u 3 | 3 . In particular, b 3,3 ̸ = 0.
Without loss of generality, we will assume that b 3,3 > 0. We may also write

Reϕ j • U -1 (u) = 1 -Q j (u) + K j (u)
where Q j is still the quadratic form introduced in Section 2 (the change of variables does not affect it since dU (0) = I 3 ) and

K j (u) = O |u 1 | 3 + |u 2 | 3 + |u 3 | 3 .
In particular, Q 1 +Q 2 only depends on (u 1 , u 2 ) and has signature (2, 0). By Lemma 3.9, there exists κ > 0 so that

R : u → κ(Q 1 + Q 2 )(u) + κ 1 κ 2 1≤k≤l≤3 b k,l u k u l
is positive definite. On the other hand, writing η j = x j + iy j , for all u ∈ U (E(δ) ∩ O 1 ),

R(u) -κ(K 1 + K 2 )(u) + κ(-2 + x 1 + x 2 ) + κ 2 H(u) + κ 1 y 2 -κ 2 y 1 ≤ κ|Reϕ 1 • U -1 (u) -x 1 | + κ|Reϕ 2 • U -1 (u) -x 2 | + κ 1 |Imϕ 1 • U -1 (u) -y 1 | + κ 2 |Imϕ 2 • U -1 (u) -y 2 | ≤ (2κ + κ 1 + κ 2 )δ 2 .
We now argue exactly as in the proof of the case s(ϕ, I, ξ) = 3, by considering

F (u) = R(u) -κ(K 1 + K 2 )(u) + κ 1 κ 2 H(u)
and by applying the parametrized Morse lemma to it. Details are left to the reader.

We end the proof of Theorem 5.1 by solving the case s(ϕ, I, ξ) = 1 and r(ϕ, I, ξ) = (2, 0) (or (0, 2)). We use again the notations of Section 2 and we assume that L(θ) = θ 1 . We know that

κ 2 Imϕ 1 (θ) -κ 1 Imϕ 2 (θ) = aθ 2 1 + bθ 1 θ 2 + cθ 1 θ 3 + Q(θ 2 , θ 3 ) + H(θ)
where Q is a quadratic form with signature (2, 0) (the case (0, 2) is symmetric) and

H(θ) = O |θ 1 | 3 + |θ 2 | 3 + |θ 3 | 3 . We also write Imϕ 1 (θ) = κ 1 θ 1 + G 1 (θ) with G 1 (θ) = O |θ 1 | 2 + |θ 2 | 2 + |θ 3 | 2 . As above, we consider the diffeomorphism U : O 1 → O 2 U (θ) = (θ 1 + G 1 (θ), θ 2 , θ 3 ).
Since dU (0) = I 3 , we still have (with the same coefficients)

(κ 2 Imϕ 1 -κ 1 Imϕ 2 ) • U -1 (u) = au 2 1 + bu 1 u 2 + cu 1 u 3 + Q(u 2 , u 3 ) + H(u) with H(u) = O |u 1 | 3 + |u 2 | 3 + |u 3 | 3 . We now apply the parametrized Morse lemma to κ 2 Imϕ 1 -κ 1 Imϕ 2 to get a diffeomorphism V : O ′ 2 ⊂ O 2 → O 3 , V (u 1 , u 2 , u 3 ) = (u 1 , v 2 (u 1 , u 2 , u 3 ), v 3 (u 1 , u 2 , u 3 )) and a C 1 -map h such that (κ 2 Imϕ 1 -κ 1 Imϕ 2 ) • (U • V ) -1 (v) = v 2 2 + v 2 3 + h(v 1 ). Setting O = U -1 (O ′
2 ), we conclude as before since, for all θ ∈ O such that |ϕ j (θ)-

η j | ≤ δ j , j = 1, 2, for v = V • U (θ), |κ 1 v 1 -x| ≤ δ 1 |v 2 2 + v 2 3 + h(v 1 )
-y| ≲ δ 2 for some real numbers x and y. □

To cover the other cases of Theorem 2.1, we shall need the following result.

Theorem 5.2. Let ϕ ∈ O(D d , D d ) ∩ C 3 (D 3 ). Let ξ ∈ T d , let I = {i 1 , .
. . , i p } ⊂ {1, . . . , d}, I ̸ = ∅ such that ϕ I (ξ) ∈ T |I| and ∇ϕ i 1 (ξ), . . . , ∇ϕ ip (ξ) are linearly independent. There exist C > 0 and a neighbourhood U of ξ such that, for all β ∈ (-1, 0], for all η ∈ T |I| , for all δ ∈ (0, +∞) |I| ,

V β z ∈ U ∩ D d : |ϕ i j (z) -η j | ≤ δ j , j = 1, . . . , p ≤ Cδ 2+β 1 • • • δ 2+β p .
The proof of this theorem is contained in [START_REF] Kosiński | Composition operators on the polydisc[END_REF]: see Lemma 12 and the proof of it.

Proof of our main theorem

This section is devoted to the proof of Theorem 2.1. Let us first prove the sufficient part. We introduce

A 1 = ξ ∈ D 3 : ϕ(ξ) ∈ T 3 . Let ξ ∈ A 1 .
We first observe that ξ belongs to T 3 . Otherwise if we assumed that ξ 1 ∈ D, we would obtain that ϕ does not depend on z 1 which would prevent dϕ(ξ) to be invertible. By Theorem 5.2, there exist a neighbourhood U(ξ) of ξ and C(ξ) > 0 such that, for all η ∈ T 3 , for all β ∈ (-1, 0], for all δ ∈ (0, 2] 3 ,

V β ϕ -1 (S(η, δ)) ∩ U(ξ) ≤ C(ξ)δ 2+β 1 δ 2+β 2 δ 2+β 3 .
Since A 1 is compact, it can be covered by a finite number of sets U(ξ 1,1 ), . . . , U(ξ 1,p ). We also get the existence of ε 1 ∈ (0, 1) such that, for all z ∈ B 1 :

= D 3 \ k U(ξ 1,k ), |ϕ j (z)| < 1 -2ε 1 for some j ∈ {1, 2, 3}.
We then set

A 2 = ξ ∈ B 1 : there exists I ⊂ {1, 2, 3}, |I| = 2, ϕ I (ξ) ∈ T 2 .
Let ξ ∈ A 2 . We first assume that ∇ϕ i 1 (ξ) and ∇ϕ i 2 (ξ) are linearly independent. Let j ∈ {1, 2, 3}\I and observe that |ϕ j (ξ)| < 1 -2ε 1 . Then we may again apply Theorem 5.2 to get a neighbourhood U(ξ) of ξ and D(ξ) > 0 such that, for all η ∈ T 3 , for all β ∈ (-1, 0], for all δ ∈ (0, 2] 3 , for all z ∈ U(ξ),

V β ϕ -1 I (S(η I , δ I )) ∩ U(ξ) ≤ D(ξ)δ 2+β i 1 δ 2+β i 2 |ϕ j (z)| < 1 -ε 1 .
Since ϕ -1 (S(η, δ)) ∩ U(ξ) is empty provided δ j < ε 1 , we get

V β ϕ -1 (S(η, δ)) ∩ U(ξ) ≤ D(ξ) ε 2+β 1 δ 2+β 1 δ 2+β 2 δ 2+β 3 ≤ C(ξ)δ 2+β 1 δ 2+β 2 δ 2+β 3
where we have set C(ξ) = D(ξ)/ε 2 1 (which does not depend on η, β ∈ (-1, 0] and δ). Assume now that ∇ϕ i 1 and ∇ϕ i 2 are not linearly independent. We claim that ξ ∈ T 3 . Otherwise, if ξ 1 ∈ D, ϕ i 1 and ϕ i 2 would not depend on z 1 . Now, let ξ ′ = (1, ξ 2 , ξ 3 ). Then it would be impossible that ϕ satisfies the assumptions of the theorem at ξ ′ , since ∇ϕ i 1 (ξ ′ ) and ∇ϕ i 2 (ξ ′ ) are linearly dependent, and since ϕ i 1 and ϕ i 2 do not depend on z 1 , conditions (2), (b), (c) and (d) cannot happen. We now argue as before, but using Theorem 5.1 to get a neighbourhood U(ξ) of ξ and C(ξ) > 0 such that, for all η ∈ T 3 , for all β ∈ (-1, 0], for all δ ∈ (0, 2] 3 ,

V β ϕ -1 (S(η, δ)) ∩ U(ξ) ≤ C(ξ)δ 2+β 1 δ 2+β 2 δ 2+β 3 .
We then cover A 2 by a finite number of these sets U(ξ 2,1 ), . . . , U(ξ 2,q ). We define ε 2 ∈ (0, ε 1 ) such that, for all z ∈ B 2 := B 1 \ k U(ξ 2,k ), |ϕ j 1 (z)| < 1 -2ε 2 and |ϕ j 2 (z)| < 1 -2ε 2 for at least two different integers j 1 , j 2 in {1, 2, 3}. We finally set A 3 = {ξ ∈ B 2 : there exists j ∈ {1, 2, 3}, ϕ j (z) ∈ T} .

Arguing as before, using Theorem 5.2, we get a finite covering U(ξ 3,1 ), . . . , U(ξ 3,r ) of A 3 and constants C(ξ 3,1 ), . . . , C(ξ 3,r ) > 0 such that, for all η ∈ T 3 , for all β ∈ (-1, 0], for all δ ∈ (0, 2] 3 ,

V β ϕ -1 (S(η, δ)) ∩ U(ξ 3,k ) ≤ C(ξ 3,k )δ 2+β 1 δ 2+β 2 δ 2+β 3 .
There also exists ε 3 ∈ (0, ε 2 ) such that, provided z ∈ B 3 := A 3 \ k U(ξ 3,k ), |ϕ j (z)| < 1-ε 3 for all j = 1, 2, 3. Let now δ ∈ (0, 2] 3 and assume that at least one of δ 1 , δ 2 , δ 3 is less than ε 3 . Then

ϕ -1 S(η, δ) =   ϕ -1 (S(η, δ)) ∩ j,k U(ξ j,k )   ∪ ϕ -1 (S(η, δ)) ∩ B 3 . Now, ϕ -1 S(η, δ) ∩ B 3 is empty, so that V β (ϕ -1 (S(η, δ)) ≤   j,k C(ξ j,k )   δ 2+β 1 δ 2+β 2 δ 2+β 3
(observe that the sum is finite). We conclude by applying Lemma 3.1.

We now prove that the conditions arising in Theorem 2.1 are necessary to ensure continuity of C ϕ on H 2 (D 3 ). For points ξ ∈ T 3 such that ϕ(ξ) ∈ T 3 , this is a consequence of [START_REF] Koo | Composition operators on the polydisc induced by smooth symbols[END_REF].

For points ξ ∈ T 3 such that there exists I ⊂ {1, 2, 3}, ϕ I (ξ) ∈ T 2 , this follows from Theorem 4.1.

Examples

We shall now exhibit examples showing the different cases which can occur with ϕ I (ξ) ∈ T 2 , |I| = 2, ∇ϕ i 1 and ∇ϕ i 2 are linearly dependent. In what follows, ε 0 (resp. ε 1 ) is the positive real number coming from Lemma 3.10 (resp. Lemma 3.12).

Example 7.1. Let ε ∈ [-ε 0 , ε 0 ] and

ϕ(z) = 1 2 z 1 z 2 + 1 + z 3 2 + iε(z 3 -1) 2 , 1 2 z 1 z 2 + 1 + z 3 2 , 0 .
Then ϕ maps D 3 into itself and C ϕ is continuous on H 2 (D 3 ) if and only if ε ̸ = 0.

Proof. Lemma 3.10 immediately implies that ϕ is a self-map of D 3 and that, for I = {1, 2}, ϕ I (ξ) ∈ T 2 =⇒ ξ = (e iα , e -iα , 1) for some α ∈ R. By circular symmetry, we may assume that ξ = e. Now,

Imϕ 1 (θ) = 1 4 (θ 1 + θ 2 + θ 3 ) -ε θ 2 3 2 + O |θ 1 | 3 + |θ 2 | 3 + |θ 3 | 3 Imϕ 2 (θ) = 1 4 (θ 1 + θ 2 + θ 3 ) + O |θ 1 | 3 + |θ 2 | 3 + |θ 3 | 3 Reϕ j (θ) = 1 - 1 4 (θ 1 + θ 2 ) 2 - 1 8 θ 2 3 + O |θ 1 | 3 + |θ 2 | 3 + |θ 3 | 3 .
Therefore, s(ϕ, I, e) = 2 and r(ϕ, I, e) = (1, 0) provided ε < 0, (0, 1) provided ε > 0, (0, 0) provided ε = 0. Hence the conclusion follows from Theorem 2.1. □ It could be observed that, for ε = 0, C ϕ is continuous on A 2 (D 3 ).

Remark 7.2. In [START_REF] Kosiński | Composition operators on the polydisc[END_REF], it is written that first order conditions, that is conditions on the first order derivative of ϕ, are not sufficient to characterize the continuity of C ϕ on H 2 (D 3 ). The argument is based on the functions ϕ(z) = (z 1 z 2 z 3 , z 1 z 2 z 3 , 0) and ψ(z) = ((z 1 + z 2 + z 3 )/3, (z 1 + z 2 + z 3 )/3, 0) and essentially reduces to the following fact: ϕ {1,2} (e) = ψ {1,2} (e) = (1, 1), ∇ϕ(e) = ∇ψ(e), C ψ is continuous on H 2 (D 3 ) whereas C ϕ is not. However, there are other points ξ ∈ T 2 such that ϕ(ξ) ∈ T 2 × D, which is not the case for ψ. We think that our example is more convincing to illustrate that the first order conditions are not sufficient to characterize the continuity of a composition operator on H 2 (D 3 ). Indeed, all maps of our family have the same points ξ ∈ D 3 such that ϕ {1,2} (ξ) ∈ T 2 × D, and the same first order derivative at these points.

Example 7.3. Let a, b, c ∈ [-ε 1 , ε 1 ] and consider ϕ defined on D 3 by ϕ(z) = (F 0 (z 1 )F 0 (z 2 )F 0 (z 3 ), F a (z 1 )F b (z 2 )F c (z 3 ), 0)
where

F ε (z) = 3 + 6z -z 2 8 + 2iε(z -1) 2 -iε(z -1) 3 .
Then ϕ is a self-map of D 3 and C ϕ is continuous on H 2 (D 3 ) if and only if ab + bc + ca > 0.

Proof. Lemma 3.12 implies that ϕ is a self-map of D 3 and that, for

I = {1, 2}, ϕ I (ξ) ∈ T 2 ⇐⇒ ξ = e. We observe that Imϕ 1 (θ) = 1 2 (θ 1 + θ 2 + θ 3 ) + O |θ 1 | 3 + |θ 2 | 3 + |θ 3 | 3 Imϕ 2 (θ) = 1 2 (θ 1 + θ 2 + θ 3 ) -aθ 2 1 -bθ 2 2 -cθ 2 3 + O |θ 1 | 3 + |θ 2 | 3 + |θ 3 | 3 Reϕ j (θ) = 1 - 1 8 (θ 1 + θ 2 + θ 3 ) 2 + O |θ 1 | 3 + |θ 2 | 3 + |θ 3 | 3 .
Hence we now have s(ϕ, I, e) = 1. To compute r(ϕ, I, e) we do the change of variables (θ 1 , θ 2 , θ 3 ) → (θ 1 -θ 2 -θ 3 , θ 2 , θ 3 ) so that, in these new coordinates,

Imϕ 2 (θ) = 1 2 θ 1 -aθ 2 1 + 2aθ 1 θ 2 + 2aθ 1 θ 3 -(a + b)θ 2 2 + (a + c)θ 2 3 + 2aθ 2 θ 3 + O |θ 1 | 3 + |θ 2 | 3 + |θ 3 | 3 .
Hence, using the notations of Section 2,

R(θ) = (a + b)θ 2 2 + 2aθ 2 θ 3 + (a + c)θ 2 3 .
The signature of R is (2, 0) or (0, 2) if and only if

4a 2 -4(a + b)(a + c) < 0 ⇐⇒ ab + bc + ca > 0.
□ It remains to provide an example with s(ϕ, I, ξ) = 3. This is easier. Indeed, consider

ϕ(z) = z 1 + z 2 + z 3 3 , z 1 + z 2 + z 3 3 , 0 .
Then ϕ is a self-map of H 2 (D 3 ). Let I = {1, 2} and ξ ∈ T 3 , so that ϕ I (ξ) ∈ T 2 . By symmetry, we may assume that ξ = e. Then for j = 1, 2,

Imϕ j (θ) = 1 3 (θ 1 + θ 2 + θ 3 ) + O |θ 1 | 2 + |θ 2 | 2 + |θ 3 | 2 Reϕ j (θ) = 1 - 1 6 θ 2 1 + θ 2 2 + θ 2 3 + O |θ 1 | 3 + |θ 2 | 3 + |θ 3 | 3 .
Therefore s(ϕ, I, e) = 3 and C ϕ defines a bounded operator on H 2 (D 3 ). Observe that since ϕ is an affine map, this was already known from the algorithm of [START_REF] Bayart | Composition operators on the polydisk induced by affine maps[END_REF].

8. Compactness issues 8.1. A general compactness theorem. Our aim now is to provide necessary and sufficient conditions for a composition operator C ϕ acting on H 2 (D d ) with d = 2 or 3 to be compact. We first state a general result implying the compactness of C ϕ on H 2 (D d ) by only estimating the V β -volume of preimages of Carleson boxes, in the spirit of Lemma 3.1. It will rely on the following precise estimate of the Carleson embedding, which is Proposition 9.3 in [START_REF] Bayart | Composition operators on the polydisk induced by affine maps[END_REF].

Lemma 8.1. Let d ≥ 1. There exists C(d) > 0 such that, for any β ∈ (-1, 0], for any µ a finite nonnegative Borel measure on D d satisfying:

∃C µ > 0, ∀ξ ∈ T d , ∀δ ∈ (0, 2] d , µ S(ξ, δ) ≤ C µ V β S(ξ, δ) , (6)
then for any f ∈ A 2 β (D d ), D d |f (z 1 , . . . , z d )| 2 dµ 1/2 ≤ C(d)C µ ∥f ∥ A 2 β (D d ) .
Theorem 8.2. Let ϕ : D d → D d be holomorphic. Let us assume that there exist a > 0 and ε : (0, a] → (0, +∞) with lim 0 + ε = 0 and such that, for all δ ∈ (0, a] d , for all β ∈ (-1, 0], for all η ∈ T d ,

(7) V β (ϕ -1 (S(η, δ))) ≤ ε(inf(δ i : i = 1, . . . , d))δ 2+β 1 • • • δ 2+β d . Then C ϕ is compact on H 2 (D d ).
Proof. Our strategy, to prove compactness is to approach C ϕ by an operator T N with rank

≤ N d , for all N ∈ N, such that, for all β ∈ (-1, 0], ∥C ϕ -T N ∥ L(A 2 β )
≤ ω N with ω N which does not depend on β and lim N →+∞ ω N = 0. Letting β to -1, we then find a sequence of operators of finite rank which converges to C ϕ in L(H 2 (D d )) leading to the compactness of C ϕ . Therefore, the proof is related to the problem of finding an upper bound for the approximation numbers of C ϕ and we are inspired by the proof of [LQRP12, Theorem 5.1]. Let us proceed with the proof. For the sake of simplicity, we shall do it only for d = 2. As explained in Section 3.1, we may assume that (7) is satisfied for W (η, δ) instead of S(η, δ). A standard argument shows that we may assume that ε is nondecreasing. We may also observe that the assumptions imply that C ϕ is continuous on A 2 β (D 2 ), β ∈ (-1, 0] as well as on H 2 (D 2 ), with sup β∈(-1,0] ∥C ϕ ∥ L(A 2 β ) < +∞ (see Section 9.2 of [START_REF] Bayart | Composition operators on the polydisk induced by affine maps[END_REF]).

Let N ∈ N and for f = α∈N 2 a α z α 1 1 z α 2 2 , let us consider P N (f ) = |α i |<N a α z α 1 1 z α 2 2 , whose rank is N d . We then set T N = C ϕ • P N . Assume for a while that f is a polynomial and let h ∈ (0, 1). We set V ϕ,β as the image of V β under ϕ and µ h,ϕ,β its restriction to

D 2 \(1 -h)D 2 . We may write ∥C ϕ (f ) -T N (f )∥ 2 β = D 2 |(f -P N (f )) • ϕ| 2 dV β = D 2 |f -P N (f )| 2 dV ϕ,β = (1-h)D 2 |f -P N (f )| 2 dV ϕ,β + D 2 |f -P N (f )| 2 dµ h,ϕ,β (8)
Let us first handle the second term. We shall apply Lemma 8.

1 with µ = µ h,ϕ,β . Let η ∈ T 2 and δ ∈ (0, 2] 2 . If δ 1 ≤ h or δ 2 ≤ h, then µ h,ϕ,β (W (η, δ)) ≤ V ϕ,β (W (η, δ)) ≤ ε(h)δ 2+β 1 δ 2+β 2 .
If δ 1 > h and δ 2 > h, then we cover W (η, δ) ∩ (D 2 \(1 -h)D 2 ) as follows. Assume for instance that η = (1, 1) and let p = ⌊ 2δ 1 h ⌋ + 1, q = ⌊ 2δ 2 h ⌋ + 1. Define (θ j ) j=1,...,p+q by

θ 1 = -δ 1 , δ 2 = -δ 1 + h, . . . , δ p = -δ 1 + (p -1)h θ p+1 = -δ 2 , δ p+2 = -δ 2 + h, . . . , δ p+q = -δ 2 + (q -1)h.
We also set, for j = 1, . . . , p, η(j) = (e iθ j , 1) and δ(j) = (h, δ 2 ), whereas, for j = p + 1, . . . , p + q, we set η j = (1, e iθ j ) and δ(j) = (δ 1 , h). We claim that

W (η, δ) ∩ (D 2 \(1 -h)D 2 ) ⊂ p+q j=1
W (η(j), δ(j)).

Indeed, pick z = (r 1 e it 1 , r 2 e it 2 ) in W (η, δ) ∩ (D 2 \(1 -h)D 2 ). We may assume r 1 ≥ 1 -h. Then there exists j ∈ {1, . . . , p} such that t 1 ∈ (θ j , θ j + h) and it is plain that z ∈ W (η(j), δ(j)). Hence, provided δ 1 > h and

δ 2 > h µ h,ϕ,β (W (η, δ)) = µ h,ϕ,β (W (η, δ) ∩ (D 2 \(1 -h)D 2 )) ≤ p+q j=1 µ h,ϕ,β (W (η(j), δ(j)) ≤ pε(h)h 2+β δ 2+β 2 + qε(h)δ 2+β 1 h 2+β ≤ ε(h) 4δ 1 h h 2+β δ 2+β 2 + 4δ 2 h δ 2+β 1 h 2+β ≤ 8ε(h)δ 2+β 1 δ 2+β 2 .
By applying Lemma 8.1, there exists C > 0 which does not depend neither on β ∈ (-1, 0] nor on h ∈ (0, 1) nor on N ∈ N such that, for all holomorphic polynomials f,

D 2 |f -P N (f )| 2 dµ h,ϕ,β ≤ Cε 2 (h)∥f -P N (f )∥ 2 β ≤ Cε 2 (h)∥f ∥ 2 β .
We turn to the first term of (8). We decompose f -P N (f ) as

(f -P N (f ))(z) = k,l≥N a k,l z k 1 z l 2 + k<N l≥N a k,l z k 1 z l 2 + k≥N l<N a k,l z k 1 z l 2 =: z N 1 z N 2 s 1 (z) + z N 2 s 2 (z) + z N 1 s 3 (z).
Therefore,

(1-h)D 2 |f -P N (f )| 2 dV ϕ,β ≤ (1 -h) 2N D 2 |s 1 | 2 dV ϕ,β + (1 -h) N D 2 |s 2 | 2 dV ϕ,β + (1 -h) N D 2 |s 3 | 2 dV ϕ,β ≤ C ′ (1 -h) 2N ∥s 1 ∥ 2 β + (1 -h) N ∥s 2 ∥ 2 β + (1 -h) N ∥s 3 ∥ 2 β where C ′ = sup ∥C ϕ ∥ 2 L(A 2 β ) : β ∈ (-1, 0] < +∞. Now, ∥s 1 ∥ 2 β = k,l≥N |a k,l | 2 (k -N + 1) β+1 (l -N + 1) β+1 = k,l≥N |a k,l | 2 (k + 1) β+1 (l + 1) β+1 × (k + 1) β+1 (l + 1) β+1 (k -N + 1) β+1 (l -N + 1) β+1 ≤ (N + 1) 2+2β ∥f ∥ 2 β .
Similarly,

∥s 2 ∥ 2 β , ∥s 3 ∥ 2 β ≤ (N + 1) 1+β ∥f ∥ 2 β .
Summarizing what we have done, since β ≤ 0, we have obtained

∥C ϕ (f ) -T N (f )∥ 2 β ≤ C ′′ ε 2 (h) + (N + 1) 2 (1 -h) 2N + (N + 1)(1 -h) N ∥f ∥ 2 β
where C ′′ > 0 only depends on C and C ′ . We choose h N = 1/ log(N + 1) and we set

ω N = C ′′ ε 2 (h N ) + (N + 1) 2 (1 -h N ) 2N + (N + 1)(1 -h N ) N
which goes to 0 as N tends to +∞. We get, for all polynomials f and all β ∈ (-1, 0],

∥C ϕ (f ) -T N (f )∥ 2 β ≤ ω N ∥f ∥ 2 β .
Letting β to -1 yields ∥C ϕ -T N ∥ L(H 2 ) ≤ √ ω N , so that C ϕ is compact as a limit of finite rank operators. □

We shall also need a necessary condition for compactness. The following corollary of [Jaf91, Theorem 2.3] will be enough for us.

δ(n)) of (0, 2] d such that δ 1 (n) • • • δ d (n) goes to 0, for any η ∈ T d , (9) σ d ({e iθ ∈ T d : ϕ(θ) ∈ S(η, δ(n))}) δ 1 (n) • • • δ d (n) n→+∞ -----→ 0.
A first application of the work of this section can be done for affine symbols. If ϕ : D d → D d is affine, for any ξ ∈ T d such that ϕ(ξ) / ∈ D d , two families of trees R -1 ξ,I,J and L -1 ξ,I,J were introduced in [START_REF] Bayart | Composition operators on the polydisk induced by affine maps[END_REF] and it was shown that C ϕ is continuous on H 2 (D d ) if and only if R -1 ξ,I,J ≤ L -1 ξ,I,J for all possible choices of ξ, I, J. Using the methods of [START_REF] Bayart | Composition operators on the polydisk induced by affine maps[END_REF] combined with Theorem 8.2 and Lemma 8.3, we can prove that C ϕ is compact on H 2 (D d ) if and only if R -1 ξ,I,J < L -1 ξ,I,J for all possible choices of ξ, I, J. Details are left to the interested reader.

8.2. Compact composition operators on H 2 (D 2 ). We shall use the results of the previous subsection to almost characterize compact composition operators on H 2 (D 2 ). Recall (see [START_REF] Kosiński | Composition operators on the polydisc[END_REF]Theorem 6

]) that for ϕ ∈ O(D 2 , D 2 ) ∩ C 1 (D 2 ), C ϕ is bounded on H 2 (D 2 ) if and only if dϕ(ξ) is invertible for all ξ ∈ T 2 such that ϕ(ξ) ∈ T 2 .
We shall need to introduce two new definitions. We start from φ ∈ O(D 2 , D) ∩ C 3 (D 2 ) and let ξ ∈ T 2 such that φ(ξ) ∈ T. Without loss of generality, we will assume ξ = e and φ(ξ) = 1. We will say that φ is truely two-dimensional at ξ provided

• ∂φ ∂z 1 (e) ̸ = 0, ∂φ ∂z 2 (e) ̸ = 0; • let us write Imφ(θ) = L(θ) + G(θ) where L is a linear form and G(θ) = O(θ 2 1 + θ 2 2 ). The map U (θ 1 , θ 2 ) = (L(θ) + G(θ), θ 2 ) defines a diffeomorphism between two neighbourhoods U and V of 0 in R 2 since ∂L ∂θ 1 (0) ̸ = 0. We then may write Reφ • U -1 (u 1 , u 2 ) = 1 + H(u 1 , u 2 ).
We shall require, for a truely two-dimensional function at ξ, that there does not exist h such that H(u 1 , u 2 ) = h(u 1 ) in a neighbourhood of 0.

We will say that φ is truely two-dimensional at ξ of order 2 provided, writing Reφ(θ) = 1-Q(θ)+o(θ 2 1 +θ 2 2 ) with Q a quadratic form, that the signature of Q is (2, 0). Observe that a truely two-dimensional function at ξ of order 2 is automatically truely two-dimensional at ξ. Indeed, ∂φ ∂z 1 (e) ̸ = 0 and ∂φ ∂z 2 (e) ̸ = 0, otherwise φ would not depend on one of the two variables and the signature of Q could not be (2, 0). Moreover, if we do the change of variables described above, we may write

Reφ • U -1 (u 1 , u 2 ) = 1 -Q(u 1 , u 2 ) + H(u 1 , u 2 ) where signature( Q) = (2, 0) and H(u 1 , u 2 ) = o(u 2 1 + u 2 2 ).
Example 8.4. Let φ(z) = z 1 z 2 . Then φ is not truely two-dimensional at e. Indeed, we may write Imφ(θ) = f (θ 1 +θ 2 ) and Reφ(θ

) = 1+g(θ 1 +θ 2 ). Let U (θ 1 , θ 2 ) = (f (θ 1 +θ 2 ), θ 2 ). A straightforward computation shows that U -1 (u 1 , u 2 ) = (f -1 (u 1 ) -u 2 , u 2 ) so that Reφ • U -1 (u 1 , u 2 ) = 1 + g(f -1 (u 1 )).
Heuristically speaking, φ is not truely two-dimensional since φ(θ) only depends on the single variable θ 1 + θ 2 .

Example 8.5. Let φ(z) = z 1 +z 2 2 . Then φ is truely two-dimensional at e of order 2 (in fact, at any ξ ∈ T 2 such that φ(ξ) ∈ T). Indeed,

Reφ(θ) = 1 - 1 4 (θ 2 1 + θ 2 2 ) + o(θ 2 1 + θ 2 2 ).
The interest of working with a truely two-dimensional function of order 2 comes from the following result, which should be compared with Theorem 5.1.

Theorem 8.6. Let φ ∈ O(D 2 , D) ∩ C 3 (D 3 ). Assume that there exists ξ ∈ T 2 such that φ(ξ) ∈ T and that φ is truely two-dimensional at ξ of order 2. Then there exist a neighbourhood U of ξ and C > 0 such that, for all β ∈ (-1, 0], for all η ∈ T, for all δ > 0,

V β z ∈ D 2 ∩ U : |φ(z) -η| < δ ≤ Cδ 2+β+ 1 2 .
Proof. We may assume ξ = e and φ(ξ) = 1. We mimic the proof of Theorem 5.1. We start with the following fact.

Fact. There exist C > 0 and a neighbourhood O of 0 in R 2 such that, for all δ > 0, for all η ∈ T,

(10) λ 2 ({θ ∈ O : |φ(θ) -η| ≤ δ}) ≤ Cδ 3 2 .
We write

Imφ(θ) = L(θ) + G(θ) Reφ(θ) = 1 -Q(θ) + o(θ 2 1 + θ 2 2 ) with G(θ) = o(|θ 1 | + |θ 2 |
) and the signature of Q equal to (2, 0). By a linear change of variables, which does not modify the signature of Q, we may assume that L(θ) = θ 1 . We then set U

(θ) = (L(θ) + G(θ), θ 2 ). There exist two neighbourhoods O 1 and O 2 of 0 in R 2 and C 1 > 0 such that U is a diffeomorphism from O 1 onto O 2 and | det(dU )| ≤ C 1 on O 1 | det(dU -1 )| ≤ C 1 on O 2 .
Since dU (0) = Id, we can still write

Reφ • U -1 (u 1 , u 2 ) = 1 -Q(u) + o(u 2 1 + u 2 2
). We then apply the parametrized Morse lemma: there exist two neighbourhoods O

′ 2 ⊂ O 2 and O 3 of 0, a constant C 2 > 0 and a diffeomorphism V : O ′ 2 → O 3 , V (u 1 , u 2 ) = (u 1 , V 2 (u 1 , u 2 )) and a C 1 -map h : R → R such that, for all v ∈ O 3 , Reφ • U -1 • V -1 (v 1 , v 2 ) = 1 -v 2 2 + h(v 1 ) whereas we have still Imφ • U -1 • V -1 (v 1 , v 2 ) = v 1 as well as | det(dV )| ≤ C 2 on O ′ 2 | det(dV -1 )| ≤ C 2 on O 3 . Therefore, for any θ ∈ U -1 (O ′ 2 ) =: O, for all η ∈ T, for all δ > 0, setting v = V • U (θ), we have |v 1 -Im(η)| < δ |1 -v 2 2 + h(v 1 ) -Re(η)| < δ. Fubini's theorem yields λ 2 V • U O ∩ φ -1 (S(η, δ)) ≲ δ 3/2
which in turn implies the fact. We finally deduce Theorem 8.6 from the fact following the method used in Theorem 5.1. □

We are now ready to almost characterize compact composition operators on H 2 (D 2 ).

Theorem 8.7. Let ϕ ∈ O(D 2 , D 2 ) ∩ C 3 (D 2 ). If C ϕ defines a compact composition operator on H 2 (D 2 ), then

(1) ϕ(T 2 ) ∩ T 2 = ∅;

(2) for all ξ ∈ T 2 , for all j ∈ {1, 2} such that ϕ j (ξ) ∈ T, ϕ j is truely two-dimensional at ξ.

Conversely, assume that

(1) ϕ(T 2 ) ∩ T 2 = ∅;

(2) for all ξ ∈ T 2 , for all j ∈ {1, 2} such that ϕ j (ξ) ∈ T, ϕ j is truely two-dimensional at ξ of order 2.

Then C ϕ is compact on H 2 (D 2 ).

Proof. We first prove the necessary condition and we start from a compact composition operator C ϕ on H 2 (D 2 ). First assume that ϕ(T 2 ) ∩ T 2 ̸ = ∅, for instance ϕ(e) = e. Let δ > 0 be small and set δ = (δ, δ). By the mean value theorem, there exists c > 0 (which does not depend on δ provided δ is small enough) such that e iθ : ϕ(θ) ∈ S(e, δ) ⊃ [-cδ, cδ] 2 .

This prevents (9) to be true and C ϕ cannot be compact. Assume now that there exists ξ ∈ T 2 and j ∈ {1, 2} such that ϕ j (ξ) ∈ T and ϕ j is not truely two-dimensional at ξ, say j = 1, ξ = e and ϕ 1 (e) = 1. If ∂ϕ 1 ∂z 2 (e) = 0, then ϕ 1 only depends on z 1 and we set δ = (δ, 2) for small values of δ. Again, there exists c > 0 such that e iθ : ϕ(θ) ∈ S(e, δ) = e iθ : |ϕ 1 (θ) -1| < δ ⊃ [-cδ, cδ] × [-1, 1] and again (9) cannot be true. If ∂ϕ 1 ∂z 1 (e) ̸ = 0 and ∂ϕ ∂z 2 (e) ̸ = 0, since ϕ 1 is not truely two-dimensional at e, let U and V be two neighbourhoods of 0, C > 0 and U a C 1diffeomorphism from U onto V such that

| det(dU )| ≤ C 1 on O 1 | det(dU -1 )| ≤ C 1 on O 2 and Imϕ 1 • U -1 (u 1 , u 2 ) = u 1 Reϕ 1 • U -1 (u 1 , u 2 ) = 1 + h(u 1 )
with h(u 1 ) = O(u 1 ). Therefore there exists c > 0 such that, for all δ > 0, for all (u 1 , u 2 ) ∈ ([-cδ, cδ] × [-1, 1]) ∩ V, |ϕ 1 • U -1 (u 1 , u 2 ) -1| < δ.

This yields U -1 ([-cδ, cδ] × [-1, 1]) ∩ V ⊂ ϕ -1 (S(e, δ))

for δ = (δ, 2) which prevents (9) to be true.

Let us turn to the positive part. It will follow by applying Theorem 8.6 with a suitable covering argument. Since ϕ(T 2 ) ∩ T 2 = ∅, there exists ε 1 ∈ (0, 1) such that, for all ξ ∈ D 2 , there exists j ∈ {1, 2} with |ϕ j (ξ)| ≤ 1 -2ε 1 . We then set A 1 = ξ ∈ D 2 : ∃j ∈ {1, 2}, ϕ j (ξ) ∈ T .

Let ξ ∈ A 1 and assume for instance that ϕ 1 (ξ) ∈ T. Let U(ξ) and C(ξ) be given by Theorem 8.6. Restricting U(ξ) if necessary, we may assume that |ϕ 2 (z)| < 1 -ε 1 for all z ∈ U(ξ). Let β ∈ (-1, 0], η ∈ T 2 and δ ∈ (0, 2] 2 . Assume first that δ 2 ≥ ε 1 . Then 1 . On the other hand, if δ 2 < ε 1 , then ϕ -1 (S(η, δ)) ∩ U(ξ) = ∅ so that (11) is verified for all δ ∈ (0, 2] 2 . By compactness, we cover A 1 by finitely many U(ξ 1 ), . . . , U(ξ p ) and we set A 2 = D 2 \ p k=1 U(ξ k ). There exists ε 2 > 0 such that, provided z ∈ B 2 , |ϕ 1 (z)|, |ϕ 2 (z)| < 1 -ε 2 . Therefore, for all η ∈ T 2 and all δ ∈ (0, 2] 2 with inf(δ 1 , δ 2 ) < ε 2 , V β ϕ -1 (S(η, δ)) ≤ touching the boundary such that C ϕ is continuous on H 2 (D 2 ) and C ϕ is compact (resp. not compact), for instance ϕ(z) = ((z 1 + z 2 )/2, 0) (resp. ϕ(z) = (z 1 z 2 , 0)).

V β ϕ -1 (S(η, δ)) ∩ U ≤ V β z ∈ D 2 ∩ U : |ϕ 1 (z) -η 1 | < δ 1 ≤ C(ξ)δ
8.3. Compact composition operators on H 2 (D 3 ). We now start a similar study of compact composition operators on H 2 (D 3 ). It turns out that the symbols which we study in depth in Section 5 give rise to noncompact composition operators. As above, we need to introduce several definitions. First of all, let φ ∈ O(D 3 , D) ∩ C 2 (D 3 ) and let ξ ∈ T 3 such that φ(ξ) ∈ T. Without loss of generality, we will assume ξ = e and φ(ξ) = 1. We will say that φ is truely at least two-dimensional at ξ provided

• at least two of ∂φ ∂z 1 (e), ∂φ ∂z 2 (e), ∂φ ∂z 3 (e) are not equal to 0; • let us assume that ∂φ ∂z 1 (e) ̸ = 0 and let us write Imφ(θ) = L(θ) + G(θ) where L is a linear form and G(θ) = O(θ 2 1 + θ 2 2 + θ 2 3 ). The map U (θ) = (L(θ) + G(θ), θ 2 , θ 3 ) defines a diffeomorphism between two neighbourhoods U and V of 0 in R 3 since ∂L ∂θ 1 (0) ̸ = 0. We then may write Reφ • U -1 (u) = 1 + H(u).

We shall require, for a truely at least two-dimensional function at ξ, that H(u) ̸ = h(u 1 ) for some function h.

Lemma 8. 3 .

 3 Let ϕ : D d → D d be holomorphic and belonging to C(D d ). Assume that C ϕ is compact on H 2 (D d ). For any sequence (

  where we have setD(ξ) = C(ξ)/ε 5/2

.

  We infer from Theorem 8.2 that C ϕ is compact. □ We can then use Examples 8.4 and 8.5 to provide examples of symbols ϕ ∈ O(D 2 , D 2 )

We will say that φ is truely at least two-dimensional at ξ of order 2 provided, writing Reφ(θ) = 1 -Q(θ) + O(θ 2 1 + θ 2 2 + θ 2 3 ) with Q a quadratic form, that the signature of Q is (2, 0) or (3, 0). As before a truely at least two-dimensional function at ξ of order 2 is automatically truely two-dimensional at ξ. We will also have to consider symbols ϕ such that there exists I ⊂ {1, 2, 3} with |I| = 2 and ϕ I (T 3 ) ∩ T 2 ̸ = ∅. Here are the relevant definitions. Let φ ∈ O(D 3 , D 2 ) ∩ C 3 (D 3 ) and let ξ ∈ T 3 such that φ(ξ) ∈ T 2 . Without loss of generality, we will assume ξ = e and φ(ξ) = (1, 1). We will say that φ is truely three-dimensional at ξ provided • ∇φ 1 (e) and ∇φ 2 (e) are linearly independent • let us assume that

is invertible and let us write Imφ j (θ) = L j (θ)+ G j (θ) where L j is a linear form and

We shall require, for a truely three-dimensional function, that H j (u) ̸ = h j (u 1 , u 2 ) for some function h j , for j = 1, 2.

We will say that φ is truely three-dimensional at ξ of order 2 provided that ∇φ 1 (e) and ∇φ 2 (e) are linearly independent and, writing Reφ j (θ) = 1-Q j (θ)+o(θ 2 1 +θ 2 2 +θ 2 3 ) with Q j a quadratic form, that the signature of both Q 1 and Q 2 is equal to (3, 0). Observe that a truely three-dimensional function at ξ of order 2 is automatically truely three-dimensional at ξ.

Example 8.8. The map φ(z) = z 1 z 2 , z 1 3 + 2z 2 3 is not truely three-dimensional at e, even if ∇φ 1 (e) and ∇φ 2 (e) are linearly independent. On the contrary, the map φ(z) =

4 z 3 is truely three-dimensional of order 2 at e. We now give an analogue of Theorem 8.6 in dimension 3. Theorem 8.9. (a) Let φ ∈ O(D 3 , D) ∩ C 3 (D 3 ). Assume that there exists ξ ∈ T 3 such that φ(ξ) ∈ T and that φ is truely at least two-dimensional at ξ of order 2. Then there exist a neighbourhood U of ξ and C > 0 such that, for all β ∈ (-1, 0], for all η ∈ T, for all δ > 0,

). Assume that there exists ξ ∈ T 3 such that φ(ξ) ∈ T 2 and that φ is truely three-dimensional at ξ of order 2. Then there exist a neighbourhood U of ξ and C > 0 such that, for all β ∈ (-1, 0], for all η ∈ T 2 , for all δ ∈ (0, 2] 2 ,

Proof. We shall be very squetchy. For (a), we just need to prove the following: there exist C > 0 and a neighbourhood O of 0 in R 2 such that, for all δ > 0, for all η ∈ T,

We may and shall assume that Imφ(θ) = θ 1 and write Reφ(θ

3 ) where signature(Q) = (2, 0) or (3, 0). In both cases we can do a linear change of variables to assume that Q(θ) = θ 2 2 + R(θ), with signature(R) = (1, 0) or (2, 0). We then apply the parametrized Morse lemma to get new variables

). We finally conclude as in Theorem 8.6. Observe that if the signature of Q is (3, 0) we could even get a better estimate by using the parametrized Morse lemma in order to write

), but this is not useful for our purpose. Regarding (b), we start by writing

with L 1 , L 2 two independent linear forms. We may assume L 1 (θ) = θ 1 , L 2 (θ) = θ 2 and using the change of variables U (θ) = (L 1 (θ) + G 1 (θ), L 2 (θ) + G 2 (θ), θ 3 ), we will assume

Let us assume that δ 1 ≤ δ 2 . Since the signature of Q 1 is (3, 0), we may apply the parametrized Morse lemma and assume, up to a change of variables, that Reφ 1 (θ) = 1 -θ 2 3 + h 1 (θ 1 , θ 2 ) with h 1 a C 1 -function defined in a neighbourhood of 0. Arguing as in the proof of Theorem 8.6, we get

□

We deduce a result giving a large number of examples of compact and noncompact composition operators on H 2 (D 3 ).

Theorem 8.10.

(2) for all ξ ∈ T 3 , for all I ⊂ {1, 2} with ϕ I (ξ) ∈ T 2 , ϕ I is truely three-dimensional at ξ; (3) for all ξ ∈ T 3 , for all j ∈ {1, 2, 3} such that ϕ j (ξ) ∈ T, ϕ j is truely at least two-dimensional at ξ.

Conversely, assume that

(1) ϕ(T 3 ) ∩ T 3 = ∅;

(2) for all ξ ∈ T 3 , for all I ⊂ {1, 2} with ϕ I (ξ) ∈ T 2 , ϕ I is truely three-dimensional at ξ of order 2;

(3) for all ξ ∈ T 3 , for all j ∈ {1, 2, 3} such that ϕ j (ξ) ∈ T, ϕ j is truely at least two-dimensional at ξ of order 2.

Then C ϕ is compact on H 2 (D 3 ).

Proof. We shall only consider the necessary condition, since the sufficient condition can be proved exactly like the sufficient condition of Theorem 8.7. That (1) is mandatory follows by the mean value theorem like in the proof of Theorem 8.7. More generally, for any d ≥ 1 and any symbol

That (3) is necessary also follows from the arguments of Theorem 8.7 and thus we just deal with (2). Assume that for I = {1, 2} and ξ = e, ϕ I (e) = (1, 1) and ϕ I is not truely three-dimensional at ξ. If ∇ϕ 1 (e) and ∇ϕ 2 (e) are not linearly independent, we may assume

3 ), j = 1, 2. Then, for some c > 0, for all δ > 0, setting δ = (δ, δ, 1),

which prevents (9) to be true. If ∇ϕ 1 (e) and ∇ϕ 2 (e) are linearly independent, we can assume

2 ), j = 1, 2. Then, for some c > 0, for all δ > 0, setting δ = (δ, δ, 1),

Then C ϕ is a compact operator on H 2 (D 3 ).

Remark 8.12. To characterize completely regular symbols leading to compact composition operators on H 2 (D 2 ) or H 2 (D 3 ), it seems that we would need to inspect high order terms in the expansion of Reϕ j and Imϕ j . In that case, some tools are missing (the signature, the parametrized Morse lemma). However, for a specific map, a detailed study using the arguments developed here should be possible. It is likely that the study of continuity on H 2 (D d ) with d ≥ 4 also requires to consider more terms in the expansion of Reϕ j and Imϕ j .