Vincent Beck

Ésar Lecoutre

RANK TWO ARTIN-SCHELTER REGULAR ALGEBRAS AND NON COMMUTING DERIVATIONS

If ∆ and Γ are two derivations of a commutative algebra A such that ∆Γ -Γ∆ = ∆ is locally nilpotent, one can endow A with a new product * whose filtered semiclassical limit is the Poisson structure ∆ ∧ Γ. In this article we first study theses (Poisson) algebras from an algebraic point of view, and when A is a polynomial algebra, we investigate their homological properties. In particular, if the derivations ∆ and Γ are linear, we prove that the algebras (A, * ) are Artin-Schelter regular algebras. Assuming furthermore a technical condition on Γ, we show that the algebra (A, * ) is Calabi-Yau if and only if the trace of Γ is equal to 1 if and only if the Poisson algebra (A, ∆ ∧ Γ) is unimodular.

Introduction

The philosophy behind noncommutative projective algebraic geometry [START_REF] Rogalski | An introduction to noncommutative projective geometry[END_REF] is to study noncommutative version of the projective spaces P n by describing suitable noncommutative algebras that would be thought of as their homogeneous coordinates rings: the so-called Artin-Schlelter regular algebras (AS regular algebras for shorts) [START_REF] Artin | Graded algebras of global dimension 3[END_REF]. Numerous example of AS regular algebras are known but a compete classification is known only up to projective dimension 2. The work of Pym [START_REF] Pym | Quantum deformations of projective three-space[END_REF] provides a partial answer in dimension 3 by focusing on a classification of graded Calabi-Yau algebras [START_REF] Ginzburg | Calabi-Yau algebras[END_REF] that are flat deformations of C[x 0 , x 1 , x 2 , x 3 ]. In [START_REF] Center | A new family of Poisson algebras and their deformations[END_REF], Sierra and the second author introduced a family of AS regular algebras R(n, a) of Gel'fand-Kirillov dimension n + 1 indexed by a scalar a. The authors proved that for any integer n 0 there exists a unique scalar a such that the algebra R(n, a) is a Calabi-Yau. This is of particular interest since it explains a specific choice of parameter made in the exceptional component E(3) of Pym's classification.

Back to the algebra R(n, a), the parameter n can be interpreted as the index of nilpotency of a maximal Jordan block seen as the restriction to degree 1 polynomials of a locally nilpotent derivation ∆ Date: February 26, 2024.

of a polynomial algebra in n + 1 variables. The construction of R(n, a) then relies on a second derivation Γ such that ∆Γ -Γ∆ = ∆ is nilpotent together with a deformation formula of Coll, Gerstenhaber, and Giaquinto [START_REF] Coll | An explicit deformation formula with noncommuting derivations, Ring theory 1989[END_REF]. The aim of this article is to investigate the following natural generalization: what are the possible algebras arising from this construction by relaxing the conditions on ∆? For instance, the case of homogeneous locally nilpotent derivations ∆ of a polynomial rings whose canonical Jordan normal form acting on the set of degree 1 polynomial is not a maximal block will be of particular interest. In this situation we prove that we obtain AS regular algebras and, among them, we describe a family of Calabi-Yau algebras indexed by r -1 scalar parameters, where r is the number of blocks in the canonical Jordan normal form of ∆ (the algebras studied in [START_REF] Center | A new family of Poisson algebras and their deformations[END_REF] correspond to the case r = 1).

This article is divided into eight sections. In Section 1 we present the general construction of a Poisson algebra A (∆,Γ) and of an associative algebra R (∆,Γ) associated to a pair of derivations (∆, Γ) of a commutative algebra A such that ∆Γ -Γ∆ = ∆ is nilpotent. Such a pair of derivation will be called a solvable pair. Section 2 is devoted to the study of general algebraic properties of A (∆,Γ) and R (∆,Γ) . A lots of example of solvable pairs are given. Moreover we compute the (Poisson) center in many cases (Lemma 2.9) and exhibit examples of important Poisson derivations of A (∆,Γ) and automorphisms of R (∆,Γ) (see Lemma 2.15 and Theorem 2.17). Furthermore we remark that the natural filtration associated to the locally nilpotent derivation ∆ is compatible with both the Poisson structure of A (∆,Γ) and the associative structure of R (∆,Γ) . This filtration is presented and studied in Section 2.4, it will be one of our main tool to prove a lot of results in this article. In the general situation not much can be said about normal elements. However, among them, we identify and study a significant subset consisting of the so-called strongly normal elements (Definition 2.29) which will be of particular interest latter on in this article. To end Section 2 we observe that up to localization the Poisson algebra A (∆,Γ) and the algebra R (∆,Γ) are particularly simple: they are isomorphic to (Poisson) Ore extensions over (a localization of) the kernel of ∆ thanks to the local slice construction.

The rest of the article is devoted to the case where the solvable pair is defined on polynomial algebra A. In Section 3 the rank of a non abelian Poisson algebra A (∆,Γ) is shown to be equal to 2. This implies that a Poisson algebra A (∆,Γ) cannot be isomorphic non trivially to tensor products of Poisson algebras of the same type. In that sense we obtain new Poisson algebras with our construction. Section 4 deals with the case of homogeneous derivations ∆ and Γ. In this situation both the Poisson center of A (∆,Γ) and the center of R (∆,Γ) are completely determined (Corollary 4.9). Moreover we construct a finer filtration than the previous one, having the benefit that the associated graded algebra is a polynomial algebra. In Section 5 we show that the algebra R (∆,Γ) is AS regular and we completely determine the (Poisson) normal elements: they are precisely the strongly normal elements introduced in Section 2. In Sections 6 and 7 we focus on the case where Γ is diagonalizable. In this situation, we are able to determine the Poisson derivation when Γ is generic, as well as we are able to give a presentation by generators and relations of R (∆,Γ) . Our last section is devoted to characterize when our algebras (resp. Poisson algebras) are Calabi-Yau (resp. unimodular), see Corollary 8.12. Finally the article ends with Appendices which recap some of the results frequently used in the article, in particular it explains the structure of the graded ring for the finer of the filtrations we study. In this article we always assume that k is a field of characteristic zero.

A deformation formula for non commuting derivations

Let A be a commutative k-algebra and ∆, Γ be two derivations of A such that [∆, Γ] = ∆Γ -Γ∆ = ∆. It is easily verified that the biderivation {-, -} = ∆ ∧ Γ satisfies the Jacobi identity, hence is a Poisson bracket on A. More precisely we have (1.1) {f, g} = ∆(f )Γ(g) -∆(g)Γ(f ) for any f, g ∈ A. Moreover, if ∆ is locally nilpotent, by setting for any f, g ∈ A

(1.2) f * g = i 0 ∆ i (f ) Γ i (g) = f g + ∆(f )Γ(g) + i 2 ∆ i (f ) Γ i (g), where Γ i = 1 i! Γ • (Γ -id) • • • • • (Γ -(i -1
)id), we define an associative product on A, see [START_REF] Coll | An explicit deformation formula with noncommuting derivations, Ring theory 1989[END_REF] for the formal version and [Pym15, Lemma 3.3] for the algebraic version. We set

• A = A (∆,Γ) = (A, {-, -}), • R = R (∆,Γ) = (A, * ).
The noncommutative algebra R is a deformation of the Poisson algebra A in the following sense. Consider A as an algebra over k[t] and define

f * t g = i 0 t i ∆ i (f ) Γ i (g).
Then A t = (A, * t ) is an associative and noncommutative algebra over k[t] such that Example 1.3. For A = k[X 0 , . . . , X n ] and the derivations ∆(X i ) = X i-1 (X -1 = 0) and Γ(X i ) = (a + i)X i (a ∈ k), the (Poisson) algebras obtained by the formulae (1.1) and (1.2) have been studied in [START_REF] Center | A new family of Poisson algebras and their deformations[END_REF]. The Poisson algebra is denoted by A(n, a) and the noncommutative algebra R(n, a). In this article, our aim is to generalize results of [START_REF] Center | A new family of Poisson algebras and their deformations[END_REF] to other pairs of derivations.

• A t /(t -1)A t ∼ = R, • A t /

Solvable pairs

2.1. Generalities. A pair (∆, Γ) of derivations of A such that [∆, Γ] = ∆ is locally nilpotent will be called a solvable pair. Since [∆, Γ] = ∆ we observe thanks to equations (1.1) and (1.2) that

Γ = 0 =⇒ ∆ = 0 =⇒ A (∆,Γ) is Poisson commutative, R (∆,Γ) is commutative.
The aim of this section is to study general facts about solvable pairs. The following result study the influence of a change of generators of A on the algebras A (∆,Γ) and R (∆,Γ) .

Lemma 2.1. Let (∆, Γ) be a solvable pair on A and σ and automorphism of the algebra A. Then (i) ( ∆, Γ) := (σ∆σ -1 , σΓσ -1 ) is a solvable pair on A, (ii) the Poisson algebras A (∆,Γ) and A ( ∆, Γ) are isomorphic, (iii) the algebras R (∆,Γ) and R ( ∆, Γ) are isomorphic.

Proof. (i) is a straightforward computation.

(ii) and (iii). In both case we show that σ is an isomorphism between the appropriate algebras. Since σ∆ = ∆σ and σΓ = Γσ we have for all f, g ∈ A σ({f, g} (∆,Γ) ) = σ∆(f )σΓ(g) -σ∆(g)σΓ(f ) = ∆σ(f ) Γσ(g) -∆σ(g) Γσ(f ) = {σ(f ), σ(g)} ( ∆, Γ) .

By induction we obtain easily that σ∆ = ∆ σ and σ Γ = Γ σ for all ∈ N. The equality σ(f * (∆,Γ) g) = σ(f ) * ( ∆, Γ) σ(g) follows from the definition of the product * given by (1.2).

Remark 2.2. When ∆ = 0 the Poisson algebra A is Poisson commutative and the algebra R is commutative for every choice of derivation Γ. Therefore there is no hope in general that a Poisson isomorphism between A (∆,Γ) and A (∆ ,Γ ) implies that there exists σ ∈ Aut(A) such that (∆ , Γ ) = (σ∆σ -1 , σΓσ -1 ) (the same remark also applies to R (∆,Γ) ). In other words, the Poisson isomorphism class of A (∆,Γ) or the k-algebra isomorphism class of R (∆,Γ) does not determine the isomorphism class of the representation of the 2-dimensional solvable Lie algebra. For a less trivial example showing that even the conjugacy class of ∆ is not determined, see Remark 2.13.

We conclude this section with some examples.

Proposition 2.3. Let A = k[X 0 , X 1 ]. Up to isomorphism any solvable pair (∆, Γ) on A with ∆ = 0 is of the form ∆ = P ∂ X1 , Γ = Q∂ X0 + R∂ X1 where P, Q ∈ k[X 0 ], P = 0 and R = R 0 + R 1 X 1 for some R 0 , R 1 ∈ k[X 0 ] are such that (1) if P ∈ k × or Q = 0, then R 1 = 1 (2) if deg X0 (P ) > 0 and Q = 0, then R 1 = 1 and we have P (R 1 -1) = Q∂ X0 (P ). In particular if deg X0 (P ) > 0, Q = 0 and R 1 ∈ k \ {1}, then there exists Λ ∈ k × , α ∈ k and an integer n > 0 such that Q = R1-1 n (X 0 -α) and P = Λ(X 0 -α) n .
Proof. Let (δ, γ) be a solvable pair on A. Thanks to Rentschler's theorem, see [START_REF] Freudenburg | Algebraic theory of locally nilpotent derivations[END_REF]Theorem 4.1], there exists α ∈ Aut(A) such that ∆ := αγα -1 = P ∂ X1 for some nonzero polynomial

P ∈ k[X 0 ]. Set Γ := αδα -1 = Q∂ X0 + R∂ X1 for some Q, R ∈ A. Then [∆, Γ] = ∆ implies that 0 = ∆(X 0 ) = ∆Γ(X 0 ) -Γ∆(X 0 ) = ∆(Q) so that Q ∈ ker ∆ = k[X 0
] since P = 0. Moreover we have

P = ∆(X 1 ) = ∆Γ(X 1 ) -Γ∆(X 1 ) = ∆(R) -Γ(P ) = P ∂ X1 (R) -Q∂ X0 (P )
so that P (∂ X1 (R) -1) = Q∂ X0 (P ). Since P = 0 two cases can happened

(1) (Q = 0 or P ∈ k × ) ⇐⇒ ∂ X1 (R) = 1 ⇐⇒ R = X 1 + R 0 with R 0 ∈ k[X 0 ] (2) (Q = 0 and deg X0 P > 0) ⇐⇒ ∂ X1 (R) = 1.
In the second case, every factors in P (∂ X1 (R) -1) = Q∂ X0 (P ) is nonzero so that the degree in

X 1 of ∂ X1 (R)-1 is equal to 0. Hence R = R 0 +R 1 X 1 with R 0 , R 1 ∈ k[X 0 ] and R 1 = 1. When R 1 = µ ∈ k\{1}
we have (µ -1)P = (-Q)P which classically implies that deg X0 (Q) = 1 and thus Q = λ(X 0 -α) for some λ ∈ k × and some α ∈ k and P = Λ(X 0 -α) n for some Λ ∈ k × and some integer n and the relation (µ -1)P = (-Q)P gives λ = (µ -1)/n.

Corollary 2.4. Every Poisson structure on A = k[X 0 , X 1 ] arising from a solvable pair is, up to isomorphism, of the form {X 0 , X 1 } = S, where S ∈ k[X 0 ]. The deformed algebra (R, * ) is given by two generators X 0 , X 1 and one relation

X 0 * X 1 -X 1 * X 0 = S.
In particular, the Poisson field Frac(A) is either Poisson commutative or a Poisson Weyl field. Similarly, the skewfield Frac(R) is either commutative or is a Weyl skewfield.

Proof. We have the result for A with S = -P Q thanks to Proposition 2.3 and by setting, if S = 0, X 1 = X 1 * S -1 so that {X 0 , X 1 } = 1. Note that the powers X * i 1 for i 0 form a basis of R = (A, * ) over k[X 0 ] so that R is isomorphic to the Ore extension k[X 0 ][X 1 ; S∂ X0 ] (see Proposition 2.32 for a proof in a more general situation). In particular R is a Noetherian domain, hence admits a skewfield of fraction Frac(R). If S = 0, set X 1 = X 1 * S -1 so that X 0 * X 1 -X 1 * X 0 = 1 and R is isomorphic to a Weyl skewfield.

Example 2.5. The Poisson structure on A = k[X 0 , X 1 ] given by {X 0 , X 1 } = X 0 X 1 cannot be obtained by a solvable pair. It follows from Corollary 2.4 since Frac(A) (resp. Frac(R)) cannot contains any Poisson bracket (resp. commutator) equal to 1, see [START_REF] Goodearl | The Dixmier-Moeglin equivalence and a Gel'fand-Kirillov problem for Poisson polynomial algebras[END_REF] (resp. [START_REF] Alev | Sur le corps des fractions de certaines algèbres quantiques[END_REF]).

Example 2.6. Let A = k[X 0 , X 1 ] and set (∆, Γ) = (∂ X0 , X 0 ∂ X0 + X i 1 ∂ X1 ) for some integer i ∈ N. One easily verify that (∆, Γ) is a solvable pair on A and that

{X 0 , X 1 } = X i 1 and X 0 * X 1 -X 1 * X 0 = X i 1 = X * i 1
where X * i 1 denote the i th power of X 1 with respect to the product * . We retrieve the following classical (Poisson) algebras: i = 0 the (Poisson) Weyl algebra, i = 1 the (symmetric) enveloping algebra of the two dimensional non abelian Lie algebra, i = 2 the (Poisson) Jordan plane.

Note that the (Poisson) Weyl algebra is classically obtained via the commuting pair of derivations

(∂ X0 , ∂ X1 ). Example 2.7. Let A = k[X, Y 1 , . . . , Y n ] and set ∆ = ∂ X and Γ = X∂ X + λ 1 Y 1 ∂ Y1 + • • • + λ n Y n ∂ Yn for some scalars λ 1 , . . . , λ n . Then (∆, Γ) is a solvable pair on A and A (∆,Γ) (resp. R (∆,Γ)
) is isomorphic to the symmetric (resp. enveloping) algebra of the Lie algebra g with basis {x, y 1 , . . . , y n } and non zero Lie bracket [x, y i ] = λ i y i , i = 1, . . . , n.

Center.

In this section, we study the elementary and general properties of solvable pairs related to the center of A (∆,Γ) and R (∆,Γ) .

Definition 2.8. Let A be a Poisson algebra. The Poisson center of A is the set

Z P (A) = {z ∈ A | {z, a} = 0 for all a ∈ A} of Poisson central elements. It is a Poisson-commutative subalgebra of A.
Lemma 2.9. Assume that ∆ = 0 and that A = A (∆,Γ) is a domain. Set R = R (∆,Γ) .

(1) Then ker ∆ ∩ ker Γ ⊆ Z P (A) with equality if ker ∆ = ker Γ.

(2) We have ker ∆ ∩ ker Γ ⊆ Z(R). Moreover, Z(R) ∩ ker ∆ ⊆ ker Γ.

(3) Assume that ker ∆ = ker Γ. Then Z(R) = ker ∆ ∩ ker Γ.

Proof.

(1) The inclusion ker ∆ ∩ ker Γ ⊆ Z P (A) is clear. Let z ∈ Z P (A). First assume that there exists x ∈ ker ∆ \ ker Γ. Then 0 = {z, x} = ∆(z)Γ(x), which implies z ∈ ker ∆. For y / ∈ ker ∆ we have 0 = {y, z} = ∆(y)Γ(z) and so z ∈ ker Γ. Now assume that there exists x ∈ ker Γ \ ker ∆. Then 0 = {x, z} = ∆(x)Γ(z) so that z ∈ ker Γ. Since ker Γ = A (otherwise Γ = 0 which implies ∆ = 0) we have for any y / ∈ ker Γ that 0 = {z, y} = ∆(z)Γ(y) and so z ∈ ker ∆.

(2) The inclusion ker ∆ ∩ ker Γ ⊆ Z(R) is clear. Let us now consider z ∈ Z(R) ∩ ker ∆. Since ∆ = 0 and ∆ is locally nilpotent, there exists x ∈ ker ∆ 2 \ ker ∆. We then have z * x = zx since z ∈ ker ∆ and

x * z = zx + ∆(x)Γ(z) since x ∈ ker ∆ 2 . But z ∈ Z(R) hence ∆(x)Γ(z) = 0 and ∆(x) = 0. Thus Γ(z) = 0.
(3) We first show that ker ∆ ⊆ ker Γ. If ker ∆ ⊆ ker Γ then ker ∆ ker Γ since ker ∆ = ker Γ. Consider x ∈ ker Γ \ ker ∆ and n ≥ 0 such that ∆ n (x) = 0 and ∆ n+1 (x) = 0. In particular ∆ n (x) ∈ ker ∆ ⊆ ker Γ. Thus [∆ n , Γ](x) = n∆ n (x) (see Lemma 2.20) implies ∆ n (Γ(x)) = n∆ n (x). But x ∈ ker Γ thus n∆ n (x) = 0 thus n = 0 and ∆(x) = 0 which is absurd. Hence we can consider x ∈ ker ∆ such that Γ(x) = 0 and choose z ∈ Z(R).

Then x * z = xz since x ∈ ker ∆. Moreover (2.10) 0 = z * x -x * z = i≥1 ∆ i (z) Γ i (x)
But [∆, Γ] = ∆ implies that ker ∆ is stable by Γ, hence stable by Γ i for all i ≥ 1 (see Lemma A.6). Therefore we have Γ i (x) ∈ ker ∆ for all i ≥ 1. Let N ≥ 1 be the smallest positive integer such that ∆ N (z) = 0. Assume that N ≥ 2. By applying ∆ N -2 to equation (2.10) we obtain that Γ(x)∆ N -1 (z) = 0. Since Γ(x) = 0, we get a contradiction. Hence N = 1 i.e. z ∈ ker ∆ and assertion (2) provides us with z ∈ ker Γ.

Example 2.11. Recall the derivations ∆ = ∂ X0 and Γ = X 0 ∂ X0 + X i 1 ∂ X1 from Example 2.6. One easily verify that ker ∆ = k[X 1 ] and ker Γ = k so that Z

P (A) = Z(R) = k. Example 2.12. Recall the derivations ∆ = ∂ X and Γ = X∂ X + λ 1 Y 1 ∂ Y1 + • • • + λ n Y n ∂ Yn from Exam- ple 2.7. Then ker ∆ = k[Y 1 , . . . , Y n ] =
ker Γ (as long as not all the λ i are zeros) so that Z P (A) = Z(R) = ker ∆ ∩ ker Γ (take x = Y i for λ i = 0 in the proposition). Note that to understand

ker ∆ ∩ ker Γ = ker Γ| k[Y1,...,Yn] = Vect{Y α1 1 • • • Y αn n | λ 1 α 1 + • • • + λ n α n = 0, α i ∈ N}
one needs to study the structure of the submonoid

n i=1 λ i N of (k, +) (or the Z-module n i=1 λ i Z if we allow for localization of the Y i 's). For instance if the λ i are N-linearly independent then Z P (A) = Z(R) = k.
The following example illustrate the fact that when ker ∆ = ker Γ the inclusion ker ∆ ∩ ker Γ ⊆ Z P (A) can be strict.

Example 2.13. Let A = k[X 0 , . . . , X n ] and consider the solvable pair (∆ (2) The set P.Der(A) of Poisson derivations of A is a Lie subalgebra of the Lie algebra of derivations of the associative commutative algebra A.

, Γ) = (X 0 ∂ X1 , X 1 ∂ X1 ). We have A = Z P (A) but ker ∆ = ker Γ = k[X 0 , X 2 , X 3 , . . . , X n ] A (see also Corollary 4.9). Similarly we have Z(R) = R so Z(R) ⊆ ker ∆ = ker Γ.
Let A = A (∆,Γ) . When ∆ = 0, the Lie algebra P.Der(A) is not reduced to 0 thanks to the following easy lemma.

Lemma 2.15. The derivation ∆ is a Poisson derivation of A.

Remark 2.16. Poisson center and Poisson derivation are compatible in the following sense. Let A be a Poisson algebra. The Poisson center of A is stable by any Poisson derivation of A: for δ ∈ P.Der(A) and a ∈ Z P (A), an easy computation shows that δ(a) ∈ Z P (A).

The following proposition can be seen as a deformation formula for the Poisson derivation ∆. Note that the map ∆ is not a derivation of R (∆,Γ) in general.

Theorem 2.17. Let (∆, Γ) be a solvable pair. For a ∈ k, the map

φ a := (id + ∆) a = k≥0 a k ∆ k is a (graded if ∆ is graded) k-algebra automorphism of R = R (∆,Γ) . The linear map δ = r 1 (-1) r-1 r ∆ r is a (graded if ∆ is graded) derivation of R.
Strictly speaking δ is an element of the formal power series ring k[[∆]] but note that since ∆ is locally nilpotent δ(f ) is a well-defined element of R for any f ∈ R.

Proof. We first prove that id + ∆ is an automorphism of R. [START_REF] Vidal | Automorphismes et dérivations de "petites normes" sur un corps valué non archimédien[END_REF] allows us to conclude that δ is a derivation of R. Hence aδ is a locally nilpotent derivation of R. So exp(aδ) = (id + ∆) a is an automorphism of R.

(id + ∆)(f ) * (id + ∆)(g) = i≥0 (∆ i (f ) + ∆ i+1 (f ))( Γ i (g) + Γ i ∆(g)) = f * g + i≥0 ∆ i+1 (f ) Γ i (g) + ∆ i (f ) Γ i ∆(g) + ∆ i+1 (f ) Γ i ∆(g) (A.6) = f * g + i≥0 ∆ i+1 (f ) Γ i (g) + f * ∆(g) + i≥0 ∆ i+1 (f ) ∆, Γ i+1 (g) = f * g + i≥0 ∆ i+1 (f ) Γ i (g) + f * ∆(g) + i≥0 ∆ i (f ) ∆, Γ i (g) = f * g + i≥0 ∆ i+1 (f ) Γ i (g) + i≥0 ∆ i (f )∆ Γ i (g) = f * g + ∆(f * g) Proposition 2 of
This last property can be shown by a direct computation using the combinatorial relation

u i=0 a v+i v+i i v u-i = u i=0 a(a-1)•••(a-v-i+1) i!(v-u+i)!(u-i)! = a v u i=0 (a-v)•••(a-v-i+1)v! i!(v-u+i)!(u-i)! = a v u i=0 a-v i v u-i = a v a u
where the last equality is the Chu-Vandermonde identity (see Appendix A).

Lemma 2.18. For any a, b ∈ k we have φ a • φ b = φ a+b . Moreover if ∆ = 0, φ a = id if and only if a = 0.

Proof. Let f ∈ R. By using equation (A.4) we have

φ a • φ b (f ) = k, ≥0 a k b ∆ k+ (f ) = u≥0 k+ =u a k b ∆ u (f ) = u≥0 a + b u ∆ u (f ) = φ a+b (f ).
When ∆ = 0, there exists f ∈ R such that ∆ 2 (f ) = 0 and ∆(f ) = 0. Then φ a (f ) = f + a∆(f ).

A filtration on R (∆,Γ)

. It is a classical fact that locally nilpotent derivations induce filtrations on commutative algebras. In this section we show that the sequence (ker ∆ i+1 ) i∈N provides us with a filtration of both the Poisson algebra A (∆,Γ) and the noncommutative algebra R (∆,Γ) .

Definition 2.19. Let ∆ be a locally nilpotent derivation on A. Set ε(0) = -∞ and for any nonzero

element f ∈ A define ε(f ) = min{i ∈ N, ∆ i+1 (f ) = 0} and set A i = {f ∈ A, ε(f ) i} = ker ∆ i+1 . By convention set A -1 = {0}.
Note that ε = deg ∆ in the notation of [Fr06, Section 1.1.8] which is a degree function thanks to [No94, Proposition 6.1.1] since ∆ is locally nilpotent.

Lemma 2.20. For any polynomial P ∈ k[T ] we have [P (∆), Γ] = Q(∆) where Q = XP . In particular, for any integer i 0 we have

∆ i Γ = Γ∆ i + i∆ i .
Proof. By linearity, it is enough to prove the relation for P = X i where i ∈ N which is easily obtained by induction.

Lemma 2.21. Let (∆, Γ) be a solvable pair on A and fix integers i, j 0.

(1) We have ∆(A i ) ⊆ A i-1 and more precisely ε(∆(f

)) = ε(f ) -1 for every f ∈ A. (2) We have Γ(A i ) ⊆ A i . Hence ε(P (Γ)(f )) ε(f ) for every f ∈ A and P ∈ k[T ]. (3) If f, g ∈ A are such that ε(f ) = i and ε(g) = j, then ε(f g) = i + j. (4) For any f, g ∈ A we have f * g -g * f = ∆(f )Γ(g) -∆(g)Γ(f ) + i 2 ∆ i (f ) Γ i (g) -∆ i (g) Γ i (f ) In particular, if ε(f ) = i and ε(g) = j then ε(f * g -g * f ) i + j -1.
Proof. Assertion (1) is obvious since ∆(f ) ∈ ker ∆ i-1 if and only if f ∈ ker ∆ i . Assertion (2) follows from Lemma 2.20 showing that ker ∆ i is stable by Γ and hence by every polynomial in Γ. Assertion (3) is true since from Leibniz formula we get ∆ i+j (f g) = i+j i ∆ i (f )∆ j (g) = 0 because A is a domain and char k = 0. Assertion (4) follows from (1.2) and then assertions (1) and (2). Recall that if (T i ) i∈N is a filtration of a ring T , its associated graded ring gr(T ) =

T i /T i-1 is an N-graded ring whose homogeneous elements of degree i are denoted by x + T i-1 for an element x ∈ T i \ T i-1 . Proposition 2.22. Assume that A is a domain.

(1) The family (A i ) i 0 is a Poisson algebra filtration of A (∆,Γ) of degree -1, meaning that it is an algebra filtration of A together with {A i , A j } ⊆ A i+j-1 for all integers i, j 0. Moreover, the associated graded algebra gr(A) is a domain.

(2) The family (A i ) i 0 is an algebra filtration of R (∆,Γ) . Moreover, the associated graded algebra gr(R) is equal to gr(A).

(3) The commutative algebra gr(R) = gr(A) can be endowed with the following three Poisson brackets (a) {f , g}

:= (f * g -g * f ) + A i+j-2 (b) {f , g} := {f, g} + A i+j-2 (c) {f , g} := ∆(f )Γ(g) -Γ(f )∆(g) ∈ A i+j-1 /A i+j-2
for homogeneous elements f and g of respective ε-degree i and j. In (c) the pair of maps (∆, Γ) is the solvable pair of homogeneous derivations of gr(A) of respective degree -1 and 0 which is induced by the solvable pair of filtered derivations ∆ and Γ of A. (4) The Poisson structures defined in (3) are all equal and make gr(A) = gr(R) into a graded Poisson algebra of degree -1.

Proof.

(1). The fact that (A i ) i 0 is an algebra filtration of A is an easy consequence of Leibniz formula, see [START_REF] Freudenburg | Algebraic theory of locally nilpotent derivations[END_REF] (2). The fact that (A i ) i 0 is an algebra filtration of R (∆,Γ) follows from equation (1.2) since the degree ε is decreased by ∆ and preserved by Γ thanks to Lemma 2.21. More precisely, if f, g ∈ A are such that ε(f ) = i and ε(g) = j, then f * g = f g + u for some u ∈ A i+j-1 . This also implies that the associated graded algebras gr(A) and gr(R) have the same multiplication, so that they are indeed equal.

(3). The fact that {-, -} is a Poisson bracket on gr(R) follows the filtered version of the semiclassical limit construction, see [START_REF] Goodearl | Semiclassical limits of quantized coordinate rings[END_REF]Section 2.4]. The fact that {-, -} is a well-defined biderivation satisfying the Jacobi identity on gr(A) follows by tedious but straightforward computation from the fact that {-, -} is a filtered Poisson bracket on A. Finally {-, -} is a Poisson bracket since (∆, Γ) is a solvable pair of derivations of gr(A).

(4). The Poisson brackets {-, -} and {-, -} are the same since both f * g -g * f and {f, g} belong to A i+j-1 combined with f * g -g * f -{f, g} ∈ A i+j-2 thanks to assertion 4 of Lemma 2.21. Finally, for homogeneous elements f and g of ε-degree i and j we have

{f , g} = ∆(f )Γ(g) -Γ(f )∆(g) = (∆(f ) + A i-2 )(Γ(g) + A i-1 ) -(∆(g) + A i-2 )(Γ(f ) + A i-1 ) = (∆(f )Γ(g) + A i+j-2 ) -(∆(g)Γ(f ) + A i+j-2 ) = (∆(f )Γ(g) -∆(g)Γ(f )) + A i+j-2 = {f, g} + A i+j-2 = {f , g}
Hence {-, -} and {-, -} are the same.

In fact it is enough to assume that ker ∆ is a domain to obtain the previous result. More precisely we have the following corollary.

Corollary 2.23. The following assertions are equivalent:

(1) ker ∆ is a domain, (2) A is a domain, (3) gr(A) = gr(R) is a domain, (4) R is a domain.

Proof.

(2) =⇒ (3) is part of assertion (1) of Proposition 2.22.

(3) =⇒ (4) is classical.

(4) =⇒ (1) since ker ∆ is a subring of R.

(1) =⇒ (2). With the notation of the proof of assertion (3) of Lemma 2.21, we have ∆ i (f ) ∈ ker ∆ and ∆ j (g) ∈ ker ∆. So i+j i ∆ i (f )∆ j (g) = 0 implies that ∆ i+j (f g) = 0, hence f g = 0.

Example 2.24.

Consider ∆ = X 0 ∂ X1 and Γ = aX 0 ∂ X0 + (a + 1)X 1 ∂ X1 on A = k[X 0 , X 1 ]. We have A i = ker ∆ i+1 = i k[X 0 ]X 1 . Then gr(A) = i 0 A i /A i-1 ∼ = i 0 k[X 0 ]X i 1 = A. Observe that X 0 ∈ A 0 and X 1 ∈ A 1 and X 0 * X 1 -X 1 * X 0 = -aX 0 * X 0 ∈ A 0
so that the induced Poisson bracket on gr(A) is given by

{X 0 + A -1 , X 1 + A 0 } = -aX 0 * X 0 + A -1 .
To lighten notation in the following example, we will denote by gr(f

) = f + A i-1 the image in gr(A) of any element f ∈ A with (f ) = i ∈ N. Example 2.25. Consider the derivation ∆ = X 0 ∂ X1 + X 1 ∂ X2 of A = k[X 0 , X 1 , X 2 ]. It is classical that A 0 = ker ∆ = k[X 0 , F 2 ] where F 2 = 2X 0 X 2 -X 2 1 , see for instance [No94, Example 6.7.1.]. Recall that ε(f g) = ε(f ) + ε(g) and that ε(X k ) = k for k ∈ {0, 1, 2}. Therefore, for all i 1, we have ε(X 1 X 2 i-1 ) = 2i -1 and ε(X 2 i ) = 2i. Hence the family B = (1, X 1 X 2 i-1 , X 2 2i
) i 1 is free over A 0 . Moreover observe that the Hilbert series of

A 0 i 1 (A 0 X 1 X 2 i-1 ⊕ A 0 X 2 i ) is the same that the one of k[X 0 , X 1 , X 2 ]. Hence B is a basis of k[X 0 , X 1 , X 2 ] over A 0 . which implies that gr(A) = A 0 ⊕ i 1 A 0 gr(X 1 X 2 i-1 )⊕ A 0 gr(X 2 i ).
Observe that gr(X 2 ) ∈ gr(A) generates a polynomial ring over A 0 and that we have the

relation X 2 1 = 2X 0 X 2 -F 2 . Therefore gr(A) ∼ = k[X 0 , F 2 , X 2 ][T ]/(T 2 -(2X 0 X 2 -F 2 )) ∼ = k[X 0 , X 2 , T ] ∼ = A since T 2 -(2X 0 X 2 -F 2 ) is a degree one polynomial in F 2 .
2.5. Normal elements. The filtration introduced in the previous section allow us to study Poisson automorphisms and Poisson derivations of gr(R) = gr(A) induced by (strongly) normal elements of R. This will be useful in Section 8. Recall that if σ and τ are endomorphisms of a ring T a (σ, τ )-derivation of T is an endomorphism δ of T such that δ(rs) = σ(r)δ(s) + δ(r)τ (s) for all r, s ∈ T .

Lemma 2.26. Let (Γ, ∆) be a solvable pair on A and consider an automorphism

Φ of R = (A, * ) that is compatible with the filtration ε, i.e. such that ε(Φ(f )) = ε(f ) for all f ∈ R. Then Φ induces a Poisson automorphism Φ of gr(R) = gr(A) given by Φ(f + A i ) = Φ(f ) + A i for all f ∈ R with ε(f ) = i. Moreover if Φ = id, then Φ -id of R induces a Poisson derivation of gr(R) = gr(A) of degree -1. Proof. Since ε(Φ(f )) = ε(f ) for all f ∈ R it is clear that Φ is a well-defined automorphism of gr(R).
Consider f, g ∈ R such that ε(f ) = i and ε(g) = j and denote by f , g their images in gr(R). We have

Φ({f , g}) = Φ(f * g -g * f + A i+j-2 ) = Φ(f ) * Φ(g) -Φ(g) * Φ(f ) + A i+j-2 = {Φ(f ), Φ(g)} hence Φ

is a Poisson automorphism of gr(A).

The equality Φ = id means that if ε(f

) = i then ε(Φ(f )-f ) i-1. Hence Φ-id induces a well-defined vector space endomorphism Φ -id of gr(R) of degree -1 by setting (Φ -id)(f ) = Φ(f ) -f + A i-2 for all f ∈ gr(R) where f ∈ R is such that ε(f ) = i. By definition we have (Φ -id)({f , g}) = (Φ -id)(f * g -g * f ) + A i+j-3 . Since Φ -id is both a (Φ, id)-derivation and a (id, Φ)-derivation of R we obtain (Φ -id)(f * g -g * f ) = Φ(f ) * (Φ(g) -g) -(Φ(g) -g) * Φ(f ) + (Φ(f ) -f ) * g -g * (Φ(f ) -f ) . Hence (Φ -id)({f , g}) = {Φ(f ), (Φ -id)(g)} + {(Φ -id)(f )
, g} and we get the result using the fact that Φ = id. Definition 2.27. Let T be a k-algebra that is a domain and let N ∈ T be a normal element. For every u ∈ T there exists a unique element Φ(u) ∈ T such that uN = N Φ(u). Then Φ is a k-algebra automorphism of T and is called the automorphism associated to N and denoted by Φ N .

Lemma 2.28. Let (Γ, ∆) be a solvable pair on a domain A and consider R = (A, * ). Let N be a normal element in R and consider Φ N its associated automorphism. We denote by N the image of N in gr(R).

Then Φ N is compatible with ε. Moreover Φ N = id and (Φ N -id) verifies that N (Φ N -id) = { • , N }. In particular, N is Poisson normal in gr(R). Proof. Let f ∈ R and set i = ε(f ) and j = ε(N ). Since f * N = N * Φ N (f ) and A is a domain, we have i + j = ε(Φ N (f )) + j. Therefore Φ N is compatible with ε.
The definition of the product

* implies that f * N = f N +u where u ∈ A i+j-1 . Similarly N * Φ N (f ) = N Φ N (f ) + v where v ∈ A i+j-1 . Hence N (Φ N (f ) -f ) ∈ A i+j-1 . Since A is a domain, we obtain that Φ N (f ) -f ∈ A i-1 i.e. Φ N = id. Moreover, we have the following equalities 0 = N * Φ N (f ) -f * N = N (Φ N (f ) -f ) + ∆(N )Γ(Φ N (f )) -∆(f )Γ(N ) + u for some u ∈ A i+j-2 Hence N (Φ N (f ) -f ) = ∆(f )Γ(N ) -∆(N )Γ(Φ N (f )) -u with ∆(f )Γ(Φ N (f )) - ∆(N )Γ(f ) ∈ A i+j-1 since Φ N is compatible with ε. This implies the desired equality since Φ N = id.
Definition 2.29 (Strongly normal element). Let (∆, Γ) be a solvable pair on A. An element f ∈ A is said to be strongly normal when ∆(f ) = 0 and there exists α ∈ k such that Γ(f ) = αf .

Lemma 2.30. Let (∆, Γ) be a solvable pair on A. If f is strongly normal in A then f is normal in R and Poisson normal in A. Moreover if A is a domain, the automorphism of R associated to N is φ α and (φ α -id) = α∆, where α is the scalar given by Γ

(f ) = αf . Proof. Since f ∈ ker ∆, we have f * g = f g for all g ∈ R, hence f * R = f A. Also we have g * f = k≥0 ∆ k (g) Γ k (f ) = k≥0 ∆ k (g) α k f = f k≥0 α k ∆ k (g) = f φ α (g) .
Recall from Theorem 2.17 that φ α is an automorphism of R. Hence the equality g

* f = f Φ α (g) implies that R * f = Af . Therefore we have f * R = f A = R * f and f is normal in R. Moreover, for every g ∈ A, we have {f, g} = ∆(f )Γ(g) -Γ(f )∆(g) = -αf ∆(g) ∈ f A and we conclude that f is Poisson normal. Assume that A is a domain. Since f ∈ ker ∆ and Γ(f ) = αf , we have ∆(f ) = 0 and Γ(f ) = αf . Hence {g, f } = α∆(g)f . Thanks to Lemma 2.28 we have {g, f } = (Φ α -id)(g)f . Since gr(A) is a domain we obtain the relation (Φ α -id) = α∆.
The converse of the preceding lemma will be proved in section 5 : in the case of a linear solvable pair, every normal element in R or Poisson normal element in A is strongly normal (Proposition 5.7).

2.6.

A canonical form up to localization. It is well know that locally nilpotent derivations with slices feature very interesting properties. For instance, if A is a k-algebra that is a domain and ∆ is a nilpotent derivation of A admitting an element s ∈ A such that ∆(s) = 1 (s is called a slice for ∆), then A is a polynomial ring in the variable s over the base ring A ∆ = ker ∆, see [START_REF] Deveney | Finston Field of Ga invariants are ruled[END_REF] and [START_REF] Van Den | Essen Locally nilpotent derivations and their applications, III[END_REF]. In this section we exploit this fact to show that, up to localization, the algebras A (∆,Γ) and R (∆,Γ) are (Poisson-) Ore extensions of the (localized) kernel of ∆ by the (Poisson) derivation Γ.

Let (∆, Γ) be a solvable pair on A with ∆ = 0. Since ∆ = 0 there exists r ∈ A such that ∆(r) = 0 and ∆ 2 (r) = 0 (any element of A 1 \ A 0 ). We consider the localization A

• = S -1 r A where S r = {∆(r) i | i 0}.

The pair of derivations (∆, Γ) extends uniquely by localization into a solvable pair of derivations on A • again denoted by (∆, Γ). Note that ∆ is locally nilpotent on

A • since ∆(r) ∈ A ∆ . Observe that the element s := r∆(r) -1 ∈ A • is a slice for ∆ since: ∆(s) = ∆(r)∆(r) -1 -r∆(r) -2 ∆ 2 (r) = 1.
2.6.1. The Poisson case. We consider the Poisson algebra (A • ) (∆,Γ) arising from the solvable pair (∆, Γ) on A • . Note that the kernel (A • ) ∆ is a Poisson commutative subalgebra of A • that is stable by Γ since for any a ∈ (A • ) ∆ we have ∆Γ(a) = Γ∆(a) + ∆(a) = 0. Hence Γ induces a Poisson derivation of (A • ) ∆ . This allows us to define the so-called Poisson-Ore extension (A • ) ∆ [X; Γ] P which is equal to the polynomial ring (A • ) ∆ [X] as a commutative algebra and whose Poisson bracket is defined to be the unique extension of the Poisson bracket of (A

• ) ∆ such that {X, a} = Γ(a) for all a ∈ (A • ) ∆ , see [Oh06]. Proposition 2.31. The map φ : (A • ) ∆ [X; Γ] P → (A • ) (∆,Γ) given by φ n i=0 a i X i = n i=0 a i s i where a i ∈ (A • ) ∆ for all i, is a Poisson algebra isomorphism. Proof. The inclusion (A • ) ∆ → A • is a Poisson algebras homomorphism. Moreover for a ∈ (A • ) ∆ , we have {s, a} = ∆(s)Γ(a) -Γ(s)∆(a) = Γ(a) since ∆(s) = 1. Thanks to the universal property of the Poisson-Ore extension ((A • ) ∆ )[X; Γ] P (see [Lec15, Proposition 1.1.15]), there exists a unique Poisson algebra homomorphism from ((A • ) ∆ )[X; Γ] P to (A • ) (∆,Γ) sending a ∈ (A • ) ∆ to a and X to s. It is given by φ. Since A • = (A • ) ∆ [s] is a polynomial ring in the slice s over (A • ) ∆ ([VdE95, Corollary 1.2]), it is clear that φ is an isomorphism.
2.6.2. The noncommutative case. We consider the algebra R

• = (R • ) (∆,Γ) = (A • , * ) arising from the solvable pair (∆, Γ) on A • . Recall that a * b = ab + i 1 ∆ i (a) Γ i (b) for all a, b ∈ A • . In particular (A • ) ∆ is a commutative subalgebra of R • . Since Γ leaves (A • ) ∆ invariant, we can consider the Ore extension (A • ) ∆ [X; Γ] which is, by definition [GW04,
Chapter 2], a free left (A • ) ∆ -module with basis {X i | i 0} and whose multiplication extends the multiplication of (A • ) ∆ by the rule Xa = aX + Γ(a) for all a ∈ (A • ) ∆ . For all integer i 0 let us denote by s * i the i th power of s with respect to the product * .

Proposition 2.32. The map φ :

(A • ) ∆ [X; Γ] → R • given by φ n i=0 a i X i = n i=0 a i s * i
where a i ∈ (A • ) ∆ for all i, is an algebra isomorphism.

Proof. The inclusion map (A • ) ∆ ⊆ R • is an algebra homomorphism. Moreover for a ∈ (A • ) ∆ , we have a * s = as and s * a = sa + Γ(a) since ∆(s) = 1 and ∆ 2 (s) = 0. Hence from the universal property (see

[GW04, Exercise 2.F]) of (A • ) ∆ [X; Γ] there exists a unique algebra homomorphism from (A • ) ∆ [X; Γ] to R • sending a ∈ (A • ) ∆ to a ∈ R • and X to s.
It is given by φ. So it remains to show that φ is bijective.

For, we will show by induction on i 0 that there exist elements a j,i ∈ (A • ) ∆ with 0 j < i such that

s * i = s i + i-1
j=0 a j,i s j . Cases i = 0 and i = 1 are trivial. Recall that ∆(s) = 1 and ∆ 2 (s) = 0 in order to compute s * (i+1) :

s * (i+1) = s * s * i =s * s i + s * i-1 j=0 a j,i s j = s i+1 + Γ(s i ) + i-1 j=0 a j,i s j+1 + i-1 j=0 Γ(a j,i s j ) =s i+1 + i(Γ(s) -s)s i-1 + is i + i j=1 a j-1,i s j + i-1 j=0 Γ(a j,i )s j + i-1 j=0 a j,i Γ(s j ) =s i+1 + i(Γ(s) -s)s i-1 + i j=0 b j,i s j + i-1 j=0 ja j,i (Γ(s) -s)s j-1 + i-1 j=0 ja j,i s j
where the b j,i for 0 j i are elements of (A • ) ∆ expressed in terms of i, a j-1,i and Γ(a j,i ). Moreover we have ∆(Γ(s) -s) = Γ∆(s) = Γ(1) = 0. Hence Γ(s) -s ∈ (A • ) ∆ and s * (i+1) has the desired form. We then deduce that (

s * i ) i 0 is a basis of R • over (A • ) ∆ since (s i ) i 0 is ([VdE95, Corollary 1.2]
) and we conclude that φ is bijective. 2.7. Skewfield of fractions. When A is a polynomial ring and the action of Γ is diagonal, the results of the previous section allow us to relate R • to a particular Lie algebra and to described the skewfield of fractions of R (∆,Γ) . Recall that, given a field F , the Weyl skewfield D 1 (F ) is the skewfield of fractions of the algebra generated over F by two elements u and v such that uv -vu = 1. Linear diagonalizable solvable pairs will be defined in Section 6.

Proposition 2.33. Assume that A is a polynomial ring and that (∆, Γ) is a nonzero linear diagonalizable solvable pair on A. Then R (∆,Γ) is, up to localization, isomorphic to the enveloping algebra of a completely solvable Lie algebra g. In particular, if g is algebraic and non abelian, the skewfield of fractions of R (∆,Γ) is isomorphic to the Weyl skewfield D 1 (F ) for a purely transcendental extension F of k.

Proof. Since ∆(X 1 ) = X 0 and ∆(X 0 ) = 0 we can apply the slice construction with the element s = X 1 X -1

0 of A • = A[X -1 0 ].
In particular, the map π : [START_REF] Van Den | Essen Locally nilpotent derivations and their applications, III[END_REF]. Therefore the algebra (A • ) ∆ is generated by elements Y ±1 0 , Y 2 , . . . , Y n where Y i denote the image of X i by π (note that π(X 0 ) = X 0 and π(X 1 ) = 0). Observe that Y 0 , Y 2 , . . . , Y n is a set of algebraically independent elements of A • since, for i 2, we can write

A • → A • given by π(a) = p 0 (-s) p p! ∆ p (a) is an algebra homomorphism such that (A • ) ∆ = π(A • ), see
Y i = π(X i ) = X i + F i where F i ∈ k[X ±1 0 , X 1 , . . . , X i-1 ].
Moreover each Y i is an eigenvector for Γ because one can verify that Γ and π commute thanks to the equality ∆Γ -Γ∆ = ∆.

Denote by g the completely solvable Lie algebra with basis {y 0 , y 2 , . . . , y n , x} and nonzero brackets [x, y i ] = λ i y i for all i, where λ i ∈ k is the eigenvalue of Y i (equivalently X i ) for Γ. The element y 0 is normal in U (g) and is its clear that (A

• ) ∆ [X; Γ] is isomorphic to the localization U (g)[y -1 0 ]. The first part of the proof is now complete since R • ∼ = (A • ) ∆ [X; Γ] thanks to Proposition 2.32.
The second assertion follows firstly from [START_REF] Mcconnell | Representations of solvable Lie algebras and the Gelfand-Kirillov conjecture[END_REF]: skewfields of fractions of enveloping algebras of algebraic completely solvable Lie algebras are isomorphic to Weyl skewfields. Moreover, the transcendence degree of the center of the enveloping skewfield of g is equal to the index of g whose value can be easily computed to be n -1 thanks to [START_REF]Ooms The Gelfand-Kirillov conjecture for semi-direct products of Lie algebras[END_REF]. The result follows.

Remark 2.34. With similar arguments (using [START_REF] Tauvel | Algèbres de Poisson et algèbres de Lie résolubles[END_REF]) the following Poisson version can be obtained: under the assumption of Proposition 2.33, the Poisson field of A (∆,Γ) is isomorphic to a Poisson Weyl field over a purely transcendental extension F of k, that is, a rational functions field F (U, V ) with {U, V } = 1.

Solvable pairs over polynomial rings

From this section included and until the end of this article we focus on the case of solvable pairs on polynomial algebras.

The rank of A (∆,Γ)

. We recall the notion of rank of an affine Poisson variety following [START_REF] Vanhaecke | Integrable systems in the realm of algebraic geometry[END_REF]. Let V = k n+1 be a Poisson variety, that is, the variety V whose coordinate ring A = O(k n+1 ) = k[X 0 , . . . , X n ] is endow with a Poisson structure. The Poisson matrix of A is the skew-symmetric matrix Π(A) = {X i , X j } 0 i,j n . For any p ∈ V the rank Rk p (A) of the Poisson bracket at p is the rank of the Poisson matrix Π(A) evaluated at p. Finally, the rank of A, denoted by Rk(A), is the maximum of the Rk p (A) for p ∈ V . Note that A is Poisson commutative if and only of Rk(A) = 0.

If (∆, Γ) is a solvable pair on A then we have

Π(A (∆,Γ) ) =       ∆(X 0 ) ∆(X 1 ) . . . ∆(X n )             Γ(X 0 ) Γ(X 1 ) . . . Γ(X n )       T -       Γ(X 0 ) Γ(X 1 ) . . . Γ(X n )             ∆(X 0 ) ∆(X 1 ) . . . ∆(X n )       T .
In particular the rank of A (∆,Γ) is at most 2.

∆ = ∆ 1 ⊗ id C and Γ = Γ 1 ⊗ id C .

Linear solvable pairs

The polynomial algebra A = k[X 0 , . . . , X n ] is a connected graded algebra once endowed with the grading d in which d(X k ) = 1 for every 0 k n. We have A = i 0 A i where, for all i 0, we denote by A i the vector space of homogeneous polynomials of degree i. A Poisson structure on A is called homogeneous if {A i , A j } ⊆ A i+j and in that case A is said to be an homogeneous polynomial Poisson algebra. We denote by P.Der gr (A) the Lie algebra of linear Poisson derivations of A, that is, Poisson derivations stabilizing A 1 . Note that linear derivations are in one to one correspondence with endomorphisms of the k-vector space A 1 or equivalently with square matrices of order n + 1 (acting on A 1 by left multiplication). Moreover, locally nilpotent derivations of A correspond to nilpotent endomorphisms/matrices. We now provide a couple of invariants for homogeneous polynomial Poisson algebras.

Proposition 4.1. Let A and B be two isomorphic homogeneous polynomial Poisson algebras.

(1) There exists a Lie algebra isomorphism between P.Der gr (A) and P.Der gr (B) that sends locally nilpotent derivations to locally nilpotent derivations.

(2) There exists a vector space isomorphism between A 1 ∩ Z P (A) and B 1 ∩ Z P (B).

Proof. Thanks to [LPV12, Proposition 8.8] a Poisson isomorphism between homogeneous polynomial

Poisson structures can always be realized by a linear isomorphism. We denote by ϕ : A → B such an isomorphism. For (1), the map Ψ : P.Der gr (A) → P.Der gr (B) given by Ψ = ϕδϕ -1 is the desired isomorphism and for (2), the restriction of ϕ to A 1 ∩ Z P (A) provides us with the desired isomorphism since ϕ(A 1 ) = B 1 and ϕ(Z P (A)) = Z P (B). Definition 4.2. A solvable pair (∆, Γ) is said to be linear if ∆ and Γ are linear derivation of A.

Note that a linear solvable pair is completely determined by its action on A 1 . Hence in the rest of the article, when considering a linear solvable pair we will focus on its action on A 1 . The article [START_REF] Center | A new family of Poisson algebras and their deformations[END_REF] consider the case where ∆ is the maximal Jordan block. In the following we will investigate algebras A (∆,Γ) and R (∆,Γ) arising from linear derivations ∆ with more than one Jordan block.

Remark 4.3. In a solvable pair, the derivation ∆ is locally nilpotent. It implies, when the solvable pair is linear, that ∆ is a nilpotent endomorphism of A 1 . We remark that when ∆ and Γ are linear derivations of A verifying [∆, Γ] = ∆ then ∆ is automatically locally nilpotent. The proof goes as follows. We have [P (∆), Γ] = (T P )[∆] for every polynomial P ∈ k[T ] by assertion (2) of Lemma 2.21. If P is the minimal polynomial of ∆ acting on the finite dimensional space A 1 , we obtain that T P (T ) is collinear to P . Hence P = T for some and ∆ acting on A 1 is nilpotent. Therefore ∆ is locally nilpotent on A and (∆, Γ) a linear solvable pair.

The following example illustrate the facts that not all homogeneous Poisson algebras of rank 2 can be obtain by a linear solvable pair.

Example 4.4. Let A = k[X, Y ] with {X, Y } = XY . As seen in Example 2.5, this Poisson bracket cannot be realized by any solvable pair, thus any linear solvable pair. Note that the linear case can be treated by a direct computation (see also 4.6) since XY is not collinear to the square of an element of A 1 .

So let (∆, Γ) be a linear solvable pair on A. Thanks to Lemma 2.1 one can always assume that ∆ is in canonical Jordan form, up to a linear automorphism of A. Recall that in canonical Jordan form, a nilpotent matrix is made of diagonal blocks of the form J ni (0), i.e. blocks of size n i with 0s everywhere except on the upper diagonal, where the entries are 1. It can be assumed that the size of the diagonal blocks J ni (0) is decreasing. So, given such a ∆, what are the possible choices for Γ ? Lemma 4.5. The set of endomorphism Γ ∈ End k (A 1 ) such that [∆, Γ] = ∆ is given by

Γ 0 + C(∆) where C(∆) = {f ∈ End k (A 1 ) ; [f, ∆] = 0} is the commutant of ∆ in End k (A 1 ) and where Γ 0 ∈ End k (A 1 ) is such that [∆, Γ 0 ] = ∆.
Proof. Follows from the fact that for two solvable pairs (∆, Γ) and (∆, Γ) we have [∆, Γ -Γ] = 0. 4.1. The case of ∆ maximal. When ∆ is a maximal Jordan block one can choose Γ 0 = diag(0, 1, . . . , n). Recall the algebras A(n, a) and R(n, a) from Example 1.3. Theorem 4.6. Given a solvable pair (∆, Γ) with ∆ a maximal Jordan block, there exists u ∈ GL(A 1 ) ∩ C(∆) such that uΓu -1 = aId A1 + Γ 0 , where a ∈ k is such that tr(Γ) = (n + 1)(a + n/2). In particular A (∆,Γ) ∼ = A(n, a) and R (∆,Γ) ∼ = R(n, a).

The proof relies on the following classical facts. Recall that

A 1 = kX 0 ⊕ • • • ⊕ kX n is a k-vector space of dimension n + 1.
(1) We have C(∆) = k[∆] = Span(id, ∆, ∆ 2 , . . . , ∆ n ) as a subalgebra of End k (A 1 ).

(2) Let u ∈ C(∆) and P ∈ k[X] such that u = P (∆). Then u ∈ GL(A 1 ) if and only if P (0) = 0.

Proof of Theorem 4.6. The relation [∆ i , Γ] = i∆ i shows that ker ∆ i is stable by Γ. So let us consider a basis (e 0 , . . . , e n ) of A 1 in which ∆ has canonical Jordan form. In such a basis Γ is upper triangular. Let us denote by (λ 0 , . . . , λ n ) the diagonal part of Γ. We have

Γ(e i ) = Γ(∆ n-i (e n )) = ∆ n-i Γ(e n ) -(n -i)∆ n-i (e n ) .
Thus by computing modulo vect(e 0 , . . . , e i-1 ), we get that λ i = λ n -(n -i) for all 0 i < n. In particular Γ has n + 1 distinct eigenvalues and thus a basis of eigenvectors. We then deduce that there exists an eigenvector e n of Γ which is not in ker ∆ n . Using the relation [∆ i , Γ] = i∆ i , we obtain that (∆ n (e n ), ∆ n-1 (e n ), . . . , e n ) is a basis of A 1 in which ∆ has canonical Jordan form and Γ is diagonal (a, a + 1, . . . , a + n) where a = λ 0 = λ n -n. Finally we get tr(Γ) = a(n + 1) + n(n + 1) 2 showing that a is uniquely determined by Γ.

Example 4.7. Let n = 1. There exists only two isomorphism classes for the algebras A = A (∆,Γ) and R = R (∆,Γ) with non necessarily maximal linear solvable pairs. If ∆ = 0 then A = R = k[X 0 , X 1 ] is a (Poisson) commutative algebra. If ∆ = 0 then it is necessarily a maximal Jordan block since n = 1. Therefore, thanks to Theorem 4.6, we have A ∼ = A(1, a) and R ∼ = R(1, a) with a ∈ k. If a = 0 we retrieve the commutative case (this is the example of Remark 2.13) and when a = 0, A is isomorphic to the Poisson algebra k[X 0 , X 1 ] with Poisson bracket {X 0 , X 1 } = X 2 0 and R is isomorphic to the Jordan plane, an algebra given by two generators X 0 , X 1 and the relation X

0 X 1 -X 1 X 0 = X 2 0 .
4.2. Center in the linear case. The following result is a generalization of Lemma 2.9 in the linear case. We describe the Poisson center of A and the center of R which allow us to determine exactly when A is Poisson commutative and R is commutative. The results of this section will be improved by Proposition 5.7 but are stated here since they will be needed in the proof of Proposition 5.7. In the next proposition the rank of ∆ seen as an endomorphism of A 1 is denoted by rank(∆).

Proposition 4.8. Let (∆, Γ) be a linear solvable pair on A. Assume that rank(∆) ≥ 2. Then we have

A 1 ∩ Z P (A) = ker ∆ ∩ ker Γ ∩ A 1 and A 1 ∩ Z(R) = ker ∆ ∩ ker Γ ∩ A 1 .
Proof. Since in this proof we only consider restrictions of ∆ and Γ to A 1 , we simply denote by ker ∆ the set ker ∆ ∩ A 1 and similarly for Γ. Since we have ker Γ ∩ ker ∆ ⊆ Z P (A) and ker Γ ∩ ker ∆ ⊆ Z(R) with equality if ker Γ = ker ∆ thanks to Lemma 2.9, it only remains to study the case ker Γ = ker ∆.

Consider z ∈ Z P (A) ∩ A 1 . Let us first show that z ∈ ker ∆ 2 . If not, z, ∆(z) and ∆ 2 (z) are linearly independent over k. Then 0 = {z, ∆(z)} = ∆(z)Γ∆(z) -∆ 2 (z)Γ(z). Since ker ∆ = ker Γ and z / ∈ ker ∆ 2 we get that Γ∆(z) = 0. Moreover ∆(z), Γ∆(z), ∆ 2 (z) and Γ(z) are homogeneous element of A of degree 1, hence are irreducible element of the unique factorization domain A. Since ∆(z) and ∆ 2 (z) are not collinear, they are non associated irreducible elements of A. The unique factorization of ∆(z)Γ∆(z) = ∆ 2 (z)Γ(z) ensures us that there exists α ∈ k such that Γ∆(z) = α∆ 2 (z) and Γ(z) = α∆(z). Computing ∆(z) = [∆, Γ](z) = α∆ 2 (z) -αΓ∆(z) = 0, we get a contradiction. Hence z ∈ ker ∆ 2 .

Before pursuing the computation of the Poisson center of A, we switch for a moment to the study of the centre of R. Consider z ∈ Z(R) ∩ A 1 . As in the Poisson case, we first show that z ∈ ker ∆ 2 . If not, the integer such that ∆ n (z) = 0 et ∆ n+1 (z) = 0 is greater than 1. We now compute z *

∆ n-1 (z) = ∆ n-1 (z) * z. For ∆((Γ -id)(∆ n-1 (z)) = Γ∆ n (z) + ∆ n (z) -∆ n (z) = 0. Indeed ∆ n (z) ∈ ker ∆ = ker Γ. So (Γ -id)(∆ n-1 (z)) ∈ ker ∆ = ker Γ. Hence ∆ n-1 (z) ∈ ker Γ 2 . Hence the relation z * ∆ n-1 (z) = ∆ n-1 (z) * z becomes ∆(z)Γ(∆ n-1 (z)) = Γ(z)∆ n (z).
Since n > 1 the element ∆(z) and ∆ n (z) are linearly independents. As in the preceding paragraph, we deduce that there exists α ∈ k such that α∆(z) = Γ(z) and α∆ n (z) = Γ(∆ n-1 (z)). The relation [∆ n-1 , Γ](z) = (n -1)∆ n-1 (z) can then be written as α∆ n (z) -α∆ n (z) = (n -1)∆ n-1 (z). Hence (n -1)∆ n-1 (z) = 0, which is absurd since n > 1 and ∆ n-1 (z) = 0. We deduce that z ∈ ker ∆ 2 and therefore that z = ∆ n-1 (z) ∈ ker Γ 2 . We now finish the proof by showing that in both cases z ∈ ker ∆ when rank∆ ≥ 2. If z / ∈ ker ∆ then there exists y such that ∆(y) is not collinear to ∆(z). For the Poisson case, we have 0 = {z, y} = ∆(z)Γ(y) -∆(y)Γ(z). For the case of R, we have z * y = zy + ∆(z)Γ(y) since z ∈ ker ∆ 2 . But y * z = yz + ∆(y)Γ(z) since z ∈ ker Γ 2 . Finally, we also have the equality ∆(z)Γ(y) = ∆(y)Γ(z). Since ker ∆ = ker Γ, we get that Γ(y) = 0 and Γ(z) = 0. Using the fact that ∆(y) and ∆(z) are not collinear, we can mimic the argument of the two preceding paragraphs to get the relation Γ(z) = α∆(z) for some α ∈ k. We then obtain ∆(z) = [∆, Γ](z) = α∆ 2 (z) -Γ∆(z). But z ∈ ker ∆ 2 and hence ∆(z) ∈ ker ∆ = ker Γ. We then conclude that ∆(z) = 0 and z ∈ ker ∆ ∩ ker Γ.

Corollary 4.9. Let (∆, Γ) be a linear solvable pair on A.

Then A 1 ∩ Z P (A) = ker ∆ ∩ ker Γ ∩ A 1 implies that A is Poisson commutative and A 1 ∩ Z(R) = ker ∆ ∩ ker Γ ∩ A 1 implies that R is commutative. Moreover A is Poisson commutative if and only if R is commutative if and only if ∆ = 0 or (∆, Γ) is conjugated to (∆ 0 , Γ 0 ) where ∆ 0 =       0 1 0 . . . 0       = X 0 ∂ X1 and Γ 0 =         0 1 0 . . . 0         = X 1 ∂ X1 Proof. Assume that A 1 ∩ Z P (A) = ker ∆ ∩ ker Γ ∩ A 1 or A 1 ∩ Z(R) = ker ∆ ∩ ker Γ ∩ A 1 ,
Lemma 2.9 and Proposition 4.8 show that ∆ = 0 (and Γ = 0) or ∆ is of Jordan type (2, 1, . . . , 1) and ker Γ = ker ∆. In the first case A is clearly Poisson commutative and R is commutative. In the second case, we can choose a basis (e 0 , . . . , e n ) of A 1 such that ∆(e 1 ) = e 0 and (e 0 , e 2 . . . , e n ) is a basis of ker ∆ = ker Γ. In such a basis, the matrix of ∆ and Γ are given by

      0 1 0 . . . 0       and        0 α 0 0 • • • 0 . . . . . . . . . . . . . . . . . . . . . . . . 0 α n 0 • • • 0       
The relation [∆, Γ] = ∆ ensures that α 1 = 1. Changing the basis (e 0 , . . . , e n ) to the basis (e 0 , e 1 + α 0 e 0 + α 2 e 2 + • • • + α n e n , e 2 , . . . , e n ) we get the matrices ∆ 0 and Γ 0 . In such a basis, we have that {X i , X j } = 0 and X i * X j = X i X j as soon as i or j belongs to {0, 2, . . . , n} (since ker Γ = ker ∆). In particular R is commutative and

A is Poisson commutative. It remains so show if A is Poisson commutative or R is commutative then ∆ = 0 or (∆, Γ) is conjugated to (∆ 0 , Γ 0 ). If A 1 ∩ Z P (A) = ker ∆ ∩ ker Γ ∩ A 1 or A 1 ∩ Z(R) = ker ∆ ∩ ker Γ ∩ A 1 then the first part of the proof allows us to conclude. If A 1 ∩ Z P (A) = ker ∆ ∩ ker Γ ∩ A 1 or A 1 ∩ Z(R) = ker ∆ ∩ ker Γ ∩ A 1 then A 1 = ker ∆ ∩ ker Γ ∩ A 1 and ∆ = 0.

4.3.

A finer filtration for the linear case. In section 2.4, we defined a filtration ε on any A with a solvable pair. When A is a polynomial ring and (∆, Γ) a linear solvable pair on A, we can define a finer filtration ε on A. The new filtration is defined as the extension to A of the restriction to A 1 of the filtration ε. Recall that A 1 denote the set of homogeneous polynomial of degree 1 which is stable by ∆ and Γ since the pair is linear (∆, Γ). Appendix C is devoted to the construction of this filtration ε and the study of its properties, see in particular Example C.3. When ∆ is a maximal Jordan block, then ε is the filtration used in [LS19, Section 3].

For this filtration ε, analogs of Lemma 2.21 and Proposition 2.22 are valid. Precisely, we have the following proposition.

Proposition 4.10. Let (∆, Γ) be a linear solvable pair on A = k[X 0 , . . . , X n ] and fix integers i, j 0. Set A i, ε = {x ∈ A, ε(x) i}.

(1) We have ∆(A i, ε ) ⊆ A i-1, ε and so ε(∆(f )) ε(f ) -1 for every f ∈ A.

(2) We have Γ(A i, ε ) ⊆ A i, ε . Hence ε(P (Γ)(f )) ε(f ) for every P ∈ k[T ] and f ∈ A.

(3) If f, g ∈ A are such that ε(f ) = i and ε(g) = j, then ε(f g) = i + j. (4) The family (A i, ε ) i∈N is a Poisson algebra filtration of A (∆,Γ) of degree -1, meaning that it is an algebra filtration of A together with {A i, ε , A j, ε } ⊆ A i+j-1, ε for all i, j 0. (5) The family (A i ) i∈N is an algebra filtration of R (∆,Γ) . Moreover, the associated graded algebra gr ε (R) is equal to gr ε (A). (6) For any f, g ∈ A we have

f * g -g * f = ∆(f )Γ(g) -∆(g)Γ(f ) + k 2 ∆ k (f ) Γ k (g) -∆ k (g) Γ k (f ) In particular, if ε(f ) = i and ε(g) = j then ε(f * g -g * f ) i + j -1. (7)
The commutative algebra gr ε (R) = gr ε (A) can be endowed with the following three Poisson brackets (a) {f , g}

:= (f * g -g * f ) + A i+j-2, ε (b) {f , g} := {f, g} + A i+j-2, ε (c) {f , g} := ∆(f )Γ(g) -Γ(f )∆(g) ∈ A i+j-1, ε /A i+j-2, ε
for homogeneous elements f and g of respective ε-degree i and j. In (c) the pair of maps (∆, Γ) is the solvable pair of homogeneous derivations of gr ε (A) of respective degree -1 and 0 which is induced by the solvable pair of filtered derivations ∆ and Γ of A (see Proposition C.2). (8) The Poisson structures defined in (3) are all equal and make gr ε (A) = gr ε (R) into a graded Poisson algebra of degree -1.

Proof. The proof is very close from the one of Lemma 2.21 and of Proposition 2.22 but since the construction of the filtration is more complicated the arguments should be given in a somewhat different order. We start with assertion (3). Proposition C.1 shows that gr ε (A) is a domain. This shows (3). To prove (1) and (2), we start with x ∈ A 1 . By definition of ε we have ε(∆(x)) ε(x) -1. Moreover, Lemma 2.20 shows that ker ∆ i ∩ A 1 is stable by Γ, hence by P (Γ) showing that ε(P (Γ)(x)) ε(x). We write now f ∈ A as a sum of products of elements of A 1 and (1) and (2) follow from (3) and the fact that ∆ and Γ are derivations. assertion (4) follows from 1.1 and then assertions (1) and ( 2) and (3). Assertion (5) and (6) follow from (1.2) and then we get assertions (1) and (2) and (3).

(7). The fact that {-, -} is a Poisson bracket on gr(R) follows the filtered version of the semiclassical limit construction, see [Goo08, Section 2.4]. The fact that {-, -} is a well-defined biderivation satisfying the Jacobi identity on gr(A) follows by tedious but straightforward computation from the fact that {-, -} is a filtered Poisson bracket on A. Finally {-, -} is a Poisson bracket since (∆, Γ) is a solvable pair of derivations of gr ε (A).

(8). The Poisson brackets {-, -} and {-, -} are the same since both f * g -g * f and {f, g} belong to A i+j-1, ε combined with f * g -g * f -{f, g} ∈ A i+j-2, ε thanks to Assertion (1), (2), (3) and (6). Finally {-, -} and {-, -} are the same since for homogeneous elements f and g of ε-degree i and j we have

{f , g} = ∆(f )Γ(g) -Γ(f )∆(g) = (∆(f ) + A i-2, ε )(Γ(g) + A j-1, ε ) -(∆(g) + A j-2, ε )(Γ(f ) + A i-1, ε ) = (∆(f )Γ(g) + A i+j-2, ε ) -(∆(g)Γ(f ) + A i+j-2, ε ) = (∆(f )Γ(g) -∆(g)Γ(f )) + A i+j-2, ε = {f, g} + A i+j-2, ε = {f , g} .
Remark 4.11 (Matrix version). Assertion (8) says that the Poisson structure on gr ε (A) is associated to the solvable pair (∆, Γ). On A 1 , the filtrations ε and ε coincide, hence by an appropriate choice of basis of A 1 and by using the Jordan reduction theorem adapted to this filtration, we obtain the following block decompositions

∆ =        0 I m0,m1 . . . . . . . . . I mr-2,mr-1 0        and Γ =        Γ 1 Γ 1,2 • • • Γ 1,r . . . . . . . . . . . . Γ r-1,r Γ r       
where for p ≥ r, we denote by

I p,r =        1 . . . 1 0 • • • 0 . . . . . . 0 • • • 0        ∈ M p,r (k)

Trigonalizable linear solvable pair

We continue with the notations of Section 4 : A = k[X 0 , . . . , X n ] is a polynomial algebra with standard grading and, for k 0, A k is the set of degree k polynomials. Definition 5.1. A linear solvable pair (∆, Γ) is said to be trigonalizable if the linear map induced by Γ on A 1 is. Proposition 5.2. Let (∆, Γ) be a linear solvable pair. Then it is trigonalizable if and only if the linear map Γ 1 induced by Γ on A 1 ∩ ker ∆ is.

Proof. Using the notations of Remark 4.11, the linear map induced by Γ on

A 1 ∩ ker ∆ is Γ 1 . Hence if Γ is trigonalizable then Γ 1 is.
Reciprocally, computing the superdiagonal blocks of the relation ∆Γ -Γ∆ = ∆ shows that for every 1 i r -1 we have the following triangular block decomposition

Γ i = Γ i+1 -I C i B i
where I denote the identity matrix of the appropriate size. Hence χ Γi+1 (X + 1) divides χ Γi (X), where χ M denote the characteristic polynomial of the square matrix M . We then deduce that when χ Γ1 is trigonalizable, every χ Γi is. Hence Γ acts on A 1 as a trigonalizable linear map.

5.1. The Artin-Schelter regular property. In this section we prove that the algebra R = R (∆,Γ) is Artin-Schelter regular when (∆, Γ) is a linear solvable pair. Recall that Artin-Schelter regular algebras are thought to be noncommutative analogue of commutative polynomial rings in the following sense [AS87]:

Definition 5.3. A connected N-graded k-algebra R is called Artin-Schelter regular or AS-regular if:

(1) gldim R < ∞;

(2) R has finite Gelfand-Kirillov dimension;

(3

) Ext i R (k R , R R ) ∼ =    0 if i = gldim R R k[ ] if i = gldim R.
where k[ ] means that the module is degree-shifted by some amount ∈ Z.

Since both ∆ and Γ are linear derivations we have ∆(A k ) ⊆ A k and Γ(A k ) ⊆ A k for all integer k 0. Therefore A k * A ⊆ A k+ for any k, 0 thanks to equation (1.2). Thus the algebra R = (A, * ) is d-graded, generated in degree 1 and with Hilbert series given by

hilb(R) = hilb(A) = 1 (1 -t) n+1 .
The following result is the key argument for our inductive proof of Theorem 5.5. Since this result relates algebras R (∆,Γ) (resp. A (∆,Γ) ) of different dimensions we will specify the (projective) dimension of the underlying polynomial ring by using the notation R n (∆,Γ) (resp. A n (∆,Γ) ).

Proposition 5.4. Let (∆, Γ) be a linear solvable pair on A = k[X 0 , . . . , X n ] such that ∆(X 0 ) = 0 and there exists α ∈ k verifying Γ(X 0 ) = αX 0 . Set A = A n (∆,Γ) and R = R n (∆,Γ) and denote by ∆ and Γ the matrices of ∆ and Γ acting on A 1 with respect to a basis starting by X 0 .

(1) The quotient algebra R/ X 0 is isomorphic to R n-1 (∆ ,Γ ) for the solvable pair (∆ , Γ ) of A n-1 = k[Y 0 , . . . , Y n-1 ] obtained from the solvable pair (∆, Γ) by deleting the first rows and first columns.

(2) X 0 is Poisson normal in A and the Poisson algebra A/ X 0 is isomorphic to A n-1 (∆ ,Γ ) for the same solvable pair (∆ , Γ ) of A n-1 .

In particular, R/ X 0 is a deformation of the Poisson algebra A/ X 0 .

Proof. (1) First note that X 0 is strongly normal in R and Poisson normal in A. The algebras R/ X 0 , A/X 0 A and R n-1 (∆ ,Γ ) can all be identified as graded vector spaces. Let π : R → R/ X 0 be the quotient map and set X i = π(X i ) for all 0 i n. Then the product in R/ X 0 is given by

X i * X j = 0 ∆ (X i ) Γ (X j )
where ∆ and Γ denote the derivations induced by∆ and Γ on the quotient

A/ X 0 ∼ = k[Y 0 , . . . , Y n-1 ].
It is now a simple verification to check that (∆, Γ) is a solvable pair on A/ X 0 that agrees with the solvable pair (∆ , Γ ) of k[Y 0 , . . . , Y n-1 ] under the identification X i+1 → Y i for all 0 i < n.

(2) X 0 is Poisson normal since it is in the kernel of ∆ and is an eigenvector of Γ. It is direct that the Poisson bracket on A/X 0 A is given by the solvable pair (∆, Γ) and we conclude as in (1).

We can now prove now prove that R (∆,Γ) is Artin-Schelter regular.

Theorem 5.5. Let (∆, Γ) be a linear solvable trigonalizable pair. The algebra R = R n (∆,Γ) is Artin-Schelter regular of global dimension n + 1.

Proof. The proof follows by induction on n, exactly as in the proof of [LS19, Theorem 3.8] since the only hypotheses used by the authors are that R is connected graded generated in degree 1 with polynomial growth and admits a normal sequence Ω of homogeneous regular elements with R/ΩR ∼ = k. Here we can also use Ω = (X 0 , X 1 , . . . , X n ) where (X 0 , . . . , X n ) is a basis of A 1 as in Remark 4.11 and Γ is triangular (Proposition 5.2).

Corollary 5.6. Let (∆, Γ) be a linear solvable pair. The algebra R = R n (∆,Γ) is Artin-Schelter regular of global dimension n + 1.

Proof. Consider k a finite field extension of k such that the characteristic polynomial of Γ acting on A 1 is split. Thanks to Theorem 5.5 the algebra R = k ⊗ R is Artin-Schelter regular since the unique extension of (∆, Γ) to R is a trigonalizable solvable pair.

Since k ⊗ is an exact functor which sends projective modules over R to projective modules over R, for any R-modules M, N , we have Proposition 5.7. Let (∆, Γ) be linear solvable pair. Assume that A is not Poisson commutative (see Corollary 4.9). Then

Ext * R ( k ⊗ M, k ⊗ N ) ∼ = Ext * R (M, k ⊗ N ) ∼ = Ext * R (M, N ) [ k:k]
* R ( k, R) ∼ = Ext * R (k, R) [ k:k] . In particular, if Ext * R ( k, R) ∼ = k then Ext * R (k, R) ∼ = k
(1) N is normal in R iff N is strongly normal iff N is Poisson normal in A. (2) N is central in R iff N ∈ ker ∆ ∩ ker Γ iff N is Poisson central in A.
Proof. The proof splits in two cases according to the value of the rank of ∆ acting on A 1 . When this rank is equal to 1, the proof relies on an explicit computation. When this rank is strictly greater than one, the proof can be mimicked from the one of [LS19, Proposition 3.21] by noticing that this proof only relies on the existence of an ε-homogeneous, irreducible, strongly normal element G which is not in ker Γ. If this is the case, it can verify that the ideal G * R = R * G of R is completely prime since G is an irreducible element of the unique factorization domain A. Notice that the existence of such a G shows that assertion (2) can also be seen as a consequence of Lemma 2.9.

First assume that (∆, Γ) is a trigonalizable solvable pair. In the following, we only consider actions of (∆, Γ) on A 1 and its subspaces. Using the notations of Remark 4.11 and Proposition 5.2, we see that Γ 1 is trigonalizable. If Γ 1 has a nonzero eigenvalue λ, then an eigenvector G of Γ 1 with respect to λ satisfies the desired properties. Hence we can assume that Γ 1 is nilpotent. As in the proof of Proposition 5.2, the relation ∆Γ -Γ∆ = ∆ shows that for every 1 i r -1 we have the following triangular block decomposition

Γ i = Γ i+1 -I C i B i
where I denote the identity matrix of the appropriate size. Hence i -1 is the only eigenvalue of Γ i and thus Γ i -(i -1)id is nilpotent. The triangular shape of the block decomposition of Γ then shows that ker ∆ i = i-1 j=0 N j where N j is the generalized eigenspace of Γ with respect to j. The relation ∆Γ -Γ∆ = ∆ can be rewritten ∆(Γ -jid) = (Γ -(j -1)id)∆ and then ∆(Γ -jid) k = (Γ -(j -1)id) k ∆ for all integer k 0. Hence ∆ maps N j into N j-1 . Moreover the relation ker ∆ i = i-1 j=0 N j shows that for all integer j 1, ∆ maps injectively N j into N j-1 .

Assume now that ∆ admits a Jordan block of size at least 3. Then 2 is an eigenvalue of Γ. If we consider X 2 such that Γ(X 2 ) = 2X 2 , then X 1 = ∆(X 2 ) ∈ ker(Γ -id), X 0 = ∆(X 1 ) ∈ ker(Γ), the family (X 0 , X 1 , X 2 ) is linearly independent and then G = 2X 2 X 0 -X 1 2 = 0 verifies ∆(G) = 0 and Γ(G) = 2G

as desired.

Hence we can now assume that ∆ admits only Jordan blocks of size 2. If ∆ admits at least two such blocks then dim(ker

∆ 2 / ker ∆) ≥ 2. Hence dim N 1 ≥ 2. If dim ker(Γ -id) ≥ 2 then choose (X 1 , Y 1 ) ∈ ker(Γ -id) linearly independent then G = X 1 ∆(Y 1 ) -Y 1 ∆(X 1 ) = 0 verifies ∆(G) = 0 and Γ(G) = G as wanted. For ∆(Y 1 ), ∆(X 1 ) ∈ ker ∆ ∩ ker Γ. If dim ker(Γ -id) = 1, since dim N 1 ≥ 2, there exists (X 1 , Y 1 ) ∈ N 1
2 such that Γ(X 1 ) = X 1 and Γ(Y 1 ) = Y 1 + X 1 (apply canonical Jordan form to Γ acting on N 1 ). Set then X 0 = ∆(X 1 ) and Y 0 = ∆(Y 1 ). We have X 0 , Y 0 ∈ ker ∆ and Γ(X 0 ) = 0 and

Γ(Y 0 ) = X 0 . Hence G = X 1 Y 0 -X 0 Y 1 = 0 verifies Γ(G) = G and ∆(G) = 0.
It remains to consider the case where ∆ admits only one Jordan block which is of size 2 that is to say the rank of ∆ is 1. In this case, we have

∆ =    0 0 1 0 0 0    and Γ =    0 L 1 0 Γ 1 0 1   
where the second column of the matrices ∆ and Γ represents matrices with (n -1) columns. By choosing the corresponding basis (X 0 , . . . , X n ) for A 1 , we have ∆ = X 0 ∂ Xn and Γ(X 0 ) = 0, Γ(X j ) ∈ k[X 0 , . . . , X n-1 ] for j ∈ {1, . . . , n -1} and Γ(X n ) = X n . We will show that every Poisson normal element Q lies in ker ∆ ∩ ker Γ and that every normal element Q in R also lies in ker ∆ ∩ ker Γ. For, write

Q = s i=0 P i X n i with P i ∈ k[X 0 , . . . , X n-1
] and an integer s 0. We have

∆(Q) = s i=0 iP i X 0 X n i-1 and Γ(Q) = s i=0 (Γ(P i ) + iP i )X n i .
Moreover there exists j 0 ∈ {1, . . . , n -1} such that Γ(X j0 ) = 0, otherwise, from Corollary 4.9, we have that A is Poisson commutative. Assume that Q is Poisson normal and compute {Q,

X j0 } = ∆(Q)Γ(X j0 ).
Since Q is Poisson normal there exists F ∈ A such that ∆(Q)Γ(X j0 ) = F Q. If F = 0 then by comparing the degree in X n of the two sides of the equality leads to something impossible. Hence F = 0 and ∆(Q) = 0 since Γ(X j0 ) = 0. So

Q = P 0 ∈ k[X 0 , . . . , X n-1 ]. By computing {Q, X n } = -Γ(Q)X 0 = -Γ(P 0 )X 0 ,
from the Poisson normality of Q, we obtain that there exists F ∈ A such that Γ(P 0 )X 0 = F P 0 . We then write

P 0 = X 0 R with X 0 R. Since X 0 ∈ ker Γ, we obtain the relation Γ(R)X 0 = F R. Hence X 0 | F , so Γ(R) = F 1 R for some F 1 ∈ A.
By comparing the total degree we obtain that F 1 ∈ k, and by multiplying by X 0 we obtain that Q = P 0 verifies Γ(P 0 ) = F 1 P 0 . But Γ is locally nilpotent on k[X 0 , . . . , X n-1 ] since it acts as a nilpotent linear map on the span of X 0 , . . . , X n-1 , hence F 1 = 0 and Q = P 0 ∈ ker Γ ∩ ker ∆ is strongly normal.

Let us now consider that

Q = 0 is normal in R and compute X j0 * Q = X j0 Q. Since Q is normal, there exists f ∈ R such that X j0 * Q = Q * f
. By applying the ε-degree, we deduce that ε(f ) = ε(X j0 ) = 0, that is to say f ∈ ker ∆. By sending the preceding equality in gr(R) which is a domain thanks to Corollary 2.23, we obtain that X j0 = f ∈ ker(A 0 /{0}) and then X j0 = f . The relation

X j0 * Q = Q * f can then be rewritten as i 1 ∆ i (Q) Γ i (X j0 ) = 0 with Γ i (X j0 ) ∈ ker ∆
By considering the smallest integer N 1 such that ∆ N (Q) = 0 and by composing the preceding relation with ∆ N -2 , we obtain that ∆

N -1 (Q)Γ(X j0 ) = 0. But Γ(X j0 ) = 0, hence N = 1 and Q ∈ ker ∆ = k[X 0 , . . . , X n-1 ]. We now compute X n * Q = X n Q + X 0 Γ(Q). Since Q is normal, there exists f such that X n * Q = Q * f = Qf . Hence Qf = X n Q + X 0 Γ(Q)
and we deduce that the total degree of f is not greater than 1 and that deg Xn (f ) = 1. By writing f = aX n + f 1 with f 1 ∈ k[X 0 , . . . , X n-1 ] and a ∈ k, we obtain a = 1 and then Qf 1 = X 0 Γ(Q). By applying the same argument that the one in the Poisson normal case, we obtain that Γ(Q) = 0. So Q ∈ ker Γ = ker ∆ is strongly normal.

To conclude, it remains to consider the case where the pair (∆, Γ) is not trigonalizable. Consider an element Q that is either Poisson normal in A or normal in R. Let k be the algebraic closure of k and consider the scalar extensions A = k ⊗ A and R = k ⊗ R of A and R respectively. The pair (∆, Γ) is a trigonalizable linear pair on A and 1

⊗ Q is Poisson normal in A or normal in R. Hence 1 ⊗ Q is strongly normal. Hence 1 ⊗ ∆(Q) = 0 and 1 ⊗ Γ(Q) = λ ⊗ Q for some λ ∈ k. But 1 ⊗ Γ(Q) ∈ 1 ⊗ A so λ ∈ k and Γ(Q) = λQ. So Q is strongly normal.
6. Diagonalizable linear solvable pair : Poisson derivations Definition 6.1. A linear solvable pair (∆, Γ) is said to be diagonalizable if there exists a basis of A consisting of eigenvectors for Γ.

Remark 6.2. The existence of a basis of A consisting of eigenvectors for Γ is equivalent to the fact that the restriction of Γ to A 1 is diagonalizable.

Indeed, if we consider a basis of eigenvectors of Γ acting on A 1 then the set of monomials formed with elements of this basis of A 1 is a basis of A consisting of eigenvectors for Γ.

Reciprocally, assume that there is a basis B of A consisting of eigenvectors for Γ. Let us show that the eigenvectors of Γ contained in A 1 generates A 1 . For this, let us consider v ∈ A 1 . We express v as a linear combination of elements of B and gather the eigenvectors associated to the same eigenvalue to write v = w 1 + • • • + w r where w i is an eigenvector for Γ associated to the eigenvalue λ i ∈ k and where λ i = λ j when i = j. By considering the family (v, Γ(v), . . . , Γ r-1 (v)) ∈ A 1 r and inverting the Vandermonde matrix associated to (λ 1 , . . . , λ r ), we get that w i ∈ A 1 for all i. This conclude the proof.

For a diagonalizable linear solvable pair (∆, Γ) we denote by n 1 n 2 • • • n r the size of the Jordan blocks of ∆ acting on A 1 . Such a ∆ is said to be of Jordan type n = (n 1 , . . . , n r ) where we set

n + 1 = n 1 + • • • + n r .
Proposition 6.3. Let (∆, Γ) be a diagonalizable linear solvable pair. Assume that ∆ is of Jordan type n = (n 1 , . . . , n r ). Then there exists a basis of A 1 and a = (a 1 , . . . , a r ) ∈ k r such that ∆ is in canonical Jordan form and Γ is diagonal of the form Γ = diag (a 1 , . . . , a 1 +n 1 -1, a 2 , . . . , a 2 +n 2 -1, . . . , a r , . . . , a r + n r -1).

Proof. This is an adaptation of the proof of the existence of the canonical Jordan form. For any integer i 0 we get that F i := ker ∆ i is stable by Γ by using the relation [∆ i , Γ] = i∆ i . There exists s 0 such that F s = A 1 since ∆ is nilpotent on A 1 . We construct by a decreasing induction subspaces

(G i ) 1 i s such that (1) F i-1 ⊕ G i = F i ; (2) ∆ maps injectively G i+1 into G i ; (3) G i is stable by Γ.
Since F s-1 is stable by Γ and Γ is diagonalizable, there exists G s which is stable by Γ such that

F s = A 1 = F s-1 ⊕ G s . Assume that G s , . . . , G s+1 have been constructed satisfying (1), (2) and (3). Then clearly ∆ maps G i+1 into F i . Moreover ∆(G i+1 ) ∩ F i-1 = {0} (since G i+1 ∩ F i = {0}) and G i+1 ∩ ker ∆ = {0}. Moreover ∆(G i+1 ) is stable by Γ. Hence F i-1 ⊕ ∆(G i+1 ) too and there exists a subspace W i of F i stable by Γ (since Γ is diagonalizable) such that F i-1 ⊕ ∆(G i+1 ) ⊕ W i = F i . It suffices now to consider G i = ∆(G i+1 ) ⊕ W i to get the result.
Let us now remark that if x is an eigenvector for Γ (associated to λ) and ∆(x) = 0 then ∆(x) is an eigenvector for Γ (associated to λ -1).

Thus if we consider a basis of G s composed of eigenvectors of Γ, then we consider the image of this basis by ∆ and complete into a basis of G s-1 (adding eigenvector of Γ belonging to W s-1 ). We proceed by induction to produce the basis we are looking for. Definition 6.4. Thanks to Lemma 2.1 and Proposition 6.3 one can assume without loss of generalities that for any diagonalizable solvable pair (∆, Γ) the Poisson algebra A (∆,Γ) is presented in the basis, denoted by (X 0 , . . . , X n ), given by Proposition 6.3. We denoted by A(n, a) the Poisson algebra A (∆,Γ) . Remark 6.5. It will sometimes be convenient to denote by λ the eigenvalue for Γ of X in A(n, a). The Poisson bracket of A(n, a) is then given by

{X i , X j } = λ j ∆(X i )X j -λ i ∆(X j )X i
for all 0 i, j n. Remark 6.6. Note that when (∆, Γ) is a linear solvable pair with ∆ maximal, then (∆, Γ) is automatically a diagonalizable linear solvable pair by Theorem 4.6. 6.1. Poisson derivations for diagonalizable linear solvable pairs. In this section, we study Poisson derivations for diagonalizable linear solvable pairs. We have seen in Lemma 2.15 that ∆ is always a Poisson derivation. In the linear case, the so-called Eulerian derivation E sending X i to itself for all 0 i n is also a Poisson derivation. Our aim is to recover ∆, or more precisely the conjugacy class of ∆ (acting on A 1 ), from the algebraic structure of A. We start with the case of a maximal Jordan block and then generalizes to every ∆ under certain generic conditions on Γ. 6.1.1. Poisson derivation for maximal Jordan block. In this paragraph we suppose that A = A (∆,Γ) , where (∆, Γ) is a linear solvable pair with ∆ a maximal Jordan block. We will show that there are no other linear Poisson derivations apart from those in the k-linear span of ∆ and E. Lemma 6.7. Assume that n 2 or that n = 1 and a = 0. For any δ ∈ P.Der gr (A) there exists a scalar λ such that δ(X 0 ) = λX 0 ∈ X 0 .

Proof. First if a = 0 (so n 2) then X 0 is, up to scalar multiple, the unique Poisson central element in A 1 thanks to [LS19, Proposition 3.1(2)]. The result follows since δ(X 0 ) is also a Poisson central element in A 1 , see Remark 2.16. Now assume a = 0 and let δ = (u ij ) ∈ M n+1 (k) so that for all 0 n we have δ

(X ) = n i=0 u i X i . Set X -1 = 0. By applying δ to the Poisson bracket {X 0 , X n } = -aX 0 X n-1 we obtain δ({X 0 , X n }) = -aδ(X 0 )X n-1 -aX 0 δ(X n-1 ) = -aX n-1 n i=0 u i0 X i -aX 0 n i=0 u i,n-1 X i , and
{δ(X 0 ), X n } + {X 0 , δ(X n )} = n i=0 u i0 {X i , X n } + n i=0 u in {X 0 , X i } = n i=0 u i0 (a + n)X i-1 X n -(a + i)X i X n-1 -a n i=0 u in X 0 X i-1 .
For 0 < i < n by comparing the coefficients of X i X n-1 we have iu i0 = 0, i.e. u i0 = 0. By comparing the coefficients of X n-1 X n we obtain au n0 = 0 so that u n0 = 0 since a = 0. Thus δ(X 0 ) = u 00 X 0 ∈ X 0 . Theorem 6.8. Assume that n 2 or that n = 1 and a = 0. Then P.Der gr (A) = kE ⊕ k∆. In particular the set of locally nilpotent derivations of P.Der gr (A) is the one dimensional subspace spanned by ∆.

Proof. We proceed by induction on n 1. First assume that a = -(n -1). Let n = 1 (so that a = 0) and δ ∈ P.Der gr (A). By Lemma 6.7 we know that δ(X 0 ) = λX 0 for a scalar λ. If δ(X 1 ) = µX 0 + νX 1 for some scalars µ, ν ∈ k, we have δ({X 0 , X 1 }) = -2aλX 2 0 and {δ(X 0 ), X 1 } + {X 0 , δ(X 1 )} = -2a(λ + ν)X 2 0 so that λ = ν. The result is proved since we have δ = λE + µ∆.

Let n > 1 and assume true the statement of the theorem for all algebras A(n -1, b) with b = -(n -2). Thanks to Lemma 6.7 there exists λ ∈ k such that δ(X 0 ) = λX 0 ∈ X 0 . Therefore δ induces a graded Poisson derivation of the Poisson algebra A/ X 0 ∼ = A(n -1, a + 1), see [LS19, Proposition 3.7(1)] for the isomorphism. Since a + 1 = -(n -2) we can apply the induction hypothesis to get that there exist µ, µ , ν i ∈ k such that

δ(X i ) = µX i + µ X i-1 + ν i X 0 2 i n, µX 1 + ν 1 X 0 i = 1.
To conclude the proof it remains to show that µ = λ, that ν 1 = µ and that ν i = 0 for all 2 i n since then δ = µE + µ ∆. If a = 0 then by applying δ to the Poisson brackets {X 0 , X 1 } (resp. {X 0 , X 2 }) a straightforward computation yields µ = λ (resp. ν 1 = µ ). If a = 0 then by applying δ to the Poisson brackets {X 1 , X 2 } a straightforward computation yields µ = λ and ν 1 = µ . Let 2 < i n. On one hand we have

δ({X 1 , X i }) = δ (a + i)X 0 X i -(a + 1)X 1 X i-1 = 2(a + i)µX 0 X i + (i -1)µ X 0 X i-1 -(a + 1)ν i-1 X 0 X 1 -2(a + 1)µX 1 X i-1 -(a + 1)µ X 1 X i-2 + (a + i)ν i X 2 0 . On the other hand {δ(X 1 ), X i } + {X 1 , δ(X i )} = 2µ (a + i)µX 0 X i -(a + 1)X 1 X i-1 + (i -1)µ X 0 X i-1 -(a + 1)µ X 1 X i-2 + aν i X 2 0 .
By comparing the coefficients of X 2 0 we get iν i = 0, i.e. ν i = 0. If i = 2 a similar but simpler computation provides ν 2 = 0.

It remains to deal with the case a = -(n -1). We proceed again by induction on n 2. It is only necessary to prove the initialization (for A(2, -1)) because the induction step proceeds exactly as in the case a = -(n -1).

We have δ(X 0 ) = λX 0 and there exists u, v, w, t, ν 1 , ν 2 ∈ k such that δ(X 1 ) = uX 2 + vX 1 + ν 1 X 0 and δ(X 2 ) = wX 2 + tX 1 + ν 2 X 0 . Direct computations from the Poisson brackets {X 0 , X 1 } = X 2 0 , {X 0 , X 2 } = X 0 X 1 and {X 1 , X 2 } = X 0 X 2 yield u = w = λ, u = ν 2 = 0 and t = ν 1 so that δ = λE + ν 1 ∆. 6.1.2. When ∆ is not necessarily maximal. We now assume that (∆, Γ) is a diagonalizable linear solvable pair but we do not assume that ∆ is a maximal Jordan block. This situation is more complicated as the following example illustrate. Example 6.9. Let ∆ be of Jordan type (2, 2) and Γ = diag(a, a + 1, b, b + 1). We obtain the Poisson brackets

{X 0 , X 1 } = -aX 2 0 , {X 0 , X 2 } = 0, {X 0 , X 3 } = -aX 0 X 2 {X 1 , X 2 } = bX 0 X 2 , {X 1 , X 3 } = (b + 1)X 0 X 3 -(a + 1)X 1 X 2 , {X 2 , X 3 } = -bX 2 2 .
It is easy to verify that P.Der gr (A (∆,Γ) ) = k∆⊕kE 01 ⊕kE 23 , where E 01 is the derivation sending X 0 to X 0 and X 1 to X 1 and both X 2 , X 3 to 0 (a similar definition applies to E 23 ). Therefore dim P.Der gr (A (∆,Γ) ) = 3 which is 1 more the number of blocks of ∆. Definition 6.10. Let (∆, Γ) be a linear solvable pair. We say that (∆, Γ) is generic or that Γ is generic if all the eigenvalues of Γ acting on A 1 are distinct and non zero. In particular a generic linear solvable pair is a diagonalisable linear solvable pair.

Recall from Definition 6.4 that without loss of generality we have A (∆,Γ) = A(n, a) where the derivation ∆ can be chosen to be in canonical Jordan from (with n 1 n 2 • • • n r denoting the size of the blocks of ∆) form and that Γ can be chosen diagonal (with the notation of Remark 6.5 for its eigenvalues). Theorem 6.11. Assume that (∆, Γ) is a generic linear solvable pair with r 2.

(1) If n 1 = 1 (i.e ∆ = 0), then P.Der gr A(n, a) = gl n+1 (k).

(2) If n = (2, 1, . . . , 1), then P.Der gr A(n, a) is the 2n-dimensional solvable non nilpotent Lie algebra L given by

L =              δ u,v,c2,...,cn,d2,...,dn :=        u v 0 u 0 c 2 d 2 . . . . . . . . . 0 c n d n        , u, v, c 2 , . . . , c n , u 2 , . . . , u n ∈ k             
Moreover, we have ZL = kid ⊕ k∆ and [L , L ] = {δ 0,0,c2,...,cn,0,...,0 , c 2 , . . . , c n ∈ k}. The ascending and descending central series of L are stationary from rank 1. (3) In the other case, P.Der gr A(n, a) is the (r + 1)-dimensional commutative Lie algebra given by k∆ ⊕ r i=1 kE i where E i is the block diagonal matrix where the diagonal block are 0 except the i th which is id ni .

In particular, up to a scalar factor, ∆ is the only locally nilpotent derivation in the center of P.Der gr A, except when ∆ = 0.

Proof. (1). The case where ∆ = 0 is trivial.

(2) and (3). We now assume that n 1 2. Let δ ∈ P.Der gr A(n, a). We first prove the following claim: if n i 2 for some i, then δ stabilizes all the blocks except possibly the i th .

It is enough to consider the case i = 1. To prove the claim we compute δ({X 1 , X j }) for j n 1 in two different ways and compare the coefficients of X 0 X . We set δ(X j ) = n i=0 u ij X i for all j. It clearly suffices to consider the case of the second block i.e. n 1 j < n 1 + n 2 . Let us prove by induction on j ∈ {n 1 , . . . , n 1 + n 2 -1} that δ(X j ) ∈ n1+n2-1 j=n1 kX i . With j = n 1 we have {X 1 , X n1 } = a 2 X 0 X n1 . On one hand we obtain δ({X 1 , X n1 }) = a 2 δ(X 0 )X n1 + a 2 X 0 δ(X n1 ). On the other hand we have

{δ(X 1 ), X n1 } + {X 1 , δ(X n1 )} = a 2 X n1 ∆(δ(X 1 )) + X 0 Γ(δ(X n1 )) -(a 1 + 1)X 1 ∆(δ(X n1 )).
By comparing the coefficients of X 0 X for = 1, n 1 we obtain a 2 u n1 = λ u n1 . Since Γ is generic λ = a 2 which is the eigenvalue of Γ associated to X n1 (we have = n 1 ). The coefficient of X 0 X 1 is a 2 u 1n1 = (a 1 + 1)u 1n1 -(a 1 + 1)u 1n1 . Thus δ(X n1 ) is collinear to X n1 (which is much more than needed) since a 2 = 0.

We now assume that the result is true for δ(X j-1 ) ∈ n1+n2-1 i=n1 kX i (with n 1 < j < n 1 + n 2 ). We have {X 1 , X j } = λ j X 0 X j -λ 1 X 1 X j-1 . By applying δ to this Poisson bracket we obtain on one hand δ({X 1 , X j }) = λ j δ(X 0 )X j + λ j X 0 δ(X j ) -λ 1 δ(X 1 )X j-1 -λ 1 X 1 δ(X j-1 ).

On the other hand we have {δ(X 1 ),

X j } + {X 1 , δ(X j )} = λ j X j ∆(δ(X 1 )) -Γ(δ(X 1 ))X j-1 + X 0 Γ(δ(X j )) -λ 1 X 1 ∆(δ(X j )).
By computing the coefficient of X 0 X for / ∈ {1, n 1 , . . . , n 1 + n 2 -1}, the induction hypothesis gives that λ j u j = λ u j . Hence u j = 0 since = j and Γ is generic. The coefficient of X 0 X 1 gives the relation λ j u 1j = λ 1 u 1j -λ 1 u 1j = 0 and the claim is proved.

We now split the argument in two cases:

(i) there are at least two blocks of size larger or equal to 2, (ii) n i = 1 for all i > 1 and n 1 2.

In case (i) by applying the claim to two different blocks of size larger or equal than 2 we see that δ must stabilize every blocks. In particular δ induces a Poisson derivation on each Poisson subalgebra B i generated by the variables X j corresponding to a block. It is clear that B i ∼ = A(n i -1, a i ) for each 1 i r, so that the induced derivation has a form δ| Bi = uE ni + v∆| Bi thanks to Theorem 6.8. Hence δ = r i=1 u i E ni +v i ∆| Bi for some u i , v i ∈ k. We will show that the v i are all equal, thus proving assertion (3) in case (i). First note that there are nothing to check if the corresponding block is of size 1 since in that case ∆ |Bi is the zero map. Now if n i 2 then we compare the coefficients of X 0 X ni after applying δ to the Poisson bracket {X 1 , X ni+1 }. On one hand we have

δ({X 1 , X ni+1 }) = δ (a ni + 1)X 0 X ni+1 -(a 1 + 1)X 1 X ni = v i (a ni + 1) -v 1 (a 1 + 1) X 0 X ni + other terms without X 0 X n1
whereas on the the other hand {δ(X 1 ),

X ni+1 } + {X 1 , δ(X ni+1 )} = (u 1 + u i ){X 1 , X ni+1 } + v 1 {X 0 , X ni+1 } + v i {X 1 , X ni } = (v i a ni -v 1 a 1 )X 0 X ni + other terms without X 0 X n1 .
Thus we have v i = v 1 , as desired.

In case (ii) we first show that δ(X i ) ∈

n1-1 j=0 kX j for all 0 i n 1 -2. Let 0 < i < n 1 n. We then have ∆(X ) = 0 and δ(X ) = u X . Hence

δ({X i , X }) = λ n k=0 u k,i-1 X k X + λ u X i-1 X and {δ(X i ), X } + {X i , δ(X )} = n1-2 k=0 λ u k+1,i X k X + λ u X i-1 X .
By comparing the coefficients of X k X for all k n 1 -1 we obtain that λ u k,i-1 = 0 so that u k,i-1 = 0 as desired.

When n 1 = 2 the previous computation provides us u 10 = 0 so that δ(X 0 ) = u 00 X 0 . Moreover by looking at the Poisson bracket {X 0 , X 1 } we obtain that u 00 = u 11 . No further restriction can be imposed on δ and we indeed obtain the form given in (2). The final assertions of (2) follow from simple computations.

When n 1 3 we can moreover prove that δ(X n1-1 ) ∈ n1-1 j=0 kX j . We showed that δ sends the Poisson subalgebra generated by X 0 , . . . , X n1-2 into the Poisson subalgebra generated by X 0 , . . . , X n1-1 , hence

δ({X 1 , X n1-1 }) =δ(λ n1-1 X 0 X n1-1 -λ 1 X 1 X n1-2 ) ∈ λ n1-1 X 0 δ(X n1-1 ) + k[X 0 , . . . , X n1-1 ] and {δ(X 1 ), X n1-1 } + {X 1 , δ(X n1-1 )} ∈ n k=0 u k,n1-1 λ k X 0 X k -a 1 X 1 ∆(δ(X n1-1 )) + k[X 0 , . . . , X n1-1 ].
By comparing the coefficients of X 0 X k for k n 1 we found that (λ k -λ n1-1 )u k,n1-1 = 0 so that u k,n1-1 = 0 thanks to our genericity hypothesis. In conclusion, δ stabilizes the subalgebra generated by X 0 , . . . , X n1-1 which is isomorphic to A(n 1 -1, a 1 ) and the result follows from Theorem 6.8. This concludes the proof of (3). Corollary 6.12. Let (∆, Γ) is a diagonalizable linear solvable pair with Γ generic and ∆ = 0. Then ∆ may be recovered from the algebraic structure of A since it is, up to a scalar, the only locally nilpotent derivation contained in the center of P.Der gr A.

6.1.3.

A remark for non necessarily diagonal Γ. In general the commutant of ∆ is not necessarily made of blocks corresponding to commutant of the Jordan blocks of ∆. We illustrate this fact in the case where ∆ has Jordan blocks of size (2, 1). Then the general form for Γ is

   a b c 0 a + 1 0 0 d e   
In A (∆,Γ) and R (∆,Γ) we have

{X 0 , X 1 } = -aX 2 0 , [X 0 , X 1 ] = -aX 0 * X 0 , {X 0 , X 2 } = 0, [X 0 , X 2 ] = 0, {X 1 , X 2 } = eX 0 X 2 + cX 2 0 , [X 1 , X 2 ] = eX 0 * X 2 + cX 0 * X 0 .
Since both X 0 and X 2 are in the kernel of ∆ we remark that Γ(X 1 ) does not appear in the formulae of the Poisson bracket and of the product * . Therefore, without loss of generality we can assume that b = d = 0. In particular Γ is triangular and its eigenvalues are a, a + 1 and e.

Case 1. If e = a and e = a + 1 then Γ is diagonalizable and one can assume c = 0 by Proposition 6.3. Case 2. Assume that e = a + 1. The solvable pair (∆, Γ) is conjugate to the diagonal solvable pair (∆, diag(a, a+1, a+1)) via the automorphism of A fixing X 0 and X 1 and sending X 2 to X 2 +cX 0 . Again we can assume Γ diagonalizable. Case 3. Assume then that e = a. If c = 0 then Γ is not diagonalizable so that the solvable pair (∆, Γ) is not diagonalizable (Remark 6.2). Note that the case c = 0 is isomorphic to the case c = 1 via the change of variable X 1 = 1 c X 1 . As a corollary of the following lemma we obtain that Case 3 above with c = 0 cannot be realized by a solvable diagonalizable pair (∆ , Γ ). Recall that the Lie algebra of linear Poisson derivations and its subset of locally nilpotent derivations is an invariant of homogeneous Poisson polynomial algebras by Proposition 4.1.

(A) =         u 00 u 01 u 02 0 u 11 u 12 0 u 21 2u 00 -u 11        
Proof. The computational proof is omitted. Now if A (∆,Γ) is realized by the solvable diagonalizable pair (∆ , Γ ), then by Theorem 6.8, we obtain that ∆ is of Jordan type (2, 1), since otherwise P.der gr (A) is of dimension 2 or 9. Consider Γ = diag (α, α + 1, ε) (with α = 0 or ε = 0 otherwise A would be Poisson commutative). If α = ε we compute that solvable pair is a consequence of the fact that Γ(X 1 ) never appears in the formulae. This is because all the other variables are in the kernel of ∆. Therefore we should expect more non diagonalizable cases when the rank of ∆ is a least 2.

7. Diagonalizable linear solvable pair: the algebra R (∆,Γ)

The aim of this section is to give a presentation for the algebra R in the case of a diagonalizable linear solvable pair (∆, Γ). According to Proposition 6.3 we can choose a basis (X 0 , . . . , X n ) of A 1 such that ∆ is in canonical Jordan form (we denote its Jordan type by n = (n 1 , . . . , n r ) with n 1 n 2 • • • n r and n + 1 = r i=1 n i ) and Γ is diagonal of the form Γ = diag (a 1 , . . . , a 1 + n 1 -1, a 2 , . . . , a 2 + n 2 -1, . . . , a r , . . . , a r + n r -1).

For any k ∈ {1, . . . , r} and 0 j < n k we denote by Y j,k = X n1+•••+n k-1 +j the j th variable of the k th block. Fix k ∈ {1, . . . , r} and 0 j < n k . Thanks to equation (1.2) we have for any

f ∈ R that f * Y j,k = ≥0 ∆ (f ) Γ (Y j,k ) = 0 ∆ (f ) a k + j Y j,k = φ a k +j (f )Y j,k .
In particular, for any k ∈ {1, . . . , r} and 0 j < n k we obtain the relations

φ -a k -j (Y j ,k ) * Y j,k = Y j ,k Y j,k = φ -a k -j (Y j,k ) * Y j ,k .
These relations can be rewritten as

(7.1) j =0 -a k -j Y j -,k * Y j,k = j =0 -a k -j Y j-,k * Y j ,k
and we have the following proposition.

Proposition 7.2. The algebra R is given by generators Y j,k for 1 k r and 1 j n k , and the homogeneous relations (7.1).

Proof. We first show that the set G = {Y j,k | 1 k r, 1 j n k } is a generating set for R. From Proposition C.1, gr ε (A) is generated by the image of the Y j,k in gr ε (A). But from Proposition 4.10, gr ε (A) = gr ε (R). Hence G is a set of generators for R as desired.

Let us denote by T the algebra given by generators Y j,k for 1 k r and 1 j n k , and relations (7.1). From the preceding argument R is a quotient of T . Moreover note that relations (7.1) are homogeneous with respect to the degree deg(Y j,k ) = 1 for all j, k. Hence the natural map from T onto R is graded. We endow the set G with the order given by Y j,k Y j ,k if k < k , or if k = k and j j . This induces an order on the monomials in the Y j,k 's. The relations (7.1) can be rewritten as

Y j,k * Y j ,k = Y j ,k * Y j,k + lower monomials
for all (j, k), (j , k ). Hence every element of T is a linear combination of monomials Y j1,k1 • • • Y js,ks in the Y j,k 's where Y j ,k Y j +1 ,k +1 for all ∈ {1, . . . , s -1}. In particular, the dimension of the homogeneous component of degree d of T is smaller than the one of R. But since R is a graded quotient of T the canonical map from T onto R must be one-to-one.

(2) there exists a basis of A 1 and a = (a 1 , . . . , a r ) ∈ k r such that ∆ is in canonical Jordan form and Γ is diagonal of the form Γ = diag (a 1 , . . . , a 1 + n 1 -1, a 2 , . . . , a 2 + n 2 -1, . . . , a r , . . . , a r + n r -1) (3) we denote by A(n, a) the corresponding A (∆,Γ) ; (4) we set n +

1 = n 1 + • • • + n r .
The parameter space of unimodular Poisson algebras A(n, a) is a (r -1)-dimensional affine subspace of k r since the Poisson algebra A(n, a) is unimodular if and only if tr(Γ) = 1 if and only if We denote by R e = R ⊗ k R op the enveloping algebra algebra of R.

Definition 8.5. We say that R is skew Calabi-Yau (or skew CY) if (i) R is homologically smooth: R has a finite projective resolution as a left R e -module such that each term is finitely generated; (ii) There are an algebra automorphism µ of R and an integer d such that

Ext i R e (R, R e ) ∼ =    0 if i = 0 1 R µ if i = d
where 1 R µ is the R-bimodule which is isomorphic to R as a k-vector space and such that r • s • t = rsµ(t).

If R is skew CY, the automorphism µ is called the Nakayama automorphism of R. If µ is inner, then R is Calabi-Yau or CY.

By [RRZ14, Lemma 1.2], any AS-regular connected graded algebra is skew CY. In particular, the algebras R (∆,Γ) are skew CY in the linear case thanks to Theorem 5.5. Proposition 8.6. Let (∆, Γ) be a linear solvable pair such that A is not Poisson commutative (see Corollary 4.9). Let Φ be an automorphism of R compatible with ε such that Φ = id (see Lemma 2.26). Assume that there exists a strongly normal element N of R whose eigenvalue with respect to Γ is nonzero. Then Φ commute with ∆ and φ c for every c ∈ k.

Proof. The proof decomposes into three steps. The first step consists to show that Φ commutes with one Φ λ by considering the normal element N ∈ R. In the second step, we deduce from the first that Φ commutes with ∆ by considering that Φ commutes with Φ kλ for k ∈ N. The third step is easy : since Φ commute with ∆, it commutes with every formal power series in ∆. Hence with φ c for every c ∈ k.

First step. Consider N ∈ ker ∆ ∩ ker(Γ -λid) with λ = 0. Then N is a normal element, hence Φ(N ) is too. Since every normal element in R is strongly normal thanks to Proposition 5.7, we deduce that there exists µ ∈ k such that Γ(Φ(N )) = µΦ(N ). Since Lemma 2.28 and Lemma 2.30 extends with the same proof to the filtration and the automophism associated to the normal element Φ(N ) is Φ • φ λ • Φ -1 = φ µ , we then deduce that the Poisson derivation associated to N is µ∆ = Φ • (λ∆) • Φ -1 (where the • notation is related to the filtration ). However we have Φ = id, hence µ∆ = λ∆. Since ∆ = 0 and then ∆ = 0, we obtain µ = λ. This shows that Φ and φ λ commute.

Second step. For f ∈ R, set n such that ∆ n+1 (f ) = 0 and ∆ n+1 (Φ(f )) = 0. The aim is to write ∆(f ) as a linear combination of (φ kα (f )) 0 k n . Since λ = 0, the matrix of binomial coefficients Remark 8.7. When the solvable pair (∆, Γ) is not necessarily linear, the preceding proposition can be adapted with the following statement. Let Φ be an automorphism of R compatible with ε and such that Φ = id. Assume that every normal element of R is strongly normal and that there exists a strongly normal element N of R whose eigenvalue with respect to Γ is nonzero. Then Φ commute with ∆ and φ c for every c ∈ k.

The preceding proposition applies in particular to the Nakayama automorphism of R as detailed in next example.

Example 8.8. Let (∆, Γ) be a linear solvable pair. From Proposition C.1, gr (A) is a polynomial algebra, hence we can apply [WZ01, Theorem 5.7] to get that the Nakayama automorphism µ of R is compatible with and verifies µ = id and (µ -id) = (1 -tr(Γ))∆ = (1 -tr(Γ))∆.

Theorem 8.9. Assume that (∆, Γ) is a generic diagonalizable linear solvable pair on A = k[X 0 , . . . , X n ] (see Definition 6.10). Then the Nakayama automorphism of R is φ 1-tr(Γ) .

Proof. Let µ be the Nakayama automorphism of R. It is a graded automorphism for the standard degree. Since ker ∆ is stable by Γ and Γ is diagonalizable, it induces a diagonalizable endomorphism of ker ∆. Hence there exists a strongly normal element N whose eigenvalue with respect to Γ is nonzero since 0 is not an eigenvalue of Γ. From Proposition 8.6 and Example 8.8, µ commute with ∆ and with φ c for all c ∈ k. In particular, Proposition 5.7 shows that every strongly normal element of A 1 is sent by µ to a strongly element of A 1 with the same eigenvalue with respect to Γ.

Let us consider a basis (X 0 , . . . , X n ) of A 1 such that ∆ is in canonical Jordan form of type n = (n 1 , . . . , n r ) and Γ is diagonal (see Proposition 6.3). For 1 k r and 0 i n r -1 we denote by Y i,k = X i+n1+•••+n k-1 the i th element of the k th block. In particular ker ∆ is the linear span of the Y 0,k for 1 k r.

Since the family G = {Y j,k | 1 k r, 1 j n k } is a generating set of R, it is enough to prove that µ and φ 1-tr(Γ) coincide on G. Therefore we have the desired equalities for every k such that n k 2. Assume now that n k 3. We prove by induction on i 2 that µ(Y i,k ) = φ 1-tr(Γ) (Y i,k ). Assume that i 1 and µ(Y j,k ) = φ 1-tr(Γ) (Y j,k ) for j i. Since both µ and φ 1-tr(Γ) commutes with ∆, we have ∆((µ -φ 1-tr(Γ) )(Y i+1,k )) = (µ -φ 1-tr(Γ) )(∆(Y i+1,k )) = (µ -φ 1-tr(Γ) )(Y i,k ) = 0 .

Hence there exist scalars α 0 , . . . , α r such that

φ -λ-1 (Y i+1,k ) * Y 1,k = Y i+1,k Y 1,k = φ -λ-i-1 (Y 1,k ) * Y i+1,k
For the sake of simplicity we set c = 1 -tr(Γ). By applying µ and φ c to the relation (8.11) and by using the commutation of µ with φ α , of φ β with φ α for every α, β ∈ k, and the relation (8.10), we obtain

0 =φ -λ-1 (µ(Y i+1,k )) * µ(Y 1,k ) -φ -λ-i-1 (µ(Y 1,k )) * µ(Y i+1,k ) =φ -λ-1 φ c (Y i+1,k ) + r =0 α Y 0, * φ c (Y 1,k ) -φ -λ-i-1 (φ c (Y 1,k )) * φ c (Y i+1,k ) + r =0 α Y 0, =φ -λ-1 r =0 α Y 0, * φ c (Y 1,k ) -φ -λ-i-1 (φ c (Y 1,k )) * r =0 α Y 0, .
By applying φ -c to the last equality, we obtain the relation

r =0 α Y 0, * Y 1,k = φ -λ-i-1 (Y 1,k ) * r =0 α Y 0,
since f = r =0 α Y 0, ∈ ker ∆ and so φ α (f ) = f for all α ∈ k. By using the expression of the product * , this last relation can be rewritten as Y 0,k r =0 α (λ -(λ + i + 1))Y 0, = 0 where λ is the eigenvalue for Γ associated to the eigenvector Y 0, . Since Γ has distinct eigenvalues, we obtain that α = 0 for all and µ(Y i+1,k ) = φ 1-tr(Γ) (Y i+1,k ). This concludes the proof.

We need a better understanding of commutation relations in the enveloping algebra of the two dimensional solvable Lie algebra spanned by ∆ and Γ.

Lemma A.6. Let B be a k-algebra (associative) and Γ, ∆ ∈ B such that [∆, Γ] = ∆. For any i 0 and k 0 we have

∆, Γ k = Γ k -1 ∆, ∆ i , Γ k = i =1 i Γ k - ∆ i
Proof. The first assertion is clear for k = 0. Let k 1. Notice that ∆(Γ -u) = (Γ -(u -1))∆ for any u ∈ k. Hence one has

k! ∆, Γ k = ∆Γ(Γ -1) • • • (Γ -(k -1)) -Γ(Γ -1) • • • (Γ -(k -1))∆ = (Γ + 1)Γ • • • (Γ -(k -2))∆ -Γ(Γ -1) • • • (Γ -(k -1))∆ = Γ • • • (Γ -(k -2))(Γ + 1 -(Γ -(k -1)))∆ = kΓ • • • (Γ -(k -2))∆ = k(k -1)! Γ k -1 ∆ = k! Γ k -1 ∆
and the result is proved. We now prove the second assertion by induction on i 0. The case i = 0 is clear. The initialization i = 1 has just been proved. Assume that the result is true for some i 1 and all k 0. Then

∆ i+1 , Γ k = ∆ i ∆ Γ k - Γ k ∆ i+1 = ∆ i Γ k ∆ + Γ k -1 ∆ - Γ k ∆ i+1 = ∆ i , Γ k ∆ + ∆ i Γ k -1 ∆ = i =1 i Γ k - + i =0 i Γ k --1 ∆ i+1 = i+1 =1 i + 1 Γ k - ∆ i+1 .
Remark A.7. The second relation of Lemma A.6 can be rewritten

∆ i Γ k = i =0 i Γ k - ∆ i = min(i,k) =0 i Γ k - ∆ i = k =0 i Γ k - ∆ i = k =0 i k - Γ ∆ i
Indeed, for > k, Γ k-= 0 and for > i, i = 0.

We finish this combinatorial appendix by the computation of the following determinant which will be useful for the determination of the Nakayama automorphism of R.

2. 3 .

 3 Derivation and automorphism. Definition 2.14. Let A be a Poisson algebra. (1) A Poisson derivation is a derivation δ of A such that for all a, b ∈ A δ({a, b}) = {δ(a), b} + {a, δ(b)}.

  Therefore we have gldim R gldim R < ∞. Since R and R are N-graded connected algebras, the case M = N = k and [Ei56, Theorem 11, Theorem 13 and Proposition 15] show that gldim R = gldim R. By considering the case M = k and N = R, we obtain that Ext

Lemma 6. 13 .

 13 Consider the Poisson algebra A = A (∆,Γ) from Case 3. If a = 0 then the space of linear Poisson derivations is of dimension 4 P.der gr (A) = In particular P.der gr (A) contains a 3-dimensional subspace of locally nilpotent derivations. If a = 0 then the space of linear Poisson derivations is of dimension 6 P.der gr

  P.der gr (A) = but contains a 2 dimensional subspace of locally nilpotent Poisson derivation, contrary to A (∆,Γ) . If α = ε then we compute that P.der gr (A) = 5, contrary to A (∆,Γ) as desired.Remark 6.14. The fact that the case Γ = d = 0 can be realized by a diagonal

.

  is invertible (see Lemma A.8). Hence there exists (α 0 , . . . , α n ) ∈ k n+1 such that We then deduce that ∆(f ) = n k=0 α k φ kλ (f ) and ∆(Φ(f )) = n k=0 α k φ kλ (Φ(f )). But Φ commutes with φ λ , therefore it commutes with φ kλ = φ λ k for any k ∈ {0, . . . , n}. And the preceding equalities show that Φ(∆(f )) = ∆(Φ(f )).

  Since µ = id (Example 8.8) and ε(Y 0,k ) = 0, we obtain µ(Y 0,k ) = Y 0,k = φ 1-tr(Γ) (Y 0,k ). Since ε(Y 1,k ) = 1, [WZ01, Theorem 5.7] and Proposition 8.1 show that µ(Y 1,k ) = Y 1,k + (1 -tr(Γ))∆(Y 1,k ) = φ 1-tr(Γ) (Y 1,k ).

=0α Y 0 ,

 0 (8.10) µ(Y i+1,k ) = φ 1-tr(Γ) (Y i+1,k ) + r If f ∈ R is an eigenvector for Γ associated to the eigenvalue λ, then for every g ∈ R we have g * f = φ λ (g)f . Set Γ(Y 0,k ) = λY 0,k where λ ∈ k. Then Γ(Y 1,k ) = (λ + 1)Y 1,k and we obtain Γ(Y i+1,k ) = (λ + i + 1)Y i+1,k by an easy induction on i ∈ {0, . . . , n k -1}. Hence we obtain that (8.11)

Lemma A. 8 .

 8 Let k be a field of characteristic 0 and A be a k-algebra. For a ∈ A and n ∈ N setM (a) = ka 0 k, n ∈ M n+1 (A)and m(a) = det(M (a)). Then m(a) = a n(n+1)/2 .

  Thanks to[START_REF] Vanhaecke | Integrable systems in the realm of algebraic geometry[END_REF] Proposition 2.17] the rank of A is invariant under isomorphism. This shows that not every Poisson bracket on A can be obtained from a solvable pair since Poisson structures with higher ranks do exist.3.2. Application: non isomorphism theorem. The notion of rank recalled in the previous section allow us to prove that a polynomial Poisson algebra arising from a solvable pair cannot be obtained non trivially by tensor products of algebras of the same type. See Appendix B for the definition of the Poisson structure on the tensor product of Poisson algebras. product, then only one factor has nonzero rank, thus making A into a Poisson central extension of a rank two Poisson algebra of the form A ( ∆, Γ) . Let s 1 be an integer and for 1 i s let A ni (∆i,Γi) be a polynomial Poisson algebra in n i indeterminates for some solvable pairs (∆ i , Γ i ) on A ni . Set B =

	The above proposition is a corollary of the following lemma.
	Lemma 3.2. s i=1 A ni (∆i,Γi) and assume that B
	has rank 2. Set n =
	Proof. Thanks to [Van01, Proposition 2.21] we have Rk(B) =

Proposition 3.1. Let A = A (∆,Γ) be non Poisson commutative. Then A is not isomorphic to a non trivial tensor product of the form s i=1 A (∆i,Γi) in the sense that if A is isomorphic to such a tensor s i=1 n i . Then there exists a solvable pair (∆ , Γ ) on a B such that the Poisson structure of B is obtained by the solvable pair (∆ , Γ ). s i=1 Rk(A ni (∆i,Γi) ). In particular the rank of all A ni (∆i,Γi) but one, say A n1 (∆1,Γ1) , should be zero, i.e. A ni (∆i,Γi) is Poisson commutative for all i > 1. Hence B = A n1 (∆1,Γ1) ⊗ C where C = s i=2 A ni (∆i,Γi) is Poisson commutative. One easily verify the conclusion of the lemma holds by setting

  since the field extension is finite. It follows that the algebra R is Artin-Schelter regular. 5.2. Normal elements. The next proposition completely describes normal elements of R and Poisson normal elements in A in the case of a linear solvable pair: they are the strongly normal element (see Definition 2.29). It generalises the results of Lemma 2.9, Proposition 4.8 and is a converve of Lemma 2.30 in the linear case.

  LS19], since n + 1 = n 1 + • • • + n r = n 1 . Calabi-Yau property for R (∆,Γ)in the generic diagonalizable case. The aim of this section is to compute the Nakayama automorphism of R in order to determine when R is a Calabi-Yau algebra.

					r-1 t=0	n t+1 a t+1 +	n t+1 2	= 1.
	Example 8.3. If r = 1 (a single Jordan block) then A is unimodular if and only if
					(n + 1)a +	n + 1 2	-1 = 0 ⇐⇒ a =	-(n + 2)(n -1) 2(n + 1)
	as expected from [Example 8.4. Let ∆ be of Jordan type (2, 1) and Γ non necessarily diagonalizable. The general form
		 a	b	c	
	for Γ is	 	0 a + 1 0	 , hence the Poisson algebra A (∆,Γ) is unimodular if and only if 2a + e = 0.
			0	d	e
	8.2.

Acknowledgment. The authors thank Patrick Le Meur for fruitful discussions on Artin-Schelter regular algebras.

Example 7.3. Let ∆ be of Jordan type (2, 2) and let Γ = diag(a, a + 1, b, b + 1) be as in Example 6.9. Thanks to Proposition 7.2 we obtain the following complete set of relations between the generators of R = R (∆,Γ) :

= (b + 1)X 0 * X 3 -(a + 1)X 1 * X 2 + (a + 1)bX 0 * X 2

Unimodarity and Calabi-Yau property

In this section we study the unimodularity of the Poisson algebra A = A (∆,Γ) in the polynomial case and determine when the algebra R = R (∆,Γ) is Calabi-Yau. Whereas the unimodularity of A is completely characterized by Proposition 8.1, we only compute the Nakayama automorphism of R in the case of a generic diagonalizable linear solvable pair (Theorem 8.9). This result generalizes [LS19, Theorem 4.16].

8.1. Unimodularity of A (∆,Γ) in the polynomial case. Unimodular Poisson algebras are thought to be Poisson analogous of Calabi-Yau algebras, see [START_REF] Dolgushev | The Van den Bergh duality and the modular symmetry of a Poisson variety[END_REF]. Recall that a polynomial Poisson algebra A = k[X 0 , . . . , X n ] is called unimodular if its modular derivation is zero, where the modular derivation m of A is given in the polynomial case by

Proposition 8.1. Let A = k[X 0 , . . . , X n ] and (∆, Γ) be a solvable pair on A. The modular derivation m of the Poisson algebra A (∆,Γ) is equal to m = (1 -div(Γ))∆. In particular, as long as ∆ = 0, the Poisson algebra A (∆,Γ) is unimodular if and only if div(Γ) = 1. When Γ is linear this is equivalent to say that tr(Γ) = 1.

Proof.

as desired. For the assertion in the linear case, recall that div(Γ) = n i=0

∂Γ(X i ) ∂X i .

We illustrate this result with several examples in the linear case.

Example 8.2. If (∆, Γ) is a diagonalizable linear solvable pair (∆, Γ), recall that

(1) its Jordan type is n = (n 1 , . . . , n r ) where n 1 n 2 • • • n r are the size of the Jordan blocks of ∆ acting on A 1 ;

Corollary 8.12. Let (∆, Γ) be a generic diagonalizable linear solvable pair on

Proof. Since R is a connected graded algebra, every inner automorphism is trivial. Hence R is Calabi-Yau if and only if its Nakayama automorphism is trivial. The result follows from Proposition 8.1 and Theorem 8.9 since Φ c = id if and only if c = 0.

and for k ∈ Z with k < 0, we set b k = 0.

Lemma A.2. Let B be a k-algebra, a, b ∈ B with ab = ba. For any n 0 we have Appendix C. Symmetric Algebra of a filtered vector space Let us consider V = ∪ i∈N V i a filtration of the vector space V , that is to say, for every i ∈ N, V i is a subspace of V and

This filtration on V defines an algebra filtration on T (V ) (the tensor algebra of V ) and S(V ) (the symmetric algebra of V ) given by

where σ : T (V ) → S(V ) is the canonical map.

Let us set M 0 = V 0 and for i 1, we consider M i a supplementary space of V i-1 in V i . We thus get that V i = j i M j and V = i∈N M i can be seen as a graded vector space which we denote by gr(V ). Using this graduation, we get a bigraduation on T (V ) =

(n,α)∈N 2 T n,α and S(V ) = (n,α)∈N 2 S n,α given by

See [Bo07, Algèbre chap III, §5.5 Proposition 7, §6.6 Proposition 10]. We set T α = n∈N T n α and S α = n∈N S n α .

Proposition C.1. The graded rings associated to the filtration T (V ) = ∪ α∈N T α and S(V ) = ∪ α∈N S α are isomorphic to T (gr(V )) and S(gr(V )).

Proof. For (α 1 , . . . , α n ) ∈ N n , let us define π β1,...,βn α1,...,αn :

Note that if there exists i such that β i > α i then π β1,...,βn α1,...,αn = 0. When (α 1 , . . . , α n ) = (β 1 . . . , β n ), we simply denote π α1,...,αn instead of π α1,...,αn α1,...,αn

Let us now define π = ⊕π α1,...,αn :

We want to show that π induces a map from T n α to T n,α . It suffices to show that if w = (w α1,...,αn ) (α1,...,αn) ∈

verifies (α1,...,αn) w α1,...,αn = 0 ∈ T n (V ) then π(w) = 0. But using the direct sum decomposition of the V i we get that

Hence the hypothesis (α1,...,αn) w α1,...,αn = 0 ∈ T n (V ) may be written for all (β 1 , . . . , β n ) such that

we get (α1...,αn) π β1,...,βn α1,...,αn (w α1,...,αn ) = 0. In particular, when β 1 + • • • + β n = α and (α 1 , . . . , α n ) = (β 1 , . . . , β n ) there exists i such that β i > α i . Hence 0 = (α1...,αn) π β1,...,βn α1,...,αn (w α1,...,αn ) = π α1,...,αn (w α1,...,αn ) as wanted. We still denote by π : T n α → T n,α the induced map. We clearly have ker π = T n α-1 showing the first result since the kernel of π α1,...,αn is included into T n α-1 (see [Bo07, Algèbre, chap.2, §3, Proposition 6]). Let us now consider the case of the symmetric algebra. We denote by σ n α : T n α → S n α and by σ n,α : T n,α → S n,α the maps induced by σ : T (V ) → S(V ).

The surjective map π : T n α → T n,α sends ker σ n α onto ker σ n,α . Hence π induces a surjective map also denoted by π : S n α → S n,α . To conclude, it remains to show that ker π = S n α-1 . This is clear that S n α-1 ⊆ ker π. The reverse inclusion follows from diagram chasing in the following diagram (where the column are exact and the second row too)

Proposition C.2. Assume that there exists linear maps ∆ :

and Γ(V i ) ⊆ V i for every i ∈ N, and [∆, Γ] = ∆. The extensions of ∆ as derivation of T (V ) and S(V ) verify ∆(T n α ) ⊆ ∆(T n α-1 ) and ∆(S n α ) ⊆ ∆(S n α-1 ). The extensions of Γ as derivation of T (V ) and S(V ) stabilise the preceding filtrations on T (V ) and S(V ). Hence they induces graded derivations ∆ of degree -1 and Γ of degree 0 on T (gr(V )) and S(gr(V )). Moreover [∆, Γ] = ∆.

In addition, ∆ induces a degree -1 map on gr(V ) and Γ induces a degree 0 map on gr(V ) and their extensions as derivations on T (gr(V )) and S(gr(V )) coincide with ∆ and Γ.

Proof. The extension of a linear map

α is mapped into T n α showing that they induces graded endomorphisms ∆ of degree -1 and Γ of degree 0 of T (gr(V )) and S(gr(V )). Moreover these endomorphisms are derivations of the graded associated rings and verify [∆, Γ] = ∆.

For the last part of the proof, since the maps considered are derivations of T (gr(V )) and S(gr(V )) it suffices to show the coincidences on gr(V ). But they follow readily from the commutative diagrams

Example C.3. Let us consider V a finite dimensional vector space and ∆ and Γ two linear endomorphisms of V verifying [∆, Γ] = ∆. Since ∆ is nilpotent and ker ∆ i is stable by Γ (see the proof of Theorem 4.6). The family of subspaces V i = ker ∆ i is a filtration of V satisfying the hypothesis of Proposition C.2.