Double Poisson Cohomology of Path Algebras of Quivers

Home , Einstein notation

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Throughout this paper we will work over an algebraically closed ﬁeld of char- acteristic 0 which we denote by C. Unadorned tensor products will be over C. We will use Sweedler notation to write down elements in the tensor product A ⊗ A for A an algebra over C.

A double Poisson algebra A is an associative unital algebra equipped with a linear map {{−, −}} : A ⊗ A → A ⊗ A that is a derivation in its second argument for the outer A-bimodule structure on A ⊗ A, where the outer action of A on A ⊗ A is deﬁned as a.(a′ ⊗ a′′).b := (aa′) ⊗ (a′′b). Furthermore, we must have that {{a, b}} = −{{b, a}}o and that the double Jacobi identity holds for all a, b, c ∈ A:

{{a, {{b, c}}′}}⊗ {{b, c}}′′ + {{c, a}}′′ ⊗ {{b, {{c, a}}′}} + {{c, {{a, b}}′}}′′ ⊗ {{a, b}}′′ ⊗ {{c, {{a, b}}′}}′ =0, where we used Sweedler notation, that is {{x, y}} = {{x, y}}′ ⊗ {{x, y}}′′ for all x, y ∈ A. Such a map is called a double Poisson bracketP on A. Double Poisson algebras were introduced in [12] as a generalization of classical Poisson geometry to the setting of noncommutative geometry. More specifically, a double Poisson bracket on an algebra A induces a Poisson structure on all finite dimensional representation spaces repn(A) of this algebra. Recall C that the coordinate ring [repn(A)] is generated as a commutative algebra by the generators aij for a ∈ A and 1 ≤ i, j ≤ n, subject to the relations j aijbjk = (ab)ik. For each n, the Poisson bracket on the coordinate ring C P[repn(A)] of the variety of n-dimensional representations of A is defined as ′ ′′ {aij, bkℓ} := {{a, b}}kj{{a, b}}iℓ. This bracket restricts to a Poisson bracket on GL C n [repn(A)] , the coordinate ring of the quotient variety issn(A) under the action of the natural symmetry group GLn of repn(A).

In case the algebra is formally smooth (i.e. quasi-free in the sense of [4]), double Poisson brackets are completely determined by double Poisson tensors, that is, degree two elements in the tensor algebra TADer(A, A ⊗ A). For example, the classical double Poisson bracket on the double Q of a quiver Q is the bracket corresponding to the double Poisson tensor ∂ ∂ P = sym ∂a ∂a∗ aX∈Q and its Poisson bracket corresponds to the symplectic form on the representation space of the double of a quiver used in the study of (deformed) preprojective algebras (see [3] and references therein for further details on deformed

2 preprojective algebras).

We will denote by Der(A) the space Der(A, A ⊗ A) of all derivations of A, with value in A ⊗ A, for the outer A-bimodule structure on A ⊗ A. This space Der(A) becomes a A-bimodule, by using the inner A-bimodule structure on A ⊗ A: if δ ∈ Der(A) and a, b, c ∈ A, then (aδb)(c)= δ(c)′b ⊗ aδ(c)′′.

As in the classical case, it is possible to define Poisson cohomology for a double Poisson bracket. This was briefly mentioned in [11] and will be formalized and illustrated in this note. More specifically, in Section 2, we will recall and formalize the definition of the double Poisson cohomology from [11]. We will then give, in Section 3, an explicit formulation of the Gerstenhaber algebra of poly-vectorfields and its noncommutative Schouten bracket for the path algebra of a quiver in terms of its graded necklace Lie algebra equipped with a graded version of the Kontsevich bracket. This description will first of all be used to define and classify linear double Poisson structures on path algebras and quivers in Section 4. On the free algebra in n variables, treated in Section 5, this classification becomes

Proposition 1 (Prop. 10, Section 5) There is a one-to-one correspondence between linear double Poisson brackets on Chx1,...,xni and associative algebra structures on V = Cx1 ⊕···⊕ Cxn. Explicitly, consider the associative algebra structure on V determined by

n k xixj := cijxk, i,j,kX=1 k C where cij ∈ , for all 1 ≤ i, j, k ≤ n, then the corresponding double Poisson bracket is given by

n k k {{xi, xj}} = (cij xk ⊗ 1 − cji 1 ⊗ xk), kX=1 which corresponds to the Poisson tensor:

n k ∂ ∂ P = cijxk . ∂xi ∂xj i,j,kX=1 Next we show there is a connection between the Hochschild cohomology of ﬁnite dimensional algebras and the double Poisson cohomology of linear double Poisson structures. We obtain

Theorem 1 (Thm. 3, Section 5) Let A = Cx1⊕· · ·⊕Cxn be an n-dimensional vector space and let n k ∂ ∂ P = cijxk ∂xi ∂xj i,j,kX=1

3 be a linear double Poisson structure on TCA = Chx1,...,xni. Consider A as an algebra through the product induced by the structure constants of P (the k C • cij ∈ ) and let HH (A) denote the Hochschild cohomology of this algebra, then • ∼ • (HP (TCA))1 = HH (A).

• TTCA Here the grading on (H (TCA)) is induced by the grading on , P [TTCA,TTCA] i which is deﬁned through deg(xi)=1.

From the appendix in [12] we know that the double Poisson cohomology of a double Poisson bracket corresponding to a bi-symplectic form (as defined in [2]) is equal to the noncommutative de Rham cohomology computed in [1], which is a translation of a similar result in classical Poisson geometry. In gen- eral, little is known about the classical Poisson cohomology and it is known to be hard to compute. In Section 6, we will compute, using the description of the algebra of poly-vectorfields in Section 3, the low-dimensional double Poisson cohomology groups for the free algebra Chx, yi, equipped with three different types of non-symplectic double Poisson brackets. This will in particular allow us to develop some tools (including a noncommutative Euler formula, Propo- sition 12) and techniques, that seem to be useful for the determination of the double Poisson cohomology.

2 Double Poisson Cohomology

In [8], Lichnerowicz observed that dπ = {π, −} with π a Poisson tensor for a Poisson manifold M ({−, −} is the Schouten-Nijenhuis bracket) is a square zero diﬀerential of degree +1, which yields a complex

0 →dπ O(M) →dπ Der(O(M)) →dπ ∧2Der(O(M)) →dπ ..., the homology of which is called the Poisson-Lichnerowicz cohomology. In this section, we show there is an analogous cohomology on TADer(A) that descends to the classical Poisson-Lichnerowicz cohomology on the quotient spaces of the representation spaces of the algebra.

In [12, §4], the notion of a diﬀerentiable double Poisson algebra was introduced. For an algebra A, the noncommutative analogue of the classical graded Lie algebra ( O(M) Der(O(M)), {−, −}) of polyvector ﬁelds on a manifold M, where {−, −}Vis the Schouten-Nijenhuis bracket, is the graded Lie algebra

TADer(A)/[TADer(A), TADer(A)][1] with graded Lie bracket {−, −} := µA ◦{{−, −}} where µA is the multiplication map on A and {{−, −}} is the double Schouten bracket deﬁned in [12, §3.2].

4 The classical notion of a Poisson tensor in this new setting becomes

Proposition 1 ([12, §4.4]) Let P ∈ (TADer(A))2 such that {P,P } = 0, then P determines a double Poisson bracket on A. We call such elements double Poisson tensors.

In case A is formally smooth (for example if A is a path algebra of a quiver), there is a one-to-one correspondence between double Poisson tensors on A and double Poisson brackets on A. For a double Poisson tensor P = δ∆, the corresponding double Poisson bracket is, for a, b ∈ A, determined by

′ ′′ ′ ′′ {{a, b}}P = δ(a) ∆δ(a) (b) − ∆(a) δ∆(a) (b).

In order to obtain the noncommutative analogue of the classical Poisson- Lichnerowicz cohomology, we observe that TA/[TA, TA][1] is a graded Lie algebra, so it is a well-known fact that if P satisﬁes {P,P } = 0 the map

dP := {P, −} : TA/[TA, TA][1] → TA/[TA, TA][1] is a square zero diﬀerential of degree +1. This leads to

Deﬁnition 2 Let A be a diﬀerentiable double Poisson algebra with double • Poisson tensor P , then the homology HP (A) of the complex

dP dP dP dP 0 → TA/[TA, TA][1]0 → TA/[TA, TA][1]1 → TA/[TA, TA][1]2 → ... is called the double Poisson-Lichnerowicz cohomology of A.

Analogous to the classical interpretation of the ﬁrst Poisson cohomology groups, we have the following interpretation of the double Poisson cohomology groups:

0 HP (A)= {double Casimir functions} := {a ∈ A | a mod [A, A] ∈ Z(A/[A, A])} 1 HP (A)= {double Poisson vector fields}/{double Hamiltonian vector fields}, where in analogy to the classical definitions, a double Poisson vector field is a degree 1 element δ ∈ TA/[TA, TA] satisfying {P, δ} = 0 and a double Hamiltonian vector field is a degree 1 element of the form {P, f} with f ∈ A/[A, A]. Indeed, let us illustrate the first claim. We have for a ∈ A that

{δ∆, a} = +∆(a)′δ∆(a)′′ − δ(a)′∆δ(a)′′ whence for any P ∈ (TADer(A))2 we get

{P, a}(b)= −{{a, b}}P

5 so if P is a double Poisson tensor and this expression is zero modulo commutators then a mod [A, A] is indeed a central element of the Lie algebra (A/[A, A], {−, −}P ).

Let A be an associative algebra with unit. From [12, §7] we know that the

Poisson bracket on repn(A) and issn(A) induced by a double Poisson tensor P corresponds to the Poisson tensor tr(P ). We furthermore know that the map tr : TA/[TA, TA][1] → Der(O(repn(A)) is a morphism of graded Lie algebras, so we have a morphismV of complexes

dP dP 0 / (TA/[TA, TA])0 / (TA/[TA, TA])1 / (TA/[TA, TA])2 / ...

tr tr tr d d / tr(P )/ tr(P )/ 2 / ... 0 / O(repn(A)) / Der(O(repn(A))) / ∧ Der(O(repn(A))) / which restricts to a morphism of complexes

dP dP 0 / (TA/[TA, TA])0 / (TA/[TA, TA])1 / (TA/[TA, TA])2 / ...

tr tr tr d d tr(P ) tr(P ) 2 0 / O(issn(A)) / Der(O(issn(A))) / ∧ Der(O(issn(A))) / ...

So there is a map from the double Poisson-Lichnerowicz cohomology to the classical Poisson-Lichnerowicz cohomology on repn(A) and issn(A).

Remark 1 It is a well-known fact that in classical Poisson cohomology, because of the biderivation property satisﬁed by the Poisson bracket, the Casimir elements form an algebra. The higher order cohomology groups all are modules over this algebra. Note that in case of double Casimir elements, this no longer is the case, as for two double Casimir elements f and g of a double Poisson tensor P , we have

{{P,fg}} = f{{P,g}} + {{P, f}}g whence {P,fg} ∈ f[A, A] + [A, A]g 6⊆ [A, A].

It is a natural and interesting question to ask whether the map from the double Poisson cohomology to the classical double Poisson cohomology is onto or not. For ﬁnite-dimensional semi-simple algebras it is onto (see [11]), but for some of the double Poisson brackets considered in Section 6, the map is not onto.

3 NC Multivector Fields and the graded Necklace Lie Algebra

For a quiver Q, the necklace Lie algebra was introduced in [1] in order to gen- eralize the classical Karoubi-De Rham complex to noncommutative geometry.

6 • • • • v HH v HH •vv H• •vv H• )) )) x ) x ) u u •) w1 • •) w2 • )) )) a a •HH • •HH • H vvv H vvv •[ •H − •[ •H a∈Q1 v [ HH v [ HH vv ∗ vv ∗ P •v a • •v a • )) )) x ) x ) •) w2 • •) w1 • )) )) • • • • HH v HH v H• •vv H• •vv

Fig. 1. Lie bracket [w1, w2] in NQ.

We will brieﬂy recall the notions from [1] needed for the remainder of this section.

Deﬁnition 3 For a quiver Q, deﬁne its double quiver Q as the quiver obtained by adding for each arrow a in Q an arrow a∗ in the opposite direction of a to Q.

Now recall that the necklace Lie algebra was deﬁned as

Definition 4 The necklace Lie algebra NQ is defined as NQ := CQ/[CQ, CQ] equipped with the Kontsevich bracket which is defined on two necklaces w1 and w2 as illustrated in Figure 2. That is, for each arrow a in w1, look for all ∗ ∗ occurrences of a in w2, remove a from w1 and a from w2 and connect the corresponding open ends of both necklaces. Next, sum all the necklaces thus obtained. Now repeat the process with the roles of w1 and w2 reversed and deduct this sum from the first.

Now consider the following grading on CQ: arrows a in the original quiver are given degree 0 and the starred arrows a∗ in Q are given degree 1. We now can consider the graded necklace Lie algebra CQ/[CQ, CQ]super equipped with a graded version of the Kontsevich bracket, as introduced in [7].

Deﬁnition 5 The graded necklace Lie algebra is deﬁned as

nQ := CQ/[CQ, CQ]super equipped with the graded Kontsevich bracket deﬁned in Figure 3. Monomials in nQ are depicted as ornate necklaces, where the beads represent arrows in the necklace and where one bead is encased, indicating the starting point of the necklace.

7 1 2 pa pa

• • C • • C ~~ CC ~~ CC ~~ C ~~ C • w1 • • w2 • @@ { @@ { @@ @@ 1 { 2 { qa • __ a qa • __ a 1 2 (−1)da +(−1)da 2 ∗ 1 ∗ pa∗ • __ a pa∗ • __ a a∈Q1 A A X ÐÐ ÐÐ ÐÐ A ÐÐ A • w2 • • w1 • @@ { @@ { @@ {{ @@ {{ • • { • • {

2 1 qa∗ qa∗

Fig. 2. the graded Kontsevich bracket {w1, w2} in nQ. The dashed links, the beads ∗ ∂ inbetween these links (denoted as a and a = ∂a and the round ornamentation are removed and the open necklaces are connected as indicated. The exponents used 1 2 1 1 2 2 1 1 are da = (|w1|− 1)|pa∗ | + |pa||qa| +1 and da = (|w2|− 1)|pa| + |pa∗ ||qa∗ |. An example of an ornate necklace is

◦ g Ó ;; ÓÓ ;; f • ;; ÓÓ ;; ÓÓ • h representing the element fδg∆h∆ if we let ◦ represent δ and • represents ∆. The identities coming from dividing out supercommutators then look like ◦ g ◦ g Ó ;; Ó ;; ÓÓ ;; ÓÓ ;; f • = f • . ;; ÓÓ ;; ÓÓ ;; ÓÓ ;; ÓÓ • h • h

This graded necklace Lie algebra is the noncommutative equivalent of the classical graded Lie algebra of multivector ﬁelds.

Theorem 1 Let Q be a quiver, then

TCQDer(CQ)/[TCQDer(CQ), TCQDer(CQ)] ∼= nQ as graded Lie algebras.

Proof. From [12] we know that the module of double derivations Der(CQ) is C ∂ generated as a Q-bimodule by the double derivations ∂a , a ∈ Q1, deﬁned as

∂ et(a) ⊗ eh(a) b = a (b)= . ∂a   0 b =6 a

 8 ∂ ∗ Now note that we may identify ∂a with an arrow a in Q in the opposite ∂ ∂ D C ∼ C ∗ direction of a: ∂a = eh(a) ∂a et(a), so TCQ er( Q) = Q. The arrows a correspond to the degree 1 elements in the tensor algebra and the original arrows to the degree 0 arrows. That is, supercommutators in the algebra on the left correspond to supercommutators in the algebra on the right.

Now note that the NC Schouten bracket on a path in Q becomes

n−1 ∗ |a1|+···+|ai| ∗ {{a , a1. . . . .an}} = (−1) a1 ...ai{{a , ai+1}}ai+2 ...an i=0 X n |a1|+···+|ai| = (−1) a1 ...ai ⊗ ai+2 ...an, i=0,aXi+1=a where a0 = t(a1). But this is the graded version of the necklace Loday algebra considered in [12]. This becomes the graded necklace Lie algebra when restricting to closed paths and modding out commutators.

An immediate corollary of the Theorem above is

Corollary 6 A nonzero double bracket on the path algebra CQ is completely determined by a linear combination of necklaces of degree 2.

For the remainder of the paper, we will assume the brackets to be nonzero.

4 Linear Double Poisson Structures

In classical Poisson geometry, linear Poisson structures are deﬁned on Cn through a Poisson tensor of the form

k ∂ ∂ π = cij xk ∧ , ∂xi ∂xj where we use Einstein notation, that is, we sum over repeated indices. For this k expression to be a Poisson tensor, the constant factors cij must satisfy

h p h p h p cjkchi + ckichj + cijchk =0,

k i.e. cij are the structure constants of an n-dimensional Lie algebra.

In order to translate this setting to NC Poisson geometry, we first of all note that the role of affine space is assumed by the representation spaces of quivers and the bivector ∂ ∧ ∂ is replaced by the degree 2 part of a necklace, ∂ ∂ , ∂xi ∂xj ∂a ∂b by Corollary 6. We define

9 Deﬁnition 7 Let Q be a quiver. A linear double bracket on CQ is a double bracket determined by a double tensor of the form ∂ ∂ P := ca a , lin bc ∂b ∂c a,b,c∈Q1 ∂ ∂X n a ∂b ∂c 6=0∈ Q a C with all cbc ∈ .

We can characterize the linear double Poisson brackets as follows.

Theorem 2 A linear double bracket ∂ ∂ P := ca a , lin bc ∂b ∂c a,b,c∈Q1 ∂ ∂X n a ∂b ∂c 6=0∈ Q

∂ ∂ ∂ Poisson bracket if and only if for all p,q,r,s ∈ Q1 such that p ∂q ∂r ∂s =06 ∈ nQ we have x p p y crscqx − cyscqr, (p,q,rs) (qr,p,s) x∈QX1 y∈QX1 where ∂ ∂ ∂ ∂ Q(p,q,rs) = {a ∈ Q | a ,p =06 ∈ n } 1 1 ∂r ∂s ∂q ∂a Q and ∂ ∂ ∂ ∂ Q(qr,p,s) = {a ∈ Q | a ,p =06 ∈ n }. 1 1 ∂q ∂r ∂a ∂s Q

Proof. We have to verify when {Plin,Plin} = 0 modulo commutators. First of all observe that a straightforward computation yields that for any x,y,z,u,v,w ∈ Q1

∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ {x ,u } = δ u + δ x ∂y ∂z ∂v ∂w xw ∂v ∂y ∂z uz ∂y ∂v ∂w ∂ ∂ ∂ ∂ ∂ ∂ −δ x − δ u uy ∂v ∂w ∂z xv ∂y ∂z ∂w modulo commutators. Next, note that this equality implies that {Plin,Plin} lies in the sub vector space of nQ that has as basis B all ornate necklaces of ∂ ∂ ∂ the form p ∂q ∂r ∂s with p,q,r,s ∈ Q1 and where p is the encased bead. We now write

∂ ∂ ∂ ∂ {Plin,Plin} = {x ,u } x,y,z u,v,w ∂y ∂z ∂v ∂w X X x u ∂ ∂ ∂ x z ∂ ∂ ∂ = cyzcvxu + cyzcvwx x,y,z u,v ∂v ∂y ∂z v,w ∂y ∂v ∂w X X X

10 x y ∂ ∂ ∂ x u ∂ ∂ ∂ − cyzcvwx − cyzcxwu , v,w ∂v ∂w ∂z u,w ∂y ∂z ∂w ! X X where in order to lighten notation, we do not explicitly write down the addi- x w tional restrictions on x,y,z,u,v,w for a cyz and cuv to be deﬁned. Regrouping ∂ ∂ ∂ this expression with respect to the basis p ∂q ∂r ∂s , we get

x p p y ∂ ∂ ∂ {Plin,Plin} = 2  crscqx − cyscqr p , ∂ ∂ ∂ ∂q ∂r ∂s p ∈B x∈Q(p,q,rs) y∈Q(qr,p,s) ∂q ∂rX∂s  X1 X1    where the summation sets simply list which coeﬃcients are deﬁned. Equating this last expression to zero concludes the proof.

We have the following immediate corollary for the induced Poisson bracket on the representation and quotient spaces.

Corollary 8 Let Plin be a linear double Poisson bracket, then the induced C C bracket tr(Plin) on repn( Q) and on issn( Q) is a linear Poisson bracket.

Proof. This is immediate from the deﬁnition of the induced bracket.

Note that this proof immediately indicates that not all linear Poisson structures are induced by linear double Poisson structures, as the condition on the coeﬃcients of the latter is more restrictive than the condition on the coeﬃ- cients of the former. We even have

Corollary 9 With notations as above, let Plin be a linear double Poisson bracket and v ∈ Q0, then Plin induces an associative algebra structure on the vector space generated by all loops in v.

Proof. It suﬃces to observe that the condition in Theorem 2 exactly determines the structure constants of an associative algebra structure on the loops in a vertex.

5 Cohomology of linear double Poisson brackets and Hochschild cohomology

In this section, we will ﬁrst of all specialize the description of linear double Poisson brackets obtained in the previous section to the free algebra in n variables. Then, we will give a link between the double Poisson cohomology

11 of such linear double Poisson brackets and the Hochschild cohomology of the corresponding associative algebra.

First of all observe that Corollary 9 becomes much stronger for the free algebra in n variables.

Proposition 10 There is a one-to-one correspondence between linear double Poisson brackets on Chx1,...,xni and associative algebra structures on V = Cx1 ⊕···⊕ Cxn. Explicitly, consider the associative algebra structure on V determined by n k xixj := cijxk, i,j,kX=1 then the corresponding double Poisson bracket is given by

n k k {{xi, xj}} = (cijxk ⊗ 1 − cji1 ⊗ xk). i,j,kX=1

Proof. This holds as the ﬁrst condition in Theorem 2 vanishes because the free algebra in n variables is the path algebra of a quiver with one vertex and n loops. As a corollary we obtain for the noncommutative aﬃne plane

Corollary 11 Up to aﬃne transformation, there are only 7 diﬀerent linear double Poisson brackets on Chx, yi. Their double Poisson brackets are

C×C ∂ ∂ ∂ ∂ Plin = x ∂x ∂x + y ∂y ∂y C×Cǫ2 ∂ ∂ Plin = x ∂x ∂x C⊕Cǫ2 ∂ ∂ ∂ ∂ ∂ ∂ Plin = x ∂x ∂x + y ∂x ∂y + y ∂y ∂x Cǫ⊕Cǫ2 ∂ ∂ Plin = y ∂x ∂x 1 B2 ∂ ∂ ∂ ∂ Plin = x ∂x ∂x + y ∂x ∂y 2 B2 ∂ ∂ ∂ ∂ Plin = x ∂x ∂y + y ∂y ∂y Cǫ2⊕Cǫ2 Plin = 0

Proof. This follows immediately from Proposition 10 and the classiﬁcation of all (non-unital) 2-dimensional associative algebras obtained in [6]. The upper indices of the brackets listed here correspond to the algebra structures they induce. Moreover, there is a direct connection between the Hochschild cohomology of the ﬁnite dimensional algebra and the double Poisson cohomology of the free algebra.

12 Theorem 3 Let A = Cx1 ⊕···⊕ Cxn be an n-dimensional vector space and let n k ∂ ∂ P = cijxk ∂xi ∂xj i,j,kX=1 be a linear double Poisson structure on TCA = Chx1,...,xni. Consider A as an algebra through the product induced by the structure constants of P and let HH•(A) denote the Hochschild cohomology of this algebra, then

• ∼ • (HP (TCA))1 = HH (A).

• TTCA Here the grading on (H (TCA)) is induced by the grading on , P [TTCA,TTCA] i which is deﬁned through deg(xi)=1.

T Proof. First of all observe that we have a basis for TCA consisting [TTCA,TTCA] i,1 of all possible elements of the form ∂ ∂ ∂ ... xℓ. ∂xk1 ∂xk2 ∂xki Now consider the linear map

∗ ∗ ∗ TTCA ϕi : A ⊗ A ⊗···⊗ A ⊗A → [TTCA, TTCA]! i factors i,1 deﬁned as | {z }

∗ ∗ ∗ ∂ ∂ ∂ ϕi(xk1 ⊗ xk2 ⊗···⊗ xki ⊗ xℓ)= ... xℓ, ∂xk1 ∂xk2 ∂xki then ϕ := (ϕi) is a morphism of complexes

0 / A d / A∗ ⊗ A d / A∗ ⊗ A∗ ⊗ A / ...

ϕ0 ϕ1 ϕ2 TT A d TT A d TT A / C P / C P / C / ... 0 [TTCA,TTCA] [TTCA,TTCA] [TTCA,TTCA] 0,1 1,1 2,1 where the upper complex is the Hochschild complex with d deﬁned as

n ∗ ∗ ∗ t ∗ ∗ ∗ ∗ d(xk1 ⊗ xk2 ⊗···⊗ xki ⊗ xℓ)= csℓ xs ⊗ xk1 ⊗ xk2 ⊗···⊗ xki ⊗ xt s,t X=1 i−1 n r kr ∗ ∗ ∗ ∗ + (−1) cst xk1 ⊗···⊗ xs ⊗ xt ⊗···⊗ xki ⊗ xℓ r=1 s,t=1 X X r−th factor n i+1 t ∗ ∗ | {z ∗ } ∗ +(−1) cℓs xk1 ⊗ xk2 ⊗···⊗ xki ⊗ xs ⊗ xt. s,t X=1

13 In order to see ϕ is a morphism of complexes, we compute

∂ ∂ ∂ i ∂ ∂ {{P, ... xℓ}} =(−1) ... {{P, xℓ}} ∂xk1 ∂xk2 ∂xki ∂xk1 ∂xki

T1 | ∂ {z∂ ∂ } + {{P, ... }}xℓ . ∂xk1 ∂xk2 ∂xki

T2 The ﬁrst term in this expression is mapped| by the{z multiplication} on the tensor algebra to

i ∂ ∂ ∂ µ(T1)=(−1) ... {P, xℓ} ∂xk1 ∂xk2 ∂xki n i ∂ ∂ ∂ r ∂ r ∂ =(−1) ... csℓxr − cℓs xr ∂xk1 ∂xk2 ∂xki r,s ∂xs ∂xs ! X=1 n r ∂ ∂ ∂ ∂ = csℓ ... xr r,s ∂xs ∂xk1 ∂xk2 ∂xki X=1 n i r ∂ ∂ ∂ ∂ −(−1) cℓs ... xr r,s ∂xk1 ∂xk2 ∂xki ∂xs X=1 The second term in the expression is mapped to

i−1 r ∂ ∂ ∂ ∂ ∂ µ(T2)= (−1) ... {P, } ... xℓ r ∂xk1 ∂xkr ∂xkr+1 ∂xkr+2 ∂xki X=0 i−1 n r kr+1 ∂ ∂ ∂ ∂ ∂ ∂ = − (−1) cst ...... xℓ. r s,t ∂xk1 ∂xkr ∂xs ∂xt ∂xkr+2 ∂xki X=0 X=1 Adding these two expressions together indeed yields

∂ ∂ ∂ ∗ ∗ ∗ dP ... xℓ = dP (ϕi(xk1 ⊗ xk2 ⊗···⊗ xki ⊗ xℓ)) ∂xk1 ∂xk2 ∂xki ! ∗ ∗ ∗ = ϕi(d(xk1 ⊗ xk2 ⊗···⊗ xki ⊗ xℓ)), so we have a morphism of complexes. It is easy to see this is an isomorphism when restricting to degree 1 terms in the lower complex, which ﬁnishes the proof.

Remark 2 Note that Theorem 3 can be seen as the noncommutative analogue of the fact that for a Lie-Poisson structure associated to a compact Lie group

14 the Poisson cohomology can be written as the tensor product of the Lie algebra cohomology with the Casimir elements of the Poisson bracket.

The relation between higher degree components of the double Poisson cohomology and the Hochschild cohomology of the corresponding ﬁnite dimensional algebra is less obvious. However, we have

Theorem 4 With notations as in the previous theorem, there is a canonical embedding 0 C 0 ⊗k ,( HP (TCA) ֒→ ⊕ HH (A, A kM≥1 where the action of A on A⊗k is the inner action on the two outmost copies of A in the tensor product.

Proof. We have already shown in Theorem 3 that the Hochschild cohomology 0 0 HH (A) corresponds to the degree 1 terms in HP (TCA). We will now show 0 0 ⊗k that (HP (TCA))k ֒→ HH (A, A ). Consider the map

k−1 TCA ⊗k ℓ ϕ0 : → A : xi1 ⊗···⊗ xik 7→ σ(1...k)xi1 ⊗···⊗ xik , [TCA, TCA] Xℓ=0 where for s ∈ Sk a permutation we deﬁne σs(a1 ⊗···⊗ak) := as(1) ⊗···⊗as(k). It is easy to see that this map is well-deﬁned. We will also need the map

TTCA ∗ ⊗k ϕ1 : → A ⊗ A [TTCA, TTCA]!1,k deﬁned as ∂ ∗ ϕ1( xj1 ...xjk )= xi ⊗ xj1 ⊗···⊗ xjk , ∂xi T where we ﬁxed a similar basis as in Theorem 3 for TCA . [TTCA,TTCA] 1,k

Now compute for f = xi1 ...xik (using Einstein notations)

′′ ′ ′′ ′ r ∂f ∂f r ∂f ∂f ∂ dP (f)= cpq xr − cqpxr ∂xq ∂xq ∂xq ∂xq ! ∂xp k−1 ′′ ′ r ∂xis+1 ∂xis+1 = cpq xis+2 ...xik xi1 ...xis xr s=0 ∂xq ∂xq X ′′ ′ r ∂xis+1 ∂xis+1 ∂ −cqp xr xis+2 ...xik xi1 ...xis ∂xq ∂xq ! ∂xp k−1 r = cpis+1 xis+2 ...xik xi1 ...xis xr s X=0

15 r ∂ −cis+1pxrxis+2 ...xik xi1 ...xis ∂xp

This is mapped by ϕ1 to

k−1 r ∗ (cpis+1 xp ⊗ xis+2 ⊗···⊗ xik ⊗ xi1 ⊗···⊗ xis ⊗ xr s=0 Xr ∗ −cis+1p xp ⊗ xr ⊗ xis+2 ⊗···⊗ xik ⊗ xi1 ⊗···⊗ xis ). On the other hand, we have

k−1

dϕ0(f)= d(xis+1 ⊗···⊗ xik ⊗ xi1 ⊗···⊗ xis ) s X=0 k−1 r ∗ = cpis xp ⊗ xis+1 ⊗···⊗ xik ⊗ xi1 ⊗···⊗ xis−1 ⊗ xr s X=0 r ∗ −cis+1p xp ⊗ xr ⊗ xis+2 ⊗···⊗ xik ⊗ xi1 ⊗···⊗ xis . So, after reindexing the ﬁrst term, we obtain d(ϕ0(f)) = ϕ1(dP (f)), which ﬁnishes the proof.

6 Examples of H0 and H1 for several double Poisson structures on Chx, yi

In this section, we will determine the double Poisson cohomology groups H0 and H1, of three diﬀerent double Poisson structures on the free algebra Chx, yi. To do this, we will use some tools and techniques that could be useful to compute other double Poisson cohomology groups, associated to other double Poisson algebras.

In order to compute the double Poisson cohomology of Chx, yi with the diﬀer- ent brackets we will introduce below, the following noncommutative version of the Euler formula will prove useful.

Proposition 12 (NC-Euler formula) Let Q be a quiver, p a path in Q and a an arrow of Q, then ∂ µ ◦ a (p) = (dega(p)) p. ∂a ! Whence ∂ µ ◦ a (p)= |p| p, ∂a ! aX∈Q1

16 with |p| the length of the necklace.

d Proof. From the deﬁnition of a da , for any path in Q, of the form a1·a2 ··· an−1· ∗ an (with n ∈ N and a1, a2,...,an, some arrows of Q), we have:

n ∂ ∂ai (a1 · a2 ··· an−1 · an)= a1 ··· ai−1 · · ai+1 ··· an ∂a i=1 ∂a Xn

= a1 ··· ai−1 · et(ai) ⊗ eh(ai) · ai+1 ··· an i=1 aX=ai n = a1 ··· ai−1 ⊗ ai+1 ··· an. i=1 aX=ai

So that, by deﬁnition of the inner product, we obtain:

∂ n a (a1 · a2 ··· an−1 · an)= a1 ··· ai−1 ⊗ ai · ai+1 ··· an, ∂a ! i=1 aX=ai proving the proposition.

The rest of this section will be devoted to the determination of the low- dimensional double Poisson cohomology groups H0 and H1 of the free algebra Chx, yi, equipped with the following (linear and non-linear) double Poisson tensors:

d d d d d d (1) the linear double Poisson brackets P0 = x dx dx and x dx dx + y dy dy , for C C which the corresponding Poisson bracket on rep1( hx, yi) = [x, y] is zero; d d d d d d d d (2) the linear double Poisson brackets P1 = x dx dx + y dx dy or x dx dy + y dy dy , C for which the Poisson bracket obtained with the trace on rep1( hx, yi) is a non-trivial Poisson bracket; d d (3) the quadratic double Poisson bracket P = x dx x dy , which induces a C quadratic (non-trivial) Poisson bracket on rep1( hx, yi).

d d d d d d 6.1 The linear double Poisson tensors x dx dx and x dx dx + y dy dy

Let us ﬁrst consider the linear double Poisson tensor, given by

2 d d P := P C×Cǫ = x . 0 lin dx dx

17 Our aim, in this subsection, is to give an explicit basis of the double Poisson 0 C 1 C cohomology groups HP0 ( hx, yi) and HP0 ( hx, yi), associated to this double Poisson tensor P0.

0 0 C First of all, let us consider the operator dP0 and the space HP0 ( hx, yi).

Proposition 13 For f ∈ Chx, yi, we have

df ′′ df ′ df ′′ df ′ dP0 (f)= ◦ dx dx x − dx dx x . Which leads to 0 C C C HP0 ( hx, yi) ≃ [x] ⊕ [y].

Proof. Let f ∈ Chx, yi and let recall that P0 denotes the double Poisson d d tensor P0 = x dx dx . Then, using the properties of the double Gerstenhaber bracket {{−, −}}, given in [12, §2.7], we compute the double Schouten-Nijenhuis bracket of f and P0:

d d {{f,P }} = {{f, x }} 0 dx dx d d d d = −x {{f, }} + {{f, x }} dx dx dx dx d df ′′ df ′ df ′′ df ′ d = x ⊗ − x ⊗ , dx dx dx dx dx dx so that, computing modulo the commutators, we obtain exactly df ′′ df ′ df ′′ df ′ d dP0 (f)= µ ({{P0, f}})= x − x . dx dx dx dx ! dx

Then, a 0-cocycle for the double Poisson cohomology, corresponding to P0 is C an element f ∈ hx, yi satisfying dP0 (f) = 0, which means that df ′′ df ′ df ′′ df ′ x − x =0, dx dx dx dx df ′′ df ′ that is to say, the element ∈ Chx, yi commutes with x, so is necessarily dx dx an element of C[x]. According to the NC-Euler Formula (Proposition 12), we have ′ ′′ ′′ ′ ∂ df df df df C C degx(f) f = µ ◦ x (f)= x ∈ x + [ hx, yi, hx, yi], ∂x ! dx dx dx dx C so that, modulo commutators, we either have degx(f) = 0 and hence f ∈ [y], or f ∈ C[x]. But then 0 C C C HP0 ( hx, yi) ≃ [x] ⊕ [y].

18 Next, let us determine the ﬁrst double Poisson cohomology group, related to P0. First of all, observe that an element of (TChx,yi/[TChx,yi, TChx,yi])1 can be d d uniquely written as f + g , with f,g ∈ Chx, yi. By a direct computation dx dy

(analogous to what we did for dP0 (f)), we obtain the value of the coboundary operator dP0 on such an element. We obtain that d d dP0 f + g = Φ1(f) + Φ2(g), dx dy ! where the operators Φ1 and Φ2 from Chx, yi to (TChx,yi/[TChx,yi, TChx,yi])2 are deﬁned, for f,g ∈ Chx, yi, by: ◦ ◦ ◦ Ñ << | @@ Ð DD ÑÑ < || @ ÐÐ D Φ1(f) := − + df ′ df ′′ − df ′ df ′′ , (1) f = 1 x x == ÒÒ dx B dx dx > dx Ò BB ~~ >> zz ◦ ◦ ~ ◦ z and ◦ ◦ | @@ Ð DD || @ ÐÐ D Φ (g) := dg ′ dg ′′ − dg ′ dg ′′ , (2) 2 x x dx B dx dx > dx BB ~~ >> zz • ~ • z d d 1 C where ◦ represents dx and • represents dy . Now, to compute HP0 ( hx, yi), we have to consider elements f,g ∈ Chx, yi, satisfying the two independent equations: Φ1(f)=0 and Φ2(g) = 0. We ﬁrst consider the second equation, Φ2(g) = 0. We then have

Proposition 14 The kernel of the linear map Φ2, from Chx, yi to (TChx,yi/[TChx,yi, TChx,yi])2 is ker(Φ2)= C[y].

Proof. Let g ∈ Chx, yi be polynomial in x and y, such that Φ2(g) = 0. Let us write g = xg0 + yg1 + a, where g0,g1 ∈ Chx, yi and a ∈ C. Then, we have dg dg dg =1 ⊗ g + x 0 + y 1 and the equation Φ (g) = 0 becomes: dx 0 dx dx 2

◦ B ◦ B ◦ B BB www BB www BB ′ w ′′ ′ w ′′ 0 = Φ2(g)= x g0 + dg0 dg0 + dg1 dg1 ?? | x x y x ? || dx G dx dx G dx • GGG || GGG || • | • | ◦ ◦ ◦ FF EE EE ÐÐ FF zz EE zz EE Ð ′z ′′ ′z ′′ − 1 xg0 − dg0 dg0 − dg1 dg1 . >> y x x y x > yy dx D dx dx D dx • y DD yy DD yy • yy • yy

19 ◦ FF ÐÐ FF Ð xg Then, we see that the term 1 > 0 has to cancel itself, which means that >> yy • yy xg0 = 0 and g = yg1 + a ∈ y Chx, yi + C.

k Now, we know that we can write g = a + k≥1 y gk with, for any k ≥ 1, dg dg dg ′ dg ′′ g ∈ x Chx, yi + C. We then have = yPk k = yk k ⊗ k and k dx dx dx dx kX≥1 kX≥1 the equation Φ2(g) = 0 becomes

k Φ2(g)=0= y Φ2(gk). kX≥1

So that, for each k ≥ 1, we must have Φ2(gk) = 0. But we have seen above that this implies gk ∈ y Chx, yi+C, while we have assumed that gk ∈ x Chx, yi+C, ∗ thus gk ∈ C, for all k ∈ N . We then conclude that g ∈ C[y].

Now, let us study the ﬁrst equation Φ1(f) = 0. To do this, we will need the following

∗ ∗ 2s k1 k2 k3 k2s−1 k2s Lemma 1 Let s ∈ N and (k1,k2,k3,...,ks) ∈ (N ) . Let m = x y x ··· x y ∈ Chx, yi. We have

◦ O s jjjj OO jjjj OO Φ1(m)= xk1 ··· yk2(i−1) xk(2i−1) yk2i ··· yk2s i=1 TT X TTT ooo TTT ◦ oo ◦ SS s mmm SSS mmm SS − xk1 ··· yk2(i−1) xk(2i−1) yk2i ··· yk2s . (3) i=2 Q k X QQQ kkk QQ ◦ kkk

Next, let n = yk1 xk2 yk3 ··· yk2s−1 xk2s ∈ Chx, yi, then

j ◦ O s jjj OOO jjjj OO k k k Φ1(n)= − y 1 ··· x 2(i−1) y (2i−1) xk2i ··· xk2s i=1 TT o X TTT ooo TTT ◦ oo ◦ SS s mmm SSS mmm SS + yk1 ··· xk2(i−1) yk(2i−1) xk2i ··· xk2s . (4) i=2 Q k X QQQ kkk QQ ◦ kkk

Proof. We will prove the ﬁrst statement of the lemma. The proof of the second statement is completely analogous. Let m = xk1 yk2 xk3 ··· xk2s−1 yk2s ∈ Chx, yi,

20 ∗ ∗ where s ∈ N and where k1,k2,k3,...,ks ∈ N are (non zero) integers. First of all, we compute:

k −1 dm s (2i−1) = xk1 ··· yk2(i−1) xj ⊗ x(k(2i−1)−1−j)yk2i ··· yk2s . dx Xi=1 jX=0

Then, by deﬁnition of Φ1, we have ◦ pp ;; ppp ;; Φ1(m)= − xk1 ...yk2s 1 N Ó NNN ÓÓ N ◦Ó ◦ UU s k(2i−1)−1 kkkk UUUU kkkk UUU + xk1 ··· yk2(i−1) xj+1 x(k(2i−1)−1−j)yk2i ··· yk2s i=1 j=0 SS i X X SSSS iiii SS ◦ iiii ◦ U s k(2i−1)−1 lll UUUU llll UUU − xk1 ··· yk2(i−1) xj x(k(2i−1)−j)yk2i ··· yk2s i=1 j=0 RR X X RRR iiii RR ◦ iiii ◦ pp ;; ppp ;; = − xk1 ...yk2s 1 N Ó NNN ÓÓ N ◦Ó ◦ O s jjjj OO jjjj OO + xk1 ··· yk2(i−1) xk(2i−1) yk2i ··· yk2s i=1 TT X TTT ooo TTT ◦ oo ◦ SS s mmm SSS mmm SS − xk1 ··· yk2(i−1) xk(2i−1) yk2i ··· yk2s , i=1 Q k X QQQ kkk QQ ◦ kkk which yields the formula (3). We will also need the following formula that gives a nice interpretation of dP0 (m), where m is a monomial like in the previous lemma.

∗ ∗ 2s Lemma 2 Let s ∈ N and (k1,k2,k3,...,ks) ∈ (N ) . We consider the monomial in Chx, yi, written as m = xk1 yk2 xk3 ··· xk2s−1 yk2s . Then, we have

dm ′′ dm ′ dm ′′ dm ′ s x − x = − xk2i−1 ··· xk2s−1 yk2s xk1 yk2 ··· yk2(i−1) dx dx dx dx i=1 Xs + yk2i ··· yk2s xk1 ··· yk2(i−1) xk(2i−1) . Xi=1 0 That is, dP0 (m) is obtained by considering all the cyclic permutations of the

21 blocks xj and yj in m (together with the sign of the permutation) and multi- d plying the result by dx . Proof. Similar to the proof of the previous lemma, we have

k −1 dm s (2i−1) = xk1 ··· yk2(i−1) xj ⊗ x(k(2i−1)−1−j)yk2i ··· yk2s . dx Xi=1 jX=0 From this, it is straightforward to obtain equation (5).

We are now able to determine the ﬁrst double Poisson cohomology group of the double Poisson algebra (Chx, yi,P0).

d d Proposition 15 Let us consider the linear double Poisson tensor P0 = x dx dx on Chx, yi. The ﬁrst double Poisson cohomology space, associated to P0 is given by: d d H1 ≃ C ⊕ C[y] . P0 dx dy

d d Proof. Let f dx + g dy be a 1-cocycle of the double Poisson cohomology associated to the double Poisson algebra (Chx, yi,P0). We have seen that the cocycle condition can be written as:

Φ1(f)=0,   Φ2(g)=0,

 where the operators Φ1 and Φ2 are deﬁned in (1) and (2). According to Propo- C C C d sition 14, we know that g ∈ [y]. As for any h ∈ hx, yi, dP0 (h) ∈ hx, yi dx , it C d is clear that the elements of [y] dy give non-trivial double Poisson cohomology 1 C classes in HP0 ( hx, yi).

Remains to consider the equation Φ1(f) = 0. First of all observe this equation implies f ∈ xChx, yiy+yChx, yix+C. In fact, suppose that there is a monomial in f that can be written as xf0 x, where f0 ∈ Chx, yi. Then, we have ◦ ◦ ◦ D E v < ÒÒ DD xxx EE vv << ÒÒ ′ x ′′ v Φ1(xf0 x)= x f0x + df0 df0 + xf0 x 1 << x dx x dx x H Ò << zz F HHH Ò z FFF yy Ò ◦ ◦ y ◦ ◦ ◦ { GG z << { GG z << ′{ ′′ z − df0 df0 − xf0 x . x dx x dx x D Ò C DD ÒÒ CC ww Ò ◦ ww ◦

22 ◦ v << vvv < But then Φ1(f) = 0, implies that the term xf0 x 1 has to cancel itself, so H Ò HHH Ò ◦Ò that xf0 x has to be zero. Now suppose a monomial of the form yf0 y appears in f. We have

◦ ◦ ◦ w < D GG ww << xxx DD {{ GG w ′ x ′′ ′{ ′′ Φ1(yf0 y)= − yf0 y 1 + df0 df0 − df0 df0 . G Ò y dx x dx y y dx x dx y GGG Ò F C Ò FFF zz CC ww ◦ ◦ z ◦ ww

◦ w << www < The term yf0 y 1 cannot appear in Φ1(f) in any other way and hence G Ò GGG Ò ◦Ò has to vanish, i.e. yf0 y has to be zero.

But then we know that f can be written as f = f2s + a, where a ∈ C and s∈N∗ f2s ∈ Chx, yi is of the form P

k1 k2 k3 k2s−1 k2s ℓ1 ℓ2 ℓ3 ℓ2s−1 ℓ2s f2s := cK x y x ··· x y − c˜L y x y ··· y x , K=(k1,...,k2s) L=(ℓ1,...,ℓ2s) ∈(XN∗)2s ∈(XN∗)2s

where cK andc ˜L are constants. According to Lemma 1, the equation Φ1(f)=0 ∗ implies that, for each s ∈ N , Φ1(f2s) = 0 (i.e., Φ1(f2s) can not be cancelled by another Φ1(f2s′ )).

Let us then consider the equation Φ1(f2s) = 0. According to Lemma 1, by col- ◦ ◦ s KK s KKK sss KK sss KK lecting the terms of the form x ···sx y ··· y and of the form x ··· ys x ··· y KK K KK ss KKK sss K◦ ss K◦ ss (which have to be cancelled by terms of the same form), we get the three following equations:

◦ O s jjjj OO jjjj OO 0= cK xk1 ··· yk2(i−1) xk(2i−1) yk2i ··· yk2s K=(k1,...,k2s) i=1 TTT o X∗ X TT oo ∈(N )2s TTT ◦ oo j ◦ O s jjj OOO jjjj OO k k k + c˜L y 1 ··· x 2(i−1) y (2i−1) xk2i ··· xk2s (5) L=(ℓ1,...,ℓ2s) i=1 TTT oo X∗ 2s X TT oo ∈(N ) TTT ◦ oo

23 and

◦ SS s mmm SSS mmm SS 0= cK xk1 ··· yk2(i−1) xk(2i−1) yk2i ··· yk2s . (6) K=(k1,...,k2s) i=2 QQ kk X∗ s X QQ kk ∈(N )2 QQ ◦ kkk

From Equation (5), we conclude, as the ﬁrst sum cannot cancel itself, that, for each 1 ≤ i ≤ s

◦ O jjjj OO jjjj OO cK xk1 ··· yk2(i−1) xk(2i−1) yk2i ··· yk2s K=(k1,...,k2s) TTT o X∗ TT oo ∈(N )2s TTT ◦ oo ◦ Q oo QQQ ooo QQQ = − c k2i k2s k k − K y ··· y x 1 ··· x (2i 1) K=(k1,...,k2s) OO mm X∗ OO mm ∈(N )2s O ◦ mmm j ◦ O jjj OOO jjjj OO k k k = − c˜L y 1 ··· x 2(i−1) y (2i−1) xk2i ··· xk2s L=(ℓ1,...,ℓ2s) TTT oo X∗ 2s TT oo ∈(N ) TTT ◦ oo and this can only happen if, for each 1 ≤ i ≤ s:

k2i k2s k1 k(2i−1) k1 k2s cK y ··· y x ··· x = c˜L y ··· x . (7) K=(k1,...,k2s) L=(ℓ1,...,ℓ2s) ∈(XN∗)2s ∈(XN∗)2s

Then, in the equation (6), the sum obtained for 2 ≤ i ≤ s has to be cancelled by the one obtained for the s − i + 2, i.e.,

◦ SS mmm SSS mmm SS cK xk1 ··· yk2(i−1) xk(2i−1) yk2i ··· yk2s K=(k1,...,k2s) QQ kk X∗ s QQ kk ∈(N )2 QQ ◦ kkk k ◦ Q kkk QQQ kkk QQ = − cK xk(2i−1) yk2i ··· yk2s xk1 ··· yk2(i−1) K=(k1,...,k2s) SSS m X∗ s SS mm ∈(N )2 SS ◦ mmm ◦ VV lll VVVVV llll VVV = cK xk1 ··· yk2(s−i+1) xk(2(s−i)+3) yk2(s−i+2) ··· yk2s K=(k1,...,k2s) RRR hh X∗ RR hhh ∈(N )2s RR ◦ hhhhh when written with exactly 2(s − i +1) blocks of the form xj or yj in the box. This implies, for each 2 ≤ i ≤ s, that

24 k(2i−1) k2s k1 k2(i−1) k1 k2s − cK x ··· y x ··· y = cK x ··· y .(8) K=(k1,...,k2s) K=(k1,...,k2s) ∈(XN∗)2s ∈(XN∗)2s

Now let k1 k2s h2s := cK x ··· y ∈ Chx, yi. K=(k1,...,k2s) ∈(XN∗)2s According to Lemma 2, we have:

dh ′′ dh ′ dh ′′ dh ′ 2s 2s x − x 2s 2s = dx dx dx dx s k2i−1 k2s−1 k2s k1 k2 k2(i−1) − cK x ··· x y x y ··· y i=1 K=(k1,...,k2s) X ∈(XN∗)2s s k2i k2s k1 k2(i−1) k(2i−1) + cK y ··· y x ··· y x . i=1 K=(k1,...,k2s) X ∈(XN∗)2s

From equations (7) and (8), we obtain

dh ′′ dh ′ dh ′′ dh ′ 2s 2s x − x 2s 2s = −s f . dx dx dx dx 2s According to Proposition 13, this yields

d d d 1 d f = f2s + a = dP0 − h2s + a , dx N∗ dx dx s dx sX∈ 1 C C d C d and we conclude that HP0 ( hx, yi) ≃ dx ⊕ [y] dy .

An analogous proof shows that

Proposition 16 Let us consider the linear double Poisson tensor

d d d d P˜ := P C×C = x + y 0 lin dx dx dy dy on Chx, yi. Then, we have:

H0 (Chx, yi) ≃ C[x] ⊕ C[y] P˜0 and the ﬁrst double Poisson cohomology space, associated to P˜0 is given by:

1 d d H ˜ (Chx, yi) ≃ C ⊕ C . P0 dx dy

25 C C×Cǫ2 Remark 3 On rep1( hx, yi), the double Poisson tensors P0 = Plin and ˜ C×C C P0 = Plin induce the trivial Poisson bracket. However, on repn( hx, yi) C×C (n ≥ 2) the double Poisson tensor Plin is mapped (by the trace map) to the ∗ ∗ canonical linear Poisson structure on the product gln ×gln. For the Lie algebra 1 C gln, we know [5] that the Lie algebra cohomology space HL(gln; ) is of dimen- ∗ ∗ sion 1. To obtain the ﬁrst Poisson cohomology group of gln × gln, we have to 1 C consider the tensor product of HL(gln × gln; ) which is of dimension two and ∗ ∗ the algebra of the Casimirs of gln × gln, which is an inﬁnite dimensional vector space. That is, the trace map from H1 (Chx, yi) to H1 (rep (Chx, yi)) P˜0 tr(P˜0) n is not onto.

d d d d d d d d 6.2 The linear double Poisson tensors x dx dx + y dx dy , x dx dy + y dy dy

Now we consider the linear double Poisson tensor:

B1 d d d d P := P 2 = x + y . 1 lin dx dx dx dy

0 C We will determine the double Poisson cohomology groups HP1 ( hx, yi) and 1 C HP1 ( hx, yi). We begin by observing

Lemma 3 Let us consider the free algebra Chx1,...,xni, associated to the quiver Q, with one vertex and n loops x1,...,xn. For each h ∈ Chx1,...,xni, we have

n d d ◦ xi (h) − xi ◦ (h) = h ⊗ 1 − 1 ⊗ h, dxi ! dxi ! ! Xi=1 (where ◦ denotes the inner multiplication). This can also be written as:

n dh ′ dh ′′ dh ′ dh ′′ xi ⊗ − ⊗ xi = h ⊗ 1 − 1 ⊗ h. dxi ! dxi ! dxi ! dxi ! ! Xi=1

Proof. This can easily be seen from the deﬁnition of the d , but it is also dxi a particular case of Proposition 6.2.2 of [12], which states that the gauge ele- d ment E of Q is given by E = , a . "da # aX∈Q1

0 C Now, let us ﬁrst consider the double Poisson cohomology space HP1 ( hx, yi).

26 Proposition 17 For f ∈ Chx, yi, we have

′′ ′ d0 (f)= ◦ y df df − • df ′′ df ′ . P1 dy dy y dx dx

Which means that 0 C C HP1 ( hx, yi)= .

Proof. In fact, by computing dP1 (f)= {P1, f}, one obtains exactly:

′′ ′ ′′ ′ ′′ ′ d0 (f)= ◦ df df x − x df df + df df y − • df ′′ df ′ . P1 dx dx dx dx dy dy y dx dx According to Lemma 3, we have

df ′ df ′′ df ′ df ′′ df ′ df ′′ df ′ df ′′ x ⊗ + y ⊗ +1 ⊗ f = ⊗ x + ⊗ y + f ⊗ 1. dx dx dy dy dx dx dy dy

Applying −op and µ, this last formula gives:

df ′′ df ′ df ′′ df ′ df ′′ df ′ df ′′ df ′ x + y = x + y , dx dx dy dy dx dx dy dy

which leads to the expression for dP1 (f) given above.

Suppose now that f is a 0-cocycle, that is to say dP1 (f) = 0. This is equivalent to say that df ′′ df ′ df ′′ df ′ y = y =0, dy dy dx dx which yields df ′′ df ′ df ′′ df ′ = =0 dy dy dx dx and df ′′ df ′ df ′′ df ′ y + x =0. dy dy dx dx This implies df ′ df ′′ df ′ df ′′ y + x ∈ [Chx, yi, Chx, yi], dy dy dx dx Using the NC-Euler Formula (Proposition 12), we can then write

f ∈ C ⊕ [Chx, yi, Chx, yi].

0 C 0 C C C Finally, HP1 ( hx, yi) = ker(dP1 )/[ hx, yi, hx, yi] = , which concludes the proof. 1 C Let us now determine HP1 ( hx, yi). We will ﬁrst use Lemma 3 to obtain a 1 useful expression for the coboundary operator dP1 .

27 d d Lemma 4 Let f + g ∈ (TC /[TC , TC ]) , then dx dy hx,yi hx,yi hx,yi 1 ◦ ◦ ◦ Ò == Ñ BB Ð >> 1 d d Ò = ÑÑ B ÐÐ > Ò df ′ df ′′ g dP1 f + g = − 1 f + − 1 dx dy ! < dy y dy >> Ð << ÑÑ = > ÐÐ ◦ Ñ == || • ◦ | ◦ ◦ • Ñ BB { >> Ð FFF ÑÑ B {{ > ÐÐ F + dg ′ dg ′′ + df ′′ df ′ − dg ′ dg ′′ dy y dy y dx dx dx y dx . = C >> == || CC ÐÐ > xxx • | • Ð • x

d d d d Proof. First, by computing dP1 f dx + g dy = {P1, f dx +g dy }, one can write d d dP1 f + g =(A)+(B)+(C), dx dy ! where ◦ ◦ ◦ ◦ Ñ << | @@ Ð DD } BB ÑÑ < || @ ÐÐ D }} B (A)= − f 1 + df ′ df ′′ − df ′ df ′′ + df ′ df ′′ == Ò dx x dx dx x dx dy y dy , = Ò B >> A ◦Ò BB ~~ > zz AA || ◦ ~ ◦ z ◦ |

◦ ◦ ◦ Ð >> | @@ Ð DD ÐÐ > || @ ÐÐ D (B)= − g 1 + dg ′ dg ′′ − dg ′ dg ′′ >> Ð x x > ÐÐ dx B dx dx > dx • BB ~~ >> zz • ~ • z ◦ ◦ } ?? { >> }} ? {{ > + dg ′ dg ′′ + df ′′ df ′ dy y dy y dx dx A C AA CC ÐÐ • • Ð and • Ð FFF ÐÐ F (C)= − dg ′ dg ′′ y . dx > dx >> xx • xx

According to Lemma 3, we have

df ′ df ′′ df ′ df ′′ df ′ df ′′ df ′ df ′′ x ⊗ + y ⊗ +1 ⊗ f = ⊗ x + ⊗ y + f ⊗ 1, dx dx dy dy dx dx dy dy

28 d and the same for g. Applying −◦ dx (where ◦ means the right inner multipli- d cation) and then the right (outer) multiplication by dx , we obtain:

◦ @ ◦ ? ◦ = | @@ } ? Ò = || }} ? ÒÒ = df ′ df ′′ + df ′ df ′′ + 1 f = dx x dx dy y dy << B A < ÑÑ BB ~~ AA ◦ Ñ ◦ ~ ◦ ◦ ◦ ◦ Ð DD Ñ BB Ñ @@ ÐÐ D ÑÑ B ÑÑ @ df ′ df ′′ + df ′ df ′′ + f 1 , dx x dx dy y dy == >> = = ~~ > zz == || ◦ ~ ◦ z ◦ | whence ◦ ◦ Ò == Ñ EEE Ò = ÑÑ E (A)= − Ò + df ′ df ′′ 1 < f y . << ÑÑ dy = dy ◦ Ñ == yy ◦ yy

d A similar argument for g (applying −◦ dx and then the right multiplication d by dy ) yields

◦ ◦ ◦ Ð >> Ñ BB { BB ÐÐ > ÑÑ B {{ B (B)= − 1 g + dg ′ dg ′′ + df ′′ df ′ >> Ð y y . > ÐÐ dy = dy dx C dx • == || CC || • | • |

Adding the expressions obtained for (A), (B) and (C), leads to the expression 1 d d of dP1 (f dx + g dy ) stated in the lemma.

1 C Now we can consider the double Poisson cohomology space HP1 ( hx, yi) = 1 0 d d 1 ker(dP1 )/Im(dP1 ). To do this, let f dx +g dy ∈ ker(dP1 ) be a 1-cocycle. According to Lemma 4, this means

◦ ◦ Ò == Ñ BB Ò = ÑÑ B (A)= − Ò + df ′ df ′′ =0, (9) 1 < f y << ÑÑ dy = dy ◦ Ñ == || ◦ |

◦ ◦ ◦ Ð >> Ñ BB { >> ÐÐ > ÑÑ B {{ > (B)= − 1 g + dg ′ dg ′′ + df ′′ df ′ =0 (10) >> Ð y y > ÐÐ dy = dy dx C dx • == || CC ÐÐ • | • Ð

29 and • Ð CC ÐÐ C (C)= − dg ′ dg ′′ =0. (11) y dx > dx >> {{ • {

Equation (9) yields f = yf˜ + a, with a ∈ C and f˜ ∈ Chx, yi. Indeed, write ◦ Ð >> ˜ ˜ ˜ ˜ C C ÐÐ > f = yf + xh + a, with f, h ∈ hx, yi and a ∈ . Then, as 1 > 1 = 0 (up >> ÐÐ ◦ Ð to commutators), we have

◦ A ◦ A ◦ B ÒÒ AA ~~ AA }} BB Ò ′ ~ ′′ ′ } ′′ (A)= − ˜ + df˜ df˜ + dh˜ dh˜ =0. 1 < xh x y < } y dy y dy dy dy < }} @ A ◦ @@ }} AA || ◦ } ◦ |

For this equality to hold, the ﬁrst term has to cancel itself, so that h˜ has to be zero and f = yf˜+ a.

A similar argument for Equation (10) shows that g = yg˜, withg ˜ ∈ Chx, yi ◦ Ð >> ÐÐ > (but, in contrast with f, g can not be a constant because 1 > 1 =6 0). Now, >> ÐÐ • Ð Equation (11) becomes

• | CC || C (C)= − dg˜ ′ dg˜ ′′ =0, y y dx B dx BB {{ • { so that dg˜ = m′ ⊗ m′′ + m′′ ⊗ m′, dx with m′, m′′ ∈ Chx, yi (using Sweedler notation). Using the NC-Euler Formula (Proposition 12), we can now write

1 ′ ′′ ′′ ′ g˜ = ′ ′′ (m xm + m xm )+ p(y), (12) degx(m m )+1 dg˜ where p ∈ Chyi. Then, computing again, we get dx

m′ ⊗ m′′ + m′′ ⊗ m′ =

30 ′ ′′ 1 dm ′′ ′ ′′ ′ dm ′ ′′ xm + m ⊗ m + m x degx(m m )+1 dx dx dm′′ dm′ + xm′ + m′′ ⊗ m′ + m′′x , dx dx ! that is to say

′ ′′ ′ ′′ ′′ ′ (degx(m m ))(m ⊗ m + m ⊗ m )= dm′ dm′′ dm′′ dm′ xm′′ + m′x + xm′ + m′′x . (13) dx dx dx dx Now let h = − (m′′xm′x + m′xm′′x) and k = −p(y)x and let us compute dh ′′ dh ′ dk ′′ dk ′ −y and −y . First, we have dx dx dx dx

dh dm′′ dm′ = − xm′x − m′′ ⊗ m′x − m′′x x − m′′xm′ ⊗ 1 dx dx dx dm′ dm′′ − xm′′x − m′ ⊗ m′′x − m′x x − m′xm′′ ⊗ 1, dx dx dk = −p ⊗ 1, dx so that

dh ′′ dh ′ dm′′ ′′ dm′′ ′ −y =2 y xm′x +2 ym′xm′′ dx dx dx ! dx ! dm′ ′′ dm′ ′ + 2 y xm′′x +2 ym′′xm′, dx ! dx ! dk ′′ dk ′ −y = yp(y). dx dx But, applying the left outer multiplication by x, −op and µ to Equation (13), we obtain

′ ′′ ′′ ′ ′ ′′ (degx(m m ))(m xm + m xm )= dm′ ′′ dm′ ′ dm′′ ′′ dm′′ ′ 2 xm′′x +2 xm′x . dx ! dx ! dx ! dx !

This implies,

dh ′′ dh ′ −y = (deg (m′m′′)+2) (ym′xm′′ + ym′′xm′) . dx dx x

31 In combination with (12) we obtain

−1 dh ′′ dh ′ dk ′′ dk ′ g = yg˜ = ′ ′′ ′ ′′ y − y . (degx(m m ) + 1)(degx(m m )+2) dx dx dx dx dh dk Now, we want to write f in terms of and . To do this, we will use the dy dy Equation (10). Using f = yf˜+ a and g = yg˜, this equation can be written as follows:

◦ B ◦ A }} BB || AA ′ } ′′ ′′ | ′ (B)= dg˜ dg˜ + df˜ df˜ =0. (14) y dy y dy y y A dx B dx AA || BB }} • | • } This implies df˜ dg˜ ′′ dg˜ ′ = − ⊗ . dx dy dy Using this expression and the NC-Euler Formula (Proposition 12), we get:

˜′ ˜′′ ˜ 1 df df f = ′ ′′ x + l(y) df˜ df˜ dx dx degx dx dx +1 −1 dg˜′′ dg˜ ′ = x + l(y), dg˜ ′ dg˜ ′′ dy dy degx dy dy +1 where l ∈ Chyi. Now, the expression forg ˜ obtained in (12) yields

′ ′′ ′ ′ ˜ −2 dm ′′ dm f = ′ xm x dm ′′ ′ ′′ dy ! dy ! ! degx dy m +2 (degx(m m )+1) ′′ ′′ ′′ ′ 2 dm ′ dm − ′′ xm x dm ′ ′ ′′ dy ! dy ! ! degx dy m +2 (degx(m m )+1) 1 dp ′′ dp ′ − x + l(y). dp ′ dp ′′ dy dy degx dy dy +1 dm′ dm′ ′ ′ C Now, as degx = degx(m ) (unless m ∈ hxi, that is, unless = 0), dy ! dy which also holds for m′′, we have exactly

′ ′′ ′ ′ ˜ −2 dm ′′ dm f = ′ ′′ ′ ′′ xm x (degx(m m ) + 2)(degx(m m )+1) dy ! dy ! !

32 ′′ ′′ ′′ ′ 2 dm ′ dm − ′′ ′ ′ ′′ xm x (degx(m m ) + 2)(degx(m m )+1) dy ! dy ! ! dp ′′ dp ′ − x + l(y) dy dy 1 dh ′′ dh ′ = ′′ ′ ′ ′′ (degx(m m ) + 2)(degx(m m )+1) dy dy dk ′′ dk ′ + + l(y). dy dy So that, if

1 L = ′′ ′ ′ ′′ h + k (degx(m m ) + 2)(degx(m m )+1) −1 ′′ ′ ′ ′′ = ′′ ′ ′ ′′ (m xm x + m xm x) − p(y)x, (degx(m m ) + 2)(degx(m m )+1) we have shown that dL ′′ dL ′ dL ′′ dL ′ g = −y , f = y + yl(y)+ a. dx dx dy dy

′′ ′ 1 Finally, as for every n ∈ N∗, yn = y dq dq , with q = yn ∈ Chyi, the element dy dy n dQ ′′ dQ ′ dQ ′′ dQ ′ yl(y) is of the form y , with Q ∈ Chyi (and in particular y = dy dy dx dx 0) and d d d f + g = d0 (L + Q)+ a . dx dy P1 dx d 0 As dx 6∈ ImdP1 , we have shown Proposition 18 The ﬁrst double Poisson cohomology group of Chx, yi, asso- d d d d ciated to the double Poisson tensor P = x + y is given by 1 dx dx dx dy d H1 (Chx, yi) ≃ C . P1 dx Remark 4 If we consider the double Poisson tensor

B2 ∂ ∂ ∂ ∂ P˜ := P 2 = x + y , 1 lin ∂x ∂y ∂y ∂y

1 B2 we obtain in a similar fashion to the computations above for P1 = Plin ,

0 1 d H ˜ ≃ C and H ˜ ≃ C . P1 P1 dy

33 Let us now consider the (classical) Poisson bracket on C[x, y], associated to d d P1, that is tr(P1) = π1 = y dx ∧ dy . According to [9] or [10], or by a direct computation, we have

d H0 (C[x, y]) = C, H1 (C[x, y]) = C , π1 π1 dx i C Hπ1 ( [x, y]) = 0, for all i ≥ 2. i C i C So that the map tr : HP1 ( hx, yi) → Hπ1 ( [x, y]) is bijective, for i =0, 1.

d d 6.3 The nonlinear double Poisson tensor P = x dx x dy

We conclude this section with the determination of the ﬁrst two double Poisson cohomology groups of a nonlinear double Poisson bracket on the free algebra in two variables.

Lemma 5 The double bracket {{−, −}} deﬁned on Chx, yi as

{{x, x}} = {{y,y}} = 0 and {{x, y}} = x ⊗ x is a double Poisson bracket.

Proof. First of all note that this bracket is deﬁned by the double Poisson d d d d tensor x dx x dy . Representing dx by ◦ and dy by •, this double Poisson bracket corresponds to the necklace P depicted as

◦ } CC }} C x A x. AA {{ • { The NC-Schouten bracket of P with itself now becomes ◦ x • x ◦ x • x } AA } AA } AA } AA }} A }} A }} A }} A x A • + x A ◦ − x A • − x A ◦ =0. AA }} AA }} AA }} AA }} • x } • x } • x } • x }

Remark 5 To stress the diﬀerence between double Poisson brackets and or- d d dinary Poisson brackets, note that y dx y dy is also a double Poisson tensor. However, taking higher degree monomials in x or y no longer yields double Poisson tensors. Whereas, of course, for C[x, y], any polynomial ψ in x and d d y deﬁnes a Poisson bracket ψ dx ∧ dy .

34 d d For the remainder of this section, P will be the double Poisson tensor x dx x dy . First of all, observe that

Proposition 19 For f ∈ Chx, yi, we have

◦ df ′′ df ′ • df ′′ df ′ dP (f)= x dy dy x − x dx dx x .

Which means that 0 C C HP ( hx, yi)= .

Proof. The computation of dP (f) was already done in greater generality in 0 C Section 1. To compute HP ( hx, yi), note that

◦ df ′′ df ′ • df ′′ df ′ x dy dy x − x dx dx x =0. implies df ′′ df ′ df ′′ df ′ = =0. dx dx dy dy But then df ′′ df ′ df ′′ df ′ x + y =0. dx dx dy dy This means df ′ df ′′ df ′ df ′′ x + y ∈ [Chx, yi, Chx, yi], dx dx dy dy which by the NC-Euler formula (Proposition 12) implies that

f ∈ C ⊕ [Chx, yi, Chx, yi].

0 C 0 C C Now HP ( hx, yi) = ker(dP )/[ hx, yi, hx, yi], ﬁnishing the proof.

Next, we can state that

d d Lemma 6 Let f dx + g dy ∈ (TChx,yi/[TChx,yi, TChx,yi])1, then

◦ = ◦ < ◦ C ÒÒ = Ñ << }} CC d d ÒÒ = ÑÑ < } dP f + g = − x f − f x + df ′ df ′′ dx dy ! << = Ò dy x x dy << ÑÑ == ÒÒ A • Ñ • Ò AA || ◦ | ◦ ◦ • z BB } CC | DD zz B }} C || D + df ′′ df ′ + dg ′ dg ′′ − dg ′ dg ′′ . x dx dx x dy x x dy dx x x dx D A B DD || AA || BB zz • | • | • z

We will denote this expression by (∗).

35 Proof. Straightforward. 1 Using this lemma, we can determine the kernel of dP . So assume now that d d 1 d d f dx + g dy ∈ ker(dP ). First of all note that if dP f dx + g dy = 0, the third d d term in the expression (∗) for dP f dx + g dy in Lemma 6 has to cancel itself and the sixth term in this expression has to cancel itself. This implies (using the Sweedler notations)

df = xm′ ⊗ m′′x + xm′′ ⊗ m′ x + n′ ⊗ n′′x + xn′′ ⊗ n′ + c 1 ⊗ 1 dy f f f f f f f f f

′ ′′ ′′ C ′ C with mf , mf , nf ∈ hx, yi, cf , nf ∈ and

dg = xm′ ⊗ m′′x + xm′′ ⊗ m′ x + n′ ⊗ n′′x + xn′′ ⊗ n′ + c 1 ⊗ 1 dx g g g g g g g g g ′ ′′ ′′ C ′ C with mg, mg, ng ∈ hx, yi, cg, ng ∈ .

Using the NC-Euler formula (Proposition 12), this implies

1 ′ ′′ ′′ ′ f = ′ ′′ x(mf ymf + mf ymf )x degy(mf mf )+1

1 ′ ′′ ′′ ′ + ′′ (nf ynf x + xnf ynf )+ p(x)+ cf y degy(nf )+1 and

1 ′ ′′ ′′ ′ g = ′ ′′ x(mgxmg + mg xmg)x degx(mgmg )+3 ′ 2ng ′′ + ′′ xng x + q(y)+ cgx. degx(ng)+2

Now note that for cf y, the ﬁrst two terms of (∗) yield

◦ > ◦ ? >> ÐÐ ?? y y Ð −cf x ? − cf > x . ?? ÐÐ >> • Ð • Because of the degree in x of the remaining terms, these terms cannot vanish unless cf = 0.

df Using this last remark and the expression for f above to compute dy again, we obtain

′ ′′ ′′ df 1 dmf ′′ ′ ′′ ′ dmf dmf ′ = ′ ′′ x ymf + mf ⊗ mf + mf y + ymf dy degy(mf mf )+1 dy dy dy

36 ′ ′′ ′ ′′ dmf +mf ⊗ mf + mf y x dy ! ′′ ′′ 1 ′ ′′ ′ dnf dnf ′ ′′ ′ + ′′ (nf ⊗ nf x + nf y x + x ynf + xnf ⊗ nf ). degy(nf )+1 dy dy

df This expression should be equal to the ﬁrst expression found for dy . That is,

′ ′′ ′′ ′ ′ ′′ ′′ ′ xmf ⊗ mf x + xmf ⊗ mf x + nf ⊗ nf x + xnf ⊗ nf ′ ′′ ′′ 1 dmf ′′ ′ ′′ ′ dmf dmf ′ = ′ ′′ x ymf + mf ⊗ mf + mf y + ymf degy(mf mf )+1 dy dy dy ′ ′′ ′ ′′ dmf + mf ⊗ mf + mf y x dy ! ′′ ′′ 1 ′ ′′ ′ dnf dnf ′ ′′ ′ + ′′ (nf ⊗ nf x + nf y x + x ynf + xnf ⊗ nf ). degy(nf )+1 dy dy whence, by comparing elements of the form x...x we obtain

dm′ dm′′ dm′′ dm′ deg (m′ m′′)(m′ ⊗m′′ +m′′ ⊗m′ )= f ym′′ +m′ y f + f ym′ +m′′y f y f f f f f f dy f f dy dy f f dy

′′ C and nf ∈ [x].

Letting y act on the equality obtained in the previous paragraph by the left outer action, we obtain

dm′ dm′′ deg (m′ m′′)(ym′ ⊗ m′′ + ym′′ ⊗ m′ )= y f ym′′ + ym′ y f y f f f f f f dy f f dy dm′′ dm′ + y f ym′ + ym′′y f . dy f f dy

Which yields, using −op and µ, the equality

′ ′′ ′′ ′ ′ ′′ degy(mf mf )(mf ymf + mf ymf )= ′ ′′ ′ ′ ′′ ′′ ′′ ′ dmf ′′ dmf dmf ′ dmf 2 ymf y + ymf y . dy ! dy ! dy ! dy ! !

2 ′′ ′ Now let h = ′ ′′ ′ ′′ mf ymf y, then (degy(mf mf )+2) (degy(mf mf )+1)

′′ ′ dh dh 1 ′ ′′ ′′ ′ x x = ′ ′′ (xmf ymf x + xmf ymf x) dy dy degy(mf mf )+1 and

37 ′′ ′ ′′ ′′ ′′ ′ dh dh 2 dmf ′ dmf x x = ′ ′′ ′ ′′ x ymf y dx dx (degy(mf mf ) + 2)(degy(mf mf )+1) dx ! dx ! ′ ′′ ′ ′ dmf ′′ dmf + ymf y x. dx ! dx ! !

So, writing

dh ′′ dh ′ f := f − x x := yp (x)+ p (x)y + p(x) 1 dy dy 1 1

n i m i with p1 := i=1 aix and p = i=0 bix and P P dh ′′ dh ′ g := g + x x, 1 dx dx

d d 1 we again obtain an element f1 dx + g1 dy in ker(dP ) by Proposition 19. Observe i dhi ′′ dhi ′ i−2 moreover that x for i ≥ 2 can be written as x dy dy x with hi = x y, so we may assume (modifying f and g by the image of a suitable h) p(x)= b1x + b0. Now note that b0 has to be equal to zero as only the ﬁrst two terms of (∗) contain b0 and these do not cancel each other. That is, we may assume p(x)= b1x.

1 The image under dP of this element becomes

◦< ◦ B d d n || << n ÒÒ BB | < ÒÒ dP f1 + g1 = − ai yxi x − ai x xiy dx dy ! B Ò << Xi=1 BB ÒÒ Xi=1 << || •Ò • | ◦ C ◦ ? n i−1 uu C n i sss ? uuu C ss ? + ai xi−j+1 yxj + ai xi−j+1y xj i j II i j K X=2 X=1 II {{ X=2 X=2 KKK I• { K• ◦ A ◦ D • E }} AA {{ DD zz EEE } ′ { ′′ ′ z ′′ − b1 x A x + dg1 dg1 − dg1 dg1 AA }} dy x x dy dx x x dx } C D • CC zz DD yyy • z • y

Note that we cancelled two terms

◦< ◦ B n || << n ÒÒ BB | < ÒÒ − ai xiy x − ai x yxi B Ò << Xi=1 BB ÒÒ Xi=1 << || •Ò • | against the similar terms obtained in the second row of (∗) for j = i in the ﬁrst sum and j = 1 in the second sum.

38 Now observe that for n ≥ 2, the terms in the second row of the expression above cannot be eliminated by any other term because of the location of the y factor, whence ai = 0 for i ≥ 2. That is, f1 = a(xy + yx)+ p(x). Moreover, if a =6 0, the expression

◦ ◦ { ?? CC {{ ? C −a  yx x + x xy  C ? CC ?? {{  {   • •    2 can only be cancelled if g1 = g2 + ay . That is, the image becomes

◦ A ◦ D • E d d }} AA {{ DD zz EEE } ′ { ′′ ′ z ′′ dP f1 + g1 = − b1 x A x + dg2 dg2 − dg2 dg2 dx dy ! AA }} dy x x dy dx x x dx } C D • CC zz DD yyy • z • y

Now if b1 =6 0, this expression can only be zero if g2 = g3 + b1y, and we get

◦ D • E d d {{ DD zz EEE ′ { ′′ ′ z ′′ dP f1 + g1 = dg3 dg3 − dg3 dg3 dx dy ! dy x x dy dx x x dx C D CC zz DD yyy • z • y

dg3 However, the ﬁrst term in this expression can only be zero if dy = 0, so i g3 ∈ C[x]. Finally, observe that in g3, we can cancel all monomials x with i ≥ 2 using h = xi−1 (which does not modify f in any way).

But then we have shown that

Theorem 5 For P as above, we have that

d d H1 (Chx, yi) ≃{(a(xy + yx)+ bx) + ay2 + by + cx + d | a,b,c,d ∈ C}, P dx dy 1 C so in particular dim HP ( hx, yi)=4

Let us now consider the Poisson bracket that corresponds to the double Poisson d d C tensor P = x dx x dy . We then obtain the Poisson algebra ( [x, y], π), where 2 d d π = tr(P )= x dx ∧ dy .

According to [10], as the polynomial x2 is not square-free, the ﬁrst Poisson 1 C cohomology space Hπ( [x, y]) is inﬁnite dimensional, so that

1 C 1 C Corollary 20 The map HP ( hx, yi) → Htr(P )( [x, y]) is not onto.

39 1 C Let us give an explicit basis for this vector space Hπ( [x, y]), in order to make explicit this map.

First of all, we recall that the Poisson coboundary operator is given by: k k C k+1 C dπ = {π, −} : Der( [x, y]) → Der( [x, y]), where {−, −} denotes the C C (classical) Schouten-NijenhuisV bracketV and Der( [x, y]) denotes the [x, y]- module of the derivations of the commutative algebra C[x, y].

We have , for f,g,h ∈ C[x, y],

0 2 dh d dh d dπ(h) = x − + dy dx dx dy ! 1 d d 2 df dg d d dπ(f dx + g dy ) = x + − 2xf ∧ . dx dy ! ! dx dy

So that,

dh dh 0 C C C Hπ( [x, y]) ≃ h ∈ [x, y] | = =0 ≃ , ( dy dx ) df dg (f,g) ∈ C[x, y]2 | x2 + − 2xf =0 dx dy H1(C[x, y]) ≃ ! . π dh dh x2 − , | h ∈ C[x, y] dy dx !

k It is clear that the coboundary operator dπ is an homogeneous operator, for example, if f and g are homogeneous polynomial of same degree d ∈ N, then 1 d d dπ(f dx +g dy ) is given by an homogeneous polynomial of degree d+1, in factor d d of dx ∧ dy . This implies that we can work “degree by degree” and consider only homogeneous polynomials. We recall the (commutative) Euler formula, for an homogeneous polynomial q ∈ C[x, y]:

dq dq x + y = deg(q) q. (15) dx dy

Let us consider (f,g) ∈ C[x, y]2, two homogeneous polynomials of same degree df dg d ∈ N, satisfying the 1-cocycle condition x2 + =2xf, equivalent to dx dy ! df dg x + =2f. We divide f and g by x2 and obtain: dx dy !

2 2 f = x f1 + xf2 + f3, g = x g1 + xg2 + g3,

40 with f1,g1 ∈ C[x, y] and f2, f3,g2,g3 ∈ C[y], homogeneous polynomials. Then the 1-cocycle condition becomes:

2 df1 2 dg1 dg2 dg3 x x + f2 + x + x + =2xf2 +2f3, dx dy dy dy ! that leads to f3 = 0 (because f3 ∈ C[y]) and df dg dg dg x2 1 + f + x2 1 + x 2 + 3 =2f . dx 2 dy dy dy 2 dg df dg dg We then have also f = 3 and x 1 + x 1 + 2 = 0. This equation then 2 dy dx dy dy dg df dg implies that 2 = 0, i.e., g ∈ C and also 1 = − 1 . Suppose now that dy 2 dx dy d ≥ 2 and let us consider the polynomial h := yf1 − xg1. We have, using the Euler formula (15) and the last equation above,

dh df dg dg dg = y 1 − g − x 1 = −y 1 − g − x 1 = −(d − 1)g , dx dx 1 dx dy 1 dx 1 dh df dg df df = f + y 1 − x 1 = f + y 1 + x 1 =(d − 1)f . dy 1 dy dy 1 dy dx 1

2 d d 0 We have obtained that, if d ≥ 2, then x (f1 dx + g1 dy ) = dπ(−h). Moreover, d g3 is an homogeneous polynomial of degree d, in C[y], so that g3 = c3y , with dg c ∈ C. We have also seen that f = 0, f = 3 = c dyd−1 and g = c ∈ C. 3 3 2 dy 3 2 2 It remains to consider the cases where d = 1 and d = 0. First, if d = 0, i.e., f,g ∈ C, then the 1-cocycle condition is equivalent to f = 0, second, if d = 1, we have (f,g)=(ax + by, cx + dy), with a,b,c,d ∈ C and the 1-cocycle condition says that a = d and b = 0. We ﬁnally have obtain the following

Proposition 21 The ﬁrst Poisson cohomology space associated to the Pois- C 2 d d son algebra ( [x, y], π = x dx ∧ dy ) is given by:

1 C C k−1 k C Hπ( [x, y]) ≃ ky x, y ⊕ (0, x). N kM∈

The image of the double Poisson cohomology under the canonical trace map in the classical cohomology now becomes

∂ ∂ Corollary 22 For the double Poisson tensor P = x ∂x x ∂y we have

1 C C k−1 k 1 C Htr(P )( [x, y]) = ky x, y ⊕ tr(HP ( hx, yi)), kM≥3

41 that is

1 C C 2 C C C tr(HP ( hx, yi)) = 2yx,y ⊕ (x, y) ⊕ (0, x) ⊕ (0, 1).

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