# Differential calculus over double Lie algebroids

Sophie Chemla

###### Abstract.

The notion of double Lie algebroid was defined by M. Van den Bergh [12] and
was illustrated by the double quasi Poisson case. We give new examples of
double Lie algebroids and develop a differential calculus in that context. We
recover the non commutative de Rham complex ([4]) and the double Poisson
-Lichnerowicz cohomology ([9]) as particular cases of our construction.

##  1\. Introduction

Let $k$ be a field of characteristic 0. The pair $({\mathcal{L}},\omega)$ is a
Lie Rinehart algebra ([15]) over a commutative $k$-algebra $A$ if
${\mathcal{L}}$ is endowed with a $k$-Lie algebra structure and an $A$-module
structure, these two structures being linked by a compatibility relation
involving the anchor $\omega$. There is a one to one correspondence ([6])
between Lie Rinehart algebra structures on ${\mathcal{L}}$ and Gerstenhaber
algebra structures on $\Lambda{\mathcal{L}}$. If $A$ is smooth and
${\mathcal{L}}$ is a finite projective $A$-module, then ${\mathcal{L}}$ is a
Lie algebroid. Lie Rinehart algebras generalize at the same time $A$-Lie
algebras (case where $\omega$ is 0) and the Lie algebra of derivations over
$A$ (then $\omega=id$). A Poisson smooth algebra gives rise to a Lie Rinehart
algebra structure (on ${\mathcal{L}}=\Omega^{1}$ ) and this is also true for
quasi Poisson smooth $G$-algebras ([1], then
${\mathcal{L}}=\Omega^{1}\oplus\Gamma(N\times{\mathfrak{g}}))$. Lie algebroids
have been extensively studied and used, in particular in Poisson geometry. A
Lie Rinehart algebra defines a differential $d_{\mathcal{L}}$ on the algebra
$\wedge_{A}{\mathcal{L}}^{*}$. In the case of a $A$-Lie algebra, we recover
the Cartan -Eilenberg complex. In the case where ${\mathcal{L}}=Der(A)$ (with
$A$ smooth), we recover the de Rham differential. In the case where
${\mathcal{L}}=\Omega^{1}(N)$ ($N$ being a Poisson manifold), we recover the
Lichnerowicz-Poisson cohomology. More generally a differential calculus has
been developed for Lie algebroids ([5]).

In this article, we are interested in the case where $A$ is not necessarily
commutative. As in [11] and [12], we follow a non commutative version of
differential geometry based on a Kontsevitch idea : for a property of a non
commutative $k$-algebra to have a geometric meaning, it should induce (through
a trace map) standard geometric properties on all representation spaces
$Rep(A,N)=Hom(A,Mat_{N}(k))$ for all integer $N$. The coordinate ring of
$Rep(A,N)$ is

${\mathcal{O}}_{Rep(A,N)}=\frac{k[a_{i,j}\mid a\in
A,\;(i,j)\in[1,N]^{2}]}{<(ab)_{i,j}=a_{i,k}b_{k,j}>}.$

One can see that the non commutative $k$-algebra $A$ is smooth if and only if
the commutative $k$-algebra ${\mathcal{O}}_{Rep(A,N)}$ is smooth for all $N$.
If $A=k<x_{1},\dots,x_{m}>$ is a free associative k-algebra generated by the
variables $x_{1},\dots,x_{m}$, then

$Rep_{N}(A)=Mat_{N}(k)\oplus Mat_{N}(k)\oplus\dots\oplus
Mat_{N}(k)\simeq\mathbb{R}^{mN^{2}}$

where, if $\alpha\in[1,m],\quad M_{\alpha}$ is the $N\times N$ matrix
$(x_{i,\alpha}^{j})_{i,j}$ and $Tr(x_{\alpha_{1}}\dots
x_{\alpha_{r}})=Tr(M_{\alpha_{1}}M_{\alpha_{2}}\dots M_{\alpha_{r}})$.

The non commutative notions defined following the Kontsevich principle have
often a name with the prefix ”double”. Thus derivations are replaced by double
derivations (that are derivations from $A$ to $A\otimes A$ considered with its
exterior $A^{e}$-structure), Poisson algebras are replaced by double Poisson
algebras, etc … The de Rham complex has been defined in the non commutative
setting, the contraction by a double vector field and Lie derivative as well
([4]). Double Poisson cohomology has been defined ([16], [9]) and computed for
examples in low dimension.

In [12]111this reference was indicated to me by V. Rubtsov, double Lie
Rinehart algebras were introduced to study the notion of double quasi Poisson
algebra. A double Lie Rinehart algebra $\mathbb{L}$ being an $A$-bimodule, one
can construct $T_{A}({\mathbb{L}})$. In [12], the double Lie Rinehart algebra
structure has been defined by the Gerstenhaber double structure it induces on
$T_{A}{\mathbb{L}}$.

We give a direct definition of a double Lie Rinehart algebras and show that
there is a correspondence between double Lie Rinehart algebras structures on
an $A^{e}$-module ${\mathbb{L}}$ and double Gerstenhaber algebra structure on
$T_{A}({\mathbb{L}})$. We give new examples of double Lie Rinehart algebras.
In particular, any associative $k$ algebra has a natural double Lie Rinehart
structure. Then, we develop a differential calculus for double Lie Rinehart
algebras : definition of a differential $d_{\mathbb{L}}$, (for which we give
an explicit formula), Lie derivative, contraction. In the case where
${\mathbb{L}}={\mathbb{D}}er(A)$, we recover the definitions of [4]. In the
case where $A$ is a double Poisson algebra and ${\mathbb{L}}=\Omega_{A}^{1}$,
the differential $d_{L}$ coincides with the double Poisson cohomology
([16],[9]). Thus we recover the classical picture : The theory of double Lie
algebroids encompasses several theories.

Acknowledgement

I am most grateful to Y. Kosmann and V. Rubtsov for their help and
encouragement, which were essential for the completion of this article. They
both gave me several references that were instrumental for me to complete this
article. They explained to me several aspects of the theory of double Poisson
algebra and commented on earlier versions of this preprint. I am happy to
extend to them my most sincere thanks.

Convention :

We will use the same notation as [11]

If $(V_{i})_{i=1,\dots,n}$ are $k$ vector spaces and $s\in S_{n}$, then for
$a=a_{1}\otimes\dots\otimes a_{n}\in V_{1}\otimes\dots\otimes V_{n}$

$\tau_{s}(a)=a_{s^{-1}(1)}\otimes\dots\otimes a_{s^{-1}(n)}.$

If $(V_{i})_{i=1,\dots,n}$ are $k$ graded vector spaces and $s\in S_{n}$, then
for $a=a_{1}\otimes\dots\otimes a_{n}\in V_{1}\otimes\dots\otimes V_{n}$

$\sigma_{s}(a)=(-1)^{t}a_{s^{-1}(1)}\otimes\dots\otimes a_{s^{-1}(n)}.$

where
$t={\displaystyle\sum_{i<j,s^{-1}(i)>s^{-1}(j)}}|a_{s^{-1}(i)}||a_{s^{-1}(j)}|$.

$\tau_{(12)}$ (respectively $\sigma_{(12)}$) will be also denoted $(x\otimes
y)^{\circ}=y\otimes x$ (respectively $(x\otimes
y)^{\circ}=(-1)^{|x||y|}y\otimes x$.)

Let $B$ be a fixed $k$-algebra that will be, most of the time, semisimple
commutative of the form $B=ke_{1}\oplus\dots\oplus ke_{n}$ with
$e_{i}^{2}=e_{i}$. A $B$-algebra is a $k$-algebra $A$ equipped with a morphism
of $k$-algebras $B\to A$. The notion of $B$-algebra allows to define relative
versions of the notions introduced.

##  2\. Definitions and generalities

Most of the definition and results of this section are in [11].

######  Definition 2.1.

An $n$-bracket is a linear map

$\\{\\!\\{-,\dots,-\\}\\!\\}:A^{\otimes n}\to A^{\otimes n}$

which is a derivation $A\to A^{\otimes n}$ in its last argument for the outer
bimodule structure on $A^{\otimes n}$ i. e

$\\{\\!\\{a_{1},a_{2},\dots,a_{n-1},a_{n}a_{n}^{\prime}\\}\\!\\}=a_{n}\\{\\!\\{a_{1},a_{2},\dots,a_{n-1},a_{n}^{\prime}\\}\\!\\}+\\{\\!\\{a_{1},a_{2},\dots,a_{n-1},a_{n}\\}\\!\\}a_{n}^{\prime}$

and which is cyclically antisymmetric in the sense that

$\tau_{(1\dots n)}\circ\\{\\!\\{-,\dots,-\\}\\!\\}\circ\tau_{(1\dots
n)}^{-1}=(-1)^{n+1}\\{\\!\\{-,\dots,-\\}\\!\\}.$

If $A$ is a $B$-algebra, then an $n$-bracket is $B$-linear if it vanishes when
its last argument is in the image of $B$.

As in [11], we set

$\\{\\!\\{a,b\\}\\!\\}_{L}=\\{\\!\\{a,b_{1}\\}\\!\\}\otimes
b_{2}\otimes\dots\otimes b_{n}$

Associated to a double bracket $\\{\\!\\{-,-\\}\\!\\}$, we define a tri-ary
operation $\\{\\!\\{-,-,-\\}\\!\\}$ as follows :

$\\{\\!\\{a,b,c\\}\\!\\}=\\{\\!\\{a,\\{\\!\\{b,c\\}\\!\\}\\}\\!\\}_{L}+\tau_{(123)}\\{\\!\\{b,\\{\\!\\{c,a\\}\\!\\}\\}\\!\\}_{L}+\tau_{(132)}\\{\\!\\{c,\\{\\!\\{a,b\\}\\!\\}\\}\\!\\}_{L}$

######  Proposition 2.2.

([11]) $\\{\\!\\{-,-,-\\}\\!\\}$ is a 3-bracket.

######  Definition 2.3.

([11]). Let $A$ be a $k$-algebra. A double bracket $\\{\\!\\{-,-\\}\\!\\}$ on
A is a double Poisson bracket if $\\{\\!\\{-,-,-\\}\\!\\}=0$. An algebra with
a double Poisson bracket is a double Poisson algebra.

######  Examples 2.4.

1) ([11], [10]) One may characterize the double Poisson brackets on $k[t]$.
For $\lambda,\mu,\nu\in k$,

$\\{\\!\\{t,t\\}\\!\\}=\lambda(t\otimes 1-1\otimes t)+\mu(t^{2}\otimes
1-1\otimes t^{2})+\nu(t^{2}\otimes t-t\otimes t^{2})$

defines a double Poisson structure if and only if $\lambda\nu-\mu^{2}=0$ and
any double Poisson structure on $k[t]$ is of this form.

2) Quadratic double Poisson bracket on $\mathbb{C}<X_{1},X_{2},\dots,X_{n}>$
have been studied in [14] as well as their relation to the solution of the
Yang Baxter equation.

The following proposition was proved in [11] :

######  Proposition 2.5.

Assume that $(A,\\{\\!\\{-,-\\}\\!\\})$ is a double Poisson algebra. Then the
following holds :

(1) $\\{-,-\\}$ is a derivation in its second argument and vanishes on
commutators in its first argument.

(2) $\\{-,-\\}$ is anti-symmetric modulo commutators.

(3) $\\{-,-\\}$ makes $A$ into a left Loday algebra, i.e $\\{-,-\\}$ satisfies
the following version of Jacobi identity

$\\{a,\\{b,c\\}\\}=\\{\\{a,b\\},c\\}+\\{b,\\{a,c\\}\\}$

(4) $\\{-,-\\}$ makes $A/[A,A]$ into a Lie algebra.

######  Definition 2.6.

([3]) A Poisson structure on $A$ is a Lie bracket $\\{-,-\\}$ on
$\dfrac{A}{[A,A]}$ such that for each $a\in A$, the map
$\\{\overline{a},-\\}:\dfrac{A}{[A,A]}\to\dfrac{A}{[A,A]}$ is induced by a
derivation on $A$.

######  Remarks 2.7.

  1. (i )

In the case where $A$ is commutative, we recover the usual Poisson bracket.

  2. (ii )

It was shown in [11] (lemme 2.6.2) that a double Poisson bracket on $A$
induces a Poisson structure on $A$.

  3. (iii )

A Poisson structure on $A$ is also called a $H_{0}$ Poisson structure as it is
a structure on $\dfrac{A}{[A,A]}=HC_{0}(A)$ which is the $0$th cyclic group of
$A$. In [2], derived Poisson structure were defined on higher cyclic
cohomology group.

Let $D$ be a graded algebra. There are two commuting $D^{e}$-module structures
on $D\otimes D$ : For any homogeneous elements $\alpha,\beta,x,y$ in $D$,

$\begin{array}[]{l}\alpha(x\otimes y)\beta=\alpha x\otimes y\beta\\\
\alpha*(x\otimes
y)*\beta=(-1)^{|\alpha||\beta|+|\alpha||x|+|y||\beta|}x\beta\otimes\alpha y\\\
\end{array}$

######  Definition 2.8.

([7]) Let $d\in\mathbb{Z}$ and let $D$ be a graded algebra. $D$ is called a
double Gerstenhaber algebra of degree $d$ if it is equipped with a graded
bilinear map

$\\{\\!\\{-,-\\}\\!\\}:D\otimes D\to D\otimes D$

of degree d such that the following identities hold :

$\begin{array}[]{l}1)\\{\\!\\{\alpha,\beta\gamma\\}\\!\\}=(-1)^{(|\alpha|+d)|\beta|}\beta\\{\\!\\{\alpha,\gamma\\}\\!\\}+\\{\\!\\{\alpha,\beta\\}\\!\\}\gamma\\\
1^{\prime})\\{\\!\\{\beta\gamma,\alpha\\}\\!\\}=(-1)^{(|\alpha|+d)|\beta|}\beta*\\{\\!\\{\alpha,\gamma\\}\\!\\}+\\{\\!\\{\alpha,\beta\\}\\!\\}*\gamma\\\
2)\\{\\!\\{\alpha,\beta\\}\\!\\}=-(-1)^{(|\alpha|+d)(|\beta|+d)}\sigma_{(12)}\\{\\!\\{\beta,\alpha\\}\\!\\}\\\
3)\\{\\!\\{\alpha,\\{\\!\\{\beta,\gamma\\}\\!\\}\\}\\!\\}_{L}+(-1)^{(|\alpha|+d)(|\beta|+|\gamma|)}\sigma_{(123)}\\{\\!\\{\beta,\\{\\!\\{\gamma,\alpha\\}\\!\\}\\}\\!\\}_{L}+(-1)^{(|\gamma|+d)(|\alpha|+|\beta|)}\sigma_{(132)}\\{\\!\\{\gamma,\\{\\!\\{\alpha,\beta\\}\\!\\}\\}\\!\\}_{L}=0\end{array}$

######  Remarks 2.9.

1) The definition of double Gerstenhaber algebra is given in [11]. It is
extended to the case of double Gerstenhaber algebra of degree $d\in\mathbb{Z}$
in [7]. The case $d=-1$ corresponds to double Gerstenhaber algebras (see [11])
and the case $d=0$ corresponds to double Poisson algebras ([11]).

2) Assertions $1)$ and $1^{\prime})$ are equivalent if assertion 2) is
satisfied ([7]).

If $D$ is a double algebra, we define the associated bracket
$\\{-,-\\}:A\otimes A\to A$ by :

$\forall(\alpha,\beta)\in
D^{2},\quad\\{\alpha,\beta\\}=\\{\\!\\{\alpha,\beta\\}\\!\\}^{\prime}\\{\\!\\{\alpha,\beta\\}\\!\\}^{\prime\prime}.$

The following proposition was partly stated in [11] :

######  Proposition 2.10.

Let $D$ be a Gerstenhaber algebra of degree $d$ and let $\alpha,\beta,\gamma$
three homogeneous elements in $D$.

1)
$\\{\alpha\otimes\beta-(-1)^{|\alpha||\beta|}\beta\otimes\alpha,\gamma\\}=0.$

2)
$\\{\alpha,\beta\\}-(-1)^{(\mid\alpha\mid+d)(\mid\beta\mid+d)}\\{\beta,\alpha\\}\in[D,D]$

3)
$\\{\alpha,\\{\\!\\{\beta,\gamma\\}\\!\\}\\}-\\{\\!\\{\\{\alpha,\beta\\},\gamma\\}\\!\\}-(-1)^{(\mid\alpha\mid+d)(|\beta|+d)}\\{\\!\\{\beta,\\{\alpha,\gamma\\}\\}\\!\\}=0$

where $\\{\alpha,-\\}$ acts on tensors by

$\\{\alpha,\\{\\!\\{\beta,\gamma\\}\\!\\}\\}=\\{\alpha,\\{\\!\\{\beta,\gamma\\}\\!\\}^{\prime}\\}\otimes\\{\\!\\{\beta,\gamma\\}\\!\\}^{\prime\prime}+(-1)^{(|\alpha|+d)|\\{\\!\\{\beta,\gamma\\}\\!\\}^{\prime}|}\\{\\!\\{\beta,\gamma\\}\\!\\}^{\prime}\otimes\\{\alpha,\\{\\!\\{\beta,\gamma\\}\\!\\}^{\prime\prime}\\}$

4) $\dfrac{D}{[D,D]}[d]$ is a graded Lie algebra.

######  Remark 2.11.

Proposition 2.10 is proved in the ungraded case in [11]. In this case, $d=0$
and $D$ is a double Poisson algebra. Our proof is similar to that of [11] so
that we only sketch the main lines of it .

###### Proof.

1) and 2) are straightforward computations if one uses the relation

$\\{\\!\\{\alpha,\beta\\}\\!\\}=-(-1)^{(|\alpha|+d)(|\beta|+d)+|\\{\\!\\{\alpha,\beta\\}\\!\\}^{\prime}||\\{\\!\\{\alpha,\beta\\}\\!\\}^{\prime\prime}|}\\{\\!\\{\beta,\alpha\\}\\!\\}$

4) is a consequence of the previous statements. Let us now prove 3).

We will make use of the following lemma whose proof is left to the reader:

######  Lemma 2.12.

Set
$\\{\\!\\{\alpha,\\{\\!\\{\gamma,\beta\\}\\!\\}\\}\\!\\}_{R}=\\{\\!\\{\alpha,\\{\\!\\{\gamma,\beta\\}\\!\\}^{\prime}\\}\\!\\}\otimes\\{\\!\\{\gamma,\beta\\}\\!\\}^{\prime\prime}$.
The following equality holds

$\\{\\!\\{\alpha,\\{\\!\\{\gamma,\beta\\}\\!\\}\\}\\!\\}_{R}=-(-1)^{(|\beta|+d)(|\gamma|+d)}\tau_{(123)}\\{\\!\\{\alpha,\\{\\!\\{\beta,\gamma\\}\\!\\}\\}\\!\\}_{L}\\\
$

Let us now compute the three terms of the equality 3). Assertion 3) will
follow from these computations.

$\begin{array}[]{rcl}\\{\alpha,\\{\\!\\{\beta,\gamma\\}\\!\\}\\}&=&\\{\alpha,\\{\\!\\{\beta,\gamma\\}\\!\\}^{\prime}\\}\\{\\!\\{\beta,\gamma\\}\\!\\}^{\prime\prime}+(-1)^{(|\alpha|+d)|\\{\\!\\{\beta,\gamma\\}\\!\\}^{\prime}|}\\{\\!\\{\beta,\gamma\\}\\!\\}^{\prime}\\{\alpha,\\{\\!\\{\
\beta,\gamma\\}\\!\\}^{\prime\prime}\\}\\\ &=&(m\otimes
1)\\{\\!\\{\alpha,\\{\\!\\{\beta,\gamma\\}\\!\\}\\}\\!\\}_{L}+(1\otimes
m)\\{\\!\\{\alpha,\\{\\!\\{\beta,\gamma\\}\\!\\}\\}\\!\\}_{R}\\\ &=&(m\otimes
1)\\{\\!\\{\alpha,\\{\\!\\{\beta,\gamma\\}\\!\\}\\}\\!\\}_{L}-(1\otimes
m)\tau_{(123)}\\{\\!\\{\alpha,\\{\\!\\{\gamma,\beta\\}\\!\\}\\}\\!\\}_{L}(-1)^{(|\beta|+d)(|\gamma|+d)}\\\
\end{array}$

$\begin{array}[]{l}\\{\\!\\{\\{\alpha,\beta\\},\gamma\\}\\!\\}\\\
=-\tau_{(12)}\\{\\!\\{\gamma,\\{\alpha,\beta\\}\\}\\!\\}(-1)^{(|\gamma|+d)(|\alpha|+|\beta|)}\\\
=-\tau_{(12)}\left[\\{\\!\\{\alpha,\beta\\}\\!\\}^{\prime}\\{\\!\\{\gamma,\\{\\!\\{\alpha,\beta\\}\\!\\}^{\prime\prime}\\}\\!\\}\right](-1)^{(|\gamma|+d)(|\alpha|+|\beta|)}-\tau_{(12)}\left[\\{\\!\\{\gamma,\\{\\!\\{\alpha,\beta\\}\\!\\}^{\prime}\\}\\!\\}\\{\\!\\{\alpha,\beta\\}\\!\\}^{\prime\prime}\right](-1)^{(|\gamma|+d)(|\alpha|+|\beta|)}\\\
=-(1\otimes
m)\tau_{(132)}\\{\\!\\{\gamma,\\{\\!\\{\alpha,\beta\\}\\!\\}\\}\\!\\}_{R}(-1)^{(|\gamma|+d)(|\alpha|+|\beta|)}-(m\otimes
1)\tau_{(123)}\\{\\!\\{\gamma,\\{\\!\\{\alpha,\beta\\}\\!\\}\\}\\!\\}_{L}(-1)^{(|\gamma|+d)(|\alpha|+|\beta|)}\\\
=(1\otimes
m)\tau_{(132)}\\{\\!\\{\gamma,\\{\\!\\{\beta,\alpha\\}\\!\\}\\}\\!\\}_{L}(-1)^{(|\gamma|+d)(|\alpha|+|\beta|)}(-1)^{(|\alpha|+d)(|\beta|+d)}\\\
\quad-(m\otimes
1)\tau_{(132)}\\{\\!\\{\gamma,\\{\\!\\{\alpha,\beta\\}\\!\\}\\}\\!\\}_{L}(-1)^{(|\gamma|+d)(|\alpha|+|\beta|)}\end{array}$

$\begin{array}[]{rcl}\\{\\!\\{\beta,\\{\alpha,\gamma\\}\\}\\!\\}&=&\\{\\!\\{\beta,\\{\\!\\{\alpha,\gamma\\}\\!\\}^{\prime}\\}\\!\\}\\{\\!\\{\alpha,\gamma\\}\\!\\}^{\prime\prime}+(-1)^{(|\beta|+d)\mid\\{\\!\\{\alpha,\gamma\\}\\!\\}^{\prime}\mid}\\{\\!\\{\alpha,\gamma\\}\\!\\}^{\prime}\\{\\!\\{\beta,\\{\\!\\{\alpha,\gamma\\}\\!\\}^{\prime\prime}\\}\\!\\}\\\
&=&(1\otimes
m)\\{\\!\\{\beta,\\{\\!\\{\alpha,\gamma\\}\\!\\}\\}\\!\\}_{L}+(m\otimes
1)\\{\\!\\{\beta,\\{\\!\\{\alpha,\gamma\\}\\!\\}\\}\\!\\}_{R}\\\ &=&(1\otimes
m)\\{\\!\\{\beta,\\{\\!\\{\alpha,\gamma\\}\\!\\}\\}\\!\\}_{L}-(m\otimes
1)\tau_{(123)}\\{\\!\\{\beta,\\{\\!\\{\gamma,\alpha\\}\\!\\}\\}\\!\\}_{L}(-1)^{(|\alpha|+d)(|\gamma|+d)}\\\
\end{array}$

∎

##  3\. Double derivations

In this section, we recall results of [11].

Let $B$ be a $k$-algebra. A $B$-algebra is a pair $(A,\eta)$ where $\eta:B\to
A$ is an algebra morphism.

Denote by $m:A\otimes_{B}A\to A$ multiplication on $A$. One sets
$\Omega^{1}_{B}(A):=\rm{Ker}(m)$. It is naturally endowed with an
$A^{e}$-module structure. If $a\in A$, then $da=a\otimes 1-1\otimes a$ belongs
to $\Omega^{1}_{B}(A)$. If $A$ is finitely generated as an $B$-algebra, then
$\Omega^{1}_{B}A$ is finitely generated as a left $A^{e}$-module.

######  Definition 3.1.

A $B$-algebra $A$ is called smooth over $B$ if it is finitely generated as an
$B$-algebra and $\Omega_{B}^{1}A$ is projective as a left $A^{e}$-module.

######  Example 3.2.

Let ${\mathcal{Q}}=(Q,I)$ be a finite quiver with vertex set
$I=\\{1,\dots,n\\}$ and edge set $Q$. Denote by $e_{i}$ the idempotent
associated to the vertex $i$ and we put $B={\displaystyle\oplus_{i}}ke_{i}$.
The path algebra $A=k[Q]$ is smooth over $B$ ([11]).

We define $\mathbb{D}er_{B}(A,A\otimes A)$ the derivations from $A$ to
$A\otimes A$ where we put the outer bimodule structure on $A\otimes A$. Inner
bimodule structure on $A^{e}$ allows to endow $\mathbb{D}er_{B}(A,A\otimes A)$
with an $A^{e}$-module structure :

$\alpha\cdot D\cdot\beta(a)=D(a)^{\prime}\beta\otimes\alpha
D(a)^{\prime\prime}$

######  Example 3.3.

We assume that we are in the situation where $B=ke_{1}\oplus\dots\oplus
ke_{n}$ with $e_{i}^{2}=e_{i}$.

Let $E_{i}:A\to A\otimes A$ defined by $E_{i}(a)=ae_{i}\otimes
e_{i}-e_{i}\otimes ae_{i}$ and $E=\displaystyle\sum_{i=1}^{n}E_{i}$.

######  Proposition 3.4.

([11]) Let $\delta$, $\Delta$ in $\mathbb{D}er_{B}(A,A\otimes A)$, then

$\begin{array}[]{l}\\{\\!\\{\delta,\Delta\\}\\!\\}_{l}^{\sim}=(\delta\otimes
1)\Delta-(1\otimes\Delta)\delta\\\
\\{\\!\\{\delta,\Delta\\}\\!\\}^{\sim}_{r}:=(1\otimes\delta)\Delta-(\Delta\otimes
1)\delta=-\\{\\!\\{\Delta,\delta\\}\\!\\}^{\sim}_{l}\\\ \end{array}$

are derivation from $A$ to $A^{\otimes 3}$ where the $A^{e}$-bimodule
structure on $A^{\otimes 3}$ is the outer structure.

We define

$\begin{array}[]{rcl}\\{\\!\\{\delta,\Delta\\}\\!\\}_{l}=\tau_{(23)}\circ\\{\\!\\{\delta,\Delta\\}\\!\\}_{l}^{\sim}\in\mathbb{D}er(A,A\otimes
A)\otimes A\\\
\\{\\!\\{\delta,\Delta\\}\\!\\}_{r}=\tau_{(12)}\circ\\{\\!\\{\delta,\Delta\\}\\!\\}_{r}^{\sim}\in
A\otimes\mathbb{D}er(A,A\otimes A)\end{array}$

We write

$\begin{array}[]{l}\\{\\!\\{\delta,\Delta\\}\\!\\}_{l}=\\{\\!\\{\delta,\Delta\\}\\!\\}_{l}^{\prime}\otimes\\{\\!\\{\delta,\Delta\\}\\!\\}_{l}^{\prime\prime}\\\
\\{\\!\\{\delta,\Delta\\}\\!\\}_{r}=\\{\\!\\{\delta,\Delta\\}\\!\\}_{r}^{\prime}\otimes\\{\\!\\{\delta,\Delta\\}\\!\\}_{r}^{\prime\prime}\\\
\end{array}$

with
$\\{\\!\\{\delta,\Delta\\}\\!\\}_{r}^{\prime},\\{\\!\\{\delta,\Delta\\}\\!\\}_{l}^{\prime\prime}\in
A$ and
$\\{\\!\\{\delta,\Delta\\}\\!\\}_{l}^{\prime},\\{\\!\\{\delta,\Delta\\}\\!\\}_{r}^{\prime\prime}\in\mathbb{D}er(A,A\otimes
A)$.

######  Theorem 3.5.

([11]) For $a,b\in A$ and $\delta,\Delta\in\mathbb{D}er(A,A\otimes A)$, the
following equations

$\begin{array}[]{l}\\{\\!\\{a,b\\}\\!\\}=0\\\
\\{\\!\\{\delta,a\\}\\!\\}=\delta(a)\in A\otimes A\\\
\\{\\!\\{\delta,\Delta\\}\\!\\}=\\{\\!\\{\delta,\Delta\\}\\!\\}_{l}+\\{\\!\\{\delta,\Delta\\}\\!\\}_{r}\end{array}$

define a unique structure of double Gerstenhaber algebra on
$D_{B}A=T_{A}\mathbb{D}er_{B}(A,A\otimes A)$.

######  Proposition 3.6.

([11]) Assume that $A$ is a finitely generated $k$-algebra. The linear map

$\begin{array}[]{rcl}\mu:\quad\quad(D_{B}A)_{n}&\to&\\{$B$-linear\;$n$-brackets\;on\;$A$\\}\\\
Q=\delta_{1}\dots\delta_{n}&\mapsto&\\{\\!\\{-,\dots,-\\}\\!\\}_{Q}=\\{\\!\\{-,\dots,-\\}\\!\\}={\displaystyle\sum_{i=0}^{n-1}(-1)^{(n-1)i}}\tau_{(1\dots
n)}^{i}\circ\\{\\!\\{-,\dots,-\\}\\!\\}^{\sim}_{Q}\circ\tau_{(1\dots
n)}^{-i}\end{array}$

where

$\\{\\!\\{a_{1},\dots,a_{n}\\}\\!\\}^{\sim}_{\delta_{1}\dots\delta_{n}}=\delta_{n}(a_{n})^{\prime}\delta_{1}(a_{1})^{\prime\prime}\otimes\delta_{1}(a_{1})^{\prime}\delta_{2}(a_{2})^{\prime\prime}\otimes\dots\otimes\delta_{n-1}(a_{n-1})^{\prime}\delta_{n}(a_{n})^{\prime\prime}$

This maps factors through $\dfrac{D_{B}A}{[D_{B}A,D_{B}A]}$. The map $\mu$ is
an isomorphism if $A$ is $B$-smooth.

The following proposition is proved in [11]

######  Proposition 3.7.

For $Q\in(D_{B}A)_{n}$, the following identity holds :

$\\{\\!\\{a_{1},\dots,a_{n}\\}\\!\\}_{Q}=(-1)^{\frac{n(n-1)}{2}}\\{\\!\\{a_{1},\dots,\\{\\!\\{a_{n-1},\\{Q,a_{n}\\}\\}\\!\\}_{L}\dots\\}\\!\\}_{L}$

The isomorphism $\mu$ allows to characterize the double Poisson algebra on a
smooth algebra :

######  Proposition 3.8.

Let $A$ be a smooth $B$-algebra. Double $B$-linear Poisson structures on $A$
are in bijection with the $P\in T^{2}\mathbb{D}er_{B}(A)$ such that
$\\{P,P\\}=0$ modulo $[D_{B}A,D_{B}A]$.

######  Definition 3.9.

We assume that we are in the situation where $B=ke_{1}\oplus\dots\oplus
ke_{n}$ with $e_{i}^{2}=e_{i}$.

Let $E_{i}:A\to A\otimes A$ defined by $E_{i}(a)=ae_{i}\otimes
e_{i}-e_{i}\otimes ae_{i}$ and $E=\displaystyle\sum_{i=1}^{n}E_{i}$.
A double quasi Poisson brachet on $A$ is a $B$-linear bracket
$\\{\\!\\{,\\}\\!\\}$ such that

$\\{\\!\\{-,-,-\\}\\!\\}=\\{\\!\\{-,-,-\\}\\!\\}_{E^{3}}.$

######  Definition 3.10.

$P\in(D_{B}A)_{2}$ is a differential double quasi Poisson bracket if

$\\{P,P\\}=\dfrac{1}{6}E^{3}\;mod\;[D_{B}A,D_{B}A]$

######  Remark 3.11.

If $A$ is smooth, then quasi Poisson bracket and differential quasi Poisson
bracket are equivalent notions.

######  Examples 3.12.

Examples of Poisson double bracket and of quasi Poisson double bracket over
the path algebra of a the double of a quiver are given in [11].

Given an oriented surface $\Sigma$ with base point $*\in\partial\Sigma$, a
quasi Poisson double algebra structure is constructed on the group algebra of
$\pi_{1}(\Sigma,*)$ in [8].

Double Poisson structures on a semi-simple algebra (over an algebraically
closed field) are described in [16].

##  4\. Double Lie algebroid

######  Definition 4.1.

A double Lie algebroid is a quadruple
$(\mathbb{L},A,\omega,\\{\\!\\{-,-\\}\\!\\})$ where

  * •

$\mathbb{L}$ is an $A^{e}$-module.

  * •

$\begin{array}[]{rcl}\\{\\!\\{,\\}\\!\\}&:&\mathbb{L}\otimes\mathbb{L}\to\mathbb{L}\otimes
A\oplus A\otimes\mathbb{L}\\\
&&(D,\Delta)\mapsto\\{\\!\\{D,\Delta\\}\\!\\}^{\prime}\otimes\\{\\!\\{D,\Delta\\}\\!\\}^{\prime\prime}=\\{\\!\\{D,\Delta\\}\\!\\}_{l}^{\prime}\otimes\\{\\!\\{D,\Delta\\}\\!\\}_{l}^{\prime\prime}+\\{\\!\\{D,\Delta\\}\\!\\}_{r}^{\prime}\otimes\\{\\!\\{D,\Delta\\}\\!\\}_{r}^{\prime\prime}\\\
&&({\rm
with}\quad\\{\\!\\{D,\Delta\\}\\!\\}_{l}^{\prime},\\{\\!\\{D,\Delta\\}\\!\\}_{r}^{\prime\prime}\in\mathbb{L}\quad{\rm
and}\quad\\{\\!\\{D,\Delta\\}\\!\\}_{l}^{\prime\prime},\\{\\!\\{D,\Delta\\}\\!\\}_{r}^{\prime}\in
A)\end{array}$

is a map satisfying
$\\{\\!\\{D,\Delta\\}\\!\\}=-\\{\\!\\{\Delta,D\\}\\!\\}^{o}$.

  * •

$\omega:\mathbb{L}\to\mathbb{D}er_{B}(A)$ is a morphism of $A^{e}$-modules.

  * •

$\\{\\!\\{\omega(D),\omega(\Delta)\\}\\!\\}=\omega\left(\\{\\!\\{D,\Delta\\}\\!\\}\right)$
where we also denote by $\omega$ the extension of $\omega$ to a map from
$A\otimes\mathbb{L}\oplus\mathbb{L}\otimes A$ to
$A\otimes\mathbb{D}er_{B}(A)\oplus\mathbb{D}er_{B}(A)\otimes A$.

  * •

Jacobi identity : If $D_{1},D_{2},D_{3}$ are elements of $\mathbb{L}$, one has
:

$\\{\\!\\{D_{1},\\{\\!\\{D_{2},D_{3}\\}\\!\\}\\}\\!\\}_{L}+\tau_{(123)}\\{\\!\\{D_{2},\\{\\!\\{D_{3},D_{1}\\}\\!\\}\\}\\!\\}_{L}+\tau_{(123)}^{2}\\{\\!\\{D_{3},\\{\\!\\{D_{1},D_{2}\\}\\!\\}\\}\\}_{L}=0$

where

$\begin{array}[]{l}\\{\\!\\{D_{1},\\{\\!\\{D_{2},D_{3}\\}\\!\\}\\}\\!\\}_{L}:=\\{\\!\\{D_{1},\\{\\!\\{D_{2},D_{3}\\}\\!\\}^{\prime}\\}\\!\\}\otimes\\{\\!\\{D_{2},D_{3}\\}\\!\\}^{\prime\prime}\\\
\forall(D,a)\in\mathbb{L}\times
A,\quad\\{\\!\\{D,a\\}\\!\\}=\omega(D)(a)\end{array}$

  * •

If $(D,\Delta)\in\mathbb{L}^{2}$, one has

$\begin{array}[]{l}\\{\\!\\{D,a\Delta\\}\\!\\}=D(a)\Delta+a\\{\\!\\{D,\Delta\\}\\!\\}\\\
\\{\\!\\{D,\Delta a\\}\\!\\}=\\{\\!\\{D,\Delta\\}\\!\\}a+\Delta
D(a)\end{array}$

where $D(a)\Delta=D(a)^{\prime}\otimes D(a)^{\prime\prime}\Delta\in
A\otimes\mathbb{L}$ and $\Delta D(a)=\Delta D(a)^{\prime}\otimes
D(a)^{\prime\prime}\in\mathbb{L}\otimes A$.

######  Remarks 4.2.

  1. (i )

When there will be ambiguity, the bracket $\\{\\!\\{-,-\\}\\!\\}$ will be
denoted $\\{\\!\\{-,-\\}\\!\\}_{\mathbb{L}}$.

  2. (ii )

The definition of a double Lie algebroid is given in [12] by its
characterization in proposition 4.5 below.

  3. (iii )

The Jacobi identity is equivalent to the following : Let $D_{1},D_{2},D_{3}$
be elements of $\mathbb{L}$ :

$\\{\\!\\{D_{1},\\{\\!\\{D_{2},D_{3}\\}\\!\\}\\}\\!\\}_{R}+\tau_{(123)}\\{\\!\\{D_{2},\\{\\!\\{D_{3},D_{1}\\}\\!\\}\\}\\!\\}_{R}+\tau_{(123)}^{2}\\{\\!\\{D_{3},\\{\\!\\{D_{1},D_{2}\\}\\!\\}\\}\\}_{R}=0$

where
$\begin{array}[]{l}\\{\\!\\{D_{1},\\{\\!\\{D_{2},D_{3}\\}\\!\\}\\}\\!\\}_{R}:=\\{\\!\\{D_{2},D_{3}\\}\\!\\}^{\prime}\otimes\\{\\!\\{D_{1},\\{\\!\\{D_{2},D_{3}\\}\\!\\}^{\prime\prime}\\}\\!\\}.\\\
\end{array}$

  4. (iv )

For differential calculus, we will assume that $A$ is a smooth $B$-algebra and
that $\mathbb{L}$ is a finitely generated projective $A^{e}$-module.

Notation : We set $\\{\\!\\{D,\Delta\\}\\!\\}_{l}$ for the component of
$\\{\\!\\{D,\Delta\\}\\!\\}$ that is $\mathbb{L}\otimes A$ and
$\\{\\!\\{D,\Delta\\}\\!\\}_{r}$ for its component in $A\otimes\mathbb{L}$.
Adopting a Sweedler’s type notation, we set

$\begin{array}[]{l}\\{\\!\\{D,\Delta\\}\\!\\}_{l}=\\{\\!\\{D,\Delta\\}\\!\\}_{l}^{\prime}\otimes\\{\\!\\{D,\Delta\\}\\!\\}_{l}^{\prime\prime}\\\
\\{\\!\\{D,\Delta\\}\\!\\}_{r}=\\{\\!\\{D,\Delta\\}\\!\\}_{r}^{\prime}\otimes\\{\\!\\{D,\Delta\\}\\!\\}_{r}^{\prime\prime}\\\
\end{array}$

with
$\\{\\!\\{D,\Delta\\}\\!\\}_{l}^{\prime},\\{\\!\\{D,\Delta\\}\\!\\}_{r}^{\prime\prime}\in\mathbb{L}$
and
$\\{\\!\\{D,\Delta\\}\\!\\}_{r}^{\prime},\\{\\!\\{D,\Delta\\}\\!\\}_{l}^{\prime\prime}\in
A$. One has

$\\{\\!\\{D,\Delta\\}\\!\\}_{l}=-\\{\\!\\{\Delta,D\\}\\!\\}_{r}^{\circ}.$

######  Lemma 4.3.

Let $D_{1}$, $D_{2}$, $D_{3}$ be three elements of $\mathbb{L}$. One has :

$\begin{array}[]{l}\\{\\!\\{D_{1},\\{\\!\\{D_{2},D_{3}\\}\\!\\}^{\prime}_{l}\\}\\!\\}^{\prime}_{l}\otimes\\{\\!\\{D_{1},\\{\\!\\{D_{2},D_{3}\\}\\!\\}^{\prime}_{l}\\}\\!\\}_{l}^{\prime\prime}\otimes\\{\\!\\{D_{2},D_{3}\\}\\!\\}_{l}^{\prime\prime}\\\
-\\{\\!\\{D_{1},D_{3}\\}\\!\\}^{\prime}_{l}\otimes\\{\\!\\{D_{2},\\{\\!\\{D_{1},D_{3}\\}\\!\\}^{\prime\prime}_{l}\\}\\!\\}^{\prime}\otimes\\{\\!\\{D_{2},\\{\\!\\{D_{1},D_{3}\\}\\!\\}^{\prime\prime}_{l}\\}\\!\\}^{\prime\prime}\\\
-\\{\\!\\{\\{\\!\\{D_{1},D_{2}\\}\\!\\}^{\prime}_{l},D_{3}\\}\\!\\}^{\prime}_{l}\otimes\\{\\!\\{D_{1},D_{2}\\}\\!\\}_{l}^{\prime\prime}\otimes\\{\\!\\{\\{\\!\\{D_{1},D_{2}\\}\\!\\}^{\prime}_{l},D_{3}\\}\\!\\}^{\prime\prime}_{l}=0\par\end{array}$

###### Proof.

The lemma follows from the Jacobi identity, taking the component on
$\mathbb{L}\otimes A\otimes A$.
∎

######  Proposition 4.4.

Let $(D,\Delta)\in\mathbb{L}^{2}$ and $(\alpha,\beta)\in A^{2}$, one has :
$\begin{array}[]{rcl}\\{\\!\\{D,\Delta\cdot\beta\\}\\!\\}_{l}&=&\\{\\!\\{D,\Delta\\}\\!\\}^{\prime}_{l}\otimes\\{\\!\\{D,\Delta\\}\\!\\}^{\prime\prime}_{l}\cdot\beta+\Delta\cdot
D(\beta)^{\prime}\otimes D(\beta)^{\prime\prime}\\\
\\{\\!\\{D,\Delta\cdot\beta\\}\\!\\}_{r}&=&\\{\\!\\{D,\Delta\\}\\!\\}^{\prime}_{r}\otimes\\{\\{D,\Delta\\}\\!\\}_{r}^{\prime\prime}\cdot\beta\\\
\\{\\!\\{D,\alpha\cdot\Delta\\}\\!\\}_{l}&=&\alpha\cdot\\{\\!\\{D,\Delta\\}\\!\\}_{l}\\\
\\{\\!\\{D,\alpha\cdot\Delta\\}\\!\\}_{r}&=&\omega(D)(\alpha)^{\prime}\otimes\omega(D)(\alpha)^{\prime\prime}\Delta+\alpha\\{\\!\\{D,\Delta\\}\\!\\}_{r}\\\
\\{\\!\\{a\cdot
D,\Delta\\}\\!\\}_{r}&=&\\{\\!\\{D,\Delta\\}\\!\\}^{\prime}_{r}\otimes
a\cdot\\{\\!\\{D,\Delta\\}\\!\\}^{\prime\prime}_{r}\\\ \\{\\!\\{a\cdot
D,\Delta\\}\\!\\}_{l}&=&-\\{\\!\\{\Delta,D\\}\\!\\}^{\prime\prime}_{r}\otimes
a\\{\\!\\{\Delta,D\\}\\!\\}^{\prime}_{r}-\\{\\!\\{\Delta,a\\}\\!\\}^{\prime\prime}D\otimes\\{\\!\\{\Delta,a\\}\\!\\}^{\prime}\\\
&=&\\{\\!\\{D,\Delta\\}\\!\\}_{l}^{\prime}\otimes
a\\{\\!\\{D,\Delta\\}\\!\\}_{l}^{\prime\prime}-\\{\\!\\{\Delta,a\\}\\!\\}^{\prime\prime}D\otimes\\{\\!\\{\Delta,a\\}\\!\\}^{\prime}\\\
\\{\\!\\{D\cdot\beta,\Delta\\}\\!\\}_{r}&=&-\\{\\!\\{\Delta,D\\}\\!\\}^{\prime\prime}_{l}\beta\otimes\\{\\!\\{\Delta,D\\}\\!\\}^{\prime}_{l}-\\{\\!\\{\Delta,\beta\\}\\!\\}^{\prime\prime}\otimes
D\\{\\!\\{\Delta,\beta\\}\\!\\}^{\prime}\\\
\\{\\!\\{D\cdot\beta,\Delta\\}\\!\\}_{l}&=&-\\{\\!\\{\Delta,D\\}\\!\\}^{\prime\prime}_{r}\beta\otimes\\{\\!\\{\Delta,D\\}\\!\\}^{\prime}_{r}=\\{\\!\\{D,\Delta\\}\\!\\}^{\prime}_{l}\beta\otimes\\{\\!\\{D,\Delta\\}\\!\\}^{\prime\prime}_{l}\par\par\par\end{array}$

###### Proof.

The proof follows from the properties of the double Lie algebroid bracket and
easy computations. ∎

######  Proposition 4.5.

Let $(\mathbb{L},A,\omega,\\{\\!\\{,\\}\\!\\}_{\mathbb{L}})$ be a quadruple
such that $\\{\\!\\{,\\}\\!\\}_{\mathbb{L}}$ is a map from
$\mathbb{L}\otimes\mathbb{L}$ to $A\otimes\mathbb{L}\oplus\mathbb{L}\otimes A$
and $\omega$ a map from $\mathbb{L}$ to $\mathbb{D}er(A)$. There is a unique
graded bilinear map $\\{\\!\\{,\\}\\!\\}:T_{A}(\mathbb{L})\otimes
T_{A}(\mathbb{L})\to T_{A}(\mathbb{L})$ of degree $-1$ satisfying the
following conditions :

  * •

For all $(\alpha,\beta,D,\Delta)\in A^{2}\otimes\mathbb{L}^{2}$,

$\begin{array}[]{l}\\{\\!\\{\alpha,\beta\\}\\!\\}=0\\\
\\{\\!\\{D,\alpha\\}\\!\\}=\omega(D)(\alpha)^{\prime}\otimes\omega(D)(\alpha)^{\prime\prime}\\\
\\{\\!\\{D,\Delta\\}\\!\\}=\\{\\!\\{D,\Delta\\}\\!\\}_{\mathbb{L}}\end{array}$

  * •

for all $(a,b,c)\in T_{A}(\mathbb{L})^{3}$

$\begin{array}[]{l}\\{\\!\\{a,bc\\}\\!\\}=(-1)^{(|a|-1)|b|}b\\{\\!\\{a,c\\}\\!\\}+\\{\\!\\{a,b\\}\\!\\}c\\\
\\{\\!\\{a,b\\}\\!\\}=-(-1)^{(|a|-1)(|b|-1)}\sigma_{(12)}\\{\\!\\{b,a\\}\\!\\}\end{array}$

$(\mathbb{L},A,\omega)$ is a double Lie Rinehart algebra over $A$ if and only
if $T_{A}(\mathbb{L})$ is a double Gerstenhaber algebra.

###### Proof.

Adopting a Sweedler type notation for $\\{\\!\\{,\\}\\!\\}$, one has for any
$a_{1},\dots,a_{m},b_{1},\dots,b_{n}$ in $T_{A}(\mathbb{L})$ :

$\\{\\!\\{a_{1}\dots a_{m},b_{1}\dots
b_{n}\\}\\!\\}={\displaystyle\sum_{p,q}}b_{1}\dots
b_{q-1}\\{\\!\\{a_{p},b_{q}\\}\\!\\}^{\prime}a_{p+1}\dots a_{m}\otimes
a_{1}\dots a_{p-1}\\{\\!\\{a_{p},b_{q}\\}\\!\\}^{\prime\prime}b_{q+1}\dots
b_{n}$

One sets

$\\{\\!\\{a,b,c\\}\\!\\}=\\{\\!\\{a,\\{\\!\\{b,c\\}\\!\\}\\}\\!\\}_{l}+(-1)^{(|a|-1)(|b|+|c|)}\tau_{(123)}\\{\\!\\{b,\\{\\!\\{c,a\\}\\!\\}\\}\\!\\}_{l}+(-1)^{|c|-1)(|a|+|b|)}\tau_{(123)}^{2}\\{\\!\\{c,\\{\\!\\{a,b\\}\\!\\}\\}\\!\\}_{l}.$

$\\{\\!\\{-,-,-\\}\\!\\}$ is a triple bracket. We want to show that it is
zero. We need to check it on generators.

First case : If two of the $a,b,c$ are in $A$ then all the terms are zero.

Second case : If all the $a$, $b$, $c$ are in $\mathbb{L}$, it is $0$ by
hypothesis.

Third case : We check Jacobi identity in the case of a triple
$(\alpha,D,\Delta)$ in $A\times\mathbb{L}\times\mathbb{L}$. We need to show
the equality :

$\\{\\!\\{\alpha,\\{\\!\\{D,\Delta\\}\\!\\}\\}\\!\\}_{L}+\sigma_{(123)}\\{\\!\\{D,\\{\\!\\{\Delta,\alpha\\}\\!\\}\\}\\!\\}_{L}+\sigma_{(123)}^{2}\\{\\!\\{\Delta,\\{\\!\\{\alpha,D\\}\\!\\}\\}\\!\\}_{L}=0$

Before computing each of these terms, let us remark that

$\omega\left(\\{\\!\\{D,\Delta\\}\\!\\}_{l}\right)=\\{\\!\\{\omega(D),\omega(\Delta)\\}\\!\\}_{l}.$

This equality can be written more simply as follows :

$\omega\left(\\{\\!\\{D,\Delta\\}\\!\\}_{l}^{\prime}\right)\otimes\\{\\!\\{D,\Delta\\}\\!\\}^{\prime\prime}_{l}=\\{\\!\\{\omega(D),\omega(\Delta)\\}\\!\\}_{l}^{\prime}\otimes\\{\\!\\{\omega(D),\omega(\Delta)\\}\\!\\}^{\prime\prime}_{l}.$

one has

$\begin{array}[]{rcl}\\{\\!\\{\alpha,\\{\\!\\{D,\Delta\\}\\!\\}\\}\\!\\}_{L}&=&\\{\\!\\{\alpha,\\{\\!\\{D,\Delta\\}\\!\\}_{l}^{\prime}\\}\\!\\}\otimes\\{\\!\\{D,\Delta\\}\\!\\}_{l}^{\prime\prime}\\\
&=&-\tau_{(12)}\omega\left(\\{\\!\\{D,\Delta\\}\\!\\}^{\prime}_{l}\right)(\alpha)\otimes\\{\\!\\{D,\Delta\\}\\!\\}^{\prime\prime}_{l}\\\
&=&-\tau_{(12)}\\{\\!\\{\omega(D),\omega(\Delta)\\}\\!\\}^{\prime}_{l}(\alpha)\otimes\\{\\!\\{\omega(D),\omega(\Delta)\\}\\!\\}^{\prime\prime}_{l}\\\
&=&-\tau_{(123)}\left[(\omega(D)\otimes
1)\omega(\Delta)-(1\otimes\omega(\Delta))\omega(D)\right](\alpha)\end{array}$
$\begin{array}[]{l}\sigma_{(123)}\\{\\!\\{D,\\{\\!\\{\Delta,\alpha\\}\\!\\}\\}\\!\\}_{L}=\sigma_{(123)}\left(\omega(D)\left[\omega(\Delta)(\alpha)^{\prime}\right]\otimes\omega(\Delta)(\alpha)^{\prime\prime}\right)=\tau_{(123)}\left((\omega(D)\otimes
1)\omega(\Delta)\right)(\alpha)\end{array}$
$\begin{array}[]{rcl}\sigma_{(23)}\\{\\!\\{\Delta,\\{\\!\\{\alpha,D\\}\\!\\}\\}\\!\\}_{L}&=&-\tau_{(132)}\left[\Delta\left(\omega(D)(\alpha)^{\prime\prime}\right)^{\prime}\otimes\Delta\left(\omega(D)(\alpha)^{\prime\prime}\right)^{\prime\prime}\otimes\omega(D)(\alpha)^{\prime}\right]\\\
&=&-\tau_{(123)}\left[\left(1\otimes\omega(\Delta)\right)\omega(D)(\alpha)\right]\par\end{array}$

∎

In the next proposition, we study the case where the anchor $\omega$ is zero.

######  Proposition 4.6.

Let $(\mathbb{L},\\{\\!\\{-,-\\}\\!\\},\omega)$ be a double Lie Rinehart
algebra over $A$ such that $\omega=0$. Set
$X\bullet_{r}Y=-\\{X,Y\\}_{r}=-\\{\\!\\{X,Y\\}\\!\\}_{r}^{\prime}\\{\\!\\{X,Y\\}\\!\\}_{r}^{\prime\prime}$
and
$X\bullet_{l}Y=\\{X,Y\\}_{l}=\\{\\!\\{X,Y\\}\\!\\}_{l}^{\prime}\\{\\!\\{X,Y\\}\\!\\}_{l}^{\prime\prime}$.
The laws $\bullet_{l}$ and $\bullet_{r}$ are associative so that
$\\{-,-\\}:{\mathbb{L}}\times\mathbb{L}\to{\mathbb{L}}$ is the difference of
two associative products. The laws induces by $\bullet_{l}$ and $\bullet_{r}$
are opposite from each other on $\dfrac{\mathbb{L}}{[A,{\mathbb{L}}]}$. The
bracket $\\{-,-\\}$ induces a Lie bracket on
$\dfrac{\mathbb{L}}{[A,{\mathbb{L}}]}$ that comes from an associative product.

###### Proof.

The associativity of $\bullet_{l}$ follows from lemma 4.3 and proposition 4.4.
The proof of the associativity of $\bullet_{l}$ is similar. Let $X$ and $Y$ in
$\mathbb{L}$.

$\begin{array}[]{rcl}X\bullet_{r}Y&=&-\\{\\!\\{X,Y\\}\\!\\}_{r}^{\prime}\\{\\!\\{X,Y\\}\\!\\}_{r}^{\prime\prime}=\\{\\!\\{Y,X\\}\\!\\}_{l}^{\prime\prime}\\{\\!\\{Y,X\\}\\!\\}_{l}^{\prime}=\\{\\!\\{Y,X\\}\\!\\}_{l}^{\prime}\\{\\!\\{Y,X\\}\\!\\}_{l}^{\prime\prime}\;{\rm
mod}\;[A,{\mathbb{L}}]\\\ &=&Y\bullet_{l}X\;{\rm
mod}\;[A,{\mathbb{L}}].\end{array}$

∎

######  Examples 4.7.

Example 1

$(\mathbb{D}er(A),A,id)$ is a double Lie algebroid.

Example 2 If $A=k$, double Lie algebroid structures on $\mathbb{L}$ over $k$
are in bijection with associative $k$-algebras structures over $\mathbb{L}$.
Indeed, If $\mathbb{L}$ is a double Lie algebroid over $k$, then the Jacobi
identities asserts that $\\{\\!\\{-,-\\}\\!\\}_{l}$ is an associative product.
Conversely, an associative product over $k$ gives rise to a double Lie
Rinehart algebra over $k$ as follows : $\\{\\!\\{X,Y\\}\\!\\}_{l}=XY\otimes
1$, $\\{\\!\\{X,Y\\}\\!\\}_{r}=-1\otimes YX$ $\\{\\!\\{X,Y\\}\\!\\}=XY\otimes
1-1\otimes YX$. The bracket $\\{\\!\\{-,-\\}\\!\\}$ identifies to the Lie
bracket coming from the associative structure.

Exemple 3:

Let us recall the definition of a double Lie algebra :

######  Definition 4.8.

Let $\mathfrak{g}$ be a vector space of finite dimension. A double Lie bracket
over $\mathfrak{g}$ is a map
$\\{\\!\\{-,-\\}\\!\\}:{\mathfrak{g}}\otimes{\mathfrak{g}}\to{\mathfrak{g}}\otimes{\mathfrak{g}}$
satisfying the Jacobi identity.

######  Remark 4.9.

Let $V$ be a $m$ dimensional $\mathbb{C}$-vector space. Double Lie algebra
structures on $V$ are in bijection ([13]) with operators $r\in End(V\otimes
V)$ such that $r(v\otimes u)=-r(u\otimes v)^{\circ}$ (skew-summetry) and
solution of the following equation

$r^{23}r^{12}+r^{31}r^{23}+r^{12}r^{31}=0$

where $r^{ij}$ acts on $V^{\otimes 3}$ non trivially on $(i,j)$ spaces and as
identity elsewhere.

Let $\mathfrak{g}$ be a double Lie algebra with double bracket
$\\{\\!\\{-,-\\}\\!\\}_{\mathfrak{g}}$. This latter induces a Poisson bracket
on $T({\mathfrak{g}})$. There is a unique double Lie algebroid structure on
$\mathbb{L}=T({\mathfrak{g}})\otimes{\mathfrak{g}}\otimes T({\mathfrak{g}})$
such that

$\begin{array}[]{l}\forall(X,Y)\in{\mathfrak{g}},\quad\forall a\in
T({\mathfrak{g}}),\\\
\\{\\!\\{X,Y\\}\\!\\}_{\mathbb{L}}=\\{\\!\\{X,Y\\}\\!\\}_{\mathfrak{g}}\\\
\omega(X)(a)=\\{\\!\\{X,a\\}\\!\\}_{T({\mathfrak{g}})}\end{array}$

Then $T({\mathfrak{g}})\otimes{\mathfrak{g}}\otimes T({\mathfrak{g}})$ is a
Lie double algebroid over $T({\mathfrak{g}})$.

Example 4: Let $V$ be an $A^{e}$-module of finite type. For $\lambda$, $\mu$
in $Hom_{A^{e}}(V,V\otimes V)$ (the exterior $A^{e}$-module structure on
$V\otimes V$ is used for the $Hom_{A^{e}}$ , one sets :

$\begin{array}[]{l}\\{\\!\\{\lambda,\mu\\}\\!\\}_{l}^{\sim}=(\lambda\otimes
1)\mu-(1\otimes\mu)\lambda\\\
\\{\\!\\{\lambda,\mu\\}\\!\\}_{r}^{\sim}=(1\otimes\lambda)\mu-(\mu\otimes
1)\lambda\\\ \end{array}$

One sets

$\begin{array}[]{l}\\{\\!\\{\lambda,\mu\\}\\!\\}_{l}=\tau_{(23)}\\{\\{\lambda,\mu\\}\\!\\}^{\sim}_{l}\\\
\\{\\!\\{\lambda,\mu\\}\\!\\}_{r}=\tau_{(12)}\\{\\{\lambda,\mu\\}\\!\\}^{\sim}_{r}\\\
\\{\\!\\{\lambda,\mu\\}\\!\\}=\\{\\!\\{\lambda,\mu\\}\\!\\}_{l}+\\{\\!\\{\lambda,\mu\\}\\!\\}_{r}\end{array}$

Then

$\begin{array}[]{rcl}\\{\\!\\{\lambda,\mu\\}\\!\\}\in Hom_{A^{e}}(V,V\otimes
V)\otimes_{k}V+V\otimes_{k}Hom_{A^{e}}(V,V\otimes V)\\\ \end{array}$

Then

$\begin{array}[]{rcl}\\{\\!\\{\lambda,\mu\\}\\!\\}\in Hom_{A^{e}}(V,V\otimes
V)\otimes_{k}T_{A}(V)\oplus T_{A}(V)\otimes_{k}Hom_{A^{e}}(V,V\otimes
V)\end{array}$

There is a unique Lie algebroid structure on
$T_{A}(V)\otimes_{k}Hom(V,V\otimes V)\otimes_{k}T_{A}(V)$ with anchor map

$\begin{array}[]{rcl}\omega:Hom_{A^{e}}(V,V\otimes V)\otimes T_{A}(V)\oplus
T_{A}(V)\otimes Hom_{A^{e}}(V,V\otimes V)&\to&\mathbb{D}er(T(V))\\\
\alpha\otimes\lambda\otimes\beta&\mapsto&\alpha\\{\\!\\{\lambda,-\\}\\!\\}\beta\end{array}$

where $\\{\\!\\{\lambda,-\\}\\!\\}$ is the unique derivation of $T_{A}(V)$
such that for all $v\in V$, $\\{\\!\\{\lambda,v\\}\\!\\}=\lambda(v)$ and, for
all $a\in A$, $\\{\\!\\{\lambda,a\\}\\!\\}=0$.

Example 5 :

Let $A$ be a $B$-algebra endowed with a $B$-linear double Poisson bracket. The
quadruple $\left(\Omega_{B}^{1}A,A,\omega,\\{\\!\\{,\\}\\!\\}\right)$ is a
double Lie Rinehart algebra with $\omega$ and $\\{\\!\\{,\\}\\!\\}$ defined as
follows :

$\begin{array}[]{rcl}\Omega_{B}^{1}A&\to&{\mathbb{D}}er_{B}(A)\\\
da&\mapsto&\\{\\!\\{a,-\\}\\!\\}\end{array}$
$\\{\\!\\{da,db\\}\\!\\}=d\left(\\{\\!\\{a,b\\}\\!\\}^{\prime}\right)\otimes\\{\\!\\{a,b\\}\\!\\}^{\prime\prime}+\\{\\!\\{a,b\\}\\!\\}^{\prime}\otimes
d\left(\\{\\!\\{a,b\\}\\!\\}^{\prime\prime}\right)$

Example 6 : ([12]) Let $(A,P)$ a double quasi Poisson algebra. Then
$\tilde{\Omega}_{A}=\Omega_{A}\oplus AEA$ has the structure of a double Lie
algebroid where the double bracket is defined as follows

$\begin{array}[]{l}\\{\\!\\{da,b\\}\\!\\}_{\tilde{\Omega}A}=\\{\\!\\{a,b\\}\\!\\}\\\
\\{\\!\\{da,db\\}\\!\\}_{\tilde{\Omega}A}=d\\{\\!\\{a,b\\}\\!\\}+\dfrac{1}{4}\left[b,\left[a,E\otimes
1-1\otimes E\right]_{*}\right]\\\
\\{\\!\\{E,X\\}\\!\\}_{\tilde{\Omega}A}=X\otimes 1-1\otimes X\end{array}$

for $a,b\in A,\quad X\in T_{A}\tilde{\Omega}_{A}$ where $[-,-]_{*}$ denotes
the commutator for the inner $A$-bimodule structure on $AEA\otimes AEA$.
Futhermore the anchor is the Ae-bimodule morphism defined by :

$\begin{array}[]{rcl}\Omega_{A}\oplus AEA&\to&D_{A}\\\
(du,\delta)&\mapsto&\\{\\!\\{u,-\\}\\!\\}+\delta\end{array}$

is surjective.

##  5\. Differential calculus

$A\otimes A$ is endowed with a $A^{e}\otimes A^{e}$-module. Let $M$ be an
$A^{e}$ -module. We choose to set

$M^{*}=\\{\lambda:M\to A\otimes A\mid\lambda(\alpha\cdot
D\cdot\beta)=\lambda(D)^{\prime}\beta\otimes\alpha\lambda(D)^{\prime\prime}\\}$

$M^{*}$ is itself an $A^{e}$-module as follows :

$\forall\lambda\in M^{*},\quad\forall(a,b)\in A^{2},\quad\forall D\in
M,\quad(a\cdot\lambda\cdot
b)(D)=a\lambda(D)^{\prime}\otimes\lambda(D)^{\prime\prime}b.$

######  Remark 5.1.

We can exchange the role of the two $A^{e}$-module structures on $A\otimes A$
and define

$M_{*}=\\{\lambda:M\to A\otimes A\mid\lambda(\alpha\cdot
D\cdot\beta)=\alpha\lambda(D)^{\prime}\otimes\lambda(D)^{\prime\prime}\beta\\}.$

Composition by $\tau$ is an isomorphism of $A^{e}$-module from $M^{*}$ to
$M_{*}$.

Let $M$ is an $A^{e}$\- module. Let us endow $A^{\otimes n+1}$ with the
$A^{e}$-module structure defined as follows :

$a^{\prime}\otimes a^{\prime\prime}\cdot(a_{1}\otimes\dots\otimes
a_{n+1})=a_{1}\otimes\dots\otimes a_{i}a^{\prime\prime}\otimes
a^{\prime}a_{i+1}\otimes\dots\otimes a_{n+1},$

then the map ([11]) 222our map is slightly different from that of VdB due to
different conventions for the dual of $\mathbb{L}$ : [11] makes use of
$M_{*}$, we make use of $M^{*}$.

$\Psi:M^{*\otimes_{A}n}\to Hom_{(A^{e})^{\otimes n}}\left(M^{\otimes
n},A^{\otimes n+1}\right)$

$\Psi(\lambda_{1}\otimes\lambda_{2}\otimes\dots\otimes\lambda_{n})(m_{1}\otimes\dots\otimes
m_{n})=\lambda_{1}(m_{1})^{\prime}\otimes\lambda_{1}(m_{1})^{\prime\prime}\lambda_{2}(m_{2})^{\prime}\otimes\dots\otimes\lambda_{n-1}(m_{n-1})^{\prime\prime}\lambda_{n}(m_{n})^{\prime}\otimes\lambda_{n}(m_{n})^{\prime\prime}$

is well defined.

If $M$ is a finitely generated $A^{e}$-projective module, then $\Psi$ is an
isomorphism of $A^{e}$-modules.

The cyclic group $C_{n}$ acts on $Hom_{(A^{e})^{\otimes n}}\left(M^{\otimes
n},A^{\otimes n}\right)$ as follows :

$\forall\omega\in Hom_{(A^{e})^{\otimes n}}\left(M^{\otimes n},A^{\otimes
n}\right),\quad\tau_{(1\dots n)}\cdot\omega=\tau_{(1\dots
n)}\circ\omega\circ\tau_{(1\dots n)}^{-1}$

The set of signed invariants of $Hom_{(A^{e})^{\otimes n}}\left(M^{\otimes
n},A^{\otimes n}\right)$ under the action of $C_{n}$ is

$s-inv\;Hom_{(A^{e})^{\otimes n}}\left(M^{\otimes n},A^{\otimes
n}\right):=\\{\omega\in Hom_{(A^{e})^{\otimes n}}\left(M^{\otimes
n},A^{\otimes n}\right)\mid\tau_{(1\dots n)}\cdot\omega=(-1)^{n-1}\omega\\}$

If $M$ is a finitely generated projective $A^{e}$-module, M. van den Bergh
showed ([11]) that $\dfrac{T^{*}(M^{*})}{[T^{*}(M^{*}),T^{*}(M^{*})]}$ is
isomorphic to $s-invHom_{(A^{e})^{\otimes n}}\left(M^{\otimes n},A^{\otimes
n}\right)$. He constructed the following isomorphism $\mu$ between the two
spaces :

$\mu:\dfrac{T^{*}(M^{*})}{[T^{*}(M^{*}),T^{*}(M^{*})]}\to
s-inv\;Hom_{(A^{e})^{\otimes n}}\left(M^{\otimes n},A^{\otimes n}\right)$

$\mu(\lambda_{1}\otimes\dots\otimes\lambda_{n})=\\{\\!\\{-,\dots,-\\}\\!\\}_{\lambda_{1}\dots\lambda_{n}}=\sum_{i}(-1)^{(n-1)i}\tau_{(1\dots
n)}^{i}\circ\Phi\circ\tau^{-i}_{(1\dots n)}$

with

$\Phi(\lambda_{1}\otimes\lambda_{2}\otimes\dots\otimes\lambda_{n})(m_{1}\otimes\dots\otimes
m_{n})=\lambda_{n}(m_{n})^{\prime\prime}\lambda_{1}(m_{1})^{\prime}\otimes\lambda_{1}(m_{1})^{\prime\prime}\lambda_{2}(m_{2})^{\prime}\otimes\dots\otimes\lambda_{n-1}(m_{n-1})^{\prime\prime}\lambda_{n}(m_{n})^{\prime}.$

Notation : In the computation,
$\Phi\left(\lambda_{1}\otimes\dots\otimes\lambda_{n}\right)$ will be denoted
$\widetilde{\lambda_{1}\otimes\dots\otimes\lambda_{n}}$.

Let $C$ be a graded algebra. We have in mind $C=T_{A}(\mathbb{L}^{*})$ or
$C=T_{A}(\mathbb{L})$ where $\mathbb{L}$ is a double Lie Rinehart algebra. Let

$\begin{array}[]{rcl}\Theta:C&\to&C\otimes C\\\
c&\mapsto&\Theta(c)^{\prime}\otimes\Theta(c)^{\prime\prime}\end{array}$

be a double derivation. One sets

$\begin{array}[]{rcl}{}^{\circ}\Theta:C&\to&C\\\
c&\mapsto&\Theta(c)^{\prime\prime}\Theta(c)^{\prime}.\end{array}$

${}^{\circ}\Theta$ is an endomorphism of $C$ and induces an endomorphism of
$\dfrac{C}{[C,C]}$.

######  Proposition 5.2.

1) Let $d:C\to C$ be a derivation of degree $\mid d\mid$. It induces a
derivation still denoted $d:C\otimes C\to C\otimes C$ :

$\forall c_{1}\otimes c_{2}\in C\otimes C,\quad d(c_{1}\otimes
c_{2})=d(c_{1})\otimes c_{2}+(-1)^{\mid c_{1}\mid}c_{1}\otimes d(c_{2})$

Let $\underline{i}:C\to C\otimes C$ be a double derivation. If $c\in C$, we
will write
$\underline{i}(c)=\underline{i}(c)^{\prime}\otimes\underline{i}(c)^{\prime\prime}$.
Then $(d\circ\underline{i})^{\circ}=d\circ\iota$ and $\left(\underline{i}\circ
d\right)^{\circ}=\iota\circ d$.

2) Let $\delta:C\to C\otimes C$ and $\Delta:C\to C\otimes C$ be two double
derivations.

$\delta\circ^{\circ}\Delta-\tau_{(12)}\Delta\circ^{\circ}\delta=^{\circ,l}\\{\\!\\{\delta,\Delta\\}\\!\\}_{l}+^{\circ,r}\\{\\!\\{\delta,\Delta\\}\\!\\}_{r}$

where
${}^{\circ,l}(\epsilon^{\prime}\otimes\epsilon^{\prime\prime})=^{\circ}\epsilon^{\prime}\otimes\epsilon^{\prime\prime}$
and
${}^{\circ,r}(\epsilon^{\prime}\otimes\epsilon^{\prime\prime})=\epsilon^{\prime}\otimes^{\circ}\epsilon^{\prime\prime}$

###### Proof.

1)
$(d\circ\underline{i})(c)=d\left(\underline{i}(c)^{\prime}\right)\otimes\underline{i}(c)^{\prime\prime}+(-1)^{\left|\underline{i}(c)^{\prime}\right|}\underline{i}(c)^{\prime}\otimes
d\left(\underline{i}(c)^{\prime\prime}\right)$.
Hence

$\begin{array}[]{rcl}(d\circ\underline{i})^{\circ}(c)&=&\underline{i}(c)^{\prime\prime}d\underline{i}(c)^{\prime}(-1)^{(|\underline{i}(c)^{\prime}|+\mid
d\mid)|\underline{i}(c)^{\prime\prime}|}+\left(d\underline{i}(c)^{\prime\prime}\right)\underline{i}(c)^{\prime}(-1)^{\left|\underline{i}(c)^{\prime\prime}\right|\left|\underline{i}(c)^{\prime}\right|}\\\
&=&d\left[\underline{i}(c)^{\prime\prime}\underline{i}(c)^{\prime}(-1)^{|\underline{i}(c)^{\prime}||\underline{i}(c)^{\prime\prime}|}\right]\\\
&=&(d\circ\iota)(c)\end{array}$

The equality $(\underline{i}\circ d)^{\circ}=\iota\circ d$ is obvious.

2) is stated in [11]. ∎

Let us now see examples of this situation.

From now on, we will assume that the $k$-algebra A is smooth and that
${\mathbb{L}}$ is a finitely generated and projective $A^{e}$-module

The contraction

Set $\mathbb{L}^{*}=Hom_{A^{e}}(\mathbb{L},A\otimes A)$ and let
$D\in\mathbb{L}$. The element $D$ defines a degree 1 double derivation
$\underline{i}_{D}:T_{A}(\mathbb{L}^{*})\to T_{A}(\mathbb{L}^{*})\otimes
T_{A}(\mathbb{L}^{*})$

$\begin{array}[]{l}\forall\alpha\in\mathbb{L}^{*},\quad\underline{i}_{D}(\alpha)=\alpha(D)^{\prime}\otimes\alpha(D)^{\prime\prime}.\end{array}$

More explicitely, the map $\underline{i}_{D}:T^{\bullet}(\mathbb{L}^{*})\to
T^{\bullet}(\mathbb{L}^{*})\otimes T^{\bullet}(\mathbb{L}^{*})$ is given by

$\underline{i}_{D}(\alpha_{1}\dots\alpha_{n})=\displaystyle\sum_{i=1}^{n}(-1)^{k-1}\alpha_{1}\dots\alpha_{k-1}\alpha_{k}(D)^{\prime}\otimes\alpha_{k}(D)^{\prime\prime}\alpha_{k+1}\dots\alpha_{n}$

######  Lemma 5.3.

([4]) For all $\Phi,\Theta$ in $\mathbb{L}$, one has :

$\begin{array}[]{l}\underline{i}_{\Theta}(\alpha\beta)=\underline{i}_{\Theta}(\alpha)\beta+(-1)^{deg\;\alpha}\alpha\underline{i}_{\Theta}(\beta)\\\
\underline{i}_{\Phi}\circ\underline{i}_{\Theta}+\underline{i}_{\Theta}\circ\underline{i}_{\Phi}=0\end{array}$

The differential

######  Theorem 5.4.

One defines $\underline{d}_{\mathbb{L}}:T_{A}(\mathbb{L}^{*})\to
T_{A}(\mathbb{L}^{*})$ by

  * •

$\underline{d}_{\mathbb{L}}:T_{A}^{k}(\mathbb{L}^{*})\to
T_{A}^{k+1}(\mathbb{L}^{*})$

  * •

$\forall
D\in\mathbb{L},\quad\underline{d}_{\mathbb{L}}a(D)=D(a)^{\prime}\otimes
D(a)^{\prime\prime}$

  * •

$\forall\lambda\in\mathbb{L}^{*},\quad\underline{d_{\mathbb{L}}}\lambda\in\mathbb{L}^{*}\otimes_{A}\mathbb{L}^{*}\simeq
Hom_{A^{e}}(\mathbb{L}\otimes\mathbb{L},A^{\otimes 3})$ is given by for all :
$D,\Delta\in{\mathbb{L}}$

$\begin{array}[]{rcl}\underline{d}_{\mathbb{L}}(\lambda)(D,\Delta)&=&D\left(\lambda(\Delta)^{\prime}\right)\otimes\lambda(\Delta)^{\prime\prime}-\lambda(D)^{\prime}\otimes\Delta\left(\lambda(D)^{\prime\prime}\right)-\tau_{(23)}\left[\lambda(\\{\\!\\{D,\Delta\\}\\!\\}^{\prime}_{L})\otimes\\{\\!\\{D,\Delta\\}\\!\\}^{\prime\prime}_{L}\right]\\\
\end{array}$

  * •

$\forall\psi\in Hom_{A^{e}}(\mathbb{L}^{\otimes n},A^{\otimes n+1})$

$\begin{array}[]{rcl}\underline{d}_{\mathbb{L}}(\psi)(D_{1},\dots,D_{n+1})&=&{\displaystyle\sum_{i=1}^{n+1}}(-1)^{i-1}\left(id^{i-1}\circ
D_{i}\circ id^{n-i}\right)\psi(D_{1},\dots,\widehat{D_{i}}\dots D_{n+1})\\\
&+&{\displaystyle\sum_{i=1}^{n}}(-1)^{i}\psi\left(D_{1}\otimes\dots\otimes\\{\\!\\{D_{i},D_{i+1}\\}\\!\\}_{L}\otimes\dots\otimes
D_{n+1}\right)\end{array}$

where if $\psi=\psi_{1}\otimes\psi_{2}\otimes\dots\otimes\psi_{n}$ with
$\psi_{i}:\mathbb{L}\to A^{\otimes 2}$, then

$\psi_{i}(\\{\\!\\{D_{i},D_{i+1}\\}\\!\\}_{L})=\tau_{(23)}\left[\psi_{i}(\\{\\!\\{D_{i},D_{i+1}\\}\\!\\}_{L}^{\prime})\otimes\\{\\!\\{D_{i},D_{i+1}\\}\\!\\}_{L}^{\prime\prime}\right]$

As $\underline{d}_{\mathbb{L}^{*}}$ is a derivation whose square is zero (that
is to say a differential).

Proof : In the proof, we will make use of the following notation : If
$\lambda\in\mathbb{L}^{*}$, then
$\underline{d}_{\mathbb{L}}(\lambda)=\underline{d}_{\mathbb{L}}(\lambda)^{(1)}\otimes\dots\otimes\underline{d}_{\mathbb{L}}(\lambda)^{(n+1)}$
where $\underline{d}_{\mathbb{L}}(\lambda)^{(i)}$ takes values in $A$.

If $D_{1}$, $D_{2}$, $D_{3}$ are three elements of $\mathbb{L}$, one has :

$\begin{array}[]{rcl}(\underline{d}_{\mathbb{L}}\circ\underline{d}_{\mathbb{L}})(\lambda)(D_{1},D_{2},D_{3})&=&D_{1}\left[D_{2}\left(\lambda(D_{3})^{\prime}\right)^{\prime}\right]\otimes
D_{2}\left(\lambda(D_{3})^{\prime}\right)^{\prime\prime}\otimes\lambda(D_{3})^{\prime\prime}-D_{1}\left(\lambda(D_{2})^{\prime}\right)\otimes
D_{3}\left(\lambda(D_{2})^{\prime\prime}\right)\\\
&-&D_{1}\left[\lambda\left(\\{\\!\\{D_{2},D_{3}\\}\\!\\}_{l}^{\prime}\right)^{\prime}\right]\otimes\\{\\!\\{D_{2},D_{3}\\}\\!\\}_{l}^{\prime\prime}\otimes\lambda\left(\\{\\!\\{D_{2},D_{3}\\}\\!\\}_{l}^{\prime}\right)^{\prime\prime}\\\
&-&D_{1}\left(\lambda(D_{3})^{\prime}\right)^{\prime}\otimes
D_{2}\left[D_{1}\left(\lambda(D_{3})^{\prime}\right)^{\prime\prime}\right]\otimes
D_{3}\left(\lambda(D_{1})^{\prime\prime}\right)^{\prime\prime}\\\
&+&\lambda(D_{1})^{\prime}\otimes
D_{2}\left[D_{3}\left(\lambda(D_{1})^{\prime\prime}\right)^{\prime}\right]\otimes
D_{3}\left(\lambda(D_{1})^{\prime\prime}\right)^{\prime\prime}\\\
&+&\lambda\left(\\{\\!\\{D_{1},D_{3}\\}\\!\\}_{l}^{\prime}\right)^{\prime}\otimes
D_{2}\left[\\{\\!\\{D_{1},D_{3}\\}\\!\\}_{l}^{\prime\prime}\right]\otimes\lambda\left(\\{\\!\\{D_{1},D_{3}\\}\\!\\}_{l}^{\prime}\right)^{\prime\prime}\\\
&+&D_{1}\left(\lambda(D_{2})^{\prime}\right)\otimes
D_{3}\left(\lambda(D_{2})^{\prime\prime}\right)\\\
&-&\lambda(D_{1})^{\prime}\otimes
D_{2}\left(\lambda(D_{1})^{\prime\prime}\right)^{\prime}\otimes
D_{3}\left[D_{2}\left(\lambda(D_{1})^{\prime\prime}\right)^{\prime\prime}\right]\\\
&-&\lambda\left(\\{\\!\\{D_{1},D_{2}\\}\\!\\}_{l}^{\prime}\right)^{\prime}\otimes\\{\\!\\{D_{1},D_{2}\\}\\!\\}_{l}^{\prime\prime}\otimes
D_{3}\left[\lambda\left(\\{\\!\\{D_{1},D_{2}\\}\\!\\}_{l}^{\prime}\right)^{\prime\prime}\right]\\\
&-&(\underline{d}_{\mathbb{L}}\lambda)^{(1)}\left(\\{\\!\\{D_{1},D_{2}\\}\\!\\}^{\prime}_{l},D_{3}\right)\otimes\\{\\!\\{D_{1},D_{2}\\}\\!\\}^{\prime\prime}_{l}\otimes(\underline{d}_{\mathbb{L}}\lambda)^{(2)}(-)\otimes(\underline{d}_{\mathbb{L}}\lambda)^{(3)}(-)\\\
&+&(\underline{d}_{\mathbb{L}}\lambda)^{(1)}\left(D_{1},\\{\\!\\{D_{2},D_{3}\\}\\!\\}^{\prime}_{l}\right)\otimes(\underline{d}_{\mathbb{L}}\lambda)^{(2)}(-)\otimes\\{\\!\\{D_{2},D_{3}\\}\\!\\}^{\prime\prime}_{l}\otimes(\underline{d}_{\mathbb{L}}\lambda)^{(3)}(-)\end{array}$

But

$\begin{array}[]{l}-(\underline{d}_{\mathbb{L}}\lambda)^{(1)}\left(\\{\\!\\{D_{1},D_{2}\\}\\!\\}^{\prime}_{l},D_{3}\right)\otimes\\{\\!\\{D_{1},D_{2}\\}\\!\\}_{l}^{\prime\prime}\otimes(\underline{d}_{\mathbb{L}}\lambda)^{(3)}(-)\otimes(\underline{d}_{\mathbb{L}}\lambda)^{(4)}(-)\\\
+(\underline{d}_{\mathbb{L}}\lambda)^{(1)}\left(D_{1},\\{\\!\\{D_{2},D_{3}\\}\\!\\}^{\prime}_{l}\right)\otimes(\underline{d}_{\mathbb{L}}\lambda)^{(2)}(-)\otimes\\{\\!\\{D_{2},D_{3}\\}\\!\\}^{\prime\prime}_{l}\otimes(\underline{d}_{\mathbb{L}}\lambda)^{(4)}(-)\\\
=-\\{\\!\\{D_{1},D_{2}\\}\\!\\}^{\prime}_{l}\left(\lambda(D_{3})^{\prime}\right)\otimes\\{\\!\\{D_{1},D_{2}\\}\\!\\}^{\prime\prime}_{l}\otimes\lambda(D_{3})^{\prime\prime}\\\
+\lambda\left(\\{\\!\\{D_{1},D_{2}\\}\\!\\}_{l}^{\prime}\right)^{\prime}\otimes\\{\\!\\{D_{1},D_{2}\\}\\!\\}^{\prime\prime}\otimes
D_{3}\left[\lambda\left(\\{\\!\\{D_{1},D_{2}\\}\\!\\}_{l}^{\prime}\right)^{\prime}\right]\\\
+\lambda\left[\\{\\!\\{\\{\\!\\{D_{1},D_{2}\\}\\!\\}^{\prime}_{l},D_{3}\\}\\!\\}_{l}^{\prime}\right]^{\prime}\otimes\\{\\!\\{D_{1},D_{2}\\}\\!\\}_{l}^{\prime\prime}\otimes\\{\\!\\{\\{\\!\\{D_{1},D_{2}\\}\\!\\}^{\prime}_{l},D_{3}\\}\\!\\}_{l}^{\prime\prime}\otimes\lambda\left[\\{\\!\\{\\{\\!\\{D_{1},D_{2}\\}\\!\\}^{\prime}_{l},D_{3}\\}\\!\\}_{l}^{\prime}\right]^{\prime\prime}\\\
+D_{1}\left[\lambda\left(\\{\\!\\{D_{2},D_{3}\\}\\!\\}_{l}^{\prime}\right)^{\prime}\right]\otimes\\{\\!\\{D_{2},D_{3}\\}\\!\\}_{l}^{\prime\prime}\otimes\lambda\left(\\{\\!\\{D_{2},D_{3}\\}\\!\\}_{l}^{\prime}\right)^{\prime\prime}\\\
-\lambda(D_{1})^{\prime}\otimes\\{\\!\\{D_{2},D_{3}\\}\\!\\}_{l}^{\prime}\left[\lambda(D_{1})^{\prime\prime}\right]^{\prime}\otimes\\{\\!\\{D_{2},D_{3}\\}\\!\\}_{l}^{\prime\prime}\otimes\\{\\!\\{D_{2},D_{3}\\}\\!\\}_{l}^{\prime}\left[\lambda(D_{1})^{\prime\prime}\right]^{\prime\prime}\\\
-\lambda\left[\\{\\!\\{D_{1},\\{\\!\\{D_{2},D_{3}\\}\\!\\}^{\prime}_{l}\\}\\!\\}^{\prime}_{l}\right]^{\prime}\otimes\\{\\!\\{D_{1},\\{\\!\\{D_{2},D_{3}\\}\\!\\}^{\prime}_{l}\\}\\!\\}_{l}^{\prime\prime}\otimes\\{\\!\\{D_{2},D_{3}\\}\\!\\}_{l}^{\prime\prime}\otimes\lambda\left[\\{\\!\\{D_{1},\\{\\!\\{D_{2},D_{3}\\}\\!\\}^{\prime}_{l}\\}\\!\\}^{\prime}_{l}\right]^{\prime\prime}\end{array}$

A lot of terms cancel, and we are left with

$\begin{array}[]{l}(\underline{d}_{\mathbb{L}}\circ\underline{d}_{\mathbb{L}})(\lambda)(D_{1},D_{2},D_{3})=\lambda\left(\\{\\!\\{D_{1},D_{3}\\}\\!\\}_{l}^{\prime}\right)^{\prime}\otimes
D_{2}\left[\\{\\!\\{D_{1},D_{3}\\}\\!\\}_{l}^{\prime\prime}\right]\otimes\lambda\left(\\{\\!\\{D_{1},D_{3}\\}\\!\\}_{l}^{\prime}\right)^{\prime\prime}\\\
+\lambda\left[\\{\\!\\{\\{\\!\\{D_{1},D_{2}\\}\\!\\}^{\prime}_{l},D_{3}\\}\\!\\}_{l}^{\prime}\right]^{\prime}\otimes\\{\\!\\{D_{1},D_{2}\\}\\!\\}^{\prime\prime}\otimes\\{\\!\\{\\{\\!\\{D_{1},D_{2}\\}\\!\\}^{\prime}_{l},D_{3}\\}\\!\\}_{l}^{\prime\prime}\otimes\lambda\left[\\{\\!\\{\\{\\!\\{D_{1},D_{2}\\}\\!\\}^{\prime}_{l},D_{3}\\}\\!\\}_{l}^{\prime}\right]^{\prime\prime}\\\
-\lambda\left[\\{\\!\\{D_{1},\\{\\!\\{D_{2},D_{3}\\}\\!\\}^{\prime}_{l}\\}\\!\\}^{\prime}_{l}\right]^{\prime}\otimes\\{\\!\\{D_{1},\\{\\!\\{D_{2},D_{3}\\}\\!\\}^{\prime}_{l}\\}\\!\\}_{l}^{\prime\prime}\otimes\\{\\!\\{D_{2},D_{3}\\}\\!\\}_{l}^{\prime\prime}\otimes\lambda\left[\\{\\!\\{D_{1},\\{\\!\\{D_{2},D_{3}\\}\\!\\}^{\prime}_{l}\\}\\!\\}^{\prime}_{l}\right]^{\prime\prime}\par\end{array}$

The equality
$\underline{d}_{\mathbb{L}}\circ\underline{d}_{\mathbb{L}}(\lambda)(D_{1},D_{2},D_{3})=0$
follows now from the lemma 4.3.$\Box$

######  Remark 5.5.

In [4], it is shown that for $\mathbb{L}=\mathbb{D}er(A)$, the complex
$\left(T_{A}^{\bullet}(\mathbb{L}),\underline{d}_{\mathbb{L}}\right)$ is
acyclic in strictly positive degree.

######  Definition 5.6.

The differential $\underline{d}_{\mathbb{L}}$ induces a differential
$d_{\mathbb{L}}:\dfrac{T_{A}(\mathbb{L}^{*})}{[T_{A}(\mathbb{L}^{*}),T_{A}(\mathbb{L}^{*})]}\to\dfrac{T_{A}(\mathbb{L}^{*})}{[T_{A}(\mathbb{L}^{*}),T_{A}(\mathbb{L}^{*})]}$.
We will write
$DR(\mathbb{L})^{\bullet}:=\dfrac{T_{A}(\mathbb{L}^{*})}{[T_{A}(\mathbb{L}^{*}),T_{A}(\mathbb{L}^{*})]}$
and $d_{\mathbb{L}}$ will be called ”the differential of the double Lie
algebroid” $\mathbb{L}$.

######  Remarks 5.7.

1) If $\mathbb{L}={\mathbb{D}}er_{B}(A)$, the complex
$\left(DR^{\bullet}(\mathbb{L}),d_{\mathbb{L}}\right)$ is the non commutative
De Rham complex ([4]).

2) If $A$ is a double Poisson algebra and $\mathbb{L}=\Omega^{1}_{k}(A)$, we
will see that the complex
$\left(DR^{\bullet}(\mathbb{L}),d_{\mathbb{L}}\right)$ is the complex
computing the non commutative Poisson cohomology ([9]).

In low dimensions, the expression of $d_{\mathbb{L}}$ is the following : For
all $a\in A$, $\phi\in\mathbb{L}$ and $\widetilde{\phi}\in
s-invHom_{(A^{e})}\left(\mathbb{L},A\right)$

$\begin{array}[]{l}d_{\mathbb{L}}(a)(D)=D(a)^{\prime\prime}D(a)^{\prime}\\\
d_{\mathbb{L}}(\widetilde{\phi})(D,\Delta)=D\left(\widetilde{\phi}(\Delta)\right)-\Delta\left(\widetilde{\phi}(D)\right)^{\circ}-\left[\widetilde{\phi}\left(\\{\\!\\{D,\Delta\\}\\!\\}_{l}^{\prime}\right)\otimes\\{\\!\\{D,\Delta\\}\\!\\}_{l}^{\prime\prime}\right]-\left[\\{\\!\\{D,\Delta\\}\\!\\}_{r}^{\prime}\otimes\widetilde{\phi}\left(\\{\\!\\{D,\Delta\\}\\!\\}_{r}^{\prime\prime}\right)\right]\\\
\end{array}$

Let us now give an expression for $d_{L}$.

######  Proposition 5.8.

Let $\mu_{1},\dots,\mu_{n}\in\mathbb{L}^{*}$. We have the following formula
where indices should be understood modulo $n+1$ :

$\begin{array}[]{l}d_{\mathbb{L}}\left(\\{\\!\\{-,\dots,-\\}\\!\\}_{\mu_{1}\dots\mu_{n}}\right)(D_{1}\otimes\dots\otimes
D_{n+1})\\\ ={\displaystyle\sum_{i=1}^{n+1}}(-1)^{n(i-1)}\tau_{(1\dots
n+1)}^{i-1}\\{\\!\\{D_{i},\\{\\!\\{D_{i+1},\dots,D_{n+1},\dots,D_{i-1}\\}\\!\\}_{\mu_{1}\dots\mu_{n}}\\}\\!\\}_{L}\\\
+{\displaystyle\sum_{i=1}^{n+1}}(-1)^{n(i+1)}\tau_{(12\dots
n+1)}^{i}\\{\\!\\{D_{i+1},D_{i+2},\dots,D_{n+1},D_{1},\dots,\\{\\!\\{D_{i-1},D_{i}\\}\\!\\}\\}\\!\\}_{\mu_{1}\dots\mu_{n},L}\end{array}$

###### Proof.

$\Phi\left(\underline{d}_{\mathbb{L}}(\mu_{1}\dots\mu_{n})\right)(D_{1}\otimes\dots\otimes
D_{n+1})=(\underline{I})+(\underline{II})$

where

$\begin{array}[]{l}(\underline{I})=\mu_{n}(D_{n+1})^{\prime\prime}D_{1}\left(\mu_{1}(D_{2})^{\prime}\right)\otimes\mu_{1}(D_{2})^{\prime\prime}\mu_{2}(D_{3})^{\prime}\otimes\otimes\dots\otimes\mu_{n-1}(D_{n})^{\prime\prime}\mu_{n}(D_{n+1})^{\prime}\\\
+\dots+{\displaystyle\sum_{i=2}^{n}}(-1)^{i-1}\mu_{n}(D_{n+1})^{\prime\prime}\mu_{1}(D_{1})^{\prime}\otimes\dots\otimes
D_{i}\left(\mu_{i-1}(D_{i-1})^{\prime\prime}\mu_{i}(D_{i+1})^{\prime}\right)\otimes\dots\otimes\mu_{n-1}(D_{n})^{\prime\prime}\mu_{n}(D_{n+1})^{\prime}\\\
(-1)^{n}D_{n+1}\left(\mu_{n}(D_{n})^{\prime\prime}\right)^{\prime\prime}\mu_{1}(D_{1})^{\prime}\otimes\mu_{1}(D_{1})^{\prime\prime}\mu_{2}(D_{2})^{\prime}\otimes\dots\otimes\mu_{n-1}(D_{n-1})^{\prime\prime}\mu_{n}(D_{n})^{\prime}\otimes
D_{n+1}\left(\mu_{n}(D_{n})^{\prime\prime}\right)^{\prime}\\\ \end{array}$

and

$\begin{array}[]{l}(\underline{II})={\displaystyle\sum_{i=1}^{n}}(-1)^{i}\mu_{n}(D_{n+1})^{\prime\prime}\mu_{1}(D_{1})^{\prime}\otimes\mu_{1}(D_{1})^{\prime\prime}\mu_{2}(D_{2})^{\prime}\otimes\dots\otimes\\\
\mu_{i-1}(D_{i-1})^{\prime\prime}\mu_{i}\left(\\{\\!\\{D_{i},D_{i+1}\\}\\!\\}_{l}^{\prime}\right)^{\prime}\otimes\\{\\!\\{D_{i},D_{i+1}\\}\\!\\}_{l}^{\prime\prime}\otimes\mu_{i}\left(\\{\\!\\{D_{i},D_{i+1}\\}\\!\\}_{l}^{\prime}\right)^{\prime\prime}\mu_{i+1}(D_{i+2})^{\prime}\otimes\dots\otimes\mu_{n-1}(D_{n})^{\prime\prime}\mu_{n}(D_{n+1})^{\prime}\end{array}$

For simplicity, we write $\sigma:=\tau_{(1\dots n)}$ and $\tau:=\tau_{(1\dots
n+1)}$. Then

$\\{\\!\\{-,\dots,-\\}\\!\\}_{\mu_{1}\dots\mu_{n}}={\displaystyle\sum_{i=1}^{n}}(-1)^{(n-1)(i-1)}\mu_{\sigma^{i}(1)}\otimes\dots\otimes\mu_{\sigma^{i}(n)}.$

Using the formulas of $\underline{I}$ and $\underline{II}$, one writes
$d_{E}(\\{\\!\\{-,\dots,-\\}\\!\\}_{\mu_{1}\dots\mu_{n}})(D_{1}\otimes\dots\otimes
D_{n+1})=(I)+(II)$, In the computation of (I), the terms of the form
$D_{1}()\otimes...$ gives
$\\{\\!\\{D_{1},\\{\\!\\{D_{2},\dots,D_{n+1}\\}\\!\\}_{\mu_{1}\dots\mu_{n}}\\}\\!\\}_{L}$.
More precisely :

$\begin{array}[]{lll}(I)&=&D_{1}\left[\mu_{n}(D_{n+1})^{\prime\prime}\mu_{1}(D_{2})^{\prime}\right]\otimes\mu_{1}(D_{2})^{\prime\prime}\mu_{2}(D_{3})^{\prime}\otimes\dots\otimes\mu_{n-1}(D_{n})^{\prime\prime}\mu_{n}(D_{n+1})^{\prime}+\dots\\\
&+&{\displaystyle\sum_{i=1}^{n}\sum_{j=1}^{n+1}}(-1)^{(n-1)(i-1)}(-1)^{n(j-1)}\tau^{j-1}\left(\mu_{k}\longleftarrow\mu_{\sigma^{i-1}(k)},D_{l}\longleftarrow
D_{\tau^{j-1}(l)}\right)\\\
&=&\\{\\!\\{D_{1},\\{\\!\\{D_{2},\dots,D_{n+1}\\}\\!\\}_{\mu_{1}\dots\mu_{n}}\\}\\!\\}_{L}+\dots\\\
&+&(-1)^{n(i-1)}\tau_{(1\dots
n+1)}^{i-1}\\{\\!\\{D_{i},\\{\\!\\{D_{i+1},\dots,D_{n+1},\dots,D_{i-1}\\}\\!\\}_{\mu_{1}\dots\mu_{n}}\\}\\!\\}_{L}+\dots\\\
&=&{\displaystyle\sum_{i=1}^{n+1}}(-1)^{n(i-1)}\tau_{(1\dots
n+1)}^{i-1}\\{\\!\\{D_{i},\\{\\!\\{D_{i+1},\dots,D_{n+1},\dots,D_{i-1}\\}\\!\\}_{\mu_{1}\dots\mu_{n}}\\}\\!\\}_{L}\end{array}$

where the notation
$\left(\mu_{k}\longleftarrow\mu_{\sigma^{i}(k)},D_{l}\longleftarrow
D_{\tau^{j}(l)}\right)$ means that we reproduce the term before replacing
$\mu_{k}$ by $\mu_{\sigma^{i}(k)}$ and $D_{l}$ by $D_{\tau^{j}(l)}$.

When computing (II), the terme finishing by
$\\{\\!\\{D_{n},D_{n+1}\\}\\!\\}^{\prime\prime}$ gives

$(-1)^{n}\\{\\!\\{D_{1},\dots,D_{n-1},\\{\\!\\{D_{n},D_{n+1}\\}\\!\\}^{\prime}\\}\\!\\}_{\mu_{1}\dots\mu_{n}}\otimes\\{\\!\\{D_{n},D_{n+1}\\}\\!\\}^{\prime\prime}$

More precisely :

$\begin{array}[]{l}(II)=\\\
(-1)^{n}\mu_{n}\left(\\{\\!\\{D_{n},D_{n+1}\\}\\!\\}^{\prime}\right)^{\prime\prime}\mu_{1}(D_{1})^{\prime}\otimes\mu_{1}(D_{1})^{\prime\prime}\mu_{2}(D_{2})^{\prime}\otimes\dots\otimes\mu_{n-1}(D_{n-1})^{\prime\prime}\mu_{n}\left(\\{\\!\\{D_{n},D_{n+1}\\}\\!\\}^{\prime}\right)\otimes\\{\\!\\{D_{n},D_{n+1}\\}\\!\\}^{\prime\prime}\\\
+\dots+{\displaystyle\sum_{i=1}^{n}\sum_{j=1}^{n+1}}(-1)^{(n-1)(i-1)}(-1)^{n(j-1)}\tau^{j-1}\left(\mu_{k}\longleftarrow\mu_{\sigma^{i-1}(k)},D_{l}\longleftarrow
D_{\tau^{j-1}(l)}\right)\\\
=(-1)^{n}\\{\\!\\{D_{1},\dots,D_{n-1},\\{\\!\\{D_{n},D_{n+1}\\}\\!\\}\\}\\!\\}_{\mu_{1}\dots\mu_{n}}+\dots+(-1)^{2n}\tau^{n}\left(D_{k}\longleftarrow
D_{\tau^{n+1}(D_{k})}\right)\\\
={\displaystyle\sum_{i=1}^{n+1}}(-1)^{n(i+1)}\tau_{(12\dots
n+1)}^{i}\\{\\!\\{D_{i+1},D_{i+2},\dots,\\{\\!\\{D_{i-1},D_{i}\\}\\!\\}\\}\\!\\}_{\mu_{1}\dots\mu_{n},L}\end{array}$

∎

We will study more in detail the case where $A$ is a smooth Poisson algebra
with double bracket defined by the double vector $P=\delta\Delta$ and
$\mathbb{L}=\Omega^{1}_{A}$. Note that
$\Omega^{1}_{A}=\mathbb{D}er_{B}(A)^{*}$ and
$\mathbb{D}er_{B}(A)=(\Omega_{A}^{1})_{*}$ so that if $D$ is in
$\mathbb{D}er_{B}(A)$, then $\sigma D\in(\Omega^{1}_{B}A)^{*}$.

######  Theorem 5.9.

1) If $a,f\in A$, then
$\underline{d}_{\mathbb{L}}(a)(df)=-\\{\\!\\{f,\\{P,a\\}\\}\\!\\}$ .

2) If $D_{1},\dots,D_{n}$ are in $\mathbb{L}$, then

$\underline{d}_{\mathbb{L}}(\sigma D_{1},\dots,\sigma
D_{n})(da_{1}\otimes\dots\otimes
da_{n+1})=-\\{\\!\\{a_{1},\\{\\!\\{a_{2},\dots\\{\\!\\{a_{n+1},\\{P,D_{1}\dots
D_{n}\\}\\}\\!\\}_{\mathbb{L}}\dots\\}\\!\\}_{\mathbb{L}}$

###### Proof.

Let us prove that
$\underline{d}_{\mathbb{L}}(a)(df)=\\{\\!\\{f,\\{\delta\Delta,a\\}\\}\\!\\}$.
With our definition, we get

$\begin{array}[]{rcl}\underline{d}_{\mathbb{L}}(a)(df)&=&\\{\\!\\{f,a\\}\\!\\}^{\prime}\otimes\\{\\!\\{f,a\\}\\!\\}^{\prime\prime}\\\
&=&\Delta(a)^{\prime}\delta(f)^{\prime\prime}\otimes\delta(f)^{\prime}\Delta(a)^{\prime\prime}-\delta(a)^{\prime}\Delta(f)^{\prime\prime}\otimes\Delta(f)^{\prime}\delta(a)^{\prime\prime}\end{array}$

On the other hand :

$\begin{array}[]{rcl}\\{\delta\Delta,a\\}&=&-\\{\\!\\{a,\delta\Delta\\}\\!\\}^{\prime\prime}\\{\\!\\{a,\delta\Delta\\}\\!\\}^{\prime}\\\
&=&-\\{\\!\\{a,\Delta\\}\\!\\}^{\prime\prime}\delta\\{\\!\\{a,\Delta\\}\\!\\}^{\prime}+\\{\\!\\{a,\delta\\}\\!\\}^{\prime\prime}\Delta\\{\\!\\{a,\delta\\}\\!\\}^{\prime}\\\
&=&\Delta(a)^{\prime}\delta\Delta(a)^{\prime\prime}-\delta(a)^{\prime}\Delta\delta(a)^{\prime\prime}\end{array}$

Hence
$\\{\delta\Delta,a\\}(df)=\delta(f)^{\prime}\Delta(a)^{\prime\prime}\otimes\Delta(a)^{\prime}\delta(f)^{\prime\prime}-\Delta(f)^{\prime}\delta(a)^{\prime\prime}\otimes\delta(a)^{\prime}\Delta(f)^{\prime\prime}$.

The proof of 2) is a consequence of the two following lemmas.

######  Lemma 5.10.

$\underline{d}_{\mathbb{L}}(\sigma D)(da\otimes
db)=-\\{\\!\\{a,\\{\\!\\{b,\\{P,D\\}\\}\\!\\}_{L}$

###### Proof.

$\begin{array}[]{l}\underline{d}_{\mathbb{L}}(\sigma D)(da\otimes db)\\\
=-D(a)^{\prime\prime}\otimes\\{\\!\\{b,D(a)^{\prime}\\}\\!\\}_{P}+\\{\\!\\{a,D(b)^{\prime\prime}\\}\\!\\}_{P}\otimes
D(b)^{\prime}-\tau_{(23)}\\{\\!\\{\sigma
D,\\{\\!\\{a,\\{P,b\\}\\}\\!\\}\\}\\!\\}_{L}\\\
=-D(a)^{\prime\prime}\otimes\\{\\!\\{b,D(a)^{\prime}\\}\\!\\}_{P}+\\{\\!\\{a,D(b)^{\prime\prime}\\}\\!\\}_{P}\otimes
D(b)^{\prime}-D\left(\\{\\!\\{a,b\\}\\!\\}^{\prime}\right)^{\prime\prime}\otimes\\{\\!\\{a,b\\}\\!\\}^{\prime\prime}\otimes
D\left(\\{\\!\\{a,b\\}\\!\\}^{\prime}\right)^{\prime}\\\
=-D(a)^{\prime\prime}\otimes\\{\\!\\{b,D(a)^{\prime}\\}\\!\\}_{P}+\\{\\!\\{a,D(b)^{\prime\prime}\\}\\!\\}_{P}\otimes
D(b)^{\prime}-\tau_{(132)}\\{\\!\\{D,\\{\\!\\{a,\\{P,b\\}\\}\\!\\}\\}\\!\\}_{L}\\\
=-D(a)^{\prime\prime}\otimes\\{\\!\\{b,D(a)^{\prime}\\}\\!\\}_{P}+\\{\\!\\{a,D(b)^{\prime\prime}\\}\\!\\}_{P}\otimes
D(b)^{\prime}-\\{\\!\\{a,\\{\\!\\{\\{P,b\\},D\\}\\!\\}\\}\\!\\}_{L}+\tau_{(123)}\\{\\!\\{\\{P,b\\},\\{\\!\\{D,a\\}\\!\\}\\}\\!\\}_{L}\\\
=-D(a)^{\prime\prime}\otimes\\{\\!\\{b,D(a)^{\prime}\\}\\!\\}_{P}+\\{\\!\\{a,D(b)^{\prime\prime}\\}\\!\\}_{P}\otimes
D(b)^{\prime}-\\{\\!\\{a,\\{\\!\\{\\{P,b\\},D\\}\\!\\}\\}\\!\\}_{L}+\tau_{(123)}\\{\\!\\{\\{b,P\\},\\{\\!\\{D,a\\}\\!\\}\\}\\!\\}_{L}\\\
=-D(a)^{\prime\prime}\otimes\\{\\!\\{b,D(a)^{\prime}\\}\\!\\}_{P}+\\{\\!\\{a,D(b)^{\prime\prime}\\}\\!\\}_{P}\otimes
D(b)^{\prime}-\\{\\!\\{a,\\{\\!\\{\\{P,b\\},D\\}\\!\\}\\}\\!\\}_{L}+\tau_{(123)}\\{\\!\\{b,\\{P,\\{\\!\\{D,a\\}\\!\\}\\}\\!\\}_{L}\\\
=-\\{\\!\\{a,D(b)^{\prime\prime}\\}\\!\\}_{P}\otimes
D(b)^{\prime}+\\{\\!\\{a,\\{\\!\\{\\{P,b\\},D\\}\\!\\}\\}\\!\\}_{L}\\\
=-\\{\\!\\{a,D(b)^{\prime\prime}\\}\\!\\}_{P}\otimes
D(b)^{\prime}+\\{\\!\\{a,\\{P,\\{\\!\\{b,D\\}\\!\\}\\}\\}\\!\\}_{L}-\\{\\!\\{a,\\{\\!\\{b,\\{P,D\\}\\}\\!\\}\\}\\!\\}_{L}\\\
=-\\{\\!\\{a,\\{\\!\\{b,\\{P,D\\}\\}\\!\\}\\}\\!\\}_{L}\end{array}$

where we used the proposition 2.10∎

######  Lemma 5.11.

If $\delta_{1},\dots,\delta_{n}$ are in $\mathbb{D}er(A)$ and
$a_{1},\dots,a_{n}\in A$, then

$\begin{array}[]{l}\\{\\!\\{a_{1},\\{\\!\\{a_{2},\dots,\\{\\!\\{a_{n},\delta_{1}\dots\delta_{n}\\}\\!\\}_{L}\\}\\!\\}_{L}\dots\\}\\!\\}_{L}=\\\
\delta_{1}(a_{1})^{\prime\prime}\otimes\delta_{1}(a_{1})^{\prime}\delta_{2}(a_{2})^{\prime\prime}\otimes\delta_{2}(a_{2})^{\prime}\delta_{3}(a_{3})^{\prime\prime}\otimes\dots\otimes\delta_{n-1}(a_{n-1})^{\prime}\delta_{n}(a_{n})^{\prime\prime}\otimes\delta_{n}(a_{n})^{\prime}\end{array}$

Proof by induction on $n$.

∎

######  Proposition 5.12.

Let $A$ be a smooth algebra. If $\mathbb{L}=\Omega^{1}_{A}$, the differential
$d_{\mathbb{L}}$ coincides with the double Poisson cohomology defined by
Pichereau and Van Weyer ([9]).

This proposition follows from the previous theorem.

The Lie derivative :

  * •

$\underline{L}_{D}(a)=D(a)$

  * •

If $\lambda\in\mathbb{L}^{*}$,

$\begin{array}[]{rcl}\underline{L}_{D}(\lambda)(\Delta)&=&\lambda(\Delta)^{\prime}\otimes
D\left(\lambda(\Delta)^{\prime\prime}\right)-\tau_{(23)}\left[\lambda\left(\\{\\!\\{D,\Delta\\}\\!\\}_{r}^{\prime\prime}\right)\otimes\\{\\!\\{D,\Delta\\}\\!\\}_{r}^{\prime}\right]\\\
&+&D\left(\lambda(\Delta)^{\prime}\right)\otimes\lambda(\Delta)^{\prime\prime}-\tau_{(32)}\left[\lambda\left(\\{\\!\\{D,\Delta\\}\\!\\}_{l}^{\prime}\right)\otimes\\{\\!\\{D,\Delta\\}\\!\\}_{l}^{\prime\prime}\right]\\\
\end{array}$

$\underline{L}_{D}(\lambda)\in
A\otimes\mathbb{L}^{*}\oplus\mathbb{L}^{*}\otimes A$ as the map

$\Delta\mapsto\lambda(\Delta)^{\prime}\otimes
D\left(\lambda(\Delta)^{\prime\prime}\right)-\tau_{(32)}\left[\lambda\left(\\{\\!\\{D,\Delta\\}\\!\\}_{r}^{\prime\prime}\right)\otimes\\{\\!\\{D,\Delta\\}\\}_{r}^{\prime}\right]$

belongs to $\mathbb{L}^{*}\otimes A$ and the map

$\Delta\mapsto
D\left(\lambda(\Delta)^{\prime}\right)\otimes\lambda(\Delta)^{\prime\prime}-\tau_{(32)}\left[\lambda(\\{\\!\\{D,\Delta\\}\\!\\}^{\prime}_{l})\otimes\\{\\!\\{D,\Delta\\}\\!\\}_{l}^{\prime\prime})\right]$

is in $A\otimes\mathbb{L}^{*}$

  * •

$\underline{L}_{D}$ is a degree preserving double derivation from
$T_{A}(\mathbb{L}^{*})$ to $T_{A}(\mathbb{L}^{*})\otimes
T_{A}(\mathbb{L}^{*})$

Standard formulas involving the differential $\underline{d}_{\mathbb{L}}$,
contraction and Lie derivative hold in the double Lie algebroid context. In
particular, one has the Cartan identity :

######  Proposition 5.13.

The map $\underline{d}_{\mathbb{L}}$ is a degree one derivation of
$T_{A}(\mathbb{L}^{*})$ and can be extended to a degree one derivation of
$T_{A}(\mathbb{L}^{*})\otimes T_{A}(\mathbb{L}^{*})$ as follows :

$\forall(\alpha,\beta)\in
T_{A}(\mathbb{L}^{*}),\quad\underline{d}_{\mathbb{L}}(\alpha\otimes\beta)=\underline{d}_{\mathbb{L}}(\alpha)\otimes\beta+(-1)^{|\alpha|}\alpha\otimes\underline{d}_{\mathbb{L}}(\beta).$

One has the following properties : For any $D$ and $\Delta$ in $\mathbb{L}$ :

1)
$\underline{d}_{\mathbb{L}}\circ\underline{i}_{D}+\underline{i}_{D}\circ\underline{d}_{\mathbb{L}}=\underline{L}_{D}$

2)
$\\{\\!\\{\underline{i}_{D},\underline{i}_{\Delta}\\}\\!\\}_{l}=\\{\\!\\{\underline{i}_{D},\underline{i}_{\Delta}\\}\\!\\}_{r}=0$.

######  Remark 5.14.

Consequences of these formulas will be seen later.

###### Proof.

2) It is enough to prove the relation
$\underline{d}_{\mathbb{L}}\circ\underline{i}_{D}+\underline{i}_{D}\circ\underline{d}_{\mathbb{L}}=\underline{L}_{D}$
on elements of $A$ and $\mathbb{L}^{*}$.

On elements of $A$, it is obvious. On elements $\lambda\in\mathbb{L}^{*}$, we
give the main steps of the computation :

$\begin{array}[]{rcl}(\underline{i}_{D}\circ\underline{d}_{\mathbb{L}}+\underline{d}_{\mathbb{L}}\circ\underline{i}_{D})(\lambda)(\Delta)&=&\underline{d}_{\mathbb{L}}\lambda(D,\Delta)-\underline{d}_{\mathbb{L}}\lambda(\Delta,D)+\underline{d}_{\mathbb{L}}\left(\lambda(D)\right)(\Delta)\\\
&=&\underline{d}_{\mathbb{L}}\lambda(D,\Delta)-\underline{d}_{\mathbb{L}}\lambda(\Delta,D)+\underline{d}_{\mathbb{L}}\left(\lambda(D)^{\prime}\right)\otimes\lambda(D)^{\prime\prime}+\lambda(D)^{\prime}\otimes\underline{d}_{\mathbb{L}}\left(\lambda(D)^{\prime\prime}\right)\\\
&=&\underline{L}_{D}(\lambda)(\Delta)\end{array}$

3) $\\{\\!\\{\underline{i}_{D},\underline{i}_{\Delta}\\}\\!\\}_{l}^{\prime}=0$
and
$\\{\\!\\{\underline{i}_{D},\underline{i}_{\Delta}\\}\\!\\}_{r}^{\prime\prime}=0$
are easy to check on elements of $A$ and $\mathbb{L}$. ∎

##  6\. From double to classical

In the section, we assume by simplicity that $B=k$.

Let $A$ be a $k$-algebra an $N\in\mathbb{N}$. Denote by $Rep_{N}(A)$ the
representation space $Hom_{alg}\left(A,M_{N}(k)\right)$ The coordinate ring of
$Rep_{N}(A)$ is

${\mathcal{O}}_{N}(A):=k[Rep_{N}(A)]=\dfrac{k\left[a_{p,q},(p,q)\in[1,N],a\in
A\right]}{<a_{p,q}b_{q,r}-(ab)_{p,r}>}$

For any element $x\in Rep_{N}(A)$ one has

$a_{p,q}(x)=x(a)_{p,q}$

######  Examples 6.1.

1) If $A=\dfrac{k[t]}{(t^{n})}$, then $Rep_{N}(A)=\\{Q\in M_{N}(k)\mid
Q^{n}=(0)\\}.$

2) If $A=k<x_{1},\dots,x_{q}>$, then $Rep_{N}(A)=M_{N}(k)\oplus\dots\oplus
M_{N}(k).$

######  Theorem 6.2.

([11]) If $(A,\\{\\!\\{-,-\\}\\!\\}$) is a double Poisson algebra, then
${\mathcal{O}}_{N}(A)$, endowed with the bracket determined by

$\\{a_{i,j},b_{u,v}\\}=\\{\\!\\{a,b\\}\\!\\}^{\prime}_{u,j}\\{\\!\\{a,b\\}\\!\\}^{\prime\prime}_{i,v}$

is a Poisson algebra.

If $a\in A$, one introduces $X(a)$ the $M_{n}(k)$ valued function on
$Rep_{N}(A)$ defined by $X(a)_{i,j}=a_{i,j}$. One has the relation
$X(ab)=X(a)X(b)$ and one defines
$Tr:\dfrac{A}{[A,A]}\to{\mathcal{O}}_{N}(A),\quad\overline{a}\to
Tr(X(a))={\displaystyle\sum_{i}}a_{i,i}$.

$GL_{N}(k)$ acts by conjugation on $Rep_{N}(A)$ and the following theorem is
shown in ([11])

######  Theorem 6.3.

The map

$\begin{array}[]{rcl}Tr:\dfrac{A}{[A,A]}&\to&{\mathcal{O}}_{N}(A)^{GL_{N}(k)}\\\
\overline{a}&\to&Tr(X(a))={\displaystyle\sum_{i}}a_{i,i}\end{array}$

is an isomorphism of Lie algebras.

If $M$ is an $A^{e}$-module, one defines ([12]) the
${\mathcal{O}}_{N}(A)$-module $(M)_{N}$

$(M)_{N}=\dfrac{k\left[m_{i,j},\;m\in
M\right]}{<a_{i,u}m_{u,j}-(am)_{i,j},a_{u,j}m_{i,u}-(ma)_{i,j}>}$

If $m\in M$, one introduces
$Tr(m)={\displaystyle\sum_{i=1}^{N}m_{i,i}\in(M)_{N}}$.

We will be interested in the case where $M=\mathbb{L}$ is a double Lie
Rinehart algebra.

If $\delta\in\mathbb{D}er(A,A\otimes A)$, one defines ([11]) the corresponding
derivation on ${\mathcal{O}}_{N}(A)$ by

$\delta_{i,j}(a_{u,v})=\delta(a)^{\prime}_{u,j}\delta(a)^{\prime\prime}_{i,v}.$

If $\delta=\delta_{1}\dots\delta_{n}$, one sets
$\delta_{i,j}=\delta_{1,ia_{1}}\delta_{2,a_{1}a_{2}}\dots\delta_{n,a_{n-1}j}\in\wedge
Der({\mathcal{O}}_{N}(A))$. In other words, this can be rewritten by the
relation $X(\delta)=X(\delta_{1})\dots X(\delta_{n})$.

######  Proposition 6.4.

([11]) If $P,Q\in DA$, then the following relation holds :

$\\{P_{i,j},Q_{u,v}\\}=\\{\\!\\{P,Q\\}\\!\\}^{\prime}_{u,j}\\{\\!\\{P,Q\\}\\!\\}_{i,v}^{\prime\prime}$

where $\\{\\!\\{-,-\\}\\!\\}$ denotes the Schouten bracket on $DA$ and
$\\{-,-\\}$ the Schouten bracket between poly-vector fields on $Rep_{N}(A)$.

######  Proposition 6.5.

The Trace map $Tr$ ([11])

$\begin{array}[]{rcl}Tr\;:\dfrac{DA}{[DA,DA]}&\to&\wedge
Der\left({\mathcal{O}}_{N}(A)\right)\\\ \delta&\mapsto&TrX(\delta)\end{array}$

is a Lie algebra homomorphism if both side are equipped with the Schouten
bracket.

######  Theorem 6.6.

([12])

There exists a unique Lie algebroid structure
$\left((\mathbb{L})_{N},{\mathcal{O}}_{N}(A),[,],\omega\right)$ with bracket
$[,]$ and anchor $\omega$ determined below by the equalities below :

$\begin{array}[]{l}\omega(X_{i,j})(a_{u,v})=\omega(X)(a)_{u,j}^{\prime}\omega(X)(a)_{i,v}^{\prime\prime}\\\
\left[X_{i,j},Y_{u,v}\right]=\\{\\!\\{X,Y\\}\\!\\}^{\prime}_{u,j}\\{\\!\\{X,Y\\}\\!\\}^{\prime\prime}_{i,v}\end{array}$

###### Proof.

The only think that is not obvious is the Jacobi identity and the fact that
$\omega$ is a Lie algebra morphism. We need to prove the identities

$\begin{array}[]{l}\left[X_{i,j},[Y_{u,v},Z_{k,m}]\right]+\left[Y_{u,v},[Z_{k,m},X_{i,j}]\right]+\left[Z_{k,m},[X_{i,j},Y_{u,v}]\right]=0\\\
\omega\left([X_{i,j},Y_{u,v}\right])=\left[\omega(X_{i,j}),\omega(Y_{u,v})\right]\end{array}$

These two identities follows from the double Jacobi identities by a
straightforward computation.∎

If $D\in\mathbb{L}$, we consider the matrix $X(D)=\left(D_{i,j}\right)$ as
being with values in $\wedge_{{\mathcal{O}}_{N}(A)}(\mathbb{L})_{N}$ and we
set

$\forall D_{1}\otimes\dots\otimes D_{n},\quad X(D_{1}\dots
D_{n})=X(D_{1})\dots
X(D_{n})\in\wedge_{{\mathcal{O}}_{N}(A)}(\mathbb{L})_{N}.$

One defines the trace map (as in [11]) by
$Tr:T_{A}(\mathbb{L})\to\wedge_{\mathcal{O}}(\mathbb{L})_{N}$ by
$Tr(D_{1}\otimes\dots\otimes D_{n})=TrX(D)$.

######  Proposition 6.7.

If $D$ and $\Delta$ are in $\mathbb{L}$, one has the following equality
$[Tr(D),Tr(\Delta)]=Tr\left(\\{\\!\\{D,\Delta\\}\\!\\}\right)$ where the left
hand side involves the Schouten bracket on
$\wedge_{{\mathcal{O}}_{N}(A)}(\mathbb{L})_{N}$ and the right hand side
involves the double Schouten bracket.

###### Proof.

it is a straightforward computation. ∎

Let $M$ be an $A^{e}$-module. To the $A^{e}$-module

$M^{*}=\\{\lambda:M\to A\otimes A\mid\lambda(amb)=\lambda(m)^{\prime}b\otimes
a\lambda(m)^{\prime\prime}\\},$

one associates

$(M^{*})_{N}=\dfrac{k\left[\lambda_{i,j},\;\lambda\in
M^{*}\right]}{<a_{i,u}\lambda_{u,j}-(a\lambda)_{i,j},a_{u,j}\lambda_{i,u}-(\lambda
a)_{i,j}>}$

If $\lambda\in M^{*}$ and $m\in M$, $\lambda_{i,j}$ defines an element of
$Hom_{\mathcal{O}_{N}(A)}\left((M)_{N},\mathcal{O}_{N}(A)\right)$ by

$\lambda_{i,j}(m_{u,v})=\lambda(m)^{\prime}_{i,v}\lambda(m)^{\prime\prime}_{u,j}$

For us, $M$ will be a double Lie Rinehart algebra $\mathbb{L}$ or its dual
${\mathbb{L}^{*}}$.

######  Remark 6.8.

If $M=\mathbb{D}er(A)$, one recovers the equality
$(\underline{d}a)_{i,j}=da_{i,j}$ ([11]).

We go on mimicking VdB’s construction :

If $\Lambda=\lambda_{1}\otimes\lambda_{2}\otimes\dots\otimes\lambda_{n}\in
T^{n}(\mathbb{M}^{*})$, we define

$\Lambda_{i,j}=\lambda_{1,ii_{1}}\lambda_{2,i_{1}i_{2}}\dots\lambda_{n,i_{n-1}j}\in\wedge\left((\mathbb{M})_{N}^{*}\right)$

and $X(\Lambda)=(\Lambda_{i,j})_{i,j}$. The latter is a matrix with values in
$\wedge\left(\mathbb{M})_{N}^{*}\right)$.

######  Lemma 6.9.

Let us identify
$\Lambda=\lambda_{1}\otimes\lambda_{2}\otimes\dots\otimes\lambda_{n}$ to
$\Phi\in Hom_{A^{e}}(\mathbb{L}^{\otimes n},A^{\otimes n+1})$. If we write
$\Phi(D_{1}\otimes\dots\otimes D_{n})$

$\Phi(D_{1}\otimes\dots\otimes D_{n})=\Phi^{(1)}(D_{1}\otimes\dots\otimes
D_{n})\otimes\dots\otimes\Phi^{(n+1)}(D_{1}\otimes\dots\otimes D_{n}),$

then

$\begin{array}[]{rcl}\Phi_{i,j}\left(D_{1,u_{1}v_{1}}\otimes\dots\otimes
D_{n,u_{n}v_{n}}\right)&=&\displaystyle\sum_{\sigma\in
S_{n}}(-1)^{n}\Phi^{(1)}(D_{\sigma(1)}\otimes\dots\otimes
D_{\sigma(n)})_{iv_{1}}\Phi^{(2)}(D_{\sigma(1)}\otimes\dots\otimes
D_{\sigma(n)})_{u_{1}v_{2}}\dots\\\
&&\Phi^{(n-1)}(D_{\sigma(1)}\otimes\dots\otimes
D_{\sigma(n)})_{u_{n-1}v_{n}}\Phi^{(n)}(D_{\sigma(1)}\otimes\dots\otimes
D_{\sigma(n)})_{u_{n}j}\end{array}$

###### Proof.

$\begin{array}[]{l}\Phi^{(1)}(D_{\sigma(1)}\dots
D_{\sigma(n)})_{iv_{1}}\Phi^{(2)}(D_{\sigma(1)}\dots
D_{\sigma(n)})_{u_{1}v_{2}}\Phi^{(n-1)}(D_{\sigma(1)}\dots
D_{\sigma(n)})_{u_{n-1}v_{n}}\Phi^{(n)}(D_{\sigma(1)}\dots
D_{\sigma(n)})_{u_{n}j}\\\
=\lambda_{1}\left(D_{\sigma(1)}\right)^{\prime}_{iv_{1}}\left[\lambda_{1}(D_{\sigma(1)})^{\prime\prime}\lambda_{2}(D_{\sigma(2)})^{\prime}\right]_{u_{1}v_{2}}\dots\left[\lambda_{n-1}(D_{\sigma(n-1)})^{\prime\prime}\lambda_{n}(D_{\sigma(n)})^{\prime}\right]_{u_{n},v_{n}}\lambda(D_{\sigma(n)})^{\prime\prime}_{u_{n},j}\\\
=\lambda_{1,ia_{1}}(D_{\sigma(1),u_{1}v_{1}})\lambda_{2,a_{1}a_{2}}(D_{\sigma(2),u_{2}v_{2}})\dots\lambda_{n,a_{n-1}j}(D_{\sigma(n),u_{n}v_{n}})\\\
=(\lambda_{1}\otimes\lambda_{2}\otimes\dots\otimes\lambda_{n})_{i,j}\left(D_{\sigma(1)u_{1}v_{1}}\otimes\dots\otimes
D_{\sigma(n),u_{n}v_{n}}\right)\end{array}$

∎

######  Lemma 6.10.

(i) If $a\in A$, then $d_{\mathbb{L}}(a)_{i,j}=d_{(\mathbb{L})_{N}}(a_{i,j})$.

(ii) If $\lambda\in\mathbb{L}^{*}$, then
$\underline{d}_{\mathbb{L}}(\lambda)_{i,j}=d_{(\mathbb{L})_{N}}(\lambda_{i,j})$

###### Proof.

Let us prove (i). On one hand, one has :

$\underline{d}_{\mathbb{L}}(a)_{i,j}(D_{u,v})=\underline{d}_{\mathbb{L}}(a)(D)^{\prime}_{i,v}\underline{d}_{\mathbb{L}}(a)(D)_{u,j}^{\prime\prime}=D(a)^{\prime}_{i,v}D(a)^{\prime\prime}_{u,j}$

On the other hand :

$\begin{array}[]{rcl}d_{(\mathbb{L})_{N}}(a_{i,j})(D_{u,v})=\omega(D_{u,v})(a_{i,j})=D(a)^{\prime}_{i,v}D(a)^{\prime\prime}_{u,j}.\end{array}$

Let us now prove (ii). In the following computation
$\left(D\leftrightarrow\Delta,u\leftrightarrow k,v\leftrightarrow p\right)$
means the same expression as before exchanging $D$ with $\Delta$, $u$ with
$k$, $v$ with $p$.

$\begin{array}[]{rcl}\underline{d}_{\mathbb{L}}(\lambda)_{i,j}(D_{k,p},\Delta_{u,v})&=&D\left(\lambda(\Delta)^{\prime}\right)^{\prime}_{i,p}D\left(\lambda(\Delta)^{\prime}\right)^{\prime\prime}_{k,v}\lambda(\Delta)^{\prime\prime}_{u,j}-\lambda(D)^{\prime}_{i,p}\Delta\left(\lambda(D)^{\prime\prime}\right)^{\prime}_{k,v}\Delta\left(\lambda(D)^{\prime\prime}\right)^{\prime\prime}_{u,j}\\\
&-&\lambda\left(\\{\\!\\{D,\Delta\\}\\!\\}^{\prime}_{l}\right)^{\prime}_{i,p}(\\{\\!\\{D,\Delta\\}\\!\\}^{\prime\prime}_{l})_{k,v}\lambda\left(\\{\\!\\{D,\Delta\\}\\!\\}^{\prime}_{l}\right)^{\prime\prime}_{u,j}\\\
&-&\left(D\leftrightarrow\Delta,u\leftrightarrow k,v\leftrightarrow
p\right)\\\
&=&D_{k,p}\left(\lambda(\Delta)^{\prime}_{i,v}\right)\lambda(\Delta)^{\prime\prime}_{u,j}-\lambda(D)^{\prime}_{i,p}\Delta_{u,v}\left(\lambda(D)^{\prime\prime}_{k,j}\right)^{\prime}\\\
&-&\lambda_{i,j}\left((\\{\\!\\{D,\Delta\\}\\!\\}^{\prime}_{l})_{u,p}\right)(\\{\\!\\{D,\Delta\\}\\!\\}^{\prime\prime}_{l})_{,k,v}\\\
&-&\left(D\leftrightarrow\Delta,u\leftrightarrow k,v\leftrightarrow
p\right)\\\
&=&D_{k,p}\left(\lambda_{i,j}(D_{u,v})\right)-\Delta_{u,v}\left(\lambda_{i,j}(D_{k,p})\right)\\\
&-&\lambda_{i,j}\left((\\{\\!\\{D,\Delta\\}\\!\\}^{\prime}_{l})_{u,p}\right)(\\{\\!\\{D,\Delta\\}\\!\\}^{\prime\prime}_{l})_{k,v}-\lambda_{i,j}\left((\\{\\!\\{D,\Delta\\}\\!\\}^{\prime\prime}_{r})_{k,v}\right)(\\{\\!\\{D,\Delta\\}\\!\\}^{\prime}_{r})_{u,p}\\\
&=&D_{k,p}\left(\lambda_{i,j}(D_{u,v})\right)-\Delta_{u,v}\left(\lambda_{i,j}(D_{k,p})\right)-\lambda_{i,j}\left([D_{k,p},\Delta_{u,v}]\right)\\\
&=&d_{(\mathbb{L)}_{N}}(\lambda_{i,j})(D_{k,p},\Delta_{u,v})\par\par\end{array}$

∎

######  Proposition 6.11.

Let $Tr$ be the Trace map : if $\Phi\in T_{A}(\mathbb{L}^{*})$,
$Tr(\Phi)=Tr\left[X(\Phi)\right]$

a) If $\lambda\in\mathbb{L}^{*}$ and $D\in\mathbb{L}$, then
$Tr\left(\lambda(D)\right)=Tr(\lambda)\left(Tr(D)\right)$.

b) If $\Phi\in T_{A}(\mathbb{L}^{*})$, one has
$Trd_{\mathbb{L}}(\Phi)=d_{(\mathbb{L})_{N}}\left((Tr\Phi)\right)$

###### Proof.

a) is an easy computation :

$Tr(\lambda)\left(TrD)\right)={\displaystyle\sum_{i,j}}\lambda_{i,i}(D_{j,j})=\lambda(D)^{\prime}_{i,j}\lambda(D)^{\prime\prime}_{j,i}=Tr\left(\lambda(D)\right).$

Let us now prove b).

Let us first prove by induction on the degree of $\Phi$ that
$\underline{d}_{\mathbb{L}}(\Phi)_{i,j}=d_{(\mathbb{L})_{N}}(\Phi_{i,j})$.

For $deg(\Phi)=1$, we have already proved it in a previous lemma.

Assume that it is proved for $\deg(\Phi)=n$ and let us prove it for
$\Phi\lambda$ if $\lambda\in\mathbb{L}^{*}$.

$\begin{array}[]{rcl}\underline{d}_{\mathbb{L}}(\Phi\lambda)_{i,j}&=&\left(\underline{d}_{\mathbb{L}}(\Phi)\lambda+\Phi\underline{d}(\lambda)\right)_{i,j}\\\
&=&\underline{d}_{\mathbb{L}}(\Phi)_{i,i_{1}}\lambda_{i_{1},j}+\Phi_{i,i_{1}}\underline{d}_{\mathbb{L}}(\lambda)_{i_{1},j}\\\
&=&d_{(\mathbb{L})_{N}}(\Phi_{i,i_{1}})\lambda_{i_{1},j}+\Phi_{i,i_{1}}d_{(\mathbb{L})_{N}}(\lambda_{i_{1},j})\\\
&=&d_{(\mathbb{L})_{N}}\left((\Phi\lambda)_{i,j}\right)\end{array}$

Then
$Tr\left[d_{\mathbb{L}}(\Phi)\right]={\displaystyle\sum_{i}}d_{\mathbb{L}}(\Phi)_{i,i}={\displaystyle\sum_{i}}d_{(\mathbb{L})_{N}}(\Phi_{i,i})=d_{(\mathbb{L})_{N}}(Tr(\Phi))$.
∎

######  Theorem 6.12.

The Tr induces a map from
$\dfrac{T_{A}(\mathbb{L}^{*})}{[T_{A}(\mathbb{L}^{*}),T_{A}(\mathbb{L}^{*})]}$
to $\bigwedge_{{\mathcal{O}}_{N}(A)}(\mathbb{L}^{*})_{N}$ that sends
$d_{\mathbb{L}}$ to $d_{(\mathbb{L})_{N}}$

##  7\. Reduced contraction and Lie derivative

######  Definition 7.1.

Let $\mathbb{L}$ be a double Lie algebroid. If $\Theta$ is in $\mathbb{L}$,
one defines the reduced Lie derivative and the reduced contraction by :

$\begin{array}[]{l}\iota_{\Theta}:T_{A}^{n}({\mathbb{L}^{*}})\to
T_{A}^{n-1}(\mathbb{L}^{*}),\quad\alpha\mapsto\left(\underline{i}_{\Theta}\alpha\right)^{\circ}\\\
{\mathcal{L}}_{\Theta}:T_{A}^{n}(\mathbb{L}^{*})\to
T_{A}^{n}(\mathbb{L}^{*}),\quad\alpha\mapsto\left(\underline{L}_{\Theta}\alpha\right)^{\circ}\\\
\end{array}$

Explicitely, if $\alpha_{1},\alpha_{2},\dots,\alpha_{n}$ are in
$\mathbb{L}^{*}$, one has

$\begin{array}[]{l}\iota_{\Theta}(\alpha_{1}\alpha_{2}\dots\alpha_{n})={\displaystyle\sum_{k=1}^{n}(-1)^{k(n-k+1)}}\alpha_{k}(\Theta)^{\prime\prime}\cdot\alpha_{k+1}\dots\alpha_{n}\alpha_{1}\dots\alpha_{k-1}\cdot\alpha_{k}(\Theta)^{\prime}\\\
{\mathcal{L}}_{\Theta}(\alpha_{1}\alpha_{2}\dots\alpha_{n})={\displaystyle\sum_{k=1}^{n}(-1)^{k(n-k+1)}}{\mathcal{L}}_{\Theta}(\alpha_{k})^{\prime\prime}\cdot\alpha_{k+1}\dots\alpha_{n}\alpha_{1}\dots\alpha_{k-1}\cdot{\mathcal{L}}_{\Theta}(\alpha_{k})^{\prime}\\\
\end{array}$

######  Proposition 7.2.

1) For any $\Theta\in\mathbb{L}$, we have the following equalities of
endomorphisms of $T_{A}(\mathbb{L}^{*})$:

$\underline{d}_{\mathbb{L}}\circ\iota_{\Theta}+\iota_{\Theta}\circ\underline{d}_{\mathbb{L}}={\mathcal{L}}_{\Theta},\quad\underline{d}_{\mathbb{L}}\circ{\mathcal{L}}_{\Theta}={\mathcal{L}}_{\Theta}\circ\underline{d}_{\mathbb{L}}$

2) The maps $\underline{d}_{\mathbb{L}}$, ${\mathcal{L}}_{\Theta}$ and
$\iota_{\Theta}$ descend to maps from $DR^{\bullet}(\mathbb{L})$ to
$DR^{\bullet}(\mathbb{L})$ denoted respectively $d_{\mathbb{L}}$,
${\mathcal{L}}_{\Theta}$ and $\iota_{\Theta}$. One has

$d_{\mathbb{L}}\circ\iota_{\Theta}+\iota_{\Theta}\circ
d_{\mathbb{L}}={\mathcal{L}}_{\Theta},\quad
d_{\mathbb{L}}\circ{\mathcal{L}}_{\Theta}={\mathcal{L}}_{\Theta}\circ
d_{\mathbb{L}}$

3) For any $\delta$ and $\Delta$ in $\mathbb{L}$, one has
$\underline{i}_{D}\iota_{\Delta}+\sigma_{(12)}\underline{i}_{\Delta}\iota_{\delta}=0$
(as maps from $T_{A}(\mathbb{L}^{*})$ to $T_{A}(\mathbb{L}^{*})\otimes
T_{A}(\mathbb{L}^{*})$.

######  Remark 7.3.

The previous proposition is proved for $\mathbb{L}=\mathbb{D}er(A)$ in [4] but
our proof is different.

###### Proof.

1) follows by applying ${}^{\circ}()$ to the relation

$\underline{L}_{\Theta}=\underline{d}_{\mathbb{L}}\circ\underline{i}_{\Theta}+\underline{i}_{\Theta}\circ\underline{d}_{\mathbb{L}}$

and using proposition 5.2.

2) followed from proposition 5.2 and from proposition 5.13.

∎

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