# Cohomology and Deformations of left-symmetric Rinehart Algebras

A. Ben Hassine  Department of Mathematics, Faculty of Science and Arts at
Belqarn, University of Bisha, Kingdom of Saudi Arabia & Faculty of Sciences,
University of Sfax, Tunisia<EMAIL_ADDRESS>,  T. Chtioui
Faculty of Sciences, University of Sfax, Tunisia<EMAIL_ADDRESS>,
M. Elhamdadi  Department of Mathematics, University of South Florida, Tampa,
FL 33620, U.S.A<EMAIL_ADDRESS>and  S. Mabrouk  Faculty of Sciences,
University of Gafsa, BP 2100, Gafsa, Tunisia<EMAIL_ADDRESS>
###### Abstract.

We introduce a notion of left-symmetric Rinehart algebras, which is a
generalization of a left-symmetric algebras. The left multiplication gives
rise to a representation of the corresponding sub-adjacent Lie-Rinehart
algebra. We construct left-symmetric Rinehart algebra from
$\mathcal{O}$-operators on Lie-Rinehart algebra. We extensively investigate
representations of a left-symmetric Rinehart algebras. Moreover, we study
deformations of left-symmetric Rinehart algebras, which is controlled by the
second cohomology class in the deformation cohomology. We also give the
relationships between $\mathcal{O}$-operators and Nijenhuis operators on left-
symmetric Rinehart algebra.

Keywords: left-symmetric Rinehart algebras, representations, cohomology,
deformations, Nijenhuis operators.

MSC(2020): Primary 17E05, 53D17; Secondary 14B12, 06B15

###### Contents

  1. 1 Introduction
  2. 2 Preliminaries
  3. 3 Some basic properties of a left-symmetric Rinehart algebra
  4. 4 Representations of left-symmetric Rinehart algebras
  5. 5 Deformation of left-symmetric Rinehart algebra
    1. 5.1 Formal deformations
    2. 5.2 Obstructions to the extension theory of deformations
    3. 5.3 Trivial deformation
  6. 6 $\mathcal{O}$-operators and Nijenhuis operators
    1. 6.1 Relationships between $\mathcal{O}$-operators and Nijenhuis operators
    2. 6.2 Compatible $\mathcal{O}$-operators and Nijenhuis operators

##  1\. Introduction

Left-symmetric algebras are algebras for which the associator
$(x,y,z):=(x\cdot y)\cdot z-x\cdot(y\cdot z)$ satisfies the identity
$(x,y,z)=(y,x,z)$. These algebras appeared as early as 1896 in the work of
Cayley [7] as rooted tree algebras. In the 1960s, they also arose from the
study of several topics in geometry and algebra, such as convex homogenous
cones [31], affine manifolds and affine structures on Lie groups [19, 28] and
deformation of associative algebras [18]. In 2006, Burde [5] wrote an
interesting survey showing the importance of left-symmetric algebras in many
areas, such as vector fields, rooted tree algebras, vertex algebras, operad
theory, deformation complexes of algebras, convex homogeneous cones, affine
manifolds and left-invariant affine structures on Lie groups [5].

Left symmetric algebras are the underlying algebraic structures of non-abelian
phase spaces of Lie algebras [20, 1], leading to a bialgebra theory of left-
symmetric algebras [3]. They can also be seen as the algebraic structures
behind the classical Yang-Baxter equations. Precisely they provide a
construction of solutions of the classical Yang-Baxter equations in certain
semidirect product Lie algebra structures (that is, over the double spaces)
induced by left-symmetric algebras [21, 2].

The notion of Lie-Rinehart algebras was introduced by J. Herz in [13] and
further developed in [29, 30]. A a notion of (Poincaré) duality for this class
of algebras was introduced in [16, 17]. Lie-Rinehart structures have been the
subject of extensive studies, in relations to symplectic geometry, Poisson
structures, Lie groupoids and algebroids and other kind of quantizations (see
[13, 15, 22, 23, 24, 25, 26]). For further details and a history of the notion
of Lie-Rinehart algebra, we refer the reader to [15]. Lie-Rinehart algebras
have been investigated furthermore in [4, 8, 9].

A left-symmetric algebroid is a geometric generalization of a left-symmetric
algebra. See [23, 24, 25] for more details and applications. The notion of a
Nijenhuis operator on a left-symmetric algebroid was introduced in [24], which
could generate a trivial deformation. More details on deformations of left-
symmetric algebras can be found in [32].

In this paper, we introduce a notion of a left-symmetric Rinehart algebra,
which is a generalization of a left-symmetric algebra and an algebraic version
of left symmetric algebroid. The following diagram shows how left-symmetric
Rinehart algebras fit in relation to Lie algebras, left-symmetric algebras and
Lie-Rinehart algebras.

Lie algebra commutatorgeneralizationLie-Rinehart algebracommutatorLeft-
symmetric algebraRota-Baxter operatorgeneralizationLeft-symmetric Rinehart
Rota-Baxter operator

The paper is organized as follows. In Section 2, we recall some definitions
concerning left-symmetric algebras and Lie-Rinehart algebra. In Section 3, we
introduce the notion of left-symmetric Rinehart algebra and give some of its
properties. As in the case of a left-symmetric algebra, one can obtain the
sub-adjacent Lie-Rinehart from a left-symmetric Rinehart by using the
commutator. The left multiplication gives rise to a representation of the sub-
adjacent Lie-Rinehart algebra. We construct left-symmetric Rinehart algebra
using $\mathcal{O}$-operators. Section 4 is devoted to the study of
representations and cohomology of left-symmetric Rinehart algebras. In Section
5, we introduce the deformation cohomology associated to a left-symmetric
Rinehart algebra, which controls the deformations. In Section 6, we introduce
the notion of a Nijenhuis operator, which could generate a trivial
deformation. In addition, we investigate some connection between
$\mathcal{O}$-operator and Nijenhuis operator.

Throughout this paper all vector spaces are over a field $\mathbb{K}$ of
characteristic zero.

##  2\. Preliminaries

In this section, we briefly recall some basics of left-symmetric algebras and
Lie-Rinehart algebras [5].

######  Definition 2.1.

A left-symmetric algebra is a vector space $L$ endowed with a linear map
$\cdot:L\otimes L\longrightarrow L$ such that for any $x,y,z\in L$,

$(x,y,z)=(y,x,z),\;\;{\rm or}\;\;{\rm equivalently,}\;\;(x\cdot y)\cdot
z-x\cdot(y\cdot z)=(y\cdot x)\cdot z-y\cdot(x\cdot z),$

where the associator $(x,y,z):=(x\cdot y)\cdot z-x\cdot(y\cdot z)$.

Let $\mathrm{ad}^{L}$ (resp. $\mathrm{ad}^{R}$) be the left multiplication
operator (resp. right multiplication operator) on $L$ that is, i.e.
$\mathrm{ad}^{L}(x)y=x\cdot y$ (resp. $\mathrm{ad}^{R}(x)y=y\cdot x$), for any
$x,y\in L$. The following lemma is given in [5].

######  Lemma 2.2.

Let $(L,\cdot)$ be a left-symmetric algebra. The commutator $[x,y]=x\cdot
y-y\cdot x$ defines a Lie algebra $L$, which is called the sub-adjacent Lie
algebra of $L$. The algebra $L$ is also called a compatible left-symmetric
algebra on the Lie algebra $L$. Furthermore, the map
$ad^{L}:L\rightarrow\mathfrak{gl}(L)$ with $x\mapsto L_{x}$ gives a
representation of the Lie algebra $(L,[\cdot,\cdot])$.

######  Definition 2.3.

Let $(L,\cdot)$ be a left-symmetric algebra and $M$ a vector space. A
representation of $L$ on $M$ consists of a pair $(\rho,\mu)$, where
$\rho:L\longrightarrow\mathfrak{gl}(M)$ is a representation of the sub-
adjacent Lie algebra $L$ on $M$ and $\mu:L\longrightarrow\mathfrak{gl}(M)$ is
a linear map satisfying:

(2.1) $\displaystyle\rho(x)\circ\mu(y)-\mu(y)\circ\rho(x)=\mu(x\cdot
y)-\mu(y)\circ\mu(x),\quad\forall~{}x,y\in L.$

The map $\rho$ is called a left representation and $\mu$ is a right
representation. Usually, we denote a representation by $(M;\rho,\mu)$. Then
$(L;\mathrm{ad}^{L},\mathrm{ad}^{R})$ is a representation of $(L,\cdot)$ which
is called adjoint representation .

The cohomology complex for a left-symmetric algebra $(L,\cdot)$ with a
representation $(M;\rho,\mu)$ is given as follows . The set of
$(n+1)$-cochains is given by

(2.2) $C^{n+1}(L,M)=\mathrm{Hom}(\wedge^{n}L\otimes L,M),\quad\forall n\geq
0.$

For all $\omega\in C^{n}(L,M)$, the coboundary operator
$\delta:C^{n}(L,M)\longrightarrow C^{n+1}(L,M)$ is given by

$\displaystyle\delta\omega(x_{1},x_{2},\dots,x_{n+1})$ $\displaystyle=$
$\displaystyle\sum_{i=1}^{n}(-1)^{i+1}\rho(x_{i})\omega(x_{1},\dots,\widehat{x_{i}},\dots,x_{n+1})$
$\displaystyle+\sum_{i=1}^{n}(-1)^{i+1}\mu(x_{n+1})\omega(x_{1},\dots,\widehat{x_{i}},\dots,x_{n},x_{i})$
$\displaystyle-\sum_{i=1}^{n}(-1)^{i+1}\omega(x_{1},\dots,\widehat{x_{i}},\dots,x_{n},x_{i}\cdot
x_{n+1})$ $\displaystyle+\sum_{1\leq i<j\leq
n}(-1)^{i+j}\omega([x_{i},x_{j}],x_{1},\dots,\widehat{x_{i}},\dots,\widehat{x_{j}},\dots,x_{n+1}).$

We then have the following lemma whose proof comes from a direct computation
using identity (2.1).

######  Lemma 2.4.

[6] The map $\delta$ satisfies $\delta^{2}=0$.

######  Definition 2.5.

A Lie-Rinehart algebra $L$ over an associative commutative algebra $A$ is a
Lie algebra over $\mathbb{K}$ with an $A$-module structure and a linear map
$\rho:L\rightarrow Der(A)$, such that the following conditions hold:

  1. (1)

$\rho$ is a representation of $(L,[\cdot,\cdot])$ on $A$.

  2. (2)

$\rho(ax)=a\rho(x)$ for all $a\in A$ and $x\in L$.

  3. (3)

The compatibility condition:

(2.3) $[x,ay]=\rho(x)ay+a[x,y],~{}~{}~{}\forall a\in A,x,y\in L.$

Let $(L,A,[\cdot,\cdot]_{L},\rho)$ and
$(L^{\prime},A^{\prime},[\cdot,\cdot]_{L^{\prime}},\rho^{\prime})$ be two Lie-
Rinehart algebras, then a Lie-Rinehart algebra homomorphism is defined as a
pair of maps $(g,f)$, where the maps $f:L\rightarrow L^{\prime}$ and
$g:A\rightarrow A^{\prime}$ are two $\mathbb{K}$-algebra homomorphisms such
that:

  * (1)

$f(ax)=g(a)f(x)$ for all $x\in L$ and $a\in A,$

  * (2)

$g(\rho(x)a)=\rho^{\prime}(f(x))g(a)$ for all $x\in L$ and $a\in A.$

Now, we recall the definition of module over a Lie-Rinehart algebra(for more
details see [10]).

######  Definition 2.6.

Let $M$ be an $A$-module. Then $M$ is a module over a Lie-Rinehart algebra
$(L,A,[\cdot,\cdot],\rho)$ if there exits a map $\theta:L\otimes M\rightarrow
M$ such that:

  1. (1)

$\theta$ is a representation of the Lie algebra $(L,[\cdot,\cdot])$ on $M$.

  2. (2)

$\theta(ax,m)=a\theta(x,m)$ for all $a\in A,x\in L,m\in M.$

  3. (3)

$\theta(x,am)=a\theta(x,m)+\rho(x)am$ for all $x\in L,a\in A,m\in M.$

We have the following lemma giving a characterization of when $\theta$ is a
representation.

######  Lemma 2.7.

$\theta$ is representation if and only if $L\oplus M$ is Lie-Rinehart algebra
over $A$.

##  3\. Some basic properties of a left-symmetric Rinehart algebra

In this section, we introduce a notion of left-symmetric Rinehart algebras
illustrated by some examples. As in the case of a left-symmetric algebra, we
obtain the sub-adjacent Lie-Rinehart from a left-symmetric Rinehart using the
commutator. In addition, we construct left-symmetric Rinehart algebra using
$\mathcal{O}$-operators.

######  Definition 3.1.

A left-symmetric Rinehart algebra is a quadruple $(L,A,\cdot,\ell)$ where
$(L,\cdot)$ is a left-symmetric algebra, $A$ is an associative commutative
algebra and $\ell:L\rightarrow Der(A)$ a linear map such that the following
conditions hold:

  1. (1)

$L$ is an $A$-module.

  2. (2)

$\ell$ is a left representation of $(L,\cdot)$ on $A$.

  3. (3)

$\ell(ax)=a\ell(x)$ for all $a\in A$ and $x\in L$.

  4. (4)

The compatibility conditions: for all $a\in A$ and $x,y\in L$

(3.1) $\displaystyle x\cdot(ay)=$ $\displaystyle\ell(x)ay+a(x\cdot y),$ (3.2)
$\displaystyle(ax)\cdot y=$ $\displaystyle a(x\cdot y).$

######  Remark 3.2.

If $\ell$ is surjective (respectively bijective) then the left-symmetric
Rinehart algebra $(L,A,\cdot,\ell)$ is called transitive (respectively
regular) left-symmetric Rinehart algebra.

######  Example 3.3.

It is clear that any left-symmetric algebra is a left-symmetric Rinehart
algebra.

######  Example 3.4.

A Novikov Poisson algebra is a left-symmetric Rinehart algebra see [34]).

######  Example 3.5.

Let $(L,A,\cdot,\ell)$ be a left-symmetric Rinehart algebra and let $L\oplus
A=\\{x+a~{}|~{}x\in L,a\in A\\}$ be the direct sum of $L$ and $A$. Then
$(L\oplus A,A,\cdot_{L\oplus A},\ell_{L\oplus A})$ is a left-symmetric
Rinehart algebra, where the $\cdot_{L\oplus A}$ is defined by, for all
$x_{1},x_{2}\in L$, $a_{1},a_{2}\in A$;

$\begin{split}(x_{1}+a_{1})\cdot_{L\oplus A}(x_{2}+a_{2})=&x_{1}\cdot
x_{2}+\ell(x_{1})(a_{2});\end{split}$

and $\ell_{L\oplus A}:L\oplus A\rightarrow Der(A)$ is defined by
$\ell_{L\oplus A}(a_{1}+x_{1})=\ell(x_{1})$. Indeed, it obvious to show that
$(L\oplus A,\cdot_{L\oplus A})$ is a left-symmetric algebra, $\ell_{L\oplus
A}$ is a representation of left-symmetric algebra $L\oplus A$ and
$\ell_{L\oplus A}\in Der(A)$.
By direct calculation, we have $\ell_{L\oplus
A}(b(x_{1}+a_{1}))=b\ell_{L\oplus A}(x_{1}+a_{1})$ for all $b,a_{1}\in A$ and
$x_{1}\in L$. On the other hand, let $x_{1},x_{2}\in L$ and $b,a_{1},a_{2}\in
A$ we have

$\displaystyle(x_{1}+a_{1})\cdot_{L\oplus A}b(x_{2}+a_{2})=$
$\displaystyle(a_{1}+x_{1})\cdot_{L\oplus A}(bx_{2}+ba_{2})$ $\displaystyle=$
$\displaystyle x_{1}\cdot(bx_{2})+\ell(x_{1})(ba_{2})$ $\displaystyle=$
$\displaystyle b(x_{1}\cdot
x_{2})+\ell(x_{1})b(x_{2})+\ell(x_{1})(b)a_{2}+b\ell(x_{1})(a_{2})$
$\displaystyle=$ $\displaystyle b\big{(}x_{1}\cdot
x_{2}+\ell(x_{1})(a_{2})\big{)}+\ell(x_{1})b(x_{2})+\ell(x_{1})(b)a_{2}$
$\displaystyle=$ $\displaystyle b\big{(}(x_{1}+a_{1})\cdot_{L\oplus
A}(x_{2}+a_{2})\big{)}+\ell_{L\oplus A}(x_{1}+a_{1})b(x_{2}+a_{2}).$

Moreover,

$\displaystyle b(x_{1}+a_{1})\cdot_{L\oplus A}(x_{2}+a_{2})=$
$\displaystyle(bx_{1}+ba_{1})\cdot_{L\oplus A}(x_{2}+a_{2})$ $\displaystyle=$
$\displaystyle(bx_{1})\cdot x_{2}+\ell(bx_{1})(a_{2})$ $\displaystyle=$
$\displaystyle b(x_{1}\cdot x_{2})+b\ell(x_{1})(a_{2})$ $\displaystyle=$
$\displaystyle b\big{(}x_{1}\cdot x_{2}+\ell(x_{1})a_{2}\big{)}$
$\displaystyle=$ $\displaystyle b\big{(}(x_{1}+a_{1})\cdot_{L\oplus
A}(x_{2}+a_{2})\big{)}.$

######  Definition 3.6.

A subalgebra $(S,A)$ of $(L,A,\cdot,\ell)$, $S$ for short, is a left-symmetric
subalgebra of $L$ such that $AS\subset S$ and $S$ acts on $A$ via the
composition $S\hookrightarrow L\rightarrow Der(A)$.
A subalgebra $(I,A)$, $I$ for short, of $L$ is called an ideal if $I$ is a
left-symmetric ideal of $L$ such that $\ell(I)AL\subset I$.

Now we have the following theorem

######  Theorem 3.7.

Let $(L,A,\cdot,\ell)$ be a left-symmetric Rinehart algebra. Then,
$(L,A,[\cdot,\cdot],\ell)$ is a Lie-Rinehart algebra, denoted by $L^{C}$,
called the sub-adjacent Lie-Rinehart algebra of $(L,A,\cdot,\ell)$.

Proof. Since $(L,\cdot)$ is a left-symmetric algebra, we have that
$(L,[\cdot,\cdot])$ is a Lie algebra. For any $a\in A$, by direct
computations, we have

$\displaystyle[x,ay]$ $\displaystyle=$ $\displaystyle x\cdot(ay)-(ay)\cdot
x=a(x\cdot y)+\ell(x)ay-a(y\cdot x)$ $\displaystyle=$ $\displaystyle
a[x,y]+\ell(x)(a)y,$

which implies that $(L,A,[\cdot,\cdot],\ell)$ is a Lie-Rinehart algebra.

To see that the linear map $\ell:L\longrightarrow Der(A)$ is a representation,
we only need to show that $\ell_{[x,y]}=[\ell_{x},\ell_{y}]_{Der(A)}$, which
follows directly from the fact that $(L,\cdot)$ is a left-symmetric algebra.
This ends the proof.

######  Definition 3.8.

Let $(L_{1},A_{1},\cdot_{1},\ell_{1})$ and $(L_{2},A_{2},\cdot_{2},\ell_{2})$
be two left-symmetric Rinehart algebras. A homomorphism of left-symmetric
Rinehart algebras is a pair of two algebra homomorphisms $(f,g)$ where
$f:L_{1}\longrightarrow L_{2}$ and $g:A_{1}\longrightarrow A_{2}$ such that:

$f(ax)=g(a)f(x),~{}~{}~{}\ \ \ \
~{}~{}~{}~{}~{}~{}g(\ell_{1}(x)a)=\ell_{2}(f(x))g(a),\quad\forall x,y\in
L_{1},a\in A_{1}.$

The following proposition is immediate.

######  Proposition 3.9.

Let $(f,g)$ be a homomorphism of left-symmetric Rinehart algebras from
$(L_{1},A_{1},\cdot_{1},\ell_{1})$ to $(L_{2},A_{2},\cdot_{2},\ell_{2})$. Then
$(f,g)$ is also a Lie-Rinehart algebra homomorphism of the corresponding sub-
adjacent Lie-Rinehart algebras.

Now we give the definition of an $\mathcal{O}$-operator.

######  Definition 3.10.

Let $(L,A,[\cdot,\cdot],\rho)$ be a Lie-Rinehart algebra and
$\theta:L\longrightarrow End(M)$ be a representation over $M$. A linear map
$T:M\longrightarrow L$ is called an $\mathcal{O}$-operator if for all $u,v\in
M$ and $a\in A$ we have

(3.3) $T(au)=aT(u),$ (3.4) $[T(u),T(v)]=T(\theta(T(u))(v)-\theta(T(v))(u)).$

######  Remark 3.11.

Consider the semidirect product Lie-Rinehart algebra
$(L\ltimes_{\theta}M,A,[\cdot,\cdot]_{L\ltimes_{\theta}M},\rho_{L\ltimes_{\theta}M})$,
where $\rho_{L\ltimes_{\theta}M}(x+u):=\rho(x)(u)$ and the bracket
$[\cdot,\cdot]_{L\ltimes_{\theta}M}$ is given by

$[x+u,y+v]_{L\ltimes_{\theta}M}=[x,y]+\theta(x)(v)-\theta(y)(u).$

Any $\mathcal{O}$-operator $T:M\longrightarrow L$ gives a Nijenhuis operator
$\tilde{T}=\left(\begin{array}[]{cc}0&T\\\ 0&0\end{array}\right)$ on the Lie-
Rinehart algebra $L\ltimes_{\theta}M$. More precisely, we have

$[\tilde{T}(x+u),\tilde{T}(y+v)]_{L\ltimes_{\theta}M}=\tilde{T}([\tilde{T}(x+u),y+v]_{L\ltimes_{\theta}M}+[x+u,\tilde{T}(y+v)]_{L\ltimes_{\theta}M}-\tilde{T}[x+u,y+v]_{L\ltimes_{\theta}M}).$

Fore more details on Nijenhuis operators and their applications the reader
should consult [11].

Let $T:M\longrightarrow L$ be an $\mathcal{O}$-operator. Define the
multiplication $\cdot_{T}$ on $M$ by

$u\cdot_{T}v=\theta(T(u))(v),\forall u,v\in M.$

We then have the following proposition

######  Proposition 3.12.

With the above notations, $(M,A,\cdot_{T},\ell_{T}=\ell\circ T)$ is a left-
symmetric Rinehart algebra, and the map $T$ is Lie-Rinehart algebra
homomorphism from $(M,[\cdot,\cdot])$ to $(L,[\cdot,\cdot])$.

Proof. It is easy to see that $(M,\cdot_{T})$ is a left-symmetric algebra. For
any $a\in A$, using Definition 3.1 and Eq.(3.3) we have

$\ell_{M}(au)=\ell(T(au))=a\ell(T(u))=a\ell_{M}(u),$

Similarly, using Definition 2.6 we obtain

$\displaystyle(au)\cdot_{T}v$ $\displaystyle=$
$\displaystyle\theta(T(au))(v)=\theta aT(u))(v)=a\theta(T(u))(v),$
$\displaystyle u\cdot_{T}(av)$ $\displaystyle=$
$\displaystyle\theta(T(u))(av)=a\theta(T(u))(v)+\ell\circ T(u)(a)v.$

Thus, $(M,A,\cdot_{T},\ell_{M})$ is a left-symmetric Rinehart algebra. Let
$[\cdot,\cdot]$ be the sub-adjacent Lie bracket on $M$. Then we have

$T[u,v]=T(u\cdot_{T}v-v\cdot_{T}u)=T(\theta(T(u))(v)-\theta(T(v))(u))=[T(u),T(v)].$

So $T$ is a homomorphism of Lie algebra.

##  4\. Representations of left-symmetric Rinehart algebras

In this section, we develop the notion of representations of a left-symmetric
Rinehart algebra and give a cohomology theory with the coefficient in a
representation.

######  Definition 4.1.

Let $(L,A,\cdot,\ell)$ be a left-symmetric Rinehart algebra and $M$ be an
$A$-module. A representation of $A$ on $M$ consists of a pair $(\rho,\mu)$,
where $\rho:L\otimes M\rightarrow M$ is a left representation of $(L,\cdot)$
on $M$ and $\mu:M\otimes L\rightarrow M$ is a linear map, such that for all
$x,y\in L$ and $\ m\in M$, we have

$\displaystyle\mu(ax)m=a\mu(x)m=\mu(x)(am)$ (4.1)
$\displaystyle\rho(x)\mu(y)-\mu(y)\rho(x)=\mu(x\cdot y)-\mu(y)\mu(x).$

We will denote this representation by $(M;\rho,\mu)$.

Let $(L,A,\cdot,\ell)$ be a left-symmetric Rinehart algebra. It is obvious
that if $(M,\rho)$ is a representation of the sub-adjacent Lie-Rinehart
algebra $(L,A,[\cdot,\cdot]^{C},\ell)$, then $(M;\rho,0)$ is a representation
of the left-symmetric Rinehart algebras $(L,A,\cdot,\ell)$.

For a left-symmetric Rinehart algebra $(L,A,\cdot,\ell)$ and a representation
$(M;\rho,\mu)$, the following proposition gives a construction of a left-
symmetric Rinehart algebra called semidirect product and denoted by
$L\ltimes_{\rho,\mu}M$.

######  Proposition 4.2.

Let $(L,A,\cdot,\ell)$ be a left-symmetric Rinehart algebra and $(M;\rho,\mu)$
a representation. Then, $(L\oplus M,A,\cdot_{L\oplus M},\ell_{L\oplus M})$ is
a left-symmetric Rinehart algebra, where $\cdot_{L\oplus M}$ and
$\ell_{L\oplus M}$ are given by

(4.2) $\displaystyle(x_{1}+m_{1})\cdot_{L\oplus M}(x_{2}+m_{2})$
$\displaystyle=$ $\displaystyle x_{1}\cdot
x_{2}+\rho(x_{1})m_{2}+\mu(x_{2})m_{1},$ (4.3) $\displaystyle\ell_{L\oplus
M}(x_{1}+m_{1})$ $\displaystyle=$ $\displaystyle\ell(x_{1}),$

for all $x_{1},x_{2}\in L$ and $\ m_{1},m_{2}\in M$.

Proof. Let $(M;\rho,\mu)$ be a representation. It is straightforward to see
that $(L\oplus M,A,\cdot_{L\oplus M})$ is a left-symmetric algebra. For any
$x_{1},x_{2}\in L$ and $m_{1},m_{2}\in M$, we have

$\displaystyle(x_{1}+m_{1})\cdot_{L\oplus M}(a(x_{2}+m_{2}))$ $\displaystyle=$
$\displaystyle x_{1}\cdot(ax_{2})+\rho(x_{1})am_{2}+\mu(ax_{2})m_{1}$
$\displaystyle=$ $\displaystyle a(x_{1}\cdot
x_{2})+\ell(x_{1})(ax_{2})+a\rho(x_{1})m_{2}+\ell(x_{1})(am_{2})+a\mu(x_{2})m_{1}$
$\displaystyle=$ $\displaystyle a((x_{1}+m_{1})\cdot_{L\oplus
M}(x_{2}+m_{2}))+\ell_{L\oplus M}(x_{1})(a)(x_{2}+m_{2}).$

On the other hand, we have

$\displaystyle(a(x_{1}+m_{1}))\cdot_{L\oplus M}(x_{2}+m_{2})$ $\displaystyle=$
$\displaystyle(ax_{1})\cdot x_{2}+\rho(ax_{1})m_{2}+\mu(x_{2})(am_{1})$
$\displaystyle=$ $\displaystyle a((x_{1}+m_{1})\cdot_{L\oplus
M}(x_{2}+m_{2})).$

Therefore, $(L\oplus M,A,\cdot_{L\oplus M},\ell_{L\oplus M})$ is a left-
symmetric Rinehart algebra.

Let $(L,A,\cdot,\ell)$ be a left-symmetric Rinehart algebra and $(M;\rho,\mu)$
be a representation.
Let $\rho^{*}:L\otimes M^{*}\rightarrow M^{*}$ and $\mu^{*}:M^{*}\otimes
L\longrightarrow M^{*}$ be defined by

$\displaystyle\langle\rho^{*}(x)\xi,m\rangle=-\langle\rho(x)m,\xi\rangle\ \
\textrm{and}\ \ \langle\mu^{*}(x)\xi,m\rangle=-\langle\mu(x)m,\xi\rangle.$

We then have the following proposition.

######  Proposition 4.3.

With the above notations, we obtain that

  * $\rm(i)$

$(M,\rho-\mu)$ is a representation of the sub-adjacent Lie-Rinehart algebra
$(L,A,[\cdot,\cdot],\ell)$.

  * $\rm(ii)$

$(M^{*},\rho^{*}-\mu^{*},-\mu^{*})$ is a representation be a left-symmetric
Rinehart algebra $(L,A,\cdot,\ell)$.

Proof. Since $(M;\rho,\mu)$ is a representation of the left-symmetric algebra
$(L,A,\cdot,\ell)$, using Proposition 4.2 we have that $(L\oplus
M,A,\cdot_{L\oplus M},\ell_{L\oplus M})$ is a left-symmetric Rinehart algebra.
Consider its sub-adjacent Lie-Rinehart algebra $(L\oplus M,A,\cdot_{L\oplus
M},[\cdot,\cdot]_{L\oplus M},\ell_{L\oplus M})$, we have

$\displaystyle[(x_{1}+m_{1}),(x_{2}+m_{2})]_{L\oplus M}$ $\displaystyle=$
$\displaystyle(x_{1}+m_{1})\cdot_{L\oplus
M}(x_{2}+m_{2})-(x_{2}+m_{2})\cdot_{L\oplus M}(x_{1}+m_{1})$ $\displaystyle=$
$\displaystyle x_{1}\cdot x_{2}+\rho(x_{1})m_{2}+\mu(x_{2})m_{1}-x_{2}\cdot
x_{1}-\rho(x_{2})m_{1}-\mu(x_{1})m_{2}$ $\displaystyle=$
$\displaystyle[x_{1},x_{2}]^{C}+(\rho-\mu)(x_{1})(m_{2})+(\rho-\mu)(x_{2})(m_{1}).$

From Lemma 2.7 we deduce that $(M,\rho-\mu)$ is a representation of Lie-
Rinehart algebra $L$ on $M$. This finishes the proof of (i).

For item (ii), it is clear that $\rho^{*}-\mu^{*}$ is just the dual
representation of the representation $\theta=\rho-\mu$ of the sub-adjacent
Lie-Rinehart algebra of $L$. We can directly check that
$-\mu^{*}(ax)\xi=-a\mu^{*}(x)\xi=-\mu^{*}(x)(a\xi)$. For any $x,y\in L$,
$\xi\in M^{*}$ and $m\in M$ we have

$\displaystyle-\langle(\rho^{*}-\mu^{*})(x)\mu^{*}(y)\xi,m\rangle+\langle\mu^{*}(y)(\rho^{*}-\mu^{*})(x)\xi,m\rangle$
$\displaystyle=$
$\displaystyle\langle\mu^{*}(y)\xi,(\rho-\mu)(x)m\rangle-\langle(\rho^{*}-\mu^{*})(x)\xi,\mu(y)m\rangle$
$\displaystyle=$
$\displaystyle-\langle\xi,\mu(y)(\rho-\mu)(x)m\rangle+\langle\xi,(\rho-\mu)(x)\mu(y)m\rangle$
$\displaystyle=$
$\displaystyle\langle\xi,(\rho-\mu)(x)\mu(y)m-\mu(y)(\rho-\mu)(x)m\rangle$
$\displaystyle=$ $\displaystyle\langle\xi,\mu(x\cdot y)m-\mu(x)\mu(y)m\rangle$
$\displaystyle=$ $\displaystyle\langle-\mu^{*}(x\cdot
y)\xi-\mu^{*}(y)\mu^{*}(x)\xi,m\rangle.$

Therefore $(M^{*},\rho^{*}-\mu^{*},-\mu^{*})$ is a representation of $L$.

The following corollary is obvious.

######  Corollary 4.4.

With the above notations, we have

  * $\rm(i)$

The left-symmetric Rinehart algebra $L\ltimes_{\rho,\mu}M$ and
$L\ltimes_{\rho-\mu,0}M$ have the same sub-adjacent Lie-Rinehart algebra
$L\ltimes_{\rho-\mu}M.$

  * $\rm(ii)$

The left-symmetric Rinehart algebras $L\ltimes_{\rho^{*},0}M^{*}$ and
$L\ltimes_{\rho^{*}-\mu^{*},-\mu^{*}}M^{*}$ have the same sub-adjacent Lie-
Rinehart algebra $L\ltimes_{\rho^{*}}M^{*}$.

Let $(M;\rho,\mu)$ be a representation of a left-symmetric Rinehart algebra
$(L,A,\cdot,\ell)$. In general, $(M^{*},\rho^{*},\mu^{*})$ is not a
representation. But we have the following proposition.

######  Proposition 4.5.

Let $(L,A,\cdot,\ell)$ be a left-symmetric Rinehart algebra and $(M;\rho,\mu)$
be a representation. Then the following conditions are equivalent:

  * $\rm(1)$

$(M;\rho-\mu,-\mu)$ is a representation of $(L,A,\cdot,\ell)$.

  * $\rm(2)$

$(M^{*},\rho^{*},\mu^{*})$ is a representation of $(L,A,\cdot,\ell)$.

  * $\rm(3)$

$\mu(x)\mu(y)=\mu(y)\mu(x)$ for all $x,y\in L.$

Proof. $(1)\Rightarrow(2)$ Using Proposition 4.3, since $(M;\rho,\mu)$ is
representation of $L$ then $(M^{*},\rho*,\mu^{*})$ is representation of $L$.
$(2)\Rightarrow(3)$ For any $x,y\in L$, $\xi\in M^{*}$ and $m\in M$ we have

$\displaystyle\langle\xi,\mu(x)\mu(y)m\rangle-\langle\xi,\mu(y)\mu(x)m\rangle$
$\displaystyle=$
$\displaystyle\langle\xi,\mu(x)\mu(y)m\rangle-\langle\xi,\mu(x\cdot
y)m\rangle+\langle\xi,\rho(x)\mu(y)m\rangle$
$\displaystyle-\langle\xi,\mu(y)\rho(x)m\rangle$ $\displaystyle=$
$\displaystyle\langle\mu^{*}(y)\mu^{*}(x)\xi,m\rangle+\langle\mu^{*}(x\cdot
y)\xi,m\rangle+\langle\mu^{*}(y)\rho^{*}(x)\xi,m\rangle$
$\displaystyle-\rho^{*}(x)\langle\mu^{*}(y)\xi,m\rangle$ $\displaystyle=$
$\displaystyle 0.$

Using Eq. (4.1) it is easy to show that $(3)\Rightarrow(1)$.

######  Example 4.6.

Let $(L,A,\cdot,\ell)$ be a transitive left-symmetric Rinehart algebra and let
$K=\ker(\ell)$. Then we have

  * $\rm(1)$

$(L,\mathrm{ad}^{L},0)$ is a representation and $(L^{*},ad^{L^{*}},0)$ is a
representation.

  * $\rm(2)$

$(K,\mathrm{ad}^{L}-\mathrm{ad}^{R},0)$ is a representation of $L$ on
$K=\ker(\ell)$ and $(K^{*},{\mathrm{ad}}^{L^{*}},0)$ is a representation of
$L$ on $K^{*}$.

  * $\rm(3)$

If $K$ is an ideal of $L$, then $(K,\mathrm{ad}^{L},\mathrm{ad}^{R})$ is a
representation of $L$ on $K$ and
$(K^{*},\mathrm{ad}^{L^{*}}-\mathrm{ad}^{R^{*}},-\mathrm{ad}^{R^{*}})$ is also
a representation on $K^{*}$ by Proposition 4.3.

Now we discuss a cohomology theory for a left-symmetric Rinehart
$(L,A,\cdot,\ell)$ with coefficients in a representation $(M;\rho,\mu)$.
Define the set of $(n+1)$-cochains by

$C^{n+1}(L,M)=\mathrm{Hom}(\Lambda^{n}L\otimes L,M)),\ n\geq 0,$

such that for any $\omega\in C^{n+1}(L,M)$ and $x_{i}\in
L,i=1,\dots,n+1,~{}~{}a\in A$

(4.4)
$\omega(x_{1},\dots,ax_{i},\dots,x_{n},x_{n+1})=a\omega(x_{1},\dots,x_{i},\dots,x_{n},x_{n+1}).$

Define the coboundary map $\delta:C^{n}(L,M)\longrightarrow C^{n+1}(L,M)$ by

$\displaystyle\delta\omega(x_{1},x_{2},\ldots,x_{n+1})$ $\displaystyle=$
$\displaystyle\sum_{i=1}^{n}(-1)^{i+1}\rho(x_{i})\omega(x_{1},x_{2},\ldots,\hat{x_{i}},\ldots,x_{n+1})$
$\displaystyle+\sum_{i=1}^{n}(-1)^{i+1}\mu(x_{n+1})\omega(x_{1},x_{2},\ldots,\hat{x_{i}},\ldots,x_{n},x_{i})$
$\displaystyle-\sum_{i=1}^{n}(-1)^{i+1}\omega(x_{1},x_{2},\ldots,\hat{x_{i}},\ldots,x_{n},x_{i}\ldots
x_{n+1})$ $\displaystyle+\sum_{1\leq i<j\leq
n}(-1)^{i+j}\omega([x_{i},x_{j}],x_{1},\ldots,\hat{x_{i}},\ldots,\hat{x_{j}},\ldots,x_{n+1}),$

for all $x_{i}\in L,i=1,2,\ldots,n+1$.

######  Proposition 4.7.

The map $\delta:C^{n+1}(L,M)\rightarrow C^{n+1}(L,M)$ satisfies
$\delta^{2}=0$. Thus giving a cochain complex $(C^{*+1}(L,M)=\bigoplus_{n\geq
0}C^{n+1}(L,M),\delta)$.

Proof.

For all $\ x_{1},\ldots,x_{n+1}\in L,\ a\in A$, we have

$\displaystyle\delta\omega(ax_{1},x_{2},\ldots,x_{n},x_{n+1})$
$\displaystyle=$
$\displaystyle\rho(ax_{1})\omega(x_{2},\ldots,x_{n},x_{n+1})-\sum_{i=2}^{n}(-1)^{i+1}\rho(x_{i})\omega(ax_{1},x_{2},\ldots,\hat{x_{i}},\ldots,x_{n+1})$
$\displaystyle+\mu(x_{n+1})\omega(x_{2},\ldots,x_{n},ax_{1})+\sum_{i=2}^{n}(-1)^{i+1}\mu(x_{n+1})\omega(ax_{1},x_{2},\ldots,\hat{x_{i}},\ldots,x_{n},x_{i})$
$\displaystyle-\omega(x_{2},\ldots,x_{n},(ax_{1})\ldots
x_{n+1})-\sum_{i=2}^{n}(-1)^{i+1}\omega(ax_{1},x_{2},\ldots,\hat{x_{i}},\ldots,x_{n},x_{i}\ldots
x_{n+1})$
$\displaystyle+\sum_{j=2}^{n}(-1)^{1+j}\omega([ax_{1},x_{j}],x_{2},\ldots,\hat{x_{j}},\ldots,x_{n+1})$
$\displaystyle+\sum_{1<i<j\leq
n}(-1)^{i+j}\omega([x_{i},x_{j}],ax_{1},\ldots,\hat{x_{i}},\ldots,\hat{x_{j}},\ldots,x_{n+1})$
$\displaystyle=$ $\displaystyle
a(\rho(x_{1})\omega(x_{2},\ldots,x_{n},x_{n+1}))-a\sum_{i=2}^{n}(-1)^{i+1}\rho(x_{i})\omega(x_{1},x_{2},\ldots,\hat{x_{i}},\ldots,x_{n+1})$
$\displaystyle+\sum_{i=2}^{n}(-1)^{i+1}\ell(x_{i})(a)\omega(x_{1},x_{2},\ldots,\hat{x_{i}},\ldots,x_{n+1})+a\mu(x_{n+1})\omega(x_{2},\ldots,x_{n},x_{1})$
$\displaystyle+a\sum_{i=2}^{n}(-1)^{i+1}\mu(x_{n+1})\omega(x_{1},x_{2},\ldots,\hat{x_{i}},\ldots,x_{n},x_{i})-a\omega(x_{2},\ldots,x_{n},x_{1}\ldots{x_{n+1}})$
$\displaystyle-\sum_{i=2}^{n}(-1)^{i+1}a\omega(x_{1},x_{2},\ldots,\hat{x_{i}},\ldots,x_{n},x_{i}\ldots
x_{n+1})$
$\displaystyle+a\sum_{j=2}^{n}(-1)^{j+1}\omega([x_{1},x_{j}],x_{2},\ldots,\hat{x_{j}},\ldots,x_{n+1})$
$\displaystyle-\sum_{j=2}^{n}(-1)^{j+1}\ell(x_{j})(a)\omega(x_{1},x_{2},\ldots,\hat{x_{j}},\ldots,x_{n+1})$
$\displaystyle+a\sum_{1\leq i<j\leq
n}(-1)^{i+j}\omega([x_{i},x_{j}],x_{1},\ldots,\hat{x_{i}},\ldots,\hat{x_{j}},\ldots,x_{n+1})$
$\displaystyle=$ $\displaystyle
a\delta\omega(x_{1},x_{2},\ldots,x_{n},x_{n+1}).$

Since $\delta\omega$ is skew-symmetric with respect to the first $n$-terms,
then $\delta\omega$ satsfies Eq.(4.4), that is $\delta\omega$ is $A$-linear
with respect to the first $n$ terms. Similarly, we can prove that
$\delta\omega$ is $A$-linear with respect to the last term. Furthermore, since
the coboundary operator $\delta$ is also a left-symmetric algebra coboundary
operator, we have $\delta^{2}=0$. This completes the proof.

For $m\in M$, define $\delta(m)(x)=\mu(x)m-\rho(x)m$. Then we have

$\delta^{2}(m)(x,y)=\rho(x)\rho(y)m-\rho(x\cdot y)m.$

From this fact, we define the set of $0$-cochains by

$C^{0}(L,M):=\\{m\in M\mid\rho(x)\rho(y)m-\rho(x\cdot y)m=0,\ \forall x,y\in
L.$

Then, we obtain a cochain complex $(C(L,M)=\bigoplus_{n\geq
0}C^{n}(L,M),\delta)$.

##  5\. Deformation of left-symmetric Rinehart algebra

We investigate in this section a deformation theory of left-symmetric Rinehart
algebra. But first let introduce some notation. For a left-symmetric Rinehart
algebra $(L,A,\cdot,\ell)$ we will denote the left-symmetric multiplication
$"\cdot"$ by $\mathfrak{m}$ in the sequel of the paper. Let $\mathbb{K}[[t]]$
be the formal power series ring in one variable $t$ and coefficients in
$\mathbb{K}$. Let $L[[t]]$ be the set of formal power series whose
coefficients are elements of $L$ (note that $L[[t]]$ is obtained by extending
the coefficients domain of $L$ from $\mathbb{K}$ to $\mathbb{K}[[t]]$). Then
$L[[t]]$ is $\mathbb{K}[[t]]$-module.

###  5.1. Formal deformations

######  Definition 5.1.

A deformation of a left-symmetric Rinehart algebra $(L,A,\cdot,\ell)$ is a
$\mathbb{K}[[t]]$-bilinear map

$\mathfrak{m}_{t}:L[[t]]\otimes L[[t]]\rightarrow L[[t]]$

which is given by $\mathfrak{m}_{t}(x,y)=\sum_{i\geq
0}t^{i}m_{i}(x,y)~{}~{}~{}~{}~{}\mbox{where}~{}~{}~{}~{}~{}~{}\mathfrak{m}_{0}=\mathfrak{m}$
and

$\mathfrak{m}_{i}(x,ay)=\mathfrak{m}_{i}(ax,y)=a~{}\mathfrak{m}_{i}(x,y),~{}~{}\forall
i\geq 1,$

such that $(L,A,\mathfrak{m}_{t},\ell)$ is left-symmetric Rinehart algebra.

Let $\mathfrak{m}_{t}$ be a deformation of $\mathfrak{m}$. Then for all $a\in
A,x,y,z\in L$

$\mathfrak{m}_{t}(\mathfrak{m}_{t}(x,y),z)-\mathfrak{m}_{t}(x,\mathfrak{m}_{t}(y,z))=\mathfrak{m}_{t}(\mathfrak{m}_{t}(y,x),z)-\mathfrak{m}_{t}(y,\mathfrak{m}_{t}(x,z)).$

Comparing the coefficients of $t^{n}$ for $n\geq 0$, we get the following
equations:

(5.1)
$\sum_{i+j=n}\mathfrak{m}_{i}(\mathfrak{m}_{j}(x,y),z)-\mathfrak{m}_{i}(x,\mathfrak{m}_{j}(y,z))-\mathfrak{m}_{i}(\mathfrak{m}_{j}(y,x),z)+\mathfrak{m}_{i}(y,\mathfrak{m}_{j}(x,z))=0.$

For $n=1$, Eq. (5.1) implies

$\displaystyle\mathfrak{m}_{1}(\mathfrak{m}(x,y),z)+\mathfrak{m}(\mathfrak{m}_{1}(x,y),z)-\mathfrak{m}_{1}(x,\mathfrak{m}(y,z))-\mathfrak{m}(x,\mathfrak{m}_{1}(y,z))$
$\displaystyle-$
$\displaystyle\mathfrak{m}_{1}(\mathfrak{m}(y,x),z)-\mathfrak{m}(\mathfrak{m}_{1}(y,x),z)+\mathfrak{m}_{1}(y,\mathfrak{m}(x,z))+\mathfrak{m}(y,\mathfrak{m}_{1}(x,z))=0.$

Or equivalently $\delta(\mathfrak{m}_{1})=0$.

The 2-cochain $\mathfrak{m}_{1}$ is called the infinitesimal of the
deformation $\mathfrak{m}_{t}$. More generally, if $m_{i}=0$ for $1\leq
i\leq(n-1)$ and $\mathfrak{m}_{n}$ is non zero cochain then $\mathfrak{m}_{n}$
is called the $n$-infinitesimal of the deformation $\mathfrak{m}_{t}$. By the
above discussion the following proposition follows immediately.

######  Proposition 5.2.

The infinitesimal of the deformation $\mathfrak{m}_{t}$ is a 2-cocycle in
$C^{2}(L,L)$. More generally, the $n$-infinitesimal is a 2-cocycle.

Now we give a notion of equivalence of two deformations

######  Definition 5.3.

Two deformations $\mathfrak{m}_{t}$ and $\tilde{\mathfrak{m}}_{t}$ are said to
be equivalent if there exists a formal automorphism

$\Phi_{t}:L[[t]]\rightarrow L[[t]]~{}~{}\mbox{ defined
as}~{}~{}\Phi_{t}=id_{L}+\sum_{i\geq 1}t^{i}\phi_{i}$

where for each $i\geq 1$, $\phi_{i}:L\rightarrow L$ is a $\mathbb{K}$-linear
map such that

$\tilde{\mathfrak{m}}_{t}(x,y)=\Phi_{t}^{-1}\mathfrak{m}_{t}(\Phi_{t}x,\Phi_{t}y).$

######  Definition 5.4.

Any deformation that is equivalent to the deformation
$\mathfrak{m}_{0}=\mathfrak{m}$ is said to be a trivial deformation.

######  Theorem 5.5.

The cohomology class of the infinitesimal of a deformation $\mathfrak{m}_{t}$
is determined by the equivalence class of $\mathfrak{m}_{t}$.

Proof. Let $\Phi_{t}$ be an equivalence of deformation between
$\mathfrak{m}_{t}$ and $\tilde{\mathfrak{m}}_{t}$. Then we get,

$\tilde{\mathfrak{m}}_{t}(x,y)=\Phi_{t}^{-1}\mathfrak{m}_{t}(\Phi_{t}x,\Phi_{t}y).$

Comparing the coefficients of $t$ from both sides of the above equation we
have

$\displaystyle\tilde{\mathfrak{m}}_{1}(x,y)+\Phi_{1}(\mathfrak{m}_{0}(x,y))=\mathfrak{m}_{1}(x,y)+\mathfrak{m}_{0}(\Phi_{1}(x),y)+\mathfrak{m}_{0}(x,\Phi_{1}(y))$

Or equivalently,

$\mathfrak{m}_{1}-\tilde{\mathfrak{m}}_{1}=\delta(\phi_{1}).$

So, cohomology class of infinitesimal of the deformation is determined by the
equivalence class of deformation of $\mathfrak{m}_{t}$.

######  Definition 5.6.

A left-symmetric Rinehart algebra is said to be rigid if and only if every
deformation of it is trivial.

######  Theorem 5.7.

A non-trivial deformation of a left-symmetric Rinehart algebra is equivalent
to a deformation whose n-infinitesimal is not a coboundary for some $n\geq 1$.

Proof. Let $\mathfrak{m}_{t}$ be a deformation of left-symmetric Rinehart
algebra with n-infinitesimal $m_{n}$ for some $n\geq 1$. Assume that there
exists a 2-cochain $\phi\in C^{1}(L,L)$ with $\delta(\phi)=m_{n}$. Then set

$\Phi_{t}=id_{L}+\phi t^{n}~{}~{}\mbox{and
define}~{}~{}\bar{\mathfrak{m}_{t}}=\Phi_{t}\circ\mathfrak{m}_{t}\circ\Phi_{t}^{-1}.$

Then by computing the expression and comparing coefficients of $t^{n}$, we get

$\bar{\mathfrak{m}_{n}}-\mathfrak{m}_{n}=-\delta(\phi).$

So, $\bar{m_{n}}=0$. We can repeat the argument to kill off any infinitesimal,
which is a coboundary.

######  Corollary 5.8.

If $H^{2}(L,L)=0$, then all deformations of $L$ are equivalent to a trivial
deformation.

###  5.2. Obstructions to the extension theory of deformations

Let $(L,A,\cdot,\ell)$ be a left-symmetric Rinehart algebra. Now we consider
the problem of extending a deformation of $\mathfrak{m}$ of order $n$ to a
deformation of $\mathfrak{m}$ of order $(n+1)$. Let $m_{t}$ be a deformation
of $\mathfrak{m}$ of order $n$. Then

${\mathfrak{m}}_{t}=\sum^{n}_{i=0}\mathfrak{m}_{i}t^{i}=\mathfrak{m}+\sum^{n}_{i=1}m_{i}t^{i}$

where $m_{i}\in C^{2}(L,L)$ for each $1\leq i\leq n$ such that

(5.2)
$\displaystyle\mathfrak{m}_{i}(x,ay)=\mathfrak{m}_{t}(ax,y)=a~{}\mathfrak{m}_{i}(x,y),~{}~{}\forall
i\geq 1,$ (5.3)
$\displaystyle\mathfrak{m}_{i}(\mathfrak{m}_{j}(x,y),z)-\mathfrak{m}_{i}(x,\mathfrak{m}_{j}(y,z))=\mathfrak{m}_{i}(\mathfrak{m}_{j}(y,x),z)-\mathfrak{m}_{i}(y,\mathfrak{m}_{j}(x,z)).$

If there exists a 2-cochain $m_{n+1}\in C^{2}(L,L)$ such that
$\tilde{\mathfrak{m}_{t}}=\mathfrak{m}_{t}+\mathfrak{m}_{n+1}t^{n+1}$ is a
deformation of $\mathfrak{m}$ of order $n+1$. Then we say that
$\mathfrak{m}_{t}$ extends to a deformation of order $(n+1)$. In this case
$\mathfrak{m}_{t}$ is called extendable.

######  Definition 5.9.

Let $\mathfrak{m}_{t}$ be a deformation of $\mathfrak{m}$ of order $n$.
Consider the cochain $Obs_{L}\in C^{3}(L,L)$, where

(5.4)
$Obs_{L}(x,y,z)=\sum\limits_{i+j=n+1;~{}i,j>0}\Big{(}\mathfrak{m}_{i}(\mathfrak{m}_{j}(x,y),z)-\mathfrak{m}_{i}(x,\mathfrak{m}_{j}(y,z))-\mathfrak{m}_{i}(\mathfrak{m}_{j}(y,x),z)+\mathfrak{m}_{i}(y,\mathfrak{m}_{j}(x,z)))\Big{)},$

for $x,y,z\in L$. The 3-cochain $Obs_{L}$ is called an obstruction cochain for
extending the deformation of $\mathfrak{m}$ of order $n$ to a deformation of
order $n+1$.

A straightforward computation gives the following

######  Proposition 5.10.

The obstructions are left-symmetric Rinehart algebra 3-cocycles.

######  Theorem 5.11.

Let $\mathfrak{m}_{t}$ be a deformation of $\mathfrak{m}$ of order $n$. Then
$\mathfrak{m}_{t}$ extends to a deformation of order $n+1$ if and only if the
cohomology class of $Obs_{L}$ vanishes.

Proof. Suppose that a deformation $\mathfrak{m}_{t}$ of order $n$ extends to a
deformation of order $n+1$. Then

$\sum_{i+j=n+1;~{}i,j\geq
0}\Big{(}\mathfrak{m}_{i}(\mathfrak{m}_{j}(x,y),z)-\mathfrak{m}_{i}(x,\mathfrak{m}_{j}(y,z))-\mathfrak{m}_{i}(\mathfrak{m}_{j}(y,x),z)+\mathfrak{m}_{i}(y,\mathfrak{m}_{j}(x,z)))\Big{)}=0.$

As a result, we get $Obs_{L}=\delta(m_{n+1})$. So, the cohomology class of
$Obs_{L}$ vanishes.

Conversely, let $Obs_{L}$ be a coboundary. Suppose that

$Obs_{L}=\delta(m_{n+1})$

for some 2-cochain $m_{n+1}$. Define a map
$\tilde{\mathfrak{m}_{t}}:L[[t]]\times L[[t]]\rightarrow L[[t]]$ as follows

$\tilde{\mathfrak{m}_{t}}=\mathfrak{m}_{t}+m_{n+1}t^{n+1}.$

Then for any $x,y,z\in L$, the map $\tilde{\mathfrak{m}_{t}}$ satisfies the
following identity

$\sum_{i+j=n+1;~{}i,j\geq
0}\Big{(}\mathfrak{m}_{i}(\mathfrak{m}_{j}(x,y),z)-\mathfrak{m}_{i}(x,\mathfrak{m}_{j}(y,z))-\mathfrak{m}_{i}(\mathfrak{m}_{j}(y,x),z)+\mathfrak{m}_{i}(y,\mathfrak{m}_{j}(x,z)))\Big{)}=0.$

This, in turn, implies that $\tilde{\mathfrak{m}_{t}}$ is a deformation of
$\mathfrak{m}$ extending $\mathfrak{m}_{t}$.

######  Corollary 5.12.

If $H^{3}(L,L)=0,$ then every 2-cocycle in $C^{2}(L,L)$ is an infinitesimal of
some deformation of $\mathfrak{m}$.

###  5.3. Trivial deformation

We study deformations of left-symmetric Rinehart algebra using the deformation
cohomology. Let $(L,A,\cdot,\ell)$ be a left-symmetric Rinehart algebra, and
$\mathfrak{m}\in C^{2}(L,L)$. Consider a $t$-parameterized family of
multiplications $\mathfrak{m}_{t}:L[[t]]\otimes L[[t]]\rightarrow L[[t]]$
given by

$\displaystyle\mathfrak{m}_{t}(x,y)$ $\displaystyle=$ $\displaystyle x\cdot
y+t\mathfrak{m}(x,y).$

If $L_{t}=(L,A,\mathfrak{m}_{t},\ell)$ is a left-symmetric Rinehart algebra
for all $t$, we say that $\mathfrak{m}$ generates a $1$-parameter
infinitesimal deformation of $(L,A,\cdot,\ell)$

Since $\mathfrak{m}$ is a $2$-cochain, we have

$\mathfrak{m}(ax,y)=a\mathfrak{m}(x,y)=\mathfrak{m}(x,ay),$

which implies that conditions (3.1) and (3.2) in Definition 3.1 are satisfied
for the multiplication $\mathfrak{m}_{t}$. Then we can deduce that
$(L,A,\mathfrak{m}_{t},\ell)$ is a deformation of $(L,A,\cdot,\ell)$ if and
only if

$\displaystyle
x\cdot\mathfrak{m}(y,z)-y\cdot\mathfrak{m}(x,z)+\mathfrak{m}(y,x)\cdot
z-\mathfrak{m}(x,y)\cdot z$ (5.5) $\displaystyle=\mathfrak{m}(y,x\cdot
z)-\mathfrak{m}(x,y\cdot z)-\mathfrak{m}([x,y],z),$

and

(5.6)
$\displaystyle\mathfrak{m}(\mathfrak{m}(x,y),z)-\mathfrak{m}(x,\mathfrak{m}(y,z))$
$\displaystyle=$
$\displaystyle\mathfrak{m}(\mathfrak{m}(y,x),z)-\mathfrak{m}(y,\mathfrak{m}(x,z)).$

Eq. $(\ref{2-closed})$ means that $\mathfrak{m}$ is a $2$-cocycle, and Eq.
$(\ref{omega bracket})$ means that $(L,A,\mathfrak{m},\ell)$ is a left-
symmetric Rinehart algebra.

Recall that a deformation is said to be trivial if there exists a family of
left-symmetric Rinehart algebra isomorphisms
$\mathrm{Id}+tN:L_{t}\longrightarrow L$.

By direct computations, $L_{t}$ is trivial if and only if

(5.7) $\displaystyle\mathfrak{m}(x,y)$ $\displaystyle=$ $\displaystyle x\cdot
N(y)+N(x)\cdot y-N(x\cdot y),$ (5.8) $\displaystyle N\mathfrak{m}(x,y)$
$\displaystyle=$ $\displaystyle N(x)\cdot N(y),$

It follows from $(\ref{Nij1})$ and $(\ref{Nij2})$ that $N$ must satisfy the
following condition:

(5.9) $\displaystyle N(x)\cdot N(y)-x\cdot N(y)-N(x)\cdot y+N^{2}(x\cdot
y)=0.$

Now we give the following definition.

######  Definition 5.13.

A linear map $N:L\longrightarrow L$ is called a Nijenhuis operator on a left-
symmetric Rinehart algebra $(L,A,\cdot,\ell)$ if

$N(ax)=aN(x),\forall x\in L,a\in A$

and the Nijenhuis condition $(\ref{integral condition of Nij})$ holds.

Obviously, any Nijenhuis operator on a left-symmetric Rinehart algebra is also
a Nijenhuis operator on the corresponding sub-adjacent Lie-Rinehart algebra.

We have seen that a trivial deformation of a left-symmetric Rinehart algebra
give rise to a Nijenhuis operator. In fact, the converse is also true as can
be seen from the following theorem.

######  Theorem 5.14.

Let $N$ be a Nijenhuis operator on a left-symmetric Rinehart algebra
$(L,A,\cdot,\ell)$. Then a deformation of $(L,A,\cdot,\ell)$ can be obtained
by putting

$\mathfrak{m}(x,y)=\delta N(x,y).$

Furthermore, this deformation is trivial.

Proof. Since $\mathfrak{m}$ is exact, $\mathfrak{m}$ is closed naturally, i.e.
Eq. (5.3) holds. To see that $\mathfrak{m}$ generates a deformation, we only
need to show that (5.6) holds, which follows from the Nijenhuis condition
(5.9). We omit the details. At last, it is obvious that

$(\mathrm{Id}+tN)(x\cdot_{t}y)=(\mathrm{Id}+tN)(x)\cdot(\mathrm{Id}+tN)(y),\quad\ell\circ(\mathrm{Id}+tN)=a_{t},$

which implies that the deformation is trivial.

######  Theorem 5.15.

Let $N$ be a Nijenhuis operator on a left-symmetric Rinehart algebra
$(L,A,\cdot,\ell)$. Then $(L,A,\cdot_{N},\ell_{N}=\ell\circ N)$ is a left-
symmetric Rinehart algebra, where

$x\cdot_{N}y=x\cdot N(y)+N(x)\cdot y-N(x\cdot y),\forall x,y\in L.$

Proof. It is obvious to show that $(L,\cdot_{N})$ is a left-symmetric algebra
and $\ell_{N}$ is a representation of L on $Der(A)$. Evidently, we have

$\ell_{N}(ax)=a\ell_{N}(x),\forall x\in L,a\in A..$

Furthermore, for any $x,y\in L$ and $a\in A$ we have

$\displaystyle x\cdot_{N}(ay)=$ $\displaystyle x\cdot
N(ay)+N(x)\cdot(ay)-N(x\cdot(ay))$ $\displaystyle=$ $\displaystyle a(x\cdot
N(y))+\ell(x)aN(y)+a(N(x)\cdot y)+\ell(N(x))ay-aN(x\cdot y)-N(\ell(x)ay)$
$\displaystyle=$ $\displaystyle a(x\cdot N(y)+N(x)\cdot y-N(x\cdot
y))+\ell_{N}(x)ay+N(\ell(x)ay)-N(\ell(x)ay).$ $\displaystyle=$ $\displaystyle
a(x\cdot_{N}y)+\ell_{N}(x)ay$

Moreover,

$\displaystyle(ax)\cdot_{N}y=$ $\displaystyle(ax)\cdot N(y)+N(ax)\cdot
y-N((ax)\cdot y)$ $\displaystyle=$ $\displaystyle a(x\cdot N(y))+a(N(x)\cdot
y)-aN(x\cdot y)$ $\displaystyle=$ $\displaystyle a(x\cdot N(y)+N(x)\cdot
y-N(x\cdot y))$ $\displaystyle=$ $\displaystyle a(x\cdot_{N}y).$

Then, $(L,A,\cdot_{N},\ell_{N}=\ell\circ N)$ is a a left-symmetric Rinehart
algebra.

Immediately, we have the following result.

######  Lemma 5.16.

Let $N$ be a Nijenhuis operator on a left-symmetric Rinehart algebra
$(L,A,\cdot,\ell)$. Then for arbitrary positive $j,k\in\mathbb{N}$, the
following equation holds

(5.10) $N^{j}(x)\cdot N^{k}(y)-N^{k}(N^{j}(x)\cdot y)-N^{j}(x\cdot
N^{k}(y))+N^{j+k}(x\cdot y)=0,\quad\forall~{}x,y\in L.$

If $N$ is invertible, this formula becomes valid for arbitrary
$j,k\in\mathbb{Z}$.

By direct calculations, we have the following corollary.

######  Corollary 5.17.

Let $(L,A,\cdot,\ell)$ left-symmetric Rinehart algebra and $N$ a Nijenhuis
operator.

  * $\rm(i)$

For all $k\in\mathbb{N}$, $(L,A,\cdot_{N^{k}},\ell_{N^{k}}=\ell\circ N^{k})$
is a left-symmetric Rinehart algebra.

  * $\rm(ii)$

For all $l\in\mathbb{N}$, $N^{l}$ is a Nijenhuis operator on the left-
symmetric Rinehart algebra $(L,A,\cdot_{N^{k}},\ell_{N^{k}})$.

  * $\rm(iii)$

The left-symmetric Rinehart algebras
$(L,A,(\cdot_{N^{k}})_{N^{l}},\ell_{N^{k+l}})$ and
$(L,A,\cdot_{N^{k+l}},\ell_{N^{k+l}})$ are the same.

  * $\rm(iv)$

$N^{l}$ is a left-symmetric Rinehart algebra homomorphism from
$(L,A,\cdot_{N^{k+l}},\ell_{N^{k+l}})$ to $(L,A,\cdot_{N^{k}},\ell_{N^{k}})$.

######  Theorem 5.18.

Let $N$ be a Nijenhuis operator on a left-symmetric Rinehart algebra
$(L,A,\cdot,\ell)$. Then the operator $P(N)=\sum_{i=0}^{n}c_{i}N^{i}$ is a
Nijenhuis operator. If $N$ is invertible, then
$Q(N)=\sum_{i=-m}^{n}c_{i}N^{i}$ is also a Nijenhuis operator.

Proof. According to Lemma 5.16, we obtain

$\displaystyle P(N)(x)\cdot P(N)(y)-P(N)(P(N)(x)\cdot y)-P(N)(x\cdot
P(N)(y))+P^{2}(N)(x\cdot y)$ $\displaystyle=$
$\displaystyle\sum_{i,j=0}^{n}c_{j}c_{k}\Big{(}N^{j}(x)\cdot
N^{k}(y)-N^{k}(N^{j}(x)\cdot y)-N^{j}(x\cdot N^{k}(y))+N^{j+k}(x\cdot
y)\Big{)}=0,\ \ \ \forall x,y\in L,$

which implies that $P(N)$ is a Nijenhuis operator. Similarly we can easy check
the second statement.

##  6\. $\mathcal{O}$-operators and Nijenhuis operators

In this section, we highlight the relationships between
$\mathcal{O}$-operators and Nijenhuis operators on left-symmetric Rinehart
algebra. Moreover, we illustrate some conections between Nijenhuis operators
and compatible $\mathcal{O}$-operators on left-symmetric Rinehart algebra.

###  6.1. Relationships between $\mathcal{O}$-operators and Nijenhuis
operators

We first give the definitions of an $\mathcal{O}$-operator and of Rota-Baxter
opeartor.

######  Definition 6.1.

An $\mathcal{O}$-operator on a left-symmetric Rinehart algebra
$(L,A,\cdot,\ell)$ associated to a representation $(M;\rho,\mu)$ is a linear
map $T:M\longrightarrow L$ satisfying

(6.1) $\displaystyle T(au)$ $\displaystyle=$ $\displaystyle aT(u),$ (6.2)
$\displaystyle T(u)\cdot T(v)$ $\displaystyle=$ $\displaystyle
T\Big{(}\rho(T(u))(v)+\mu(T(v))(u)\Big{)},\quad\forall u,v\in M,a\in A.$

######  Definition 6.2.

Let $(L,A,\cdot,\ell)$ be a left-symmetric Rinehart algebra and
$\mathcal{R}:L\longrightarrow L$ a linear operator. If $\mathcal{R}$ satisfies

(6.3) $\displaystyle\mathcal{R}(ax)$ $\displaystyle=$ $\displaystyle
a\mathcal{R}(x),$ (6.4) $\displaystyle\mathcal{R}(x)\cdot\mathcal{R}(y)$
$\displaystyle=$ $\displaystyle\mathcal{R}(\mathcal{R}(x)\cdot
y+x\cdot\mathcal{R}(y))+\lambda\mathcal{R}(x\cdot y),\quad\forall x,y\in
L,a\in A,$

then $\mathcal{R}$ is called a Rota-Baxter operator of weight $\lambda$ on
$L$.

Notice that a Rota-Baxter operator of weight zero on a left-symmetric Rinehart
algebra $L$ is exactly an $\mathcal{O}$-operator associated to the adjoint
representation $(L;ad^{L},ad^{R})$.

The following proposition gives connections between Nijenhuis operators and
Rota-Baxter operators.

######  Proposition 6.3.

Let $(L,A,\cdot,\ell)$ be a left-symmetric Rinehart algebra and
$N:L\longrightarrow L$ a linear operator.

  1. (i)

If $N^{2}={\rm Id}$, then $N$ is a Nijenhuis operator if and only if $N\pm{\rm
Id}$ is a Rota-Baxter operator of weight $\mp 2$ on $(L,A,\cdot,\ell)$.

  2. (ii)

If $N^{2}=0$, then $N$ is a Nijenhuis operator if and only if $N$ is a Rota-
Baxter operator of weight zero on $(L,A,\cdot,\ell)$.

  3. (iii)

If $N^{2}=N$, then $N$ is a Nijenhuis operator if and only if $N$ is a Rota-
Baxter operator of weight $-1$ on $(L,A,\cdot,\ell)$.

Proof. For Item (i), for all $x,y\in L$ then we have

$\displaystyle(N-Id)(x)\cdot(N-Id)(y)-(N-Id)\big{(}(N-Id)(x)\cdot
y+x\cdot(N-Id)(y)\big{)}+2(N-Id)(x\cdot y)$ $\displaystyle=$ $\displaystyle
N(x)\cdot N(y)-N\big{(}N(x)\cdot y+x\cdot N(y)\big{)}+x\cdot y$
$\displaystyle=$ $\displaystyle N(x)\cdot N(y)-N\big{(}N(x)\cdot y+x\cdot
N(y)\big{)}+N^{2}(x\cdot y).$

So $N$ is a Nijenhuis operator if and only if $N-Id$ is a Rota-Baxter operator
of weight $2$ on $L$. Similarly, we obtain that $N$ is a Nijenhuis operator if
and only if $N+Id$ is a Rota-Baxter operator of weight $-2$ on $L$.
Items (ii) and (iii) are obvious from the definitions of Nijenhuis operators
and Rota-Baxter operator.

######  Proposition 6.4.

Let $(L,A,\cdot,\ell)$ be a left-symmetric Rinehart algebra and $(M;\rho,\mu)$
be a representation on $L$. Let $T:M\rightarrow L$ be a linear map. For any
$\lambda$, $T$ is an $\mathcal{O}$-operator on $L$ associated to
$(M;\rho,\mu)$ if and only if the linear map
$\mathcal{R}_{T,\lambda}:=\left(\begin{array}[]{cc}0&T\\\ 0&-\lambda{\rm
Id}\end{array}\right)$ is a Rota-Baxter operator of weight $\lambda$ on the
semidirect product left-symmetric Rinehart algebra $({L\oplus
M},\cdot_{L\oplus M})$, where the multiplication $\cdot_{L\oplus M}$ is given
by (4.2).

Proof. It is easy to check the Eq.(6.3). Let $x_{1},x_{2}\in L$ and
$m_{1},m_{2}\in M$,

$\displaystyle\mathcal{R}_{T,\lambda}(x_{1}+m_{1})\cdot_{L\oplus
M}\mathcal{R}_{T,\lambda}(x_{2}+m_{2})$ $\displaystyle=(T(m_{1})-\lambda
m_{1})\cdot_{L\oplus M}(T(m_{2})-\lambda m_{2})$ (6.5)
$\displaystyle=T(m_{1})\cdot
T(m_{2})-\lambda\rho(T(m_{1})m_{2}-\lambda\mu(T(m_{2}))m_{1}.$

On the other hand,

$\displaystyle\mathcal{R}_{T,\lambda}\big{(}\mathcal{R}_{T,\lambda}(x_{1}+m_{1})\cdot_{L\oplus
M}(x_{2}+m_{2})+(x_{1}+m_{1})\cdot_{L\oplus
M}\mathcal{R}_{T,\lambda}(x_{2}+m_{2})\big{)}+$
$\displaystyle\lambda\mathcal{R}_{T,\lambda}((x_{1}+m_{1})\cdot_{L\oplus
M}(x_{2}+m_{2}))$ $\displaystyle=$
$\displaystyle\mathcal{R}_{T,\lambda}\big{(}(T(m_{1})-\lambda
m_{1})\cdot_{L\oplus M}(x_{2}+m_{2})+(x_{1}+m_{1})\cdot_{L\oplus
M}(T(m_{2})-\lambda m_{2})\big{)}+$
$\displaystyle\lambda\mathcal{R}_{T,\lambda}(x_{1}\cdot
x_{2}+\rho(x_{1})m_{2}+\mu(x_{2})m_{1})$ $\displaystyle=$
$\displaystyle\mathcal{R}_{T,\lambda}\big{(}T(m_{1})\cdot
x_{2}+\rho(T(m_{1}))m_{2}-\lambda\mu(x_{2})m_{1}+x_{1}\cdot
T(m_{2})-\lambda\rho(x_{1})m_{2}+\mu(T(m_{2}))m_{1}\big{)}$
$\displaystyle+\lambda\big{(}T(\rho(x_{1})m_{2})+T(\mu(x_{2})m_{1})-\lambda(\rho(x_{1})m_{2}+\mu(x_{2})m_{1})\big{)}$
$\displaystyle=$ $\displaystyle
T\big{(}\rho(T(m_{1}))m_{2}+\mu(T(m_{2}))m_{1}\big{)}-\lambda
T\big{(}\mu(x_{2})m_{1}+\rho(x_{1})m_{2}\big{)}-\lambda\Big{(}\rho(T(m_{1}))m_{2}+\mu(T(m_{2}))m_{1}\Big{)}$
$\displaystyle+\lambda^{2}\Big{(}\mu(x_{2})m_{1}+\rho(x_{1})m_{2}\big{)}+\lambda\big{(}T(\rho(x_{1})m_{2})+T(\mu(x_{2})m_{1})\big{)}-\lambda^{2}\big{(}\rho(x_{1})m_{2}+\mu(x_{2})m_{1})\big{)}$
(6.6) $\displaystyle=$ $\displaystyle
T\big{(}\rho(T(m_{1}))m_{2}+\mu(T(m_{2}))m_{1}\big{)}-\lambda\rho(T(m_{1}))m_{2}-\lambda\mu(T(m_{2}))m_{1}.$

According to Eqs. (6.5) and (6.6), $\mathcal{R}_{T,\lambda}$ is a Rota-Baxter
operator of weight $\lambda$ on the semidirect product left-symmetric Rinehart
algebra $({L\oplus M},\cdot_{L\oplus M})$ if and only if $T$ is an
$\mathcal{O}$-operator on $(L,A,\cdot,\ell)$ associated to $(M;\rho,\mu)$.

######  Proposition 6.5.

Let $(L,A,\cdot,\ell)$ be a left-symmetric Rinehart algebra and let
$(M;\rho,\mu)$ be a representation on $L$. Let $T:M\rightarrow L$ be a linear
map. Then the following statements are equivalent.

  1. (i)

$T$ is an $\mathcal{O}$-operator on the left-symmetric Rinehart algebra
$(L,A,\cdot,\ell)$.

  2. (ii)

$\mathcal{N}_{T}:=\left(\begin{array}[]{cc}0&T\\\ 0&{\rm
Id}\end{array}\right)$ is a Nijenhuis operator on left-symmetric Rinehart
algebra $({L\oplus M},\cdot_{L\oplus M})$.

  3. (iii)

$N_{T}:=\left(\begin{array}[]{cc}0&T\\\ 0&0\end{array}\right)$ is a Nijenhuis
operator on the left-symmetric Rinehart algebra $({L\oplus M},\cdot_{L\oplus
M})$.

Proof. Note that $\mathcal{N}_{T}=\mathcal{R}_{T,-1}$ and
$(\mathcal{N}_{T})^{2}=\mathcal{N}_{T}$ then $\mathcal{N}_{T}$ is a Nijenhuis
operator on the left-symmetric Rinehart algebra $({L\oplus M},\cdot_{L\oplus
M})$ using (iii) in Proposition 6.3.
Similarly $N_{T}=\mathcal{R}_{T,0}$ and $(N_{T})^{2}=0$, then $N_{T}$ is a
Nijenhuis operator on the left-symmetric Rinehart algebra $({L\oplus
M},\cdot_{L\oplus M})$ according (ii) in Proposition 6.3.

###  6.2. Compatible $\mathcal{O}$-operators and Nijenhuis operators

In this subsection we study compatibility of $\mathcal{O}$-operators and
Nijenhuis operators. First we start with the following definition.

######  Definition 6.6.

Let $(L,A,\cdot,\ell)$ be a left-symmetric Rinehart algebra and let
$(M;\rho,\mu)$ be a representation. Let $T_{1},T_{2}:M\longrightarrow L$ be
two $\mathcal{O}$-operators associated to $(M;\rho,\mu)$. Then $T_{1}$ and
$T_{2}$ are called compatible if $T_{1}+T_{2}$ is an $\mathcal{O}$-operator
associated to $(M;\rho,\mu)$.

Let $T_{1},T_{2}:M\longrightarrow L$ be two $\mathcal{O}$-operators on a left-
symmetric Rinehart algebra $(L,A,\cdot,\ell)$ associated to a representation
$(M;\rho,\mu)$ such that

(6.7) $\displaystyle T_{1}(u)\cdot T_{2}(v)+T_{2}(u)\cdot T_{1}(v)$
$\displaystyle=T_{1}\Big{(}\rho(T_{2}(u))(v)+\mu(T_{2}(v))(u)\Big{)}+T_{2}\Big{(}\rho(T_{1}(u))(v)+\mu(T_{1}(v))(u)\Big{)},$

for all $u,v\in M.$

######  Lemma 6.7.

Two operators $T_{1}$ and $T_{2}$ are compatible if and only if the equation
(6.7) holds.

Proof. For all $u,v\in M$, $a\in A$, we have

$\displaystyle(T_{1}+T_{2})(au)$ $\displaystyle=T_{1}(au)+T_{2}(au)$
$\displaystyle=aT_{1}(u)+aT_{2}(u)$ $\displaystyle=a(T_{1}+T_{2})(u).$

Furthermore,

$\displaystyle(T_{1}+T_{2})(u)\cdot(T_{1}+T_{2})(v)-(T_{1}+T_{2})\Big{(}\rho((T_{1}+T_{2})(u))(v)+\mu((T_{1}+T_{2})(v))(u)\Big{)},$
$\displaystyle=$ $\displaystyle T_{1}(u)\cdot T_{1}(v)+T_{1}(u)\cdot
T_{2}(v)+T_{2}(u)\cdot T_{1}(v)+T_{2}(u)\cdot T_{2}(v)$
$\displaystyle-(T_{1}+T_{2})\Big{(}\rho(T_{1}(u))(v)+\rho(T_{2}(u))(v)+\mu(T_{1}(v))(u)+\mu(T_{2}(v))(u)\Big{)}$
$\displaystyle=$ $\displaystyle T_{1}(u)\cdot T_{1}(v)+T_{1}(u)\cdot
T_{2}(v)+T_{2}(u)\cdot T_{1}(v)+T_{2}(u)\cdot T_{2}(v)$ $\displaystyle-
T_{1}\big{(}\rho(T_{1}(u))(v)+\mu(T_{1}(v))(u)\big{)}-T_{1}\big{(}\rho(T_{2}(u))(v)+\mu(T_{2}(v))(u)\big{)}$
$\displaystyle-
T_{2}\big{(}\rho(T_{1}(u))(v)+\mu(T_{1}(v))(u)\big{)}-T_{2}\big{(}\rho(T_{2}(u))(v)+\mu(T_{2}(v))(u)\big{)}$
$\displaystyle=$ $\displaystyle T_{1}(u)\cdot T_{2}(v)+T_{2}(u)\cdot
T_{1}(v)-T_{1}\Big{(}\rho(T_{2}(u))(v)+\mu(T_{2}(v))(u)\Big{)}-T_{2}\Big{(}\rho(T_{1}(u))(v)+\mu(T_{1}(v))(u)\Big{)}$

Then $T_{1}+T_{2}$ is an $\mathcal{O}$-operator associated to $(M;\rho,\mu)$
if and only if Eq.(6.7) holds.

######  Remark 6.8.

The equation (6.7) implies that for any $k_{1},k_{2}$ the linear combination
$k_{1}T_{1}+k_{2}T_{2}$ is an $\mathcal{O}$-operator.

There is a close relationship between a Nijenhuis operator and a pair of
compatible $\mathcal{O}$-operators as can be seen from the following
proposition.

######  Proposition 6.9.

Let $T_{1},T_{2}:M\longrightarrow L$ be two $\mathcal{O}$-operators on a left-
symmetric Rinehart algebra $(L,A,\cdot,\ell)$ associated to a representation
$(M;\rho,\mu)$. Suppose that $T_{2}$ is invertible. If $T_{1}$ and $T_{2}$ are
compatible, then $N=T_{1}\circ T_{2}^{-1}$ is a Nijenhuis operator on the
left-symmetric Rinehart algebra $(L,A,\cdot,\ell)$.

Proof. For all $x,y\in L$, since $T_{2}$ is invertible, there exist $u,v\in M$
such that $T_{2}(u)=x,T_{2}(v)=y$. Hence $N=T_{1}\circ T_{2}^{-1}$ is a
Nijenhuis operator if and only if the following equation holds:

(6.8) $NT_{2}(u)\cdot NT_{2}(v)=N(NT_{2}(u)\cdot T_{2}(v)+T_{2}(u)\cdot
NT_{2}(v))-N^{2}(T_{2}(u)\cdot T_{2}(v)).$

Since $T_{1}=N\circ T_{2}$ is an $\mathcal{O}$-operator, the left hand side of
the above equation is

$NT_{2}(\rho(NT_{2}(u))(v)+\mu(NT_{2}(v))(u)).$

Using the fact that $T_{2}$ and $T_{1}=N\circ T_{2}$ are two compatible
$\mathcal{O}$-operators, we get

$\displaystyle NT_{2}(u)\cdot T_{2}(v)+T_{2}(u)\cdot NT_{2}(v)$
$\displaystyle=$ $\displaystyle
T_{2}(\rho(NT_{2}(u))(v)+\mu(NT_{2}(v))(u))+NT_{2}(\rho(T_{2}(u))(v)+\mu(T_{2}(v))(u))$
$\displaystyle=$ $\displaystyle
T_{2}(\rho(NT_{2}(u))(v)+\mu(NT_{2}(v))(u))+N(T_{2}(u)\cdot T_{2}(v)).$

Hence Eq.(6.8) holds by acting $N$ on both sides of the last equality.

Using an $\mathcal{O}$-operator and a Nijenhuis operator, we can construct a
pair of compatible $\mathcal{O}$-operators.

######  Proposition 6.10.

Let $T:M\longrightarrow L$ be an $\mathcal{O}$-operator on a left-symmetric
Rinehart algebra $(L,A,\cdot,\ell)$ associated to a representation
$(M;\rho,\mu)$ and let $N$ be a Nijenhuis operator on $(L,A,\cdot,\ell)$. Then
$N\circ T$ is an $\mathcal{O}$-operator on the left-symmetric Rinehart algebra
$(L,A,\cdot,\ell)$ associated to $(M;\rho,\mu)$ if and only if for all $u,v\in
M$, the following equation holds:

$\displaystyle N\Big{(}NT(u)\cdot T(v)+T(u)\cdot NT(v)\Big{)}$ (6.9)
$\displaystyle=$ $\displaystyle
N\Big{(}T\big{(}\rho(NT(u))(v)+\mu(NT(v))(u)\big{)}+NT\big{(}\rho(T(u))(v)+\mu(T(v))(u)\big{)}\Big{)}.$

If in addition $N$ is invertible, then $T$ and $NT$ are compatible. More
explicitly, for any $\mathcal{O}$-operator $T$, if there exists an invertible
Nijenhuis operator $N$ such that $NT$ is also an $\mathcal{O}$-operator, then
$T$ and $NT$ are compatible.

Proof. Let $u,v\in M$ and $a\in A$, we have

$\displaystyle NT(au)=N(T(au))=N(aT(u))=aNT(u).$

In addition, since $N$ is a Nijenhuis operator and $T$ is an
$\mathcal{O}$-operator we have

$\displaystyle NT(u)\cdot NT(v)$ $\displaystyle=N\Big{(}NT(u)\cdot
T(v)+T(u)\cdot NT(v)\Big{)}-N^{2}(T(u)\cdot T(v))$
$\displaystyle=NT\Big{(}\rho(NT(u))(v)+\mu(NT(v))(u)\Big{)}$

if and only if (6.9) holds.

If $NT$ is an $\mathcal{O}$-operator and $N$ is invertible, then we have

$NT(u)\cdot T(v)+T(u)\cdot
NT(v)=T\big{(}\rho(NT(u))(v)+\mu(NT(v))(u)\big{)}+NT\big{(}\rho(T(u))(v)+\mu(T(v))(u)\big{)},$

which is exactly the condition that $NT$ and $T$ are compatible.

The following results are an immediate consequence of the last two
propositions.

######  Corollary 6.11.

Let $T_{1},T_{2}:M\longrightarrow L$ be two $\mathcal{O}$-operators on a left-
symmetric Rinheart algebra $(L,A,\cdot,\ell)$ associated to a representation
$(M;\rho,\mu)$. Suppose that $T_{1}$ and $T_{2}$ are invertible. Then $T_{1}$
and $T_{2}$ are compatible if and only if $N=T_{1}\circ T_{2}^{-1}$ is a
Nijenhuis operator.

Since a Rota-Baxter operator of weight zero is an $\mathcal{O}$-operator on a
left-symmetric Rinheart algebra associated to the adjoint representation, we
have the following corollary.

######  Corollary 6.12.

Let $(L,A,\cdot,\ell)$ be a left-symmetric Rinheart algebra. Suppose that
$\mathcal{R}_{1}$ and $\mathcal{R}_{2}$ are two invertible Rota-Baxter
operators of weight zero. Then $\mathcal{R}_{1}$ and $\mathcal{R}_{2}$ are
compatible in the sense that any linear combination of $\mathcal{R}_{1}$ and
$\mathcal{R}_{2}$ is still a Rota-Baxter operator of weight zero if and only
if $N=\mathcal{R}_{1}\circ\mathcal{R}_{2}^{-1}$ is a Nijenhuis operator.

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