# Cohomology of Leibniz Triple Systems and its applications

Xueru Wu  X. Wu: School of Mathematics and Statistics, Northeast Normal
University, Changchun 130024, China<EMAIL_ADDRESS>,  Liangyun Chen  L.
Chen: School of Mathematics and Statistics, Northeast Normal University,
Changchun 130024, China<EMAIL_ADDRESS>and  Yao Ma∗ Y. Ma: School of
Mathematics and Statistics, Northeast Normal University, Changchun 130024,
China<EMAIL_ADDRESS>
###### Abstract.

In this paper, we introduce the first and third cohomology groups on Leibniz
triple systems, which can be applied to extension theory and $1$-parameter
formal deformation theory. Specifically, we investigate the central extension
theory for Leibniz triple systems and show that there is a one-to-one
correspondence between equivalent classes of central extensions of Leibniz
triple systems and the third cohomology group. We study the $T^{*}$-extension
of a Leibniz triple system and we determined that every even-dimensional
quadratic Leibniz triple system $(\mathfrak{L},B)$ is isomorphic to a
$T^{*}$-extension of a Leibniz triple system under a suitable condition. We
also give a necessary and sufficient condition for a quadratic Leibniz triple
system to admit a symplectic form. At last, we develop the $1$-parameter
formal deformation theory of Leibniz triple systems and prove that it is
governed by the cohomology groups.

*Corresponding author.

_MSC_(2020). 17A32, 17A40, 17B56, 16S20.

_Key words and phrases_. Leibniz triple systems, cohomology, central
extension, $T^{*}$-extension, deformation.

The second author is supported by NNSF of China (No.11771069). The third
author is supported by NNSF of China (Nos. 11801066, 11771410).

##  1\. Introduction

The notion of Leibniz triple systems was introduced by Bremner and Sánchez-
Ortega [4] in 2014, which were obtained by using Kolesnikov-Pozhideav
algorithm that takes the defining identities for a variety of algebras and
produces the defining identities for the corresponding variety of dialgebras
[8], to Lie triple systems. Besides, Leibniz triple systems can also be
obtained from any Malcev dialgebra [5]. Furthermore, Leibniz triple systems
are related to Leibniz algebras in the same way that Lie triple systems are
related to Lie algebras. So it is natural to prove the analogous results from
the theory of Lie triple systems to Leibniz triple systems. We list here some
known results on this kind of algebraic structures: a Leibniz triple system
decomposes as the orthogonal direct sum of well-described ideals if it admits
a multiplicative basis [1]; Levi’s theorem for Leibniz triple systems is
determined [11]; the theory of centroid on Leibniz triple systems refers to
[6]. In this paper, we will introduce the cohomology theory of Leibniz triple
systems and use it to study the extension theory and $1$-parameter formal
deformation theory, which generalize partial results in [2, 3, 7, 9, 10, 12,
13, 14, 15, 16].

The paper is organized as follows. In Section 2, first, we will recall some
basic definitions of Leibniz triple system and then we introduce the first and
third cohomology groups of Leibniz triple systems. Section 3 concerns central
extensions and we prove that there is a one-to-one correspondence between
equivalent classes of central extensions of a Leibniz triple system and the
third cohomology group. Section 4 is to introduce the $T^{*}$-extension theory
of Leibniz triple systems. We prove that every even-dimensional quadratic
Leibniz triple system containing an isotropic ideal is isomorphic to the
$T^{*}$-extension of a Leibniz triple system. We also give the definition of
symplectic form for a Leibniz triple system and get a sufficient and necessary
condition for a quadratic Leibniz triple system to admit a symplectic
structure. In Section 5, we study the $1$-parameter formal deformation theory
of Leibniz triple systems. We show that the cohomology group defined in
Section 2 is suitable for this $1$-parameter formal deformation theory.

Throughout this paper, all Leibniz triple system $\mathfrak{L}$ are defined
over a fixed but arbitrary field $\mathbb{F}$.

##  2\. Cohomology of Leibniz triple systems

In this section, we first recall some basic definitions and give the dual
modules of Leibniz triple systems. Then we will give the first and third
cohomology group of Leibniz triple systems.

######  Definition 2.1.

[4] A Leibniz triple system is a vector space $\mathfrak{L}$ endowed with a
trilinear operation
$\\{\cdot,\cdot,\cdot\\}:\mathfrak{L}\times\mathfrak{L}\times\mathfrak{L}\longrightarrow\mathfrak{L}$
satisfying

(2.1)
$\displaystyle\\{a,b,\\{c,d,e\\}\\}=\\{\\{a,b,c\\},d,e\\}-\\{\\{a,b,d\\},c,e\\}-\\{\\{a,b,e\\},c,d\\}+\\{\\{a,b,e\\},d,c\\},$
(2.2)
$\displaystyle\\{a,\\{b,c,d\\},e\\}=\\{\\{a,b,c\\},d,e\\}-\\{\\{a,c,b\\},d,e\\}-\\{\\{a,d,b\\},c,e\\}+\\{\\{a,d,c\\},b,e\\},$

for all $a,b,c,d,e\in\mathfrak{L}$.

Note that a Leibniz triple system can be given by a Lie triple system with the
same ternary product. A Leibniz algebra $L$ with product $[\cdot,\cdot]$
becomes a Leibniz triple system when $\\{x,y,z\\}:=[[x,y],z],$ for all
$x,y,z\in L.$ More examples refer to [4]. Denote by End$(\mathfrak{L})$ the
set consisting of all linear maps on a Leibniz triple system
$(\mathfrak{L},\\{\cdot,\cdot,\cdot\\})$.

######  Definition 2.2.

[1] Let $I$ be a subspace of a Leibniz triple system $\mathfrak{L}.$ Then $I$
is called a subsystem of $\mathfrak{L},$ if $\\{I,I,I\\}\subset I;$ $I$ is
called an ideal of $\mathfrak{L},$ if
$\\{I,\mathfrak{L},\mathfrak{L}\\}+\\{\mathfrak{L},I,\mathfrak{L}\\}+\\{\mathfrak{L},\mathfrak{L},I\\}\subset
I.$ Moreover, if
$\\{I,I,\mathfrak{L}\\}=\\{I,\mathfrak{L},I\\}=\\{\mathfrak{L},I,I\\}=0,$ then
$I$ is called an abelian ideal of $\mathfrak{L}.$

######  Definition 2.3.

[11] Let $\mathfrak{L}$ be a Leibniz triple system and $V$ a vector space. $V$
is called an $\mathfrak{L}$-module, if $\mathfrak{L}\dot{+}V$ is a Leibniz
triple system such that $(1)$ $\mathfrak{L}$ is a subsystem, $(2)$
$\\{a,b,c\\}\in V$ if any one of $a,b,c\in V;$ $(3)$ $\\{a,b,c\\}=0$ if any
two of $a,b,c\in V.$

######  Definition 2.4.

[11] Let $\mathfrak{L}$ be a Leibniz triple system and $V$ a vector space.
Suppose $l,m,r:\mathfrak{L}\times\mathfrak{L}\longrightarrow$ End$(V)$ are
bilinear maps such that

(2.3) $\displaystyle l(a,\\{b,c,d\\})$
$\displaystyle\\!=\\!l(\\{a,b,c\\},d)\\!-\\!l(\\{a,c,b\\},d)\\!-\\!l(\\{a,d,b\\},c)\\!+\\!l(\\{a,d,c\\},b),$
(2.4) $\displaystyle m(a,d)l(b,c)$
$\displaystyle=m(\\{a,b,c\\},d)-m(\\{a,c,b\\},d)-r(c,d)m(a,b)+r(b,d)m(a,c),$
(2.5) $\displaystyle m(a,d)m(b,c)$
$\displaystyle=r(c,d)l(a,b)-r(c,d)m(a,b)-m(\\{a,c,b\\},d)+r(b,d)l(a,c),$ (2.6)
$\displaystyle m(a,d)r(b,c)$
$\displaystyle=r(c,d)m(a,b)-r(c,d)l(a,b)-r(b,d)l(a,c)+m(\\{a,c,b\\},d),$ (2.7)
$\displaystyle r(\\{a,b,c\\},d)$
$\displaystyle=r(c,d)r(a,b)-r(c,d)r(b,a)-r(b,d)r(c,a)+r(a,d)r(c,b),$ (2.8)
$\displaystyle l(a,b)l(c,d)$
$\displaystyle=l(\\{a,b,c\\},d)-l(\\{a,b,d\\},c)-r(c,d)l(a,b)+r(d,c)l(a,b),$
(2.9) $\displaystyle l(a,b)m(c,d)$
$\displaystyle=m(\\{a,b,c\\},d)-r(c,d)l(a,b)-l(\\{a,b,d\\},c)+m(\\{a,b,d\\},c),$
(2.10) $\displaystyle l(a,b)r(c,d)$
$\displaystyle=r(c,d)l(a,b)-m(\\{a,b,c\\},d)-m(\\{a,b,d\\},c)+l(\\{a,b,d\\},c),$
(2.11) $\displaystyle m(a,\\!\\{b,\\!c,\\!d\\})$
$\displaystyle\\!=\\!r(c,\\!d)m(a,\\!b)\\!-\\!r(b,\\!d)m(a,\\!c)\\!-\\!r(b,\\!c)m(a,\\!d)\\!+\\!r(c,\\!b)m(a,\\!d),$
(2.12) $\displaystyle r(a,\\{b,c,d\\})$
$\displaystyle=r(c,d)r(a,b)-r(b,d)r(a,c)-r(b,c)r(a,d)+r(c,b)r(a,d),$

for all $a,b,c,d\in\mathfrak{L}.$ Then $(r,m,l)$ is called a representation of
$\mathfrak{L}$ on $V$.

######  Remark 2.5.

In [11], the authors proved that the notions of an $\mathfrak{L}$-module and a
representation of $\mathfrak{L}$ are equivalent in the following sense. If
$(r,m,l)$ is a representation of $\mathfrak{L}$ on $V,$ then
$\mathfrak{L}\dot{+}V$ endowed with

(2.13)
$\displaystyle\\{x+u,y+v,z+w\\}_{\mathfrak{L}\dot{+}V}=\\{x,y,z\\}+(r(y,z)u+m(x,z)v+l(x,y)w)$

becomes a Leibniz triple system. If $V$ is an $\mathfrak{L}$-module then it
induces a representation of $\mathfrak{L}$ on $V$ by setting
$l(a,b)(v)=\\{a,b,v\\},$ $m(a,b)(v)=\\{a,v,b\\}$ and $r(a,b)(v)=\\{v,a,b\\},$
for any $a,b\in\mathfrak{L}$ and $v\in V.$ In particular, $\mathfrak{L}$
together with $l(a,b)(c)=\\{a,b,c\\},$ $m(a,b)(c)=\\{a,c,b\\}$ and
$r(a,b)(c)=\\{c,a,b\\},$ for all $a,b,c\in\mathfrak{L},$ gives a
representation of $\mathfrak{L}$ on $\mathfrak{L},$ called the adjoint
$\mathfrak{L}$-module. Next, we give the example of the dual
$\mathfrak{L}$-module.

######  Example 2.6.

Let $(\mathfrak{L},\\{\cdot,\cdot,\cdot\\})$ be a Leibniz triple system and
$\mathfrak{L}^{*}$ the dual space of $\mathfrak{L}.$ Define the following
bracket on $\mathfrak{L}\dot{+}\mathfrak{L}^{*}$ as

(2.14)
$\displaystyle\\{a+f,b+g,c+h\\}_{\mathfrak{L}\dot{+}\mathfrak{L}^{*}}=\\{a,b,c\\}+\\{f,b,c\\}+\\{a,g,c\\}+\\{a,b,h\\},$

for all $a,b,c\in\mathfrak{L}$ and $f,g,h\in\mathfrak{L}^{*},$ where
$\\{f,b,c\\}(n)=f(\\{n,c,b\\}),$ $\\{a,g,c\\}(n)=g(\\{c,n,a\\})$ and
$\\{a,b,h\\}(n)=h(\\{b,a,n\\}),$ for all $n\in\mathfrak{L}.$ Then
$(\mathfrak{L}\dot{+}\mathfrak{L}^{*},\\{\cdot,\cdot,\cdot\\}_{\mathfrak{L}\dot{+}\mathfrak{L}^{*}})$
is a Leibniz triple system, which can be checked straightforwardly, here we
only check Eq. (2.1) as an example. Note that, for all
$a,b,c,d,e\in\mathfrak{L}$ and $f,g,h,s,t\in\mathfrak{L}^{*},$ we have

$\displaystyle\\{a+f,b+g,\\{c+h,d+s,e+t\\}_{\mathfrak{L}\dot{+}\mathfrak{L}^{*}}\\}_{\mathfrak{L}\dot{+}\mathfrak{L}^{*}}$
$\displaystyle=$
$\displaystyle\\{a+f,b+g,\\{c,d,e\\}+\\{h,d,e\\}+\\{c,s,e\\}+\\{c,d,t\\}\\}_{\mathfrak{L}\dot{+}\mathfrak{L}^{*}}$
$\displaystyle=$
$\displaystyle\\{a,b,\\{c,d,e\\}\\}+\\{a,b,\\{h,d,e\\}+\\{c,s,e\\}+\\{c,d,t\\}\\}+\\{f,b,\\{c,d,e\\}\\}+\\{a,g,\\{c,d,e\\}\\},$
$\displaystyle\\{\\{a+f,b+g,c+h\\}_{\mathfrak{L}\dot{+}\mathfrak{L}^{*}},d+s,e+t\\}_{\mathfrak{L}\dot{+}\mathfrak{L}^{*}}$
$\displaystyle=$
$\displaystyle\\{\\{a,b,c\\}+\\{a,b,h\\}+\\{f,b,c\\}+\\{a,g,c\\},d+s,e+t\\}_{\mathfrak{L}\dot{+}\mathfrak{L}^{*}}$
$\displaystyle=$
$\displaystyle\\{\\{a,b,c\\},d,e\\}+\\{\\{a,b,c\\},d,t\\}+\\{\\{a,b,c\\},s,e\\}+\\{\\{a,b,h\\}+\\{f,b,c\\}+\\{a,g,c\\},d,e\\},$
$\displaystyle\\{\\{a+f,b+g,d+s\\}_{\mathfrak{L}\dot{+}\mathfrak{L}^{*}},c+h,e+t\\}_{\mathfrak{L}\dot{+}\mathfrak{L}^{*}}$
$\displaystyle=$
$\displaystyle\\{\\{a,b,d\\}+\\{a,b,s\\}+\\{f,b,d\\}+\\{a,g,d\\},c+h,e+t\\}_{\mathfrak{L}\dot{+}\mathfrak{L}^{*}}$
$\displaystyle=$
$\displaystyle\\{\\{a,b,d\\},c,e\\}+\\{\\{a,b,d\\},c,t\\}+\\{\\{a,b,d\\},h,e\\}+\\{\\{a,b,s\\}+\\{f,b,d\\}+\\{a,g,d\\},c,e\\},$
$\displaystyle\\{\\{a+f,b+g,e+t\\}_{\mathfrak{L}\dot{+}\mathfrak{L}^{*}},c+h,d+s\\}_{\mathfrak{L}\dot{+}\mathfrak{L}^{*}}$
$\displaystyle=$
$\displaystyle\\{\\{a,b,e\\}+\\{a,b,t\\}+\\{f,b,e\\}+\\{a,g,e\\},c+h,d+s\\}_{\mathfrak{L}\dot{+}\mathfrak{L}^{*}}$
$\displaystyle=$
$\displaystyle\\{\\{a,b,e\\},c,d\\}+\\{\\{a,b,e\\},c,s\\}+\\{\\{a,b,e\\},h,d\\}+\\{\\{a,b,t\\}+\\{f,b,e\\}+\\{a,g,e\\},c,d\\},$
$\displaystyle\\{\\{a+f,b+g,e+t\\}_{\mathfrak{L}\dot{+}\mathfrak{L}^{*}},d+s,c+h\\}_{\mathfrak{L}\dot{+}\mathfrak{L}^{*}}$
$\displaystyle=$
$\displaystyle\\{\\{a,b,e\\}+\\{a,b,t\\}+\\{f,b,e\\}+\\{a,g,e\\},d+s,c+h\\}_{\mathfrak{L}\dot{+}\mathfrak{L}^{*}}$
$\displaystyle=$
$\displaystyle\\{\\{a,b,e\\},d,c\\}+\\{\\{a,b,e\\},d,h\\}+\\{\\{a,b,e\\},s,c\\}+\\{\\{a,b,t\\}+\\{f,b,e\\}+\\{a,g,e\\},d,c\\},$

If $\mathfrak{L}\dot{+}\mathfrak{L}^{*}$ becomes a Leibniz triple system, we
must have

$\displaystyle\\{a+f,b+g,\\{c+h,d+s,e+t\\}_{\mathfrak{L}\dot{+}\mathfrak{L}^{*}}\\}_{\mathfrak{L}\dot{+}\mathfrak{L}^{*}}$
$\displaystyle=$
$\displaystyle\\{\\!\\{a+f,\\!b+g,\\!c+h\\}_{\mathfrak{L}\dot{+}\mathfrak{L}^{*}\\!},\\!d+s,\\!e+t\\}_{\mathfrak{L}\\!\dot{+}\\!\mathfrak{L}^{*}}-\\{\\!\\{\\!a+f,\\!b+g,\\!d+s\\}_{\mathfrak{L}\dot{+}\mathfrak{L}^{*}\\!},\\!c+h,\\!e+t\\}_{\mathfrak{L}\\!\dot{+}\\!\mathfrak{L}^{*}}$
$\displaystyle-\\{\\!\\{\\!a+f,\\!b+g,\\!e+t\\}_{\mathfrak{L}\dot{+}\mathfrak{L}^{*}\\!},\\!c+h,\\!d+s\\}_{\mathfrak{L}\\!\dot{+}\\!\mathfrak{L}^{*}}+\\{\\!\\{\\!a+f,\\!b+g,\\!e+t\\}_{\mathfrak{L}\dot{+}\mathfrak{L}^{*}\\!},\\!d+s,\\!c+h\\}_{\mathfrak{L}\\!\dot{+}\\!\mathfrak{L}^{*}}.$

Since

$\displaystyle\\{f,b,\\{c,d,e\\}\\}(n)=$ $\displaystyle
f(\\{n,\\{c,d,e\\},b\\}),$ $\displaystyle\\{\\{f,b,c\\},d,e\\}(n)=$
$\displaystyle\\{f,b,c\\}(\\{n,e,d\\})=f(\\{\\{n,e,d\\},c,b\\}),$
$\displaystyle\\{\\{f,b,d\\},c,e\\}(n)=$
$\displaystyle\\{f,b,d\\}(\\{n,e,c\\})=f(\\{\\{n,e,c\\},d,b\\}),$
$\displaystyle\\{\\{f,b,e\\},c,d\\}(n)=$
$\displaystyle\\{f,b,e\\}(\\{n,d,c\\})=f(\\{\\{n,d,c\\},e,b\\}),$
$\displaystyle\\{\\{f,b,e\\},d,c\\}(n)=$
$\displaystyle\\{f,b,e\\}(\\{n,c,d\\})=f(\\{\\{n,c,d\\},e,b\\}),$

and

$\displaystyle\\{n,\\{c,d,e\\},b\\}=\\{\\{n,c,d\\},e,b\\}-\\{\\{n,d,c\\},e,b\\}-\\{\\{n,e,c\\},d,b\\}+\\{\\{n,e,d\\},c,b\\},$

it follows that

$\displaystyle\\{f,b,\\{c,d,e\\}\\}=\\{\\{f,b,c\\},d,e\\}-\\{\\{f,b,d\\},c,e\\}-\\{\\{f,b,e\\},c,d\\}+\\{\\{f,b,e\\},d,c\\}.$

Similarly, we have

$\displaystyle\\{a,g,\\{c,d,e\\}\\}=$
$\displaystyle\\{\\{a,g,c\\},d,e\\}-\\{\\{a,g,d\\},c,e\\}-\\{\\{a,g,e\\},c,d\\}+\\{\\{a,g,e\\},d,c\\},$
$\displaystyle\\{a,b,\\{h,d,e\\}\\}=$
$\displaystyle\\{\\{a,b,h\\},d,e\\}-\\{\\{a,b,d\\},h,e\\}-\\{\\{a,b,e\\},h,d\\}+\\{\\{a,b,e\\},d,h\\},$
$\displaystyle\\{a,b,\\{c,s,e\\}\\}=$
$\displaystyle\\{\\{a,b,c\\},s,e\\}-\\{\\{a,b,s\\},c,e\\}-\\{\\{a,b,e\\},s,c\\}+\\{\\{a,b,e\\},c,s\\},$
$\displaystyle\\{a,b,\\{c,d,t\\}\\}=$
$\displaystyle\\{\\{a,b,c\\},d,t\\}-\\{\\{a,b,d\\},c,t\\}-\\{\\{a,b,t\\},c,d\\}+\\{\\{a,b,t\\},d,c\\}.$

So we have proved that $\mathfrak{L}\dot{+}\mathfrak{L}^{*}$ satisfies Eq.
(2.1).

Similarly, $\mathfrak{L}\dot{+}\mathfrak{L}^{*}$ satisfies Eq. (2.2), hence
$(\mathfrak{L}\dot{+}\mathfrak{L}^{*},\\{\cdot,\cdot,\cdot\\}_{\mathfrak{L}\dot{+}\mathfrak{L}^{*}})$
is a Leibniz triple system, which implies that $\mathfrak{L}^{*}$ becomes an
$\mathfrak{L}$-module, which is called the dual $\mathfrak{L}$-module.

######  Definition 2.7.

Let $V$ be an $\mathfrak{L}$-module. A $(2n+1)$-linear map
$f:\underbrace{\mathfrak{L}\times\cdots\times\mathfrak{L}}_{(2n+1)~{}{\rm
times}}\rightarrow V$ is called a $(2n+1)$-cochain of $\mathfrak{L}$ on $V.$
Denote by $C^{2n+1}(\mathfrak{L},V)$ the set of all $(2n+1)$-cochains, for
$n\geq 0.$

######  Definition 2.8.

Let $\mathfrak{L}$ be a Leibniz triple system and $(r,m,l)$ a representation
of $\mathfrak{L}$ on $V.$ For $n=1,2$ the coboundary operator
$\delta^{2n-1}:C^{2n-1}(\mathfrak{L},V)\rightarrow C^{2n+1}(\mathfrak{L},V)$
is defined as follows.
$(1)$ A $1$-coboundary operator of $\mathfrak{L}$ on $V$ is defined by

$\displaystyle\delta^{1}:C^{1}(\mathfrak{L},V)$ $\displaystyle\rightarrow
C^{3}(\mathfrak{L},V)$ $\displaystyle f$ $\displaystyle\mapsto\delta^{1}f$

for $f\in C^{1}(\mathfrak{L},V)$ and

$\displaystyle\delta^{1}f(x_{1},x_{2},x_{3})=r(x_{2},x_{3})f(x_{1})+m(x_{1},x_{3})f(x_{2})+l(x_{1},x_{2})f(x_{3})-f(\\{x_{1},x_{2},x_{3}\\}).$

$(2)$ A $3$-coboundary operator of $\mathfrak{L}$ on $V$ consists a pair of
maps $(\delta^{3}_{1},\delta^{3}_{2}),$ where

$\displaystyle\delta^{3}_{i}:C^{3}(\mathfrak{L},V)$ $\displaystyle\rightarrow
C^{5}(\mathfrak{L},V)$ $\displaystyle g$ $\displaystyle\mapsto\delta^{3}_{i}g$

for $g\in C^{3}(\mathfrak{L},V)$ and

(2.15) $\displaystyle\begin{aligned}
&\delta^{3}_{1}g(x_{1},x_{2},x_{3},x_{4},x_{5})\\\
=&g(x_{1},x_{2},\\{x_{3},x_{4},x_{5}\\})-g(\\{x_{1},x_{2},x_{3}\\},x_{4},x_{5})+g(\\{x_{1},x_{2},x_{4}\\},x_{3},x_{5})\\\
&+g(\\{x_{1},x_{2},x_{5}\\},x_{3},x_{4})-g(\\{x_{1},x_{2},x_{5}\\},x_{4},x_{3})+l(x_{1},x_{2})g(x_{3},x_{4},x_{5})\\\
&-r(x_{4},x_{5})g(x_{1},x_{2},x_{3})+r(x_{3},x_{5})g(x_{1},x_{2},x_{4})+r(x_{3},x_{4})g(x_{1},x_{2},x_{5})\\\
&-r(x_{4},x_{3})g(x_{1},x_{2},x_{5}),\end{aligned}$ (2.16)
$\displaystyle\begin{aligned}
&\delta^{3}_{2}g(x_{1},x_{2},x_{3},x_{4},x_{5})\\\
=&g(x_{1},\\{x_{2},x_{3},x_{4}\\},x_{5})-g(\\{x_{1},x_{2},x_{3}\\},x_{4},x_{5})+g(\\{x_{1},x_{3},x_{2}\\},x_{4},x_{5})\\\
&+g(\\{x_{1},x_{4},x_{2}\\},x_{3},x_{5})-g(\\{x_{1},x_{4},x_{3}\\},x_{2},x_{5})+m(x_{1},x_{5})g(x_{2},x_{3},x_{4})\\\
&-r(x_{4},x_{5})g(x_{1},x_{2},x_{3})+r(x_{4},x_{5})g(x_{1},x_{3},x_{2})+r(x_{3},x_{5})g(x_{1},x_{4},x_{2})\\\
&-r(x_{2},x_{5})g(x_{1},x_{4},x_{3}).\end{aligned}$

######  Remark 2.9.

Similarly to the definition of $(r,m,l)$ in Remark 2.5, we can rewrite Eqs.
$(\ref{222.221})$-$(\ref{222.222})$ as follows

(2.17) $\displaystyle\begin{aligned}
&\delta^{3}_{1}g(x_{1},x_{2},x_{3},x_{4},x_{5})\\\
=&g(x_{1},x_{2},\\{x_{3},x_{4},x_{5}\\})-g(\\{x_{1},x_{2},x_{3}\\},x_{4},x_{5})+g(\\{x_{1},x_{2},x_{4}\\},x_{3},x_{5})\\\
&+g(\\{x_{1},x_{2},x_{5}\\},x_{3},x_{4})-g(\\{x_{1},x_{2},x_{5}\\},x_{4},x_{3})+\\{x_{1},x_{2},g(x_{3},x_{4},x_{5})\\}\\\
&-\\{g(x_{1},x_{2},x_{3}),x_{4},x_{5}\\}+\\{g(x_{1},x_{2},x_{4}),x_{3},x_{5}\\}+\\{g(x_{1},x_{2},x_{5}),x_{3},x_{4}\\}\\\
&-\\{g(x_{1},x_{2},x_{5}),x_{4},x_{3}\\},\end{aligned}$ (2.18)
$\displaystyle\begin{aligned}
&\delta^{3}_{2}g(x_{1},x_{2},x_{3},x_{4},x_{5})\\\
=&g(x_{1},\\{x_{2},x_{3},x_{4}\\},x_{5})-g(\\{x_{1},x_{2},x_{3}\\},x_{4},x_{5})+g(\\{x_{1},x_{3},x_{2}\\},x_{4},x_{5})\\\
&+g(\\{x_{1},x_{4},x_{2}\\},x_{3},x_{5})-g(\\{x_{1},x_{4},x_{3}\\},x_{2},x_{5})+\\{x_{1},g(x_{2},x_{3},x_{4}),x_{5}\\}\\\
&-\\{g(x_{1},x_{2},x_{3}),x_{4},x_{5}\\}+\\{g(x_{1},x_{3},x_{2}),x_{4},x_{5}\\}+\\{g(x_{1},x_{4},x_{2}),x_{3},x_{5}\\}\\\
&-\\{g(x_{1},x_{4},x_{3}),x_{2},x_{5}\\}.\end{aligned}$

Recall that, a derivation of a Leibniz triple system $\mathfrak{L}$ is a
linear map $D$ $:$ $\mathfrak{L}$ $\longrightarrow$ $\mathfrak{L}$ satisfying

(2.19) $\displaystyle
D(\\{a,b,c\\})=\\{D(a),b,c\\}+\\{a,D(b),c\\}+\\{a,b,D(c)\\},$

for all $a,b,c\in\mathfrak{L}$. Now we introduce the first and third
cohomology groups of a Leibniz triple system. We will see that the derivation
is a $1$-cocycle.

Let $\mathfrak{L}$ be a Leibniz triple system and $V$ an
$\mathfrak{L}$-module. We call the set

$\displaystyle Z^{1}(\mathfrak{L},V)=\\{f\in
C^{1}(\mathfrak{L},V)~{}|~{}\delta^{1}f=0\\}$

the space of $1$-cocycles of $\mathfrak{L}$ on $V$. We can see that for any
$f\in Z^{1}(\mathfrak{L},\mathfrak{L}),$ the space of $1$-cocycles of
$\mathfrak{L}$ on its adjoint module, $f$ is a derivation of
$(\mathfrak{L},\\{\cdot,\cdot,\cdot\\}).$ The set

$\displaystyle Z^{3}(\mathfrak{L},V)=\\{g\in
C^{3}(\mathfrak{L},V)~{}|~{}\delta^{3}_{1}g=\delta^{3}_{2}g=0\\}$

is called the space of $3$-cocycles of $\mathfrak{L}$ on $V$. The set

$\displaystyle B^{3}(\mathfrak{L},V)=\\{\delta^{1}f~{}|~{}f\in
C^{1}(\mathfrak{L},V)\\}$

is called the space of $1$-coboundaries of $\mathfrak{L}$ on $V$.

######  Example 2.10.

Let $\mathfrak{L}=\mathbb{K}\langle x,y\rangle$ be a Leibniz triple system
with $\\{x,y,y\\}=\\{y,y,y\\}=x$ and $\mathfrak{L}^{*}$ its dual space with
the dual basis $\\{x^{*},y^{*}\\}.$ Define a trilinear map
$\theta:\mathfrak{L}\times\mathfrak{L}\times\mathfrak{L}\rightarrow\mathfrak{L}^{*}$
by $\theta(x,y,y)=\theta(y,y,y)=x^{*}+y^{*},$ the rest are $0.$ We can easily
prove that $\theta$ is a $3$-cocycle of $\mathfrak{L}$.

######  Proposition 2.11.

$Z^{3}(\mathfrak{L},V)$ and $B^{3}(\mathfrak{L},V)$ are defined as above. Then
$B^{3}(\mathfrak{L},V)\subset Z^{3}(\mathfrak{L},V).$

###### Proof.

For any $\delta^{1}f\in B^{3}(\mathfrak{L},V),$ we need to prove that
$\delta^{3}_{1}\delta^{1}f=0$ and $\delta^{3}_{2}\delta^{1}f=0$. In fact, by
the definitions of $1$-coboundary operator, $3$-coboundary operator,
$\mathfrak{L}$-module and Eq. (2.17), we have

$\displaystyle\delta^{3}_{1}\delta^{1}f(x_{1},x_{2},x_{3},x_{4},x_{5})$
$\displaystyle=$
$\displaystyle\delta^{1}f(x_{1},x_{2},\\{x_{3},x_{4},x_{5}\\})-\delta^{1}f(\\{x_{1},x_{2},x_{3}\\},x_{4},x_{5})+\delta^{1}f(\\{x_{1},x_{2},x_{4}\\},x_{3},x_{5})$
$\displaystyle+\delta^{1}f(\\{x_{1},x_{2},x_{5}\\},x_{3},x_{4})-\delta^{1}f(\\{x_{1},x_{2},x_{5}\\},x_{4},x_{3})+\\{x_{1},x_{2},\delta^{1}f(x_{3},x_{4},x_{5})\\}$
$\displaystyle-\\{\delta^{1}f(x_{1},x_{2},x_{3}),x_{4},x_{5}\\}+\\{\delta^{1}f(x_{1},x_{2},x_{4}),x_{3},x_{5}\\}+\\{\delta^{1}f(x_{1},x_{2},x_{5}),x_{3},x_{4}\\}$
$\displaystyle-\\{\delta^{1}f(x_{1},x_{2},x_{5}),x_{4},x_{3}\\}$
$\displaystyle=$
$\displaystyle\\{f(x_{1}),x_{2},\\{x_{3},x_{4},x_{5}\\}\\}+\\{x_{1},f(x_{2}),\\{x_{3},x_{4},x_{5}\\}\\}+\\{x_{1},x_{2},f(\\{x_{3},x_{4},x_{5}\\})\\}$
$\displaystyle-f(\\{x_{1},x_{2},\\{x_{3},x_{4},x_{5}\\}\\})-\\{f(\\{x_{1},x_{2},x_{3}\\}),x_{4},x_{5}\\}-\\{\\{x_{1},x_{2},x_{3}\\},f(x_{4}),x_{5}\\}$
$\displaystyle-\\{\\{x_{1},x_{2},x_{3}\\},x_{4},f(x_{5})\\}+f(\\{\\{x_{1},x_{2},x_{3}\\},x_{4},x_{5}\\})+\\{f(\\{x_{1},x_{2},x_{4}\\}),x_{3},x_{5}\\}$
$\displaystyle+\\{\\{x_{1},x_{2},x_{4}\\},f(x_{3}),x_{5}\\}+\\{\\{x_{1},x_{2},x_{4}\\},x_{3},f(x_{5})\\}-f(\\{\\{x_{1},x_{2},x_{4}\\},x_{3},x_{5}\\})$
$\displaystyle+\\{f(\\{x_{1},x_{2},x_{5}\\}),x_{3},x_{4}\\}+\\{\\{x_{1},x_{2},x_{5}\\},f(x_{3}),x_{4}\\}+\\{\\{x_{1},x_{2},x_{5}\\},x_{3},f(x_{4})\\}$
$\displaystyle-f(\\{\\{x_{1},x_{2},x_{5}\\},x_{3},x_{4}\\})-\\{f(\\{x_{1},x_{2},x_{5}\\}),x_{4},x_{3}\\}-\\{\\{x_{1},x_{2},x_{5}\\},f(x_{4}),x_{3}\\}$
$\displaystyle-\\{\\{x_{1},x_{2},x_{5}\\},x_{4},f(x_{3})\\}+f(\\{\\{x_{1},x_{2},x_{5}\\},x_{4},x_{3}\\})+\\{x_{1},x_{2},\\{f(x_{3}),x_{4},x_{5}\\}\\}$
$\displaystyle+\\{x_{1},x_{2},\\{x_{3},f(x_{4}),x_{5}\\}\\}+\\{x_{1},x_{2},\\{x_{3},x_{4},f(x_{5})\\}\\}-\\{x_{1},x_{2},f(\\{x_{3},x_{4},x_{5}\\})\\}$
$\displaystyle-\\{\\{f(x_{1}),x_{2},x_{3}\\},x_{4},x_{5}\\}-\\{\\{x_{1},f(x_{2}),x_{3}\\},x_{4},x_{5}\\}-\\{\\{x_{1},x_{2},f(x_{3})\\},x_{4},x_{5}\\}$
$\displaystyle+\\{f(\\{x_{1},x_{2},x_{3}\\}),x_{4},x_{5}\\}+\\{\\{f(x_{1}),x_{2},x_{4}\\},x_{3},x_{5}\\}+\\{\\{x_{1},f(x_{2}),x_{4}\\},x_{3},x_{5}\\}$
$\displaystyle+\\{\\{x_{1},x_{2},f(x_{4})\\},x_{3},x_{5}\\}-\\{f(\\{x_{1},x_{2},x_{4}\\}),x_{3},x_{5}\\}+\\{\\{f(x_{1}),x_{2},x_{5}\\},x_{3},x_{4}\\}$
$\displaystyle+\\{\\{x_{1},f(x_{2}),x_{5}\\},x_{3},x_{4}\\}+\\{\\{x_{1},x_{2},f(x_{5})\\},x_{3},x_{4}\\}-\\{f(\\{x_{1},x_{2},x_{5}\\}),x_{3},x_{4}\\}$
$\displaystyle-\\{\\{f(x_{1}),x_{2},x_{5}\\},x_{4},x_{3}\\}-\\{\\{x_{1},f(x_{2}),x_{5}\\},x_{4},x_{3}\\}-\\{\\{x_{1},x_{2},f(x_{5})\\},x_{4},x_{3}\\}$
$\displaystyle+\\{f(\\{x_{1},x_{2},x_{5}\\}),x_{4},x_{3}\\}$ $\displaystyle=$
$\displaystyle 0.$

On the other hand, it follows from the definitions of $1$-coboundary operator,
$3$-coboundary operator, $\mathfrak{L}$-module and Eq. (2.18) that

$\displaystyle\delta^{3}_{2}\delta^{1}f(x_{1},x_{2},x_{3},x_{4},x_{5})$
$\displaystyle=$
$\displaystyle\delta^{1}f(x_{1},\\{x_{2},x_{3},x_{4}\\},x_{5})-\delta^{1}f(\\{x_{1},x_{2},x_{3}\\},x_{4},x_{5})+\delta^{1}f(\\{x_{1},x_{3},x_{2}\\},x_{4},x_{5})$
$\displaystyle+\delta^{1}f(\\{x_{1},x_{4},x_{2}\\},x_{3},x_{5})-\delta^{1}f(\\{x_{1},x_{4},x_{3}\\},x_{2},x_{5})+\\{x_{1},\delta^{1}f(x_{2},x_{3},x_{4}),x_{5}\\}$
$\displaystyle-\\{\delta^{1}f(x_{1},x_{2},x_{3}),x_{4},x_{5}\\}+\\{\delta^{1}f(x_{1},x_{3},x_{2}),x_{4},x_{5}\\}+\\{\delta^{1}f(x_{1},x_{4},x_{2}),x_{3},x_{5}\\}$
$\displaystyle-\\{\delta^{1}f(x_{1},x_{4},x_{3}),x_{2},x_{5}\\}$
$\displaystyle=$
$\displaystyle\\{f(x_{1}),\\{x_{2},x_{3},x_{4}\\},x_{5}\\}+\\{x_{1},f(\\{x_{2},x_{3},x_{4}\\}),x_{5}\\}+\\{x_{1},\\{x_{2},x_{3},x_{4}\\},f(x_{5})\\}$
$\displaystyle-f(\\{x_{1},\\{x_{2},x_{3},x_{4}\\},x_{5}\\})-\\{f(\\{x_{1},x_{2},x_{3}\\}),x_{4},x_{5}\\}-\\{\\{x_{1},x_{2},x_{3}\\},f(x_{4}),x_{5}\\}$
$\displaystyle-\\{\\{x_{1},x_{2},x_{3}\\},x_{4},f(x_{5})\\}+f(\\{\\{x_{1},x_{2},x_{3}\\},x_{4},x_{5}\\})+\\{f(\\{x_{1},x_{3},x_{2}\\}),x_{4},x_{5}\\}$
$\displaystyle+\\{\\{x_{1},x_{3},x_{2}\\},f(x_{4}),x_{5}\\}+\\{\\{x_{1},x_{3},x_{2}\\},x_{4},f(x_{5})\\}-f(\\{\\{x_{1},x_{3},x_{2}\\},x_{4},x_{5}\\})$
$\displaystyle+\\{f(\\{x_{1},x_{4},x_{2}\\}),x_{3},x_{5}\\}+\\{\\{x_{1},x_{4},x_{2}\\},f(x_{3}),x_{5}\\}+\\{\\{x_{1},x_{4},x_{2}\\},x_{3},f(x_{5})\\}$
$\displaystyle-f(\\{\\{x_{1},x_{4},x_{2}\\},x_{3},x_{5}\\})-\\{f(\\{x_{1},x_{4},x_{3}\\}),x_{2},x_{5}\\}-\\{\\{x_{1},x_{4},x_{3}\\},f(x_{2}),x_{5}\\}$
$\displaystyle-\\{\\{x_{1},x_{4},x_{3}\\},x_{2},f(x_{5})\\}+f(\\{\\{x_{1},x_{4},x_{3}\\},x_{2},x_{5}\\})+\\{x_{1},\\{f(x_{2}),x_{3},x_{4}\\},x_{5}\\}$
$\displaystyle+\\{x_{1},\\{x_{2},f(x_{3}),x_{4}\\},x_{5}\\}+\\{x_{1},\\{x_{2},x_{3},f(x_{4})\\},x_{5}\\}-\\{x_{1},f(\\{x_{2},x_{3},x_{4}\\}),x_{5}\\}$
$\displaystyle-\\{\\{f(x_{1}),x_{2},x_{3}\\},x_{4},x_{5}\\}-\\{\\{x_{1},f(x_{2}),x_{3}\\},x_{4},x_{5}\\}-\\{\\{x_{1},x_{2},f(x_{3})\\},x_{4},x_{5}\\}$
$\displaystyle+\\{f(\\{x_{1},x_{2},x_{3}\\}),x_{4},x_{5}\\}+\\{\\{f(x_{1}),x_{3},x_{2}\\},x_{4},x_{5}\\}+\\{\\{x_{1},f(x_{3}),x_{2}\\},x_{4},x_{5}\\}$
$\displaystyle+\\{\\{x_{1},x_{3},f(x_{2})\\},x_{4},x_{5}\\}-\\{f(\\{x_{1},x_{3},x_{2}\\}),x_{4},x_{5}\\}+\\{\\{f(x_{1}),x_{4},x_{2}\\},x_{3},x_{5}\\}$
$\displaystyle+\\{\\{x_{1},f(x_{4}),x_{2}\\},x_{3},x_{5}\\}+\\{\\{x_{1},x_{4},f(x_{2})\\},x_{3},x_{5}\\}-\\{f(\\{x_{1},x_{4},x_{2}\\}),x_{3},x_{5}\\}$
$\displaystyle-\\{\\{f(x_{1}),x_{4},x_{3}\\},x_{2},x_{5}\\}-\\{\\{x_{1},f(x_{4}),x_{3}\\},x_{2},x_{5}\\}-\\{\\{x_{1},x_{4},f(x_{3})\\},x_{2},x_{5}\\}$
$\displaystyle+\\{f(\\{x_{1},x_{4},x_{3}\\}),x_{2},x_{5}\\}$ $\displaystyle=$
$\displaystyle 0.$

Thus the proposition has been proved. ∎

Hence we can define $1$-cohomology space and $3$-cohomology space of
$\mathfrak{L}$ as the factor spaces

$\displaystyle H^{1}(\mathfrak{L},V):=$ $\displaystyle Z^{1}(\mathfrak{L},V).$
$\displaystyle H^{3}(\mathfrak{L},V):=$ $\displaystyle
Z^{3}(\mathfrak{L},V)/B^{3}(\mathfrak{L},V).$

######  Remark 2.12.

Recall that $\mathfrak{L}\dot{+}V$ endowed with Eq. $(\ref{222.123})$ becomes
a Leibniz triple system. Using the $Z^{3}(\mathfrak{L},V)$ defined above, we
have a new Leibniz triple system structure on $\mathfrak{L}\dot{+}V$ by a
$3$-cocycle.

For any $\theta\in Z^{3}(\mathfrak{L},V),$ set

$\displaystyle\\{(x,a),(y,b),(z,c)\\}_{\theta}=(\\{x,y,z\\},\theta(x,y,z)).$

Note that

$\displaystyle\\{(x,a),(y,b),\\{(u,c),(v,d),(w,e)\\}_{\theta}\\}_{\theta}$
$\displaystyle=$
$\displaystyle\\{(x,a),(y,b),(\\{u,v,w\\},\theta(u,v,w))\\}_{\theta}=(\\{x,y,\\{u,v,w\\}\\},\theta(x,y,\\{u,v,w\\})$
$\displaystyle=$
$\displaystyle\Big{(}\\{\\{x,y,u\\},v,w\\},\theta(\\{x,y,u\\},v,w)\Big{)}-\Big{(}\\{\\{x,y,v\\},u,w\\},\theta(\\{x,y,v\\},u,w)\Big{)}$
$\displaystyle-\Big{(}\\{\\{x,y,w\\},u,v\\},\theta(\\{x,y,w\\},u,v)\Big{)}+\Big{(}\\{\\{x,y,w\\},v,u\\},\theta(\\{x,y,w\\},v,u)\Big{)}$
$\displaystyle=$
$\displaystyle\\{\\{(x,a),(y,b),(u,c)\\}_{\theta},(v,d),(w,e)\\}_{\theta}-\\{\\{(x,a),(y,b),(v,d)\\}_{\theta},(u,c),(w,e)\\}_{\theta}$
$\displaystyle-\\{\\{(x,a),(y,b),(w,e)\\}_{\theta},(u,c),(v,d)\\}_{\theta}+\\{\\{(x,a),(y,b),(w,e)\\}_{\theta},(v,d),(u,c)\\}_{\theta},$

and

$\displaystyle\\{(x,a),\\{(y,b),(u,c),(v,d)\\}_{\theta},(w,e)\\}_{\theta}$
$\displaystyle=$
$\displaystyle\\{(x,a),(\\{y,u,v\\},\theta(y,u,v)),(w,e)\\}_{\theta}=(\\{x,\\{y,u,v\\},w\\},\theta(x,\\{y,u,v\\},w)$
$\displaystyle=$
$\displaystyle\Big{(}\\{\\{x,y,u\\},v,w\\},\theta(\\{x,y,u\\},v,w)\Big{)}-\Big{(}\\{\\{x,u,y\\},v,w\\},\theta(\\{x,u,y\\},v,w)\Big{)}$
$\displaystyle-\Big{(}\\{\\{x,v,y\\},u,w\\},\theta(\\{x,v,y\\},u,w)\Big{)}+\Big{(}\\{\\{x,v,u\\},y,w\\},\theta(\\{x,v,u\\},y,w)\Big{)}$
$\displaystyle=$
$\displaystyle\\{\\{(x,a),(y,b),(u,c)\\}_{\theta},(v,d),(w,e)\\}_{\theta}-\\{\\{(x,a),(u,c),(y,b)\\}_{\theta},(v,d),(w,e)\\}_{\theta}$
$\displaystyle-\\{\\{(x,a),(v,d),(y,b)\\}_{\theta},(u,c),(w,e)\\}_{\theta}+\\{\\{(x,a),(v,d),(u,c)\\}_{\theta},(y,b),(w,e)\\}_{\theta}.$

Then, $(\mathfrak{L}_{\theta},\\{\cdot,\cdot,\cdot\\}_{\theta})$ is a Leibniz
triple system.

In the following sections, we will apply the cohomology theory of Leibniz
triple systems to discuss the central extension, $T^{*}$-extension and
$1$-parameter formal deformation of a Leibniz triple system.

##  3\. Central extension of Leibniz triple system

In this section, we give the definition of central extension of a Leibniz
triple system $(\mathfrak{L},\\{\cdot,\cdot,\cdot\\})$ by an
$\mathfrak{L}$-module $V$ and prove that there is a one-to-one correspondence
between equivalent classes of central extensions of
$(\mathfrak{L},\\{\cdot,\cdot,\cdot\\})$ and $H^{3}(\mathfrak{L},V).$

######  Definition 3.1.

Suppose that $(\mathfrak{L},\\{\cdot,\cdot,\cdot\\})$ is a Leibniz triple
system and $V$ is an $\mathfrak{L}$-module.
$(\mathfrak{L}_{c},\\{\cdot,\cdot,\cdot\\}_{c})$ is called a central extension
of $(\mathfrak{L},\\{\cdot,\cdot,\cdot\\})$ by $V$ if there is an exact row of
Leibniz triple systems

$\textstyle{0\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{V\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\iota}$$\textstyle{\mathfrak{L}_{c}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\pi}$$\textstyle{\mathfrak{L}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{0}$

such that $\iota(V)$ is contained in the center of $\mathfrak{L}_{c},$ which
is defined as

$\displaystyle
Z(\mathfrak{L}_{c})=\\{x\in\mathfrak{L}_{c}~{}|~{}\\{x,\mathfrak{L}_{c},\mathfrak{L}_{c}\\}_{c}=\\{\mathfrak{L}_{c},x,\mathfrak{L}_{c}\\}_{c}=\\{\mathfrak{L}_{c},\mathfrak{L}_{c},x\\}_{c}=0\\}.$

######  Definition 3.2.

If $(\mathfrak{L}_{c}^{\prime},\\{\cdot,\cdot,\cdot\\}_{c}^{\prime})$ is
another central extension of $(\mathfrak{L},\\{\cdot,\cdot,\cdot\\})$ by $V$
such that there is an isomorphism
$\phi:(\mathfrak{L}_{c},\\{\cdot,\cdot,\cdot\\}_{c})\rightarrow(\mathfrak{L}_{c}^{\prime},\\{\cdot,\cdot,\cdot\\}_{c}^{\prime}),$
and the diagram

$\setcounter{MaxMatrixCols}{11}\begin{CD}0@>{}>{}>V@>{\iota}>{}>\mathfrak{L}_{c}@>{\pi}>{}>\mathfrak{L}@>{}>{}>0\\\
\Big{\|}@V{}V{\phi}V\Big{\|}\\\ 0@>{}>{}>V\
@>{\iota^{\prime}}>{}>\mathfrak{L}_{c}^{\prime}@>{\pi^{\prime}}>{}>\mathfrak{L}@>{}>{}>0\end{CD}$

commutes, then the two central extensions
$(\mathfrak{L}_{c},\\{\cdot,\cdot,\cdot\\}_{c})$ and
$(\mathfrak{L}_{c}^{\prime},\\{\cdot,\cdot,\cdot\\}_{c}^{\prime})$ are said to
be equivalent.

######  Theorem 3.3.

There is a one-to-one correspondence between equivalent classes of central
extensions of $(\mathfrak{L},\\{\cdot,\cdot,\cdot\\})$ by $V$ and
$H^{3}(\mathfrak{L},V).$

###### Proof.

Step 1, construct a map $\sigma_{1}$ from Ext${}_{c}(\mathfrak{L},V)$ the set
of equivalent classes of central extensions of $\mathfrak{L}~{}by~{}V$ to
$H^{3}(\mathfrak{L},V).$

We first show that there exists a map from the set of central extensions of
$(\mathfrak{L},\\{\cdot,\cdot,\cdot\\})$ by $V$ to $Z^{3}(\mathfrak{L},V).$
Suppose that $(\mathfrak{L}_{c},\\{\cdot,\cdot,\cdot\\}_{c})$ is a central
extension of $(\mathfrak{L},\\{\cdot,\cdot,\cdot\\})$ by $V.$ Then we have the
following exact sequence

$\lx@xy@svg{\hbox{\raise 0.0pt\hbox{\kern
5.5pt\hbox{\ignorespaces\ignorespaces\ignorespaces\hbox{\vtop{\kern
0.0pt\offinterlineskip\halign{\entry@#!@&&\entry<EMAIL_ADDRESS>0.0pt\hbox{\hbox{\kern 0.0pt\raise 0.0pt\hbox{\hbox{\kern 3.0pt\raise
0.0pt\hbox{$\textstyle{0\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$}}}}}}}\ignorespaces\ignorespaces\ignorespaces\ignorespaces{}{\hbox{\lx@xy@droprule}}\ignorespaces{\hbox{\kern
29.5pt\raise 0.0pt\hbox{\hbox{\kern 0.0pt\raise
0.0pt\hbox{\lx@xy@tip{1}\lx@xy@tip{-1}}}}}}{\hbox{\lx@xy@droprule}}{\hbox{\lx@xy@droprule}}{\hbox{\kern
29.5pt\raise 0.0pt\hbox{\hbox{\kern 0.0pt\raise 0.0pt\hbox{\hbox{\kern
3.0pt\raise
0.0pt\hbox{$\textstyle{V\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$}}}}}}}\ignorespaces\ignorespaces\ignorespaces\ignorespaces{}{\hbox{\lx@xy@droprule}}\ignorespaces\ignorespaces\ignorespaces{\hbox{\kern
51.47124pt\raise 4.50694pt\hbox{{}\hbox{\kern 0.0pt\raise
0.0pt\hbox{\hbox{\kern 3.0pt\hbox{\hbox{\kern
0.0pt\raise-1.50694pt\hbox{$\scriptstyle{\iota}$}}}\kern
3.0pt}}}}}}\ignorespaces{\hbox{\kern 67.55554pt\raise 0.0pt\hbox{\hbox{\kern
0.0pt\raise
0.0pt\hbox{\lx@xy@tip{1}\lx@xy@tip{-1}}}}}}{\hbox{\lx@xy@droprule}}{\hbox{\lx@xy@droprule}}{\hbox{\kern
67.55554pt\raise 0.0pt\hbox{\hbox{\kern 0.0pt\raise 0.0pt\hbox{\hbox{\kern
3.0pt\raise
0.0pt\hbox{$\textstyle{\mathfrak{L}_{c}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$}}}}}}}\ignorespaces\ignorespaces\ignorespaces\ignorespaces{}{\hbox{\lx@xy@droprule}}\ignorespaces\ignorespaces\ignorespaces{\hbox{\kern
88.62802pt\raise 4.50694pt\hbox{{}\hbox{\kern 0.0pt\raise
0.0pt\hbox{\hbox{\kern 3.0pt\hbox{\hbox{\kern
0.0pt\raise-1.50694pt\hbox{$\scriptstyle{\pi}$}}}\kern
3.0pt}}}}}}\ignorespaces{\hbox{\kern 106.22897pt\raise 0.0pt\hbox{\hbox{\kern
0.0pt\raise
0.0pt\hbox{\lx@xy@tip{1}\lx@xy@tip{-1}}}}}}{\hbox{\lx@xy@droprule}}{\hbox{\lx@xy@droprule}}{\hbox{\kern
106.22897pt\raise 0.0pt\hbox{\hbox{\kern 0.0pt\raise 0.0pt\hbox{\hbox{\kern
3.0pt\raise
0.0pt\hbox{$\textstyle{\mathfrak{L}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$}}}}}}}\ignorespaces\ignorespaces\ignorespaces\ignorespaces{}{\hbox{\lx@xy@droprule}}\ignorespaces{\hbox{\kern
142.47897pt\raise 0.0pt\hbox{\hbox{\kern 0.0pt\raise
0.0pt\hbox{\lx@xy@tip{1}\lx@xy@tip{-1}}}}}}{\hbox{\lx@xy@droprule}}{\hbox{\lx@xy@droprule}}{\hbox{\kern
142.47897pt\raise 0.0pt\hbox{\hbox{\kern 0.0pt\raise 0.0pt\hbox{\hbox{\kern
3.0pt\raise 0.0pt\hbox{$\textstyle{0}$}}}}}}}\ignorespaces}}}}\ignorespaces.$

Since $\pi$ is surjective, there exists a linear map
$s:\mathfrak{L}\rightarrow\mathfrak{L}_{c},$ such that $\pi
s=\rm{id}_{\mathfrak{L}}.$ Then for any $x,y,z\in\mathfrak{L},$

$\displaystyle\pi(\\{s(x),s(y),s(z)\\}_{c})-\pi s(\\{x,y,z\\})=\\{\pi s(x),\pi
s(y),\pi s(z)\\}-\\{x,y,z\\}=0.$

It follows that $\\{s(x),s(y),s(z)\\}_{c}-s(\\{x,y,z\\})\in{\rm
Ker}\pi=\iota(V)\subset Z(\mathfrak{L}_{c}),$ hence, for all
$x_{1},x_{2},x_{3},x_{4},$ $x_{5}\in\mathfrak{L},$

$\displaystyle\\{\\{s(x_{1}),s(x_{2}),s(x_{3})\\}_{c},x_{4},x_{5}\\}_{c}$
$\displaystyle=\\{s(\\{x_{1},x_{2},x_{3}\\}),x_{4},x_{5}\\}_{c},$
$\displaystyle\\{x_{4},\\{s(x_{1}),s(x_{2}),s(x_{3})\\}_{c},x_{5}\\}_{c}$
$\displaystyle=\\{x_{4},s(\\{x_{1},x_{2},x_{3}\\}),x_{5}\\}_{c},$
$\displaystyle\\{x_{4},x_{5},\\{s(x_{1}),s(x_{2}),s(x_{3})\\}_{c}\\}_{c}$
$\displaystyle=\\{x_{4},x_{5},s(\\{x_{1},x_{2},x_{3}\\})\\}_{c}.$

Consider a map $g:\mathfrak{L}\times\mathfrak{L}\times\mathfrak{L}\rightarrow
V,$

$\displaystyle\iota g(x,y,z)=\\{s(x),s(y),s(z)\\}_{c}-s(\\{x,y,z\\}),$

which is well-defined, since $\iota$ is injective. Clearly, $g\in
C^{3}(\mathfrak{L},V).$ Moreover, it is straightforwardly to check
$\delta^{3}_{1}g=\delta^{3}_{2}g=0,$ and so $g\in Z^{3}(\mathfrak{L},V).$ Here
we only check that $\delta^{3}_{1}g=0$ as an example. Note that it is
sufficient to show that $\iota\delta^{3}_{1}g=0,$ since $\iota$ is injective.
In fact,

$\displaystyle\iota\delta^{3}_{1}g(x_{1},x_{2},x_{3},x_{4},x_{5})$
$\displaystyle=$
$\displaystyle\iota\Big{(}g(\\{x_{1},x_{2},x_{3}\\},x_{4},x_{5})-g(\\{x_{1},x_{2},x_{4}\\},x_{3},x_{5})-g(\\{x_{1},x_{2},x_{5}\\},x_{3},x_{4})$
$\displaystyle+g(\\{x_{1},x_{2},x_{5}\\},x_{4},x_{3})-g(x_{1},x_{2},\\{x_{3},x_{4},x_{5}\\})\Big{)}$
$\displaystyle=$
$\displaystyle\\{s(\\{x_{1},x_{2},x_{3}\\}),s(x_{4}),s(x_{5})\\}_{c}-s(\\{\\{x_{1},x_{2},x_{3}\\},x_{4},x_{5}\\})$
$\displaystyle-\\{s(\\{x_{1},x_{2},x_{4}\\}),s(x_{3}),s(x_{5})\\}_{c}+s(\\{\\{x_{1},x_{2},x_{4}\\},x_{3},x_{5}\\})$
$\displaystyle-\\{s(\\{x_{1},x_{2},x_{5}\\}),s(x_{3}),s(x_{4})\\}_{c}+s(\\{\\{x_{1},x_{2},x_{5}\\},x_{3},x_{4}\\})$
$\displaystyle+\\{s(\\{x_{1},x_{2},x_{5}\\}),s(x_{4}),s(x_{3})\\}_{c}-s(\\{\\{x_{1},x_{2},x_{5}\\},x_{4},x_{3}\\})$
$\displaystyle-\\{s(x_{1}),s(x_{2}),s(\\{x_{3},x_{4},x_{5}\\}\\}_{c}+s(\\{x_{1},x_{2},\\{x_{3},x_{4},x_{5}\\}\\})$
$\displaystyle=$
$\displaystyle\\{\\{s(x_{1}),s(x_{2}),s(x_{3})\\}_{c},s(x_{4}),s(x_{5})\\}_{c}-\\{\\{s(x_{1}),s(x_{2}),s(x_{4})\\}_{c},s(x_{3}),s(x_{5})\\}_{c}$
$\displaystyle-\\{\\{s(x_{1}),s(x_{2}),s(x_{5})\\}_{c},s(x_{3}),s(x_{4})\\}_{c}+\\{\\{s(x_{1}),s(x_{2}),s(x_{5})\\}_{c},s(x_{4}),s(x_{3})\\}_{c}$
$\displaystyle-\\{s(x_{1}),s(x_{2}),\\{s(x_{3}),s(x_{4}),s(x_{5})\\}_{c}\\}_{c}=0.$

We claim that the above arguments induce a map from
Ext${}_{c}(\mathfrak{L},V)$ to $H^{3}(\mathfrak{L},V).$ Indeed, suppose that
$(\mathfrak{L}_{c},\\{\cdot,\cdot,\cdot\\}_{c})$ and
$(\mathfrak{L}_{c}^{\prime},\\{\cdot,\cdot,\cdot\\}_{c}^{\prime})$ are
equivalent central extensions of $(\mathfrak{L},\\{\cdot,\cdot,\cdot\\})$ by
$V.$ Then we have the following diagram

$\textstyle{0\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{V\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\iota}$$\textstyle{\mathfrak{L}_{c}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\pi}$$\scriptstyle{\phi}$$\textstyle{\mathfrak{L}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{s}$$\textstyle{0}$$\textstyle{0\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{V\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\iota^{\prime}}$$\textstyle{\mathfrak{L}_{c}^{\prime}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\pi^{\prime}}$$\textstyle{\mathfrak{L}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{s^{\prime}}$$\textstyle{0}$

such that $\phi\iota=\iota^{\prime}$ and $\pi=\pi^{\prime}\phi,$ with $\phi$
being an isomorphism and $\pi
s=\pi^{\prime}s^{\prime}=\rm{id}_{\mathfrak{L}}.$ Let $g,g^{\prime}$ be their
corresponding $3$-cocycles constructed as above, i.e.,

$\displaystyle\iota g(x,y,z)=$
$\displaystyle\\{s(x),s(y),s(z)\\}_{c}-s(\\{x,y,z\\}),$
$\displaystyle\iota^{\prime}g^{\prime}(x,y,z)=$
$\displaystyle\\{s^{\prime}(x),s^{\prime}(y),s^{\prime}(z)\\}_{c}-s^{\prime}(\\{x,y,z\\}).$

Then

(3.1) $\displaystyle\iota^{\prime}g(x,y,z)=$ $\displaystyle\phi\iota
g(x,y,z)=\phi(\\{s(x),s(y),s(z)\\}_{c})-\phi s(\\{x,y,z\\}).$

Now define $f\in C^{1}(\mathfrak{L},V)$ by

$\iota^{\prime}f(x)=s^{\prime}(x)-\phi s(x),~{}\forall x\in\mathfrak{L},$

which is reasonable since Im$\iota^{\prime}=$Ker$\pi^{\prime}$ and

$\displaystyle\pi^{\prime}s^{\prime}(x)-\pi^{\prime}\phi s(x)=x-\pi s(x)=0,$

Since $s^{\prime}(x)-\phi s(x)=\iota^{\prime}f(x)\in
Z(\mathfrak{L}_{c}^{\prime}),$

(3.2)
$\displaystyle\\{s^{\prime}(x),s^{\prime}(y),s^{\prime}(z)\\}_{c}^{\prime}=\\{\phi
s(x),\phi s(y),\phi s(z)\\}_{c}^{\prime}=\phi(\\{s(x),s(y),s(z)\\}_{c}).$

Combing Eqs. (3.1) and (4.6),

$\displaystyle\iota^{\prime}(g^{\prime}-g)(x,y,z)=-\iota^{\prime}f(\\{x,y,z\\})=\iota^{\prime}(\delta^{1}f)(x,y,z).$

So $g^{\prime}-g=\delta^{1}f\in B^{3}(\mathfrak{L},V).$ Hence we give a map
from Ext${}_{c}(\mathfrak{L},V)$ to $H^{3}(\mathfrak{L},V)$ denote by
$\sigma_{1}.$

_Step 2, construct a map $\sigma_{2}$ from $H^{3}(\mathfrak{L},V)$ to
$Ext_{c}(\mathfrak{L},V).$_

Let $g\in Z^{3}(\mathfrak{L},V)$ and $\mathfrak{L}_{c}=\mathfrak{L}+V$ with

$\displaystyle\\{(x,a),(y,b),(z,c)\\}_{c}=(\\{x,y,z\\},g(x,y,z)).$

Then Remark 2.12 says that $(\mathfrak{L}_{c},\\{\cdot,\cdot,\cdot\\}_{c})$ is
a Leibniz triple system. Define $\iota:V\rightarrow\mathfrak{L}_{c}$ by
$\iota(a)=(0,a)$ and $\pi:\mathfrak{L}_{c}\rightarrow\mathfrak{L}$ by
$\pi(x,a)=x.$ Clearly, $(\mathfrak{L}_{c},\\{\cdot,\cdot,\cdot\\}_{c})$ is a
central extension of $(\mathfrak{L},\\{\cdot,\cdot,\cdot\\})$ by $V.$ Then we
have construct a map from $Z^{3}(\mathfrak{L},V)$ to
Ext${}_{c}(\mathfrak{L},V).$

Suppose $g,g^{\prime}\in Z^{3}(\mathfrak{L},V)$ and $g-g^{\prime}\in
B^{3}(\mathfrak{L},V)$, i.e., there exist $f\in C^{1}(\mathfrak{L},V)$
satisfying $(g^{\prime}-g)=\delta^{1}f.$ Then
$(g^{\prime}-g)(x,y,z)=-f(\\{x,y,z\\}).$ Let
$(\mathfrak{L}_{c},\\{\cdot,,\cdot,\cdot\\}_{c})$
and$(\mathfrak{L}_{c}^{\prime},\\{\cdot,,\cdot,\cdot\\}_{c}^{\prime})$ be two
central extensions of $(\mathfrak{L},\\{\cdot,\cdot,\cdot\\})$ by $V$ defined
as above with respect to $g$ and $g^{\prime},$ respectively. Then
$\iota(a)=\iota^{\prime}(a)=(0,a)$ and $\pi(x,a)=\pi^{\prime}(x,a)=x.$
Consider a linear map

$\displaystyle\phi:(\mathfrak{L}_{c},\\{\cdot,,\cdot,\cdot\\}_{c})$
$\displaystyle\rightarrow(\mathfrak{L}_{c}^{\prime},\\{\cdot,,\cdot,\cdot\\}_{c}^{\prime})$
$\displaystyle(x,a)$ $\displaystyle\mapsto(x,a-f(x)).$

Then $\phi\iota(a)=\iota^{\prime}(a)$ and
$\pi^{\prime}\phi(x,a)=\pi^{\prime}(x,a-f(x))=x=\pi(x,a).$ Next, we prove that
$\phi$ is an isomorphism.

If $\phi(x,a)=\phi(\tilde{x},\tilde{a}),$ it follows that
$(x,a-f(x))=(\tilde{x},\tilde{a}-f(\tilde{x})),$ that is, $x=\tilde{x}$ and
$a-f(x)=\tilde{a}-f(\tilde{x}),$ then $a=\tilde{a},$ and so $\phi$ is
injective; $\phi$ is obviously surjective. Note that

$\displaystyle\phi(\\{(x,a),(y,b),(z,c)\\}_{c})$ $\displaystyle=$
$\displaystyle\phi(\\{x,y,z\\},g(x,y,z))=(\\{x,y,z\\},g(x,y,z)-f(\\{x,y,z\\}))$
$\displaystyle=$
$\displaystyle(\\{x,y,z\\},g^{\prime}(x,y,z))=\\{(x,a-f(x)),(y,b-f(y)),(z,c-f(z))\\}_{c}^{\prime}$
$\displaystyle=$
$\displaystyle\\{\phi(x,a),\phi(y,b),\phi(z,c)\\}_{c}^{\prime}.$

Therefore, $(\mathfrak{L}_{c},\\{\cdot,\cdot,\cdot\\}_{c})$ and
$(\mathfrak{L}_{c}^{\prime},\\{\cdot,\cdot,\cdot\\}_{c}^{\prime})$ are
equivalent central extensions of $(\mathfrak{L},\\{\cdot,\cdot,\cdot\\})$ by
$V.$ Hence we find a map from $H^{3}(\mathfrak{L},V)$ to
Ext${}_{c}(\mathfrak{L},V)$ by $\sigma_{2}.$

It is obvious that $\sigma_{1}\sigma_{2}={\rm id}_{H^{3}(\mathfrak{L},V)}$ and
$\sigma_{2}\sigma_{1}={\rm id}_{{\rm Ext}_{c}(\mathfrak{L},V)}.$ We get the
assertion of the theorem. ∎

##  4\. $T^{*}$-extension of Leibniz triple systems

At the beginning of this section we will study the quadratic structure of
Leibniz triple systems, and then we introduce the theory of $T^{*}$-extension
to Leibniz triple systems, at last, we study the symplectic form of a Leibniz
triple system.

###  4.1. Quadratic Leibniz triple systems

In this subsection, we will introduce the quadratic Leibniz triple system
$(\mathfrak{L},B)$ and prove that $\mathfrak{L}$ can be decomposed into the
direct sum of non-degenerate ideals. We also give the relation between central
descending series $(C^{n}(\mathfrak{L}))_{n\geq 0}$ and central ascending
series $(C_{n}(\mathfrak{L}))_{n\geq 0}.$

######  Definition 4.1.

Let $\mathfrak{L}$ be a Leibniz triple system. A bilinear form $B$ on
$\mathfrak{L}$ is said to be symmetric if $B(x,y)=B(y,x)$, for all
$x,y\in\mathfrak{L}$. $B$ is said to be non-degenerate if
$\mathfrak{L}^{\bot}=0,$ where
$\mathfrak{L}^{\bot}=\\{x\in\mathfrak{L}~{}|~{}B(x,y)=B(y,x)=0,\forall
y\in\mathfrak{L}\\}.$

For any $x,y\in\mathfrak{L},$ denote by $R_{(x,y)},$ $M_{(x,y)}$ and
$L_{(x,y)}$ the right, middle and left multiplication operator on
$\mathfrak{L},$ respectively, i.e.,
$R_{(x,y)},~{}M_{(x,y)},~{}L_{(x,y)}:\mathfrak{L}\times\mathfrak{L}\rightarrow$
End$(\mathfrak{L})$ are defined as $R_{(x,y)}z:=\\{z,x,y\\},$
$M_{(x,y)}z:=\\{x,z,y\\}$ and $L_{(x,y)}z:=\\{x,y,z\\},$ which can be used to
characterize the invariance of $B.$

######  Definition 4.2.

A symmetric bilinear form $B$ on a Leibniz triple system $\mathfrak{L}$ is
said to be right-invariant $($resp. left-invariant or middle-invariant $)$ if

$\displaystyle B(R_{(a,b)}x,y)$ $\displaystyle=B(x,R_{(b,a)}y)$
$\displaystyle\Big{(}{\rm resp}.~{}B(L_{(a,b)}x,y)=B(x,L_{(b,a)}y),~{}$
$\displaystyle{\rm or}~{}B(M_{(a,b)}x,y)=B(x,M_{(b,a)}y)\Big{)},$

for all $x,y,a,b\in\mathfrak{L}$.

######  Remark 4.3.

If $B$ is symmetric then any two invariance of $B$ lead to the third
invariance. Here, we only check that the middle-invariant can be deduced by
the right-invariant and left-invariant as an example, which follows from for
all $a,b,x,y\in\mathfrak{L},$

$\displaystyle B(M_{(a,b)}x,y)=$ $\displaystyle
B(L_{(a,x)}b,y)=B(b,L_{(x,a)}y)=B(b,R_{(a,y)}x)$ $\displaystyle=$
$\displaystyle B(R_{(y,a)}b,x)=B(M_{(b,a)}y,x)=B(x,M_{(b,a)}y).$

######  Definition 4.4.

Let $\mathfrak{L}$ be a Leibniz triple system together with a bilinear form
$B$. We say that $(\mathfrak{L},B)$ is a quadratic Leibniz triple system if
$B$ is non-degenerate symmetric invariant. An ideal $I$ of $\mathfrak{L}$ is
called non-degenerate if $B|_{I\times I}$ is non-degenerate. A quadratic
Leibniz triple system $(\mathfrak{L},B)$ is said to be $B$-irreducible if
$\mathfrak{L}$ contains no nontrivial non-degenerate ideal.

######  Lemma 4.5.

Let $(\mathfrak{L},B)$ be a quadratic Leibniz triple system.

$(1)$ If $\mathfrak{L}$ is finite dimensional and $I$ is a subspace of
$\mathfrak{L},$ then $\mathfrak{L}=I\dot{+}I^{\bot},$ $(I^{\bot})^{\bot}=I.$

$(2)$ If $I$ is an ideal of $\mathfrak{L},$ then $I^{\bot}$ is an ideal of
$\mathfrak{L}.$

$(3)$ If $\mathfrak{L}$ is finite dimensional and $I$ is a non-degenerate
subspace of $\mathfrak{L},$ then $I^{\bot}$ is non-degenerate.

###### Proof.

(1) Let $\\{x_{1},x_{2},...,x_{l}\\}$ be a basis of $I$ and extend it to a
basis of $\mathfrak{L}$ say $\\{x_{1},x_{2},...,x_{n}\\}.$ Then $x\in
I^{\bot},$ if and only if $B(x_{i},x)=0,$ $1\leq i\leq l.$ Suppose
$x=\sum\limits_{i=1}^{n}k_{i}x_{i}.$ Then $B(x_{i},x)=0,$ if and only if
$B(x_{i},\sum\limits_{j=1}^{n}k_{j}x_{j})=\sum\limits_{j=1}^{n}B(x_{i},x_{j})k_{j}=0,$
$1\leq i\leq l,$ i.e.,

$\left(\begin{array}[]{cccc}B(x_{1},x_{1})&\cdots&B(x_{1},x_{n})\\\
\vdots&&\vdots\\\ B(x_{l},x_{1})&\cdots&B(x_{l},x_{n})\\\
\end{array}\right)\left(\begin{array}[]{cccc}k_{1}\\\ \vdots\\\ k_{n}\\\
\end{array}\right)=\left(\begin{array}[]{cccc}0\\\ \vdots\\\ 0\\\
\end{array}\right),$

which implies that $I^{\bot}$ is isomorphism to
$\\{X\in\mathbb{F}^{n}~{}|~{}AX=0\\}$ as vector space. Then
dim$I^{\bot}=n-$rank$A.$ Note that rank$A=l=$dim$I$ and $I\cap I^{\bot}=0$ by
non-degenerate of $B.$ It follows that $\mathfrak{L}=I\dot{+}I^{\bot}.$
$(I^{\bot})^{\bot}=I$ follows from the facts that $I\subset(I^{\bot})^{\bot}$
and dim$I=$dim$(I^{\bot})^{\bot}=$dim$\mathfrak{L}-$dim$I^{\bot}.$

(2) It follows the invariance of $B.$

(3) Follows from the non-degenerate of $B$ and $B|_{I\times I}$ and
$B(I,I^{\bot})=0.$ ∎

######  Proposition 4.6.

Let $(\mathfrak{L},B)$ be a finite dimensional quadratic Leibniz triple
system. Then $\mathfrak{L}=\oplus_{i=1}^{r}\mathfrak{L}_{i}$ such that for all
$1\leq i\leq r,$ we have

$(1)$ $\mathfrak{L}_{i}$ is a non-degenerate ideal;

$(2)$ $\mathfrak{L}_{i}$ is $B$-irreducible.

###### Proof.

By induction on dim$\mathfrak{L}=n$ and using the facts that
$\mathfrak{L}=I\dot{+}I^{\bot}$ and any ideal of $I$ is also an ideal of
$\mathfrak{L},$ the proposition follows.

If $n=1,$ the assertion is true.

Assume that dim$\mathfrak{L}<n$ is true.

When dim$\mathfrak{L}=n,$ we will do it in two cases, if $\mathfrak{L}$ is
$B$-irreducible, then the assertion is true; if not, let $I$ be a
$B$-irreducible ideal of $\mathfrak{L}.$ By Lemma 4.5,
$\mathfrak{L}=I\dot{+}I^{\bot}.$ Applying the induction hypothesis to $I$ and
$I^{\bot}.$ This ends the proof of the proposition. ∎

Now we are ready to study central descending series
$(C^{n}(\mathfrak{L}))_{n\geq 0}$ and central ascending series
$(C_{n}(\mathfrak{L}))_{n\geq 0}.$ We will prove that
$(C^{r}(\mathfrak{L}))^{\bot}=C_{r}(\mathfrak{L})$ for a quadratic Leibniz
triple system $(\mathfrak{L},B).$ Also we conclude that the centralizer
$Z_{\mathfrak{L}}(I)$ of a quadratic Leibniz triple system $(\mathfrak{L},B)$
contains $I^{\bot}$ if and only if $I$ is an ideal of $\mathfrak{L}.$ This
work is inspired by [14].

######  Definition 4.7.

Let $\mathfrak{L}$ be a Leibniz triple system, the central descending series
$(C^{n}(\mathfrak{L}))_{n\geq 0}$ is defined by
$C^{0}(\mathfrak{L}):=\mathfrak{L}$ and
$C^{n+1}(\mathfrak{L}):=\\{C^{n}(\mathfrak{L}),\mathfrak{L},\mathfrak{L}\\}+\\{\mathfrak{L},C^{n}(\mathfrak{L}),\mathfrak{L}\\}+\\{\mathfrak{L},\mathfrak{L},C^{n}(\mathfrak{L})\\}.$
And the central ascending series $(C_{n}(\mathfrak{L}))_{n\geq 0}$ is defined
by $C_{0}(\mathfrak{L}):=0$ and
$C_{n+1}(\mathfrak{L}):=C(C_{n}(\mathfrak{L}))$ where
$C(C_{r}(\mathfrak{L})):=\\{x\in\mathfrak{L}~{}|~{}\\{x,\mathfrak{L},\mathfrak{L}\\}+\\{\mathfrak{L},x,\mathfrak{L}\\}+\\{\mathfrak{L},\mathfrak{L},x\\}\subseteq
C_{r}(\mathfrak{L})\\}.$

######  Theorem 4.8.

Let $(\mathfrak{L},B)$ be a quadratic Leibniz triple system. Then for all
integer $r\geq 0,$ we have $(C^{r}(\mathfrak{L}))^{\bot}=C_{r}(\mathfrak{L}).$

###### Proof.

We proceed by induction on $r.$

If $r=0,$
$(C^{0}(\mathfrak{L}))^{\bot}=\mathfrak{L}^{\bot}=0=C_{0}(\mathfrak{L}),$ the
equality holds.

Suppose the theorem is true for $r$ and we will show that it is true for
$r+1.$ Let $w\in C_{r+1}(\mathfrak{L}).$ Note that $u\in
C^{r+1}(\mathfrak{L})$ is a finite sum
$u=\Sigma\Big{(}\\{x_{i},y_{i},z_{i}\\}+\\{y_{i}^{\prime},x_{i}^{\prime},z_{i}^{\prime}\\}+\\{y_{i}^{\prime\prime},z_{i}^{\prime\prime},x_{i}^{\prime\prime}\\}\Big{)},$
for all $x_{i},x_{i}^{\prime},x_{i}^{\prime\prime}\in C^{r}(\mathfrak{L})$ and
$y_{i},z_{i},y_{i}^{\prime},z_{i}^{\prime},y_{i}^{\prime\prime},z_{i}^{\prime\prime}\in\mathfrak{L}.$
Then

$\displaystyle B(u,w)=$ $\displaystyle\Sigma
B(\\{x_{i},y_{i},z_{i}\\}+\\{y_{i}^{\prime},x_{i}^{\prime},z_{i}^{\prime}\\}+\\{y_{i}^{\prime\prime},z_{i}^{\prime\prime},x_{i}^{\prime\prime}\\},w)$
$\displaystyle=$ $\displaystyle\Sigma B(\\{x_{i},y_{i},z_{i}\\},w)+\Sigma
B(\\{y_{i}^{\prime},x_{i}^{\prime},z_{i}^{\prime}\\},w)+\Sigma
B(\\{y_{i}^{\prime\prime},z_{i}^{\prime\prime},x_{i}^{\prime\prime}\\},w)$
$\displaystyle=$ $\displaystyle\Sigma B(x_{i},\\{w,z_{i},y_{i}\\})+\Sigma
B(x_{i}^{\prime},\\{z_{i}^{\prime},w,y_{i}^{\prime}\\})+\Sigma
B(x_{i}^{\prime\prime},\\{z_{i}^{\prime\prime},y_{i}^{\prime\prime},w\\}).$

Since $(C^{r}(\mathfrak{L}))^{\bot}=C_{r}(\mathfrak{L})$ and

$\displaystyle\\{w,z_{i},y_{i}\\}$
$\displaystyle\in\\{C_{r+1}(\mathfrak{L}),\mathfrak{L},\mathfrak{L}\\}\subseteq
C_{r}(\mathfrak{L}),$ $\displaystyle\\{z_{i}^{\prime},w,y_{i}^{\prime}\\}$
$\displaystyle\in\\{\mathfrak{L},C_{r+1}(\mathfrak{L}),\mathfrak{L}\\}\subseteq
C_{r}(\mathfrak{L}),$
$\displaystyle\\{z_{i}^{\prime\prime},y_{i}^{\prime\prime},w\\}$
$\displaystyle\in\\{\mathfrak{L},\mathfrak{L},C_{r+1}(\mathfrak{L})\\}\subseteq
C_{r}(\mathfrak{L}),$

we have
$B(x_{i},\\{w,z_{i},y_{i}\\})=B(x_{i}^{\prime},\\{z_{i}^{\prime},w,y_{i}^{\prime}\\})=B(x_{i}^{\prime\prime},\\{z_{i}^{\prime\prime},y_{i}^{\prime\prime},w\\})=0,$
hence $B(u,w)=0.$ That means
$C_{r+1}(\mathfrak{L})\subseteq(C^{r+1}(\mathfrak{L}))^{\bot}.$

Conversely, for any $x\in(C^{r+1}(\mathfrak{L}))^{\bot}$ and for any
$y,z\in\mathfrak{L},$ $w\in C^{r}(\mathfrak{L}),$ since $\\{w,z,y\\},$
$\\{z,w,y\\},$ $\\{z,y,w\\}\in C^{r+1}(\mathfrak{L}),$ we have

$\displaystyle B(\\{x,y,z\\},w)$ $\displaystyle=B(x,\\{w,z,y\\})=0,$
$\displaystyle B(\\{y,x,z\\},w)$ $\displaystyle=B(x,\\{z,w,y\\})=0,$
$\displaystyle B(\\{y,z,x\\},w)$ $\displaystyle=B(x,\\{z,y,w\\})=0,$

then
$\\{x,y,z\\},\\{y,x,z\\},\\{y,z,x\\}\in(C^{r}(\mathfrak{L}))^{\bot}=C_{r}(\mathfrak{L}),$
i.e., $x\in C_{r+1}(\mathfrak{L}),$ that means
$(C^{r+1}(\mathfrak{L}))^{\bot}\subset C_{r+1}(\mathfrak{L}).$ This ends the
proof of the theorem. ∎

######  Definition 4.9.

[11] Let $\mathfrak{L}$ be a Leibniz triple system and $I$ a non-empty subset
of $\mathfrak{L}.$ We call

$\displaystyle
Z_{\mathfrak{L}}(I)\\!=\\!\\{x\in\mathfrak{L}~{}|~{}\\{x,I,\mathfrak{L}\\}=\\{x,\mathfrak{L},I\\}=\\{\mathfrak{L},x,I\\}=\\{\mathfrak{L},I,x\\}=\\{I,x,\mathfrak{L}\\}=\\{I,\mathfrak{L},x\\}=0\\}$

the centralizer of $I$ in $\mathfrak{L}$.

Combing Definition 4.9 and Lemma 4.5, one has

######  Proposition 4.10.

For a finite dimensional quadratic Leibniz triple system $(\mathfrak{L},B),$ a
subspace $I$ of $\mathfrak{L}$ is an ideal if and only if
$Z_{\mathfrak{L}}(I)$ contains $I^{\bot}.$

###  4.2. $T^{*}$-extension of a quadratic Leibniz triple system

In this section, we give the method of $T^{*}$-extension for a Leibniz triple
system $\mathfrak{L}.$ We will prove that every $2n$-dimensional Leibniz
triple system $\mathfrak{L}$ is isomorphic to a $T^{*}$-extension of another
Leibniz triple system if $\mathfrak{L}$ contains an $n$-dimensional isotropic
ideal.

######  Lemma 4.11.

Let $(\mathfrak{L},\\{\cdot,\cdot,\cdot\\})$ be a Leibniz triple system and
$\mathfrak{L}^{*}$ the dual space of $\mathfrak{L}.$ Consider a trilinear map
$\theta:\mathfrak{L}\times\mathfrak{L}\times\mathfrak{L}\rightarrow\mathfrak{L}^{*}$
and the linear space $\mathfrak{L}\dot{+}\mathfrak{L}^{*}$ endowed with the
triple product

(4.1)
$\displaystyle\\{x+f,y+g,z+h\\}_{\theta}=\\{x,y,z\\}+\theta(x,y,z)+\\{f,y,z\\}+\\{x,g,z\\}+\\{x,y,h\\},$

for all $x,y,z\in\mathfrak{L}$ and $f,g,h\in\mathfrak{L}^{*},$ $(f,g,h)$ is a
dual representation of $\mathfrak{L}.$ The linear space
$\mathfrak{L}\dot{+}\mathfrak{L}^{*}$ endowed with
$\\{\cdot,\cdot,\cdot\\}_{\theta}$ is a Leibniz triple system if and only if
the map $\theta$ is a $3$-cocycle.

###### Proof.

Example 2.6 says that $\mathfrak{L}\dot{+}\mathfrak{L}^{*}$ is a Leibniz
triple system for $\theta=0$. Then one observes that
$\mathfrak{L}\dot{+}\mathfrak{L}^{*}$ in the assertion becomes a Leibniz
triple system if and only if $\theta$ satisfies

(4.2) $\displaystyle\begin{aligned}
&\theta(x,y,\\{z,u,v\\})+\\{x,y,\theta(z,u,v)\\}\\\
=&\theta(\\{x,y,z\\},u,v)-\theta(\\{x,y,u\\},z,v)-\theta(\\{x,y,v\\},z,u)+\theta(\\{x,y,v\\},u,z)\\\
&+\\{\theta(x,y,z),u,v\\}-\\{\theta(x,y,u),z,v\\}-\\{\theta(x,y,v),z,u\\}+\\{\theta(x,y,v),u,z\\},\end{aligned}$

and

(4.3) $\displaystyle\begin{aligned}
&\theta(x,\\{y,z,u\\},v\\})+\\{x,\theta(y,z,u),v\\}\\\
=&\theta(\\{x,y,z\\},u,v)-\theta(\\{x,z,y\\},u,v)-\theta(\\{x,u,y\\},z,v)+\theta(\\{x,u,z\\},y,v)\\\
&+\\{\theta(x,y,z),u,v\\}-\\{\theta(x,z,y),u,v\\}-\\{\theta(x,u,y)z,v\\}+\\{\theta(x,u,z),y,v\\},\end{aligned}$

i.e., $\theta$ is a $3$-cocycle. ∎

######  Definition 4.12.

Let $\mathfrak{L},$ $\mathfrak{L}^{*},$ $\theta$ and
$\\{\cdot,\cdot,\cdot\\}_{\theta}$ be as above. Then the Leibniz triple system
$\mathfrak{L}\dot{+}\mathfrak{L}^{*}$ endowed with
$\\{\cdot,\cdot,\cdot\\}_{\theta}$ is called a $T^{*}$-extension of
$\mathfrak{L},$ denoted by $T^{*}_{\theta}\mathfrak{L}.$ In particular, when
$\theta=0,$ we denote it by $T^{*}\mathfrak{L}$ called the trivial
$T^{*}$-extension of $\mathfrak{L}.$

######  Remark 4.13.

For any $T^{*}$-extension of $\mathfrak{L},$ it is easy to check that
$\mathfrak{L}^{*}$ is an ideal of $T^{*}_{\theta}\mathfrak{L}$ and
$\mathfrak{L}\simeq T^{*}_{\theta}\mathfrak{L}/\mathfrak{L}^{*}.$

######  Proposition 4.14.

Consider a bilinear form $B$ on $\mathfrak{L}\dot{+}\mathfrak{L}^{*}$ by

(4.4) $\displaystyle B(x+f,y+g)=g(x)+f(y),$

for all $x,y\in\mathfrak{L},$ $f,g\in\mathfrak{L}^{*},$ if

(4.5) $\displaystyle\theta(x,y,z)(u)=\theta(y,x,u)(z)=\theta(u,z,y)(x),$

for all $x,y,z,u\in\mathfrak{L},$ then $B$ is a non-degenerate symmetric
invariant bilinear form on $\mathfrak{L}\dot{+}\mathfrak{L}^{*}.$ The pair
$(\mathfrak{L}\dot{+}\mathfrak{L}^{*},\\{\cdot,\cdot,\cdot\\}_{\theta},B)$ is
a quadratic Leibniz triple system.

###### Proof.

It is sufficiency to prove the symmetric, non-degenerate and invariance of
$B.$

For the left-invariance of $B,$ note that, for any $x,y,z,u\in\mathfrak{L}$
and $f,g,h,l\in\mathfrak{L}^{*},$

$\displaystyle B(L_{(x+f,y+g)}(z+h),u+l)=B(\\{x+f,y+g,z+h\\}_{\theta},u+l)$
$\displaystyle=$ $\displaystyle
B(\\{x,y,z\\}+\theta(x,y,z)+\\{f,y,z\\}+\\{x,g,z\\}+\\{x,y,h\\},u+l)$
$\displaystyle=$ $\displaystyle
l(\\{x,y,z\\})+\theta(x,y,z)(u)+\\{f,y,z\\}(u)+\\{x,g,z\\}(u)+\\{x,y,h\\}(u)$
$\displaystyle=$ $\displaystyle
l(\\{x,y,z\\})+f(\\{u,z,y\\})+g(\\{z,u,x\\})+h(\\{y,x,u\\})+\theta(x,y,z)(u),$
$\displaystyle B(z+h,L_{(y+g,x+f)}(u+l))=B(z+h,\\{y+g,x+f,u+l\\}_{\theta})$
$\displaystyle=$ $\displaystyle
B(z+h,\\{y,x,u\\}+\theta(y,x,u)+\\{g,x,u\\}+\\{y,f,u\\}+\\{y,x,l\\})$
$\displaystyle=$ $\displaystyle
h(\\{y,x,u\\})+\\{g,x,u\\}(z)+\\{y,f,u\\}(z)+\\{y,x,l\\}(z)+\theta(y,x,u)(z)$
$\displaystyle=$ $\displaystyle
h(\\{y,x,u\\})+g(\\{z,u,x\\})+f(\\{u,z,y\\})+l(\\{x,y,z\\})+\theta(y,x,u)(z),$

which coincide Eq. (4.5). Similarly, one can prove that right-invariance of
$B.$ So we deduce the invariance of $B$.

As for the non-degenerate of $B$, let
$x+f\in(\mathfrak{L}\dot{+}\mathfrak{L}^{*})^{\bot}.$ For any
$y\in\mathfrak{L}$ we have $B(x+f,y+0)=f(y)+0(x)=0,$ which implies $f=0.$ For
any $g\in\mathfrak{L}^{*},$ $B(x+f,0+g)=f(0)+g(x)=0,$ we have $x=0,$ hence $B$
is non-degenerate. ∎

Now, we are ready to coincide that a quadratic Leibniz triple system
$(\mathfrak{L},B)$ and $T^{*}$-extension of a Leibniz triple system.

######  Lemma 4.15.

Let $(\mathfrak{L},B)$ be a quadratic Leibniz triple system of $2n$-dimension
and $I$ an isotropic $({\rm i.e.},I\subseteq I^{\bot})$ $n$-dimensional
subspace of $\mathfrak{L}.$ Then $I$ is an ideal of $\mathfrak{L}$ if and only
if $I$ is abelian ideal of $\mathfrak{L}$, that is
$\\{I,\mathfrak{L},\mathfrak{L}\\}=\\{\mathfrak{L},I,\mathfrak{L}\\}=\\{\mathfrak{L},\mathfrak{L},I\\}=0.$

###### Proof.

Note that dim$I$ \+ dim$I^{\bot}$ = $n$ \+ dim$I^{\bot}$ = $2n,$
dim$I^{\bot}=$ dim$I=n$ and $I\subset I^{\bot},$ we have $I=I^{\bot}.$

If $I$ is an ideal of $\mathfrak{L},$ then by right-invariance of $B$

$B(\mathfrak{L},\\{I,I,\mathfrak{L}\\})=B(\\{\mathfrak{L},\mathfrak{L},I\\},I)\subset
B(I,I)=B(I,I^{\bot})=0,$

which implies $\\{I,I,\mathfrak{L}\\}\subset\mathfrak{L}^{\bot}=0.$ Similarly,
$\\{I,\mathfrak{L},I\\}=\\{\mathfrak{L},I,I\\}=0.$

Conversely, it follows from

$B(I,\\{I,\mathfrak{L},\mathfrak{L}\\})=B(\\{\mathfrak{L},I,I\\},I)=B(0,I)=0$

that $\\{I,\mathfrak{L},\mathfrak{L}\\}\subset I^{\bot}=I.$ Similarly,
$\\{\mathfrak{L},I,\mathfrak{L}\\}\subset I$ and
$\\{\mathfrak{L},\mathfrak{L},I\\}\subset I.$ This implies that $I$ is an
ideal of $\mathfrak{L}.$ ∎

######  Theorem 4.16.

Let $(\mathfrak{L},B)$ be a quadratic Leibniz triple system of $2n$-dimension.
Then $(\mathfrak{L},B)$ is isometric to a $T^{*}$-extension
$(T^{*}_{\theta}J,B_{J})$ if and only if $\mathfrak{L}$ contains an isotropic
ideal $I$ of $n$-dimensional. Furthermore, $J$ is isomorphic to
$\mathfrak{L}/I.$

###### Proof.

The necessity is clear by Remark 4.13, since $J^{*}$ is an ideal of
$T^{*}_{\theta}J$ and $J\simeq T^{*}_{\theta}J/J^{*}$ and $B(J^{*},J^{*})=0,$
$J^{*}\subset(J^{*})^{\bot},$ i.e., $J^{*}$ is isotropic.

For the sufficiency, let $I$ be an $n$-dimensional isotropic ideal of
$\mathfrak{L}.$ Suppose $J=\mathfrak{L}/I$ and $p:\mathfrak{L}\rightarrow J$
be the canonical projection. We will show that $\mathfrak{L}$ is some
$T^{*}$-extension of $J$ by two steps.

_Step 1, construct a $T^{*}$-extension $T^{*}_{\theta}J$ of $J.$_

First, by the non-degenerate of $B$ we can find an isotropic complement
subspace $J_{0}$ to $I$ in $\mathfrak{L},$ i.e., $\mathfrak{L}=J_{0}\dot{+}I$
and $J_{0}\subset J_{0}^{\bot}.$ Then dim$\mathfrak{L}-$dim$I=n$ and by Lemma
4.15. we have $J_{0}^{\bot}=J_{0}.$

Denote $p_{0}$ $($resp. $p_{1})$ the projection $\mathfrak{L}\rightarrow
J_{0}$ $($resp. $\mathfrak{L}\rightarrow I)$ and define a linear map

$\displaystyle\mu^{B}:I$ $\displaystyle\rightarrow J^{*}$ $\displaystyle i$
$\displaystyle\mapsto\mu^{B}(i):\mu^{B}(i)(p(x)):=B(i,x).$

It is easily seen that $\mu^{B}$ is well-defined, since for $p(x)=p(y)$ then
$x-y\in I,$ hence $B(i,x-y)\in B(i,I)=0,$ so $B(i,x)=B(i,y),$ which implies
$\mu^{B}$ is well-defined and it is easily seen that $\mu^{B}$ is linear. If
$\mu^{B}(i)=\mu^{B}(j)$ then
$\mu^{B}(i)(p(x))-\mu^{B}(j)(p(x))=B(i,x)-B(j,x),$ for all $x\in\mathfrak{L},$
which implies $i-j\in\mathfrak{L}^{\bot}=0,$ hence $\mu^{B}$ is injective. As
dim$I=$ dim$J^{*},$ $\mu^{B}$ is also surjective.

In addition, $\mu^{B}$ has the following properties:

$\displaystyle\mu^{B}(\\{a_{1},a_{2},i\\})(pa_{3})=$ $\displaystyle
B(\\{a_{1},a_{2},i\\},a_{3})=B(i,\\{a_{2},a_{1},a_{3}\\})$ $\displaystyle=$
$\displaystyle\mu^{B}(i)(\\{pa_{2},pa_{1},pa_{3}\\})=\\{pa_{1},pa_{2},\mu^{B}(i)\\}(pa_{3}),$

i.e., $\mu^{B}(\\{a_{1},a_{2},i\\})=\\{pa_{1},pa_{2},\mu^{B}(i)\\},$ for any
$a_{1},a_{2},a_{3}\in\mathfrak{L}$ and $i\in I.$ Further, a similar
computation shows that,

$\displaystyle\mu^{B}(\\{i,a_{1},a_{2}\\})=$
$\displaystyle\\{\mu^{B}(i),pa_{1},pa_{2}\\}$
$\displaystyle\mu^{B}(\\{a_{1},i,a_{2}\\})=$
$\displaystyle\\{pa_{1},\mu^{B}(i),pa_{2}\\}.$

Define a trilinear map

$\displaystyle\theta:J\times J\times J$ $\displaystyle\rightarrow J^{*}$
$\displaystyle(p(b_{1}),p(b_{2}),p(b_{3}))$
$\displaystyle\mapsto\mu^{B}(p_{1}(\\{b_{1},b_{2},b_{3}\\})),$

for all $b_{1},b_{2},b_{3}\in J_{0}.$ Then $\theta$ is well-defined, since
$p|_{J_{0}}:J_{0}\rightarrow J_{0}/I\simeq\mathfrak{L}/I=J$ is a linear
isomorphism and $\theta\in C^{3}(J,J^{*})$ it is easy to check that $\theta\in
Z^{3}(J,J^{*}).$ Now, define the bracket on $J\dot{+}J^{*}$ by (4.1). Then
$J\dot{+}J^{*}$ is a Leibniz triple system.

_Step 2, construct a isomorphism from $\mathfrak{L}$ to $J\dot{+}J^{*}.$_

Let $\phi:\mathfrak{L}\rightarrow J\dot{+}J^{*}$ be a linear map, via
$\phi(b_{0}+i)=pb_{0}+\mu^{B}(i),$ for any $b_{0}\in J_{0},i\in I.$ Since the
restriction of $p$ to $J_{0}$ and $\mu^{B}$ are linear isomorphisms, the map
$\phi$ is also a linear isomorphism. Note that for any $b_{1},b_{2},b_{3}\in
J_{0}$ and $i_{1},i_{2},i_{3}\in I,$ we have

$\displaystyle\phi(\\{b_{1}+i_{1},b_{2}+i_{2},b_{3}+i_{3}\\})$
$\displaystyle=$
$\displaystyle\phi\Big{(}\\{b_{1},b_{2},b_{3}\\}+\\{b_{1},b_{2},i_{3}\\}+\\{i_{1},b_{2},b_{3}\\}+\\{b_{1},i_{2},b_{3}\\}\Big{)}$
$\displaystyle=$
$\displaystyle\phi\Big{(}p_{0}(\\{b_{1},b_{2},b_{3}\\})+p_{1}(\\{b_{1},b_{2},b_{3}\\})+\\{b_{1},b_{2},i_{3}\\}+\\{i_{1},b_{2},b_{3}\\}+\\{b_{1},i_{2},b_{3}\\}\Big{)}$
$\displaystyle=$ $\displaystyle
p(p_{0}(\\{b_{1},b_{2},b_{3}\\}))+\mu^{B}\Big{(}p_{1}(\\{b_{1},b_{2},b_{3}\\})+\\{b_{1},b_{2},i_{3}\\}+\\{i_{1},b_{2},b_{3}\\}+\\{b_{1},i_{2},b_{3}\\}\Big{)}$
$\displaystyle=$ $\displaystyle
p(\\{b_{1},b_{2},b_{3}\\})+\Omega(pb_{1},pb_{2},pb_{3})+\\{pb_{1},pb_{2},\mu^{B}(i_{3})\\}+\\{\mu^{B}(i_{1}),pb_{2},pb_{3}\\}+\\{pb_{1},\mu^{B}(i_{2}),pb_{3}\\}$
$\displaystyle=$
$\displaystyle\\{pb_{1}+\mu^{B}(i_{1}),pb_{2}+\mu^{B}(i_{2}),pb_{3}+\mu^{B}(i_{3})\\}$
$\displaystyle=$
$\displaystyle\\{\phi(b_{1}+i_{1}),\phi(b_{2}+i_{2}),\phi(b_{3}+i_{3})\\},$

where we have made use of the definition of $\phi$ and $\theta,$ and the above
properties of $\mu^{B}.$ Then $\phi$ is a Leibniz triple system isomorphism.
Furthermore, we have

$\displaystyle B_{J}(\phi(x+i),\phi(x^{\prime}+i^{\prime}))=$ $\displaystyle
B_{J}(p(x)+\mu^{B}(i),p(x^{\prime})+\mu^{B}(i^{\prime}))$ $\displaystyle=$
$\displaystyle\mu^{B}(i^{\prime})(p(x))+\mu^{B}(i)(p(x^{\prime}))$
$\displaystyle=$ $\displaystyle B(i^{\prime},x)+B(i,x^{\prime})$
$\displaystyle=$ $\displaystyle B(x+i,x^{\prime}+i^{\prime}).$

Then $\phi$ is isometric. We also have $B_{J}$ is non-degenerate following
from the non-degenerate property of $B.$ So $(J\dot{+}J^{*},B_{J})$ is a
quadratic Leibniz triple system. In this way, we get a $T^{*}$-extension
$T^{*}_{\theta}J$ of $J,$ and consequently, $(\mathfrak{L},B)$ and
$(T^{*}_{\theta}J,B_{J})$ are isometric as required. ∎

###  4.3. Symplectic form of a Leibniz triple system

In this subsection, we will introduce the symplectic form of a Leibniz triple
system, and we give a necessary and sufficient condition for a quadratic
Leibniz triple system to admit a symplectic form.

######  Definition 4.17.

Let $(\mathfrak{L},\\{\cdot,\cdot,\cdot\\})$ be a Leibniz triple system. A
symplectic form is a non-degenerate skewsymmetric bilinear form $\omega$
satisfying

(4.6)
$\displaystyle\omega(u,\\{x,y,z\\})+\omega(x,\\{u,z,y\\})+\omega(y,\\{z,u,x\\})+\omega(z,\\{y,x,u\\})=0,$

for all $x,y,z,u\in\mathfrak{L}.$ A Leibniz triple system endowed with a
symplectic form is called symplectic Leibniz triple system, denoted by
$(\mathfrak{L},\omega).$

######  Definition 4.18.

A symmetric $($resp. skewsymmetric$)$ linear map $f$ on $\mathfrak{L}$ with
respect to $B$ satisfying $B(f(x),y)=B(x,f(y))$ $({\rm
resp}.~{}B(f(x),y)=-B(x,f(y))),$ for all $x,y\in\mathfrak{L}.$

######  Theorem 4.19.

Let $(\mathfrak{L},B)$ be a quadratic Leibniz triple system. There is a
symplectic form $\omega$ on $\mathfrak{L}$ if and only if there is an
invertible derivation of $\mathfrak{L}$ which is skewsymmetric with respect to
$B.$

###### Proof.

For the sufficiency, let $D$ be an invertible and skewsymmetric derivation on
$\mathfrak{L}$ with respect to $B.$ Define $\omega(x,y)=B(D(x),y).$ In the
following we prove that $\omega$ is skewsymmetric, non-degenerate and
satisfies Eq. (4.6). Since

$\omega(x,y)=B(D(x),y)=-B(x,D(y))=-B(D(y),x)=-\omega(y,x),$

and $D$ is invertible, $B$ is non-degenerate, we have $\omega$ is
skewsymmetric and non-degenerate. Furthermore, by the skewsymmetric of $D$
with respect to $B$ and the invariance of $B,$

$\displaystyle\omega(u,\\{x,y,z\\})=$ $\displaystyle B(D(u),\\{x,y,z\\})$
$\displaystyle=$ $\displaystyle-B(u,D(\\{x,y,z\\}))$ $\displaystyle=$
$\displaystyle-B(u,\\{D(x),y,z\\})-B(u,\\{x,D(y),z\\})-B(u,\\{x,y,D(z)\\})$
$\displaystyle=$
$\displaystyle-B(\\{u,z,y\\},D(x))-B(\\{z,u,x\\},D(y))-B(\\{y,x,u\\},D(z))$
$\displaystyle=$
$\displaystyle-B(D(x),\\{u,z,y\\})-B(D(y),\\{z,u,x\\})-B(D(z),\\{y,x,u\\})$
$\displaystyle=$
$\displaystyle-\omega(x,\\{u,z,y\\})-\omega(y,\\{z,u,x\\})-\omega(z,\\{y,x,u\\}),$

then we have

$\displaystyle\omega(u,\\{x,y,z\\})+\omega(x,\\{u,z,y\\})+\omega(y,\\{z,u,x\\})+\omega(z,\\{y,x,u\\})=0,$

for all $x,y,z,u\in\mathfrak{L}.$ So $\omega$ is a symplectic form.

For the necessity, if $\omega$ is a symplectic form on $\mathfrak{L}.$ Define
$D$ by $\omega(x,y)=B(D(x),y),$ for all $x,y\in\mathfrak{L},$ which is well-
defined since the non-degenerate of $B$ and $B,$ $\omega$ are bilinear form
deduce that $D$ is a linear map. By the definition of $\omega$ we have

$\displaystyle
B(D(u),\\{x,y,z\\})+B(D(x),\\{u,z,y\\})+B(D(y),\\{z,u,x\\})+B(D(z),\\{y,x,u\\})=0.$

Since $D$ is skewsymmetric, $B$ is invariant and symmetric,

$\displaystyle-B(D(\\{x,y,z\\}),u)+B(\\{D(x),y,z\\},u\\})+B(\\{x,D(y),z\\},u\\})+B(\\{x,y,D(z)\\},u)=0,$

i.e.,

$\displaystyle
B(D(\\{x,y,z\\}),u)=B(\\{D(x),y,z\\},u\\})+B(\\{x,D(y),z\\},u\\})+B(\\{x,y,D(z)\\},u).$

Again $B$ is non-degenerate $D$ becomes a derivation on $\mathfrak{L}.$ $D$ is
skewsymmetric, since

$\displaystyle B(D(x),y)=\omega(x,y)=-\omega(y,x)=-B(D(y),x)=-B(x,D(y)).$

For the invertible of $D,$ define $D^{\prime}$ by
$\omega(D^{\prime}(x),y)=B(x,y)$ for any $x,y\in\mathfrak{L},$ which is well-
defined since the non-degenerate of $B,$ note that

$\displaystyle\omega(D^{\prime}D(x),y)=B(D(x),y)=\omega(x,y),$ $\displaystyle
B(DD^{\prime}(x),y)=\omega(D^{\prime}(x),y)=B(x,y),$

it follows that $D^{\prime}D=DD^{\prime}=$id. Thus $D$ is invertible. So we
have finished the proof of this theorem. ∎

######  Remark 4.20.

The skewsymmetric derivation $D$ of a quadratic Leibniz triple system
$(\mathfrak{L},B)$ in the above theorem is also skewsymmetric with respect to
the symplectic form $\omega.$ In fact, for any $x,y\in\mathfrak{L},$

$\displaystyle\omega(D(x),y)=B(D^{2}(x),y)=-B(D(x),D(y))=-\omega(x,D(y)).$

######  Corollary 4.21.

Let $T^{*}\mathfrak{L}$ be a $T^{*}$-extension of a Leibniz triple system
$\mathfrak{L}$ admitting a non-degenerate symmetric invariant bilinear form
$B$ defined by Eq. $(\ref{224.55})$, if $T^{*}\mathfrak{L}$ has an invertible
skewsymmetric derivation $D,$ then there exists a symplectic form for
$T^{*}\mathfrak{L}.$

By the conclusion of Lemma 4.21, we give the following example for a
symplectic Leibniz triple system.

######  Example 4.22.

Let $(\mathfrak{L},\\{\cdot,\cdot,\cdot\\}_{\mathfrak{L}})$ be a Leibniz
triple system. Consider the $G_{n}=t\mathbb{F}[t]/t^{n}\mathbb{F}[t],$ where
$n$ is a nonzero positive integer. Define the product on
$\mathfrak{L}_{n}=\mathfrak{L}\otimes G_{n},$ as follows,

$\displaystyle\\{x\otimes\overline{t^{p}},y\otimes\overline{t^{q}},z\otimes\overline{t^{r}}\\}=\\{x,y,z\\}_{\mathfrak{L}}\otimes\overline{t^{p+q+r}},$

where $x,y,z\in\mathfrak{L},$ $p,q,r\in\mathbb{N}/\\{0\\}.$ Then
$\mathfrak{L}_{n}$ is a nilpotent Leibniz triple system. By the universal
property of tensor product define a linear map $D$ on $\mathfrak{L}_{n}$ by
$D(x\otimes\overline{t^{p}})=p(x\otimes\overline{t^{p}}),$ for any
$x\in\mathfrak{L},$ and $p\in\\{1,...,n-1\\}$ it is easy to check that $D$ is
an invertible derivation of $\mathfrak{L}_{n}.$

Consider the trivial $T^{*}$-extension deduce the quadratic Leibniz triple
system on $\mathfrak{L}_{n},$ which is denoted by
$\widetilde{\mathfrak{L}_{n}}:=\mathfrak{L}_{n}\dot{+}\mathfrak{L}_{n}^{*},$
when $\widetilde{\mathfrak{L}_{n}}$ endowed with the derivation
$\widetilde{D}(x+f)=D(x)-f\circ D,$ for $x\in\mathfrak{L}_{n},$
$f\in\mathfrak{L}_{n}^{*},$ then $\widetilde{D}$ is invertible with
$\widetilde{D}^{-1}=D^{-1}(x)-f\circ D^{-1},$ so $\widetilde{D}$ is an
invertible derivation on $\widetilde{\mathfrak{L}_{n}}$ which is skewsymmetric
with respect to $B,$ since $B(\widetilde{D}(x+f),y+h)=h\circ D(x)-f\circ
D(y),$ hence the quadratic Leibniz triple system
$(\widetilde{\mathfrak{L}_{n}},B)$ admits a symplectic structure.

##  5\. $1$-parameter formal deformation of Leibniz triple systems

At last, we use the cohomology theory defined in Section 2 to study the
$1$-parameter formal deformation of Leibniz triple systems. Let
$(\mathfrak{L},\\{\cdot,\cdot,\cdot\\})$ be a Leibniz triple system and
$\mathbb{F}[[t]]$ be the formal power series over $\mathbb{F}.$ Suppose that
$\mathfrak{L}[[t]]$ is the set of formal power series over $\mathfrak{L}.$
Then for an $\mathbb{F}$-trilinear map
$f:\mathfrak{L}\times\mathfrak{L}\times\mathfrak{L}\rightarrow\mathfrak{L},$
it is natural to extend it to an $\mathbb{F}[[t]]$-trilinear map
$f:\mathfrak{L}[[t]]\times\mathfrak{L}[[t]]\times\mathfrak{L}[[t]]\rightarrow\mathfrak{L}[[t]],$
by

$\displaystyle f\Big{(}\sum\limits_{i\geq 0}x_{i}t^{i},\sum\limits_{j\geq
0}y_{j}t^{j},\sum\limits_{k\geq 0}z_{k}t^{k}\Big{)}=\sum\limits_{i,j,k\geq
0}f(x_{i},y_{j},z_{k})t^{i+j+k}.$

######  Definition 5.1.

Let $(\mathfrak{L},\\{\cdot,\cdot,\cdot\\})$ be a Leibniz triple system over
$\mathbb{F}.$ A $1$-parameter formal deformation of
$(\mathfrak{L},\\{\cdot,\cdot,\cdot\\})$ is a formal power series
$d_{t}:\mathfrak{L}[[t]]\times\mathfrak{L}[[t]]\times\mathfrak{L}[[t]]\rightarrow\mathfrak{L}[[t]]$
of the form $d_{t}(x,y,z)=\sum\limits_{i\geq 0}d_{i}(x,y,z)t^{i}$ where each
$d_{i}$ is an $\mathbb{F}$-trilinear map
$d_{i}:\mathfrak{L}\times\mathfrak{L}\times\mathfrak{L}\rightarrow\mathfrak{L}$
$($extended to be $\mathbb{F}[[t]]$-trilinear$)$ and
$d_{0}(x,y,z)=\\{x,y,z\\},$ such that the following identities hold

(5.1) $\displaystyle\begin{aligned}
d_{t}(u,v,d_{t}(x,y,z))=&d_{t}(d_{t}(u,v,x),y,z)-d_{t}(d_{t}(u,v,y),x,z)-d_{t}(d_{t}(u,v,z),x,y)\\\
&+d_{t}(d_{t}(u,v,z),y,x),\\\ \end{aligned}$ (5.2)
$\displaystyle\begin{aligned}
d_{t}(u,d_{t}(x,y,z),v)=&d_{t}(d_{t}(u,x,y),z,v)-d_{t}(d_{t}(u,y,x),z,v)-d_{t}(d_{t}(u,z,x),y,v)\\\
&+d_{t}(d_{t}(u,z,y),x,v).\end{aligned}$

Eqs.$(\ref{226.1})$ and $(\ref{226.2})$ are called the deformation equations
and give a Leibniz triple system structure on
$\mathfrak{L}_{t}=(\mathfrak{L}[[t]],d_{t}).$

Note that Eqs. $(\ref{226.1})$ and $(\ref{226.2})$ can be expressed as

(5.3) $\displaystyle\begin{aligned} \sum\limits_{i,j\geq
0}d_{i}(u,v,d_{j}(x,y,z))=&\sum\limits_{i,j\geq
0}d_{i}(d_{j}(u,v,x),y,z)-\sum\limits_{i,j\geq 0}d_{i}(d_{j}(u,v,y),x,z)\\\
&-\sum\limits_{i,j\geq 0}d_{i}(d_{j}(u,v,z),x,y)+\sum\limits_{i,j\geq
0}d_{i}(d_{j}(u,v,z),y,x),\end{aligned}$ (5.4) $\displaystyle\begin{aligned}
\sum\limits_{i,j\geq 0}d_{i}(u,d_{j}(x,y,z),v)=&\sum\limits_{i,j\geq
0}d_{i}(d_{j}(u,x,y),z,v)-\sum\limits_{i,j\geq 0}d_{i}(d_{j}(u,y,x),z,v)\\\
&-\sum\limits_{i,j\geq 0}d_{i}(d_{j}(u,z,x),y,v)+\sum\limits_{i,j\geq
0}d_{i}(d_{j}(u,z,y),x,v).\end{aligned}$

Then

(5.5) $\displaystyle\begin{aligned}
\sum\limits_{i+j=r}&\Big{(}d_{i}(d_{j}(u,v,x),y,z)-d_{i}(d_{j}(u,v,y),x,z)-d_{i}(d_{j}(u,v,z),x,y)+d_{i}(d_{j}(u,v,z),y,x)\\\
&-d_{i}(u,v,d_{j}(x,y,z))\Big{)}=0,\end{aligned}$ (5.6)
$\displaystyle\begin{aligned}
\sum\limits_{i+j=r}&\Big{(}d_{i}(d_{j}(u,x,y),z,v)-d_{i}(d_{j}(u,y,x),z,v)-d_{i}(d_{j}(u,z,x),y,v)+d_{i}(d_{j}(u,z,y),x,v)\\\
&-d_{i}(u,d_{j}(x,y,z),v)\Big{)}=0.\end{aligned}$

######  Definition 5.2.

Suppose that $d_{t}(x,y,z)=\sum\limits_{i\geq 0}d_{i}(x,y,z)t^{i}$ is a
$1$-parameter formal deformation of $(\mathfrak{L},\\{\cdot,\cdot,\cdot\\}).$
Then $d_{1}$ is called an infinitesimal deformation. In general, if $d_{i}=0,$
for $1\leq i\leq n-1,$ $d_{n}\neq 0,$ then $d_{n}$ is called an
$n$-infinitesimal deformation.

######  Proposition 5.3.

The infinitesimal deformation $d_{1}$ is a $3$-cocycle in
$C^{3}(\mathfrak{L},\mathfrak{L}).$ In general, $n$ infinitesimal deformation
$d_{n}$ is also a $3$-cocycle in $C^{3}(\mathfrak{L},\mathfrak{L}).$

###### Proof.

Taking $r=1$ in Eqs. (5.5) and (5.6), we obtain two equations for $d_{1}:$

(5.7) $\displaystyle\begin{aligned}
&\\{d_{1}(u,v,x),y,z\\}-\\{d_{1}(u,v,y),x,z\\}-\\{d_{1}(u,v,z),x,y\\}+\\{d_{1}(u,v,z),y,x\\}\\\
&-\\{u,v,d_{1}(x,y,z)\\}+d_{1}(\\{u,v,x\\},y,z)-d_{1}(\\{u,v,y\\},x,z)-d_{1}(\\{u,v,z\\},x,y)\\\
&+d_{1}(\\{u,v,z\\},y,x)-d_{1}(u,v,\\{x,y,z\\})=0,\end{aligned}$ (5.8)
$\displaystyle\begin{aligned}
&\\{d_{1}(u,x,y),z,v\\})-\\{d_{1}(u,y,x),z,v\\}-\\{d_{1}(u,z,x),y,v\\}+\\{d_{1}(u,z,y),x,v\\}\\\
&-\\{u,d_{1}(x,y,z),v\\}+d_{1}(\\{u,x,y\\},z,v)-d_{1}(\\{u,y,x\\},z,v)-d_{1}(\\{u,z,x\\},y,v)\\\
&+d_{1}(\\{u,z,y\\},x,v)-d_{1}(u,\\{x,y,z\\},v)=0.\end{aligned}$

So $d_{1}$ is a $3$-cocycle with respect to adjoint module $\mathfrak{L}.$
Similarly, one can check that $d_{n}$ is a $3$-cocycle. ∎

Now, we characterize the $1$-parameter formal deformation through cohomology
of $\mathfrak{L}.$

######  Definition 5.4.

Let $(\mathfrak{L},\\{\cdot,\cdot,\cdot\\})$ be a Leibniz triple system.
Suppose that $d_{t}(x,y,z)=\sum\limits_{i\geq 0}d_{i}(x,y,z)t^{i}$ and
$d_{t}^{\prime}(x,y,z)=\sum\limits_{i\geq 0}d_{i}^{\prime}(x,y,z)t^{i}$ are
two $1$-parameter formal deformations of
$(\mathfrak{L},\\{\cdot,\cdot,\cdot\\}).$ They are called equivalent, denote
by $d_{t}\sim d_{t}^{\prime},$ if there is a formal isomorphism of
$\mathbb{F}[[t]]$-modules

$\displaystyle\phi_{t}(x)=\sum\limits_{i\geq
0}\phi_{i}(x)t^{i}:(\mathfrak{L}[[t]],d_{t})\rightarrow(\mathfrak{L}[[t]],d_{t}^{\prime}),$

where each $\phi_{i}:\mathfrak{L}\rightarrow\mathfrak{L}$ is an
$\mathbb{F}$-linear map $($extended to be $\mathbb{F}[[t]]$-linear$)$ and
$\phi_{0}=$id${}_{\mathfrak{L}},$ satisfying

$\displaystyle\phi_{t}d_{t}(x,y,z)=d_{t}^{\prime}(\phi_{t}(x),\phi_{t}(y),\phi_{t}(z)).$

If $d_{1}=d_{2}=...=0,$ $d_{t}=d_{0}$ is said to be the null deformation. A
$1$-parameter formal deformation $d_{t}$ is called trivial if $d_{t}\sim
d_{0}.$ A Leibniz triple system $(\mathfrak{L},\\{\cdot,\cdot,\cdot\\})$ is
called analytically rigid, if every $1$-parameter formal deformation $d_{t}$
is trivial.

######  Theorem 5.5.

Let $d_{t}(x,y,z)=\sum\limits_{i\geq 0}d_{i}(x,y,z)t^{i}$ and
$d_{t}^{\prime}(x,y,z)=\sum\limits_{i\geq 0}d_{i}^{\prime}(x,y,z)t^{i}$ be
equivalent $1$-parameter formal deformations of
$(\mathfrak{L},\\{\cdot,\cdot,\cdot\\}).$ Then $d_{1}$ and $d_{1}^{\prime}$
belong to the same cohomology class in $H^{3}(\mathfrak{L},\mathfrak{L}).$

###### Proof.

Suppose that $\phi_{t}(x)=\sum\limits_{i\geq 0}\phi_{i}(x)t^{i}$ is the formal
$\mathbb{F}[[t]]$-module isomorphism such that

$\displaystyle\sum\limits_{i\geq 0}\phi_{i}\Big{(}\sum\limits_{j\geq
0}d_{j}(x,y,z)\Big{)}t^{i+j}=\sum\limits_{i\geq
0}d_{i}^{\prime}\Big{(}\sum\limits_{k\geq
0}\phi_{k}(x)t^{k},\sum\limits_{l\geq 0}\phi_{l}(y)t^{l},\sum\limits_{m\geq
0}\phi_{m}(z)t^{m}\Big{)}t^{i}.$

It follows that

$\displaystyle\sum\limits_{i+j=n}\phi_{i}(d_{j}(x,y,z))t^{i+j}=\sum\limits_{i+k+l+m=n}d_{i}^{\prime}(\phi_{k}(x),\phi_{l}(y),\phi_{m}(z))t^{i+k+l+m}.$

In particular,

$\displaystyle\sum\limits_{i+j=1}\phi_{i}(d_{j}(x,y,z))=\sum\limits_{i+k+l+m=1}d_{i}^{\prime}(\phi_{k}(x),\phi_{l}(y),\phi_{m}(z)),$

that is,

$\displaystyle
d_{1}(x,y,z)+\phi_{1}(\\{x,y,z\\})=\\{\phi_{1}(x),y,z\\}+\\{x,\phi_{1}(y),z\\}+\\{x,y,\phi_{1}(z)\\}+d_{1}^{\prime}(x,y,z),$

we have

$\displaystyle
d_{1}(x,y,z)-d_{1}^{\prime}(x,y,z)=\\{\phi_{1}(x),y,z\\}+\\{x,\phi_{1}(y),z\\}+\\{x,y,\phi_{1}(z)\\}-\phi_{1}(\\{x,y,z\\}).$

Then $d_{1}-d_{1}^{\prime}\in B^{3}(\mathfrak{L},\mathfrak{L}).$ ∎

######  Theorem 5.6.

Suppose that $(\mathfrak{L},\\{\cdot,\cdot,\cdot\\})$ is a Leibniz triple
system such that $H^{3}(\mathfrak{L},\mathfrak{L})=0.$ Then
$(\mathfrak{L},\\{\cdot,\cdot,\cdot\\})$ is analytically rigid.

###### Proof.

Let $d_{t}$ be a $1$-parameter formal deformation of
$(\mathfrak{L},\\{\cdot,\cdot,\cdot\\}).$ Suppose that
$d_{t}=d_{0}+\sum\limits_{i\geq n}d_{i}t^{i},$ and by Proposition 5.3
$d_{n}\in Z^{3}(\mathfrak{L},V)=B^{3}(\mathfrak{L},\mathfrak{L}).$ It follows
that there exists $f_{n}\in C^{1}(\mathfrak{L},\mathfrak{L}),$ such that

(5.9) $\displaystyle
d_{n}(x,y,z)=\\{f_{n}(x),y,z\\}+\\{x,f_{n}(y),z\\}+\\{x,y,f_{n}(z)\\}-f_{n}(\\{x,y,z\\}).$

Let $\phi_{t}=$id${}_{\mathfrak{L}}-f_{n}t^{n}$ and
$d_{t}^{\prime}(x,y,z)=\phi_{t}^{-1}d_{t}(\phi_{t}(x),\phi_{t}(y),\phi_{t}(z)).$
It is straightforward to prove that $d_{t}^{\prime}$ is a $1$-parameter formal
deformation of $(\mathfrak{L},\\{\cdot,\cdot,\cdot\\})$ and $d_{t}\sim
d_{t}^{\prime}.$ Suppose that $d_{t}^{\prime}=\sum\limits_{i\geq
0}d_{i}^{\prime}t^{i},$ we express the coefficient in front of the
$d_{t}^{\prime}$ in terms of $d_{i}^{\prime}.$ Then

$\displaystyle({\rm id}_{\mathfrak{L}}-f_{n}t^{n})\Big{(}\sum\limits_{i\geq
0}d_{i}^{\prime}(x,y,z)t^{i}\Big{)}=\Big{(}d_{0}+\sum\limits_{i\geq
n}d_{i}t^{i}\Big{)}(x-f_{n}(x)t^{n},y-f_{n}(y)t^{n},z-f_{n}(z)t^{n}),$

i.e.,

$\displaystyle\sum\limits_{i\geq
0}d_{i}^{\prime}(x,y,z)t^{i}-\sum\limits_{i\geq
0}f_{n}d_{i}^{\prime}(x,y,z)t^{i+n}$ $\displaystyle=$
$\displaystyle\\{x,y,z\\}-(\\{f_{n}(x),y,z\\}+\\{x,f_{n}(y),z\\}+\\{x,y,f_{n}(z)\\})t^{n}+(\\{f_{n}(x),f_{n}(y),z\\}$
$\displaystyle+\\{x,f_{n}(y),f_{n}(z)\\}+\\{f_{n}(x),y,f_{n}(z)\\})t^{2n}-\\{f_{n}(x),f_{n}(y),f_{n}(z)\\}t^{3n}+\sum\limits_{i\geq
n}d_{i}(x,y,z)t^{i}$ $\displaystyle-\sum\limits_{i\geq
n}(d_{i}(f_{n}(x),y,z)+d_{i}(x,f_{n}(y),z)+d_{i}(x,y,f_{n}(z)))t^{i+n}$
$\displaystyle+\sum\limits_{i\geq
n}(d_{i}(f_{n}(x),f_{n}(y),z)+d_{i}(x,f_{n}(y),f_{n}(z))+d_{i}(f_{n}(x),y,f_{n}(z)))t^{i+2n}$
$\displaystyle-\sum\limits_{i\geq
n}(d_{i}(f_{n}(x),f_{n}(y),f_{n}(z))t^{i+3n}.$

Then we have $d_{1}^{\prime}=...=d_{n-1}^{\prime}=0$ and

$\displaystyle
d_{n}^{\prime}(x,y,z)-f_{n}(\\{x,y,z\\})=-(\\{f_{n}(x),y,z\\}+\\{x,f_{n}(y),z\\}+\\{x,y,f_{n}(z)\\})+d_{n}(x,y,z).$

By Eq. (5.9), $d_{n}^{\prime}=0$ and $d_{t}^{\prime}=d_{0}+\sum\limits_{i\geq
n+1}d_{i}^{\prime}t^{i}.$ Repeat this procedure, one observe that there is no
nontrivial deformation of $(\mathfrak{L},\\{\cdot,\cdot,\cdot\\}),$ i.e.,
$(\mathfrak{L},\\{\cdot,\cdot,\cdot\\})$ is analytically rigid. ∎

At last, we define two $5$-cochains Ob${}_{n+1}^{i}(i=1,2)$ by

$\displaystyle{\rm Ob}_{n+1}^{1}(u,v,x,y,z)=$
$\displaystyle\sum\limits_{i+j=n+1\atop
i,j>0}\Big{(}d_{i}(d_{j}(u,v,x),y,z)-d_{i}(d_{j}(u,v,y),x,z)$ $\displaystyle-
d_{i}(d_{j}(u,v,z),x,y)+d_{i}(d_{j}(u,v,z),y,x)-d_{i}(u,v,d_{j}(x,y,z))\Big{)},$
$\displaystyle{\rm Ob}_{n+1}^{2}(u,v,x,y,z)=$
$\displaystyle\sum\limits_{i+j=n+1\atop
i,j>0}\Big{(}d_{i}(d_{j}(u,v,x),y,z)-d_{i}(d_{j}(u,x,v),y,z)$ $\displaystyle-
d_{i}(d_{j}(u,y,v),x,z)+d_{i}(d_{j}(u,y,x),v,z)-d_{i}(u,d_{j}(v,x,y),z)\Big{)}.$

######  Definition 5.7.

The $5$-cochains Ob${}_{n+1}^{i}(\mathfrak{L})\in
C^{5}(\mathfrak{L},\mathfrak{L})(i=1,2)$ are called $(n+1)$-th obstruction
cochain for extending a given deformation of order $n$ (i.e.
$d_{n}=\sum\limits_{i=0}^{n}d^{i}t^{i}$) to a deformation of $\mathfrak{L}$ of
order $(n+1).$

######  Theorem 5.8.

Let $d_{t}$ be a deformation of $\mathfrak{L}$ of order $n.$ Then $d_{t}$
extends to a deformation of order $n+1,$
$d_{n+1}=\sum\limits_{i=0}^{n+1}d^{i}t^{i}$ if and only if
Ob${}_{n+1}^{i}(\mathfrak{L})=\delta^{3}_{i}d_{n+1}(i=1,2).$

###### Proof.

For the necessity, it is obvious by the definition of
Ob${}_{n+1}^{i}(\mathfrak{L})$ and Eqs. (5.5) and (5.6).

For the sufficiency, suppose
Ob${}_{n+1}^{i}(\mathfrak{L})=\delta^{3}_{i}d_{n+1}(i=1,2)$, for some
$3$-cochain $d_{n+1}\in C^{3}(\mathfrak{L},\mathfrak{L}),$ that is
Ob${}_{n+1}^{i}(\mathfrak{L})$ is a coboundary and
$\tilde{d_{t}}=d_{t}+d_{n+1}t^{n+1}.$ Observe that $\tilde{d_{t}}$ satisfies
Eqs. (5.5) and (5.6) for $0\leq r\leq n+1$. So $\tilde{d_{t}}$ is an extension
of $d_{t}$ and is of order $n+1.$ ∎

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