# Classification, Derivations and Centroids of Low-Dimensional Complex BiHom-
Trialgebras

Ahmed Zahari
Université de Haute Alsace,
IRIMAS-Département de Mathématiques,
18, rue des Frères Lumière F-68093 Mulhouse, France.
Bouzid Mosbahi
University of Sfax, Faculty of Sciences of Sfax, BP 1171, 3000 Sfax, Tunisia.
Imed Basdouri
University of Gafsa, Faculty of Sciences of Gafsa, 2112 Gafsa, Tunisia.
e-mail address<EMAIL_ADDRESS>address<EMAIL_ADDRESS>mail address<EMAIL_ADDRESS>
Abstract. The basic objective of this research work is to investigate the
stucture of BiHom-associative trialgebras. In this respect we build up one
important class of BiHom-trialgebras and determine properties of right, left
and middle operations in BiHom-associative trialgebras. We provide a
classification of $n$-dimensional BiHom-trialgebras for $n\leq 3$. Relying
upon the classification result of BiHom-associative trialgebras, the
derivations and centroids of low-dimensional BiHom-associative trialgebras are
characterized. Certain properties of the centroids in light of BiHom-
associative trialgebras are reviwed the centroids of low-dimensional BiHom-
associative trialgebras are computed.

AMS Subject Classification : 17A30, 17A32, 16D20, 16W25, 17B63 .

Keywords: BiHom-associative trialgebra, Classification, Centroid, Derivation,
Central derivation.

##  1 Introduction

A BiHom-associative trialgebra $(\mathcal{A},\dashv,\vdash,\bot,\alpha,\beta)$
involves a vector space, three multiplications and two linear self maps. It
may be regarded as a deformation of an associative algebras, where the
associativity condition is twisted by a linear map $\alpha$ and $\beta$, such
that when $\alpha=id$ (rep. $\beta=id$), the BiHom-associative trialgebra
degenerate to exactly BiHom-triassociative algebras. In the current work, our
central focus is upon the structure of BiHom-associative trialgebras. Let
$\mathcal{A}$ be an $n$-dimensional $\mathbb{F}$-linear space and let
$\left\\{e_{1},e_{2},\cdots,e_{n}\right\\}$ be a basis of $\mathcal{A}$. A
BiHom-associative trialgebra structure on $\mathcal{A}$ with products
$\mu,\,\lambda$ and $\xi$ being determined by $3n^{3}$ structure constants
$\mathcal{\gamma}_{ij}^{k},\,\mathcal{\delta}_{ij}^{k}$ and
$\mathcal{\xi}_{ij}^{k}$, are $e_{i}\dashv
e_{j}=\sum_{k=1}^{n}\gamma_{ij}^{k}e_{k},\quad e_{i}\vdash
e_{j}=\sum_{k=1}^{n}\delta_{ij}^{k}e_{k}$ and $e_{i}\bot
e_{j}=\sum_{k=1}^{n}\xi_{ij}^{k}e_{k}$. They are also determined by $\alpha$
(resp. $\beta$) which is provided by ${2n^{2}}$ structure constants $a_{ij}$
(resp. $b_{ij}$), where $\alpha(e_{i})=\sum_{j=1}^{n}a_{ji}e_{j}$ (resp.
$\beta(e_{j})=\sum_{k=1}^{n}b_{kj}e_{k}$). Requiring the algebra structure to
be BiHom-associative trialgebra and unital gives rise to a sub-variety of
$\mathcal{BT}as$ of $\mathbb{F}^{3n^{3}+2n^{2}}$. Base changes in
$\mathcal{A}$ result in the natural transport of structure action of
$GL_{n}(k)$ on $\mathcal{BT}as$. From this perspective, isomorphism classes of
$n$-dimensional BiHom-algebras go in one-to-one correspondence with the orbits
of the action of $GL_{n}(\mathbb{F})$ on $\mathcal{BT}as.$

Classification problems of the BiHom-associative trialgebras using the
algebraic and geometric technique have triggered the widest interest in the
derivations and centroids of BiHom-associative trialgebras. The BiHom-
associative trialgebras introduced by Loday [5] with a motivation to provide
dual dialgebras, have been further investigated with regard to several areas
in mathematics and physics. The classification of Hom-associative algebras was
set forward by [1] and A. Zahari and I. Bakayoko who studied the
classification of BiHom-associative dialgebras [2].

This paper centers around examining the derivations and centroids of finite
dimensional associative trialgebras. The algebra of derivations and centroids
are highly needed and extremely useful in terms of algebraic and geometric
classification problems of algebras.

The current paper is laid out as follows. In the fist section, we introduce
the topic alongside with some previously obtained results. The target of this
paper is to introduce and classify derivation as well as centroids of BiHom-
associative trialgebras. In section 2, we explore the structure of BiHom-
associative trialgebras. In section 3, we display the algebraic varieties of
BiHom-associative trialgebras, and we exhibit classifications, up to
isomorphism, of two-dimensional and three-dimensional BiHom-associative
trialgebras.
In Section 4, we identify the classification of the derivations. Eventually,
in Section 5, we present the classification of the centroids. In this case,
the concepts of derivations and centroids is largely inspired and whetted
significant scientific concern from that of finite-dimensional algebras. The
algebra of centroids plays an important role in the classification problems as
well as in different applications of algebras.In this study, we elaborated
classification results of two and three-dimensional BiHom-associative
trialgebras. All considered algebras and vectors spaces are supposed over a
field $\mathbb{F}$ of characteristic zero.

##  2 Structure of BiHom-associative trialgebras

######  Definition 2.1.

A BiHom-associative trialgebra is a $6$-tuple
$(\mathcal{A},\dashv,\vdash,\bot,\alpha,\beta)$ consisting of a linear space
$\mathcal{A}$, linear maps
$\dashv,\vdash,\bot:\mathcal{A}\otimes\mathcal{A}\longrightarrow\mathcal{A}$
and $\alpha,\beta:\mathcal{A}\longrightarrow\mathcal{A}$ satisfying, for all
$x,y,z\in\mathcal{A}$, the following conditions :

$\displaystyle\alpha\circ\beta$ $\displaystyle=$
$\displaystyle\beta\circ\alpha,$ (2.1) $\displaystyle(x\dashv
y)\dashv\beta(z)$ $\displaystyle=$ $\displaystyle\alpha(x)\dashv(y\dashv z),$
(2.2) $\displaystyle(x\dashv y)\dashv\beta(z)$ $\displaystyle=$
$\displaystyle\alpha(x)\dashv(y\vdash z)=\alpha(x)\dashv(y\bot z),$ (2.3)
$\displaystyle(x\dashv y)\dashv\beta(z)$ $\displaystyle=$
$\displaystyle\alpha(x)\vdash(y\dashv z),$ (2.4) $\displaystyle(x\dashv
y)\vdash\beta(z)$ $\displaystyle=$ $\displaystyle\alpha(x)\vdash(y\vdash
z)=(x\bot y)\vdash\beta(z),$ (2.5) $\displaystyle(x\vdash y)\vdash\beta(z)$
$\displaystyle=$ $\displaystyle\alpha(x)\vdash(y\vdash z),$ (2.6)
$\displaystyle(x\bot y)\dashv\beta(z)$ $\displaystyle=$
$\displaystyle\alpha(x)\bot(y\dashv z),$ (2.7) $\displaystyle(x\dashv
y)\bot\beta(z)$ $\displaystyle=$ $\displaystyle\alpha(x)\bot(y\vdash z),$
(2.8) $\displaystyle(x\vdash y)\bot\beta(z)$ $\displaystyle=$
$\displaystyle\alpha(x)\vdash(y\bot z),$ (2.9) $\displaystyle(x\bot
y)\bot\beta(z)$ $\displaystyle=$ $\displaystyle\alpha(x)\bot(y\bot z),$ (2.10)

for all $x,y,z\in\mathcal{A}.$ We call $\alpha$ and $\beta$ ( in this order )
the structure maps of $\mathcal{A}.$

######  Remark 2.2.

In addition, $\alpha$ and $\beta$ are endomorphisms with respect to
$\dashv,\vdash$ and $\bot$. Then, $\mathcal{A}$ is said to be a multiplicative
BiHom-associative trialgebra

$\displaystyle\alpha(x\dashv y)=\alpha(x)\dashv\alpha(y)$ ;
$\displaystyle\beta(x\dashv y)=\beta(x)\dashv\beta(y),$
$\displaystyle\alpha(x\vdash y)=\alpha(x)\vdash\alpha(y)$ ;
$\displaystyle\beta(x\vdash y)=\beta(x)\vdash\beta(y),$
$\displaystyle\alpha(x\bot y)=\alpha(x)\bot\alpha(y)$ ;
$\displaystyle\beta(x\bot y)=\beta(x)\bot\beta(y),$

for all $x,y\in\mathcal{A}.$

######  Definition 2.3.

A morphism of BiHom-associative trialgebra is a linear map

$\psi:(\mathcal{A},\dashv,\vdash,\bot,\alpha,\beta)\rightarrow(\mathcal{A}^{\prime},\dashv^{\prime},\vdash^{\prime},\bot^{\prime},\alpha^{\prime},\beta^{\prime})$

such that

$\alpha^{\prime}\circ\psi=\psi\circ\alpha,\;\;\beta^{\prime}\circ\psi=\psi\circ\beta$

and

$\displaystyle\psi(x\dashv y)$ $\displaystyle=$
$\displaystyle\psi(x)\dashv^{\prime}\psi(y),$ $\displaystyle\psi(x\vdash y)$
$\displaystyle=$ $\displaystyle\psi(x)\vdash^{\prime}\psi(y),$
$\displaystyle\psi(x\bot y)$ $\displaystyle=$
$\displaystyle\psi(x)\bot^{\prime}\psi(y),$

for all $x,y\in\mathcal{A}.$

######  Remark 2.4.

A bijective homomorphism is an isomorphism of $\mathcal{A}_{1}$ and
$\mathcal{A}_{2}$.

######  Definition 2.5.

Let $(\mathcal{A},\ast,\alpha,\beta)$ be a BiHom-associative trialgebra. If
there is an associative trialgebra $(\mathcal{A},\ast^{\prime})$ such that
$x\ast^{\prime}y=\alpha^{-1}(x)\ast\beta^{-1}(y),\,\forall x,y\in A$, we say
that $(\mathcal{A},\ast,\alpha,\beta)$ is of an associative type and
$({A},\ast^{\prime})$ is its compatible associative algebra or the untwist of
$(\mathcal{A},\ast,\alpha,\beta)$.

######  Proposition 2.6.

Let $(\mathcal{A},\dashv,\vdash,\bot,\alpha,\beta)$ be an $n$-dimensional
BiHom-associative trialgebra and let $\psi:\mathcal{A}\rightarrow\mathcal{A}$
be an invertible linear map. Therefore, there is an isomorphism with an
n-dimensional BiHom-associative trialgebra
$(\mathcal{A},\dashv^{\prime},\vdash^{\prime},\bot^{\prime},\psi\alpha\psi^{-1},\psi\beta\psi^{-1})$,
where
$(\dashv^{\prime},\vdash^{\prime},\bot^{\prime})=\psi\circ(\dashv,\vdash,\bot)\circ(\psi^{-1}\otimes\psi^{-1}$).
Furthermore, if
$\left\\{\phi^{k}_{ij},\gamma^{k}_{ij},\delta^{k}_{ij}\right\\}$ are the
structure constants of $(\dashv,\vdash,\bot)$ with respect to the basis
$\left\\{e_{1},\dots,e_{n}\right\\}$, then
$(\dashv^{\prime},\vdash^{\prime},\bot^{\prime})$ has the same structure
constants with respect to the basis
$\left\\{\psi(e_{1}),\dots,\psi(e_{n})\right\\}$.

###### Proof.

We prove, for any invertible linear map
$\psi:\mathcal{A}\rightarrow\mathcal{A},that\,(\mathcal{A},\dashv^{\prime},\vdash^{\prime},\bot^{\prime},\psi\alpha\psi^{-1},\psi\beta\psi^{-1})$
is a BiHom-associative trialgebra.

$\begin{array}[]{ll}(x\dashv^{\prime}y)\dashv^{\prime}\psi\beta\psi^{-1}(z)&=\psi\dashv(\psi^{-1}\otimes\psi^{-1})(x,y)\dashv^{\prime}\psi\beta\psi^{-1}(z)\\\
&=(\psi(\psi^{-1}\otimes\psi^{-1})(x\dashv
y))\dashv^{\prime}\psi\beta\phi^{-1}(z)\\\
&=\psi(\psi^{-1}\otimes\psi^{-1})(\psi(\psi^{-1}\otimes\psi^{-1})(x\dashv
y)\dashv\phi^{-1}\beta\psi(z)\\\
&=\psi(\psi^{-1}(x)\dashv\psi^{-1}(y))\dashv\psi^{-1}\beta\psi(z)\\\
&=\psi(\alpha\psi^{-1}(x)\dashv(\psi^{-1}(y)\vdash\psi^{-1}(z)))\\\
&=\psi(\psi^{-1}\otimes\psi^{-1})(\psi\otimes\psi)\alpha\psi^{-1}(x)\dashv(\psi^{-1}\otimes\psi^{-1}(y\vdash
z))\\\
&=\psi(\psi^{-1}\otimes\psi^{-1})(\psi\alpha\psi^{-1}(x)\dashv(\psi(\psi^{-1}\otimes\psi^{-1})(y\vdash
z)))\\\
&=\psi\alpha\psi^{-1}(x)\dashv^{\prime}(y\vdash^{\prime}z).\end{array}$

Hence,
$(\mathcal{A},\dashv^{\prime},\vdash^{\prime},\bot^{\prime},\psi\alpha\psi^{-1},\psi\beta\psi^{-1})$
is a BiHom-associative trialgebra.
It is also multiplicative. Indeed, for $\alpha$

$\begin{array}[]{ll}\psi\alpha\psi^{-1}(x\dashv^{\prime}y)&=\psi\alpha\psi^{-1}\psi\dashv(\psi^{-1}\otimes\psi^{-1})(x,y)\\\
&=\psi\alpha\dashv(\psi^{-1}\otimes\psi^{-1})(x,y)\\\
&=\psi(\alpha\psi^{-1}(x)\dashv\alpha\psi^{-1}(y))\\\
&=\psi((\psi^{-1}\otimes\psi^{-1})(\psi\otimes\phi)\alpha\psi^{-1}(x)\dashv\alpha\psi^{-1}(y))=(\psi\alpha\psi^{-1}(x)\dashv^{\prime}\psi\alpha\psi^{-1}(y)).\end{array}$

We have also for $\beta$

$\begin{array}[]{ll}\psi\beta\psi^{-1}(x\dashv^{\prime}y)&=\psi\beta\psi^{-1}\psi\dashv(\psi^{-1}\otimes\psi^{-1})(x,y)\\\
&=\psi\beta\dashv(\psi^{-1}\otimes\psi^{-1})(x,y)\\\
&=\psi(\beta\psi^{-1}(x)\dashv\beta\psi^{-1}(y))\\\
&=\psi((\psi^{-1}\otimes\psi^{-1})(\psi\otimes\phi)\beta\psi^{-1}(x)\dashv\beta\psi^{-1}(y))=(\psi\beta\psi^{-1}(x)\dashv^{\prime}\psi\beta\psi^{-1}(y)).\end{array}$

Thus,
$\psi:(\mathcal{A},\dashv,\vdash,\bot,\alpha,\beta)\rightarrow(\mathcal{A},\dashv^{\prime},\vdash^{\prime},\bot^{\prime},\psi\alpha\psi^{-1},\psi\beta\psi^{-1})$
is a BiHom-associative trialgebra morphism, since
$\psi\circ\dashv=\psi\circ\dashv\circ(\psi^{-1}\otimes\psi^{-1})\circ(\psi\otimes\psi)=\mu^{\prime}\circ(\psi\otimes\psi)$
and $(\psi\alpha\phi^{-1})\circ\psi=\psi\circ\alpha$ and
$(\psi\beta\psi^{-1})\circ\psi=\psi\circ\beta.$
It is easy to infer that $\left\\{\psi(e_{i}),\cdots,\psi(e_{n})\right\\}$ is
a basis of $\mathcal{A}$. For $i,j=1,\cdots,n$, we have

$\begin{array}[]{ll}(\psi(e_{i})\dashv\psi(e_{j}))&=\psi(\psi^{-1}(e_{i})\dashv\psi^{-1}(e_{j}))=\psi(e_{i}\dashv
e_{j})=\sum_{k=1}^{n}\gamma^{k}_{ij}\psi(e_{k}).\end{array}$ ∎

######  Proposition 2.7.

Let $(\mathcal{A},\dashv,\vdash,\bot,\alpha,\beta)$ be a BiHom-associative
trialgebra over $\mathbb{F}$.
Let
$(\mathcal{A},\dashv^{\prime},\vdash^{\prime},\bot^{\prime},\phi\alpha\psi^{-1},\psi\beta\psi^{-1})$
be its isomorphic BiHom-associative trialgebra described in Proposition 2.6.
If $\psi$ is an automorphism of
$(\mathcal{A},\dashv,\vdash,\bot,\alpha,\beta)$, then $\psi\phi\psi^{-1}$ is
an automorphism of
$(\mathcal{A},\dashv^{\prime},\vdash^{\prime},\bot^{\prime},\psi\alpha\psi^{-1},\psi\beta\psi^{-1})$.

###### Proof.

Note that $\gamma=\psi\alpha\psi^{-1}$. It follows that

$\psi\phi\psi^{-1}\gamma=\psi\phi\psi^{-1}\psi\alpha\psi^{-1}=\psi\phi\alpha\psi^{-1}=\psi\alpha\phi\psi^{-1}=\psi\alpha\psi^{-1}\psi\phi\psi^{-1}=\gamma\psi\phi\psi^{-1}.$

For $\beta$, we assume $\lambda=\psi\beta\psi^{-1}$. As a result, we get

$\psi\phi\psi^{-1}\lambda=\psi\phi\psi^{-1}\psi\beta\psi^{-1}=\psi\phi\beta\psi^{-1}=\psi\beta\phi\psi^{-1}=\psi\beta\psi^{-1}\psi\phi\psi^{-1}=\lambda\psi\phi\psi^{-1}.$

For any $x,y\in T,$

$\begin{array}[]{ll}\psi\phi\psi^{-1}(\psi(x)\dashv^{\prime}\psi(y))&=\psi\phi\psi^{-1}\psi(x\dashv
y)=\psi\phi(x\dashv y)=\psi(\phi(x)\dashv\phi(y))\\\
&=(\psi\phi(x)\dashv^{\prime}\psi\phi(y))=(\psi\phi\psi^{-1}(\psi(x)\dashv^{\prime}\phi\psi\phi^{-1}(\phi(y))).\end{array}$

By Definition, $\psi\phi\psi^{-1}$ is an automorphism of
$(A,\dashv^{\prime},\vdash^{\prime},\bot^{\prime},\psi\alpha\psi^{-1},\psi\beta\psi^{-1})$.
∎

In the below proposition, we prove that BiHom-associative trialgebras are
closed under a direct sum, and provide a condition for which a linear map
becomes a morphism of direct sum of BiHom-associative trialgebras.

######  Proposition 2.8.

Let
$(\mathcal{A},\dashv_{\mathcal{A}},\vdash_{\mathcal{A}},\bot_{\mathcal{A}},\alpha_{\mathcal{A}},\beta_{\mathcal{A}})$
and
$(\mathcal{B},\dashv_{\mathcal{B}},\vdash_{\mathcal{B}},\bot_{\mathcal{B}},\alpha_{\mathcal{B}},\beta_{\mathcal{A}})$
be two BiHom-associative trialgebras. Then, there exists a BiHom-associative
trialgebra structure on $\mathcal{A}\oplus\mathcal{B}$ with the bilinear maps
$\triangleleft,\triangleright,\ast:(\mathcal{A}\oplus\mathcal{B})^{\otimes
2}\rightarrow\mathcal{A}\oplus\mathcal{B}$ indicated by

$(a_{1}+b_{2})\dashv(a_{2}+b_{2}):=a_{1}\dashv_{\mathcal{A}}a_{2}+b_{2}\dashv_{\mathcal{B}}b_{2},$
$(a_{1}+b_{1})\vdash(a_{2}+b_{2}):=a_{1}\vdash_{\mathcal{A}}a_{2}+b_{\mathcal{A}}\vdash_{\mathcal{B}}b_{2}$
$(a_{1}+b_{1})\bot(a_{2}+b_{2}):=a_{1}\bot_{\mathcal{A}}a_{2}+b_{\mathcal{A}}\bot_{\mathcal{B}}b_{2}$

and the linear maps
$\alpha=\alpha_{\mathcal{A}}+\alpha_{\mathcal{B}},\,\beta=\beta_{\mathcal{A}}+\beta_{\mathcal{B}}:\mathcal{A}\oplus\mathcal{B}\rightarrow\mathcal{A}\oplus\mathcal{B}$
expressed by

$(\alpha_{\mathcal{A}}+\alpha_{\mathcal{B}})(a+b):=\alpha_{\mathcal{A}}(a)+\alpha_{\mathcal{B}}(b),\,(\beta_{\mathcal{A}}+\beta_{\mathcal{B}})(a+b):=\beta_{\mathcal{A}}(a)+\beta_{\mathcal{B}}(b),\,\forall(a,b)\in\mathcal{A}\times\mathcal{B}.$

Moreover, if $\xi:\mathcal{A}\rightarrow\mathcal{B}$ is a linear map, then,

$\xi:(\mathcal{A},\dashv_{\mathcal{A}},\vdash_{\mathcal{A}},\alpha_{\mathcal{A}},\beta_{\mathcal{A}})\rightarrow(\mathcal{B},\dashv_{\mathcal{B}},\vdash_{\mathcal{B}},\alpha_{\mathcal{B}},\beta_{\mathcal{B}})$

is a morphism if and only if its graph
$\Gamma_{\xi}=\\{(x,\xi(x)),x\in\mathcal{A}\\}$ is a BiHom-subalgebra of
$(\mathcal{A}\oplus\mathcal{B},\triangleleft,\triangleright,\ast,\alpha,\beta)$.

###### Proof.

The proof of the first part derives from a direct computation. For this
reason, we omit it. Now, let us suppose that
$\xi:(\mathcal{A},\dashv_{\mathcal{A}},\vdash_{\mathcal{A}},\alpha_{\mathcal{A}},\beta_{\mathcal{A}})\rightarrow(\mathcal{B},\dashv_{\mathcal{B}},\vdash_{\mathcal{B}},\alpha_{\mathcal{B}},\beta_{\mathcal{B}})$
is a morphism of BiHom-associative trialgebras. Consequently

$(u+\xi(u))\dashv(v+\xi(v))=(u\dashv_{\mathcal{A}}v+\xi(u)\dashv_{B}\xi(v))=u\dashv_{\mathcal{A}}v+\xi(u\dashv_{\mathcal{A}}v),$
$(u+\xi(u))\vdash(v+\xi(v))=(u\vdash_{\mathcal{A}}v+\xi(u)\vdash_{B}\xi(v))=u\vdash_{\mathcal{A}}v+\xi(u\vdash_{\mathcal{A}}v).$
$(u+\xi(u))\bot(v+\xi(v))=(u\bot_{\mathcal{A}}v+\xi(u)\vdash_{B}\xi(v))=u\bot_{A}v+\xi(u\bot_{\mathcal{A}}v).$

Thus, the graph $\Gamma_{\xi}$ is closed under the operations $\dashv,\vdash$
and $\bot$.
Additionally, since
$\xi\circ\alpha_{\mathcal{A}}=\alpha_{\mathcal{B}}\circ\xi,$ and
$\xi\circ\beta_{\mathcal{A}}=\beta_{\mathcal{B}}\circ\xi,$ we have

$(\alpha_{\mathcal{A}}\oplus\alpha_{\mathcal{B}})(u,\xi(u))=(\alpha_{\mathcal{A}}(u),\alpha_{\mathcal{B}}\circ\xi(u))=(\alpha_{\mathcal{A}}(u),\xi\circ\alpha_{\mathcal{A}}(u)),$

and

$(\beta_{\mathcal{A}}\oplus\beta_{\mathcal{B}})(u,\xi(u))=(\beta_{\mathcal{A}}(u),\beta_{\mathcal{B}}\circ\xi(u))=(\beta_{\mathcal{A}}(u),\xi\circ\beta_{\mathcal{A}}(u));$

which implies that $\Gamma_{\xi}$ is closed
$\alpha_{\mathcal{A}}\oplus\alpha_{\mathcal{B}}$ and
$\beta_{\mathcal{A}}\oplus\beta_{\mathcal{B}}.$ Thus, $\Gamma_{\xi}$ is a
BiHom-subalgebra of
$(\mathcal{A}\oplus\mathcal{B},\dashv,\vdash,\alpha,\beta).$
Conversely, if the graph $\Gamma_{\xi}\subset\mathcal{A}\oplus\mathcal{B}$ is
a BiHom-subalgebra of
$(\mathcal{A}\oplus\mathcal{B},\dashv,\vdash,\alpha,\beta)$, then we get

$(u+\xi(u))\dashv(v+\xi(v))=(u\dashv_{\mathcal{A}}v+\xi(u)\dashv_{\mathcal{B}}\xi(v))\in\Gamma_{\xi},$
$(u+\xi(u))\vdash(v+\xi(v))=(u\vdash_{\mathcal{A}}v+\xi(u)\vdash_{\mathcal{B}}\xi(v))\in\Gamma_{\xi},$
$(u+\xi(u))\bot(v+\xi(v))=(u\bot_{\mathcal{A}}v+\xi(u)\bot_{\mathcal{B}}\xi(v))\in\Gamma_{\xi}.$

Furthermore,
$(\alpha_{\mathcal{A}}\oplus\alpha_{\mathcal{B}})(\Gamma_{\xi})\subset\Gamma_{\xi},\,and(\beta_{\mathcal{A}}\oplus\beta_{\mathcal{B}})(\Gamma_{\xi})\subset\Gamma_{\xi},$
imply that

$(\alpha_{\mathcal{A}}\oplus\alpha_{\mathcal{B}})(u,\xi(u))=(\alpha_{\mathcal{A}}(u),\alpha_{\mathcal{B}}\circ\xi(u))\in\Gamma_{\xi},\,(\beta_{\mathcal{A}}\oplus\beta_{\mathcal{B}})(u,\xi(u))=(\beta_{\mathcal{A}}(u),\beta_{B}\circ\xi(u))\in\Gamma_{\xi},$

which is equivalent to the condition
$\alpha_{\mathcal{B}}\circ\xi(u)=\xi\circ\alpha_{\mathcal{A}}(u),$ i.e.,
$\alpha_{\mathcal{B}}\circ\xi=\xi\circ\alpha_{\mathcal{A}}.$
Similarly, $\beta_{\mathcal{B}}\circ\xi=\xi\circ\beta_{\mathcal{A}}$.
Therefore, $\xi$ is a morphism of BiHom-associative trialgebras. ∎

######  Definition 2.9.

Let $(\mathcal{A},\dashv,\vdash,\bot,\alpha,\beta)$ be a BiHom-associative
trialgebra endowed with a linear map $R:\mathcal{A}\longrightarrow\mathcal{A}$
such that $\alpha\circ R=R\circ\alpha$, $\beta\circ R=R\circ\beta$ and

$\begin{array}[]{ll}R(x)\vdash R(y)&=R(R(x)\dashv y+x\dashv
R(y)+\lambda(x\dashv y)),\end{array}$ $\begin{array}[]{ll}R(x)\dashv
R(y)&=R(R(x)\vdash y+x\vdash R(y)+\lambda(x\vdash y)),\end{array}$
$\begin{array}[]{ll}R(x)\bot R(y)&=R(R(x)\bot y+x\bot R(y)+\lambda(x\bot
y)),\end{array}$

with $\lambda\in\mathbb{K},x,y\in\mathcal{A}$, is called a Rota-Baxter BiHom-
associative trialgebra, and R is called a Rota-Baxter operator of weight
$\lambda$ on $\mathcal{A}$.

######  Example 2.10.

We consider the $2$-dimensional BiHom-associative trialgebra with a basis
$\left\\{e_{1},e_{2}\right\\}$. For $e_{1}\dashv e_{2}=e_{1},\,e_{2}\dashv
e_{1}=e_{1},\,e_{1}\vdash e_{2}=e_{1},\,e_{1}\bot e_{2}=e_{1},e_{2}\bot
e_{2}=e_{1},\,\alpha(e_{2})=e_{1},\,\beta(e_{2})=e_{1}$ and $R(e_{1})=-\lambda
e_{1},\,R(e_{2})=-\lambda e_{2}.$

######  Proposition 2.11.

Let $(\mathcal{A},\ast,\alpha,\beta,R)$ be some Rota-Baxter BiHom-associative
trialgebras of weight $\lambda$, and three new operations $\bot,\dashv$ and
$\vdash$ on $\mathcal{A}$ are defined by $x\dashv y=x\ast R(y),\,x\vdash
y=R(x)\ast y$ and $x\bot y=\lambda x\ast y$. Then,
$(\mathcal{A},\dashv,\vdash,\bot,\alpha,\beta)$ is a BiHom-associative
trialgebra.

###### Proof.

We prove only one axiom, as others are proven similarly. Indeed, for any
$x,y,z\in\mathcal{A},$

$\begin{array}[]{ll}(x\vdash y)\dashv\beta(z)&=(R(x)\ast y)\dashv\beta(z)\\\
&=(R(x)\ast y)\ast R(\beta(z))\\\ &=\alpha(R(x))\ast(y\dashv y)\\\
&=\alpha(x)\vdash(y\dashv y),\end{array}$

which ends the proof. ∎

######  Proposition 2.12.

Let $(\mathcal{A},\dashv,\vdash,\bot,\alpha,\beta)$ be a BiHom-associative
trialgebra and suppose
$\alpha^{2}=\beta^{2}\alpha\circ\beta=\beta\circ\alpha=id.$ Then,
$(A,\dashv,\vdash,\bot,\alpha,\beta)\cong(\mathcal{A},\dashv,\vdash,\bot,\beta,\alpha).$

###### Proof.

We prove only one axiom, as others are proven similarly. For any
$x,y,z\in\mathcal{A},$

$\begin{array}[]{ll}\alpha(x)\dashv(y\vdash z)&=\alpha(x)\dashv(y\bot z)\\\
&\Leftrightarrow\alpha(\alpha\beta(x))\dashv(y\vdash
z)=\alpha(\alpha\beta(x))\dashv(y\bot z)\\\
&\Leftrightarrow\alpha^{2}\beta(x)\dashv(y\vdash
z)=\alpha^{2}\beta(x)\dashv(y\bot z)\\\ &\Leftrightarrow\beta(x)\dashv(y\vdash
z)=\beta(x)\dashv(y\bot z).\end{array}$

Hence,
$(\mathcal{A},\bot,\dashv,\vdash,\alpha,\beta)\cong(\mathcal{A},\bot,\dashv,\vdash,\beta,\alpha).$
∎

######  Proposition 2.13.

Let $(\mathcal{A},\dashv,\vdash,\bot,\alpha,\beta)$ be a BiHom-associative
trialgebra. Therefore, $(\mathcal{A},\dashv,\bot,\ast,\alpha,\beta)$ is a
BiHom-associative trialgebra, where for any $x,y,z\in\mathcal{A},$ $x\ast
y=x\vdash y+x\bot y.$

###### Proof.

We prove only one axiom, as others are proven similarly. For any
$x,y,z\in\mathcal{A},$

$\begin{array}[]{ll}(x\ast y)\dashv\beta(z)&=(x\vdash y+x\bot
y)\dashv\beta(z)\\\ &=(x\vdash y)\dashv\beta(z)+(x\bot y)\dashv\beta(z)\\\
&=\alpha(x)\vdash(y\dashv z)+\alpha(x)\bot(y\dashv z)=\alpha(x)\ast(y\dashv
z),\end{array}$

which ends the proof. ∎

######  Proposition 2.14.

Let $(\mathcal{A},\dashv,\vdash,\bot,\alpha,\beta)$ be BiHom-associative
trialgebra. Define now binany operationqs by

$\begin{array}[]{ll}x\ast y&=x\dashv y-y\vdash x\\\ \left[x,y\right]&=x\bot
y-y\bot x.\end{array}$

Then $(\mathcal{A},\ast,\left[,\right],\alpha,\beta)$ becames a BiHom-
associative trialgebra.

###### Proof.

By define, for any $x,y,z\in\mathcal{A}$ we have

$\begin{array}[]{ll}\left[x,y\right]\ast\beta(z)&=\left[x,y\right]\dashv\beta(z)-\alpha(z)\vdash\left[x,y\right]\\\
&=(x\vdash y-y\bot x)\dashv\beta(z)-\alpha(z)\vdash(x\bot y-y\bot
y).\end{array}$

and

$\begin{array}[]{ll}\left[x\ast z,\beta(y)\right]+\left[\alpha(x),y\ast
z\right]&=\left[x\dashv z-z\vdash,\beta(y)\right]+\left[\alpha(x),y\dashv
z-z\vdash y\right]\\\ &=(x\vdash z-z\vdash
x)\bot\beta(y)-\alpha(y)\bot(x\dashv z-z\vdash x)\\\ &+\alpha(x)\bot(y\dashv
z-z\vdash y)-(y\dashv z-z\vdash y)\bot\beta(x).\\\ \end{array}$

Since

$\begin{array}[]{ll}(x\bot y)\dashv\beta(z)&=\alpha(x)\bot(y\vdash z),\\\
(y\bot x)\dashv\beta(z)&=\alpha(y)\bot(x\vdash z),\\\ \alpha(z)\vdash(x\bot
y)&=(z\vdash x)\bot\beta(y),\\\ \alpha(z)\vdash(y\bot x)&=(z\vdash
y)\bot\beta(x),\\\ (x\bot z)\bot\beta(y)&=\alpha(x)\bot(z\vdash y),\\\
\alpha(y)\bot(z\vdash x)&=(y\dashv z)\bot\beta(x),\\\ \end{array}$

in a BiHom-associative trialgebra, we have
$\left[x,y\right]\ast\alpha\beta(z)=\left[x\ast
z,\beta(y)\right]+\left[\alpha(x),y\ast z\right].$ This ends the proof. ∎

######  Proposition 2.15.

Let $(\mathcal{A},\dashv,\vdash,\bot,\alpha,\beta)$ be a BiHom-associative
trialgebra and $x\ast y=x\vdash y+x\dashv y+x\bot y.$ Then,
$(\mathcal{A},\ast,\alpha,\beta)$ is a BiHom-associative algebra.

###### Proof.

For any $x,y,z\in\mathcal{A},$

$\begin{array}[]{ll}(x\ast y)\ast\beta(z)&-\alpha(x)\ast(y\ast z)\\\
&=(x\vdash y+x\dashv y+x\bot y)\ast\beta(z)-\alpha(x)\ast(y\vdash z+y\dashv
z+y\bot z)\\\ &=(x\vdash y)\ast\beta(z)+(x\dashv y)\ast\beta(z)+(x\bot
y)\ast\beta(z)-\alpha(x)\ast(y\vdash z)-\alpha(x)\ast(y\dashv
z)-\alpha(x)\ast(y\bot z)\\\ &=(x\vdash y)\vdash\beta(z)+(x\vdash
y)\dashv\beta(z)+(x\vdash y)\bot\beta(z)+(x\dashv y)\vdash\beta(z)+(x\dashv
y)\dashv\beta(z)+(x\dashv y)\bot\beta(z)\\\ &+(x\bot y)\vdash\beta(z)+(x\bot
y)\dashv\beta(z)+(x\bot y)\bot\beta(z)-\alpha(x)\vdash(y\vdash
z)-\alpha(x)\dashv(y\vdash z)-\alpha(x)\bot(y\vdash z)\\\
&-\alpha(x)\vdash(y\dashv z)-\alpha(x)\dashv(y\dashv z)-\alpha(x)\bot(y\dashv
z)-\alpha(x)\vdash(y\bot z)-\alpha(x)\dashv(y\bot z)-\alpha(x)\bot(y\bot
z).\end{array}$

This confirms that $(\mathcal{A},\ast,\alpha,\beta)$ is a BiHom-associative
algebra. ∎

######  Definition 2.16.

An averaging operator over a BiHom-associative trialgebra
$(\mathcal{A},\dashv,\bot,\ast,\alpha,\beta)$ is a linear map
$\xi:\mathcal{A}\longrightarrow\mathcal{A}$ such that
$\alpha\circ\xi=\xi\circ\alpha$ and $\beta\circ\xi=\xi\circ\beta$, and for all
$x,y\in\mathcal{A},$

$\begin{array}[]{ll}\xi(\xi(x)\bot
y)&=\xi(x)\bot\xi(y)=\xi(x\bot\xi(y)),\end{array}$
$\begin{array}[]{ll}\xi(\xi(x)\dashv
y)&=\xi(x)\dashv\xi(y)=\xi(x\dashv\xi(y)),\end{array}$
$\begin{array}[]{ll}\xi(\xi(x)\vdash
y)&=\xi(x)\vdash\xi(y)=\xi(x\vdash\xi(y)).\end{array}$

######  Proposition 2.17.

Let $(\mathcal{A},\cdot)$ be an associative algebra and let
$\alpha,\beta:\mathcal{A}\longrightarrow\mathcal{A}$ be two averaging
operators such that $(\mathcal{A},\cdot,\alpha,\beta)$ is a BiHom-associative
algebra. For any $x,y\in\mathcal{A},$ define new operators on $\mathcal{A}$ by
$x\dashv y=\alpha(x)\cdot y,\quad x\vdash y=x\cdot\beta(y),\quad x\bot
y=\alpha(x)\cdot\beta(y).$

###### Proof.

We prove only one axiom, as others are proven similarly. For any
$x,y,z\in\mathcal{A},$

$\begin{array}[]{ll}(x\dashv y)\bot\beta(z)&=(\alpha(x)\cdot y)\bot\beta(z)\\\
&=\alpha(\alpha(x)\cdot y)\cdot\beta(\beta(y))\\\
&=(\alpha(x)\cdot\alpha(y))\cdot\beta(\beta(z))=\alpha(\alpha(x))\cdot(\alpha(y)\cdot\beta(z)),\end{array}$

which ends the proof. ∎

##  3 Classification of low-dimensional BiHom-associative trialgebras

This section provides a detailed description of $\alpha\beta$-derivation of
BiHom-associative trialgebras in dimension two and three over the field
$\mathbb{F}.$ Let $\left\\{e_{1},e_{2},e_{3},\cdots,e_{n}\right\\}$ be a basis
of an n-dimensional BiHom-associative trialgebra $\mathcal{A}.$ The product of
basis is expressed in terms of

$e_{i}\dashv e_{j}=\sum_{k=1}^{n}\gamma_{ij}^{k}e_{k};\quad e_{i}\vdash
e_{j}=\sum_{k=1}^{n}\delta_{ij}^{k}e_{k};\quad e_{i}\bot
e_{j}=\sum_{k=1}^{n}\xi_{ij}^{k}e_{k};\,\,\alpha(e_{i})=\sum_{j=1}^{n}a_{ji}e_{j};\quad\beta(e_{j})=\sum_{k=1}^{n}b_{kj}e_{k}.$

We have

$\displaystyle\sum_{j=1}^{n}b_{ji}a_{kj}$ $\displaystyle=$
$\displaystyle\sum_{j=1}^{n}a_{ji}b_{kj},$ (3.1)
$\displaystyle\sum_{p=1}^{n}\sum_{q=1}^{n}\gamma_{ij}^{p}b_{qk}\gamma_{pq}^{r}$
$\displaystyle=$
$\displaystyle\sum_{p=1}^{n}\sum_{q=1}^{n}a_{pi}\gamma_{jk}^{q}\gamma_{pq}^{r},$
(3.2)
$\displaystyle\sum_{p=1}^{n}\sum_{q=1}^{n}\gamma_{ij}^{p}b_{qk}\gamma_{pq}^{r}$
$\displaystyle=$
$\displaystyle\sum_{p=1}^{n}\sum_{q=1}^{n}a_{pi}\delta_{jk}^{q}\gamma_{pq}^{r}=\sum_{p=1}^{n}\sum_{q=1}^{n}a_{pi}\xi_{jk}^{q}\gamma_{pq}^{r},$
(3.3)
$\displaystyle\sum_{p=1}^{n}\sum_{q=1}^{n}\delta_{ij}^{p}b_{qk}\gamma_{pq}^{r}$
$\displaystyle=$
$\displaystyle\sum_{p=1}^{n}\sum_{q=1}^{n}a_{pi}\gamma_{jk}^{q}\delta_{pq}^{r},$
(3.4)
$\displaystyle\sum_{p=1}^{n}\sum_{q=1}^{n}\gamma_{ij}^{p}b_{qk}\delta_{pq}^{r}$
$\displaystyle=$
$\displaystyle\sum_{p=1}^{n}\sum_{q=1}^{n}a_{pi}\gamma_{jk}^{q}\delta_{pq}^{r}=\sum_{p=1}^{n}\sum_{q=1}^{n}\xi_{ij}^{p}b_{qk}\delta_{pq}^{r},$
(3.5)
$\displaystyle\sum_{p=1}^{n}\sum_{q=1}^{n}\delta_{ij}^{p}b_{qk}\delta_{pq}^{r}=\sum_{p=1}^{n}\sum_{q=1}^{n}a_{pi}\delta_{jk}^{q}\delta_{pq}^{r}$
;
$\displaystyle\quad\sum_{p=1}^{n}\sum_{q=1}^{n}\xi_{ij}^{p}b_{qk}\gamma_{pq}^{r}=\sum_{p=1}^{n}\sum_{q=1}^{n}a_{pi}\gamma_{jk}^{q}\xi_{pq}^{r},$
(3.6)
$\displaystyle\sum_{p=1}^{n}\sum_{q=1}^{n}\gamma_{ij}^{p}b_{qk}\xi_{pq}^{r}=\sum_{p=1}^{n}\sum_{q=1}^{n}a_{pi}\delta_{jk}^{q}\xi_{pq}^{r}$
;
$\displaystyle\sum_{p=1}^{n}\sum_{q=1}^{n}\delta_{ij}^{p}b_{qk}\xi_{pq}^{r}=\sum_{p=1}^{n}\sum_{q=1}^{n}a_{pi}\xi_{jk}^{q}\delta_{pq}^{r},$
(3.7)
$\displaystyle\sum_{p=1}^{n}\sum_{q=1}^{n}\xi_{ij}^{p}b_{qk}\xi_{pq}^{r}$
$\displaystyle=$
$\displaystyle\sum_{p=1}^{n}\sum_{q=1}^{n}a_{pi}\xi_{jk}^{q}\xi_{pq}^{r},$
(3.8)

where ${(a_{ij})}_{1\leq i,j\leq n}$ refers to the matrix of $\alpha$,
${(b_{ij})}_{1\leq i,j\leq n}$ corresponds to the matrix of $\beta$ and
$(\gamma_{ij}^{k})$, $(\delta_{ij}^{k})$ and $(\xi_{ij}^{k})$ stand for the
structure constants of $\mathcal{A}.$

The axioms in Remark 2.2 are,respectively,equivalent to

$\displaystyle\sum_{k=1}^{n}\gamma_{ij}^{k}a_{qk}=\sum_{k=1}^{n}\sum_{p=1}^{n}a_{ki}a_{pj}\gamma_{kp}^{q},\quad\sum_{k=1}^{n}\delta_{ij}^{k}a_{qk}=\sum_{k=1}^{n}\sum_{p=1}^{n}a_{ki}a_{pj}\delta_{kp}^{q},\quad\sum_{k=1}^{n}\xi_{ij}^{k}a_{qk}=\sum_{k=1}^{n}\sum_{p=1}^{n}a_{ki}a_{pj}\xi_{kp}^{q}.$
$\displaystyle\sum_{k=1}^{n}\gamma_{ij}^{k}b_{qk}=\sum_{k=1}^{n}\sum_{p=1}^{n}b_{ki}a_{pj}\gamma_{kp}^{q},\quad\sum_{k=1}^{n}\delta_{ij}^{k}b_{qk}=\sum_{k=1}^{n}\sum_{p=1}^{n}b_{ki}a_{pj}\delta_{kp}^{q},\quad\sum_{k=1}^{n}\xi_{ij}^{k}b_{qk}=\sum_{k=1}^{n}\sum_{p=1}^{n}b_{ki}a_{pj}\xi_{kp}^{q}.$

###  3.1 Classification in two-dimensional of BiHom-associative trialgebras

Note that $\mathcal{BT}as_{n}^{m}$ denote $m^{th}$ isomorphism class of BiHom-
associative trialgebra in dimension $n.$

######  Theorem 3.1.

Any two-dimensional complex BiHom-associative trialgebras are either
associative or isomorphic to one of the following pairwise non-isomorphic
BiHom-associative trialgebras :

$\mathcal{BT}as_{2}^{1}$ : $\begin{array}[]{ll}e_{1}\dashv e_{2}=e_{1},\\\
e_{2}\dashv e_{1}=e_{1},\end{array}\quad$ $\begin{array}[]{ll}e_{2}\dashv
e_{2}=e_{1},\\\ e_{1}\vdash e_{2}=e_{1},\end{array}\quad$
$\begin{array}[]{ll}e_{2}\vdash e_{1}=e_{1},\\\ e_{2}\bot
e_{2}=e_{1},\end{array}\quad$ $\begin{array}[]{ll}\alpha(e_{2})=e_{1},\\\
\beta(e_{2})=e_{1}.\end{array}$

$\mathcal{BT}as_{2}^{2}$ : $\begin{array}[]{ll}e_{1}\dashv e_{2}=e_{1},\\\
e_{2}\dashv e_{1}=e_{1},\\\ \end{array}\quad$ $\begin{array}[]{ll}e_{2}\dashv
e_{2}=e_{1},\\\ e_{1}\vdash e_{2}=e_{1},\\\ \end{array}\quad$
$\begin{array}[]{ll}e_{2}\vdash e_{1}=e_{1},\\\ e_{2}\vdash e_{2}=e_{1},\\\
\end{array}$ $\begin{array}[]{ll}e_{2}\bot e_{2}=e_{1},\\\
\alpha(e_{2})=\beta(e_{2})=e_{1}.\end{array}$

$\mathcal{BT}as_{2}^{3}$ : $\begin{array}[]{ll}e_{2}\dashv e_{2}=e_{1},\\\
e_{2}\vdash e_{2}=e_{1},\end{array}\quad$ $\begin{array}[]{ll}e_{2}\bot
e_{1}=e_{1},\\\ e_{2}\bot e_{2}=e_{1},\\\ \end{array}\quad$
$\begin{array}[]{ll}\alpha(e_{1})=\beta(e_{1})=e_{1},\\\
\alpha(e_{2})=\beta(e_{2})=e_{1}+e_{2}.\end{array}$

$\mathcal{BT}as_{2}^{4}$ : $\begin{array}[]{ll}e_{2}\dashv e_{1}=e_{1},\\\
e_{2}\dashv e_{2}=e_{1}+e_{2},\end{array}\,$ $\begin{array}[]{ll}e_{2}\vdash
e_{1}=e_{1},\\\ e_{2}\vdash e_{2}=e_{1}+e_{2},\end{array}\,$
$\begin{array}[]{ll}e_{2}\bot e_{1}=e_{1},\\\ e_{2}\bot
e_{2}=e_{1}+e_{2},\end{array}$
$\begin{array}[]{ll}\alpha(e_{1})=\beta(e_{1})=e_{1},\\\
\alpha(e_{2})=\beta(e_{2})=e_{1}+e_{2}.\end{array}$

$\mathcal{BT}as_{2}^{5}$ : $\begin{array}[]{ll}e_{1}\dashv e_{2}=e_{1},\\\
e_{2}\dashv e_{2}=e_{2},\\\ \end{array}\quad$ $\begin{array}[]{ll}e_{1}\vdash
e_{2}=e_{1},\\\ e_{2}\vdash e_{2}=e_{2},\\\ \end{array}\,$
$\begin{array}[]{ll}e_{1}\bot e_{2}=e_{1},\\\ e_{2}\bot
e_{2}=e_{2},\end{array}$
$\begin{array}[]{ll}\alpha(e_{1})=\beta(e_{1})=e_{1},\\\
\alpha(e_{2})=e_{2},\,\beta(e_{2})=e_{1}+e_{2}.\end{array}$

$\mathcal{BT}as_{2}^{6}$ : $\begin{array}[]{ll}e_{2}\dashv e_{2}=e_{1},\\\
e_{2}\vdash e_{2}=e_{1},\end{array}\quad$ $\begin{array}[]{ll}e_{2}\bot
e_{2}=e_{1},\\\ \alpha(e_{1})=\beta(e_{1})=e_{1},\\\ \end{array}\,$
$\begin{array}[]{ll}\alpha(e_{2})=e_{2},\\\
\beta(e_{2})=e_{1}+e_{2}.\end{array}$

$\mathcal{BT}as_{2}^{7}$ : $\begin{array}[]{ll}e_{1}\dashv e_{2}=e_{1},\\\
e_{1}\dashv e_{2}=e_{1}+e_{2},\end{array}\,$ $\begin{array}[]{ll}e_{1}\vdash
e_{2}=e_{1},\\\ e_{2}\vdash e_{2}=e_{1}+e_{2},\end{array}\,$
$\begin{array}[]{ll}e_{1}\bot e_{2}=e_{1},\\\ e_{2}\bot
e_{2}=e_{1}+e_{2},\end{array}\,$ $\begin{array}[]{ll}\alpha(e_{1})=e_{1},\\\
\alpha(e_{2})=e_{1}+e_{2},\end{array}$
$\begin{array}[]{ll}\beta(e_{1})=e_{1},\\\
\beta(e_{2})=e_{1}+e_{2}.\end{array}$

###  3.2 Classification in three-dimensional of BiHom-associative trialgebras

######  Theorem 3.2.

Any three-dimensional complex BiHom-associative trialgebras are either
associative or isomorphic to one of the following pairwise non-isomorphic
BiHom-associative trialgebra :

$\mathcal{BT}as_{3}^{1}$ : $\begin{array}[]{ll}e_{1}\dashv e_{2}=e_{1},\\\
e_{2}\dashv e_{2}=e_{1},\\\ e_{2}\dashv e_{3}=e_{1},\end{array}\quad$
$\begin{array}[]{ll}e_{2}\vdash e_{1}=e_{1},\\\ e_{2}\vdash e_{2}=e_{1},\\\
e_{3}\vdash e_{2}=e_{1},\end{array}\,$ $\begin{array}[]{ll}e_{2}\bot
e_{2}=e_{1},\\\ e_{2}\bot e_{3}=e_{1},\\\ e_{3}\bot e_{2}=e_{1},\end{array}$
$\begin{array}[]{ll}\alpha(e_{3})=e_{3},\\\ \beta(e_{2})=e_{1}.\end{array}$

$\mathcal{BT}as_{3}^{2}$ : $\begin{array}[]{ll}e_{1}\dashv e_{2}=e_{1},\\\
e_{2}\dashv e_{1}=e_{1},\\\ e_{3}\dashv e_{2}=e_{1},\end{array}\quad$
$\begin{array}[]{ll}e_{2}\vdash e_{2}=e_{1},\\\ e_{2}\vdash e_{3}=e_{1},\\\
e_{3}\vdash e_{3}=e_{1},\end{array}\,$ $\begin{array}[]{ll}e_{2}\bot
e_{1}=e_{1},\\\ e_{2}\bot e_{3}=e_{1},\\\ e_{3}\bot e_{2}=e_{1},\end{array}$
$\begin{array}[]{ll}\alpha(e_{3})=e_{3},\\\ \beta(e_{2})=e_{1},\\\
\beta(e_{3})=e_{3}.\end{array}$

$\mathcal{BT}as_{3}^{3}$ : $\begin{array}[]{ll}e_{2}\dashv e_{3}=e_{1},\\\
e_{3}\dashv e_{1}=e_{1},\\\ e_{3}\dashv e_{2}=e_{1},\end{array}\quad$
$\begin{array}[]{ll}e_{1}\vdash e_{2}=e_{1}\\\ e_{2}\vdash e_{3}=e_{1},\\\
e_{3}\vdash e_{2}=e_{1},\end{array}\,$ $\begin{array}[]{ll}e_{1}\bot
e_{1}=e_{1},\\\ e_{1}\bot e_{2}=e_{1},\\\ e_{2}\bot e_{3}=e_{1},\end{array}$
$\begin{array}[]{ll}e_{3}\bot e_{1}=e_{1},\\\ \alpha(e_{2})=e_{2},\\\
\beta(e_{3})=e_{3}.\end{array}$

$\mathcal{BT}as_{3}^{4}$ : $\begin{array}[]{ll}e_{1}\dashv e_{2}=e_{1},\\\
e_{2}\dashv e_{1}=e_{3},\\\ e_{2}\dashv e_{2}=e_{1}+e_{3},\\\
\end{array}\quad$ $\begin{array}[]{ll}e_{1}\vdash e_{2}=e_{1}\\\ e_{2}\vdash
e_{1}=e_{1}\\\ e_{2}\vdash e_{2}=e_{1}+e_{3},\\\ \end{array}\,$
$\begin{array}[]{ll}e_{2}\vdash e_{3}=e_{3},\\\ e_{2}\bot e_{2}=e_{1},\\\
e_{2}\bot e_{3}=e_{1},\\\ \end{array}$ $\begin{array}[]{ll}e_{3}\bot
e_{3}=e_{3},\\\ \alpha(e_{2})=e_{1},\\\ \beta(e_{2})=e_{1}.\end{array}$

$\mathcal{BT}as_{3}^{5}$ : $\begin{array}[]{ll}e_{1}\dashv e_{2}=e_{1},\\\
e_{2}\dashv e_{1}=e_{3},\\\ e_{2}\dashv e_{2}=e_{1},\\\ \end{array}\quad$
$\begin{array}[]{ll}e_{2}\dashv e_{3}=e_{1},\\\ e_{2}\vdash e_{1}=e_{1}\\\
e_{2}\vdash e_{3}=e_{1}\end{array}\,$ $\begin{array}[]{ll}e_{3}\vdash
e_{2}=e_{1},\\\ e_{1}\bot e_{2}=e_{1},\\\ e_{2}\bot e_{2}=e_{1},\end{array}$
$\begin{array}[]{ll}e_{3}\bot e_{2}=e_{1},\\\
\alpha(e_{3})=\beta(e_{3})=e_{3},\\\ \beta(e_{2})=e_{1}.\end{array}$

$\mathcal{BT}as_{3}^{6}$ : $\begin{array}[]{ll}e_{1}\dashv e_{2}=e_{1},\\\
e_{2}\dashv e_{1}=e_{1},\\\ e_{2}\dashv e_{2}=e_{1},\end{array}\quad$
$\begin{array}[]{ll}e_{2}\vdash e_{2}=e_{1},\\\ e_{2}\vdash e_{3}=e_{1},\\\
e_{3}\vdash e_{3}=e_{1},\end{array}\,$ $\begin{array}[]{ll}e_{2}\bot
e_{1}=e_{1},\\\ e_{2}\bot e_{2}=e_{1},\\\ e_{2}\bot e_{3}=e_{1},\end{array}$
$\begin{array}[]{ll}e_{3}\bot e_{3}=e_{1},\\\ \alpha(e_{3})=e_{3},\\\
\beta(e_{2})=e_{1}.\end{array}$

$\mathcal{BT}as_{3}^{7}$ : $\begin{array}[]{ll}e_{1}\dashv e_{1}=e_{1},\\\
e_{1}\dashv e_{2}=e_{1},\\\ e_{2}\dashv e_{3}=e_{1},\end{array}\quad$
$\begin{array}[]{ll}e_{3}\dashv e_{1}=e_{1},\\\ e_{1}\vdash e_{1}=e_{1}\\\
e_{2}\vdash e_{3}=e_{1},\end{array}\,$ $\begin{array}[]{ll}e_{3}\vdash
e_{1}=e_{1},\\\ e_{1}\bot e_{2}=e_{1},\\\ e_{2}\bot e_{3}=e_{1},\end{array}$
$\begin{array}[]{ll}e_{3}\bot e_{2}=e_{1},\\\ \alpha(e_{2})=e_{2},\\\
\beta(e_{3})=e_{3}.\end{array}$

$\mathcal{BT}as_{3}^{8}$ : $\begin{array}[]{ll}e_{1}\dashv e_{1}=e_{1},\\\
e_{2}\dashv e_{3}=e_{1},\\\ e_{3}\dashv e_{1}=e_{1},\end{array}\quad$
$\begin{array}[]{ll}e_{3}\dashv e_{2}=e_{1},\\\ e_{1}\vdash e_{2}=e_{1}\\\
e_{2}\vdash e_{3}=e_{1},\end{array}\,$ $\begin{array}[]{ll}e_{3}\vdash
e_{1}=e_{1},\\\ e_{2}\bot e_{3}=e_{1},\\\ e_{3}\bot e_{1}=e_{1},\end{array}$
$\begin{array}[]{ll}e_{3}\bot e_{2}=e_{1},\\\ \alpha(e_{2})=e_{2},\\\
\beta(e_{3})=e_{3}.\end{array}$

$\mathcal{BT}as_{3}^{9}$ : $\begin{array}[]{ll}e_{1}\dashv e_{1}=e_{1},\\\
e_{1}\dashv e_{2}=e_{1},\\\ e_{3}\dashv e_{1}=e_{1},\end{array}\quad$
$\begin{array}[]{ll}e_{2}\vdash e_{1}=e_{1}\\\ e_{3}\vdash e_{1}=e_{1},\\\
e_{3}\vdash e_{2}=e_{1},\end{array}\,$ $\begin{array}[]{ll}e_{1}\bot
e_{1}=e_{1},\\\ e_{3}\bot e_{1}=e_{1},\\\ e_{3}\bot e_{2}=e_{1},\end{array}$
$\begin{array}[]{ll}\alpha(e_{2})=e_{2},\\\ \beta(e_{3})=e_{3}.\end{array}$

$\mathcal{BT}as_{3}^{10}$ : $\begin{array}[]{ll}e_{2}\dashv e_{1}=e_{2},\\\
e_{2}\dashv e_{2}=e_{2},\\\ e_{3}\dashv e_{1}=e_{2},\end{array}\quad$
$\begin{array}[]{ll}e_{3}\dashv e_{2}=e_{2},\\\ e_{2}\vdash e_{1}=e_{2}\\\
e_{3}\vdash e_{1}=e_{2},\end{array}\,$ $\begin{array}[]{ll}e_{3}\vdash
e_{2}=e_{2},\\\ e_{2}\bot e_{2}=e_{2},\\\ e_{3}\bot e_{1}=e_{2},\end{array}$
$\begin{array}[]{ll}e_{3}\bot e_{2}=e_{2},\\\ \alpha(e_{1})=e_{1},\\\
\beta(e_{3})=e_{3}.\end{array}$

$\mathcal{BT}as_{3}^{11}$ : $\begin{array}[]{ll}e_{2}\dashv e_{1}=e_{3},\\\
e_{2}\dashv e_{3}=e_{3},\\\ e_{3}\dashv e_{1}=e_{3},\end{array}\quad$
$\begin{array}[]{ll}e_{2}\vdash e_{1}=e_{3}\\\ e_{3}\vdash e_{1}=e_{3},\\\
e_{3}\vdash e_{3}=e_{3},\end{array}\,$ $\begin{array}[]{ll}e_{2}\bot
e_{1}=e_{3},\\\ e_{2}\bot e_{3}=e_{3},\\\ e_{3}\bot e_{1}=e_{3},\end{array}$
$\begin{array}[]{ll}e_{3}\bot e_{3}=e_{3},\\\ \alpha(e_{1})=e_{1},\\\
\beta(e_{3})=e_{3}.\end{array}$

$\mathcal{BT}as_{3}^{12}$ : $\begin{array}[]{ll}e_{2}\dashv e_{1}=e_{3},\\\
e_{3}\dashv e_{1}=e_{3},\\\ e_{3}\dashv e_{3}=e_{3},\end{array}\quad$
$\begin{array}[]{ll}e_{1}\vdash e_{2}=e_{3}\\\ e_{2}\vdash e_{1}=e_{3},\\\
e_{3}\vdash e_{1}=e_{3},\end{array}\,$ $\begin{array}[]{ll}e_{2}\bot
e_{1}=e_{3},\\\ e_{2}\bot e_{3}=e_{3},\\\ e_{3}\bot e_{3}=e_{3},\end{array}$
$\begin{array}[]{ll}\alpha(e_{1})=e_{1},\\\ \beta(e_{3})=e_{3}.\end{array}$

$\mathcal{BT}as_{3}^{13}$ : $\begin{array}[]{ll}e_{1}\dashv e_{2}=e_{3},\\\
e_{2}\dashv e_{1}=e_{3},\\\ e_{3}\dashv e_{1}=e_{3},\end{array}\quad$
$\begin{array}[]{ll}e_{3}\dashv e_{3}=e_{3},\\\ e_{2}\vdash e_{1}=e_{3}\\\
e_{2}\vdash e_{3}=e_{3},\end{array}\,$ $\begin{array}[]{ll}e_{3}\vdash
e_{1}=e_{3},\\\ e_{1}\bot e_{2}=e_{3},\\\ e_{2}\bot e_{3}=e_{3},\end{array}$
$\begin{array}[]{ll}e_{3}\bot e_{3}=e_{3},\\\ \alpha(e_{1})=e_{1},\\\
\beta(e_{3})=e_{3}.\end{array}$

$\mathcal{BT}as_{3}^{14}$ : $\begin{array}[]{ll}e_{1}\dashv e_{2}=e_{3},\\\
e_{2}\dashv e_{2}=e_{3},\\\ e_{2}\dashv e_{3}=e_{3},\end{array}\quad$
$\begin{array}[]{ll}e_{2}\vdash e_{1}=e_{3}\\\ e_{2}\vdash e_{2}=e_{3},\\\
e_{3}\vdash e_{2}=e_{3},\end{array}\,$ $\begin{array}[]{ll}e_{3}\vdash
e_{3}=e_{3},\\\ e_{1}\bot e_{2}=e_{3},\\\ e_{2}\bot e_{2}=e_{3},\end{array}$
$\begin{array}[]{ll}e_{3}\bot e_{2}=e_{3},\\\ \alpha(e_{1})=e_{1},\\\
\beta(e_{1})=e_{1}.\end{array}$

$\mathcal{BT}as_{3}^{15}$ : $\begin{array}[]{ll}e_{2}\dashv e_{1}=e_{3},\\\
e_{2}\dashv e_{2}=e_{3},\\\ e_{2}\dashv e_{3}=e_{3},\end{array}\quad$
$\begin{array}[]{ll}e_{2}\vdash e_{2}=e_{3}\\\ e_{2}\vdash e_{3}=e_{3},\\\
e_{3}\vdash e_{3}=e_{3},\end{array}\,$ $\begin{array}[]{ll}e_{2}\bot
e_{1}=e_{3},\\\ e_{3}\bot e_{2}=e_{3},\\\ e_{3}\bot e_{3}=e_{3},\end{array}$
$\begin{array}[]{ll}\alpha(e_{1})=e_{1},\\\ \beta(e_{1})=e_{1}.\end{array}$

$\mathcal{BT}as_{3}^{16}$ : $\begin{array}[]{ll}e_{2}\dashv
e_{1}=e_{1}+e_{3},\\\ e_{2}\dashv e_{2}=e_{1}+e_{3},\\\ e_{2}\dashv
e_{3}=e_{1}+e_{3},\end{array}\quad$ $\begin{array}[]{ll}e_{3}\dashv
e_{2}=e_{3},\\\ e_{1}\vdash e_{2}=e_{1}+e_{3}\\\ e_{2}\vdash
e_{1}=e_{1}+e_{3},\end{array}\,$ $\begin{array}[]{ll}e_{2}\bot
e_{2}=e_{1}+e_{3},\\\ e_{2}\bot e_{3}=e_{3},\\\ e_{3}\bot
e_{2}=e_{1},\end{array}$ $\begin{array}[]{ll}\alpha(e_{2})=e_{1},\\\
\alpha(e_{3})=e_{2},\\\ \beta(e_{2})=e_{1}.\end{array}$

$\mathcal{BT}as_{3}^{17}$ : $\begin{array}[]{ll}e_{1}\dashv
e_{1}=e_{1}+e_{3},\\\ e_{2}\dashv e_{1}=e_{1}+e_{3},\\\ e_{2}\dashv
e_{2}=e_{1}+e_{3},\end{array}\quad$ $\begin{array}[]{ll}e_{2}\dashv
e_{3}=e_{1}+e_{3},\\\ e_{2}\vdash e_{1}=e_{1}+e_{3}\\\ e_{2}\vdash
e_{2}=e_{1}+e_{3},\end{array}\,$ $\begin{array}[]{ll}e_{3}\vdash
e_{2}=e_{1}+e_{3},\\\ e_{2}\bot e_{2}=e_{1}+e_{3},\\\ e_{3}\bot
e_{2}=e_{1}+e_{3},\end{array}$ $\begin{array}[]{ll}\alpha(e_{2})=e_{1},\\\
\alpha(e_{3})=e_{2},\\\ \beta(e_{2})=e_{1}.\end{array}$

$\mathcal{BT}as_{3}^{18}$ : $\begin{array}[]{ll}e_{2}\dashv
e_{1}=e_{1}+e_{3},\\\ e_{2}\dashv e_{2}=e_{1}+e_{3},\\\ e_{2}\dashv
e_{3}=e_{1}+e_{3},\end{array}\quad$ $\begin{array}[]{ll}e_{3}\dashv
e_{2}=e_{1}+e_{3},\\\ e_{1}\vdash e_{2}=e_{1}+e_{3}\\\ e_{2}\vdash
e_{1}=e_{1}+e_{3},\end{array}\,$ $\begin{array}[]{ll}e_{3}\vdash
e_{2}=e_{1}+e_{3},\\\ e_{2}\bot e_{2}=e_{1}+e_{3},\\\ e_{2}\bot
e_{3}=e_{1}+e_{3},\end{array}$ $\begin{array}[]{ll}\alpha(e_{2})=e_{1},\\\
\alpha(e_{3})=e_{2},\\\ \beta(e_{2})=e_{1}.\end{array}$

$\mathcal{BT}as_{3}^{19}$ : $\begin{array}[]{ll}e_{2}\dashv e_{2}=e_{1},\\\
e_{2}\dashv e_{3}=e_{1}+e_{3},\\\ e_{3}\dashv e_{3}=e_{3},\\\
\end{array}\quad$ $\begin{array}[]{ll}e_{1}\vdash e_{2}=e_{1}\\\ e_{2}\vdash
e_{1}=e_{1}\\\ e_{3}\vdash e_{3}=e_{1}+e_{3},\\\ \end{array}\,$
$\begin{array}[]{ll}e_{2}\bot e_{1}=e_{1}+e_{3},\\\ e_{2}\bot
e_{3}=e_{1}+e_{3},\\\ e_{3}\bot e_{3}=e_{1}+e_{3},\end{array}$
$\begin{array}[]{ll}\alpha(e_{2})=e_{1},\\\ \beta(e_{2})=e_{1}.\end{array}$

$\mathcal{BT}as_{3}^{20}$ : $\begin{array}[]{ll}e_{1}\dashv
e_{2}=e_{1}+e_{3},\\\ e_{2}\dashv e_{2}=e_{1}+e_{3},\\\ e_{2}\dashv
e_{3}=e_{1}+e_{3},\end{array}\quad$ $\begin{array}[]{ll}e_{2}\vdash
e_{2}=e_{1}+e_{3}\\\ e_{2}\vdash e_{3}=e_{1}+e_{3},\\\ e_{3}\vdash
e_{2}=e_{1}+e_{3},\end{array}\,$ $\begin{array}[]{ll}e_{2}\bot e_{1}=e_{1},\\\
e_{2}\bot e_{2}=e_{1}+e_{3},\\\ e_{3}\bot e_{2}=e_{1}+e_{3},\end{array}$
$\begin{array}[]{ll}\alpha(e_{2})=e_{1},\\\ \alpha(e_{3})=e_{2},\\\
\beta(e_{2})=e_{1}.\end{array}$

$\mathcal{BT}as_{3}^{21}$ : $\begin{array}[]{ll}e_{1}\dashv e_{2}=e_{1},\\\
e_{2}\dashv e_{1}=e_{1}+e_{3},\\\ e_{2}\dashv e_{2}=e_{1}+e_{3},\\\
\end{array}\quad$ $\begin{array}[]{ll}e_{1}\vdash e_{2}=e_{1}+e_{3}\\\
e_{2}\vdash e_{1}=e_{1}+e_{3}\\\ e_{2}\vdash e_{2}=e_{1}+e_{3},\\\
\end{array}\,$ $\begin{array}[]{ll}e_{2}\bot e_{1}=e_{1},\\\ e_{2}\bot
e_{3}=e_{1}+e_{3},\\\ e_{3}\bot e_{3}=e_{1}+e_{3},\\\ \end{array}$
$\begin{array}[]{ll}\alpha(e_{2})=e_{1},\\\ \beta(e_{2})=e_{1}.\end{array}$

$\mathcal{BT}as_{3}^{22}$ : $\begin{array}[]{ll}e_{1}\dashv e_{2}=e_{3},\\\
e_{2}\dashv e_{2}=e_{1}+e_{3},\\\ e_{3}\dashv
e_{3}=e_{1}+e_{3},\end{array}\quad$ $\begin{array}[]{ll}e_{2}\vdash
e_{1}=e_{1}\\\ e_{2}\vdash e_{2}=e_{1}+e_{3},\\\ e_{3}\vdash
e_{2}=e_{1}+e_{3},\end{array}\,$ $\begin{array}[]{ll}e_{2}\bot
e_{2}=e_{1}+e_{3},\\\ e_{2}\bot e_{3}=e_{1}+e_{3},\\\ e_{3}\bot
e_{2}=e_{1},\end{array}$ $\begin{array}[]{ll}\alpha(e_{2})=e_{1},\\\
\alpha(e_{3})=e_{2},\\\ \beta(e_{2})=e_{1}.\end{array}$

$\mathcal{BT}as_{3}^{23}$ : $\begin{array}[]{ll}e_{2}\dashv e_{1}=e_{3},\\\
e_{2}\dashv e_{2}=e_{1}+e_{3},\\\ e_{2}\dashv
e_{3}=e_{1}+e_{3},\end{array}\quad$ $\begin{array}[]{ll}e_{3}\dashv
e_{2}=e_{1}+e_{3},\\\ e_{2}\vdash e_{1}=e_{3}\\\ e_{2}\vdash
e_{2}=e_{1},\end{array}\,$ $\begin{array}[]{ll}e_{2}\bot e_{1}=e_{1}+e_{3},\\\
e_{2}\bot e_{2}=e_{3},\\\ e_{2}\bot e_{3}=e_{3},\end{array}$
$\begin{array}[]{ll}\alpha(e_{2})=e_{1},\\\ \alpha(e_{3})=e_{2},\\\
\beta(e_{2})=e_{1}.\end{array}$

$\mathcal{BT}as_{3}^{24}$ : $\begin{array}[]{ll}e_{2}\dashv e_{1}=e_{3},\\\
e_{2}\dashv e_{2}=e_{3},\\\ e_{2}\dashv e_{3}=e_{1}+e_{3},\end{array}\quad$
$\begin{array}[]{ll}e_{2}\vdash e_{2}=e_{1}+e_{3}\\\ e_{2}\vdash
e_{3}=e_{3},\\\ e_{2}\vdash e_{3}=e_{1},\end{array}\,$
$\begin{array}[]{ll}e_{1}\bot e_{2}=e_{1}+e_{3},\\\ e_{2}\bot
e_{3}=e_{1}+e_{3},\\\ e_{3}\bot e_{2}=e_{3},\end{array}$
$\begin{array}[]{ll}\alpha(e_{2})=e_{1},\\\ \alpha(e_{3})=e_{2},\\\
\beta(e_{2})=e_{1}.\end{array}$

##  4 Derivations of low-dimensional BiHom-associative trialgebras

######  Definition 4.1.

An $\alpha\beta$-derivation of the BiHom-associative trialgebras $\mathcal{A}$
is a linear transformation : $\mathcal{D}:\mathcal{A}\rightarrow\mathcal{A}$
satisfying

$\displaystyle\alpha\circ d=d\circ\alpha$ , $\displaystyle\beta\circ
d=d\circ\beta$ (4.1) $\displaystyle d(x\dashv y)$ $\displaystyle=$
$\displaystyle d(x)\dashv\alpha\beta(y)+\alpha\beta(x)\dashv d(y)$ (4.2)
$\displaystyle d(x\vdash y)$ $\displaystyle=$ $\displaystyle
d(x)\vdash\alpha\beta(y)+\alpha\beta(x)\vdash d(y)$ (4.3) $\displaystyle
d(x\bot y)$ $\displaystyle=$ $\displaystyle
d(x)\bot\alpha\beta(y)+\alpha\beta(x)\bot d(y),$ (4.4)

for all $x,y\in\mathcal{A}.$

This section illustrates in depth, $\alpha\beta$-derivation of BiHom-
associative trialgebras in dimension two and three over the field
$\mathbb{F}.$ Let $\left\\{e_{1},e_{2},e_{3},\cdots,e_{n}\right\\}$ be a basis
of an n-dimensional BiHom-associative trialgebra $\mathcal{A}.$ The product of
basis is denoted by

$\displaystyle d(e_{p})=\sum_{q=1}^{n}d_{qp}e_{q}.$

We have

$\displaystyle\sum_{p=1}^{n}d_{pk}a_{qp}=\sum_{p=1}^{n}a_{pk}d_{qp}$ ;
$\displaystyle\sum_{p=1}^{n}d_{pk}b_{qp}=\sum_{p=1}^{n}b_{pk}d_{qp},$ (4.5)
$\displaystyle\sum_{k=1}^{n}\gamma_{ij}^{p}d_{rp}=\sum_{k=1}^{n}\sum_{q=1}^{n}\sum_{p=1}d_{ki}b_{pj}a_{qp}\gamma_{kq}^{r}$
$\displaystyle+$
$\displaystyle\sum_{q=1}^{n}\sum_{k=1}^{n}\sum_{p=1}b_{ki}a_{qk}d_{pj}\gamma_{qp}^{r},$
(4.6)
$\displaystyle\sum_{k=1}^{n}\delta_{ij}^{p}d_{rp}=\sum_{k=1}^{n}\sum_{q=1}^{n}\sum_{p=1}d_{ki}b_{pj}a_{qp}\delta_{kq}^{r}$
$\displaystyle+$
$\displaystyle\sum_{q=1}^{n}\sum_{k=1}^{n}\sum_{p=1}b_{ki}a_{qk}d_{pj}\delta_{qp}^{r},$
(4.7)
$\displaystyle\sum_{k=1}^{n}\xi_{ij}^{p}d_{rp}=\sum_{k=1}^{n}\sum_{q=1}^{n}\sum_{p=1}d_{ki}b_{pj}a_{qp}\xi_{kq}^{r}$
$\displaystyle+$
$\displaystyle\sum_{q=1}^{n}\sum_{k=1}^{n}\sum_{p=1}b_{ki}a_{qk}d_{pj}\xi_{qp}^{r}.$
(4.8)

######  Theorem 4.2.

The derivation of two-dimensional BiHom-associative trialgebras has the
following form :

IC | Der$(\mathcal{D})$ | $Dim(\mathcal{D})$ | IC | Der$(\mathcal{D})$ | $Dim(\mathcal{D})$
---|---|---|---|---|---
$\mathcal{BT}as_{2}^{1}$ | $\left(\begin{array}[]{cccc}0&0\\\ d_{21}&0\end{array}\right)$ | 1 | $\mathcal{BT}as_{2}^{2}$ | $\left(\begin{array}[]{cccc}0&0\\\ d_{21}&0\end{array}\right)$ | 1
$\mathcal{BT}as_{2}^{6}$ | $\left(\begin{array}[]{cccc}0&0\\\ d_{21}&0\end{array}\right)$ | 1 |  |  |

######  Theorem 4.3.

The derivation of three-dimensional BiHom-associative trialgebras has the
following form :

IC | Der$(\mathcal{D})$ | $Dim(\mathcal{D})$ | IC | Der$(\mathcal{D})$ | $Dim(\mathcal{D})$
---|---|---|---|---|---
$\mathcal{BT}as_{3}^{1}$ | $\left(\begin{array}[]{cccc}0&0&0\\\ d_{21}&0&0\\\ 0&0&d_{33}\end{array}\right)$ | 2 | $\mathcal{BT}as_{3}^{2}$ | $\left(\begin{array}[]{cccc}0&0&0\\\ d_{21}&0&0\\\ 0&0&0\end{array}\right)$ | 1
$\mathcal{BT}as_{3}^{3}$ | $\left(\begin{array}[]{cccc}0&0&0\\\ 0&d_{22}&0\\\ 0&0&d_{33}\end{array}\right)$ | 2 | $\mathcal{BT}as_{3}^{4}$ | $\left(\begin{array}[]{cccc}0&0&0\\\ d_{21}&0&d_{23}\\\ 0&0&0\end{array}\right)$ | 2
$\mathcal{BT}as_{3}^{5}$ | $\left(\begin{array}[]{cccc}0&0&0\\\ d_{21}&0&0\\\ 0&0&0\end{array}\right)$ | 1 | $\mathcal{BT}as_{3}^{6}$ | $\left(\begin{array}[]{cccc}0&0&0\\\ d_{21}&0&0\\\ 0&0&d_{33}\end{array}\right)$ | 2
$\mathcal{BT}as_{3}^{7}$ | $\left(\begin{array}[]{cccc}0&0&0\\\ 0&d_{22}&0\\\ 0&0&d_{33}\end{array}\right)$ | 2 | $\mathcal{BT}as_{3}^{8}$ | $\left(\begin{array}[]{cccc}0&0&0\\\ 0&d_{22}&0\\\ 0&0&d_{33}\end{array}\right)$ | 2
$\mathcal{BT}as_{3}^{9}$ | $\left(\begin{array}[]{cccc}0&0&0\\\ 0&d_{22}&0\\\ 0&0&d_{33}\end{array}\right)$ | 2 | $\mathcal{BT}as_{3}^{10}$ | $\left(\begin{array}[]{cccc}d_{11}&0&0\\\ 0&0&0\\\ 0&0&d_{33}\end{array}\right)$ | 2
$\mathcal{BT}as_{3}^{11}$ | $\left(\begin{array}[]{cccc}d_{11}&0&0\\\ 0&d_{22}&0\\\ 0&0&0\end{array}\right)$ | 2 | $\mathcal{BT}as_{3}^{12}$ | $\left(\begin{array}[]{cccc}d_{11}&0&0\\\ 0&d_{22}&0\\\ 0&0&0\end{array}\right)$ | 2
$\mathcal{BT}as_{3}^{13}$ | $\left(\begin{array}[]{cccc}d_{11}&0&0\\\ 0&d_{22}&0\\\ 0&0&0\end{array}\right)$ | 2 | $\mathcal{BT}as_{3}^{14}$ | $\left(\begin{array}[]{cccc}d_{11}&0&0\\\ 0&d_{22}&d_{23}\\\ 0&0&0\end{array}\right)$ | 3
$\mathcal{BT}as_{3}^{15}$ | $\left(\begin{array}[]{cccc}d_{11}&0&0\\\ 0&0&d_{23}\\\ 0&0&0\end{array}\right)$ | 2 | $\mathcal{BT}as_{3}^{19}$ | $\left(\begin{array}[]{cccc}0&0&0\\\ d_{21}&0&d_{23}\\\ 0&0&0\end{array}\right)$ | 2

###### Proof.

Departing from Theorem 4.3, we provide the proof only for one case in order to
illustrate the used approach. The other cases can be handled similarly with or
no modification(s). We refer to Theorem 4.3 and use the system of equations
(4.5), (4.6), (4.7), (4.8) of operations $\dashv,\vdash$ and $\bot$,
respectively, to obtain all derivations of BiHom-associative trialgebra in
dimension three. Let’s consider ${Trias}_{3}^{1}$. Applying the systems of
equations (4.5), (4.6), (4.7), (4.8), we therefore get the derivation for
${Trias}_{3}^{1}$ as follows $d_{1}=\left(\begin{array}[]{ccc}0&0&0\\\
d_{21}&0&0\\\ 0&0&d_{33}\end{array}\right)$. Clary, $d_{1}(e_{1})=e_{2}$,
$d_{2}(e_{3})=e_{3}$. Hence, the derivations of ${Trias}_{3}^{1}$ are
indicated as follows
$d_{1}=\left(\begin{array}[]{ccc}0&0&0\\\ 1&0&0\\\ 0&0&0\end{array}\right)$,
$d_{2}=\left(\begin{array}[]{ccc}0&0&0\\\ 0&0&0\\\ 0&0&1\end{array}\right)$ is
basis of $Der(\mathcal{D})$ and Dim$Der(\mathcal{D})=2.$ The derivations of
the remaining parts of three-dimension associative trialgebras can be handled
in a similar manner as depicted above. ∎

######  Corollary 4.4.

  * •

The dimensions of the derivations of two-dimensional BiHom-associative
trialgebras range between zero and two.

  * •

The dimensions of the derivations of three-dimensional BiHom-associative
trialgebras range between zero and three.

##  5 Centroids of low-dimensional BiHom-associative trialgebras

###  5.1 Properties of centroids of BiHom-associative trialgebras

In this section, we display the following results as far as properties of
centroids of BiHom-associative trialgebras $\mathcal{A}$ are concerned.

######  Definition 5.1.

Let $\mathcal{H}$ be a nonempty subset of $\mathcal{A}$. The subset

$Z_{\mathcal{A}}(\mathcal{H})=\left\\{x\in\mathcal{H}\,|\,\alpha\beta(x)\bullet\mathcal{H}=\mathcal{H}\bullet\alpha\beta(x)=0\right\\},$
(5.1)

is said to be centralized of $\mathcal{H}$ in $\mathcal{A}$, where the
$\bullet$ is $\dashv,\vdash$ and $\bot,$ respectively.

######  Definition 5.2.

Let $\psi\in End(\mathcal{A})$. If $\psi(\mathcal{A})\subseteq Z(\mathcal{A})$
and $\psi(\mathcal{A}^{2})=0$, then $\psi$ is called a central
$\alpha\beta$-derivation. The set of all central $\alpha\beta$-derivations of
$\mathcal{A}$ is denoted by $\mathcal{C}(\mathcal{A})$.

######  Proposition 5.3.

Consider $(\mathcal{A},\dashv,\vdash,\bot,\alpha,\beta)$ a BiHom-associative
trialgebra. Then,

  1. i)

$\Gamma(\mathcal{A})Der(\mathcal{A})\subseteq Der(\mathcal{A})$.

  2. ii)

$\left[\Gamma(\mathcal{A}),Dr(\mathcal{A})\right]\subseteq\Gamma(\mathcal{A}).$

  3. iii)

$\left[\Gamma(\mathcal{A}),\Gamma(\mathcal{A})\right](\mathcal{A})\subseteq\Gamma(\mathcal{A})$
and $\left[\Gamma(\mathcal{A}),\Gamma(\mathcal{A})\right](\mathcal{A}^{2})=0.$

###### Proof.

The proof of parts $i)-iii)$ is straightforward relying on definitions of
$\alpha\beta$-derivation and centroids. ∎

Now, we shall introduce centroid for BiHom-associative trialgebras.

######  Definition 5.4.

Let $(\mathcal{A},\dashv,\vdash,\bot,\alpha,\beta)$ be a BiHom-associative
trialgebra. A linear map $\psi:\mathcal{A}\rightarrow\mathcal{A}$ is called an
element of $(\alpha,\beta)$-element of centroids on $\mathcal{A}$ if, for all
$x,y\in\mathcal{A}$,

$\displaystyle\alpha\circ\psi$ $\displaystyle=$
$\displaystyle\psi\circ\alpha,\quad\beta\circ\psi=\psi\circ\beta,$ (5.2)
$\displaystyle\psi(x)\dashv\alpha\beta(y)$ $\displaystyle=$
$\displaystyle\psi(x)\dashv\psi(y)=\alpha\beta(x)\dashv\psi(y),$ (5.3)
$\displaystyle\psi(x)\vdash\alpha\psi(y)$ $\displaystyle=$
$\displaystyle\psi(x)\vdash\psi(y)=\alpha\beta(x)\vdash\psi(y),$ (5.4)
$\displaystyle\psi(x)\bot\alpha\beta(y)$ $\displaystyle=$
$\displaystyle\psi(x)\bot\psi(y)=\alpha\beta(x)\bot\psi(y).$ (5.5)

The set of all $(\alpha,\beta)$-elements of centroid of $\mathcal{A}$ is
denoted $Cent_{(\alpha,\beta)}(\mathcal{A})$. The centroid of $\mathcal{A}$ is
denoted $Cent(\mathcal{A})$.

######  Proposition 5.5.

Let $(\mathcal{A},\dashv,\vdash,\bot,\alpha,\beta)$ be a BiHom-associative
trialgebra and $\varphi\in Cent(\mathcal{A}),\,d\in Der(\mathcal{A}).$ Then,
$\varphi\circ d$ is a $\alpha\beta$-derivation of $\mathcal{A}.$

###### Proof.

Indeed, if $x,y\in\mathcal{A}$, then

$\begin{array}[]{ll}(\varphi\circ d)(x\bullet
y)&=\varphi(d(x)\bullet\alpha\beta(y)+\alpha\beta(x)\bullet d(y))\\\
&=\varphi(d(x)\bullet y)+\varphi(x\bullet d(y))=(\varphi\circ
d)(x)\bullet\alpha\beta(y)+\alpha\beta(x)\bullet(\varphi\circ
d)(y),\end{array}$

where $\bullet$ is $\dashv,\vdash$ and $\bot$, respectively. ∎

######  Proposition 5.6.

Let $(\mathcal{A},\dashv,\vdash,\bot,\alpha,\beta)$ be a BiHom-associative
trialgebra over a field $\mathbb{F}$. Hence,
$\mathcal{C}(\mathcal{A})=Cent(\mathcal{A})\cap Der(\mathcal{A}).$

###### Proof.

If $\psi\in Cent(\mathcal{A})\cap Der(\mathcal{A})$, then by definition of
$Cent(\mathcal{A}$ and $Der(\mathcal{A})$, we have

$\psi(x\bullet y)=\psi(x)\bullet\alpha\beta(y)+\alpha\beta(x)\bullet\psi(y)$
and $\psi(x\bullet y)=\psi(x)\circ\alpha\beta(y)=\alpha\beta(x)\circ\psi(y)$
for $x,y\in\mathcal{A}.$ The yields $\psi(\mathcal{A}\mathcal{A})=0$ and
$\psi(\mathcal{A})\subseteq Z(\mathcal{A})$ i.e $Cent(\mathcal{A})\cap
Der(\mathcal{A})\subseteq Cent(\mathcal{A}).$ The inverse is obvious since
$\mathcal{C}(\mathcal{A})$ is in both $Cent(\mathcal{A})$ and
$Der(\mathcal{A}),$ where $\bullet$ is $\dashv,\vdash$ and $\bot$,
respectively. ∎

######  Proposition 5.7.

Let $(\mathcal{A},\dashv,\vdash,\bot,\alpha,\beta)$ be a BiHom-associative
trialgebra. Therefore, for any $d\in Der(\mathcal{A})$ and $\varphi\in
Cent(\mathcal{A})$, we have

  1. (i)

The composition $d\circ\varphi$ is in $Cent(\mathcal{A})$, if and only if
$\varphi\circ d$ is a central $\alpha\beta$-derivation of $\mathcal{A}.$

  2. (ii)

The composition $d\circ\varphi$ is a $\alpha\beta$-derivation of
$\mathcal{A}$, if and only if $\left[d,\varphi\right]$ is a central
$\alpha\beta$-derivation of $\mathcal{A}.$

###### Proof.

  1. i)

For any $\varphi\in Cent(\mathcal{A}),\,d\in
Der(\mathcal{A}),\,\forall\,x,y\in\mathcal{A}$, we have

$\begin{array}[]{ll}d\circ\varphi(x\bullet y)=d\circ\varphi(x)\bullet
y&=d\circ\varphi(x)\bullet y+\varphi(x)\bullet d(y)\\\
&=d\circ\varphi(x)\bullet y+\varphi\circ d(x\bullet y)-\varphi\circ
d(x)\bullet y.\end{array}$

Thus, $(d\circ\varphi-\varphi\circ d)(x\bullet
y)=(d\bullet\varphi-\varphi\circ d)(x)\bullet y.$

  2. ii)

Let $d\circ\varphi\in Der(\mathcal{A})$. Using $\left[d,\varphi\right]\in
Cent(\mathcal{A})$, we get

$\left[d,\varphi\right](x\bullet
y)=(\left[d,\varphi\right](x))\bullet\alpha\beta(y)=\alpha\beta(x)\bullet(\left[d,\varphi\right](y))$
(5.6)

On the other side, $\left[d,\varphi\right]d\circ\varphi-\varphi\circ d$ and
$d\circ\varphi,\varphi\circ d\in Der(\mathcal{A}).$ Therefore,

$\left[d,\varphi\right](x\bullet
y)=(d(\varphi\circ(x))\bullet\alpha\beta(y)+\alpha\beta(x)\bullet(d\circ\varphi(y))-(\varphi\circ
d(x))\bullet\alpha\beta(y)-\alpha\beta(x)\bullet(\varphi\circ d(y)).$ (5.7)

Referring to (5.6) and (5.7), we get
$\alpha\beta(x)\bullet(\left[d,\varphi\right])(y)=(\left[d,\varphi\right])(x)\bullet\alpha\beta(y)=0.$

At this stage of analysis,let $\left[d,\varphi\right]$ be a central
$\alpha\beta$-derivation of $\mathcal{A}$. Then,

$\begin{array}[]{ll}d\circ\varphi(x\bullet
y)&=\left[d\circ\varphi\right](x\bullet y)+(\varphi\circ d)(x\bullet y)\\\
&=\varphi(\circ d(x)\bullet\alpha\beta(y))+\varphi(\alpha\beta(x)\bullet
d(y))\\\ &=(\varphi\circ
d)(x)\bullet\alpha\beta(y)+\alpha\beta(x)\bullet(\varphi\circ
d)(y),\end{array}$

where $\bullet$ represents the products $\dashv,\vdash$ and $\bot$,
respectively. ∎

###  5.2 Centroids of BiHom-associative trialgebras

This section adresses in details the centroid of BiHom-associative trialgebras
in dimension two and three over the field $\mathbb{F}.$ Let
$\left\\{e_{1},e_{2},e_{3},\cdots,e_{n}\right\\}$ be a basis of an
n-dimensional BiHom-associative trialgebra $\mathcal{A}.$ The product of basis
is determined by

$\displaystyle\psi(e_{i})=\sum_{j=1}^{n}c_{ji}e_{j}.$

We,therefore,have

$\displaystyle\sum_{p=1}^{n}c_{pi}a_{qp}=\sum_{p=1}^{n}a_{pi}c_{qp}$ ;
$\displaystyle\sum_{p=1}^{n}c_{pi}b_{qp}=\sum_{p=1}^{n}b_{pi}c_{qp},$ (5.8)
$\displaystyle\sum_{k=1}^{n}\sum_{q=1}\sum_{p=1}c_{ki}b_{pj}a_{qp}\gamma_{kq}^{r}=\sum_{p=1}^{n}\sum_{q=1}^{n}c_{pi}c_{qj}\gamma_{pq}^{r}$
$\displaystyle=$
$\displaystyle\sum_{q=1}^{n}\sum_{k=1}^{n}\sum_{p=1}^{n}b_{ki}a_{qk}c_{pj}\gamma_{qp}^{r},$
(5.9)
$\displaystyle\sum_{k=1}^{n}\sum_{q=1}\sum_{p=1}c_{ki}b_{pj}a_{qp}\delta_{kq}^{r}=\sum_{p=1}^{n}\sum_{q=1}^{n}c_{pi}c_{qj}\delta_{pq}^{r}$
$\displaystyle=$
$\displaystyle\sum_{q=1}^{n}\sum_{k=1}^{n}\sum_{p=1}^{n}b_{ki}a_{qk}c_{pj}\delta_{qp}^{r},$
(5.10)
$\displaystyle\sum_{k=1}^{n}\sum_{q=1}\sum_{p=1}c_{ki}b_{pj}a_{qp}\xi_{kq}^{r}=\sum_{p=1}^{n}\sum_{q=1}^{n}c_{pi}c_{qj}\xi_{pq}^{r}$
$\displaystyle=$
$\displaystyle\sum_{q=1}^{n}\sum_{k=1}^{n}\sum_{p=1}^{n}b_{ki}a_{qk}c_{pj}\xi_{qp}^{r}.$
(5.11)

######  Theorem 5.8.

The centroids of three-dimensional complex BiHom-associative trialgebras are
illustrated as follows :

IC | $Cent(\mathcal{A})$ | $Dim(Cent(\mathcal{A}))$ | IC | $Cent(\mathcal{A})$ | $Dim(Cent(\mathcal{A}))$
---|---|---|---|---|---
$\mathcal{BT}as_{3}^{1}$ | $\left(\begin{array}[]{cccc}0&0&0\\\ c_{21}&0&0\\\ 0&0&0\end{array}\right)$ | 1 | $\mathcal{BT}as_{3}^{2}$ | $\left(\begin{array}[]{cccc}0&0&0\\\ c_{21}&0&0\\\ 0&0&0\end{array}\right)$ | 1
$\mathcal{BT}as_{3}^{3}$ | $\left(\begin{array}[]{cccc}0&0&0\\\ 0&0&0\\\ 0&0&c_{33}\end{array}\right)$ | 1 | $\mathcal{BT}as_{3}^{4}$ | $\left(\begin{array}[]{cccc}0&0&0\\\ 0&0&0\\\ c_{31}&0&0\end{array}\right)$ | 1
$\mathcal{BT}as_{3}^{5}$ | $\left(\begin{array}[]{cccc}0&0&0\\\ c_{21}&0&0\\\ 0&0&0\end{array}\right)$ | 1 | $\mathcal{BT}as_{3}^{6}$ | $\left(\begin{array}[]{cccc}0&0&0\\\ c_{21}&0&0\\\ 0&0&0\end{array}\right)$ | 1
$\mathcal{BT}as_{3}^{7}$ | $\left(\begin{array}[]{cccc}0&0&0\\\ 0&0&0\\\ 0&0&c_{33}\end{array}\right)$ | 1 | $\mathcal{BT}as_{3}^{8}$ | $\left(\begin{array}[]{cccc}0&0&0\\\ 0&0&0\\\ 0&0&c_{33}\end{array}\right)$ | 1
$\mathcal{BT}as_{3}^{9}$ | $\left(\begin{array}[]{cccc}0&0&0\\\ 0&0&0\\\ 0&0&c_{33}\end{array}\right)$ | 1 | $\mathcal{BT}as_{3}^{10}$ | $\left(\begin{array}[]{cccc}0&0&0\\\ 0&0&0\\\ 0&0&c_{33}\end{array}\right)$ | 1
$\mathcal{BT}as_{3}^{11}$ | $\left(\begin{array}[]{cccc}0&0&0\\\ 0&c_{22}&0\\\ 0&0&0\end{array}\right)$ | 1 | $\mathcal{BT}as_{3}^{12}$ | $\left(\begin{array}[]{cccc}0&0&0\\\ 0&c_{22}&0\\\ 0&0&0\end{array}\right)$ | 1
$\mathcal{BT}as_{3}^{13}$ | $\left(\begin{array}[]{cccc}0&0&0\\\ 0&c_{22}&0\\\ 0&0&0\end{array}\right)$ | 1 | $\mathcal{BT}as_{3}^{15}$ | $\left(\begin{array}[]{cccc}c_{11}&0&0\\\ 0&0&0\\\ 0&0&0\end{array}\right)$ | 1
$\mathcal{BT}as_{3}^{16}$ | $\left(\begin{array}[]{cccc}0&0&0\\\ 0&0&0\\\ c_{31}&0&0\end{array}\right)$ | 1 | $\mathcal{BT}as_{3}^{18}$ | $\left(\begin{array}[]{cccc}0&0&0\\\ 0&0&0\\\ c_{31}&0&0\end{array}\right)$ | 1

IC | $Cent(\mathcal{A})$ | $Dim(Cent(\mathcal{A}))$ | IC | $Cent(\mathcal{A})$ | $Dim(Cent(\mathcal{A}))$
---|---|---|---|---|---
$\mathcal{BT}as_{3}^{19}$ | $\left(\begin{array}[]{cccc}0&0&0\\\ 0&0&0\\\ c_{13}&0&0\end{array}\right)$ | 1 | $\mathcal{BT}as_{3}^{20}$ | $\left(\begin{array}[]{cccc}0&0&0\\\ 0&0&0\\\ c_{31}&0&0\end{array}\right)$ | 1
$\mathcal{BT}as_{3}^{21}$ | $\left(\begin{array}[]{cccc}0&0&0\\\ 0&0&0\\\ c_{31}&0&0\end{array}\right)$ | 1 | $\mathcal{BT}as_{3}^{22}$ | $\left(\begin{array}[]{cccc}0&0&0\\\ 0&0&0\\\ c_{31}&0&0\end{array}\right)$ | 1
$\mathcal{BT}as_{3}^{23}$ | $\left(\begin{array}[]{cccc}0&0&0\\\ 0&0&0\\\ c_{31}&0&0\end{array}\right)$ | 1 | $\mathcal{BT}as_{3}^{24}$ | $\left(\begin{array}[]{cccc}0&0&0\\\ 0&0&0\\\ c_{31}&0&0\end{array}\right)$ | 1.

######  Corollary 5.9.

  * •

The dimensions of the centroids of two-dimensional BiHom-associative
trialgebras are zero.

  * •

The dimensions of the centroids of three-dimensional BiHom-associative
trialgebras range between zero and one.

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