# Multipliers and weak multipliers of algebras

Yuji Kobayashi and Sin-Ei Takahasi

( Laboratory of Mathematics and Games
(https://math-game-labo.com)
2020 MSC Numbers: Primary 43A22; Secondary 17A99, 46J10.
Keywords: (weak) multiplier, (non)associative algebra, Jordan algebra,
zeropotent algebra, annihilator, nihil decomposition, matrix representation.
)

###### Abstract

We study general properties of multipliers and weak multipliers of algebras.
We apply the results to determine the (weak) multipliers of associative
algebras and zeropotent algebras of dimension 3 over an algebraically closed
field.

##  1 Introduction

Multipliers of algebras, in particular, multipliers of Banach algebras, have
been discussed in analysis. In this paper we will discuss them in a purely
algebraic manner.

Let $B$ be a Banach algebra. A mapping $T:B\rightarrow B$ is called a
multiplier of $B$, if it satisfies the condition (I) $xT(y)=T(xy)=T(x)y\
(x,y\in B)$. Let $M(B)$ denote the collection of all multipliers of $B$, and
let $B(B)$ be the collection of all bounded linear operators on $B$. Then
$M(B)$ forms an algebra and $B(B)$ forms a Banach algebra. $B$ is called
without order if it has no nonzero left or right annihilator. If $B$ is
without order, then $M(B)$ forms a commutative closed subalgebra of $B(B)$
(see [2], Proposition 1.4.11). In 1953, Wendel [7] proved an important result
that the multiplier algebra of $L^{1}(G)$ on a locally compact abelian group
$G$ is isometrically isomorphic to the measure algebra on $G$. The general
theory of multipliers of Banach algebras has been developed by Johnson [1]. A
good reference to the theory of multipliers of Banach algebra is given in
Larsen [5].

When $B$ is without order, $T$ is a multiplier if it satisfies the condition
(II) $xT(y)=T(x)y\ (x,y\in B)$. Many researchers had been unaware of
difference between conditions (I) and (II) until Zivari-Kazempour [8] (see
also [9]) recently clearly stated the difference. We call a mapping $T$
satisfying (II) a weak multiplier and denote the set of weak multipliers of
$B$ by $M^{\prime}(B)$. Then, $M(B)$ is in general a proper subset of
$M^{\prime}(B)$. Furthermore, (weak) multipliers can be defined for an algebra
$A$ not necessarily associative, and they are not linear mappings in general.
We denote the spaces of linear multipliers and linear weak multipliers of $A$
by $LM(A)$ and $LM^{\prime}(A)$ respectively. $M(A)$ and $LM(A)$ are
subalgebras of the algebra $A^{A}$ consisting of all mappings from $A$ to
itself. Meanwhile, $M^{\prime}(A)$ and $LM^{\prime}(A)$ are closed under the
operation $\circ$ defined by $T\circ S=TS+ST$, and they form a Jordan algebra.

In Sections 2 - 5 we study general properties of (weak) multipliers. In
particular, in sections 3 and 4 we give a decomposition theorem (Theorem 3.1),
and a matrix equation (Theorem 4.2) for (weak) multipliers. They play an
essential role to analyze (weak) multipliers.

Complete classifications of associative algebras and zeropotent algebras of
dimension 3 over an algebraically closed field of characteristic not equal to
2 were given in Kobayashi et al, [3] and [4]. In Sections 6 and 7 we
completely determine the (linear) (weak) multipliers of those algebras.

##  2 Multipliers and weak multipliers

Let $K$ be a field and $A$ be a (not necessarily associative) algebra over
$K$. The set $A^{A}$ of all mappings from $A$ to $A$ forms an associative
algebra over $K$ in the usual manner. Let $L(A)$ denotes the subalgebra of
$A^{A}$ of all linear mappings from $A$ to $A$.

A mapping $T:A\rightarrow A$ is a weak multiplier of $A$, if

$xT(y)=T(x)y$ (1)

holds for any $x,y\in A$, and $T$ is a multiplier, if

$xT(y)=T(xy)=T(x)y$ (2)

for any $x,y\in A$. Let $M(A)$ (resp. $M^{\prime}(A)$) denote the set of all
multipliers (resp. weak multipliers) of $A$. Define

$LM(A)\overset{\text{\tiny def}}{=}M(A)\cap L(A)\ \,\mbox{and}\
\,LM^{\prime}(A)\overset{\text{\tiny def}}{=}M^{\prime}(A)\cap L(A).$

######  Proposition 2.1.

$M(A)$ (resp. $LM(A)$) is a unital subalgebra of $A^{A}$ (resp. $L(A)$), and
$M^{\prime}(A)$ (resp. $LM^{\prime}(A)$) is a Jordan subalgebra of $A^{A}$
(resp. $L(A)$).

###### Proof.

First, the zero mapping is a multiplier because all the three terms in (2) are
zero. Secondly, the identity mapping is also a multiplier because the three
terms in (2) are $xy$. Let $T,U\in M(A)$. Then we have

$x(T+U)(y)=xT(y)+xU(y)=T(xy)+U(xy)=T(x)y+U(x)y=(T+U)(x)y$ (3)

and

$x(TU)(y)=xT(U(y))=T(xU(y))=TU(xy)=T(U(x)y)=(TU)(x)y$ (4)

for any $x,y\in A$. Hence, $T+U,TU\in M(A)$. Finally let $k\in K$, then

$x(kT)(y)=kxT(y)=kT(xy)=kT(x)y=(kT)(x)y,$ (5)

and so $kT\in M(A)$. Therefore, $M(A)$ is a unital subalgebra of $A^{A}$, and
$LM(A)=M(A)\cap L(A)$ is a unital subalgebra of $L(A)$.

Next, let $T,U\in M^{\prime}(A)$. Then, the equalities in (3) and (5) hold
removing the center terms $T(xy)+U(xy)$ and $kT(xy)$ respectively. Hence,
$M^{\prime}(A)$ is a subspace of $A^{A}$. Moreover, we have

$x(TU)(y)=xT(U(y))=T(x)U(y)=U(T(x))y=UT(x)y$

and similarly $x(UT)(y)=TU(x)y$ for any $x,y\in A$. Hence,

$x(TU+UT)(y)=(TU+UT)(x)y.$

It follows that $TU+UT\in M^{\prime}(A)$.111In general, for an associative
algebra $A$ over a field $K$ of characteristic $\neq 2$, the Jordan product
$\circ$ on $A$ is defined by $x\circ y=(xy+yx)/2$ for $x,y\in A$. ∎

The opposite $A^{\rm op}$ of $A$ is the algebra with the same elements and the
addition as $A$, but the multiplication $*$ in it is reversed, that is,
$x*y=yx$ for all $x,y\in A$.

######  Proposition 2.2.

$A$ and $A^{\rm op}$ have the same multipliers and weak multiplies, that is,

$M(A^{\rm op})=M(A)\ \ \mbox{and}\ \ M^{\prime}(A^{\rm op})=M^{\prime}(A).$

###### Proof.

Let $T\in A^{A}$. Then, $T\in M^{\prime}(A)$, if and only if

$x*T(y)=T(y)x=yT(x)=T(x)*y$

for any $x,y\in A$, if and only if $T\in M^{\prime}(A^{\rm op})$. Further,
$T\in M(A)$, if and only if

$x*T(y)=T(y)x=T(yx)=T(x*y)=yT(x)=T(x)*y$

for any $x,y\in A$, if and only if $T\in M(A^{\rm op})$. ∎

Let Ann${}_{l}(A)$ (resp. Ann${}_{r}(A)$) be the left (resp. right)
annihilator of $A$ and let $A_{0}$ be their intersection, that is,

${\rm Ann}_{l}(A)=\\{a\in A\,|\,ax=0{\ \rm for\ all\ }x\in A\\},$ ${\rm
Ann}_{r}(A)=\\{a\in A\,|\,xa=0{\ \rm for\ all\ }x\in A\\}$

and

$A_{0}={\rm Ann}_{l}(A)\cap{\rm Ann}_{r}(A).$

They are all subspaces of $A$, and when $A$ is an associative algebra, they
are two-sided ideals. For a subset $X$ of $A$, $\langle X\rangle$ denotes the
subspace of $A$ generated by $X$.

######  Proposition 2.3.

A weak multiplier $T$ of $A$ such that $\langle T(A)\rangle\cap A_{0}=\\{0\\}$
is a linear mapping.

###### Proof.

Let $x,y,z\in A$ and $a,b\in K$, and let $T$ be a weak multiplier. We have

$\displaystyle T(ax+by)z$
$\displaystyle=(ax+by)T(z)=axT(z)+byT(z)=aT(x)z+bT(y)z$
$\displaystyle=(aT(x)+bT(y))z.$

Because $z$ is arbitrary, we have $w=T(ax+by)-aT(x)-bT(y)\in{\rm Ann}_{l}(A)$.
Similarly, we can show $w\in{\rm Ann}_{r}(A)$, and so $w\in A_{0}$. Hence, if
$\langle T(A)\rangle\cap A_{0}=\\{0\\}$, then $w=0$ because $w\in\langle
T(A)\rangle$. Since $a,b,x,y$ are arbitrary, $T$ is a linear mapping. ∎

######  Corollary 2.4.

If $A_{0}=\\{0\\}$, then any weak multiplier is a linear mapping over $K$,
that is, $M^{\prime}(A)=LM^{\prime}(A)$ and $M(A)=LM(A)$.

######  Proposition 2.5.

If $T$ is a weak multiplier, then $T({\rm Ann}_{l}(A))\subseteq{\rm
Ann}_{l}(A)$, $T({\rm Ann}_{r}(A))\subseteq{\rm Ann}_{r}(A)$ and
$T(A_{0})\subseteq A_{0}$ .

###### Proof.

Let $x\in{\rm Ann}_{l}(A)$, then for any $y\in A$ we have

$0=xT(y)=T(x)y.$

Hence, $T(x)\in{\rm Ann}_{l}(A)$. The other cases are similar. ∎

In this paper we denote the subset $\\{xy\,|\,x,y\in A\\}$ of $A$ by $A^{2}$,
though usually $A^{2}$ denotes the subspace of $A$ generated by this set.

######  Proposition 2.6.

Any mapping $T:A\rightarrow A$ such that $T(A)\subseteq A_{0}$ is a weak
multiplier. Such a mapping $T$ is a multiplier if and only if
$T(A^{2})=\\{0\\}$. In particular, if $A$ is the zero algebra, every mapping
$T$ is a weak multiplier, and it is a multiplier if only if $T(0)=0$.

###### Proof.

If $T(A)\subseteq A_{0}$, the both sides are 0 in (1) and $T$ is a weak
multiplier. This $T$ is a multiplier, if only if the term $T(xy)$ in the
middle of (2) is $0$ for all $x,y\in A$, that is, $T(A^{2})=\\{0\\}$. If $A$
is the zero algebra, then $A=A_{0}$ and $A^{2}$ = {0}. Hence, any $T$ is a
weak multiplier and it is a multiplier if and only if $T(0)=0$. ∎

##  3 Nihil decomposition

Let $A_{1}$ be a subspace of $A$ such that

$A=A_{1}\oplus A_{0}.$ (6)

Here, $A_{1}$ is not unique, but choosing an appropriate $A_{1}$ will become
important later. When $A_{1}$ is fixed, any mapping $T\in A^{A}$ is uniquely
decomposed as

$T=T_{1}+T_{0}$ (7)

with $T_{1}(A)\subseteq A_{1}$ and $T_{0}(A)\subseteq A_{0}$. We call (6) and
(7) a nihil decompositions of $A$ and $T$ respectively. Let $\pi:A\rightarrow
A_{1}$ be the projection and $\mu:A_{1}\rightarrow A$ be the embedding, that
is, $\pi(x_{1}+x_{0})=\mu(x_{1})=x_{1}$ for $x_{1}\in A_{1}$ and $x_{0}\in
A_{0}$.

Let $M_{1}(A)$ (resp. $M_{0}(A)$) denote the set of all multipliers $T$ of $A$
with $T(A)\subseteq A_{1}$ (resp. $T(A)\subseteq A_{0}$). Similarly, the sets
$M^{\prime}_{1}(A)$ and $M^{\prime}_{0}(A)$ of weak multipliers of $A$ are
defined. Also, set

$LM_{i}(A)=M_{i}(A)\cap L(A)\ \,\mbox{and}\
\,LM^{\prime}_{i}(A)=M^{\prime}_{i}(A)\cap L(A)$

for $i=0,1$. By Proposition 2.3 we see

$M^{\prime}_{1}(A)=LM^{\prime}_{1}(A)\ \,\mbox{and}\ \,M_{1}(A)=LM_{1}(A),$

and by Proposition 2.6 we have

$M^{\prime}_{0}(A)=A_{0}^{A},\,\,M_{0}(A)=\\{T\in
A_{0}^{A}\,|\,T(A^{2})=\\{0\\}\\}.$ (8)

######  Theorem 3.1.

Let $A$ = $A_{1}\oplus A_{0}$ and $T=T_{1}+T_{0}$ be nihil decompositions of
$A$ and $T\in A^{A}$ respectively.

(i) $T$ is a weak multiplier, if and only if $T_{1}$ is a weak multiplier. If
$T$ is a weak multiplier, $T_{1}$ is a linear mapping satisfying
$T_{1}(A_{0})=\\{0\\}$.

(ii) If $T_{1}$ is a multiplier and $T_{0}(A^{2})=\\{0\\}$, then $T$ is a
multiplier. If $A_{1}$ is a subalgebra of $A$, the converse is also true.

Suppose that $A_{1}$ is a subalgebra of $A$, and let $\Phi$ be a mapping
sending $R\in(A_{1})^{A_{1}}$ to $\mu\circ R\circ\pi\in A^{A}$. Then,

(iii) $\Phi$ gives an algebra isomorphism from $M(A_{1})$ onto $M_{1}(A)$ and
a Jordan isomorphism from $M^{\prime}(A_{1})$ onto $M^{\prime}_{1}(A)$.

###### Proof.

Let $x,y\in A$.

(i) If $T$ is a weak multiplier, then

$xT_{1}(y)=x(T(y)-T_{0}(y))=xT(y)=T(x)y=T_{1}(x)y.$

Thus, $T_{1}$ is also a weak multiplier. Moreover, $T_{1}$ is a linear mapping
by Proposition 2.3 and $T_{1}(A_{0})\subseteq A_{1}\cap A_{0}=\\{0\\}$ by
Proposition 2.5. Conversely, if $T_{1}$ is a weak multiplier, then

$xT(y)=xT_{1}(y)=T_{1}(x)y=T(x)y,$

and so $T$ is a weak multiplier.

(ii) If $T_{1}$ is a multiplier and $T_{0}(A^{2})=0$, then $T$ is a multiplier
because

$\displaystyle
xT(y)=xT_{1}(y)=T_{1}(xy)=T(xy)-T_{0}(xy)=T(xy)=T_{1}(x)y=T(x)y.$

Next suppose that $A_{1}$ is a subalgebra. If $T$ is a multiplier, then for
any $x,y\in A$ we have

$T_{1}(xy)+T_{0}(xy)=T(xy)=xT(y)=x_{1}T_{1}(y),$ (9)

where $x=x_{1}+x_{0}$ with $x_{1}\in A_{1}$ and $x_{0}\in A_{0}$. Here,
$x_{1}T_{1}(y)\in A_{1}$ because $A_{1}$ is a subalgebra, and thus, we have
$T_{0}(xy)=x_{1}T_{1}(y)-T_{1}(xy)\in A_{0}\cap A_{1}=\\{0\\}$. Since $x,y$
are arbitrary, we get $T_{0}(A^{2})=\\{0\\}$. Moreover, $T_{1}$ is a
multiplier because $T_{1}(xy)=x_{1}T_{1}(y)=xT_{1}(y)$ by (9) and similarly
$T_{1}(xy)=T_{1}(x)y$. The converse is already proved above.

(iii) Let $S\in(A_{1})^{A_{1}}$ and $x=x_{1}+x_{0},y=y_{1}+y_{0}\in A$ with
$x_{1},y_{1}\in A_{1}$ and $x_{0},y_{0}\in A_{0}$. Then,
$\pi(x)=\mu(x_{1})=x_{1}$, $\pi(y)=\mu(y_{1})=y_{1}$ and

$\Phi(S)(x)=\mu(S(\pi(x)))=\mu(S(x_{1}))=S(x_{1}).$

Thus, if $S\in M^{\prime}(A_{1})$, we have

$\displaystyle
x\Phi(S)(y)=xS(y_{1})=x_{1}S(y_{1})=S(x_{1})y_{1}=\Phi(S)(x)y_{1}=\Phi(S)(x)y.$

Hence, $\Phi(S)\in M^{\prime}_{1}(A)$. Moreover, if $S\in M(A_{1})$, then
because $\pi$ is a homomorphism, we have

$\Phi(S)(xy)=S(\pi(xy))=S(x_{1}y_{1})=x_{1}S(y_{1})=x\Phi(S)(y),$

and hence $\Phi(S)\in M_{1}(A)$. Conversely, let $T\in M^{\prime}_{1}(A)$,
then because $T$ is a linear mapping satisfying $T(A_{0})=\\{0\\}$, there is a
linear mapping $S\in L(A_{1})$ on $A_{1}$ such that $\Phi(S)=T$, that is,
$S(x_{1})=T(x)=T(x_{1})$. We have

$x_{1}S(y_{1})=x_{1}T(y_{1})=T(x_{1})y_{1}=S(x_{1})y_{1},$

and hence $S\in M^{\prime}(A_{1})$. Therefore, $\Phi$ is a linear isomorphism
from $M^{\prime}(A_{1})$ to $M^{\prime}_{1}(A)$. Similarly, $\Phi$ gives a
linear isomorphism from $M(A_{1})$ to $M_{1}(A)$. Moreover, for $T,U\in
M^{\prime}(A_{1})$, we have

$\Phi(TU)=\mu\circ T\circ U\circ\pi=\mu\circ T\circ\pi\circ\mu\circ
U\circ\pi=\Phi(T)\Phi(U).$

Thus, $\Phi$ gives an isomorphism of algebras from $M(A_{1})$ to $M_{1}(A)$
and a Jordan isomorphism from $M^{\prime}(A_{1})$ to $M^{\prime}_{1}(A)$. ∎

Theorem 3.1 implies

$M^{\prime}(A)=\ M^{\prime}_{1}(A)\oplus M^{\prime}_{0}(A),\ M_{1}(A)\oplus
M_{0}(A)\subseteq M(A),$

where $M^{\prime}_{0}(A)$ and $M_{0}(A)$ are given as (8). Moreover, if
$A_{1}$ is a subalgebra, we have

$M^{\prime}(A)\cong M^{\prime}(A_{1})\oplus(A_{0})^{A},\ \ M(A)\cong
M(A_{1})\oplus\\{T\in(A_{0})^{A}\,|\,T(A^{2})=\\{0\\}\\}.$ (10)

######  Corollary 3.2.

Any weak multiplier $T$ is written as

$T=T_{1}+R$ (11)

with $T_{1}\in LM_{1}^{\prime}(A)$ and $R\in(A_{0})^{A},$ and it is a
multiplier if and only if

$R(x_{1}y_{1})=x_{1}T_{1}(y_{1})-T_{1}(x_{1}y_{1})$ (12)

for any $x_{1},y_{1}\in A_{1}$.

###### Proof.

As stated above $T$ is written as (11). Let $x=x_{1}+x_{0},y=y_{1}+y_{0}\in A$
with $x_{1},y_{1}\in A_{1}$ and $x_{0},y_{0}\in A_{0}$ be arbitrary, then we
have

$xT(y)=x_{1}(T_{1}(y)+R(y))=x_{1}T_{1}(y)=x_{1}T_{1}(y_{1})$ (13)

because $R(A)\subseteq A_{0}$ and $T_{1}(A_{0})=\\{0\\}$. The last term in
(13) is also equal to $T_{1}(x_{1})y_{1}=T(x)y$. Hence, $T$ is a multiplier,
if and only if it is equal to
$T(xy)=T(x_{1}y_{1})=T_{1}(x_{1}y_{1})+R(x_{1}y_{1})$, if and only if (12)
holds. ∎

##  4 Linear multipliers and matrix equation

In this section, $A$ is a finite dimensional algebra over $K$. We drive a
matrix equation for a linear mapping on $A$ to be a (weak) multiplier. Suppose
that $A$ is $n$-dimensional with basis $E=\\{e_{1},e_{2},\dots,e_{n}\\}$.

######  Lemma 4.1.

A linear mapping $T:A\rightarrow A$ is a weak multiplier if and only if

$e_{i}T(e_{j})=T(e_{i})e_{j},$ (14)

and it is a multiplier if and only if

$T(e_{i}e_{j})=e_{i}T(e_{j})=T(e_{i})e_{j},$ (15)

for all $e_{i},e_{j}\in E$.

###### Proof.

The necessity of the conditions (14) and (15) is obvious. Let
$x=x_{1}e_{1}+x_{2}e_{2}+\cdots+x_{n}e_{n},y=y_{1}e_{1}+y_{2}e_{2}+\cdots+y_{n}e_{n}\in
A$ with $x_{1},x_{2},\dots,x_{n},y_{1},y_{2},\dots,y_{n}\in K$. If $T$
satisfies (14), then we have

$\displaystyle xT(y)$ $\displaystyle=$
$\displaystyle(\sum_{i}x_{i}e_{i})T(\sum_{j}y_{j}e_{j})=(\sum_{i}x_{i}e_{i})(\sum_{j}y_{j}T(e_{j}))=\sum_{i,j}x_{i}y_{j}e_{i}T(e_{j})$
$\displaystyle=$
$\displaystyle\sum_{i,j}x_{i}y_{j}T(e_{i})e_{j}=(\sum_{i}x_{i}T(e_{i}))(\sum_{j}y_{j}e_{j})=T(x)y.$

Hence, $T$ is a weak multiplier. Moreover, if $T$ satisfies (15), it is a
multiplier in a similar way ∎

Let $\bm{A}$ (we use the bold character) be the multiplication table of $A$ on
$E$. $\bm{A}$ is a matrix whose elements are from $A$ defined by

$\bm{A}=\bm{E}^{t}\bm{E},$ (16)

where $\bm{E}=(e_{1},e_{2},\dots,e_{n})$ (we again use the bold face $\bm{E}$)
is the row vector consisting the basis elements. For a linear mapping $T$ on
$A$ and a matrix $\bm{B}$ over $A$, $T(\bm{B})$ denotes the matrix obtained by
applying $T$ component-wise, that is, the $(i,j)$-element of $T(\bm{B})$ is
$T(b_{ij})$ for the $(i,j)$-element $b_{ij}$ of $\bm{B}$.222This is called a
broadcasting (cf. [6]). We use the same character $T$ for the representation
matrix of $T$ on $E$, that is,

$T(\bm{E})=\bm{E}T.$ (17)

.

######  Theorem 4.2.

A linear mapping $T$ is a weak multiplier of $A$ if and only if

$\bm{A}T=T^{t}\bm{A},$ (18)

and $T$ is a multiplier if and only if

$T(\bm{A})=\bm{A}T=T^{t}\bm{A}.$ (19)

###### Proof.

By (16) and (17) we have

$(e_{1},e_{2},\dots,e_{n})^{t}(T(e_{1}),T(e_{2}),\dots,T(e_{n}))=\bm{E}^{t}T(\bm{E})=\bm{E}^{t}\bm{E}T=\bm{A}T$
(20)

and

$(T(e_{1}),T(e_{2}),\dots,T(e_{2}))^{t}(e_{1},e_{2},\dots,e_{n})=T(\bm{E})^{t}\bm{E}=T^{t}\bm{E}^{t}\bm{E}=T^{t}\bm{A}.$
(21)

By Lemma 4.1, $T$ is a weak multiplier, if and only if (20) and (21) are
equal, if and only if (18) holds. Moreover, $T$ is multiplier if and only if,
the leftmost sides of (20) and (21) are equal to
$(T(e_{i}e_{j}))_{i,j=1,2,\dots,n}=T(\bm{A})$, if and only if (19) holds. ∎

The multiplication table of the opposite algebra $A^{\rm op}$ of $A$ is
$\bm{A}^{\rm t}$. So, the algebras with multiplication tables transposed to
each other have the same (weak) multipliers.

##  5 Associative algebras

In this section $A$ is an associative algebra over $K$.

######  Proposition 5.1.

If $A_{0}=\\{0\\}$, then we have

$M(A)=M^{\prime}(A)=LM(A)=LM^{\prime}(A).$

###### Proof.

Let $T\in M^{\prime}(A)$, then we have

$T(xy)z=xyT(z)=xT(y)z\ \ {\rm and}\ \ zT(xy)=T(z)xy=zT(x)y$

for any $x,y,z\in A$. It follows that

$T(xy)-xT(y)\in{\rm Ann}_{l}(A)\cap{\rm Ann}_{r}(A)=A_{0}=\\{0\\}.$

Hence, $T(xy)=xT(y)$ and we see $T\in M(A)$. Moreover, $T\in LM(A)$ by
Proposition 2.3. ∎

Let $a\in A$. If $xay=axy$ (resp. $xay=xya$) for any $x,y\in A$, $a$ is called
a left (resp. right) central element, and $a$ is called a central element if
$ax=xa$ for any $x\in A$. Let $Z_{l}(A)$, (resp. $Z_{r}(A)$, $Z(A)$) denotes
the set of all left central (resp. right central, central) elements.

######  Lemma 5.2.

$Z_{l}(A)$ (resp. $Z_{r}(A)$, $Z(A)$) are subalgebra of $A$ containing ${\rm
Ann}_{l}(A)$ (resp. ${\rm Ann}_{r}(A)$, $A_{0}$).

###### Proof.

Straightforward. ∎

For $a\in A$, $l_{a}$ (resp. $r_{a}$) denotes the left (resp. right)
multiplication by $a$, that is,

$l_{a}(x)=ax,\ \ \ r_{a}(x)=xa$

for $x\in A$. They are linear mappings.

######  Lemma 5.3.

For $a\in A$ the following statements are equivalent.

(i) $l_{a}$ (resp. $r_{a}$) is a multiplier,

(ii) $l_{a}$ (resp. $r_{a}$) is a weak multiplier,

(iii) $a$ is left (resp. right) central.

###### Proof.

If $l_{a}$ is a weak multiplier, then

$xay=xl_{a}(y)=l_{a}(x)y=axy$

for any $x,y\in A$. Hence, $a$ is left central. Because
$l_{a}(x)y=axy=l_{a}(xy)$ and $xl_{a}(y)=xay$ for any $x,y\in A$, $l_{a}$ is a
multiplier if $a$ is left central. The other case is similar, and we see that
the three statements are equivalent. ∎

######  Lemma 5.4.

Suppose that $A$ has a right (resp. left) identity. Then, a left (resp. right)
central element is central, and ${\rm Ann}_{l}(A)=\\{0\\}$ (resp. ${\rm
Ann}_{r}(A)=\\{0\\}$).

###### Proof.

Easy. ∎

Because ${\rm Ann}_{l}(A)\subseteq Z_{l}$ and ${\rm Ann}_{r}(A)\subseteq
Z_{r}$ by Lemma 5.2, we can make the quotient algebras
$\bar{Z}_{l}(A)=Z_{l}(A)/{\rm Ann}_{l}(A)\ \,\mbox{and}\
\,\bar{Z}_{r}(A)=Z_{r}(A)/{\rm Ann}_{r}(A).$

######  Theorem 5.5.

Suppose that $A$ has a left (resp. right) identity $e$. Then, any multiplier
is a left (resp. right) multiplication by a left (resp. right) central element
and is a linear multiplier. The mapping $\phi:Z_{l}(A)\rightarrow M(A)$
sending $a\in Z_{l}(A)$ to $l_{a}$ induces an isomorphism
$\bar{\phi}:\overline{Z_{l}}(A)\rightarrow M(A)$ of algebras. In particular,
if $A$ is unital, $M(A)$ is isomorphic to $Z(A)$.

###### Proof.

Suppose that $A$ has a left identity $e$. Let $T\in M^{\prime}(A)$ and set
$a=T(e)$. Then we have

$T(x)=eT(x)=T(e)x=ax$

for any $x\in A$. Hence, $T=l_{a}$, where $a\in Z_{l}(A)$ and $T$ is a linear
multiplier by Lemma 5.3. Therefore, $M^{\prime}(A)=LM(A)$ and $\phi$ is
surjective. Moreover, for $a\in Z_{l}(A)$, $\phi(a)=0$, if and only if $ax=0$
for any $x\in A$, if and only if $a\in{\rm Ann}_{l}(A)$. Thus we have
Ker$(\phi)={\rm Ann}_{l}(A)$, and $\phi$ induces the desired isomorphism.
Similarly, if $A$ has a right identity, $M(A)$ is isomorphic to
$\overline{Z_{r}}(A)$. Finally, if $A$ has the identity, then
Ann${}_{l}(A)=\mbox{Ann}_{r}(A)=\\{0\\}$ and hence $M(A)$ is isomorphic to
$Z(A)$. ∎

##  6 3-dimensional associative algebras

Over an algebraically closed field $K$ of characteristic not equal to 2, we
have, up to isomorphism, 24 families of associative algebras of dimension 3.
They are 5 unital algebras $U_{0},U_{1},U_{2}$, $U_{3},U_{4}$ defined on basis
$E=\\{e,f,g\\}$ by

$\small{\begin{pmatrix}e&f&g\\\ f&0&0\\\ g&0&0\end{pmatrix},\
\begin{pmatrix}e&f&g\\\ f&0&f\\\ g&-f&e\end{pmatrix},\ \begin{pmatrix}e&0&0\\\
0&f&0\\\ 0&0&g\end{pmatrix},\ \begin{pmatrix}e&0&0\\\ 0&f&g\\\
0&g&0\end{pmatrix},\ \begin{pmatrix}e&f&g\\\ f&g&0\\\ g&0&0\end{pmatrix}},$

5 curled algebras $C_{0},C_{1},C_{2},C_{3},C_{4}$ defined by

$\small{\begin{pmatrix}0&0&0\\\ 0&0&0\\\ 0&0&0\end{pmatrix},\
\begin{pmatrix}0&0&0\\\ 0&0&e\\\ 0&-e&0\end{pmatrix},\
\begin{pmatrix}{0}&{0}&{0}\\\ {e}&{f}&{0}\\\ {0}&{g}&{0}\end{pmatrix},\
\begin{pmatrix}{0}&{0}&{0}\\\ {0}&{0}&{0}\\\ {e}&{f}&{g}\end{pmatrix},\
\begin{pmatrix}{0}&{0}&{e}\\\ {0}&{0}&{f}\\\ {0}&{0}&{g}\end{pmatrix},}$

non-unital 4 straight algebras $S_{1},S_{2},S_{3},S_{4}$ defined by

${\small\begin{pmatrix}{f}\ &{g}&{0}\\\ {g}\ &{0}&{0}\\\ {0}\
&{0}&{0}\end{pmatrix},\ \begin{pmatrix}{e}\ &{0}&{0}\\\ {0}\ &{g}&{0}\\\ {0}\
&{0}&{0}\end{pmatrix},\ \begin{pmatrix}{e}\ &{0}&{0}\\\ {0}\ &{f}&{0}\\\ {0}\
&{0}&{0}\end{pmatrix},\ \begin{pmatrix}{e}\ &{f}&{0}\\\ {f}\ &{0}&{0}\\\ {0}\
&{0}&{0}\end{pmatrix},}\vspace{-1mm}$

and non-unital 10 families of waved algebras $W_{1}$, $W_{2}$, $W_{4}$,
$W_{5}$, $W_{6}$, $W_{7}$, $W_{8}$, $W_{9}$, $W_{10}$ and
$\big{\\{}W_{3}(k)\big{\\}}_{k\in H}$ defined by

${\small\begin{pmatrix}{0}&{0}&{0}\\\ {0}&{0}&{0}\\\
{0}&{0}&{e}\end{pmatrix},\ \begin{pmatrix}{0}&{0}&{0}\\\ {0}&{0}&{0}\\\
{0}&{e}&{0}\end{pmatrix},\ \begin{pmatrix}{e}&{0}&{0}\\\ {0}&{0}&{0}\\\
{0}&{0}&{0}\end{pmatrix},\ \begin{pmatrix}{0}&{0}&{0}\\\ {0}&{0}&{0}\\\
{0}&{f}&{g}\end{pmatrix},\ \begin{pmatrix}{0}&{0}&{0}\\\ {0}&{0}&{f}\\\
{0}&{0}&{g}\end{pmatrix},}$ $\small{\begin{pmatrix}e&0&0\\\ 0&0&0\\\
0&f&g\end{pmatrix},\ \begin{pmatrix}{e}&{0}&{0}\\\ {0}&{0}&{f}\\\
{0}&{0}&{g}\end{pmatrix},\ \begin{pmatrix}{0}&{e}&{0}\\\ {e}&{f}&{0}\\\
{0}&{g}&{0}\end{pmatrix},\ \begin{pmatrix}{0}&{e}&{0}\\\ {e}&{f}&{g}\\\
{0}&{0}&{0}\end{pmatrix}\ \mbox{and}\ \begin{pmatrix}{0}&{0}&{0}\\\
{0}&{e}&{0}\\\ {0}&k{e}&{e}\end{pmatrix},}$

respectively, where $H$ is a subset of $K$ such that $K=H\cup-H$ and
$H\cap-H=\\{0\\}$ (see [3] for details). We determine the (weak) multipliers
of them below.

(0) $A=C_{0}$ is the zero algebra, so by Proposition 2.6, we have

$M^{\prime}(A)={A}^{A},\ \,M(A)=\\{T\in{A}^{A}\,|\,T(0)=0\\}$

and

$LM(A)=LM^{\prime}(A)=L(A).$

(i) The unital algebras $U_{0},U_{2}$, $U_{3},U_{4}$ are commutative, so for
such $A$ we have

$M(A)=LM(A)=M^{\prime}(A)=LM^{\prime}(A)=\\{l_{x}|x\in A\\}\cong A$

by Theorem 5.5. For $A=U_{1}$, we have

$M(A)=LM(A)=M^{\prime}(A)=LM^{\prime}(A)\cong Z(A)=Ke.$

(ii) For $A=C_{1}$, We have $A_{0}=\mbox{Ann}_{l}(A)=\mbox{Ann}_{r}(A)=Ke$,
and a nihil decomposition $A=A_{1}\oplus A_{0}$ with $A_{1}=Kf+Kg$. Let
$T_{1}\in M^{\prime}_{1}(A)$, then by Theorem 3.1, $T_{1}$ is a linear mapping
such that $T_{1}(Ke)=\\{0\\}$. Let

$T_{1}=\begin{pmatrix}0&0&0\\\ 0&q&r\\\ 0&t&u\end{pmatrix}$ (22)

with $q,r,t,u\in K$ be the representation matrix of $T_{1}$ on $E$. By Theorem
4.2, $T_{1}$ is a weak multiplier, if and only if

$\begin{pmatrix}0&0&0\\\ 0&te&ue\\\
0&-qe&-re\end{pmatrix}=\bm{A}T_{1}=T_{1}^{\rm t}\bm{A}=\begin{pmatrix}0&0&0\\\
0&-te&qe\\\ 0&-ue&re\end{pmatrix},$

if and only if $r=t=0$ and $q=u$. Hence,
$M^{\prime}_{1}(A)=\\{T_{q}\,\big{|}\,q\in K\\}$, where
$T_{q}=\small{\begin{pmatrix}0&0&0\\\ 0&q&0\\\ 0&0&q\end{pmatrix}}$. By
Theorem 3.1 we see

$M^{\prime}(A)=\\{T_{q}\,\big{|}\,q\in K\\}\oplus(Ke)^{A},$

and

$LM^{\prime}(A)=\left\\{\small{\begin{pmatrix}a&b&c\\\ 0&q&0\\\
0&0&q\end{pmatrix}}\,\Big{|}\,a,b,c,q\in K\right\\}.$

By the multiplication table of $A$, we have $\alpha\beta=(xv-yz)e$ for
$\alpha=xf+yg,\beta=zf+vg\in A_{1}$ with $x,y,z,v\in K.$ By Corollary 3.2,
$T\in M^{\prime}(A)$ is given by $T=T_{q}+R$ with $R\in(Ke)^{A}$ and this $T$
is a multiplier, if and only if

$\displaystyle R((xv-yz)e)$ $\displaystyle=$ $\displaystyle R(\alpha\beta)\ =\
\alpha T_{q}(\beta)-T_{q}(\alpha\beta)$ $\displaystyle=$
$\displaystyle\alpha(q\beta)-T_{q}((xv-yz)e)\ =\ q(xv-yz)e$

for any $\alpha$ and $\beta$, if and only if $R(xe)=qxe$ for all $x\in K$. Let
$S_{q}=\small{\begin{pmatrix}q&0&0\\\ 0&q&0\\\ 0&0&q\end{pmatrix}}$ be the
scalar multiplication by $q\in K$. Then, we see $(T-S_{q})(A)\subseteq
A_{0}=Ke$ and

$(T-S_{q})(xe)=T_{q}(xe)+R(xe)-S_{q}(xe)=0+qxe-qxe=0,$

for any $x\in K$, that is, $(T-S_{q})(Ke)=\\{0\\}$. Thus, we have

$M(A)=\\{S_{q}\,\big{|}\,q\in K\\}\oplus\\{R\in(Ke)^{A}\,|\,R(Ke)=\\{0\\}\\},$

and

$LM(A)=\left\\{\small{\begin{pmatrix}a&b&c\\\ 0&a&0\\\
0&0&a\end{pmatrix}}\,\Big{|}\,a,b,c\in K\right\\}.$

(iii) $A=C_{2}$: Because Ann${}_{l}(A)=Ke$ and Ann${}_{r}(A)=Kg$, we see
$A_{0}=\\{0\\}$. Hence, any weak multiplier $T$ is a linear multiplier by
Proposition 5.1. By Theorem 4.2,

$T=\begin{pmatrix}a&b&c\\\ p&q&r\\\ s&t&u\end{pmatrix}$ (23)

is a (weak) multiplier, if and only if

$\begin{pmatrix}0&0&0\\\ ae+pf&be+qf&ce+rf\\\
pg&qg&rg\end{pmatrix}=\bm{A}T=T^{\rm t}\bm{A}=\begin{pmatrix}pe&pf+sg&0\\\
qe&qf+tg&0\\\ re&rf+ug&0\end{pmatrix},$

if and only if $b=c=p=r=s=t=0$ and $a=q=u$, that is, $T$ is the scalar
multiplication $S_{a}$ by $a$. Consequently,

$M(A)=M^{\prime}(A)=LM(A)=LM^{\prime}(A)=\\{S_{a}\,\big{|}\,a\in K\\}\cong K.$

(iv) $C_{3}$ and $C_{4}$ are opposed to each other, and have the same (weak)
multipliers by Proposition 2.2. Let $A=C_{3}$, then, $A$ has a left identity
$g$, $Z_{l}(A)=A$ and Ann${}_{l}(A)=Ke+Kf$. Hence, by Theorem 5.4,

$M(A)=M^{\prime}(A)=LM(A)=LM^{\prime}(A)=A/(Ke+Kg)=\\{S_{a}\,\big{|}\,a\in
K\\}.$

(v) $A=S_{1}$: We have $A_{0}=\mbox{Ann}_{l}(A)=\mbox{Ann}_{r}(A)=Kg$, and
$A=A_{1}\oplus A_{0}$ with $A_{1}=Ke+Kf$. Then, $T_{1}\in M^{\prime}_{1}(A)$
is a linear mapping with $T(Kg)=\\{0\\}$. Let

$T_{1}=\begin{pmatrix}a&b&0\\\ p&q&0\\\ 0&0&0\end{pmatrix}$ (24)

be its representation on $E$. $T_{1}$ is a weak multiplier, if and only if

$\begin{pmatrix}af+pg&bf+qg&0\\\ ag&bg&0\\\
0&0&0\end{pmatrix}=\bm{A}T_{1}=T_{1}^{\rm
t}\bm{A}=\begin{pmatrix}af+pg&ag&0\\\ bf+qg&bg&0\\\ 0&0&0\end{pmatrix},$

if and only if $b=0$ and $a=q$. Hence,

$M^{\prime}(A)=\left\\{T_{1}^{a,p}\,|\,a,p\in K\right\\}\oplus(Kg)^{A},$

where $T_{1}^{a,p}=\small{\begin{pmatrix}a&0&0\\\ p&a&0\\\
0&0&0\end{pmatrix}}$. So, $T\in M^{\prime}(A)$ is written as $T=T_{1}^{a,p}+R$
with $R\in(Kg)^{A}$, and this $T$ is multiplier, if and only if

$\displaystyle R(xzf+(xv+yz)g)$ $\displaystyle=$ $\displaystyle
R(\alpha\beta)\ =\ \alpha T_{1}^{a,p}(\beta)-T_{1}^{a,p}(\alpha\beta)$
$\displaystyle=$ $\displaystyle\alpha(aze+(pz+av)f)-T_{1}^{a,p}(xzf+(xv+yz)g)$
$\displaystyle=$ $\displaystyle axzf+(pxz+axv+ayz)g-axzf$ $\displaystyle=$
$\displaystyle(pxz+a(xv+yz))g$

for any $\alpha=xe+yf,\beta=ze+vf\in A_{1}$ with $x,y,z,v\in K$, if and only
if $R(xf+yg)=(px+ay)g$ for all $x,y\in K$. Let
$T^{a,p}=\small{\begin{pmatrix}a&0&0\\\ p&a&0\\\ 0&p&a\end{pmatrix}}$, then
$(T-T^{a,p})(A)\subseteq Kg$, and

$\displaystyle(T-T^{a,p})(xf+yg)$ $\displaystyle=$
$\displaystyle\left(T_{1}^{a,p}+R-T^{a,p}\right)(xf+yg)$ $\displaystyle=$
$\displaystyle axf+(px+ay)g-(axf+pxg+ayg)=0.$

for any $x,y\in K$. Thus, $(T-T^{a,p})(Kf+Kg)=\\{0\\}$, and hence

$M(A)=\\{T^{a,p}\,|\,a,p\in
K\\}\oplus\\{R\in(Kg)^{A}\,|\,R(Kf+Kg)=\\{0\\}\\}.$

Taking the intersections of $M^{\prime}(A)$ and $M(A)$ with $L(A)$, we obtain

$LM^{\prime}(A)=\left\\{\small{\begin{pmatrix}a&0&0\\\ p&a&0\\\
s&t&u\end{pmatrix}}\,\Big{|}\,a,p,s,t,u\in K\right\\}$

and

$LM(A)=\left\\{\small{\begin{pmatrix}a&0&0\\\ p&a&0\\\
s&p&a\end{pmatrix}}\,\Big{|}\,a,p,s\in K\right\\}.$

(vi) $A=S_{2}$: We have $A_{0}=\mbox{Ann}_{l}(A)=\mbox{Ann}_{r}(A)=Kg$, and
$A=A_{1}\oplus A_{0}$ with $A_{1}=Ke+Kf$. Let a linear mapping $T_{1}\in
M^{\prime}_{1}(A)$ be represented as (24), then $T_{1}$ is a weak multiplier,
if and only if

$\begin{pmatrix}ae&be&0\\\ pg&qg&0\\\ 0&0&0\end{pmatrix}=\bm{A}T=T^{\rm
t}\bm{A}=\begin{pmatrix}ae&pg&0\\\ be&qg&0\\\ 0&0&0\end{pmatrix},$

if and only if $b=p=0$. Hence,

$M^{\prime}(A)=\\{T_{1}^{a,q}\,\big{|}\,a,q\in K\\}\oplus(Kg)^{A},$

where $T_{1}^{a,q}=\small{\begin{pmatrix}a&0&0\\\ 0&q&0\\\
0&0&0\end{pmatrix}}$, and

$LM^{\prime}(A)=\left\\{\small{\begin{pmatrix}a&0&0\\\ 0&q&0\\\
s&t&u\end{pmatrix}}\,\Big{|}\,a,q,s,t,u\in K\right\\}.$

By Corollary 3.2, a weak multiplier $T$ written as $T=T_{1}^{a,q}+R$ for
$a,q\in K$ and $R\in(Kg)^{A}$ is multiplier, if and only if

$\displaystyle R(xze+yvg)$ $\displaystyle=$ $\displaystyle R(\alpha\beta)\ =\
\alpha T_{1}^{a,q}(\beta)-T_{1}^{a,q}(xze+yvg)$ $\displaystyle=$
$\displaystyle(xe+yf)(aze+qvf)-axze$ $\displaystyle=$ $\displaystyle yqvg,$

for any $\alpha=xe+yf,\beta=ze+vf\in A_{1}$ with $x,y,z,v\in K$, if only if
$R(xe+yg)=qyg$ for all $x,y\in K$. Let $T^{a,q}=\small{\begin{pmatrix}a&0&0\\\
0&q&0\\\ 0&0&q\end{pmatrix}}$, then $T^{a,q}\in M(A)$ and we have
$(T-T^{a,q})(xe+yg)=0$ for any $x,y\in K$ in the same way as (v) above. Hence,
$(T-T^{a,q})(Ke+Kg)=\\{0\\}$, and we have

$M(A)=\\{T^{a,p}\,|\,a,p\in K\\}\oplus\\{R\in(Kg)^{A}\,|\,R(Ke+Kg)=\\{0\\}\\}$

and

$LM(A)=\left\\{\small{\begin{pmatrix}a&0&0\\\ 0&p&0\\\
0&t&0\end{pmatrix}}\,\Big{|}\,a,p,t\in K\right\\}.$

(vii) $A=S_{3}$: We have $A_{0}=Kg$ and $A=A_{1}\oplus A_{0}$ with
$A_{1}=Ke+Kf$. Since $A_{1}$ is a subalgebra of $A$, by Theorem 3.1 we obtain
the equalities (10) in Section 3. Because $A_{1}$ is a commutative unital
algebra,

$M(A_{1})=M^{\prime}(A_{1})=A_{1}=\left\\{\small{\begin{pmatrix}a&0\\\
0&b\end{pmatrix}}\,\Big{|}\,a,b\in K\right\\}$

by Theorem 5.5. Hence,

$M^{\prime}(A)=A_{1}\oplus(Kg)^{A}\ \,\mbox{and}\
\,M(A)=A_{1}\oplus\\{T\in(Kg)^{A}\,|\,T(Ke+Kf)=0\\}.$

Intersecting with $L(A)$ we have

$LM^{\prime}(A)=\left\\{\small{\begin{pmatrix}a&0&0\\\ 0&b&0\\\
s&t&u\end{pmatrix}}\,\Big{|}\,a,b,s,t,u\in K\right\\}\ {\rm and}\ \
LM(A)=\left\\{\small{\begin{pmatrix}a&0&0\\\ 0&b&0\\\
0&0&u\end{pmatrix}}\,\Big{|}\,a,b,u\in K\right\\}.$

(viii) $A=S_{4}$: We have $A=A_{1}\oplus A_{0}$ with $A_{0}=Kg$ and
$A_{1}=Ke+Kf$. Because $A_{1}$ is a commutative unital subalgebra of $A$,
similarly to above we have

$M^{\prime}(A)=A_{1}\oplus(Kg)^{A}=\left\\{\small{\begin{pmatrix}a&0\\\
b&a\end{pmatrix}}\,\Big{|}\,a,b\in K\right\\}\oplus(Kg)^{A},$ $\displaystyle
M(A)$ $\displaystyle=$ $\displaystyle
A_{1}\oplus\left\\{T\in(Kg)^{A}\,|\,T(A^{2})=0\right\\}$ $\displaystyle=$
$\displaystyle\left\\{\small{\begin{pmatrix}a&0\\\
b&a\end{pmatrix}}\,\Big{|}\,a,b\in
K\right\\}\oplus\\{T\in(Kg)^{A}\,|\,T(Ke+Kf)=0\\},$
$LM^{\prime}(A)=\left\\{\small{\begin{pmatrix}a&0&0\\\ b&a&0\\\
s&t&u\end{pmatrix}}\,\Big{|}\,a,b,s,t,u\in K\right\\}\ {\rm and}\ \
LM(A)=\left\\{\small{\begin{pmatrix}a&0&0\\\ b&a&0\\\
0&0&u\end{pmatrix}}\,\Big{|}\,a,b,u\in K\right\\}.$

(ix) $A=W_{1}:$ We have $A=A_{1}\oplus A_{0}$ with $A_{0}=Ke+Kf$ and
$A_{1}=Kg$. Let $T_{1}\in M^{\prime}_{1}(A)$, then $T_{1}$ is a linear mapping
with $T_{1}(A_{0})=\\{0\\}$. So $T_{1}$ is determined by $T_{1}(g)=ag$ with
$a\in K$. Let denote this $T_{1}$ by $T^{a}_{1}$. We have

$M^{\prime}(A)=\\{T^{a}_{1}\,|\,a\in K\\}\oplus(Ke+Kf)^{A}.$

A weak multiplier $T=T^{a}_{1}+R$ with $R\in(Ke+Kf)^{A}$ is a multiplier, if
and only if

$R(xye)=R((xg)(yg))=xgT^{a}_{1}(yg)-T^{a}_{1}(xye)=axye$

for all $x,y\in K$, if and only if $R(xe)=axe$ for any $x\in K$. Let
$T_{a}=\small{\begin{pmatrix}a&0&0\\\ 0&0&0\\\ 0&0&a\end{pmatrix}}$. Then,
$(T-T_{a})(Ke)=\\{0\\}$ and it follows that

$M(A)=\\{T_{a}\,\big{|}\,a\in
K\\}\oplus\\{R\in(Ke+Kf)^{A}\,\big{|}\,R(Ke)=\\{0\\}\\}.$

Also we have

$LM^{\prime}(A)=\left\\{\small{\begin{pmatrix}a&b&c\\\ p&q&r\\\
0&0&u\end{pmatrix}}\,\Big{|}\,a,b,c,p,q,r,u\in K\right\\}$

and

$LM(A)=\left\\{\small{\begin{pmatrix}a&b&c\\\ 0&q&r\\\
0&0&a\end{pmatrix}}\,\Big{|}\,a,b,c,q,r\in K\right\\}.$

(x) $A=W_{2}$ 333This is the algebra taken up in [8]: We have $A=A_{1}\oplus
A_{0}$ with $A_{0}=Ke$ and $A_{1}=Kf+Kg$. $T\in M^{\prime}_{1}(A)$ is a linear
mapping with $T(Ke)=\\{0\\}$. Let $T$ be represented as (22), then $T$ is a
weak multiplier if and only if

$\begin{pmatrix}0&0&0\\\ 0&0&0\\\ 0&qe&re\end{pmatrix}=\bm{A}T=T^{\rm
t}\bm{A}=\begin{pmatrix}0&0&0\\\ 0&te&0\\\ 0&ue&0\end{pmatrix},$

if and only if $r=t=0$, $q=u$. Hence,

$M^{\prime}(A)=\\{T_{q}\,|\,q\in K\\}\oplus(Ke)^{A},$

where $T_{q}=\small{\begin{pmatrix}0&0&0\\\ 0&q&0\\\ 0&0&q\end{pmatrix}}$. So,

$LM^{\prime}(A)=\left\\{\small{\begin{pmatrix}a&b&c\\\ 0&q&0\\\
0&0&q\end{pmatrix}}\,\Big{|}\,a,b,c,q\in K\right\\}.$

A weak multiplier $T=T_{q}+R$ with $R\in(Ke)^{A}$ is a multiplier, if and only
if

$\displaystyle R(yze)=R(\alpha\beta)=\alpha
T_{q}(\beta)-T_{q}(yze)=\alpha(q\beta)=qyze$

for any $\alpha=xf+yg,\beta=zf+vg\in A_{1}$ with $x,y,z,v\in K$, if and only
if $R(xe)=qxe$ for all $x\in K$. Let $S_{a}$ be the scalar multiplication by
$a\in K$. Then, $(T-S_{a})(Ke)=\\{0\\}$, and hence,

$M(A)=\\{S_{a}\,|\,a\in K\\}\oplus\\{R\in(Ke)^{A}\,|\,R(Ke)=\\{0\\}\\}$

and

$LM(A)=\left\\{\small{\begin{pmatrix}a&b&c\\\ 0&a&0\\\
0&0&a\end{pmatrix}}\,\Big{|}\,a,b,c\in K\right\\}.$

(xi) $A=W_{4}$: We have $A=A_{1}\oplus A_{0}$ with $A_{0}=Kf+Kg$ and
$A_{1}=Ke$. Because $A_{1}$ is a subalgbra isomorphic to the base field $K$,
for $T_{1}^{a}\in M(A_{1})$ with $a\in K$ given by $T_{1}^{a}(e)=ae$, we see

$M^{\prime}(A)=\\{T_{1}^{a}\,|\,a\in K\\}\oplus(fK+gK)^{A}$

and

$M(A)=\\{T_{1}^{a}\,|\,a\in K\\}\oplus\\{R\in(fK+gK)^{A}\,|\,R(Ke)=0\\}$

by Theorem 3.1. Taking the intersection with $L(A)$ we have

$LM^{\prime}(A)=\left\\{\small{\begin{pmatrix}a&0&0\\\ p&q&r\\\
s&t&u\end{pmatrix}}\,\Big{|}\,a,p,q,r,s,t,u\in K\right\\}$

and

$LM(A)=\left\\{\small{\begin{pmatrix}a&0&0\\\ 0&q&r\\\
0&t&u\end{pmatrix}}\,\Big{|}\,a,q,r,t,u\in K\right\\}.$

(xii) $W_{5}$ and $W_{6}$ are opposed. Let $A=W_{5}$, then $A=A_{1}\oplus
A_{0}$ with $A_{0}=Ke$ and $A_{1}=Kf+Kg$. Since $A_{1}$ is a subalgebra of
$A$, we have the equalities (10). Because $A_{1}$ has a left identity $g$, we
have

$M(A_{1})=LM(A_{1})=M^{\prime}(A_{1})=LM^{\prime}(A_{1})\cong(A_{1})/Kf\cong
Kg$

by Theorem 5.5. So, any element in $M(A_{1})$ is a scalar multiplication
$S_{1}^{q}$ in $A_{1}$ by $q\in K$. By Theorem 3.1 we have

$M^{\prime}(A)=\\{S_{1}^{q}\,|\,q\in K\\}\oplus(Ke)^{A},$
$M(A)=\\{S_{1}^{q}\,|\,q\in K\\}\oplus\\{R\in(Ke)^{A}\,|\,R(Kf+Kg)=0\\},$
$LM^{\prime}(A)=\left\\{\small{\begin{pmatrix}a&b&c\\\ 0&q&0\\\
0&0&q\end{pmatrix}}\,\Big{|}\,a,b,c,q\in K\right\\}\ \,{\rm and}\
\,LM(A)=\left\\{\small{\begin{pmatrix}a&0&0\\\ 0&q&0\\\
0&0&q\end{pmatrix}}\,\Big{|}\,a,q\in K\right\\}.$

(xiii) $W_{7}$ and $W_{8}$ are opposed. Let $A=W_{7}$. We see
$A_{0}=\mbox{Ann}_{r}(A)=\\{0\\}$. Hence, any weak multiplier is a linear
multiplier by Proposition 5.1, and $T$ represented as (23) is a weak
multiplier, if and if

$\begin{pmatrix}ae&be&ce\\\ 0&0&0\\\
pf+sg&qf+tg&rf+ug\end{pmatrix}=\bm{A}T=T^{\rm
t}\bm{A}=\begin{pmatrix}ae&sf&sg\\\ be&tf&tg\\\ ce&uf&ug\end{pmatrix},$

if and only if $b=c=p=r=s=t=0,q=u$. Therefore,

$M(A)=LM(A)=M^{\prime}(A)=LM^{\prime}(A)=LM^{\prime}(A)=\left\\{\small{\begin{pmatrix}a&0&0\\\
0&q&0\\\ 0&0&q\end{pmatrix}}\,\Big{|}\,a,q\in K\right\\}.$

(xiv) $W_{9}$ and $W_{10}$ are opposed. Let $A=W_{9}$. Then, because
$A_{0}=\mbox{Ann}_{l}(A)=\\{0\\}$, any weak multiplier is a linear multiplier
and a linear mapping $T$ represented as (23) is a weak multiplier if and only
if

$\begin{pmatrix}pe&qe&re\\\ ae+pf&be+qf&ce+rf\\\
pg&qg&rg\end{pmatrix}=\bm{A}T=T^{\rm t}\bm{A}=\begin{pmatrix}pe&ae+pf+sg&0\\\
qe&be+qf+tg&0\\\ re&ce+rf+ug&0\end{pmatrix}$

$c=p=r=s=t=0,a=q=u$. Therefore,

$LM(A)=M(A)=LM^{\prime}(A)=M^{\prime}(A)=\left\\{\small{\begin{pmatrix}a&b&0\\\
0&a&0\\\ 0&0&a\end{pmatrix}}\,\Big{|}\,a,b\in K\right\\}.$

(xv) $A=W_{3}(k)$. We have $A=A_{1}\oplus A_{0}$ with $A_{0}=Ke$ and
$A_{1}=Kf+Kg$. $T\in M^{\prime}_{1}(A)$ is a linear mapping with
$T(Ke)=\\{0\\}$. Let $T$ be represented as (22), then $T$ is a weak multiplier
if and only if

$\begin{pmatrix}0&0&0\\\ 0&qe&re\\\
0&(kq+t)e&(kr+u)e\end{pmatrix}=\bm{A}T=T^{\rm t}\bm{A}=\begin{pmatrix}0&0&0\\\
0&(q+kt)e&te\\\ 0&(r+ku)e&ue\end{pmatrix}.$

If $k=0$, then the above holds if and only if $r=t$, and otherwise it holds if
and only if $r=t=0,q=u$. Thus,

$M^{\prime}(A)=\\{T_{1}^{q,r,u}\,\big{|}\,q,r,u\in K\\}\oplus(Ke)^{A},\
LM^{\prime}(A)=\left\\{\small{\begin{pmatrix}a&b&c\\\ 0&q&r\\\
0&r&u\end{pmatrix}}\,\Big{|}\,a,b,c,q,r,u\in K\right\\}$

if $k=0$, and

$M^{\prime}(A)=\\{T_{1}^{q}\,\big{|}\,q\in K\\}\oplus(Ke)^{A},\
\,LM^{\prime}(A)=\left\\{\small{\begin{pmatrix}a&b&c\\\ 0&q&0\\\
0&0&q\end{pmatrix}}\,\Big{|}\,a,b,c,q\in K\right\\}$

if $k\neq 0$, where

$T_{1}^{q,r,u}=\small{\begin{pmatrix}0&0&0\\\ 0&q&r\\\ 0&r&u\end{pmatrix}}\
\,{\rm and}\ \,T_{1}^{q}=\small{\begin{pmatrix}0&0&0\\\ 0&q&0\\\
0&0&q\end{pmatrix}}.$

When $k=0$, $T=T_{1}^{q,r,u}+R$ with $R\in(Ke)^{A}$ is multiplier, if and only
if

$\displaystyle R((xz+yv)e)$ $\displaystyle=$ $\displaystyle R(\alpha\beta)\ =\
\alpha T_{1}^{q,r,u}(\beta)-T_{1}^{q,r,u}((xz+yv)e)$ $\displaystyle=$
$\displaystyle\alpha((qz+rv)f+(rz+uv)g)=(qxz+r(xv+yz)+uyv)e$

for any $\alpha=xf+yg,\beta=zf+vg\in A_{1}$ with $x,y,z,v\in K$, if and only
if $q=u,r=0$ and $R(xe)=qxe$ for all $x\in K$. While, when $k\neq 0$,
$T=T_{1}^{q}+R$ with $R\in(Ke)^{A}$ is a multiplier, if and only if

$\displaystyle R((xz+y(kz+v))e)$ $\displaystyle=$ $\displaystyle
R(\alpha\beta)\ =\ \alpha T_{1}^{q}(\beta)-T_{1}^{q}(xue+y(kz+v)e)$
$\displaystyle=$ $\displaystyle\alpha(q\beta)\ =\ q(xz+y(kz+v))e$

for any $\alpha,\beta$, if and only if $R(xe)=qxe$ for any $x\in K$. In the
both cases, with the scalar multiplication $S_{a}$ by $a\in K$, we have
$(T-S_{a})(Ke)=\\{0\\}$. Therefore, for arbitrary $k$ (either $k$ is zero or
nonzero) we see

$M(A)=\\{S_{a}\,\big{|}\,a\in K\\}\oplus\\{R\in(Ke)^{A}\,|\,R(Ke)=\\{0\\}\\}$

and

$LM(A)=\left\\{\small{\begin{pmatrix}a&b&c\\\ 0&a&0\\\
0&0&a\end{pmatrix}}\,\Big{|}\,a,b,c\in K\right\\}.$

##  7 3-dimensional zeropotent algebras

$A$ is a zeropotent algebra if $x^{2}=0$ for all $x\in A$. The zeropotent
algebra $A$ over $K$ is anti-commutative, that is, $xy=-yx$ for all $x,y\in
A$. Thus we see

$A_{0}=\mbox{Ann}_{l}(A)=\mbox{Ann}_{r}(A).$

Let $A$ be a zeropotent algebras of dimension 3 over $K$ with char$(K)\neq 2$.
Let ${E}=\\{e,f,g\\}$ be a basis of $A$. Because $A$ is anti-commutative, the
multiplication table $\bm{A}$ of $A$ on $E$ is given as

$\bm{A}=\begin{pmatrix}0&\alpha&-\beta\\\ -\alpha&0&\gamma\\\
\beta&-\gamma&0\end{pmatrix},$

where

$\begin{cases}\gamma=fg=a_{11}e+a_{12}f+a_{13}g\\\
\beta=ge=a_{21}e+a_{22}f+a_{23}g\\\
\alpha=ef=a_{31}e+a_{32}f+a_{33}g\end{cases}$

for $a_{ij}\in K$. We call $A=\small{\begin{pmatrix}a_{11}&a_{12}&a_{13}\\\
a_{21}&a_{22}&a_{23}\\\ a_{31}&a_{32}&a_{33}.\end{pmatrix}}$ the structural
matrix of $A$ (we use the same symbol $A$ for the algebra and its structural
matrix).

######  Lemma 7.1.

If ${\rm rank}(A)\geq 2$, then $A_{0}=\\{0\\}$.

###### Proof.

If ${\rm rank}(A)\geq 2$, at least two of $fg,ge,ef$ are linearly independent.
Suppose that $\alpha=ef$ and $\beta=ge$ are linearly independent (the other
cases are similar). If $x=ae+bf+cg$ with $a,b.c\in K$ is in
$\mbox{Ann}_{l}(A)$, then $xe=-b\alpha+c\beta$, $xf=a\alpha-c\gamma$ and
$xg=-a\beta+b\gamma$ are all zero. It follows that $a=b=c=0$. Hence, we have
Ann${}_{l}(A)=\\{0\\}$ and $A_{0}=\mbox{Ann}_{l}(A)=\\{0\\}$. ∎

######  Theorem 7.2.

Let $A$ be a zeropotent algebra $A$ of dimension 3 with rank$(A)\geq 2$ over
$K$. Then, any weak multiplier of $A$ is the scalar multiplication $S_{a}$ for
some $a\in K$, that is,

$M(A)=M^{\prime}(A)=LM(A)=LM^{\prime}(A)=\\{S_{a}\,|\,a\in K\\}.$

###### Proof.

By Lemma 7.1 and Corollary 2.4, any weak multiplier $T$ is a linear mapping.
Let $T\in L(A)$ be represented as (23). By Theorem 4.2, $T$ is a weak
multiplier, if and only if $\bm{A}T=T^{\rm t}\bm{A}$, if and only if

$\small{\begin{pmatrix}p\alpha-s\beta&q\alpha-t\beta&r\alpha-u\beta\\\
-a\alpha+s\gamma&-b\alpha+t\gamma&-c\alpha+u\gamma\\\
a\beta-p\gamma&b\beta-q\gamma&c\beta-r\gamma\end{pmatrix}=\begin{pmatrix}-p\alpha+s\beta&a\alpha-s\gamma&-a\beta+p\gamma\\\
-q\alpha+t\beta&b\alpha-t\gamma&-b\beta+q\gamma\\\
-r\alpha+u\beta&c\alpha-u\gamma&-c\beta+r\gamma\end{pmatrix}}$ (25)

holds. Suppose that $\alpha=ef,\beta=ge$ are linearly independent (the other
cases are similar). Then, because $p\alpha-s\beta=-p\alpha+s\beta$ by
comparing the (1,1)-elements of two matrices in (25), we have $p=s=0$.
Comparing (1,2)-elements and (1,3)-elements, we have
$q\alpha-t\beta=a\alpha-s\gamma=a\alpha$ and
$r\alpha-u\beta=-a\beta+p\gamma=-a\beta$ respectively. It follows that $a=q=u$
and $r=t=0$. Comparing (2,2)-elements and (3,3)-elements, we see $b=c=0$.
Consequently, (23) holds if and only if $b=c=p=r=s=t=0\ \ \mbox{and}\ \
a=q=u$, that is, $T=S_{a}$. ∎

In [4] we classify the zeropotent algebras of dimension 3 over an
algebraically field $K$ of characteristic not equal to 2. Up to isomorphism,
we have 10 families of zeropotent algebras. They are

$Z_{0},Z_{1},Z_{2},Z_{3},\\{Z_{4}(a)\\}_{a\in
H},Z_{5},Z_{6},\\{Z_{7}(a)\\}_{a\in H},Z_{8}\mbox{ and }Z_{9}$

defined by the structural matrices

$\small{\begin{pmatrix}0&0&0\\\ 0&0&0\\\
0&0&0\end{pmatrix},\begin{pmatrix}0&0&0\\\ 0&0&0\\\
0&0&1\end{pmatrix},\begin{pmatrix}0&0&1\\\ 0&0&0\\\
0&0&1\end{pmatrix},\begin{pmatrix}0&1&0\\\ -1&0&0\\\
0&0&0\end{pmatrix},\begin{pmatrix}0&0&0\\\ 0&1&a\\\ 0&0&1\end{pmatrix},}$
$\small{\begin{pmatrix}0&1&0\\\ 0&0&0\\\
0&0&1\end{pmatrix},\begin{pmatrix}0&1&1\\\ 0&0&1\\\
0&0&1\end{pmatrix},\begin{pmatrix}1&a&0\\\ 0&1&0\\\
0&0&1\end{pmatrix},\begin{pmatrix}1&2&2\\\ 0&1&2\\\
0&0&1\end{pmatrix}\mbox{and }\begin{pmatrix}1&3&3\\\ 0&1&3\\\
0&0&1\end{pmatrix},}$

respectively.

$Z_{0}$ is the zero algebra, and $Z_{1}$ is isomorphic to the 3-dimensional
associative algebra $C_{1}$, and their (weak) multipliers are already
determined in Section 6. The algebras $Z_{3}$ to $Z_{9}$ have rank greater or
equal to 2, which are covered by Theorem 7.2.

Thus, only $A=Z_{2}$ is left to be analyzed. The multiplication table $\bm{A}$
of $A$ is $\small{\begin{pmatrix}0&g&0\\\ -g&0&g\\\ 0&-g&0\end{pmatrix}}$. We
see $A_{0}=\mbox{Ann}_{l}(A)=\mbox{Ann}_{r}(A)=K(e+g),$ and we have the nihil
decomposition $A=A_{0}\oplus A_{1}$ with $A_{1}=Ke+Kf$. A weak multiplier
$T\in M^{\prime}_{1}(A)$ is a linear mapping represented by
$\small{\begin{pmatrix}a&b&c\\\ p&q&r\\\ 0&0&0\end{pmatrix}}$ satisfying

$\begin{pmatrix}pg&qg&rg\\\ -ag&-bg&-cg\\\
-pg&-qg&-rg\end{pmatrix}=\bm{A}T=T^{\rm t}\bm{A}=\begin{pmatrix}-pg&ag&pg\\\
-qg&bg&qg\\\ -rg&cg&rg\end{pmatrix}.$

Hence, $a=-c=q$ and $b=p=r=0$. Let $T_{a}$ be this linear mapping, then by
Theorem 3.1 we have

$M^{\prime}(A)=\\{T_{a}\,|\,a\in K\\}\oplus(K(f+g))^{A}$

and

$LM^{\prime}(A)=\left\\{\small{\begin{pmatrix}a+s&t&-a+u\\\ 0&a&0\\\
s&t&u\end{pmatrix}}\,\Big{|}\,a,s,t,u\in K\right\\}.$

By Corollary 3.2, a weak multiplier $T=T_{a}+R$ with $R\in(K(e+g))^{A}$ is a
multiplier if and only if for any $\zeta=xe+yf$ and $\eta=ze+vf$ with
$x,y,z,v\in K$,

$\displaystyle R((xv-yz)g)$ $\displaystyle=$ $\displaystyle R(\zeta\eta)\ =\
\zeta T_{a}(\eta)-T_{a}((xv-yz)g)$ $\displaystyle=$
$\displaystyle\zeta(a\eta)+a(xv-yz)e\ =\ a(xv-yz)(e+g)$

holds. It follows that $R(xg)=ax(e+g)$ for all $x\in K$. Let $S_{a}$ be the
scalar multiplication by $a$, then $(T-S_{a})(A)\subseteq K(e+g)$ and
$(T-S_{a})(Kg)=\\{0\\}$. Hence, we obtain

$M(A)=\\{S_{a}\,|\,a\in K\\}\oplus\\{R\in(K(e+g))^{A}\,|\,R(Kg)={0}\\},$

and

$LM(A)=\left\\{\small{\begin{pmatrix}a+s&t&0\\\ 0&a&0\\\
s&t&a\end{pmatrix}}\,\Big{|}\,a,s,t\in
K\right\\}=\left\\{\small{\begin{pmatrix}a&b&0\\\ 0&c&0\\\
a-c&b&c\end{pmatrix}}\,\Big{|}\,a,b,c\in K\right\\}.$

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