The factorization problem for Jordan algebras. Applications]
The factorization problem for Jordan algebras. Applications

A. L. Agore
Simion Stoilow Institute of Mathematics of the Romanian Academy, P.O. Box 1-764, 014700 Bucharest, Romania and Vrije Universiteit Brussel, Pleinlaan 2, B-1050 Brussels, Belgium

<EMAIL_ADDRESS>and<EMAIL_ADDRESS>
G. Militaru
Faculty of Mathematics and Computer Science, University of Bucharest, Str.
Academiei 14, RO-010014 Bucharest 1, Romania and Simion Stoilow Institute of Mathematics of the Romanian Academy, P.O. Box 1-764, 014700 Bucharest, Romania

<EMAIL_ADDRESS>and<EMAIL_ADDRESS>[2010]16T10, 16T05, 16S40

This work was supported by a grant of the Ministry of Research, Innovation and Digitization, CNCS/CCCDI–UEFISCDI, project number
PN-III-P4-ID-PCE-2020-0458, within PNCDI III

[2020]17C10, 17C50, 17C55

We investigate the factorization problem as well as the classifying complements problem in the setting of Jordan algebras. Matched pairs of Jordan algebras and the corresponding bicrossed products are introduced. It is shown that any Jordan algebra which factorizes through two given Jordan algebras is isomorphic to a bicrossed product associated to a certain matched pair between the same two Jordan algebras. Furthermore, a new type of deformation of a Jordan algebra is proposed as the main step towards solving the classifying complements problem.

§ INTRODUCTION

The factorization problem is an old and notoriously difficult problem which stems in group theory. A weaker version of the factorization problem was first considered in [23, 27] where groups $G$ which admit two subgroups $A$ and $B$ such that $G = AB$, are studied under the name of permutable groups. If we assume, in addition, that $A$ and $B$ have trivial intersection then we say that $G$ factorizes through $A$ and $B$. The factorization problem for groups asks for the description and classification of all groups which factorize through two given groups. As explained in [3, 4], despite its elementary statement, the factorization problem is far from being an easy question. An important turning point for the factorization problem was the introduction of the matched pairs of groups [28] which consist of two groups that act on each other in a compatible way. To each matched pair of groups one can associate a so-called bicrossed product, i.e. a group structure on the direct product of the underlying sets constructed from the two actions. This leads to a new and more computational approach to the factorization problem. Consequently, the factorization problem comes down to finding all matched pairs between two given groups and classifying the corresponding bicrossed products. Therefore, the same strategy relying on matched pairs and bicrossed products was used to approach the factorization problem for various mathematical objects such as: (co)algebras [11, 12, 13], Lie
algebras and Lie groups [21, 24], Leibniz algebras [6], Hopf algebras [25], fusion categories [15] and so on. Furthermore, this new approach has also the advantage of opening the way to new classification methods as evidenced in [2, 3, 8, 19] for Hopf algebras. Since their introduction by P. Jordan in $1933$, Jordan algebras have appeared in various different fields of mathematics and mathematical physics such us the theory of superstrings, supersymmetry, projective geometry, Lie algebras and algebraic groups, representation theory or functional analysis [26]. In the setting of Jordan algebras, the factorization problem can be stated as follows:

Factorization problem. Let $A$ and $V$ be two
given Jordan algebras. Describe and classify up to an isomorphism
all Jordan algebras $E$ that factorize through $A$ and $V$, i.e.
$E$ contains $A$ and $V$ as Jordan subalgebras such that $E = A +
V$ and $A \cap V = \{0\}$.

Another closely-related problem which we will consider here was first
introduced in [5] at the level of Lie/Hopf algebras as
a converse of the factorization problem. Corresponding theories have been developed for associative algebras [1] or groups [7] where an important connection with the problem of classifying all groups of a given finite order was highlighted. For Jordan algebras it comes down to the following:

Classifying complements problem. Let $A \subseteq
E$ be a Jordan subalgebra of a Jordan algebra $E$. If a complement
of $A$ in $E$ exists (i.e. a Jordan subalgebra $V\subseteq E$ such
that $E = A + V$ and $A \cap V = \{0\}$), describe and classify up
to an isomorphism all others complements of $A$ in $E$.

The paper is devoted to the two problems listed above and is organized as follows: in prel we set the notation and recall some
basic definitions and properties of Jordan algebras. cazurispeciale focuses on the factorization problem for Jordan algebras. mpmL introduces matched pairs of Jordan algebras and bicross connects them to the bicrossed product construction. These are the
Jordan algebra counterparts of similar constructions performed for groups [28] or Lie algebras [22, 25]. bicromlver proves that the
bicrossed product of two Jordan algebras is the object responsible
for the factorization problem and is the Jordan algebra version of
<cit.>. As the examples in this section show, most low dimensional Jordan algebras can be written as bicrossed products of certain Jordan subalgebras. This suggests that achieving the classification of bicrossed products can lead to a successful strategy for classifying Jordan algebras. To this end, a general classification result for bicrossed products is presented in morfismebicross.

The last section of the paper deals with the classifying complements problem for Jordan algebras following the general strategy we developed previously in [1, 5, 7] for groups and associative/Lie/Hopf algebras. Consequently, a new type of deformation of a Jordan algebra arising from a matched pair is introduced in deforJ. More precisely, these deformations involve a certain linear map $r \colon B \to A$ associated to the canonical matched pair $(A,\, B,\, \triangleleft,\, \triangleright)$, called a deformation map. It is shown that all $A$-complements of a Jordan algebra $E$ are obtained from a given complement $B$ by this new kind of deformation. More precisely, for each deformation map $r \colon B \to A$ we can define a new $A$-complement denoted by $B_{r}$ and, conversely, for any $A$-complement $B'$ we can find a deformation map of the canonical matched pair $(A,\, B,\, \triangleleft,\, \triangleright)$ such that $B' \cong B_{r}$. Relevant examples are presented. In particular, it is proved that $4$ out of the $6$ isomorphism types of real Jordan algebras of dimension $2$ (<cit.>) appear as deformations of a given $2$-dimensional Jordan algebra. This further substantiate the importance of these objects for the classification problem of Jordan algebras.

§ PRELIMINARIES
All vector spaces, (bi)linear maps, tensor products are over a
field $k$ of characteristic $\neq 2$. A Jordan algebra is a vector space $A$ together with a
bilinear map $\cdot : A \times A \to A$, called multiplication, which is commutative and satisfies the Jordan identity, i.e. for any $a$, $b\in A$ we have:

\begin{equation*}
a\cdot b = b\cdot a, \qquad (a^2 \cdot b) \cdot a = a^2 \cdot (b
\cdot a)
\end{equation*}

A Jordan algebra $A$ with trivial multiplication, i.e. $a \cdot b := 0$ for all $a$, $b\in A$, will be called abelian and a vector space $V$ endowed with the abelian multiplication will be denoted by $V_0$. Throughout, when there is no fear of confusion, we will denote the multiplication of a Jordan algebra just by juxtaposition.
As in the case of Lie algebras, any associative algebra induces a Jordan algebra. More precisely, given an associative algebra $A$ one can define a Jordan algebra structure on the underlying vector space by $x \cdot y :=
2^{-1} (xy + yx)$, for all $x$, $y\in A$. The automorphism group of the Jordan algebra $A$ will be denoted by ${\rm Aut}_{\rm
J}(A)$. We recall from [9] the following concept:

A right action of a Jordan algebra $A$ on a vector space
$M$ is a bilinear map $\triangleleft : M \times A \to M$ such that
for any $a\in A$ and $x \in M$ we have:

\begin{eqnarray} \eqlabel{actiunedr}
(x \triangleleft a^2) \triangleleft a = (x \triangleleft a)
\triangleleft a^2
\end{eqnarray}

Similarly, a left action of $A$ on $M$ is a bilinear map
$\triangleright : A \times M \to M$ such that for any $a\in A$ and
$x \in M$ we have:

\begin{eqnarray} \eqlabel{actiunest}
a \triangleright  (a^2 \triangleright x) = a^2 \triangleright (a
\triangleright x)
\end{eqnarray}

The canonical maps $\triangleright : A \times A \to A$ and
$\triangleright : A \times A^* \to A^*$ given for any $a$, $b\in
A$ and $a^* \in A^*$ by:

\begin{equation} \eqlabel{calduramare}
a \triangleright b := a \cdot b, \qquad (a \triangleright a^*) (b)
:= a^* (a\cdot b)
\end{equation}

are left actions of $A$ on $A$ and $A^* = {\rm Hom}_k
\, (A, \, k) $, respectively.

A Jordan bimodule [14] over a Jordan algebra $A$ is a vector space
$M$ together with two linear maps:

\begin{eqnarray*}
&& M \otimes A \to M, \,\,\, x \otimes a \mapsto xa\\
&& A \otimes M \to M, \,\,\, a \otimes x \mapsto ax
\end{eqnarray*}

subject to the following compatibilities:

\begin{eqnarray}
&& xa = ax\\\
&& a(a^{2}x) = a^{2}(xa)\eqlabel{bim2}\\
&& (a^{2}b)x - a^{2}(bx) = 2 \bigl[(ab)(ax) - a(b(ax))\bigl]\eqlabel{bim3}
\end{eqnarray}

For all unexplained notions pertaining to Jordan algebra theory we refer the reader to [16, 20, 26].

§ MATCHED PAIRS AND BICROSSED PRODUCTS OF JORDAN ALGEBRAS

In this section we introduce matched pairs of Jordan algebras and the corresponding bicrossed product in order to approach the factorization problem. Part of our results regarding matched pairs will be derived as a special case from the more general theory developed in <cit.> in connection to the extending structures problem.

Let $A$ and $V$ be two Jordan algebras. Then the quadruple $(A, V, \triangleleft, \, \triangleright)$ is
called a matched pair of Jordan algebras if $\triangleleft
: V\times A \to V$ is a right action of the Jordan algebra $A$ on
$V$, $\triangleright : V \times A \to A$ is a left action of the
Jordan algebra $V$ on $A$ and the following compatibility conditions hold for all $a$, $b \in A$
and $x$, $y \in V$:

\begin{eqnarray*}
&& \hspace*{-8mm} {\rm (MP1)}\,\, a \, (x \triangleright a^2) + (x
\triangleleft a^2 )\triangleright a = a^2 \, (x \triangleright a) + (x \triangleleft a) \triangleright a^{2};\\
&& \hspace*{-8mm}{\rm (MP2)}\,\, x \triangleleft (x^2 \triangleright a) + (x^2
\triangleleft a)  \, x = x^2 \triangleleft (x
\triangleright a) + x^2  \, (x \triangleleft a);\\
&& \hspace*{-8mm} {\rm (MP3)}\,\, 2\bigl[\bigl((x \triangleleft a) \triangleleft b\bigl) \triangleleft \,a + x \triangleleft \bigl((x \triangleright a) b\bigl) + x \triangleleft \bigl((x \triangleleft a) \triangleright b\bigl) + \bigl((x \triangleleft a) \triangleleft b\bigl)x \bigl] + (x^{2} \triangleleft b) \triangleleft a +\\
&& \,\,\,\,\,\, x \triangleleft (ba^{2}) = 2\bigl[(x \triangleleft a) \triangleleft ba + (x \triangleleft a) \triangleleft (x \triangleright b) + (x \triangleleft b) \triangleleft (x \triangleright a) + (x \triangleleft a)(x \triangleleft b)\bigl] +\\
&& \,\,\,\,\,\, x^{2} \triangleleft ba + (x \triangleleft b) \triangleleft a^{2};\\
&& \hspace*{-8mm} {\rm (MP4)}\,\, 2\bigl[y \triangleright (x \triangleright a) + x \triangleright \bigl(y \triangleright (x \triangleright  a)\bigl) + \bigl(y \triangleleft (x \triangleright a)\bigl) \triangleright\, a + \bigl((x \triangleleft a) y\bigl) \triangleright a\bigl] + (x^{2}y) \triangleright a +\\
&& \,\,\,\,\,\, x \triangleright (y \triangleright a^{2}) = 2\bigl[(y \triangleright a)(x \triangleright a) + xy \triangleright (x \triangleright a) + (y \triangleleft a) \triangleright (x \triangleright a) + (x \triangleleft a) \triangleright (y \triangleright a)\bigl] +\\
&& \,\,\,\,\,\, x^{2} \triangleright (y \triangleright a) + xy \triangleright a^{2};\\
&& \hspace*{-8mm} {\rm (MP5)}\,\, 2\bigl[\bigl(y \triangleleft (x \triangleright a)\bigl) \triangleleft \,a + \bigl((x \triangleleft a)y\bigl) \triangleleft a + x \triangleleft \bigl(y \triangleright (x \triangleright a)\bigl) + \bigl(y \triangleleft (x \triangleright a)\bigl)x + \bigl((x \triangleleft a)y\bigl)x\bigl] +\\
&& \,\,\,\,\,\, x^{2}y \triangleleft a + x \triangleleft (y \triangleright a^{2}) + (y \triangleleft a^{2}) x = 2\bigl[(x \triangleleft a) \triangleleft (y \triangleright a) + (y \triangleleft a) \triangleleft (x \triangleright a) + \\
&& \,\,\,\,\,\, xy \triangleleft (x \triangleright a) + (x \triangleleft a)xy + (x \triangleleft a)(y \triangleleft  a)\bigl] + x^{2} \triangleleft (y \triangleright a) + xy \triangleleft a^{2} + x^{2}(y \triangleleft a); \\
&& \hspace*{-8mm} {\rm (MP6)}\,\, 2\bigl[\bigl((x \triangleright a)b\bigl)a + \bigl((x \triangleleft a) \triangleright b\bigl)a + \bigl((x \triangleleft a) \triangleleft b\bigl) \triangleright a + x \triangleright \bigl((x \triangleright a)b\bigl) + x \triangleright \bigl((x \triangleleft a) \triangleright b\bigl)\bigl] +\\
&& \,\,\,\,\,\, (x^{2} \triangleright b)a + (x^{2} \triangleleft b)  \triangleright a + x  \triangleright (a^{2}b) = 2 \bigl[(x \triangleright a)(ab) + (x \triangleleft a) \triangleright (ba) + (x  \triangleright a) (x  \triangleright b) +\\
&& \,\,\,\,\,\, (x  \triangleleft b) \triangleright (x \triangleright a) + (x \triangleleft a) \triangleright (x \triangleright b)\bigl] + x^{2} \triangleright (ba) + (x \triangleleft b) \triangleright a^{2} +a^{2}(x \triangleright b).
\end{eqnarray*}

We denote by $ A \bowtie V$ the
vector space $A \, \times V$ together with the bilinear map $
\circ : (A \times V) \times (A \times V) \to A \times V$ defined

\begin{equation}\eqlabel{bracbicross}
(a, x) \circ (b, y) := \bigl( a b + x \triangleright b
+ y\triangleright a , \,\,  x\triangleleft b + y\triangleleft a +
x y  \bigl)
\end{equation}

The next result provides a motivation for the matched pair axioms as introduced in mpmL and is the Jordan algebra version of <cit.>:

Let $A$ and $V$ be two Jordan algebras and $\triangleleft
\colon V\times A \to V$, $\triangleright \colon V \times A \to A$ two bilinear maps. The multiplication defined in bracbicross is a Jordan algebra structure on $A \times V$ if and only if $(A, V, \triangleleft, \, \triangleright)$ is a matched pair of Jordan algebras. In this case, the Jordan algebra $A
\bowtie V$ will be called the bicrossed product associated to the matched pair $(A, V, \triangleleft, \, \triangleright)$.

Note that the proof can be performed in a direct manner by a rather long and laborious computation. Instead, we will apply the results from <cit.>. Indeed, by <cit.>, a Jordan extending datum of $A$ through $V$ with $f := 0$ comes down precisely to the definition of a matched pair while the multiplication of the corresponding unified product reduces to the one in bracbicross. The conclusion now follows from <cit.> and <cit.>.

Our first examples of a bicrossed product are generic ones, namely the semidirect product of Jordan algebras [16] and the null split extension of a Jordan algebra by a bimodule <cit.>.

$1.$ Let $A$, $V$ be two Jordan algebras and $\triangleright : V
\times A \to A$ a left action such that the following
compatibilities hold for all $a$, $b \in A$ and $x$, $y \in V$:

\begin{eqnarray*} \eqlabel{Lsemidirectpr}
&& \hspace*{-8mm}{\rm (L1)}\,\,\,\,\,\, a \, ( x \triangleright a^2) = a^2 \, (x \triangleright a)\\
&&  \hspace*{-8mm}{\rm (L2)}\,\,\,\,\,\, 2\bigl[y \triangleright (x \triangleright a) - (y \triangleright a)(x \triangleright a) + x \triangleright \bigl(y \triangleright (x \triangleright  a)\bigl) -xy \triangleright (x \triangleright a) \bigl]\\
&&\quad =  x^{2} \triangleright (y \triangleright a) -  (x^{2}y) \triangleright a +  (xy) \triangleright a^{2} -  x \triangleright (y \triangleright a^{2})\\
&& \hspace*{-8mm}{\rm (L3)}\,\,\,\,\,\,  2\bigl[\bigl((x \triangleright a)b\bigl)a - (x \triangleright a)(ab) + x \triangleright \bigl((x \triangleright a)b\bigl) -  (x  \triangleright a) (x  \triangleright b)\bigl]\\
&&\quad = x^{2} \triangleright (ba) - (x^{2} \triangleright b)a + a^{2}(x \triangleright b) - x \triangleright (a^{2}b)
\end{eqnarray*}

Then the direct product $A \times V$ together with the multiplication given as follows $a$, $b \in A$ and $x$, $y \in V$:

\begin{equation}\eqlabel{semiL}
(a, x) \circ (b, y) := \bigl( a b + x \triangleright b
+ y\triangleright a , \,\, x y  \bigl)
\end{equation}

is a Jordan algebra called the left semidirect product and will be denoted by $A \ltimes V$.

It can be easily seen that the left semidirect product can be recovered from mpmL by considering a matched pair $(A, V, \triangleleft, \, \triangleright)$ with $\triangleleft := 0$. Then (MP2), (MP3) and (MP5) are trivially fulfilled while (MP1), (MP4) and (MP6) reduce to the compatibilities (L1), (L2) and (L3) respectively.

Similarly, the right semidirect product also appears as a special case of the bicrossed product. To this end, consider $A$, $V$ to be two Jordan algebras and $\triangleleft : V \times A \to V$ a right action such that the following compatibilities hold for all $a$, $b \in A$ and $x$, $y \in V$:

\begin{eqnarray*} \eqlabel{Rsemidirectpr}
&& \hspace*{-8mm}{\rm (R1)}\,\,\,\,\,\, x\, (x^2
\triangleleft a) = x^2  \, (x \triangleleft a)\\
&&  \hspace*{-8mm}{\rm (R2)}\,\,\,\,\,\,
2\bigl[\bigl((x \triangleleft a) \triangleleft b\bigl) \triangleleft \,a - (x \triangleleft a) \triangleleft ba + \bigl((x \triangleleft a) \triangleleft b\bigl)x - (x \triangleleft a)(x \triangleleft b)\bigl] \\
&&\quad =  x^{2} \triangleleft ba - (x^{2} \triangleleft b) \triangleleft a +  (x \triangleleft b) \triangleleft a^{2} - x \triangleleft (ba^{2})\\
&& \hspace*{-8mm}{\rm (R3)}\,\,\,\,\,\,  2\bigl[\bigl((x \triangleleft a)y\bigl) \triangleleft\, a - (x \triangleleft a)(y \triangleleft  a) + \bigl((x \triangleleft a)y\bigl)x - (x \triangleleft a)xy\bigl]\\
&&\quad = xy \triangleleft a^{2} - (y \triangleleft a^{2}) x + x^{2}(y \triangleleft a) - x^{2}y \triangleleft a
\end{eqnarray*}

Then the direct product $A \times V$ together with the multiplication given as follows $a$, $b \in A$ and $x$, $y \in V$:

\begin{equation}\eqlabel{semiR}
(a, x) \circ (b, y) := \bigl( a b, \,\, x\triangleleft b + y\triangleleft a + x y  \bigl)
\end{equation}

is a Jordan algebra called the right semidirect product and will be denoted by $A \rtimes V$.

The right semidirect product can be recovered from mpmL by considering a matched pair $(A, V, \triangleleft, \, \triangleright)$ with $\triangleright := 0$. Then (MP1), (MP4) and (MP6) are trivially fulfilled while (MP2), (MP3) and (MP5) reduce to the compatibilities (R1), (R2) and (R3) respectively.

$2.$ Let $A$ be a Jordan algebra and $M$ a Jordan bimodule over $A$. We will see $M$ as an abelian Jordan algebra, i.e. $xy = 0$ for all $x$, $y \in M$. It is straightforward to see that $\triangleleft \colon M \otimes A \to M$ defined by $x \triangleleft a = xa$ is a right action of $A$ on $M$ satisfying $(R1)-(R3)$, where we denote by juxtaposition the bimodule structure on $M$. Indeed, to start with, $\triangleleft$ is a right action by virtue of bim2. Furthermore, $(R1)$ and $(R3)$ hold trivially since we assumed $M$ to be an abelian Jordan algebra and by the same argument $(R2)$ follows from bim3. The multiplication on the corresponding right semidirect product $A \ltimes M$ is given as follows for all $a$, $b \in A$ and $x$, $y \in M$ by:

\begin{eqnarray*}\eqlabel{semiR}
(a, x) \circ (b, y) := \bigl( a b, \,\, xb + ya  \bigl)
\end{eqnarray*}

which is precisely the null split extension of a Jordan algebra by a bimodule.

A straightforward computation shows that $V \cong \{0\} \times V$ is an ideal
of $A \rtimes V$, $A \cong A  \times \{0\}$ is a subalgebra of $A
\rtimes V$ and the canonical inclusion $i_A : A \to A \rtimes V$,
$i_A (a) = (a, 0)$ is a split monomorphism of Jordan algebras: its
retraction $\pi_A : A \rtimes V \to A$, $\pi_A (a, x) := a$, for
all $a\in A$ and $x\in V$ is a Jordan algebra map.

Conversely, the right semidirect product of Jordan algebras describes
split monomorphisms in the category: more precisely, if $i: A \to E$ is a split
monomorphism of Jordan algebras with splitting map $p: E \to A$, then $E \cong A \rtimes V$, where $V = {\rm ker} p$ is obviously a Jordan subalgebra of $E$.
Indeed, $\psi \colon A \rtimes V \to E$ defined by $\psi(a,\,x) = a+x$ for all $(a,\,x) \in A \rtimes V$ is a Jordan algebra isomorphism, where $A \rtimes V$  is the right semidirect product corresponding to the right action $\triangleleft : V \times A \to V$ given by $x \triangleleft a = xa$. To this end, for all $a$, $b \in A$ and $x$, $y \in V$ we have:

\begin{eqnarray*}
\psi(a,\,x) \psi(b,\, y) &=& (a+x)(b+y) = ab+ay+xb+xy = \psi(ab,\, xb+ya+xy)\\
&=& \psi \bigl((a,\,x) \circ (b,\,y)\bigl)
\end{eqnarray*}

In what follows, when describing a matched pair of Jordan algebras, we only indicate the non-zero values of the two actions.

$1.$ <cit.> lists the $7$ isomorphism classes of non-associative unitary Jordan algebras of dimension $4$. It can be easily seen that all those Jordan algebras can be written as bicrossed products of certain subalgebras. We include here one example; $J_{5}$ is the $4$-dimensional real Jordan algebra with multiplication table defined as follows:

$J_{5}$     $a$     $b$     $u$     $v$

$a$     $a$     $0$     $\frac{1}{2}u$     $v$

$b$     $0$     $b$     $\frac{1}{2}u$     $0$

$u$     $\frac{1}{2}u$     $\frac{1}{2}u$     $0$     $0$

$v$     $v$     $0$     $0$     $0$

$J_{5}$ is a right semidirect product between the $2$-dimensional Jordan algebras $A \,=\, <a,\, b ~|~ a^{2} = a,\, b^{2} = b, \, ab =0>$ and the abelian $2$-dimensional Jordan algebra $V$ generated by $u$ and $v$ corresponding to the right action $\triangleleft \colon V \times A \to V$ defined as follows:

v \triangleleft a = v, \qquad u \triangleleft a = u \triangleleft b = \frac{1}{2} u.

$2.$ Similarly, among the $19$ isomorphism classes of non-associative unitary Jordan algebras of dimension $5$ listed in <cit.> there is only one which can not be written as a bicrossed product, namely $J_{3}$. We include below two examples: the Jordan algebras denoted in [17] by $J_{7}$ and $J_{17}$. $J_{7}$ is the $5$-dimensional real Jordan algebra with multiplication table defined as follows:

$J_{7}$     $a$     $b$     $c$     $u$     $v$

$a$     $a$     $0$     $\frac{1}{2} c$     $\frac{1}{2} u$     $\frac{1}{2} v$

$b$     $0$     $b$     $\frac{1}{2} c$     $\frac{1}{2} u$     $\frac{1}{2} v$

$c$     $\frac{1}{2} c$     $\frac{1}{2} c$     $0$     $\frac{1}{2} (a+b)$     $0$

$u$     $\frac{1}{2} u$     $\frac{1}{2} u$     $\frac{1}{2} (a+b)$     $0$     $0$

$v$     $\frac{1}{2} v$     $\frac{1}{2} v$     $0$     $0$     $0$

$J_{7}$ is a bicrossed product between the $3$-dimensional Jordan algebra $A\, =\, <a,\, b,\,c ~|~ a^{2} = a,\, b^{2} = b,\, c^{2} = 0,\, ab=0,\, ac = \frac{1}{2}c,\, bc= \frac{1}{2}c>$ and the abelian $2$-dimensional Jordan algebra $V$ generated by $u$ and $v$. The actions $(\triangleleft, \, \triangleright)$ of the associated matched pair are defined as follows:

\begin{eqnarray*}
&& u \triangleleft a = u \triangleleft b = \frac{1}{2} u,\,\,\, v \triangleleft a = v \triangleleft b =\frac{1}{2} v,\,\,\, u \triangleright c = \frac{1}{2}(a+b)
\end{eqnarray*}

$J_{17}$ is the $5$-dimensional real Jordan algebra with the following multiplication table:

$J_{17}$     $a$     $b$     $c$     $u$     $v$

$a$     $a$     $b$     $\frac{1}{2} c$     $0$     $\frac{1}{2} v$

$b$     $b$     $0$     $0$     $0$     $0$

$c$     $\frac{1}{2} c$     $0$     $0$     $\frac{1}{2} c$     $0$

$u$     $0$     $0$     $\frac{1}{2} c$     $u$     $\frac{1}{2} v$

$v$     $\frac{1}{2} v$     $0$     $0$     $\frac{1}{2} v$     $0$

$J_{17}$ is a bicrossed product between the Jordan algebras $A\, =\, <a,\, b,\,c ~|~ a^{2} = a,\, b^{2} = 0,\, c^{2} = 0,\, ab=b,\, ac = \frac{1}{2} c,\, bc= 0>$ and $V = <u,\, v ~|~ u^{2} = u,\, v^{2} = 0,\, uv = \frac{1}{2} v>$. The actions $(\triangleleft, \, \triangleright)$ of the associated matched pair are defined as follows:

u \triangleright c =  \frac{1}{2} c,\,\,\,\, v \triangleleft a = \frac{1}{2} v.

The bicrossed product of two Jordan algebras is the construction
responsible for the factorization problem for Jordan
algebras as formulated in the introduction. Indeed, we can prove the Jordan algebra version of

A Jordan algebra $E$ factorizes through two given Jordan algebras
$A$ and $V$ if and only if there exists a matched pair of Jordan
algebras $(A, V, \triangleleft, \, \triangleright)$ such that $E
\cong A \bowtie V $.

First observe that $A \cong A\times \{0\}$ and $V \cong
\{0\}\times V$ are Jordan subalgebras of $A \bowtie V $ and obviously $A \bowtie V $ factorizes through $A\times \{0\}$ and
$\{0\}\times V$.

Conversely, assume that a Jordan algebra $E$ factorizes through two Jordan subalgebras $A$ and $V$. Consider $\pi_{A} \colon E \to A$ be the canonical projection, i.e. $\pi_{A}(a) = a$ for all $a \in A$ and ${\rm ker}\, \pi_{A} = V$. We can now define the following:

\begin{eqnarray}
\triangleright \colon V \times A \to
A, \qquad x \triangleright a &:=& \pi_{A} (x a ) = \pi_{A}(a x ) \eqlabel{bala1}\\
\triangleleft \colon V \times A \to V,
\qquad x \triangleleft a &:=& x a - \pi_{A} (x a) \eqlabel{bala2}
\end{eqnarray}

We will show that $\bigl(A,\, V,\, \triangleleft, \,
\triangleright \bigl)$ is a matched pair of Jordan algebras and $ \varphi: A \bowtie V \to
E$, $\varphi(a, x) := a + x$ is an isomorphism of Jordan algebras. To this end note that $\varphi$ is a linear isomorphism between $E$ and the direct product of vector spaces $A \times V$ with the inverse given by $\varphi^{-1}(y) := \bigl(\pi_{A}(y), \, y -
\pi_{A}(y)\bigl)$, for all $y \in E$. Therefore, there exists a unique Jordan
algebra structure on $A \times V$ such that $\varphi$ becomes an
isomorphism of Jordan algebras and this unique multiplication on
$A \times V$ is given for any $a$, $b \in A$ and $x$, $y\in V$ by:

(a, x) \circ (b, y) := \varphi^{-1} \bigl( \varphi(a, x) \,
\varphi(b, y) \bigl)

The following straightforward computation shows that the above multiplication indeed coincides with the one associated to the matched pair $\bigl(A,\, V,\, \triangleleft, \,
\triangleright \bigl)$ as defined in bracbicross:

\begin{eqnarray*}
(a, x) \circ (b, y) &=& \varphi^{-1} \bigl(\varphi(a, x) \,  \varphi(b, y)\bigl) = \varphi^{-1} \bigl( (a+x) (b+y) \bigl) \\
&=& \varphi^{-1} (a b + a y + x b + x
y) \\
&=& \bigl( \pi_{A}(a b) + \pi_{A}(a y ) + \pi_{A} (x b) +
\pi_{A}(x y), \, \\
&& a b + a y + x b + x y - \pi_{A}(a b) - \pi_{A}(a y )- \pi_{A} (x b) -
\pi_{A}(x y)  \bigl)\\
&=& \Bigl(a b + x \triangleright b + y \triangleright a, \,\,  x \triangleleft b + y \triangleleft a + x\, y \Bigl)
\end{eqnarray*}

Based on bicromlver we can restate the factorization
problem as follows:

Let $A$ and $V$ be two given Jordan algebras. Describe the
set of all matched pairs $(A, V, \, \triangleleft, \,
\triangleright)$ and classify up to an isomorphism all bicrossed
products $A \bowtie V$.

In this way, the factorization problem can be divided into two
parts: the first one is a computational one which requires the
explicit description of all matched pairs $(\triangleleft, \,
\triangleright)$ between Jordan algebras $A$ and $V$ while the
second one deals with the classification of all bicrossed products
$A \bowtie V$ associated to the matched pairs previously
described. There is no general strategy for approaching the first
part of the problem which needs to be treated 'case by case' for
all specific examples of Jordan algebras (in the spirit of
[2, 3]). The tool for approaching the second part of the
problem is the next result which can be also used to describe the
automorphisms group ${\rm Aut}_J ({A \bowtie V})$ of a given
bicrossed product. In light of mpbpex, the classification of bicrossed products is an important step in the classification of all Jordan algebras of a given dimension.

Let $(A, V, \triangleleft, \triangleright)$ and $(A', V',
\triangleleft', \triangleright')$ be two matched pairs of Jordan
algebras. Then there exists a bijective correspondence between the
set of all morphisms of Jordan algebras $\psi : A \bowtie V \to A'
\bowtie ' V' $ and the set of all quadruples $(r, s, t, q)$, where
$r: A \to A'$, $s: A \to V'$, $t: V \rightarrow A'$, $q: V
\rightarrow V'$ are linear maps satisfying the following
compatibility conditions for any $a$, $b\in A$ and $x$, $y\in V$:

\begin{eqnarray}
r(a\, b) - r(a) \, r(b) &{=}&  s(a) \triangleright' r(b) + s(b) \triangleright' r(a) \eqlabel{C1}\\
s(a \ b) - s(a) \, s(b) &{=}& s(a) \triangleleft' r(b) + s(b) \triangleleft' r(a)  \eqlabel{C2}\\
t(x \, y) - t(x) \, t(y)  &{=}&  q(x) \triangleright' t(y) + q(y) \triangleright' t(x)   \eqlabel{C3}\\
q(x\, y) - q(x) \, q(y)  &{=}&  q(x)\triangleleft' t(y) + q(y) \triangleleft' t(x)  \eqlabel{C4}\\
r(x \triangleright a) + t (x \triangleleft a) &{=}& r(a) \, t(x) + s(a)  \triangleright' t(x) + q(x)  \triangleright' r(a) \eqlabel{C5}\\
s(x \triangleright a) + q (x \triangleleft a)  &{=}& s(a) \, q(x) + s(a)  \triangleleft' t(x) + q(x)
\triangleleft' r(a) \eqlabel{C6}
\end{eqnarray}

Under the above bijection the Jordan algebra map $ \psi =
\psi_{(r, s, t, q)} : A \bowtie V \to A' \bowtie ' V' $
corresponding to $(r, s, t, q)$ is given by:

\begin{equation}\eqlabel{morfbicros}
\psi \bigl((a, \, x)\bigl) = \bigl(r(a) + t(x) , \, s(a) + q(x)
\end{equation}

for all $a \in A$ and $x\in V$.

By the universal property of the direct product of vector spaces
we obtain that for any linear map $\psi : A \bowtie V \to A'
\bowtie ' V'$ there exists a unique quadruple $(r, s, t, q)$ of
linear maps $r: A \to A'$, $s: A \to V'$, $t: V \rightarrow A'$,
$q: V \rightarrow V'$ such that $\psi \bigl((a, \, x)\bigl) =
\bigl(r(a) + t(x) , \, s(a) + q(x) )\bigl)$, for all $a \in A$ and
$x\in V$. It remains to investigate when such a map $\psi =
\psi_{(r, s, t, q)}$ is a morphism of Jordan algebras between the
two bicrossed products, i.e. the following holds for all $(a, x),
(b, y) \in A\times V$:

\begin{equation}\eqlabel{morftest}
\psi \bigl( (a, \, x) \circ_{A \bowtie V} \, (b, \, y)  \bigl) =
\psi (a, x) \circ_{A' \bowtie ' V'} \, \psi (b, y)
\end{equation}

This is again a rather cumbersome but straightforward computation
which will be skipped. We only indicate the main strategy for the
proof: first, we can prove that morftest holds for the
pairs $(a, 0)$ and $(b, 0)$ if and only if C1 and
C2 are fulfilled. Secondly, it can be shown that
morftest holds for the pairs $(0, x)$ and $(0, y)$ if and
only if C3 and C4 hold. Finally,
morftest holds for the pairs $(a, 0)$ and $(0, x)$ if and
only if C5 and C6 hold and this finishes the

Let $A_{0}$ be an abelian Jordan algebra and consider $k_0$ to be the abelian Jordan algebra of dimension $1$. Then
there exists a bijection between the set of all matched pairs of Jordan algebras $(A_0, k_0, \triangleleft, \triangleright)$ and the set of
all linear endomorphisms $D \in {\rm End}_k (A)$ such that $D^3 = 0$ and the matched pair $(\triangleleft, \triangleright)$
associated to $D$ is given as follows for all $\alpha \in k$ and $a\in A$:

\begin{equation} \eqlabel{mpab1}
\alpha \triangleleft a := 0, \qquad \alpha \triangleright a := \alpha D(a)
\end{equation}

In particular, any bicrossed product $A_0 \bowtie k_0$ of Jordan algebras is a
left semidirect product $A_0 \ltimes k_0$, which we denote by $A(D) := A \times k$ and whose multiplication is given
for all $a$, $b\in A$ and $\alpha$, $\beta \in k$ by:

(a, \, \alpha) \circ (b, \, \beta) = (\alpha \, D(b) + \beta \, D(a), \, 0)

Indeed, since $V = k$, there exists a bijection between the set of all bilinear maps $(\triangleleft, \triangleright)$,
$\triangleleft : k \times A \to k$ and $\triangleright : k\times A \to A$ and the set of all linear maps
$(\lambda, \, D) \in A^* \times {\rm End}_k (A)$ given such that

\alpha \triangleleft a := \alpha \lambda (a), \qquad \alpha \triangleright a := \alpha D(a)

for all $\alpha \in k$, $a\in A$. Now since both $A$ and $k_0$ are abelian Jordan algebras, a laborious but straightforward computation shows that the axioms (MP1)$-$(MP6) from mpmL are reduced to the following two conditions: $\lambda = 0$ and $D^3 (a) = 0$, for all $a\in A$.

§ APPLICATION: CLASSIFYING COMPLEMENTS FOR JORDAN ALGEBRAS

Let $A \subseteq E$ be a Jordan subalgebra of $E$. A Jordan
subalgebra $B$ of $E$ is called a complement of $A$ in $E$
(or an $A$-complement of $E$) if $E$ factorizes through $A$
and $B$, i.e. $E = A + B$ and $A \cap B = \{0\}$.
bicromlver implies that the Jordan algebra $B \cong
\{0\}\times B$ is a complement of $A \cong A\times \{0\}$ in the
bicrossed product $A \bowtie B$. Moreover, if $B$ is an
$A$-complement of $E$, then there exists a matched pair of Jordan
algebras $(A, B, \triangleright, \triangleleft)$ such that the
corresponding bicrossed product $A \bowtie B$ is isomorphic as a
Jordan algebra with $E$. The actions of the matched pair $(A, B,
\triangleright, \triangleleft)$ arise from the unique

\begin{equation}\eqlabel{Lie456}
(0,\,b) \circ (a,\,0) = (b \triangleright a,\, b \triangleleft a)
\end{equation}

for all $a \in A$, $b \in B$. The matched pair constructed in
Lie456 will be called the canonical matched pair
associated to the $A$-complement $B$ of $E$.

For a Jordan subalgebra $A$ of $E$ we denote by ${\mathcal F} (A,
\, E)$ the isomorphism classes of $A$-complements of $E$. The
factorization index of $A$ in $E$ is defined as $[E : A]^f
:= |\, {\mathcal F} (A, \, E) \,|$.

Let $(A, B, \triangleright, \triangleleft)$ be a matched pair of
Jordan algebras. A $k$-linear map $r: B \to A$ is called a
deformation map of the matched pair $(A, B, \triangleright,
\triangleleft)$ if the following compatibility holds for all $x$,
$y \in B$:

\begin{equation}\eqlabel{factLie}
r(x y) \, - \, r(x) r(y) = x \triangleright
r(y) + y \triangleright r(x) - r \bigl( x \triangleleft r(y) + y
\triangleleft r(x) \bigl)
\end{equation}

We denote by ${\mathcal D}{\mathcal M} \, (B, A\, | \,
(\triangleright, \triangleleft) )$ the set of all deformation maps
of the matched pair $(A, B, \triangleright, \triangleleft)$.

Consider $J$ to be the $4$-dimensional real Jordan algebra with multiplication table defined as follows:

$J$     $a$     $b$     $u$     $v$

$a$     $a$     $0$     $0$     $\frac{1}{2}v$

$b$     $0$     $b$     $0$     $\frac{1}{2}v$

$u$     $0$     $0$     $u$     $0$

$v$     $\frac{1}{2}v$     $\frac{1}{2}v$     $0$     $0$

$J$ is a semidirect product between the $2$-dimensional Jordan algebras $A \,=\, <a,\, b ~|~ a^{2} = a,\, b^{2} = b, \, ab =0>$ and $V = <u,\, v ~|~ u^{2} = u,\, v^{2} = 0,\, uv = 0>$. Furthermore, the associated matched pair $(\triangleleft, \, \triangleright)$ consists of the trivial left action $\triangleright \colon V \times A \to A$ and the right action $\triangleleft \colon V \times A \to V$ defined as follows:

v \triangleleft a = v \triangleleft b = \frac{1}{2} v.

It can be easily seen by a straightforward computation that the deformation maps $r \colon V \to A$ associated to this matched pair are given as follows for some $\alpha \in \RR$:

\begin{eqnarray}
&& r(u) = 0,\,\,\, r(v) = \alpha\, b \eqlabel{def1}\\
&& r(u) = 0,\,\,\, r(v) = \alpha\, a \eqlabel{def2}\\
&& r(u) = a+b,\,\,\, r(v) = \alpha\, b \eqlabel{def3}\\
&& r(u) = a+b,\,\,\, r(v) = \alpha\, a \eqlabel{def4}\\
&& r(u) = a,\,\,\, r(v) = 0 \eqlabel{def5}\\
&& r(u) = b,\,\,\, r(v) = 0 \eqlabel{def6}
\end{eqnarray}

Using this concept the following deformation of a Jordan algebra is

Let $A$ be a Jordan subalgebra of $E$, $B$ a given $A$-complement
of $E$ and $r\colon B \to A$ a deformation map of the associated
canonical matched pair $(A, B, \triangleright, \triangleleft)$.

$(1)$ Let $f_{r}: B \to E = A \bowtie B$ be the $k$-linear map
defined for all $x \in B$ by:

$$f_{r}(x) = \bigl(r(x),\, x\bigl)$$

Then $\widetilde{B} : = {\rm Im}(f_{r})$ is an $A$-complement of

$(2)$ $B_{r} := B$, as a $k$-vector space, with the new multiplication defined for all $x$, $y \in B$ by:

\begin{equation}\eqlabel{rLiedef}
x\, \cdot_{r} \, y := x y + x \triangleleft r(y) + y
\triangleleft r(x)
\end{equation}

is a Jordan algebra called the $r$-deformation of $B$.
Furthermore, $B_{r} \cong \widetilde{B}$, as Jordan algebras.

$(1)$ To start with, we will prove that $\widetilde{B} =
\{\bigl(r(x),\, x \bigl) ~|~ x \in B\}$ is a Jordan subalgebra of
$E = A \bowtie B$. Indeed, for all $x$, $y \in B$ we have:

\begin{eqnarray*}
\bigl(r(x),\, x\bigl) \circ \bigl(r(y),\, y\bigl)
&\stackrel{\equref{bracbicross}}{=}& \bigl(\underline{r(x) r(y) + x \triangleright r(y) + y \triangleright
r(x)},\,\,x \triangleleft r(y) + y \triangleleft
r(x) + x y\bigl)\\
&\stackrel{\equref{factLie}}{=}& \Bigl( r\bigl(x \triangleleft
r(y) + y \triangleleft r(x) + x y\bigl),\, x
\triangleleft r(y) + y \triangleleft r(x) + x y\Bigl)
\end{eqnarray*}

and the latter term obviously belongs to $\widetilde{B}$. Moreover, it is straightforward to see that $A \cap \widetilde{B}
= \{0\}$ and $(a,\, b) = \bigl(a - r(b),\,0\bigl)+\bigl(r(b),\,b
\bigl) \in A + \widetilde{B}$. Therefore, $\widetilde{B}$ is a
$A$-complement of $E$.

$(2)$ We denote by $\widetilde{f_{r}}$ the $k$-linear isomorphism
from $B$ to $\widetilde{B}$ induced by $f_{r}$. We will prove that
$\widetilde{f_{r}}$ is also a Jordan algebra map if we consider on
$B=B_r$ the multiplication given by rLiedef. Indeed, for any
$x$, $y \in B$ we have:

\begin{eqnarray*}
\widetilde{f_{r}}\bigl(x\, \cdot_{r} \,
y\bigl)&\stackrel{\equref{rLiedef}}{=}& \widetilde{f_{r}}\bigl(x y + x \triangleleft r(y) + y \triangleleft
&{=}& \bigl(\underline{r(x y + x \triangleleft r(y) + y
\triangleleft r(x))},\, x y + x \triangleleft r(y) + y
\triangleleft
&\stackrel{\equref{factLie}}{=}& \bigl(r(x) r(y) + x
\triangleright r(y) + y \triangleright r(x),\, x y + x
\triangleleft r(y) + y \triangleleft
&\stackrel{\equref{bracbicross}}{=}& \bigl(r(x),\,x\bigl) \circ
\bigl( r(y),\, y\bigl) = \widetilde{f_{r}}(x)\, \circ
\widetilde{f_{r}}(y)
\end{eqnarray*}

This shows that $B_{r}$ is a Jordan algebra and the proof is
now finished.

We are now able to describe all complements of a Jordan subalgebra
$A$ of $E$.

Let $A$ be a Jordan subalgebra of $E$, $B$ a given $A$-complement
of $E$ with the associated canonical matched pair of Jordan
algebras $(A, B, \triangleright, \triangleleft)$. Then
$\overline{B}$ is an $A$-complement of $E$ if and only if there
exists an isomorphism of Jordan algebras $\overline{B} \cong
B_{r}$, for some deformation map $r \colon B \to A$ of the matched
pair $(A, B, \triangleright, \triangleleft)$.

Let $\overline{B}$ be an arbitrary $A$-complement of $E$. Since $E
= A \bowtie B = A \bowtie \overline{B}$ we can find four
$k$-linear maps:

u: B \to A, \quad v: B \to \overline{B}, \quad t:\overline{B} \to
A, \quad w: \overline{B} \to B

such that for all $x \in B$ and $y \in \overline{B}$ we have:

\begin{equation} \eqlabel{lie111}
(0,\, x) = \bigl(u(x),\, v(x)\bigl), \qquad (0,\,y) =
\bigl(t(y),\, w(y)\bigl)
\end{equation}

It can be easily proved that $v$ and $w$ are inverses to each
other and, in particular, that $v: B \to \overline{B}$ is a
$k$-linear isomorphism. We denote by $\tilde{v}: B \to A \bowtie
B$ the composition:

\tilde{v} : \, B \, \stackrel{v} {\longrightarrow} \, \overline{B}
\, \stackrel{i}{\hookrightarrow} \, E \, = \,A \bowtie B

More precisely, we have $(0,\,\tilde{v}(x)) = (0,\,v(x))
\stackrel{\equref{lie111}}{=} (-u(x),\, x)$, for all $x \in B$. Consider now $r := -u$; we will prove that $r$ is a deformation map
and $\overline{B} \cong B_{r}$. Indeed, $\overline{B} = {\rm Im}
(v) = {\rm Im} (\tilde{v})$ is a Jordan subalgebra of $E = A
\bowtie B$ and therefore we have:

\begin{eqnarray*}
\bigl(r(x),\, x\bigl) \circ \bigl(r(y),\, y\bigl)
&\stackrel{\equref{bracbicross}}{=}& \bigl(r(x) r(y) + x
\triangleright r(y) + y \triangleright r(x),\,\,x \triangleleft
r(y) + y \triangleleft r(x) + x y\bigl)\\
&=& (r(z),\, z)
\end{eqnarray*}

for some $z \in B$. Thus, we obtain:

\begin{equation}\eqlabel{lie113}
r(z) = r(x)\, r(y) + x \triangleright r(y) + y
\triangleright r(x), \qquad z = x \triangleleft r(y) + y
\triangleleft r(x) + x y
\end{equation}

Now by applying $r$ to the second part of lie113 we

\begin{eqnarray*}
r(z) = r\bigl(x \triangleleft r(y) + y \triangleleft r(x) + x y\bigl)
\end{eqnarray*}

which combined with the first part of lie113 shows that
$r$ is a deformation map of the matched pair $(A, B,
\triangleright, \triangleleft)$. Moreover, for all $x$, $y \in B$
we have:

\begin{eqnarray*}
&& \bigl(0,\, v(x) v(y)\bigl) = \bigl(0,\, v(x)\bigl)
\circ \bigl(0,\,v(y)\bigl) = \bigl(r(x),\, x\bigl)\circ
\bigl(r(y),\,
y\bigl)= \bigl(r(z),\, z\bigl)\\
&=& \bigl(0,\, v(z)\bigl) = \bigl(0,\, v(x \triangleleft r(y) + y
\triangleleft r(x) + x y)\bigl) = \bigl(0,\, v(x
\cdot_{r} y)\bigl)
\end{eqnarray*}

that is, $v: B_{r} \to \overline{B}$ is a Jordan algebra map and
the proof is now finished.

In order to classify all complements we need to introduce the

Let $(A, B, \triangleright, \triangleleft)$ be a matched pair of
Jordan algebras. Two deformation maps $r$, $s: B \to A$ are called
equivalent and we denote this by $r \sim s$ if there exists
$\sigma: B \to B$ a $k$-linear automorphism of $B$ such that for
all $x$, $y\in B$ we have:

\begin{equation}\eqlabel{equivLiemaps}
\sigma (x y) - \sigma(x)\, \sigma(y) = \sigma(x)
\triangleleft s \bigl(\sigma(y)\bigl) - \sigma\bigl(x
\triangleleft r(y)\bigl) + \sigma(y) \triangleleft s
\bigl(\sigma(x)\bigl) - \sigma \bigl(y \triangleleft r(x)\bigl)
\end{equation}

We conclude this section with the following classification result
for complements of Jordan algebras:

Let $A$ be a Jordan subalgebra of $E$, $B$ an $A$-complement of $E$
and $(A, B, \triangleright, \triangleleft)$ the associated
canonical matched pair. Then:

$(1)$ $\sim$ is an equivalence relation on ${\mathcal D}{\mathcal
M} \, (B, A \, | \, (\triangleright, \triangleleft) )$. We denote
by  ${\mathcal H}{\mathcal A}^{2} (B, A \, | \, (\triangleright,
\triangleleft) )$ the quotient set ${\mathcal D}{\mathcal M} \,
(B, A \, | \, (\triangleright, \triangleleft) )/ \sim$.

$(2)$ There exists a bijection between the isomorphism classes of
all $A$-complements of $E$ and ${\mathcal H}{\mathcal A}^{2} (B, A
\, | \, (\triangleright, \triangleleft) )$. In particular, the
factorization index of $A$ in $E$ is computed by the formula:

[E : A]^f = | {\mathcal H}{\mathcal A}^{2} (B, A \, | \,
(\triangleright, \triangleleft) )|

where $|X|$ denotes the cardinal of the set $X$.

In light of descrierecomlie, in order to classify all
$A$-complements of $E$ it suffices to consider only
$r$-deformations of $B$, for various deformation maps $r : B \to
A$. Now let $r$, $s : B \to A$ be two deformation maps. Since $B_r =
B_s : = B$ as $k$-spaces, it follows that the Jordan algebras $B_r$
and $B_s$ are isomorphic if and only if there exists a $k$-linear
isomorphism $\sigma : B_{r} \to B_{s}$ which is also a Jordan
algebra map. It can be easily seen using rLiedef that
$\sigma$ is a Jordan algebra map if and only if the compatibility
condition equivLiemaps holds, i.e. $r \sim s$. Hence, $r
\sim s$ if and only if $\sigma: B_r \to B_s$ is an isomorphism of
Jordan algebras. To conclude, $\sim$ is an equivalence relation on
${\mathcal D}{\mathcal M} (B, A \, | \, (\triangleright,
\triangleleft) )$ and the map

{\mathcal H}{\mathcal A}^{2} (B,\,A \, | \, (\triangleright,
\triangleleft) ) \to {\mathcal F} (A, E),  \qquad \overline{r}
\mapsto B_{r}

is a bijection between sets, where $\overline{r}$ is the
equivalence class of $r$ via the relation $\sim$.

We end this section with an example which computes the factorization index of a certain Jordan algebra extension.

Consider again $(A,\,V,\, \triangleright, \triangleleft)$ to be the matched pair depicted in defmap. We will prove that the factorization index $[J : A]^f = 4$, where $J = A \bowtie V$. We start by describing the Jordan algebras $V_{r}$ corresponding to the deformation maps listed in defmap. First note that the Jordan algebras $V_{r}$ corresponding to the $r$-deformations given in def1 and def2 are equal and defined by $V_{r} \,=\, \{u,\,v ~|~ u^{2} = u, \, v^{2} = \alpha v,\, uv=0\}$ for some $\alpha \in \RR$. If $\alpha = 0$ we obtain the Jordan algebra $V$ while if $\alpha \neq 0$ we have an isomorphism between $V_{r}$ and the Jordan algebra $V_{1} = \{u,\, v ~|~ u^{2} = u,\, v^{2} = 0,\, uv=0\}$. Furthermore, it can be easily seen that $V$ is not isomorphic to $V_{1}$. Next, the Jordan algebras $V_{r}$ corresponding to the $r$-deformations given in def3 and def4 are equal and defined by $V_{r} \,=\, \{u,\,v ~|~ u^{2} = u, \, v^{2} = \beta v,\, uv=v\}$ for some $\beta \in \RR$. If $\beta = 0$ we obtain the Jordan algebra $V_{2} = \{u,\, v ~|~ u^{2} = u,\, v^{2} = 0,\, uv=v\}$ while if $\beta \neq 0$ we have an isomorphism between $V_{r}$ and the Jordan algebra $V_{2}' = \{u,\, v ~|~ u^{2} = u,\, v^{2} = v,\, uv=v\}$. It can be easily seen that $V_{2}'$ is isomorphic to $V_{1}$ and that $V_{2}$ is not isomorphic to $V$ nor to $V_{1}$. Similarly,  the Jordan algebras $V_{r}$ corresponding to the $r$-deformations given in def5 and def6 are equal and defined by $V_{3} = \{u,\, v ~|~ u^{2} = u,\, v^{2} = 0,\, uv= \frac{1}{2}v\}$. As $V_{3}$ is not isomorphic to any of the Jordan algebras $V$, $V_{1}$ or $V_{2}$ we can conclude that $[J : A]^f = 4$, as desired.

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11institutetext: Kyoto University, Kyoto-shi, Kyoto 606-8501, Japan
11email<EMAIL_ADDRESS><EMAIL_ADDRESS>

# Self-Agreement: A Framework for Fine-tuning Language Models to Find
Agreement among Diverse Opinions

Shiyao Ding 11 Takayuki Ito 11

###### Abstract

Finding an agreement among diverse opinions is a challenging topic in
multiagent systems. Recently, large language models (LLMs) have shown great
potential in addressing this challenge due to their remarkable capabilities in
comprehending human opinions and generating human-like text. However, they
typically rely on extensive human-annotated data. In this paper, we propose
Self-Agreement, a novel framework for fine-tuning LLMs to autonomously find
agreement using data generated by LLM itself. Specifically, our approach
employs the generative pre-trained transformer-3 (GPT-3) to generate multiple
opinions for each question in a question dataset and create several agreement
candidates among these opinions. Then, a bidirectional encoder representations
from transformers (BERT)-based model evaluates the agreement score of each
agreement candidate and selects the one with the highest agreement score. This
process yields a dataset of question-opinion-agreements, which we use to fine-
tune a pre-trained LLM for discovering agreements among diverse opinions.
Remarkably, a pre-trained LLM fine-tuned by our Self-Agreement framework
achieves comparable performance to GPT-3 with only 1/25 of its parameters,
showcasing its ability to identify agreement among various opinions without
the need for human-annotated data.

###### Keywords:

Multiagent sytems, Consensus building Opinion summarization Large language
models.

##  1 Introduction

Consensus refers to a significant portion of a group agreeing on a specific
topic, which is essential for collaboration and a crucial element of
democracy. Consequently, consensus building as an important topic of
multiagent systems has been extensively studied, such as finding an agreement
through discussions among diverse opinions [4], which is often challenging
[1]. The recent rise and success of large language models (LLMs), such as the
generative pre-trained transformer-3 (GPT-3) [2], offer promising
opportunities for addressing this challenge, by leveraging their capabilities
in comprehending human opinions and generating human-like text.

Currently, employing LLMs for finding agreement among diverse opinions can be
divided into two kinds of approaches. The first approach involves zero-shot
learning on a pretrained LLM, such as GPT-3/4, which can output agreement
without additional training. However, accessing these models requires the use
of an application programming interface (API), which can introduce latency,
high costs, and limited accessibility for research. Moreover, while some
pretrained LLMs like Meta’s LLaMA [13] are available for local download, their
massive parameters often lead to substantial storage and computational
requirements during inference, making them less feasible for certain
applications.

The second approach focuses on few-shot learning, which entails fine-tuning a
LLM using a dataset containing opinions and agreement data. One notable
example is [1], which relies on a 70 billion parameter model and expert-
annotated data. However, this dependence on high-quality, human-generated data
and the associated costs can be a significant barrier for many researchers and
organizations seeking to leverage these models for consensus-building tasks.
These limitations highlight the need for more accessible and cost-effective
solutions in applying LLMs to find agreement among diverse opinions without
heavily relying on human resources.

In this paper, we address the above challenges by proposing Self-Agreement, a
framework designed to autonomously find agreement among diverse opinions using
LLMs. Our method consists of the following four key steps. First, we employ
GPT-3 to generate multiple opinions on each question given a dataset
containing various questions. Next, we use GPT-3 to generate several potential
agreements among these diverse opinions for each question. We then evaluate
the agreement score of each agreement candidate, choosing the one with the
highest score as the optimal agreement to construct a question-opinion-
agreement dataset. Finally, we fine-tune a pre-trained LLM using the above
dataset. This step allows the model to adapt to the specific task of finding
agreements among various opinions. In our evaluation, we fine-tuned a 7
billion pre-trained LLM using our Self-Agreement framework. Remarkably, our
model achieves a similar performance to GPT-3 with only 1/25 of its
parameters, showcasing its ability to identify areas of agreement among
conflicting views.

In summary, our contributions include: 1) a large dataset consisting of
various questions, opinions, and agreement candidates, which can serve as a
valuable resource for building and evaluating consensus-building models; 2)
the Self-Agreement framework which is designed to autonomously find agreement
among diverse opinions using LLMs without human-annotated data; 3) a
demonstration of fine-tuning a pre-trained model, with a comprehensive set of
experiments confirming its effectiveness and performance.

##  2 Related Work

The concept of agreements can vary depending on the task. In this paper, we
consider agreement as a form of opinion summarization from users, which may be
agreed upon by all the users, similar to the approaches used in [1] [12].

Opinion Summarization Most opinion summarization methods follow a three-step
process [5][6]: First, aspect extraction is performed, which involves
identifying the relevant features or aspects of the product that the user is
commenting on. Second, sentiment prediction is used to determine the sentiment
of the extracted aspects, whether it is positive, negative, or neutral. This
step helps to understand the overall opinion of the user. Finally, summary
generation is used to present the identified opinions to the user in a concise
and easily understandable manner. This step involves condensing the extracted
aspects and their corresponding sentiments into a brief summary. Most methods
rely on extractive techniques for creating textual summaries, which select
representative segments from the source text. However, this can result in loss
of information that may be useful depending on user needs.

Most current approaches for opinion summarization, as described in the
reference [11], involve encoding documents and then decoding the learned
representations into an abstractive summary. These methods leverage the
success of sequence-to-sequence neural network architectures and are trained
using sets of opinions and their corresponding summaries. In this approach,
there is no need to explicitly identify aspects and sentiment for the opinion
summarization task, as these are learned implicitly from the training data.
However, due to memory limitations, training these models end-to-end with a
large number of input reviews for each target entity is practically infeasible
[10].

LLMs for Opinion Summarization With the rapid development of LLMs, especially
transformer-based LLMs, they have shown great potential for tackling tasks
such as opinion summarization. The transformer model, proposed in 2017 [14],
has become a classic model for natural language processing (NLP) tasks, and
various Transformer-based large language models have been proposed.

Bidirectional encoder representations from transformers (BERT) was one of the
first models that pretrained on large corpora and fine-tuned for specific
tasks, enabling it to capture intricate language patterns [3]. Then, GPT have
also seen significant advancements, with GPT-1, GPT-2 [8], and GPT-3 [2],
focusing on large-scale unsupervised learning to generate human-like text.
Other notable LLM architectures include RoBERTa [7], which improves upon BERT
with enhanced pretraining and training techniques; Text-to-Text Transfer
Transformer (T5) [9], which adopts a unified text-to-text approach for various
NLP tasks; and Large Language Model Meta AI (LLaMA) [13].

Those LLMs have shown great potential in opinion summarization tasks, largely
due to their ability to process and generate natural language. Bakker et al.
[1] fine-tuned a 70 billion parameter pretrained LLM to produce statements
that maximize the expected approval for groups with diverse opinions. The
model demonstrated exceptional performance, generating consensus statements
preferred by over 70% of human users compared to prompted LLMs. However, their
approach relies on human annotations. Although the Self-Instruction framework
proposed by Wang et al. [15] utilizes pre-trained LLM outputs for fine-tuning
LLMs to follow human instructions, it still requires the preparation of seed
instructions. Our Self-Agreement framework, on the other hand, specifically
targets consensus-building tasks and does not depend on any human-generated
instructions. This distinction highlights the advantages of the Self-Agreement
framework in handling diverse opinions and generating agreement statements
without extensive reliance on human-annotated data.

##  3 Method

Annotating and generating large-scale agreement data usually require
significant human resources, as it often involves thousands of human-written
questions and agreement data. Although those data can be automatically output
in some cases, human input is still needed to score or rank the outputs,
allowing LLMs to fit human preferences. In this section, we present our Self-
Agreement framework, where the entire process, from generating to scoring
agreements, does not require human intervention. This makes it a completely
automatic process, as shown in Fig.1.

Figure 1: The framework of the Self-Agreement.

###  3.1 Automatic Opinion-Agreement Data Generation

We first define a consensus-building instance $CB^{i}$ for question $i$ as a
tuple of $CB^{i}=<q^{i},OP^{i},AC^{i}>$, where $q^{i}$ denotes question $i$,
$OP^{i}=\\{op^{i}_{1},...,op^{i}_{|O^{i}|}\\}$ is the opinion set, and
$AC^{i}=\\{ac^{i}_{1},...,ac^{i}_{|AC^{i}|}\\}$ is the set of agreement
candidates. As mentioned in the previous section, we consider the definition
of agreement as a form of opinion summarization from users, which may be
agreed upon by all users. Then, for a set $Q=\\{q^{1},q^{2},...,q^{|Q|}\\}$ of
questions, we aim to generate $|Q|$ consensus-building instances as a
question-opinion-agreement dataset.

First, for opinion generation for each question $q^{i}$, inspired by the Self-
Instruct approach, which utilizes large pretrained language models to create
new and diverse instructions in a bootstrapping fashion, our framework employs
the GPT-3 to generate several diverse opinions with the prompt “Generate $n$
opinions for the question of $q^{i}$”, as shown in Steps 1 and 2 of Fig. 1.
However, unlike Self-Instruction, which first relies on a set of seed human-
written opinions, Self-Agreement does not require any previously written human
opinions. Then, in Step 3 of agreement generation, we also employ GPT-3 to
create $m$ agreement candidates with the prompt “Find an agreement for the
following opinions $OP^{i}$,” where an instance is shown in Table 1.

###  3.2 Score Agreement Candidates

For each consensus-building instance, the goal is to choose the optimal
agreement candidate that matches all the opinions well. First, we can define a
scoring function $Mat:OP^{i}\times AC^{i}\rightarrow\mathcal{R}$ to evaluate
how well agreement $ac^{i}_{j}$ matches opinion $op^{i}_{k}$. In this paper,
we set the range of values of $Mat$ as $[0,1]$, where
$Mat(op^{i}_{k},ac^{i}_{j})=0$ represents the agreement $ac^{i}_{j}$ being
completely irrelevant to opinion $op^{i}_{k}$, and 1 means the agreement
$ac^{i}_{j}$ completely supports opinion $op^{i}_{k}$. such as,
$Mat(op^{i}_{k},ac^{i}_{j})|_{ac^{i}_{j}=op^{i}_{k}}=1$.

For each question $q^{i}$, we obtain $|OP^{i}|$ opinions and $|AC^{i}|$
agreement candidates. The model assigns a score to each candidate agreement,
reflecting its relevance and compatibility with the given opinions. By
calculating the agreement candidates according to their cumulative scores, we
can identify the optimal agreement candidate that represents the various
viewpoints. The best agreement candidate, $ac^{i}_{*}$, is the one with the
highest agreement score:

$ac^{i}_{*}=\text{argmax}_{ac^{i}_{j}\in AC^{i}}\sum_{op^{i}_{k}\in
OP^{i}}Mat(op^{i}_{k},ac^{i}_{j})$ (1)

Although various functions can be used as the agreement score function, in
this paper, we employ a BERT-based model to evaluate the similarity between
each opinion and agreement pair. This model computes contextualized embeddings
for each sentence, capturing their semantic representations. The similarity
score is then calculated by comparing these embeddings, typically using a
distance metric such as cosine similarity. Thus, we take $ac^{i}_{*}$ as the
output of the instance, as shown in Step 5 of Fig.1.

Table 1: A sample of self-agreement process with three opinions and two candidate agreements on the topic of trans fats. Topic |  What is Trans Fat? How to reduce that?
---|---
Opinion 1 |  Trans fats are a type of fat created when hydrogen is added to vegetable oil during the production process, and can increase the risk of heart disease and other health issues. The best way to reduce trans fat intake is to eat a balanced diet with plenty of fruits, vegetables, whole grains, and lean proteins.
Opinion 2 |  Trans fats are unhealthy and should be avoided whenever possible. To reduce trans fat intake, individuals should read labels carefully and opt for foods with no trans fat or low amounts of trans fat. Additionally, it is important to limit processed foods and choose items that are made with healthier fats, such as olive oil or avocado oil.
Opinion 3 |  Trans fats are an unhealthy type of fat and should be avoided. To reduce trans fat intake, individuals should limit their consumption of processed and fast foods and instead focus on eating fresh, whole foods. Additionally, they should check food labels to make sure they are not consuming hidden trans fats.
Agreement Candidate 1 |  All three opinions agree that trans fats are unhealthy and should be avoided. To reduce trans fat intake, individuals should limit their consumption of processed and fast foods, read labels carefully, and opt for foods with no trans fat or low amounts of trans fat. Additionally, it is recommended to focus on eating fresh, whole foods, and choosing items that are made with healthier fats, such as olive oil or avocado oil.
Agreement Candidate 2 |  All three opinions agree that trans fats are an unhealthy type of fat that should be avoided. To reduce trans fat intake, individuals should limit their consumption of processed and fast foods, check food labels, and opt for foods with no trans fat or low amounts of trans fat. Additionally, they should focus on eating fresh, whole foods with healthier fats, such as olive oil or avocado oil.

###  3.3 Fine-tuning the Language Model

After building the question-opinion-agreement dataset, we fine-tune a pre-
trained language model for the task of consensus building, as shown in Step 6
of Fig.1. In this paper, we employ a 7-billion-parameter LLaMA model based on
the Alpaca-LoRA architecture for fine-tuning. The Alpaca-LoRA architecture
comprises two components: 1) the original 7-billion-parameter model, and 2) an
adapter module. During the fine-tuning process, only the adapter module’s
parameters are updated, while the original model’s parameters remain
unchanged. Both components contribute to the inference process.

Each instance in the training dataset consists of an “Instruction,” an “Input”
and an “Output”, as shown in Fig.1. Focusing on the task of consensus
building, we set the instruction as “Find an agreement among the following
opinions.” This fine-tuning process equips the language model with the
capability to efficiently identify consensus among diverse opinions.

##  4 Evaluation

###  4.1 Evaluation Setting

Dataset In this paper, we use Yahoo! Answers topic classification dataset [16]
which includes 1,400,000 training samples and 60,000 testing samples. Each
sample includes following 5 parts: id, topic (class label) question title,
question content and best answer, where the table A.1 in Appendix shows some
examples. We choose 1000 question titles from training samples for generating
training dataset. We then use each question content to generate opinions by
using GPT-3. We consider both of the opinions have conflict and not, where the
corresponding prompts are as follows, prompt1: Generate three opinions for the
topic of $topic_{i}$ which do not have a conflict and prompt2: Generate three
opinions for the topic of $topic_{i}$ which have a conflict. Then we use GPT-3
to generate an agreement by inputing the prompt as Please generate an
agreement of the following opinions. independently for $m$ times. Thus, for
each question, we have $n$ opinions and $m$ agreement candidates. We set
$q=1000$, $m=3$ and $n=4$ in this paper, which corresponds to 1000 questions,
6000 opinions (with conflict:3000, without conflict:3000) and 8000 agreement
candidates.

Fine-tuning LLM For fine-tuning the LLM, we select a 7 billion model from
LLaMA [13] as a pre-trained model to be fine-tuned. We utilized the above
dataset containing opinions with and without conflicts. Additionally, we
divided it into two datasets based on the method of choosing an agreement
candidate. We considered two approaches: selecting an optimal agreement
candidate or randomly picking one for training. Consequently, this generated
four distinct cases, as shown in Table 2.

|  Random Agreement Candidate |  Optimal Agreement Candidate
---|---|---
Without conflict opinions |  Prompt: Generate three opinions for the topic which do not have a conflict Output: Randomly choosing one from all agreement candidates as Output |  Prompt: Generate three opinions for the topic which do not have a conflict Output: Optimally choosing one from all agreement candidates as Output
With conflict opinions |  Prompt: Generate three opinions for topic which have a conflict Output:Randomly choosing one from all agreement candidates as Output. |  Prompt: Generate three opinions for topic which have a conflict Output: Optimally choosing the one with maximized consensus score from all agreement candidates as Output
Table 2: Four cases to fine-tune LLMs.

Correspondingly, we employ GPT-3 as a baseline for comparison. For the test
set, we randomly select 100 questions from the same dataset, excluding those
used in the training dataset, to generate both opinions with conflicts and
without conflicts. Table A.2 provides some examples of these test cases.

Figure 2: The comparison of finding agreement results between GPT-3 and
Alpaca-LoRA-7B (Self-Instruct) in four cases. To calculate the average
agreement score, agreements are generated from the 100 test samples, and each
agreement score is calculated using the summation component in Eq. (1).

###  4.2 Evaluation Results

We compare our methods with GPT-3 and present the results for the four cases
depicted in Fig. 2. To calculate the average agreement score, agreements are
generated from the 100 test samples, and each agreement score is calculated
using the summation component in Eq. (1). First, when randomly selecting an
agreement candidate as the final opinion summarization, Alpaca-LoRA-7B (Self-
Instruct) achieves comparable results in both cases of opinions with conflicts
and without conflicts. Also, both Alpaca-LoRA-7B (Self-Instruct) and GPT-3
exhibit lower scores for opinions with conflict than for opinions without
conflict. This can be attributed to the fact that it is generally more
challenging for agreement candidates to accommodate conflicting opinions.

In the case where we select one optimal agreement candidate, the scores for
both models improve. For instance, in the case of Alpaca-LoRA-7B (Self-
Instruct), the score increases to 0.86 from 0.85 in the situation without
conflicting opinions and to 0.81 from 0.79 in the situation with conflicting
opinions. These improvements highlight the benefits of selecting the optimal
agreement candidate for opinion summarization tasks. Most notably, Alpaca-
LoRA-7B (Self-Instruct) demonstrates competitive performance in opinion
summarization when compared to GPT-3, which has 175 billion parameters. This
underscores the effectiveness of our Self-Agreement framework to fine-tune an
LM model in handling diverse opinions.

Table 3: One instance of comparing Agreement candidates of GPT-3 and Alpaca-LoRA-7B (Self-Instruct) from opinions with conflict. Topic |  Which is the country with top development indicators?
---|---
Opinion 1 |  India is the country with top development indicators because it has achieved great progress in the fields of education, health, and poverty alleviation.
Opinion 2 |  China is the country with top development indicators because it has made great strides in economic growth and infrastructure development.
Opinion 3 |  The United States is the country with top development indicators because it has the highest GDP per capita and the most advanced technology.
Agreement Candidate (GPT-3) |  We agree that India, China, and the United States are all countries with top development indicators. India has achieved great progress in the fields of education, health, and poverty alleviation; China has made great strides in economic growth and infrastructure development; and the United States has the highest GDP per capita and the most advanced technology. Together, these three countries demonstrate the highest level of development in the world.
Agreement Candidate (Self-Agreement) |  We agree that India, China, and the United States are all countries with top development indicators. India has achieved great progress in the fields of education, health, and poverty alleviation, China has made great strides in economic growth and infrastructure development, and the United States has the highest GDP per capita and the most advanced technology. However, we disagree on which country has the top development indicators, as each country has its own strengths and weaknesses.

Further, we list an instance of the above result. Table 3 shows the comparison
results of the agreement candidates generated by GPT-3 models and Alpaca-LoRA
(Self-Agreement), in response to a given topic: Which is the country with top
development indicators?. The opinions represent three different views on which
country has the top development indicators. In the GPT-3 generated agreement
candidate, the model introduces India, China, and The United States as
countries with high development indicators. Although these countries are well-
regarded for their development, the response deviates from the given opinions.
This indicates that GPT-3 may not effectively capture the essence of the
conflicting opinions in this scenario. However, the Alpaca-LoRA-7B (Self-
Instruct) generated agreement candidate acknowledges all three opinions and
successfully incorporates the key points from each. It emphasizes the
strengths of India, China, and the United States while also recognizing that
it is difficult to determine which country has the top development indicators,
as each has its own strengths and weaknesses (shown in the highlighted parts
of Table 3). This response demonstrates the effectiveness of the Self-
Agreement framework in accurately capturing and representing diverse opinions
in a conflict situation.

Both GPT-3 and the Self-Agreement framework demonstrate their ability to
synthesize diverse opinions into a coherent agreement candidate. However, the
Alpaca-LoRA-7B model fine-tuned by Self-Agreement framework provides a more
balanced perspective by acknowledging the underlying disagreement and avoiding
a definitive conclusion. This illustrates the effectiveness of the Self-
Agreement framework in handling conflicting opinions and generating a
consensus statement that fairly represents the diversity of viewpoints.

Further, catastrophic forgetting is a phenomenon in which a model trained on a
new task, significantly degrades its generalization performance on the
original task, resulting in a severe loss of previous knowledge. To test for
catastrophic forgetting, we use some instruction data from the Self-
Instruction framework and find that the Alpaca-LoRA-7B model, after fine-
tuning with the Self-Agreement framework, still demonstrates a good response
to general instructions. Some instances are listed in Table A.2.

##  5 Conclusion

In this paper, we introduced the Self-Agreement, an efficient framework to
fine-tune LLMs to autonomously find agreement among diverse opinions. Our
method eliminates the need for expensive human-generated data. We also
presented a large dataset of questions, opinions, and agreement candidates,
serving as a valuable resource for future consensus-building models. Our
experiments highlight the effectiveness of our framework in consensus-building
tasks while achieving comparable performance to GPT-3 with only 1/25 of its
parameters.

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## Appendix

The Fig.A.1 illustrates the topic distribution of questions from both train
and test samples. As can be observed from the pie charts, the questions are
distributed across various topics. Moreover, Table A.1 provides several
examples from the Yahoo dataset, showcasing the diversity of questions and
answers across different topics.

Figure A.1: The topic distribution of questions from train and test samples. Id |  Topic (Class label) |  Question Title |  Question Content |  Best Answer
---|---|---|---|---
0 |  4 (Computers Internet) |  ”why doesn’t an optical mouse work on a glass table?” |  ”or even on some surfaces?” |  ”Optical mice use an LED and a camera to rapidly capture images of the surface beneath the mouse. …
1 |  5 (Sports) |  ”What is the best off-road motorcycle trail ?” |  ”long-distance trail throughout CA” |  ”i hear that the mojave road is amazing!”
2 |  2 (Health) |  ”What is Trans Fat? How to reduce that?” |  ”I heard that tras fat is bad for the body. Why is that? Where can we find it in our daily food?” |  ”Trans fats occur in manufactured foods…
Table A.1: Some examples of Yahoo answer topic dataset

The following table demonstrates how well the model of Alpaca-LoRA-7B can
still perform general instructions after fine-tuning with the Self-Agreement
framework, indicating that catastrophic forgetting has been mitigated.

Table A.2: Sample general instruction responses from Alpaca-LoRA-7B after fine-tuning with the Self-Agreement framework. Instruction |  Input |  Output
---|---|---
Give three tips for staying healthy. |  |  1\. Eat a balanced diet of fruits, vegetables, and whole grains. 2\. Exercise regularly to keep your body in shape. 3\. Get enough sleep to recharge your body and mind.
What are the three primary colors? |  |  The three primary colors are red, blue, and yellow. These colors can be used to create any other color by mixing them together.
Describe the structure of an atom. |  |  An atom is made up of a nucleus surrounded by a cloud of electrons. The nucleus is made up of protons and neutrons, while the cloud of electrons is made up of negatively charged electrons.
Classify the following into animals, plants, and minerals |  Oak tree, copper ore, elephant. |  1.Oak tree is an example of an animal, as it is a living organism. 2.Copper ore is an example of a mineral, as it is a naturally occurring substance. 3.Elephant is an example of a plant, as it is a living organism that grows and reproduces.
Identify the odd one out. |  Twitter, Instagram, Telegram |  Telegram is the odd one out as it is a messaging app, while Twitter and Instagram are social media platforms.