数理解析研究所講究録
第 1873 巻 2014 年 106-113

106

Triple systems and structurable algebras *Noriaki Kamiya1 and Susumu Okubo2
1Department of Mathematics, University of Aizu
965-8580, Aizuwakamatsu, Japan, kamiya@u-aizu.ac.jp

2Department of Physics and Astronomy, University of Rochester
Rochester, New York, 14627, U.S.A., okubo@pas.rochester.edu

Abstract

Freudenthal-Kantor triple
We will show that we can always construct
Freudenthal-Kantor triple system. Moreover, we will
system from a given
prove that we can construct a structurable algebra from any $(-1,1)$ FreudenthalKantor triple system with a distinguished element satisfying $eex=x$ for the triple$a(-\epsilon, \delta)$$(\epsilon, \delta)$$e$

product.

AMS classification: $17C50;17A40;17B60$
Keywords:triple systems,Lie algebras, structurable algebra.

1. Introduction and summary of Main Results
Let be a vector space over a field $F$ with a triple product denoted by jaxtaposition,$xyz\in V$ for$y,$ $z\in V$ Then, the triple systems $(V, xyz)$ is called
FreudenthalFKTS), provided thatKantor triple system ([17]) which hereafter abreviated$V$$a(\epsilon, \delta)$$x,$$($$is$$as (\epsilon, \delta)$

$V$

(1.1)

$K(x, y)z=xzy-\delta yzx$ ,(1.2)$L(x, y)z=xyz$(1.3)$[L(u, v), L(x, y)]=L(uvx, y)+\epsilon L(x, vuy)$(1.4)$K(K(u, v)x, y)=L(y, x)K(u, v)-\epsilon K(u, v)L(x, y)$ .

(1.5)$K,$ $L;V\otimes Varrow$Enddefined bysatisfyandHere, and are constants with values 1 or $-1.$
ternary system $(V, xy, xyz)$ is called
space
structurable, ([2]), wherewith both binary product $xy$ and ternary
is a vector
product $xyz$ , satisfying the following conditions;$A$$\delta$$\epsilon$$V$

This paper is a survey note and the detail is described in other article. And this is talked by aworkshop in RIMS (Kyoto University) 2013, $Feb.$
$*$

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1. The triple system $(V, xyz)$ is $a(-1,1)$ FKTS.

2. The binary algebra $(V, xy)$ is a unital involutive algebra with involutioni.e.,
th $=x$ and$map:xarrow\overline{x},$$\overline{xy}=\overline{y}\overline{x}.$3. The triple product $xyz$ is expressed as(1.6)$xyz=(z\overline{y})x-(z\overline{x})y+(x\overline{y})z$

in terms of the binary products.
We may then call the binary system $(V, xy)$ to be a structurable (or Allison) algebra.
It posseses many interesting properties. First, it satisfies the triality relation ([3],[16]).
Second, its assoicated Lie algebra is a-graded Lie algebra of type([5]) and isinvariant under the symmetric group of degree 4, as are noted in ([8] and [9]). We may
also note that structurable algebras can be formulated purely as binary algebras without
any reference on the triple product ([3] and [16]). In ([14]), we have alternatively shownthat it can be derived from any $(-1,1)$ FKTS satisfying some additional conditions as in$BC_{1}$$B_{1}$$S_{4}$

the following theorem:

Theorem 1.1
Let $(V, xyz)$ be $a(-1,1)FKTS$ over the field $F$ of characteristica distinguished element $e\in V$ satisfying$\neq 2,$ $\neq 3$, possessing$eex=x$(1.7)$exe+2xee=3eex$(1.8)

andfor any $x\in V$ We then introduce a linear mapping$xarrow\overline{x}$and a binary product $xy$ in$V$

$by$

(1.9)

$\overline{x}=2x-xee$$xy=yex-\overline{x}\overline{y}e+\overline{x}ey$.

(1.10)

Then, the ternarll system $(V, xy, xyz)$ is structurable, so that $(V, xy)$ is a sturucturable
algebra with the unit element and the involution map. Conversely, for any
structurable algebra $(V, xy)$ with the unit element , the ternary system $(V, xy, xyz)$ is
structurable, satisfying $Eq.(l.7)$ and (1.8), if we define the triple product by $Eq.(l.6)$ .
$xarrow\overline{x}$

$e$

$e$

One of main results of this paper is that we can essentially dispense with the assumption of Eq.(1.8) as in the following theorem: to be proved in other paper

Theorem 1.2([18])Let $(V, xyz)$ be $a(-1,1)FKTS$ over the field of characteristic
distinguished element $e\in V$ satisfying $eex=x$ . Then there exists(1)$\sigma^{2}=1,$$\sigma e=e$(2) $\sigma(xyz)=(\sigma x)(\sigma y)(\sigma z)$ .

, possessing aEndsatisfying$\neq 2,$ $\neq 3$$\sigma\in$$V$(l.lla)$(1.11b)$

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Moreover, if we introduce a new triple product $\{x, y, z\}$ in$\{x, y, z\}=x(\sigma y)z$$V$by
(1.12),

then the triple system $(V, \{x, y, z\})$ is also $a(-1,1)FKTS$, satisfying$\{e, e, x\}=x, (1.13a)$$\{e, x, e\}+2\{x, e, e\}=3\{e, e, x\}.

Especially, if we set(1.13b)$$\overline{x}=2x-\{x, e, e\}, (1.14a)$$x\cdot y=\{y, e, x\}-\{\overline{x},\overline{y}, e\}+\{\overline{x}, e, y\}, (1.14b)$

then the ternary system $(V, x\cdot y, \{x, y, z\})$ is structurable,
where denoted by $Rx=xee,$ $Mx=exe,$ $Q=3Id-2R$ , and $\sigma=MQ^{-1}.$
The explicit formula for will be given in $([18J)$ .
$\sigma$

A similar situation exists for $(-1, -1)$ FKTS.
Theorem 1.3
Let $(V, xyz)$ be $a(-1, -1)FKTS$ over the fieldof characteristic not 2, possessing
a distinguished element $e\in V$ satisfying $eex=x$ for any $x\in V$ Then, there exists
satisfying $Eqs.(l.ll)$ . Moreover, introducing the new triple product $\{x, y, z\}$End
by $Eq.(l.12)$ , the new triple system $(V, \{x, y, z\})$ is also $a(-1, -1)FKTS$, satisfying now
$F$

$V$

$\sigma\in$(1.15)$\{e, e, x\}=\{e, x, e\}=x$ .

Especially, $(V, \{x, y, z\})$ is unitary $(-1, -1)FKTS$. (The definition of the unitary FKTS
be given by $Id\in\{K(x, y)\}_{span}.)$ .

At this point, it may be worthwhile to note that any$(1, \delta)$$FKTS$ $(i.e., \epsilon=+1)$cannot

satisfy $eex=x$ , since it contradicts Eq.(1.4) for the choice of $u=v=x=y=e$ . Also, in
FKTS, and
FKTS from any([18]), we will show that we can always constructsuggest that for $(1, 1)$ FKTS the condition $eex=x$ can be replaced by $efx=-fex=x$ fora pair $f\in V$. We can then construct a structurable algebra from any such $(1, 1)$ FKTS,and vice-versa.
$a(-\epsilon, \delta)$$(\epsilon, \delta)$$e,$Appendix (due to N.Kamiya)
Proposition A.
Under the notation of above Theorem 1.3, if we define the involution and the binary
product as follows;$\overline{x}=\{xee\}$where$Rx=\{xee\},$

and

$x$

$\bullet$$y=Ryox+Rxoy-Ry\circ Rx,$

$xoy=\{exy\}$ , then we have
$x=x, \overline{x\bullet y}=\overline{y}\bullet\overline{x}=$furthermore,$xoy=1/3\{\overline{y}\bullet(x-\overline{x})+(2x+\overline{x}) \bullet y\}$109
Proof.

From the definition of$\overline{x}$

and

$x$

$\bullet$$y$, by straightfoward calculations, we obtain$\overline{x\bullet y}=R(x\circ Ry)+R(yoRx)-xoy,$$\overline{y}\bullet\overline{x}=\overline{y}ox+\overline{x}oy-R(\overline{y}0\overline{x})$.

On the other hand, from$R(xoy)=x\circ Ry-Ry\circ x+RyoRx-Rxoy+yoRx,$
it follows that
$R(\overline{y}\circ\overline{x})=\overline{y}\circ x-x\circ\overline{y}+x\circ y-y\circ\overline{x}+\overline{x}\circ y.$$R(xo\overline{y})=xoy-y\circ x+yo\overline{x}-\overline{x}0\overline{y}+\overline{y}0\overline{x}$$R(yoRx)=yox-x\circ y+xo\overline{y}-\overline{y}0\overline{x}+\overline{x}\circ\overline{y}.$Therefore, combining these relations, we get$\overline{x\bullet y}=\overline{y}e\overline{x}$and $x==x.$From$x\bullet y=Ry\circ x+Rx\circ y-Ry\circ Rx$

it follows that for$U_{0}=\{x|Rx=-x\}, U_{2}=\{x|Rx=x\},$if
if$x\in U_{0},$$y\in U_{0}$, then $x\bullet y=-2yox-xoy,$$x\in U_{0},$$y\in U_{2}$, thenif$x\in U_{2}$, then$x\bullet$$x$$\bullet$$y=2yox-xoy,$$y=x\circ y.$Thus we have the relation of theand $xoy$ . This completes the proof.
Corollary $([2011$ Glasgow $J.Math.vol53,p727-739])$ For balanced $(-1,-1)Freudenthal-$
Kantor triple system, if we set$x$$\bullet$$y$$\overline{x}=R(x)=2<x, e>e-x, x\circ y=\{exy\}$

then we have
$x=x, \overline{xoy}=\overline{y}0\overline{x}=$In particular, also we get$x=x, \overline{x\bullet y}=\overline{y}\bullet\overline{x}=$that is, this algebra is a quadratic algebra.
Following([2006 Lec.Notes in Pure and App.Math. vo1246,Chapter 16, CRC]), we
recall the definition of the quadratic triple system defined by$\{xxy\}=\{yxx\}=<x|x>y, <x|y>=<y|x>\in\Phi.$

Remark. we note that by means of $x\circ y=\{xey\}$ , this product has the structure of aquadratic algebra with respect to$0.$

110
Proposition B. Let
$e$

$A$

be a nonassociative algebra with involution and unit element

such that$e\bullet x=x, x\bullet e=x, x=x, \overline{x\bullet y}=\overline{y}\bullet\overline{x}=.$

If

$x$

$\bullet$$\overline{x}\in\Phi e$and define$\{xyz\}=(x\bullet\overline{y})\bullet z-(z\bullet\overline{y})\bullet x+(z\bullet\overline{x})\bullet y$then$\{xyz\}^{-}$

is a quadratic triple system w.r.t. new triple product $\{xyz\}^{\sim}=\{yxz\}.$

Proof.

is denoted by $<x|x>e.$
From $\{xxy\}=\{yxx\}=<x,$ $x>y$ , it is clear, where
RemarkFor the above Prop. , the triple product{xyz} is closedly related to an anti-structurable
algebras ([2010 Bull. Aust. Math. vol.81,132-155.]).
Example (a)is theLet be a set of matrix $n\cross n$ and $O(n, \Phi)=\{x\in U|x^{t}x=Id\}$ , wheretranspose matrix of . We define a triple product as follows;$x$$\bullet$$\overline{x}$$B$$t_{X}$$U$$x$$\{xyz\}=(x^{t}y)z-(z^{t}y)x+(z^{t}x)y, x, y, z\in U$

then this triple product has a structure of $(-1,-1)Freudenthal$-Kantor triple system.
Furthermore, if we set$\{xyz\}^{\sim}=\{yxz\},$this triple system is a quadratic triple system. On the other hand, we have$\{uux\}=\{uxu\}=x, \{xuu\}=u^{t}xu, \forall u\in O(n, \Phi)$that is, is a left unit element of $U.$
If we define a binary product by $xoy=uxy,$ $u\in O(n, \Phi)$ and an involution by$\overline{x}=Rx=\{xuu\}$ then we have$u$$xoy=u^{t}xy-y^{t}xu+y^{t}ux,\overline{x}=u^{t}xu, x==x.$

However,$\overline{x\circ y}\neq\overline{y}\circ\overline{x}$and$\overline{x}\bullet$$y=\overline{y}$$\bullet$$\overline{x}.$Indeed, we have$\{xxy\}^{\sim}=<x|x>y, \{yxx\}^{\sim}=<x|x>y$$x^{t}x=<x|x>e,$$e=Id_{n},$$e$is the identity matrix element. Thus, it is clear thatis a quadratic triple system.
On the other hand, from the relations;$xoy=u^{t}xy-y^{t}xu+y^{t}ux,$$x \bullet y=Ry\circ x+Rxoy-RyoRx,$$\overline{x}=Rx=\{xuu\}=u^{t}xu$

by $u^{t}u=Id_{n}$ we get$yox=u^{t}yx-x^{t}yu+x^{t}uy,$$\{xyz\}^{\sim}$

111$Ry\circ x=y^{t}ux-x^{t}uy+x^{t}yu,$$Rx\circ y=x^{t}uy-y^{t}ux+y^{t}xu,$$Ry\circ Rx=y^{t}xu-u^{t}xy+y^{t}ux.$

Furthermore we get the relations;$x \bullet y=x^{t}yu+u^{t}xy-y^{t}ux,$$x\bullet y=y^{t}xu+u^{t}yx-u^{t}xu^{t}yu,$$y \bullet x=y^{t}xu+u^{t}yx-x^{t}uy,$$\overline{y}\bullet\overline{x}=u^{t}yx+y^{t}xu-u^{t}xu^{t}yu.$

Therefore we obtain$\overline{x\bullet y}=\overline{y}\bullet\overline{x}, x=x=.$This case has$\overline{x\circ y}\neq\overline{y}\circ\overline{x}, x=x=.$

Example (b).
Let be a set of Mat $(1, n, \Phi)$ . We define the triple product as follows;$U$$\{xyz\}=<x|y>z-<y|z>x+<z|x>y, <x|y>=<y|x>\in\Phi.$

Then this product has a structure of balanced $(-1,-1)Freudenthal$ -Kantor triple system.
Thus, as same as Example (a), the triple product $\{xyz\}^{\sim}=\{yxz\}$ is a quadratic triple
system.
In deed, it is clear that$\{xxy\}^{\sim}=\{xyx\}^{\sim}=<x|x>y.$On the other hand, for, we give an example, of left unit element from the structurable
algebra as follows.
Let$z\in Mat(n, n;\Phi)$ ,$\delta=1$$x,$ $y,$$\{xyz\}=x^{t}yz+z^{t}yx-z^{t}xy,$where is the transpose matrix of , and chose $e\in O(n, \Phi)$ .
Then we have$t_{X}$$x$$\{eex\}=x, e^{t}e=Id_{n}$We now define$\overline{x}:=e^{t}xe$and$x\cdot y:=\{\overline{x}ey\}-\{\overline{x}\sigma(\overline{y})e\}+\{yex\}$where $\sigma:=M(5-2R)/3,$ $Mx:=\{exe\},$ $Rx:=\{xee\}.$Then we have
$Mx=2e^{t}xe-x,$ $Rx=2x-e^{t}xe,$ $(5-2R)x=2x+e^{t}xe,$$\sigma x=x.$

From these relations, it follows that$\{\overline{x}ey\}=\overline{x}^{t}ey+y^{t}e\overline{x}-y^{t}\overline{x}e$$M(5-2R)x=3x$ , and so112$\{\overline{x}\sigma(\overline{y})e\}=\overline{x}^{t}\overline{y}e+e^{t}\overline{y}\overline{x}-e^{t}\overline{x}\overline{y}$$\{yex\}=y^{t}ex+x^{t}ey-x^{t}ye.$

By means of$\overline{x}=e^{t}xe$, from straightfoward calculations, we get$x\cdot y=x^{t}ey$(as usual product)$\overline{x\cdot y}=e^{t}ye^{t}xe$$\overline{y}\cdot\overline{x}=e^{t}ye^{t}ee^{t}xe=e^{t}ye^{t}xe (by e^{t}ee=e)$Hence we obtain$\overline{x\cdot y}=\overline{y}\cdot\overline{x}$.

and $x==x.$

Coccluding Remark
For left unit $(-1, \delta)$ Freudenthal-Kantor triple systems, it seems that there is a relationship as follows;$—$ –structurable algebra,$\delta=1$$\delta=-1$$—–$ –weakly quadratic algebra.

In our subjects;
N.Jacobson, H.Freudenthal, I.L.Kantor, E.Zelmanov, S.Okubo, are important persons,historically in my opinion.
. In view point of generalization of numbers;$R–>C-->H-->O-->H_{3}(O)$$–>M(H_{3}(O))-->E_{8}$ (Lie algebra of $dim248$ )$O$—-$>ImO-->G_{2},$ $H\otimes O-->E_{7},$ $O\oplus O-->F(4)$ .

. In view point of construction of Lie algebras and superalgebras:
By means of triple systems, we can construct to all simple algebras and superalgebras
without concept of root systems, and also the tangent spaces of some symmetric space orbounded domain correspond to certain triple systems.
$(i.e.,$ symmetric space
triple system $etc.)$ .
. In view point of physics application:(Clasical)Yang-Baxter equations, Dynamical Yang-Baxter equations, $M$ theory(hermitiantriple systems), String theory.
The example of hermitian triple system is as follows;$rightarrow Lie$$\{xyz\}=x^{t}\overline{y}z-z^{t}\overline{y}x+z^{t}x\overline{y}, x, y, z\in Mat(m, n;C)$is the conjugation of C.
wherThis triple system is $a(-1,-1)-\mathbb{R}eudenthal$ -Kantor triple system with$e^{-}$$\{\alpha x, \beta y, \gamma z\}=\alpha\overline{\beta}\gamma\{xyz\}$,for all$\alpha,$$\beta,$$\gamma\in C$

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$J$$BC_{1},$$S_{4}$$S_{4}$$(\alpha, \beta, \gamma)$$U(\epsilon, \delta)$