Derivations and dimensionally nilpotent derivations in Lie triple algebras

TL;DR

This paper investigates derivations in non-nilpotent Lie triple algebras, characterizing their structure based on idempotent elements and exploring their multiplicative properties, including connections to gametic and Bernstein algebras.

Contribution

It provides a detailed classification of derivation structures in non-nilpotent Lie triple algebras and links their basis structures to known algebraic forms.

Findings

Derivation algebra structure depends on the presence of idempotents.

When dimension n=2p+1, bases align with gametic algebra G(2p+2,2).

For n=2p, the algebra is either like the previous case or a Bernstein algebra.

Abstract

In this paper, we first study derivations in non nilpotent Lie triple algebras. We determine the structure of derivation algebra according to whether the algebra admits an idempotent or a pseudo-idempotent. We study the multiplicative structure of non nilpotent dimensionally nilpotent Lie triple algebras. We show that when $n=2p+1$ the adapted basis coincides with the canonical basis of the gametic algebra $G(2p+2,2)$ or this one obviously associated to a pseudo-idempotent and if $n=2p$ then the algebra is either one of the precedent case or a conservative Bernstein algebra.

Equations

$$2x(x(xy))+y{x}^3=3x(y{x}^2)$$

$$x^2(yx)=(x^{2}y)x.$$

$$d(t)=0\text{ and }d(e)\in A_e(1/2).$$

$$d(x_1{y}_1)$$

$$h_d((x_1{y}_1{)}_1)=[h_d(x_1)y_1+x_1{h}_d(y_1){]}_1\text{ et }$$

$$f_d((x_1{y}_1{)}_{1/2})=[h_d(x_1)y_1+x_1{h}_d(y_1){]}_{1/2}=2h_d(x_1{y}_1)t,\text{ and we have }(v).$$

$$g_d((x_0{y}_0{)}_0)=[g_d(x_0)y_0+x_0{g}_d(y_0){]}_0\text{ et }$$

$$f_d((x_0{y}_0{)}_{1/2})=[g_d(x_0)y_0+x_0{g}_d(y_0){]}_{1/2}=2g_d(x_0{y}_0)t,\text{and we have }(vi).$$

$$d(x_{1/2}{y}_{1/2})$$

$$h_d((x_{1/2}{y}_{1/2}{)}_1)=[f_d(x_{1/2})y_{1/2}+x_{1/2}{f}_d(y_{1/2}){]}_1\text{ et }$$

$$g_d((x_{1/2}{y}_{1/2}{)}_0)=[f_d(x_{1/2})y_{1/2}+x_{1/2}{f}_d(y_{1/2}){]}_0,\text{ and we have }(vii).$$

$$d(x_1{x}_0)$$

$$f_d((x_1{x}_0{)}_{1/2})=h_d(x_1)x_0+x_1{g}_d(x_0),\text{ and we have }(viii).$$

$$d(x_1{x}_{1/2})$$

$$f_d(x_1{x}_{1/2})=h_d(x_1)x_{1/2}+x_1{f}_d(x_{1/2}),\text{ and we have }(ix).$$

$$d(x_0{x}_{1/2})$$

$$f_d(x_0{x}_{1/2})=g_d(x_0)x_{1/2}+x_0{f}_d(x_{1/2}),\text{ and finally }(x).$$

$${\lambda }_2+{\gamma }_{112}=\frac{1}{2}\text{ et }{a}_{2,k}+{\gamma }_{11k}=0(3\le k\le n).$$

$${\lambda }_3+3{\gamma }_{123}=\frac{1}{2}\text{ et }{a}_{3,k}+3{\gamma }_{1,2,k}=0(4\le k\le n).$$

$${\lambda }_3+{\gamma }_{123}={\lambda }_2\text{ et }{a}_{3,k}+{\gamma }_{12k}=a_{2,k-1}(4\le k\le n).$$

$${\lambda }_4+4{\gamma }_{134}+3{\gamma }_{224}=\frac{1}{2}$$

$${\lambda }_4+{\gamma }_{134}={\lambda }_3\text{ et }{a}_{4,k}+{\gamma }_{13k}=a_{3,k-1}(5\le k\le n)$$

$$\{\begin{array}{ll}{e}_1{e}_{2p}+e_2{e}_{2p-1}={\epsilon }_1{e}_{2p+1},& \\ \cdots \cdots ,& \\ e_i{e}_{2p+1-i}+e_{i+1}{e}_{2p-i}={\epsilon }_i{e}_{2p+1},& \\ \cdots \cdots ,& \\ e_p{e}_{p+1}+e_{p+1}{e}_p=2e_p{e}_{p+1}={\epsilon }_p{e}_{2p+1}.& \end{array}$$

$$\{\begin{array}{ll}{e}_1{e}_{2p-1}+e_2{e}_{2p-2}=0,& \\ e_2{e}_{2p-2}+e_3{e}_{2p-3}=0,& \\ \cdots \cdots ,& \\ e_{p-1}{e}_{p+1}+e_p^2=0.& \end{array}$$

$$\{\begin{array}{ll}{e}_1{e}_{2k}+e_2{e}_{2k-1}=0,& \\ e_2{e}_{2k-1}+e_3{e}_{2k-2}=0,& \\ \cdots \cdots ,& \\ e_{k-1}{e}_{k+2}+e_k{e}_{k+1}=0,& \\ e_k{e}_{k+1}+e_{k+1}{e}_k=2e_k{e}_{k+1}=0.& \end{array}$$

Full text

Keyword: Dimensionally nilpotent Lie triple algebra, pseudo-idempotent, Jordan algebra, ascending basis, adapted basis.

2010 Mathematics Subject Classification : Primary 17A30, secondary 17D92, 17B40, 17C10

1 Introduction

A $n+1$ finite dimensional algebra $A$ is dimensionally nilpotent if there is a derivation $d:A\longrightarrow A$ such that $d^{n+1}=0$ and $d^{n}\neq 0$. This notion has been studied by G.F. Leger and P.L. Manley[8] for Lie algebras, J.M. Osborn [12] for Jordan algebras, Micali and Ouattara[9] for genetic algebras. Recently, V. Eberlin [4] has deepened the work of the authors of [8] in his thesis. Regarding Jordan algebras, Osborn shows that every dimensionally nilpotent Jordan $K$-algebra is either nilpotent or satisfies $A/Rad(A)\simeq K$.

We study the case of non nilpotent dimensionally Lie triple algebras. In an adapted basis we caracterize the multiplicative structure of these algebras relative to the parity of $n$. More precisely we show that when $n=2p+1$, the adapted basis coincides with canonical basis of the gametic algebra $G(2p+2,2)$ or this one obviously associated to a pseudo-idempotent. If $n=2p$ then this algebra is either one of the precedent case or a train algebra of rank $3$ which is a Jordan algebra [13]. Since Jordan algebras are also Lie triple ones the final corollary describes non nilpotent dimensionally nilpotent Jordan algebras.

2 Preliminaries

A Lie triple algebra is a commutative algebra satisfying

[TABLE]

while a Jordan algebra is a commutative algebra satisfying

[TABLE]

Every Jordan algebra satisfies identity (1).

Theorem 2.1** ([5]).**

Let $A$ be a Lie triple algebra and $L$ the ideal generated by the associators $(x^{2},x,x)$. Then $L^{2}=0$ and $A/L$ is a Jordan algebra.

Definition 2.2**.**

A pseudo-idempotent of $A$ is a non-zero element $e$ such that there is $t\neq 0$ in $L$ satisfying $e^{2}=e+t$ and $et=\frac{1}{2}t$.

Theorem 2.3** ([2]).**

Every Lie triple non nilalgebra contains either a non-zero idempotent, or a pseudo-idempotent.

Definition 2.4**.**

An ideal $I$ of an algebra $A$ is said to be caracteristic if $d(I)\subseteq I$ for every derivation $d$ of $A$. An ideal $I$ of an algebra $A$ is said to be d-invariant if $d(I)\subseteq I$ for a given derivation $d$ of $A$.

3 Caracterization of derivations

In this paragraph we study the derivations in Lie triple non nilalgebras . We give a caracterization, distinguishing two cases: with an idempotent or with a pseudo-idempotent.

3.1 Lie triple algebras with idempotent

Relative to the non-zero idempotent $e$, $A$ admits the following Peirce decomposition $A=A_{e}(1)\oplus A_{e}(\frac{1}{2})\oplus A_{e}(0)$. Relations between Peirce components and the products of their elements are ruled by the following lemma:

Lemma 3.1** ([2], Lemme 2.2).**

*Let $A=A_{e}(1)\oplus A_{e}(1/2)\oplus A_{e}(0)$ be the Peirce decomposition of $A$ relative to a non-zero idempotent. Then

1. $(i)$

$A_{e}(1/2)A_{e}(1/2)\subseteq A_{e}(1)+A_{e}(0)$, $A_{e}(\lambda)A_{e}(\lambda)\subseteq A_{e}(\lambda)$, $A_{e}(\lambda)A_{e}(1/2)\subseteq A_{e}(1/2)$,

$A_{e}(\lambda)A_{e}(1-\lambda)=0$, $(\lambda=0,1)$ ; 2. $(ii)$

$(x_{1}y_{1})a_{1/2}=x_{1}(y_{1}a_{1/2})+y_{1}(x_{1}a_{1/2})$,

$(x_{0}y_{0})a_{1/2}=x_{0}(y_{0}a_{1/2})+y_{0}(x_{0}a_{1/2})$* ;* 3. $(iii)$

$[x_{1}(x_{1/2}a_{1/2})]_{1}=[(x_{1}x_{1/2})a_{1/2}+(x_{1}a_{1/2})x_{1/2}]_{1}$,

$[x_{0}(x_{1/2}a_{1/2})]_{0}=[(x_{0}x_{1/2})a_{1/2}+(x_{0}a_{1/2})x_{1/2}]_{0}$* ;* 4. $(iv)$

$[(x_{1}x_{1/2})y_{1/2}]_{0}=[(x_{1}y_{1/2})x_{1/2}]_{0}$,

$[(x_{0}x_{1/2})y_{1/2}]_{1}=[(x_{0}y_{1/2})x_{1/2}]_{1}$* ;* 5. $(v)$

$x_{0}(y_{1}a_{1/2})=y_{1}(x_{0}a_{1/2})$* ;* 6. $(vi)$

$x_{1/2}(x_{1/2}^{2})_{1}=x_{1/2}(x_{1/2}^{2})_{0}=\frac{1}{2}x_{1/2}^{3}$* ;* 7. $(vii)$

$(x_{1/2}y_{1/2})_{0}z_{1/2}+(y_{1/2}z_{1/2})_{0}x_{1/2}+(z_{1/2}x_{1/2})_{0}y_{1/2}\\ {\qquad\qquad\qquad}=(x_{1/2}y_{1/2})_{1}z_{1/2}+(y_{1/2}z_{1/2})_{1}x_{1/2}+(z_{1/2}x_{1/2})_{1}y_{1/2}$.

Since $A$ is $e$-stable, i.e. $A_{e}(\lambda)A_{e}(1/2)\subseteq A_{e}(1/2)$ and $[(x_{\lambda}x_{1/2})y_{1/2}]_{1-\lambda}=[(x_{\lambda}y_{1/2})x_{1/2}]_{1-\lambda}$ with $\lambda=0,1$, calculations on derivations give results similar to [1, Corollary 2], precisely.

Theorem 3.2**.**

Every derivation $d$ of $A$ is determined and only defined by a quadruplet $(d(e),f_{d},g_{d},h_{d})$ with $f_{d}\in End_{K}(A_{e}(1/2))$, $g_{d}\in Der_{K}(A_{e}(0))$ and $h_{d}\in Der_{K}(A_{e}(1))$ satisfying the following conditions:

1. $(i)$

$d(e)\in A_{e}(1/2)$* ;* 2. $(ii)$

$d(x_{1})=h_{d}(x_{1})+2d(e)x_{1}$* ;* 3. $(iii)$

$d(x_{1/2})=f_{d}(x_{1/2})+2(d(e)x_{1/2})_{0}-2(d(e)x_{1/2})_{1}$* ;* 4. $(iv)$

$d(x_{0})=g_{d}(x_{0})-2d(e)x_{0}$* ;* 5. $(v)$

$h_{d}(x_{1}y_{1})=h_{d}(x_{1})y_{1}+x_{1}h_{d}(y_{1})$* ;* 6. $(vi)$

$g_{d}(x_{0}y_{0})=g_{d}(x_{0})y_{0}+x_{0}g_{d}(y_{0})$* ;* 7. $(vii)$

$h_{d}((x_{1/2}y_{1/2})_{1})=[f_{d}(x_{1/2})y_{1/2}+x_{1/2}f_{d}(y_{1/2})]_{1}$* ;* 8. $(viii)$

$g_{d}((x_{1/2}y_{1/2})_{0})=[f_{d}(x_{1/2})y_{1/2}+x_{1/2}f_{d}(y_{1/2})]_{0}$* ;* 9. $(ix)$

$f_{d}(x_{1}x_{1/2})=h_{d}(x_{1})x_{1/2}+x_{1}f_{d}(x_{1/2})$* ;* 10. $(x)$

$f_{d}(x_{0}x_{1/2})=g_{d}(x_{0})x_{1/2}+x_{0}f_{d}(x_{1/2}).$**

Proposition 3.3**.**

Let $A$ be a Lie triple algebra and $A=A_{e}(1)\oplus A_{e}(1/2)\oplus A_{e}(0)$ the Peirce decomposition of $A$ relative to an idempotent $e\neq 0$. Subspaces $J_{\lambda}=\{x_{\lambda}\in A_{e}(\lambda)\mid x_{\lambda}A_{e}(1/2)=0\}$ $(\lambda=0,1)$ and $J=J_{0}\oplus J_{1}$ are caracteristic ideals of $A$ and the quotient algebra $A/J$ is a Jordan algebra.

Proof.

Considering $J_{\lambda}=\ker(S_{\lambda})$, with $S_{\lambda}:A_{e}(\lambda)\rightarrow End_{K}(A_{e}(1/2)),x_{\lambda}\mapsto S_{\lambda}(x_{\lambda})$ and $S_{\lambda}(x_{\lambda}):a_{1/2}\mapsto x_{\lambda}a_{1/2}$. We know by ([10]) that $J_{\lambda}$ is an ideal of $A_{e}(\lambda)$ ($\lambda=0,1$) and since $A_{e}(\lambda)A_{e}(1/2)\subseteq A_{e}(1/2)$, then $J=J_{1}+J_{0}$ is an ideal of $A$ such that $A/J$ is a Jordan algebra ([11], Proposition 6.7).

Let’s consider $d\in Der_{K}(A)$, $x_{\lambda}\in J_{\lambda}(e)$ and $a_{1/2}\in A_{e}(1/2)$. We have $0=d(x_{\lambda}a_{1/2})=x_{\lambda}d(a_{1/2})+d(x_{\lambda})a_{1/2}$. But $d(a_{1/2})=f_{d}(a_{1/2})+2(d(e)a_{1/2})_{0}-2(d(e)a_{1/2})_{1},$ therefore we have $x_{\lambda}d(a_{1/2})=0$ because $x_{\lambda}(d(e)a_{1/2})_{\lambda}=[(x_{\lambda}d(e))a_{1/2}+(x_{\lambda}a_{1/2})d(e)]_{\lambda}$. Hence $d(x_{\lambda})a_{1/2}=0$ with $\lambda=0,1.$ But, on the one hand we have $d(x_{1})=h_{d}(x_{1})-2d(e)x_{1},$ and $0=d(x_{1})a_{1/2}=h_{d}(x_{1})a_{1/2}$ and then $h_{d}(x_{1})\in J_{1}$, on the other hand we have $d(x_{0})=g_{d}(x_{0})-2d(e)x_{0}$, with $0=d(x_{0})a_{1/2}=g_{d}(x_{0})a_{1/2}$ and then $g_{d}(x_{0})\in J_{0}$. Hence $d(J_{\lambda})\subseteq J_{\lambda}$ and we conclude that $d(J)\subseteq J$.     $\Box$

3.2 Lie triple algebras with pseudo-idempotent

Lemma 3.4** ([2], Proposition 4.3).**

Let $L=L_{e}(1)\oplus L_{e}(1/2)\oplus L_{e}(0)$ and $A=A_{e}(1)\oplus A_{e}(1/2)\oplus A_{e}(0)$ be the respective Peirce decomposition of $L$ and $A$, relative to the pseudo-idempotent $e$, satisfying $e^{2}=e+t$ with $t\in L_{1/2}$ fixed. Then

1. $(i)$

$A_{e}(0)L_{e}(1/2)\subseteq L_{e}(1/2)$*, $A_{e}(1)L_{e}(1/2)\subseteq L_{e}(1/2)$, $A_{e}(1)L_{e}(1)\subseteq L_{e}(1)$, *

$A_{e}(0)L_{e}(0)\subseteq L_{e}(0)$*, $A_{e}(0)L_{e}(1)=A_{e}(1)L_{e}(0)=0$, *

$A_{e}(1/2)L_{e}(0)=A_{e}(1/2)L_{e}(1)=A_{e}(1/2)L_{e}(1/2)=0$* ;* 2. $(ii)$

$A_{e}(1)A_{e}(0)\subseteq L_{e}(1/2)$, $A_{e}(0)A_{e}(1/2)\subseteq A_{e}(1/2)$, $A_{e}(1)A_{e}(1/2)\subseteq A_{e}(1/2)$,

$A_{e}(0)A_{e}(0)\subseteq A_{e}(0)+L_{e}(1/2)$, $A_{e}(1)A_{e}(1)\subseteq A_{e}(1)+L_{e}(1/2)$,

$A_{e}(1/2)A_{e}(1/2)\subseteq A_{e}(1)+A_{e}(0)$* ;* 3. $(iii)$

$(x_{0}y_{0})_{1/2}=4(x_{0}t)y_{0}=4(y_{0}t)x_{0}$* ;*

$(x_{1}y_{1})_{1/2}=4(x_{1}t)y_{1}=4(y_{1}t)x_{1}$* ; *

$(x_{0}y_{1})_{1/2}=4(x_{0}t)y_{1}=4(y_{1}t)x_{0}$* ;* 4. $(iv)$

$(x_{1}y_{1})a_{1/2}=x_{1}(y_{1}a_{1/2})+y_{1}(x_{1}a_{1/2})$* ;* 5. $(v)$

$(x_{0}y_{0})a_{1/2}=x_{0}(y_{0}a_{1/2})+y_{0}(x_{0}a_{1/2})$* ;* 6. $(vi)$

$x_{0}(y_{1}a_{1/2})=y_{1}(x_{0}a_{1/2})$* ;* 7. $(vii)$

$[x_{0}(x_{1/2}a_{1/2})]_{0}=[(x_{0}x_{1/2})a_{1/2}+(x_{0}a_{1/2})x_{1/2}]_{0}$* ;* 8. $(viii)$

$[x_{1}(x_{1/2}a_{1/2})]_{1}=[(x_{1}x_{1/2})a_{1/2}+(x_{1}a_{1/2})x_{1/2}]_{1}$* ;* 9. $(ix)$

$[(x_{0}x_{1/2})y_{1/2}]_{1}=[(x_{0}y_{1/2})x_{1/2}]_{1}$* ;*

$[(x_{1}x_{1/2})y_{1/2}]_{0}=[(x_{1}y_{1/2})x_{1/2}]_{0}$* ;* 10. $(x)$

$(x_{1/2}y_{1/2})_{0}z_{1/2}+(y_{1/2}z_{1/2})_{0}x_{1/2}+(z_{1/2}x_{1/2})_{0}y_{1/2}\\ {\qquad\qquad\qquad}=(x_{1/2}y_{1/2})_{1}z_{1/2}+(y_{1/2}z_{1/2})_{1}x_{1/2}+(z_{1/2}x_{1/2})_{1}y_{1/2}$.

Lemma 3.5**.**

Let $A$ be a Lie triple algebra and $e$ a pseudo-idempotent of $A$: $e^{2}=e+t,\;et=\frac{1}{2}t,\;t^{2}=0$ with $t\in L$. For every derivation $d$ of $A$, we have

[TABLE]

Proof.

Let’s consider $d\in Der_{K}(A)$. Since $e^{2}=e+t$, we have $2ed(e)=d(e)+d(t)$. Setting $d(e)=[d(e)]_{1}+[d(e)]_{1/2}+[d(e)]_{0}$, we have $d(t)=[d(e)]_{1}-[d(e)]_{0}$. Because of $2et=t$, we deduce $2ed(t)+2d(e)t=d(t)$. We have $2d(e)t=-[d(e)]_{1}-[d(e)]_{0}$. We know that $t\in L_{e}({1/2})$ and $L_{e}({1/2})$ is an ideal of $A$. It follows that $[d(e)]_{1}=[d(e)]_{0}=0$.     $\Box$

Theorem 3.6**.**

Every derivation $d$ of $A$ is determined and only defined by a quadruplet $(d(e),f_{d},g_{d},h_{d})$ with $f_{d}\in End_{K}(A_{e}(1/2))$, $g_{d}\in End_{K}(A_{e}(0))$ and $h_{d}\in End_{K}(A_{e}(1))$ satisfying the following conditions:

1. $(i)$

$d(e)\in A_{e}(1/2)$* ;* 2. $(ii)$

$d(x_{1})=h_{d}(x_{1})+2d(e)x_{1}$* ;* 3. $(iii)$

$d(x_{1/2})=f_{d}(x_{1/2})+2(d(e)x_{1/2})_{0}-2(d(e)x_{1/2})_{1}$* ;* 4. $(iv)$

$d(x_{0})=g_{d}(x_{0})-2d(e)x_{0}$* ;* 5. $(v)$

$h_{d}((x_{1}y_{1})_{1})=[h_{d}(x_{1})y_{1}+x_{1}h_{d}(y_{1})]_{1}$* ;*

$f_{d}((x_{1}y_{1})_{1/2})=[h_{d}(x_{1})y_{1}+x_{1}h_{d}(y_{1})]_{1/2}=2h_{d}((x_{1}y_{1})_{1})t$* ;* 6. $(vi)$

$g_{d}((x_{0}y_{0})_{0})=[g_{d}(x_{0})y_{0}+x_{0}g_{d}(y_{0})]_{0}$* ;*

$f_{d}((x_{0}y_{0})_{1/2})=[g_{d}(x_{0})y_{0}+x_{0}g_{d}(y_{0})]_{1/2}=2g_{d}((x_{0}y_{0})_{0})t$* ;* 7. $(vii)$

$h_{d}((x_{1/2}y_{1/2})_{1})=[f_{d}(x_{1/2})y_{1/2}+x_{1/2}f_{d}(y_{1/2})]_{1}$* ;*

$g_{d}((x_{1/2}y_{1/2})_{0})=[f_{d}(x_{1/2})y_{1/2}+x_{1/2}f_{d}(y_{1/2})]_{0}$* ;* 8. $(viii)$

$f_{d}(x_{1}x_{0})=h_{d}(x_{1})x_{0}+x_{1}g_{d}(x_{0})$* ;* 9. $(ix)$

$f_{d}(x_{1}x_{1/2})=h_{d}(x_{1})x_{1/2}+x_{1}f_{d}(x_{1/2})$* ;* 10. $(x)$

$f_{d}(x_{0}x_{1/2})=g_{d}(x_{0})x_{1/2}+x_{0}f_{d}(x_{1/2}).$**

Proof.

Let $d$ be a derivation of $A$ and $e$ a pseudo-idempotent of $A$. Since $d(e)\in A_{e}(1/2)$, we have $(i)$. Let $x_{1}\in A_{e}(1)$. We have $ex_{1}=x_{1}$, and then $d(e)x_{1}+ed(x_{1})=d(x_{1})$. Let’s set $d(x_{1})=a_{1}+a_{1/2}+a_{0}$. Then $d(e)x_{1}+a_{1}+\frac{1}{2}a_{1/2}=a_{1}+a_{1/2}+a_{0}$, and we have $a_{1/2}=2d(e)x_{1}$ and $a_{0}=0$. Hence $d(x_{1})=h_{d}(x_{1})+2d(e)x_{1}$ with $h_{d}$ an endomorphism of $A_{e}(1)$ and $(ii)$ is prooved.

By similar calculations we have $(iii)$ and $(iv)$.

with $x_{1}$, $y_{1}\in A_{e}(1)$, we have $(x_{1}y_{1})_{1/2}=4x_{1}(y_{1}t)=4y_{1}(x_{1}t)=2(x_{1}y_{1})t$

[TABLE]

because $d((x_{1}y_{1}))t=h_{d}(x_{1}y_{1})t$.

We also have,

[TABLE]

It follows that

[TABLE]

[TABLE]

We show by similar calculations that:

[TABLE]

[TABLE]

Let $x_{1/2},y_{1/2}\in A_{e}(1/2)$. We have

[TABLE]

But

[TABLE]

because of identity $(x)$ of Lemma 3.4. It follows that:

[TABLE]

and we have

[TABLE]

[TABLE]

We have

[TABLE]

[TABLE]

In a similar way

[TABLE]

[TABLE]

So we have

[TABLE]

[TABLE]

Conversely, once we have identities $(i)$ to $(xi)$, setting $x=x_{1}+x_{1/2}+x_{0}$ and $y=y_{1}+y_{1/2}+y_{0}$, we show that $d(xy)=d(x)y+xd(y)$.     $\Box$

Example 3.7**.**

Let $A$ be the four dimensional Lie triple $K$-algebra which multiplication table in the basis $\{e,t,u,r\}$ is given by : $e^{2}=e+t$, $u^{2}=u+r$, $et=\frac{1}{2}t$, $ur=\frac{1}{2}r$, all other product being zero. The Peirce decomposition of $A$ relative to pseudo-idempotent $e$ gives $A_{e}(1)=K(e+2t)$, $A_{e}(1/2)=Kt$, $A_{e}(0)=<u,r>$. Let $d$ be a derivation of $A$. Since $e$ and $u$ are pseudo-idempotents, we have $d(t)=d(r)=0$, $d(e)=\alpha t$, $d(u)=\beta r$. The derivation algebra is two dimensional.

Example 3.8**.**

Let’s consider the four dimensional Lie triple $K$-algebra $A$ which multiplication table in the basis $\{e,t_{1},t_{2},v\}$ is given by : $e^{2}=e+t_{1}$, $et_{1}=\frac{1}{2}t_{1}$, $et_{2}=\frac{1}{2}t_{2}$, $ev=v$ and $vt_{1}=t_{2}$, all other products being zero. Then we have $A_{e}(1)=<(e+2t),v>$, $A_{e}(1/2)=<t_{1},t_{2}>$, $A_{e}(0)=0$. Let $d$ be a derivation of $A$. We have $d(t_{1})=0.$ Set $d(e)=\alpha_{1}t_{1}+\beta_{1}t_{2}.$ Thus $\alpha_{1}t_{1}+\beta_{1}t_{2}=d(e+2t_{1})=h_{d}(e+2t_{1})+2d(e)(e+2t_{1})=h_{d}(e+2t_{1})+\alpha_{1}t_{1}+\beta_{1}t_{2}.$ It follows that $h_{d}(e+2t_{1})=0.$

Setting $d(v)=\alpha_{2}(e+2t_{1})+\beta_{2}v+\gamma_{2}t_{1}+\eta_{2}t_{2}$, relation $0=d(v^{2})=2d(v)v$ gives $\alpha_{2}=\gamma_{2}=0.$ It follows that $h_{d}(v)=\beta_{2}v.$ Furthermore, relation $f_{d}((vt_{1})_{1/2})=h_{d}(t_{1})+f_{d}(t_{1})$ gives $f_{d}(t_{2})=\beta_{2}t_{2}.$

We have $d(e)=\alpha_{1}t_{1}+\beta_{1}t_{2},h_{d}(e+2t_{1})=0,h_{d}(v)=\beta_{2}v,f_{d}(t_{1})=0,f_{d}(t_{2})=\beta_{2}t_{2}.$ The derivation algebra is three dimensional.

Proposition 3.9**.**

Let’s consider a pseudo-idempotent $e\neq 0$. Subspace $J_{e}(1/2)=\{x_{1/2}\in A_{e}(1/2)\mid x_{1/2}A_{e}(1/2)=0\}$ is a caracteristic ideal and $A/J_{e}(1/2)$ is a Lie triple algebra with $\overline{e}$ as idempotent.

Proof.

Let $x_{1/2}\in J_{e}(1/2)$, $a_{1/2}\in A_{e}(1/2)$ and $y_{\lambda}\in A_{e}(\lambda)$ ($\lambda=0,1$). We have $[(x_{1/2}y_{\lambda})a_{1/2}]_{\lambda}=[(a_{1/2}y_{\lambda})x_{1/2}]_{\lambda}=0$ and $[(x_{1/2}y_{\lambda})a_{1/2}]_{1-\lambda}=[(a_{1/2}y_{\lambda})x_{1/2}]_{1-\lambda}=0$, and then $(x_{1/2}y_{\lambda})a_{1/2}=0$. Hence $A_{e}(\lambda)J_{e}(1/2)\subseteq J_{e}(1/2)$, and it follows that $AJ_{e}(1/2)\subseteq J_{e}(1/2)$. $J_{e}(1/2)$ is an ideal of $A$. Since $t\in L_{1/2}\subseteq J_{e}(1/2)$, $\overline{e}$ is an idempotent of quotient algebra $A/J_{e}(1/2)$.

Let’s consider now $d\in Der_{K}(A)$, $x_{1/2}\in J_{e}(1/2)$ and $a_{1/2}\in A_{e}(1/2)$. We have $0=d(x_{1/2}a_{1/2})=x_{1/2}d(a_{1/2})+d(x_{1/2})a_{1/2}$. But $d(x_{1/2})=f_{d}(x_{1/2})\in A_{1/2}$ and $d(a_{1/2})=f_{d}(a_{1/2})+2(d(e)a_{1/2})_{0}-2(d(e)a_{1/2})_{1},$ it follows that $x_{1/2}d(a_{1/2})=0$ because $x_{1/2}(d(e)a_{1/2})_{1}=x_{1/2}(d(e)a_{1/2})_{0}$. So $d(x_{1/2})a_{1/2}=0$, and $d(x_{1/2})\in J_{e}(1/2)$. We conclude that $d(J_{e}(1/2))\subseteq J_{e}(1/2)$.     $\Box$

4 Dimensionally nilpotent Lie algebras

Definition 4.1**.**

Let $A$ be a $n+1$ finite dimensional $K$-algebra . If there is a nilpotent $K$-derivation $d$ of $A$ such that $d^{n+1}=0$ and $d^{n}\neq 0$, $d$ is said to be dimensionally nilpotent, and so is the algebra $A$, though $A$ is not necessarily nilpotent.

If so it is, there is a basis $\{e_{0},e_{1},\ldots,e_{n}\}$ of $A$ such that $d(e_{i})=e_{i+1}$ $(i=0,\ldots n-1)$ and $d(e_{n})=0$ and the basis $\{e_{0},e_{1},\ldots,e_{n}\}$ is said to be adapted to $d$.

Example 4.2**.**

[9, Exemple 2.5] Let $K$ be a commutative field of characteristic $\neq 2$ and $A=G(n+1,2)$ the gametic diploid algebra with $n+1$ alleles. Its multiplication table in the natural basis $\{a_{0},\ldots,a_{n}\}$ is given by $a_{i}a_{j}=\frac{1}{2}a_{i}+\frac{1}{2}a_{j}$. We know that the mapping $\omega:A\rightarrow K,a_{i}\mapsto 1$ is a weight function and if we set $e_{i}=a_{0}-a_{i}$ $(i\neq 0)$ then $\{e_{1},\ldots,e_{n}\}$ is a basis of the ideal $N=\ker\omega$ and $e_{0}=a_{0}$ is an idempotent of $A$ such that $\{e_{0},e_{1},\ldots,e_{n}\}$ is the canonical basis of $A$, so $e_{0}e_{i}=\frac{1}{2}e_{i}$ $(i=l,\ldots,n)$ and if $d$ is a derivation of $A$, $e_{0}d(e_{i})=\frac{1}{2}d(e_{i})$ $(i=1,\ldots,n)$, because $d(e_{0})\in N$ and $N$ is a zero algebra. So we just need to define $d:A\rightarrow A$ by $d(e_{i})=e_{i+1}$ $(i=1,\ldots,n-1)$, $d(e_{0})=e_{1}$ and $d(e_{n})=0$. It follows that $d^{n+1}=0$ and $d^{n}\neq 0$, showing the gametic algebra $A=G(n+1,2)$ is dimensionally nilpotent.

Example 4.3**.**

Let $K$ be a commutative field of characteristic $\neq 2$ and $A$ the $n+1$ dimensional commutative $K$-algebra, which multiplication table in the basis $\{e_{0},e_{1},\ldots,e_{n}\}$ is given by $e_{0}e_{i}=\frac{1}{2}e_{i}$ $(i=1,\ldots,n)$, $e_{0}^{2}=e_{0}+e_{n}$, all other product being zero. If $d$ is a derivation of $A$, $e_{0}d(e_{i})=\frac{1}{2}d(e_{i})$ $(i=1,\ldots,n)$ because $d(e_{0})\in N=<e_{1},\ldots,e_{n}>$ and $N$ is a zero algebra. Here, we just need again to define $d:A\rightarrow A$ by $d(e_{i})=e_{i+1}$ $(i=1,\ldots,n-1)$, $d(e_{0})=e_{1}$ and $d(e_{n})=0$. We have $d^{n+1}=0$ and $d^{n}\neq 0$, that shows the algebra $A$ is dimensionally nilpotent. Since $Ke_{n}$ is an ideal, the quotient algebra $A/Ke_{n}$ is isomorphic to $G(n,2)$.

4.1 Basic tools

Theorem 4.4** ([12]).**

Let $K$ be a perfect field of characteristic $\neq 2$ and $3$ and $A$ finite dimensional $K$-Jordan algebra, dimensionally nilpotent. Then either $A$ is nilpotent or $\dim_{K}(A/rad(A))=1$.

Remark 4.5*.*

Let $A$ be a dimensionally nilpotent Lie triple non nilalgebra. Because of Theorem 2.3 we consider two cases :

• $1)$

$A$ has an idempotent $e$. Since the ideal $J$ is caracteristic, the quotient algebra $\overline{A}=A/J$ is a dimensionally nilpotent Jordan algebra. Because of Theorem 4.4 we have $\dim_{K}(\overline{A}/rad(\overline{A})=1$ and since $rad(\overline{A})\simeq rad(A)/J$, the first isomorphism theorem gives $A/rad(A)\simeq\overline{A}/rad(\overline{A})$ and $\dim_{K}A/rad(A)=1$. Then we can write $A=Ke\oplus N$, with $N=rad(A)$.

• $2)$

$A$ has a pseudo-idempotent $e$. Since the ideal $J_{e}(1/2)$ is caracteristic, the quotient algebra $\overline{A}=A/J_{e}(1/2)$ is a dimensionally Lie triple algebra with $\overline{e}$ as idempotent. Because of $1)$ we can write $\overline{A}=K\overline{e}\oplus\overline{N}$ with $\overline{N}=rad(\overline{A})$. So we have $A=Ke\oplus N$ with $N=rad(A)$.

Lemma 4.6**.**

Let $x$, $y\in N$ such that $x\neq 0$ and $\alpha\in K$. If $xy=\alpha y$ then $\alpha=0$ or $y=0$.

Proof.

Since $N$ is nilpotent, there is $m\in\mathbb{N}^{*}$ such that $L_{x}^{m}(y)=\alpha^{m}y=0$, $L_{x}$ being the multiplicative operator by $x$. Then $\alpha=0$ or $y=0$.     $\Box$

From now on, throughout the paper, $A$ is a dimensionally nilpotent Lie triple non nilalgebra of dimension $n+1$, with $\{e_{0},e_{1},\ldots,e_{n}\}$ an adapted basis to the derivation $d$. We can consider $e_{0}$ either, as an idempotent, or a pseudo-idempotent. In the last case, $e_{0}^{2}=e_{0}+t$, $e_{0}t=\frac{1}{2}t$ and $t^{2}=0$ implies $d(t)=0$ (Lemma 3.5), that means $t=\alpha e_{n}$ with $\alpha\in K$. Since $t\in A_{e}(1/2)$, if $\alpha\neq 0$, then $e_{n}\in A_{e}(1/2)$.

Lemma 4.7**.**

We have:

$(i)$* $e_{0}e_{n}=\lambda_{n}e_{n}$*

$(ii)$* $e_{k}e_{n}=0$ with $1\leq k\leq n$*

Proof.

Let’s write $e_{0}e_{n}=\sum^{n}_{i=0}\lambda_{i}e_{i}$. Deriving $k$ times successively, we have $e_{k}e_{n}=\sum^{n-k}_{i=0}\lambda_{i}e_{i+k}$. With $k=n$, it follows that $e_{n}^{2}=\lambda_{0}e_{n}$ and because of Lemma 4.6 we have $\lambda_{0}=0$. Set $k=n-1$, one has $e_{n-1}e_{n}=\lambda_{1}e_{n}$. That implies $\lambda_{1}=0$. And so on, we have $\lambda_{0}=\lambda_{1}=\cdots=\lambda_{n-1}=0$, $e_{0}e_{n}=\lambda_{n}e_{n}$. Deriving successively $e_{0}e_{n}$ it follows that $e_{k}e_{n}=0$ with $1\leq k\leq n$.     $\Box$

Lemma 4.8**.**

We have :

• $(i)$

$e_{0}e_{k}=\lambda_{k}e_{k}+\sum^{n}_{i=k+1}a_{k,i}e_{i}$* with $1\leq k\leq n-1$;*

• $(ii)$

$e_{i}e_{k}=\sum^{n}_{j=k+1}\gamma_{ikj}e_{j}$* with $1\leq i\leq k\leq n-1$;*

• $(iii)$

$\lambda_{k}\in\{0,\frac{1}{2},1\}$* with $1\leq k\leq n$.*

Proof.

Reason by recurrence on $n$. With $n=1$ the multiplication table of the algebra $A$ is given by $e_{0}^{2}=e_{0}$, $e_{0}e_{1}=\frac{1}{2}e_{1}$, $e_{1}^{2}=0$ and the lemma is satisfied. Assume the lemma is true until an order $n$. Because of Lemma 4.7 the subspace $I_{n+1}=Ke_{n+1}$ is a $d$-invariant ideal of $A$. The quotient algebra $A/I_{n+1}$ is dimensionally nilpotent of dimension $n+1$. By the hypothesis, we have $\overline{e}_{0}\overline{e}_{k}=\lambda_{k}\overline{e}_{k}+\sum_{i=k+1}^{n}a_{k,i}\overline{e}_{i}$ and $\overline{e}_{i}\overline{e}_{k}=\sum_{j=k+1}^{n}\gamma_{ikj}\overline{e}_{j}$, with $1\leq i\leq k\leq n$. Otherwise $e_{0}e_{k}=\lambda_{k}e_{k}+\sum_{i=k+1}^{n}a_{k,i}e_{i}+a_{k,n+1}e_{n+1}$ and $e_{i}e_{k}=\sum_{j=k+1}^{n}\gamma_{ikj}e_{j}+\gamma_{ik,n+1}e_{n+1}$; and results $(i)$ and $(ii)$ follow.

Now we just need to show $(iii)$. Since $2L_{e_{0}}^{3}-3L_{e_{0}}^{2}+L_{e_{0}}=0$, with $L_{e_{0}}$ being the multiplicative operator by $e_{0}$, applying it to $e_{k}$ we have $2\lambda_{k}^{3}-3\lambda_{k}^{2}+\lambda_{k}=0$, soit $\lambda_{k}\in\{0,\frac{1}{2},1\}$.     $\Box$

4.2 Example of low dimensions

Here we deal with cases $1\leq n\leq 4$. Let $A$ be a dimensionally nilpotent Lie triple algebra, of dimension $n+1$ and $\{e_{0},e_{1},\ldots,e_{n}\}$ be a basis adapted to $d$. We have $\ker d=Ke_{n}$. Since $e_{0}$ is an idempotent or a pseudo-idempotent, $e_{1}=d(e_{0})\in A_{e}(1/2)$, i.e $e_{0}e_{1}=\frac{1}{2}e_{1}$. Deriving this we have $e_{0}e_{2}+e_{1}^{2}=\frac{1}{2}e_{2}$, that means

[TABLE]

We also have $e_{1}^{2}=\sum_{k=2}^{n}\gamma_{11k}e_{k}$ which derivative is $2e_{1}e_{2}=\sum_{k=2}^{n-1}\gamma_{11k}e_{k+1}=\sum_{k=3}^{n}\gamma_{11,k-1}e_{k}$, that means $2\gamma_{12k}=\gamma_{11,k-1}$ $(3\leq k\leq n)$. Let’s derive for the second time $e_{0}e_{1}=\frac{1}{2}e_{1}$. We have $e_{0}e_{3}+3e_{1}e_{2}=\frac{1}{2}e_{3}$, that means

[TABLE]

However we have, $d(e_{0}e_{2})=e_{0}e_{3}+e_{1}e_{2}=\lambda_{2}e_{3}+\sum_{4}^{n}a_{2,k-1}e_{k}$, that implies

[TABLE]

So $(\ref{Eq4})$ implies $\gamma_{123}\in\{-1,-\frac{1}{2},0,\frac{1}{2},1\}$. So it is necessary to take $\lambda_{3}=\frac{1}{2}$ in $(\ref{Eq3})$. Whence $\lambda_{3}=\lambda_{2}=\frac{1}{2}$ and $\gamma_{123}=\gamma_{112}=0$ if $3\leq n$.

Deriving $e_{0}e_{3}+3e_{1}e_{2}=\frac{1}{2}e_{3}$, one has $e_{0}e_{4}+4e_{1}e_{3}+3e_{2}^{2}=\frac{1}{2}e_{4}$, that means

[TABLE]

However, $d(e_{0}e_{3})=e_{0}e_{4}+e_{1}e_{3}=\lambda_{3}e_{4}+\sum_{k=5}^{n}a_{3,k-1}e_{k}$, which implies

[TABLE]

We also have $e_{1}e_{2}=\sum_{k=4}^{n}\gamma_{12k}e_{k}$ because $\gamma_{123}=0$ and $d(e_{1}e_{2})=e_{1}e_{3}+e_{2}^{2}=\sum_{k=5}^{n}\gamma_{12,k-1}e_{k}$, which implies $\gamma_{223}=0$ and $\gamma_{134}+\gamma_{224}=0$.

Case $\dim_{K}A=2$ i.e $n=1$.

We obviously have $e_{0}e_{1}=\frac{1}{2}e_{1}$, $e_{1}^{2}=0$, $e_{0}^{2}=e_{0}$ or $e_{0}^{2}=e_{0}+e_{1}$ all other product being zero.

[TABLE]

[TABLE]

Case $\dim_{K}A=3$ i.e $n=2$.

Because of $(\ref{Eq2})$ we have $\lambda_{2}+\gamma_{112}=\frac{1}{2}$. Let’s discuss the possible values of $\lambda_{2}$.

$\ast$ $\lambda_{2}=0$ $\Rightarrow$ $\gamma_{112}=\frac{1}{2}$, so $e_{0}^{2}=e_{0}$, $e_{0}e_{1}=\frac{1}{2}e_{1}$, $e_{1}^{2}=\frac{1}{2}e_{2}$ all other product being zero.

[TABLE]

$\ast$ $\lambda_{2}=\frac{1}{2}$ $\Rightarrow$ $\gamma_{112}=0$, so $e_{0}^{2}=e_{0}$ or $e_{0}^{2}=e_{0}+e_{2}$, $e_{0}e_{1}=\frac{1}{2}e_{1}$, $e_{0}e_{2}=\frac{1}{2}e_{2}$ all other product being zero.

[TABLE]

[TABLE]

$\ast$ $\lambda_{2}=1$ $\Rightarrow$ $\gamma_{112}=-\frac{1}{2}$, so $e_{0}^{2}=e_{0}$, $e_{0}e_{1}=\frac{1}{2}e_{1}$, $e_{0}e_{2}=e_{2}$, $e_{1}^{2}=-\frac{1}{2}e_{2}$ all other product being zero.

[TABLE]

Case $\dim_{K}A=4$ i.e $n=3$.

Because of the preliminary calculations, $\lambda_{3}=\lambda_{2}=\lambda_{1}=\frac{1}{2}$, $\gamma_{112}=\gamma_{123}=0$ and $a_{2,3}+\gamma_{113}=0$. So $e_{0}e_{3}=\frac{1}{2}e_{3}$ $\Rightarrow$ $e_{3}\in A_{\frac{1}{2}}$ and $e_{1}^{2}=\gamma_{113}e_{3}$. Since $A_{\frac{1}{2}}^{2}\subseteq A_{0}+A_{1}$ we have $\gamma_{113}=0$ implying $a_{2,3}=0$ and finally $e_{0}e_{2}=\frac{1}{2}e_{2}$. So we have the following multiplication table : $e_{0}e_{1}=\frac{1}{2}e_{1}$, $e_{0}e_{2}=\frac{1}{2}e_{2}$, $e_{0}e_{3}=\frac{1}{2}e_{3}$, $e_{0}^{2}=e_{0}$ or $e_{0}^{2}=e_{0}+e_{3}$, all other product being zero.

[TABLE]

[TABLE]

Case $\dim_{K}A=5$ i.e $n=4$.

$\ast$ $\lambda_{4}=0$ $\Rightarrow$ $\gamma_{134}=\frac{1}{2}$ because of $(\ref{Eq6})$. Since $e_{1}e_{3}=\gamma_{134}e_{4}=\frac{1}{2}e_{4}$ we have $e_{3}\in A_{\frac{1}{2}}$ because $A_{\frac{1}{2}}^{2}\subseteq A_{0}+A_{1}$, $A_{\frac{1}{2}}A_{1}\subseteq A_{\frac{1}{2}}$, $A_{\frac{1}{2}}A_{0}\subseteq A_{\frac{1}{2}}$. So $e_{0}e_{3}=\frac{1}{2}e_{3}$ and $a_{3,4}=0$ $\Rightarrow$ $\gamma_{124}=0$ because of $(\ref{Eq4})$ and finally $\gamma_{113}=a_{2,3}=0$. In the same way $e_{2}^{2}=-\frac{1}{2}e_{4}$ $\Rightarrow$ $e_{2}\in A_{e}(0)$ or $e_{2}\in A_{\frac{1}{2}}$, because $A_{e}(1/2)^{2}\subseteq A_{e}(0)+A_{e}(1)$ and $A_{e}(0)^{2}\subseteq A_{e}(0)$. But $e_{0}e_{2}=\frac{1}{2}e_{2}+a_{2,4}e_{4}$ $\Rightarrow$ $e_{2}\in A_{\frac{1}{2}}$ so $a_{2,4}=0=\gamma_{114}$ because of $(\ref{Eq2})$. Whence the following multiplication table

[TABLE]

$\ast$ $\lambda_{4}=\frac{1}{2}$ $\Rightarrow$ $\gamma_{134}=\gamma_{224}=0$ because of $(\ref{Eq5})$ and $(\ref{Eq6})$. We have $e_{1}e_{2}=\gamma_{1,2,4}e_{4}\in A_{\frac{1}{2}}$ $\Rightarrow$ $e_{2}\in A_{e}(0)$ or $e_{2}\in A_{e}(1)$ this is a contradiction (because $e_{0}e_{2}=\frac{1}{2}e_{2}+a_{2,3}e_{3}$) so $\gamma_{124}=0$ and then $a_{3,4}=\gamma_{113}=0$. Whence the following multiplication table:

[TABLE]

[TABLE]

$\ast$ $\lambda_{4}=1$ $\Rightarrow$ $e_{4}\in A_{e}(1)$ $\Rightarrow$ $\gamma_{134}=-\frac{1}{2}$

We have $e_{1}e_{3}=-\frac{1}{2}e_{4}$ $\Rightarrow$ $e_{3}\in A_{\frac{1}{2}}$ $\Rightarrow$ $e_{0}e_{3}=\frac{1}{2}e_{3}$ $\Rightarrow$ $a_{3,4}=0$. So $\gamma_{124}=\gamma_{113}=a_{2,3}=0$. In the same way $e_{2}^{2}=\frac{1}{2}e_{4}$ $\Rightarrow$ $e_{2}\in A_{\frac{1}{2}}$ (because $e_{2}$ can not be in $A_{e}(1)$), $e_{0}e_{2}=\frac{1}{2}e_{2}$ $\Rightarrow$ $a_{2,4}=\gamma_{114}=0$. Whence this table

[TABLE]

4.3 Main results in general case

Theorem 4.9** (Main theorem).**

Let $A$ be dimensionally nilpotent Lie triple non nilalgebra. Let $\{e_{0},e_{1},\ldots,e_{n}\}$ be an adapted basis of $A$. Then:

$1\textdegree)$* If $n=2p+1$, the multiplication table of $A$ is one of the two following:*

• $(i)$

$e_{0}^{2}=e_{0}$, $e_{0}e_{i}=\frac{1}{2}e_{i}$ $(1\leq i\leq 2p+1)$, all other product being zero.

• $(i^{\prime})$

$e_{0}^{2}=e_{0}+e_{2p+1}$, $e_{0}e_{i}=\frac{1}{2}e_{i}$ $(1\leq i\leq 2p+1)$, all other product being zero.

$2\textdegree)$* If $n=2p$, the multiplication table of $A$ is one of the four following :*

• $(i)$

$e_{0}^{2}=e_{0}$, $e_{0}e_{i}=\frac{1}{2}e_{i}$ $(1\leq i\leq 2p)$, all other product being zero.

• $(i^{\prime})$

$e_{0}^{2}=e_{0}+e_{2p}$, $e_{0}e_{i}=\frac{1}{2}e_{i}$ $(1\leq i\leq 2p)$, all other product being zero.

• $(ii)$

$e_{0}^{2}=e_{0}$, $e_{0}e_{i}=\frac{1}{2}e_{i}$ $(1\leq i\leq 2p-1)$, $e_{0}e_{2p}=0$, $e_{i}e_{2p-i}=\frac{1}{2}(-1)^{i-1}e_{2p}$, $(1\leq i\leq p)$, all other product being zero.

• $(iii)$

$e_{0}^{2}=e_{0}$, $e_{0}e_{i}=\frac{1}{2}e_{i}$ $(1\leq i\leq 2p-1)$, $e_{0}e_{2p}=e_{2p}$, $e_{i}e_{2p-i}=\frac{1}{2}(-1)^{i}e_{2p}$ $(1\leq i\leq p)$, all other product being zero.

Proof.

Reason by recurrence on $n$. Subsection 4.2 shows that the theorem is true when $n\leq 4$. Assume it is true until an order $n>4$ and let’s show it remains true for $n+1$. Integer $n$ being either even or odd, we consider two cases :

$\mathbf{1)}$ $n=2p$ is even. The multiplication table of $A$ has the following form $e_{0}e_{k}=\frac{1}{2}e_{k}+a_{k,2p+1}e_{2p+1}$ $(1\leq k\leq 2p-1)$, $e_{0}e_{2p}=\lambda_{2p}e_{2p}+a_{2p,2p+1}e_{2p+1}$, $e_{0}e_{2p+1}=\lambda_{2p+1}e_{2p+1}$ and $e_{i}e_{2p-i}=\varepsilon_{i}e_{2p}+\gamma_{i,2p-i,2p+1}e_{2p+1}$, with $\varepsilon_{i}=0$, $\varepsilon_{i}=\frac{1}{2}(-1)^{i-1}$ or $\varepsilon_{i}=\frac{1}{2}(-1)^{i}$ $(i=1,\ldots,p)$ according to $\lambda_{2p}=\frac{1}{2}$, $\lambda_{2p}=0$ or $\lambda_{2p}=1$, respectively. We have $d(e_{i}e_{2p-i})=e_{i}e_{2p+1-i}+e_{i+1}e_{2p-i}=\varepsilon_{i}e_{2p+1}$, and then the following system

[TABLE]

We see that $e_{p}e_{p+1}=\frac{1}{2}\varepsilon_{p}e_{2p+1}$, $e_{p-1}e_{p+2}=(\varepsilon_{p-1}-\frac{1}{2}\varepsilon_{p})e_{2p+1}=\frac{3}{2}\varepsilon_{p-1}e_{2p+1}$, $e_{p-i}e_{p+1+i}=\frac{2i+1}{2}\varepsilon_{p-i}e_{2p+1}$, $e_{1}e_{2p}=\frac{2p-1}{2}\varepsilon_{1}e_{2p+1}$. But $d(e_{0}e_{2p})=e_{0}e_{2p+1}+e_{1}e_{2p}=\lambda_{2p}e_{2p+1}$, that means $e_{1}e_{2p}=(\lambda_{2p}-\lambda_{2p+1})e_{2p+1}$, and also $\lambda_{2p}-\lambda_{2p+1}=\frac{2p-1}{2}\varepsilon_{1}$.

• If

$\lambda_{2p}=0$, we have $\varepsilon_{1}=\frac{1}{2}$ and $\lambda_{2p+1}=-\frac{2p-1}{4}\not\in\{0,\frac{1}{2},1\}$, impossible.

• If

$\lambda_{2p}=1$, we have $\varepsilon_{1}=-\frac{1}{2}$ and $\lambda_{2p+1}=1+\frac{2p-1}{4}=\frac{2p+3}{4}\not\in\{0,\frac{1}{2},1\}$, impossible.

• Hence

$\lambda_{2p}=\frac{1}{2}$, $\varepsilon_{1}=0$ and $\lambda_{2p+1}=\frac{1}{2}$.

Since all the $\lambda_{k}$ are equal to $\frac{1}{2}$ ($k\neq 2p+1$), applying $2L_{e_{0}}^{3}-3L_{e_{0}}^{2}+L_{e_{0}}=0$ to $e_{k}$, it follows that $2a_{k,2p+1}\lambda_{2p+1}(\lambda_{2p+1}-1)=0$. If $\lambda_{2p+1}=\frac{1}{2}$, then we have $a_{k,2p+1}=0$, which means $e_{0}e_{k}=\frac{1}{2}e_{k}$ for all $k$. Hence $e_{i}e_{j}=0$, with $1\leq i\leq j\leq n$.

$\mathbf{2)}$ $n=2p-1$ is odd. The multiplication table of $A$ has the following form $e_{0}e_{k}=\frac{1}{2}e_{k}+a_{k,2p}e_{2p}$ ($k=1,\ldots,2p-1$), $e_{0}e_{2p}=\lambda_{2p}e_{2p}$ and $e_{i}e_{2p-1-i}=\gamma_{i,2p-1-i,2p}e_{2p}$ ($i=1,\ldots,p-1$). Deriving this last relation, we have $e_{i}e_{2p-i}+e_{i+1}e_{2p-1-i}=0$, and the following system

[TABLE]

So $e_{1}e_{2p-1}=-e_{2}e_{2p-2}=e_{3}e_{2p-3}=\cdots=(-1)^{i-1}e_{i}e_{2p-i}=\cdots=(-1)^{p-1}e^{2}_{p}$, that means $e_{i}e_{2p-i}=(-1)^{i-1}e_{1}e_{2p-1}$. However, since $d(e_{0}e_{2p-1})=e_{0}e_{2p}+e_{1}e_{2p-1}=\frac{1}{2}e_{2p}$, we have $e_{1}e_{2p-1}=(\frac{1}{2}-\lambda_{2p})e_{2p}$. Because $\lambda_{2p}\in\{0,\frac{1}{2},1\}$, we consider three situations :

• If

$\lambda_{2p}=0$, we have $e_{1}e_{2p-1}=\frac{1}{2}e_{2p}$, $e_{i}e_{2p-i}=\frac{1}{2}(-1)^{i-1}e_{2p}$ $(i=1,\ldots,p)$ and $e_{0}^{2}=e_{0}$.

• If

$\lambda_{2p}=1$, we have $e_{1}e_{2p-1}=-\frac{1}{2}e_{2p}$, $e_{i}e_{2p-i}=\frac{1}{2}(-1)^{i}e_{2p}$ $(i=1,\ldots,p)$ and $e_{0}^{2}=e_{0}$.

• If

$\lambda_{2p}=\frac{1}{2}$, we have $e_{1}e_{2p-1}=e_{i}e_{2p-i}=0$ $(i=1\ldots p)$ and $2a_{i,2p}\lambda_{2p}(\lambda_{2p}-1)=0$ shows that $a_{i,2p-i}=0$. Hence $e_{0}e_{i}=\frac{1}{2}e_{i}$ $(i=1,\ldots,p)$. Furthermore we have, either $e_{0}^{2}=e_{0}$, or $e_{0}^{2}=e_{0}+e_{2p}$.

For cases $\lambda_{2p}=0$ and $\lambda_{2p}=1$, We just need to show $e_{i}e_{j}=0$ for $i+j<2p$. The following lemma completes the proof of the theorem. And Note 4.12 shows that all algebras defined in this theorem are Lie triple.     $\Box$

Lemma 4.10**.**

$e_{0}e_{i}=\frac{1}{2}e_{i}$* for $i=1,\ldots,n-1$.*

Proof.

One has $e_{i}e_{2k-i}=\gamma_{i,2k-i,n}e_{n}$ for $i=1,\ldots,k-1$. Deriving this we have $e_{i}e_{2k-i+1}+e_{i+1}e_{2k-i}=0$. By varying $i$ we have the following system

[TABLE]

Going up the lines of this system we see that $e_{i}e_{2k+1-i}=0$ for $i=1,\ldots,k$ in particular $e_{1}e_{2k}=0$, so $e_{0}e_{2k+1}+e_{1}e_{2k}=\frac{1}{2}e_{2k+1}$ and $e_{0}e_{2k+1}=\frac{1}{2}e_{2k+1}$, $k=0,\ldots,p-1$.

Now we make a recurrence on $n$. Assume it is true until an order $n$. We distinguish two cases :

• •

$n+1=2p+1$ is odd. We have $e_{0}e_{n+1}=\frac{1}{2}e_{n+1}$, which imposes $e_{0}e_{n}=\frac{1}{2}e_{n}$.

• •

$n+1=2p$ is even. Since $n=2p-1$ is odd, we have $e_{0}e_{n}=\frac{1}{2}e_{n}$.

$\Box$

Since every commutative Jordan algebra is a Lie triple algebra, we have the following result:

Corollary 4.11**.**

Let $A$ be a commutative Jordan non nilalgebra, dimensionally nilpotent. Let $\{e_{0},e_{1},\ldots,e_{n}\}$ be an adapted basis of $A$. Then:

$1\textdegree)$* If $n=2p+1$, the multiplication table of $A$ is:*

$e_{0}^{2}=e_{0}$, $e_{0}e_{i}=\frac{1}{2}e_{i}$ $(1\leq i\leq 2p+1)$, all other product being zero.

$2\textdegree)$* If $n=2p$, the multiplication table of $A$ is one of the following three tables :*

• $(i)$

$e_{0}^{2}=e_{0}$, $e_{0}e_{i}=\frac{1}{2}e_{i}$ $(1\leq i\leq 2p)$, all other product being zero.

• $(ii)$

$e_{0}^{2}=e_{0}$, $e_{0}e_{i}=\frac{1}{2}e_{i}$ $(1\leq i\leq 2p-1)$, $e_{0}e_{2p}=0$, $e_{i}e_{2p-i}=\frac{1}{2}(-1)^{i-1}e_{2p}$, $(1\leq i\leq p)$, all other product being zero.

• $(iii)$

$e_{0}^{2}=e_{0}$, $e_{0}e_{i}=\frac{1}{2}e_{i}$ $(1\leq i\leq 2p-1)$, $e_{0}e_{2p}=e_{2p}$, $e_{i}e_{2p-i}=\frac{1}{2}(-1)^{i}e_{2p}$ $(1\leq i\leq p)$, all other product being zero.

Proof.

The Corollary follows from Theorem 4.9 knowing that Jordan algebras do not admit pseudo-idempotent.     $\Box$

Note 4.12*.*

1. Multiplication tables in Theorem 4.9 $1)~{}(i)$ and in Corollary 4.11 $1)$ when $n=2p+1$ is odd are those of gametic algebras $G(2p+2,2)$ (Example 4.2). It is the same for those defined in Theorem 4.9 $2)~{}(i)$ and in Corollary 4.11 $2)~{}(i)$ when $n=2p$ is even. These are gametic algebras $G(2p+1,2)$. They are caracterized as elementary train algebras with equation $x^{2}-\omega(x)x=0$, in which $\omega:A\rightarrow K$, $e_{0}\mapsto 1$, $e_{i}\mapsto 0$ is a homomorphism of algebras.

2. Multiplication tables in Theorem 4.9 $2)~{}(ii)$ and in Corollary 4.11 $2)~{}(ii)$, when $n=2p$ is even, are those of normal Bernstein algebras of type $(2p,1)$. Normal Bernstein algebras are defined by equation $x^{2}y=\omega(x)xy$. These are Bernstein-Jordan algebras, caracterized by the train equation $x^{3}-\omega(x)x^{2}=0$ [13, 15].

3. Multiplication tables in Theorem 4.9 $2)~{}(iii)$ and in Corollary 4.11 $2)~{}(iii)$, when $n=2p$ is even, are those of the other class of train algebras of rank $3$ which are Jordan algebras of type $(2p,1)$. They are defined by equation $x^{3}-2\omega(x)x^{2}+\omega(x)^{2}x=0$ [13, Theorem 2.1].

4. Multiplication tables in Theorem 4.9 $1)~{}(i^{\prime})$ and $2)~{}(i^{\prime})$ are those of train algebras satisfying $x^{3}-\frac{3}{2}\omega(x)x^{2}+\frac{1}{2}\omega(x)^{2}x=0$. These are Lie triple algebras because of [2, Proposition 5.2, (iii)].