Hom-约当三系的表示与扩张 Representation and Extension of Hom-Jordan Triple System

doi:10.12677/AAM.2020.97129

Advances in Applied Mathematics
Vol. 09 No. 07 ( 2020 ), Article ID: 36753 , 11 pages
10.12677/AAM.2020.97129

Representation and Extension of Hom-Jordan Triple System

Lele Guo

●How to Cite this Article

School of Mathematics, Liaoning Normal University, Dalian Liaoning

Received: Jul. 5^th, 2020; accepted: Jul. 21^st, 2020; published: Jul. 28^th, 2020

ABSTRACT

This paper mainly studies the representation and extension of Hom-Jordan triple system. Firstly, we give the definition of representation of Hom-Jordan triple system, the method of judging the representation and an example of representation. At the same time, the condition that the dual of Hom-Jordan triple system is the representation is given. Secondly, two methods of extension of Hom-Jordan triple system are found. One is to construct an extension by a special linear function, and the other is the T^*-extension of Hom-Jordan triple system.

Keywords:Hom-Jordan Triple System, Representation, T^*-Extension

Hom-约当三系的表示与扩张

郭乐乐

辽宁师范大学数学学院，辽宁大连

收稿日期：2020年7月5日；录用日期：2020年7月21日；发布日期：2020年7月28日

摘要

本文主要介绍了Hom-约当三系的表示与扩张。首先，给出了Hom-约当三系表示的定义、判断表示的方法以及表示的例子，同时给出Hom-约当三系的表示的对偶是表示所满足的条件。其次，找到Hom-约当三系扩张的两种方法，其一是利用一个特殊的线性函数而构造出的扩张，其二是Hom-约当三系的T^*-扩张。

关键词 :Hom-约当三系，表示，T^*-扩张

This work is licensed under the Creative Commons Attribution International License (CC BY 4.0).

http://creativecommons.org/licenses/by/4.0/

1. 引言

约当代数和约当三系与对称流形有密切联系，实的紧的约当三系与对称R-空间之间存在一一对应关系 [1]。此外，约当三系与结合代数也有密切联系。在结合代数 $(A, \circ)$ 上定义运算 $a \circ (b \circ c) + b \circ (a \circ c) + (a \circ b) \circ c$ 可以得到约当三系 [2]。约当三系作为独立的代数体系已经有很多相关的结果，例如约当三系的分类 [3]、约当三系的表示 [4]、约当三系的上同调 [5] 等。Hom-约当三系是 [6] 中引入的，是约当三系的推广。本文将考虑Hom-约当三系的表示和扩张。

2. Hom-约当三系的表示

本文所指的线性空间都是数域K上的有限维线性空间。

定义1.1 [5] 设J是线性空间， $α_{1}, α_{2} \in E n d (J)$ ， ${\cdot, \cdot, \cdot} : J \times J \times J \to J$ 是三线性映射，如果对于任意的 $x, y, z, u, v \in J$ 满足下列条件

${x, y, z} = {z, y, x},$ (1.1)

$\begin{array}{l} {α_{1} (x), α_{2} (y), {z, u, v}} - {α_{1} (z), α_{2} (u), {x, y, v}} \\ = {{x, y, z}, α_{1} (u), α_{2} (v)} - {α_{1} (z), {y, x, u}, α_{2} (v)}, \end{array}$ (1.2)

则称 $(J, {\cdot, \cdot, \cdot}, α_{1}, α_{2})$ 为Hom-约当三系。

当 $α_{1} = i d, α_{2} = i d$ 时， $(J, {\cdot, \cdot, \cdot}, α_{1}, α_{2})$ 为约当三系。

定义1.2设 $(J, {\cdot, \cdot, \cdot}, α_{1}, α_{2})$ 为Hom-约当三系，V是线性空间， $A_{1}, A_{2} \in E n d (V)$ ， $θ_{12}, θ_{13} : J \times J \to E n d (V)$ 为双线性映射，如果对于任意的 $x, y, z, u \in J$ 有

$θ_{13} (x, y) = θ_{13} (y, x),$ (1.3)

$\begin{array}{l} θ_{12} (α_{2} (x), α_{1} (u)) θ_{12} (z, y) - θ_{13} (α_{1} (z), α_{2} (x)) θ_{13} (y, u) \\ - θ_{12} ({z, u, x}, α_{2} (y)) A_{1} + θ_{12} (α_{1} (z), α_{2} (u)) θ_{12} (x, y) = 0, \end{array}$ (1.4)

$\begin{array}{l} θ_{12} (α_{2} (y), α_{1} (u)) θ_{13} (x, z) - θ_{13} (α_{1} (z), α_{2} (y)) θ_{12} (u, x) \\ - θ_{13} (α_{1} (x), {z, u, y}) A_{2} + θ_{12} (α_{1} (z), α_{2} (u)) θ_{13} (x, y) = 0, \end{array}$ (1.5)

$\begin{array}{l} θ_{12} (α_{2} (z), α_{1} (u)) θ_{12} (x, y) - θ_{12} (α_{2} (z), {y, x, u}) A_{1} \\ - θ_{12} (α_{1} (x), α_{2} (y)) θ_{12} (z, u) + θ_{12} ({x, y, z}, α_{2} (u)) A_{1} = 0, \end{array}$ (1.6)

$\begin{array}{l} θ_{13} ({x, y, z}, α_{2} (u)) A_{1} - θ_{13} (α_{1} (z), α_{2} (u)) θ_{12} (y, x) \\ - θ_{12} (α_{1} (x), α_{2} (y)) θ_{13} (z, u) + θ_{13} (α_{1} (z), {x, y, u}) A_{2} = 0, \end{array}$ (1.7)

$\begin{array}{l} θ_{12} ({x, y, z}, α_{1} (u)) A_{2} - θ_{12} (α_{1} (z), {y, x, u}) A_{2} \\ - θ_{12} (α_{1} (x), α_{2} (y)) θ_{12} (z, u) + θ_{12} (α_{1} (z), α_{2} (u)) θ_{12} (x, y) = 0, \end{array}$ (1.8)

则称 $(V, θ_{12}, θ_{13}, A_{1}, A_{2})$ 是 $(J, {\cdot, \cdot, \cdot}, α_{1}, α_{2})$ 的表示。

例1.1设 $(J, {\cdot, \cdot, \cdot}, α_{1}, α_{2})$ 为Hom-约当三系， $L : J \times J \to E n d (J)$ ， $M : J \times J \to E n d (J)$ 为双线性映射，其中 $L (x, y) (z) = {x, y, z}$ ， $M (x, y) (z) = {x, z, y}$ $(\forall x, y, z \in J)$ ，则 $(J, L, M, α_{1}, α_{2})$ 为 $(J, {\cdot, \cdot, \cdot}, α_{1}, α_{2})$ 的表示，称为伴随表示。

证明：直接验证可知对于 $(J, L, M, α_{1}, α_{2})$ 有(1.3)~(1.8)成立。

定理1.1设 $(J, {\cdot, \cdot, \cdot}, α_{1}, α_{2})$ 为Hom-约当三系，V是线性空间， $A_{1}, A_{2} \in E n d (V)$ ， $θ_{12}, θ_{13} : J \times J \to E n d (V)$ 为双线性映射，在 $J \oplus V$ 上定义

${x + a, y + b, z + c} = {x, y, z} + θ_{12} (z, y) a + θ_{13} (x, z) b + θ_{12} (x, y) c,$

$(α_{1} + A_{1}) (x + a) = α_{1} (x) + A_{1} (a),$

$(α_{2} + A_{2}) (x + a) = α_{2} (x) + A_{2} (a),$

其中 $\forall x, y, z \in J, \forall a, b, c \in V$ ，则 $(J \oplus V ， {\cdot, \cdot, \cdot}, α_{1} + A_{1}, α_{2} + A_{2})$ 为Hom-约当三系当且仅当 $(V, θ_{12}, θ_{13}, A_{1}, A_{2})$ 是 $(J, {\cdot, \cdot, \cdot}, α_{1}, α_{2})$ 的表示。

证明: $(J \oplus V, {\cdot, \cdot, \cdot}, α_{1} + A_{1}, α_{2} + A_{2})$ 为Hom-约当三系当且仅当下列等式成立

${x + a, y + b, z + c} = {z + c, y + b, x + a},$ (1.9)

$\begin{array}{l} {{x + a, y + b, z + c}, (α_{1} + A_{1}) (u + d), (α_{2} + A_{2}) (v + e)} \\ - {(α_{1} + A_{1}) (z + c), {y + b, x + a, u + d}, (α_{2} + A_{2}) (v + e)} \\ - {(α_{1} + A_{1}) (x + a), (α_{2} + A_{2}) (y + b), {z + c, u + d, v + e}} \\ + {(α_{1} + A_{1}) (z + c), (α_{2} + A_{2}) (u + d), {x + a, y + b, v + e}} = 0, \end{array}$ (1.10)

其中 $\forall x + a, y + b, z + c, u + d, v + e \in J \oplus V$ 。

由于

$\begin{array}{l} {x + a, y + b, z + c} - {z + c, y + b, x + a} \\ = {x, y, z} - {z, y, x} + θ_{12} (z, y) a - θ_{12} (z, y) a + θ_{13} (x, z) b \\ - θ_{13} (z, x) b + θ_{12} (x, y) c - θ_{12} (x, y) c, \end{array}$

因此(1.9)成立当且仅当 $θ_{13}$ 对称，即(1.3)式成立。

(1.10)式左边直接计算得

$\begin{array}{l} = {{x, y, z}, α_{1} (u), α_{2} (v)} - {α_{1} (z), {y, x, u}, α_{2} (v)} \\ - {α_{1} (x), α_{2} (y), {z, u, v}} + {α_{1} (z), α_{2} (u), {x, y, v}} \\ + (θ_{12} (α_{2} (v), α_{1} (u)) θ_{12} (z, y) - θ_{13} (α_{1} (z), α_{2} (v)) θ_{13} (y, u) \\ - θ_{12} ({z, u, v}, α_{2} (y)) A_{1} + θ_{12} (α_{1} (z), α_{2} (u)) θ_{12} (v, y)) (a) \\ + (θ_{12} (α_{2} (v), α_{1} (u)) θ_{13} (x, z) - θ_{13} (α_{1} (z), α_{2} (v)) θ_{12} (u, x) \end{array}$

$\begin{array}{l} - θ_{13} (α_{1} (x), {z, u, v}) A_{2} + θ_{12} (α_{1} (z), α_{2} (u)) θ_{13} (x, v)) (b) \\ + (θ_{12} (α_{2} (v), α_{1} (u)) θ_{12} (x, y) - θ_{12} (α_{2} (v), {y, x, u}) A_{1} \\ - θ_{12} (α_{1} (x), α_{2} (y)) θ_{12} (v, u) + θ_{12} ({x, y, v}, α_{2} (u)) A_{1}) (c) \\ + (θ_{13} ({x, y, z}, α_{2} (v)) A_{1} - θ_{13} (α_{1} (z), α_{2} (v)) θ_{12} (y, x) \end{array}$

$\begin{array}{l} - θ_{12} (α_{1} (x), α_{2} (y)) θ_{13} (z, v) + θ_{13} (α_{1} (z), {x, y, v}) A_{2}) (d) \\ + (θ_{12} ({x, y, z}, α_{1} (u)) A_{2} - θ_{12} (α_{1} (z), {y, x, u}) A_{2} \\ - θ_{12} (α_{1} (x), α_{2} (y)) θ_{12} (z, u) + θ_{12} (α_{1} (z), α_{2} (u)) θ_{12} (x, y)) ( e ) \end{array}$

由 $(J, {\cdot, \cdot, \cdot}, α_{1}, α_{2})$ 为Hom-约当三系，则等式(1.2)成立。因此，(1.10)式成立当且仅当(1.4)~(1.8)成立。因此，结论成立。

设 $V, W$ 为线性空间， $φ : V \to W$ 为线性映射，定义 $φ^{*} : W^{*} \to V^{*}$ ，其中

$〈 φ^{*} (f), v 〉 = 〈 f, φ (v) 〉 (\forall f \in W^{*}, v \in V),$

则 $φ^{*}$ 为线性映射，称为 $φ$ 的对偶映射。

设 $(J, {\cdot, \cdot, \cdot}, α_{1}, α_{2})$ 为Hom-约当三系，V为线性空间， $θ : J \times J \to E n d (V)$ 为线性映射，定义 $θ^{*} : J \times J \to E n d (V^{*})$ ，其中

$〈 θ^{*} (x, y) f, v 〉 = 〈 f, θ (x, y) v 〉 (\forall x, y \in J, v \in V, f \in V^{*}),$

则 $θ^{*}$ 为线性映射。

定理1.2设 $(J, {\cdot, \cdot, \cdot}, α_{1}, α_{2})$ 为Hom-约当三系， $(V, θ_{12}, θ_{13}, A_{1}, A_{2})$ 是 $(J, {\cdot, \cdot, \cdot}, α_{1}, α_{2})$ 的表示，则 $(V^{*}, θ_{12}^{*}, θ_{13}^{*}, A_{1}^{*}, A_{2}^{*})$ 也是 $(J, {\cdot, \cdot, \cdot}, α_{1}, α_{2})$ 的表示当且仅当满足下列条件

$\begin{array}{l} θ_{12} (z, y) θ_{12} (α_{2} (x), α_{1} (u)) - θ_{13} (y, u) θ_{13} (α_{1} (z), α_{2} (x)) \\ - A_{1} θ_{12} ({z, u, x}, α_{2} (y)) + θ_{12} (x, y) θ_{12} (α_{1} (z), α_{2} (u)) = 0, \end{array}$ (1.11)

$\begin{array}{l} θ_{13} (x, z) θ_{12} (α_{2} (y), α_{1} (u)) - θ_{12} (u, x) θ_{13} (α_{1} (z), α_{2} (y)) \\ - A_{2} θ_{13} (α_{1} (x), {z, u, y}) + θ_{13} (x, y) θ_{12} (α_{1} (z), α_{2} (u)) = 0, \end{array}$ (1.12)

$\begin{array}{l} θ_{12} (x, y) θ_{12} (α_{2} (z), α_{1} (u)) - A_{1} θ_{12} (α_{2} (z), {y, x, u}) \\ - θ_{12} (z, u) θ_{12} (α_{1} (x), α_{2} (y)) + A_{1} θ_{12} ({x, y, z}, α_{2} (u)) = 0, \end{array}$ (1.13)

$\begin{array}{l} A_{1} θ_{13} ({x, y, z}, α_{2} (u)) - θ_{12} (y, x) θ_{13} (α_{1} (z), α_{2} (u)) \\ - θ_{13} (z, u) θ_{12} (α_{1} (x), α_{2} (y)) + A_{2} θ_{13} (α_{1} (z), {x, y, u}) = 0, \end{array}$ (1.14)

$\begin{array}{l} A_{2} θ_{12} ({x, y, z}, α_{1} (u)) - A_{2} θ_{12} (α_{1} (z), {y, x, u}) \\ - θ_{12} (z, u) θ_{12} (α_{1} (x), α_{2} (y)) + θ_{12} (x, y) θ_{12} (α_{1} (z), α_{2} (u)) = 0, \end{array}$ (1.15)

其中 $\forall x, y, z, u \in J$ 。

证明； $(V^{*}, θ_{12}^{*}, θ_{13}^{*}, A_{1}^{*}, A_{2}^{*})$ 为 $(J, {\cdot, \cdot, \cdot}, α_{1}, α_{2})$ 的表示当且仅当 $(V^{*}, θ_{12}^{*}, θ_{13}^{*}, A_{1}^{*}, A_{2}^{*})$ 满足等式(1.3)~(1.8)。 $\forall x, y, z, u \in J, f \in V^{*}, v \in V$ ，直接计算得

$〈 [θ_{13}^{*} (x, y) - θ_{13}^{*} (y, x)] f, v 〉 = 〈 f, [θ_{13} (x, y) - θ_{13} (y, x)] v 〉,$

$\begin{array}{l} 〈 [θ_{12}^{*} (α_{2} (x), α_{1} (u)) θ_{12}^{*} (z, y) - θ_{13}^{*} (α_{1} (z), α_{2} (x)) θ_{13}^{*} (y, u) \\ - θ_{12}^{*} ({z, u, x}, α_{2} (y)) A_{1}^{*} + θ_{12}^{*} (α_{1} (z), α_{2} (u)) θ_{12}^{*} (x, y)] f, v 〉 \\ = 〈 f, [θ_{12} (z, y) θ_{12} (α_{2} (x), α_{1} (u)) - θ_{13} (y, u) θ_{13} (α_{1} (z), α_{2} (x)) \\ - A_{1} θ_{12} ({z, u, x}, α_{2} (y)) + θ_{12} (x, y) θ_{12} (α_{1} (z), α_{2} (u))] v 〉, \end{array}$

$\begin{array}{l} 〈 [θ_{12}^{*} (α_{2} (y), α_{1} (u)) θ_{13}^{*} (x, z) - θ_{13}^{*} (α_{1} (z), α_{2} (y)) θ_{12}^{*} (u, x) \\ - (α_{1} (x), {z, u, y}) A_{2}^{*} + θ_{12}^{*} (α_{1} (z), α_{2} (u)) θ_{13}^{*} (x, y)] f, v 〉 \\ = 〈 f, [θ_{13} (x, z) θ_{12} (α_{2} (y), α_{1} (u)) - θ_{12} (u, x) θ_{13} (α_{1} (z), α_{2} (y)) \\ - A_{2} (α_{1} (x), {z, u, y}) + θ_{13} (x, y) θ_{12} (α_{1} (z), α_{2} (u))] v 〉, \end{array}$

$\begin{array}{l} 〈 [θ_{12}^{*} (α_{2} (z), α_{1} (u)) θ_{12}^{*} (x, y) - θ_{12}^{*} (α_{2} (z), {y, x, u}) A_{1}^{*} \\ - θ_{12}^{*} (α_{1} (x), α_{2} (y)) θ_{12}^{*} (z, u) + θ_{12}^{*} ({x, y, z}, α_{2} (u)) A_{1}^{*}] f, v 〉 \\ = 〈 f, [θ_{12} (x, y) θ_{12} (α_{2} (z), α_{1} (u)) - A_{1} θ_{12} (α_{2} (z), {y, x, u}) \\ - θ_{12} (z, u) θ_{12} (α_{1} (x), α_{2} (y)) + A_{1} θ_{12} ({x, y, z}, α_{2} (u))] v 〉, \end{array}$

$\begin{array}{l} 〈 [θ_{13}^{*} ({x, y, z}, α_{2} (u)) A_{1}^{*} - θ_{13}^{*} (α_{1} (z), α_{2} (u)) θ_{12}^{*} (y, x) \\ - θ_{12}^{*} (α_{1} (x), α_{2} (y)) θ_{13}^{*} (z, u) + θ_{13}^{*} (α_{1} (z), {x, y, u}) A_{2}^{*}] f, v 〉 \\ = 〈 f, [A_{1} θ_{13} ({x, y, z}, α_{2} (u)) - θ_{12} (y, x) θ_{13} (α_{1} (z), α_{2} (u)) \\ - θ_{13} (z, u) θ_{12} (α_{1} (x), α_{2} (y)) + A_{2} θ_{13} (α_{1} (z), {x, y, u})] v 〉, \end{array}$

$\begin{array}{l} 〈 [θ_{12}^{*} ({x, y, z}, α_{1} (u)) A_{2}^{*} - θ_{12}^{*} (α_{1} (z), {y, x, u}) A_{2}^{*} \\ - θ_{12}^{*} (α_{1} (x), α_{2} (y)) θ_{12}^{*} (z, u) + θ_{12}^{*} (α_{1} (z), α_{2} (u)) θ_{12}^{*} (x, y)] f, v 〉 \\ = 〈 f, [A_{2} θ_{12} ({x, y, z}, α_{1} (u)) - A_{2} θ_{12} (α_{1} (z), {y, x, u}) \\ - θ_{12} (z, u) θ_{12} (α_{1} (x), α_{2} (y)) + θ_{12} (x, y) θ_{12} (α_{1} (z), α_{2} (u))] v 〉, \end{array}$

由 $(V, θ_{12}, θ_{13}, A_{1}, A_{2})$ 是 $(J, {\cdot, \cdot, \cdot}, α_{1}, α_{2})$ 的表示可知 $θ_{13} (x, y) = θ_{13} (y, x)$ ，则 $(V^{*}, θ_{12}^{*}, θ_{13}^{*}, A_{1}^{*}, A_{2}^{*})$ 上等式(1.3)~(1.8)成立等价于等式(1.11)~(1.15)成立。因此结论成立。

推论1.1设 $(J, {\cdot, \cdot, \cdot}, α_{1}, α_{2})$ 为Hom-约当三系，则 $(J^{*}, L^{*}, M^{*}, α_{1}^{*}, α_{2}^{*})$ 为 $(J, {\cdot, \cdot, \cdot}, α_{1}, α_{2})$ 的表示当且仅当J上的运算 ${\cdot, \cdot, \cdot}$ 满足下列条件

$\begin{array}{l} {z, y, {α_{2} (x), α_{1} (u), v}} - {y, {α_{1} (z), v, α_{2} (x)}, u} \\ - α_{1} ({{z, u, x}, α_{2} (y), v}) + {x, y, {α_{1} (z), α_{2} (u), v}} = 0, \end{array}$ (1.16)

$\begin{array}{l} {x, {α_{2} (y), α_{1} (u), v}, z} - {u, x, {α_{1} (z), v, α_{2} (y)}} \\ - α_{2} ({α_{1} (x), v, {z, u, y}}) + {x, {α_{1} (z), α_{2} (u), v}, y} = 0, \end{array}$ (1.17)

$\begin{array}{l} {x, y, {α_{2} (z), α_{1} (u), v}} - α_{1} ({α_{2} (z), {y, x, u}, v}) \\ - {z, u, {α_{1} (x), α_{2} (y), v}} + α_{1} ({x, y, z}, α_{2} (u), v) = 0, \end{array}$ (1.18)

$\begin{array}{l} α_{1} ({x, y, z}, v, α_{2} (u)) - {y, x, {α_{1} (z), v, α_{2} (u)}} \\ - {z, {α_{1} (x), α_{2} (y), v}, u} + α_{2} ({α_{1} (z), v, {x, y, u}}) = 0, \end{array}$ (1.19)

$\begin{array}{l} α_{2} ({x, y, z}, α_{1} (u), v) - α_{2} ({α_{1} (z), {y, x, u}, v}) \\ - {z, u, {α_{1} (x), α_{2} (y), v}} + {x, y, {α_{1} (z), α_{2} (u), v}} = 0, \end{array}$ (1.20)

其中 $\forall x, y, z, u, v \in J$ 。

证明：利用定理1.2，取 $θ_{12} = L, θ_{13} = M$ ，通过计算可直接得出。

设 $(J, {\cdot, \cdot, \cdot}, α_{1}, α_{2})$ 为Hom-约当三系， $V, W$ 为线性空间， $θ^{'} : J \times J \to E n d (V)$ ， $θ^{″} : J \times J \to E n d (W)$ 为双线性映射， $A^{'} \in E n d (V)$ ， $A^{″} \in E n d (W)$ ，定义 $θ^{'} \otimes θ^{″} : J \times J \to E n d (V \otimes W)$ ， $A^{'} \otimes A^{″} \in E n d (V \otimes W)$ ，其中

$θ^{'} \otimes θ^{″} (x, y) (v \otimes w) = θ^{'} (x, y) v \otimes w + v \otimes θ^{″} (x, y) w,$

$A^{'} \otimes A^{″} (v \otimes w) = A^{'} (v) \otimes A^{″} (w),$

其中 $\forall x, y \in J, v \in V, w \in W$ 。

定理1.3设 $(J, {\cdot, \cdot, \cdot}, α_{1}, α_{2})$ 为Hom-约当三系， $(V, {θ^{'}}_{12}, {θ^{'}}_{13}, {A^{'}}_{1}, {A^{'}}_{2}), (W, {θ^{″}}_{12}, {θ^{″}}_{13}, {A^{″}}_{1}, {A^{″}}_{2})$ 均为 $(J, {\cdot, \cdot, \cdot}, α_{1}, α_{2})$ 的表示，则 $(V \otimes W, {θ^{'}}_{12} \otimes {θ^{″}}_{12}, {θ^{'}}_{13} \otimes {θ^{″}}_{13}, {A^{'}}_{1} \otimes {A^{″}}_{1}, {A^{'}}_{2} \otimes {A^{″}}_{2})$ 是 $(J, {\cdot, \cdot, \cdot}, α_{1}, α_{2})$ 的表示当且仅当 $(V, {θ^{'}}_{12}, {θ^{'}}_{13}, {A^{'}}_{1}, {A^{'}}_{2}), (W, {θ^{″}}_{12}, {θ^{″}}_{13}, {A^{″}}_{1}, {A^{″}}_{2})$ 满足下列条件

$\begin{array}{l} {θ^{'}}_{12} ({z, u, x}, α_{2} (y)) {A^{'}}_{1} (v) \otimes w - {θ^{'}}_{12} ({z, u, x}, α_{2} (y)) {A^{'}}_{1} (v) \otimes {A^{″}}_{1} (w) \\ + v \otimes {θ^{″}}_{12} ({z, u, x}, α_{2} (y)) {A^{″}}_{1} (w) - {A^{'}}_{1} (v) \otimes {θ^{″}}_{12} ({z, u, x}, α_{2} (y)) {A^{″}}_{1} (w) \\ + {θ^{'}}_{12} (α_{2} (x), α_{1} (u)) (v) \otimes {θ^{″}}_{12} (z, y) (w) + {θ^{'}}_{12} (z, y) (v) \otimes {θ^{″}}_{12} (α_{2} (x), α_{1} (u)) (w) \\ - {θ^{'}}_{13} (y, u) (v) \otimes {θ^{″}}_{13} (α_{1} (z), α_{2} (x)) (w) - {θ^{'}}_{13} (α_{1} (z), α_{2} (x)) (v) \otimes {θ^{″}}_{13} (y, u) (w) \\ + {θ^{'}}_{12} (x, y) (v) \otimes {θ^{″}}_{12} (α_{1} (z), α_{2} (u)) (w) + {θ^{'}}_{12} (α_{1} (z), α_{2} (u)) (v) \otimes {θ^{″}}_{13} (x, y) (w) = 0, \end{array}$ (1.21)

$\begin{array}{l} {θ^{'}}_{13} (α_{1} (x), {z, u, y}) {A^{'}}_{2} (v) \otimes w - {θ^{'}}_{13} (α_{1} (x), {z, u, y}) {A^{'}}_{2} (v) \otimes {A^{″}}_{2} (w) \\ + v \otimes {θ^{″}}_{13} (α_{1} (x), {z, u, y}) {A^{″}}_{2} (w) - {A^{'}}_{2} (v) \otimes {θ^{″}}_{13} (α_{1} (x), {z, u, y}) {A^{″}}_{2} (w) \\ + v \otimes {θ^{″}}_{13} (α_{1} (x), {z, u, y}) {A^{″}}_{2} (w) - {A^{'}}_{2} (v) \otimes {θ^{″}}_{13} (α_{1} (x), {z, u, y}) {A^{″}}_{2} (w) \\ + {θ^{'}}_{13} (x, z) (v) \otimes {θ^{″}}_{12} (α_{2} (y), α_{1} (u)) (w) + {θ^{'}}_{12} (α_{2} (y), α_{1} (u)) (v) \otimes {θ^{″}}_{13} (x, z) (w) \\ + {θ^{'}}_{13} (x, y) (v) \otimes {θ^{″}}_{12} (α_{1} (z), α_{2} (u)) (w) + {θ^{'}}_{12} (α_{1} (z), α_{2} (u)) (v) \otimes {θ^{″}}_{13} (x, y) (w) = 0, \end{array}$ (1.22)

$\begin{array}{l} {θ^{'}}_{12} (α_{2} (z), α_{1} (u)) {θ^{'}}_{12} (x, y) (v) \otimes w + {θ^{'}}_{12} (x, y) (v) \otimes {θ^{″}}_{12} (α_{2} (z), α_{1} (u)) (w) \\ + {θ^{'}}_{12} (α_{2} (z), α_{1} (u)) (v) \otimes {θ^{″}}_{12} (x, y) (w) + v \otimes {θ^{″}}_{12} (α_{2} (z), α_{1} (u)) {θ^{″}}_{12} (x, y) (w) \\ - {θ^{'}}_{12} (α_{2} (z), {y, x, u} {A^{'}}_{1} (v) \otimes {A^{″}}_{1} (w) - {A^{'}}_{1} (v) \otimes {θ^{″}}_{12} (α_{2} (z), {y, x, u}) {A^{″}}_{1} (w) \\ - {θ^{'}}_{12} (α_{1} (x), α_{2} (y)) {θ^{'}}_{12} (z, u) (v) \otimes w - {θ^{'}}_{12} (z, u) (v) \otimes {θ^{″}}_{12} (α_{1} (x), α_{2} (y)) w \\ - {θ^{'}}_{12} (α_{1} (x), α_{2} (y)) (v) \otimes {θ^{″}}_{12} (z, u) (w) - v \otimes {θ^{″}}_{12} (α_{1} (x), α_{2} (y)) {θ^{″}}_{12} (z, u) w \\ + {θ^{'}}_{12} ({x, y, z}, α_{2} (u)) {A^{'}}_{1} (v) \otimes {A^{″}}_{1} (w) + {A^{'}}_{1} (v) \otimes {θ^{″}}_{12} ({x, y, z}, α_{2} (u)) {A^{″}}_{1} (w) = 0, \end{array}$ (1.23)

$\begin{array}{l} {θ^{'}}_{13} ({x, y, z}, α_{2} (u)) {A^{'}}_{1} (v) \otimes {A^{″}}_{1} (w) + {A^{'}}_{1} (v) \otimes {θ^{″}}_{13} ({x, y, z}, α_{2} (u)) {A^{″}}_{1} (w) \\ - {θ^{'}}_{13} (α_{1} (z), α_{2} (u)) {θ^{'}}_{12} (y, x) (v) \otimes (w) - {θ^{'}}_{12} (y, x) (v) \otimes {θ^{″}}_{13} (α_{1} (z), α_{2} (u)) (w) \\ - {θ^{'}}_{13} (α_{1} (z), α_{2} (u)) (v) \otimes {θ^{″}}_{12} (y, x) (w) - v \otimes {θ^{″}}_{13} (α_{1} (z), α_{2} (u)) {θ^{″}}_{12} (y, x) (w) \\ - {θ^{'}}_{12} (α_{1} (x), α_{2} (y)) {θ^{'}}_{13} (z, u) (v) \otimes w - {θ^{'}}_{13} (z, u) (v) \otimes {θ^{″}}_{12} (α_{1} (x), α_{2} (y)) (w) \\ - {θ^{'}}_{12} (α_{1} (x), α_{2} (y)) (v) \otimes {θ^{″}}_{13} (z, u) (w) - v \otimes {θ^{″}}_{12} (α_{1} (x), α_{2} (y)) {θ^{″}}_{13} (z, u) (w) \\ + {θ^{'}}_{13} (α_{1} (z), {x, y, u}) {A^{'}}_{1} (v) \otimes {A^{″}}_{2} (w) + {A^{'}}_{2} (v) \otimes {θ^{″}}_{13} (α_{1} (z), {x, y, u}) {A^{″}}_{2} (w) = 0, \end{array}$ (1.24)

$\begin{array}{l} {θ^{'}}_{12} ({x, y, z}, α_{1} (u)) {A^{'}}_{2} (v) \otimes {A^{″}}_{2} (w) + {A^{'}}_{2} (v) \otimes {θ^{″}}_{12} ({x, y, z}, α_{1} (u)) {A^{″}}_{2} (w) \\ - {θ^{'}}_{12} (α_{1} (z), {y, x, u}) {A^{'}}_{2} (v) \otimes {A^{″}}_{2} (w) - {A^{'}}_{2} (v) \otimes {θ^{″}}_{12} (α_{1} (z), {y, x, u}) {A^{″}}_{2} (w) \\ - {θ^{'}}_{12} (α_{1} (x), α_{2} (y)) {θ^{'}}_{12} (z, u) (v) \otimes w - {θ^{'}}_{12} (z, u) (v) \otimes {θ^{″}}_{12} (α_{1} (x), α_{2} (y)) (w) \\ - {θ^{'}}_{12} (α_{1} (x), α_{2} (y)) (v) \otimes {θ^{″}}_{12} (z, u) w - v \otimes {θ^{″}}_{12} (α_{1} (x), α_{2} (y)) {θ^{″}}_{12} (z, u) (w) \\ + {θ^{'}}_{12} (α_{1} (z), α_{2} (u)) {θ^{'}}_{12} (x, y) (v) \otimes w + {θ^{'}}_{12} (x, y) (v) \otimes {θ^{″}}_{12} (α_{1} (z), α_{2} (u)) (w) \\ + {θ^{'}}_{12} (α_{1} (z), α_{2} (u)) (v) \otimes {θ^{″}}_{12} (x, y) (w) + v \otimes {θ^{″}}_{12} (α_{1} (z), α_{2} (u)) {θ^{″}}_{12} (x, y) (w) = 0, \end{array}$ (1.25)

其中 $\forall x, y, z, u \in J, v \in V, w \in W$ 。

证明：由 ${θ^{'}}_{12} \otimes {θ^{″}}_{12}, {θ^{'}}_{13} \otimes {θ^{″}}_{13}, {A^{'}}_{1} \otimes {A^{″}}_{1}, {A^{'}}_{2} \otimes {A^{″}}_{2}$ 的定义知为线性映射。 $(V \otimes W, {θ^{'}}_{12} \otimes {θ^{″}}_{12}, {θ^{'}}_{13} \otimes {θ^{″}}_{13}, {A^{'}}_{1} \otimes {A^{″}}_{1}, {A^{'}}_{2} \otimes {A^{″}}_{2})$ 是 $(J, {\cdot, \cdot, \cdot}, α_{1}, α_{2})$ 的表示当且仅当对于 $(V \otimes W, {θ^{'}}_{12} \otimes {θ^{″}}_{12}, {θ^{'}}_{13} \otimes {θ^{″}}_{13}, {A^{'}}_{1} \otimes {A^{″}}_{1}, {A^{'}}_{2} \otimes {A^{″}}_{2})$ 有(1.3)~(1.8)成立。由于

$\begin{array}{l} {θ^{'}}_{13} \otimes {θ^{″}}_{13} (x, y) (v \otimes w) - {θ^{'}}_{13} \otimes {θ^{″}}_{13} (y, x) (v \otimes w) \\ = {θ^{'}}_{13} (x, y) v \otimes w - {θ^{'}}_{13} (y, x) v \otimes w + v \otimes {θ^{″}}_{13} (x, y) w - v \otimes {θ^{″}}_{13} (y, x) w, \end{array}$

则 ${θ^{'}}_{13} \otimes {θ^{″}}_{13}$ 满足等式(1.3)当且仅当 ${θ^{'}}_{13}, {θ^{″}}_{13}$ 满足等式(1.3)。在 $V \otimes W$ 上等式(1.4)左边

$\begin{array}{l} {θ^{'}}_{12} \otimes {θ^{″}}_{12} (α_{2} (x), α_{1} (u)) {θ^{'}}_{12} \otimes {θ^{″}}_{12} (z, y) (v \otimes w) - {θ^{'}}_{13} \otimes {θ^{″}}_{13} (α_{1} (z), α_{2} (x)) {θ^{'}}_{13} \otimes {θ^{″}}_{13} (y, u) (v \otimes w) \\ - {θ^{'}}_{12} \otimes {θ^{″}}_{12} ({z, u, x}, α_{2} (y)) {A^{'}}_{1} \otimes {A^{″}}_{1} (v \otimes w) + {θ^{'}}_{12} \otimes {θ^{″}}_{12} (α_{1} (z), α_{2} (u)) {θ^{'}}_{12} \otimes {θ^{″}}_{12} (x, y) (v \otimes w) \\ = {θ^{'}}_{12} ({z, u, x}, α_{2} (y)) {A^{'}}_{1} (v) \otimes w - {θ^{'}}_{12} ({z, u, x}, α_{2} (y)) {A^{'}}_{1} (v) \otimes {A^{″}}_{1} (w) \\ + v \otimes {θ^{″}}_{12} ({z, u, x}, α_{2} (y)) {A^{″}}_{1} (w) - {A^{'}}_{1} (v) \otimes {θ^{″}}_{12} ({z, u, x}, α_{2} (y)) {A^{″}}_{1} ( w ) \end{array}$

$\begin{array}{l} + {θ^{'}}_{12} (α_{2} (x), α_{1} (u)) (v) \otimes {θ^{″}}_{12} (z, y) (w) + {θ^{'}}_{12} (z, y) (v) \otimes {θ^{″}}_{12} (α_{2} (x), α_{1} (u)) (w) \\ - {θ^{'}}_{13} (y, u) (v) \otimes {θ^{″}}_{13} (α_{1} (z), α_{2} (x)) (w) - {θ^{'}}_{13} (α_{1} (z), α_{2} (x)) (v) \otimes {θ^{″}}_{13} (y, u) (w) \\ + {θ^{'}}_{12} (x, y) (v) \otimes {θ^{″}}_{12} (α_{1} (z), α_{2} (u)) (w) + {θ^{'}}_{12} (α_{1} (z), α_{2} (u)) (v) \otimes {θ^{″}}_{12} (x, y) (w), \end{array}$

因此 $(V \otimes W, {θ^{'}}_{12} \otimes {θ^{″}}_{12}, {θ^{'}}_{13} \otimes {θ^{″}}_{13}, {A^{'}}_{1} \otimes {A^{″}}_{1}, {A^{'}}_{2} \otimes {A^{″}}_{2})$ 满足等式(1.4)当且仅当等式(1.21)式成立。同理， $(V \otimes W, {θ^{'}}_{12} \otimes {θ^{″}}_{12}, {θ^{'}}_{13} \otimes {θ^{″}}_{13}, {A^{'}}_{1} \otimes {A^{″}}_{1}, {A^{'}}_{2} \otimes {A^{″}}_{2})$ 满足等式(1.5)当且仅当等式(1.22)成立，满足等式(1.6)当且仅当

等式(1.23)成立，满足等式(1.7)当且仅当等式(1.24)成立，满足等式(1.8)当且仅当等式(1.25)成立，因此，结论成立。

3. Hom-约当三系的扩张

定理2.1设 $(J, {\cdot, \cdot, \cdot}, α_{1}, α_{2})$ 为Hom-约当三系， $φ : J \times J \times J \to F$ 是三线性函数，在

$J \oplus V = {x_{1} + λ_{1} c | x_{1} \in J, λ_{1} \in F}$ 上定义

${x_{1} + λ_{1} c, x_{2} + λ_{2} c, x_{3} + λ_{3} c} = {x_{1}, x_{2}, x_{3}} + φ (x_{1}, x_{2}, x_{3}) c,$

$(α_{1} + i d) (x_{1} + λ_{1} c) = α_{1} (x_{1}) + λ_{1} c,$

$(α_{2} + i d) (x_{1} + λ_{1} c) = α_{2} (x_{1}) + λ_{1} c,$

则 $(J \oplus V, {\cdot, \cdot, \cdot}, α_{1} + i d, α_{2} + i d)$ 为Hom-约当三系的充分必要条件是 $φ$ 满足

$φ (x_{1}, x_{2}, x_{3}) = φ (x_{3}, x_{2}, x_{1}),$ (2.1)

$\begin{array}{l} φ (α_{1} (x_{1}), α_{2} (x_{2}), {x_{3}, x_{4}, x_{5}}) - φ ({x_{1}, x_{2}, x_{3}}, α_{1} (x_{4}), α_{2} (x_{5})) \\ + φ (α_{1} (x_{3}), {x_{2}, x_{1}, x_{4}}, α_{2} (x_{5})) - φ (α_{1} (x_{3}), α_{2} (x_{4}), {x_{1}, x_{2}, x_{5}}) = 0, \end{array}$ (2.2)

其中 $\forall x_{1}, x_{2}, x_{3}, x_{4}, x_{5} \in J, λ_{1}, λ_{2}, λ_{3} \in F$ 。

证明： $\forall x_{1}, x_{2}, x_{3}, x_{4}, x_{5} \in J, λ_{1}, λ_{2}, λ_{3}, λ_{4}, λ_{5} \in F$ ，则 $(J \oplus V, {\cdot, \cdot, \cdot}, α_{1} + i d, α_{2} + i d)$ 为Hom-约当三系当且仅当下列等式成立

${x_{1} + λ_{1} c, x_{2} + λ_{2} c, x_{3} + λ_{3} c} = {x_{3} + λ_{3} c, x_{2} + λ_{2} c, x_{1} + λ_{1} c},$ (2.3)

$\begin{array}{l} {(α_{1} + i d) (x_{1} + λ_{1} c), (α_{2} + i d) (x_{2} + λ_{2} c), {x_{3} + λ_{3} c, x_{4} + λ_{4} c, x_{5} + λ_{5} c}} \\ - {(α_{1} + i d) (x_{3} + λ_{3} c), (α_{2} + i d) (x_{4} + λ_{4} c), {x_{1} + λ_{1} c, x_{2} + λ_{2} c, x_{5} + λ_{5} c}} \\ - {{x_{1} + λ_{1} c, x_{2} + λ_{2} c, x_{3} + λ_{3} c}, (α_{1} + i d) (x_{4} + λ_{4} c), (α_{2} + i d) (x_{5} + λ_{5} c)} \\ + {(α_{1} + i d) (x_{3} + λ_{3} c), {x_{2} + λ_{2} c, x_{1} + λ_{1} c, x_{4} + λ_{4} c}, (α_{2} + i d) (x_{5} + λ_{5} c)} = 0, \end{array}$ (2.4)

由 $(J, {\cdot, \cdot, \cdot}, α_{1}, α_{2})$ 为Hom-约当三系知 ${x_{1}, x_{2}, x_{3}} = {x_{3}, x_{2}, x_{1}}$ ，则等式(2.3)成立当且仅当 $φ$ 满足等式(2.1)。由 $(J, {\cdot, \cdot, \cdot}, α_{1}, α_{2})$ 为Hom-约当三系知等式(1.2)成立，则等式(2.4)成立等价于等式(2.2)成立。因此，结论成立。

定理2.2设 $(J, {\cdot, \cdot, \cdot}, α_{1}, α_{2})$ 为Hom-约当三系，且满足(1.11)~(1.15)， $ω : J \times J \times J \to J^{*}$ 为三线性映射，在 $J \oplus J^{*}$ 上定义

${x + a^{*}, y + b^{*}, z + c^{*}} = {x, y, z} + ω (x, y, z) + L^{*} (x, y) c^{*} + L^{*} (z, y) a^{*} + M^{*} (x, z) b^{*},$

$(α_{1} + α_{1}^{*}) (x + a^{*}) = α_{1} (x) + α_{1}^{*} (a^{*}),$

$(α_{2} + α_{2}^{*}) (x + a^{*}) = α_{2} (x) + α_{2}^{*} (a^{*}),$

其中 $\forall x, y, z \in J, a^{*}, b^{*}, c^{*} \in J^{*}$ ，则 $(J \oplus J^{*}, {\cdot, \cdot, \cdot}, α_{1} + α_{1}^{*}, α_{2} + α_{2}^{*})$ 为Hom-约当三系当且仅当 $ω$ 满足

$ω (x, y, z) = ω (z, y, x),$ (2.5)

$\begin{array}{l} ω (α_{1} (x), α_{2} (y), {z, u, v}) - ω (α_{1} (z), α_{2} (u), {x, y, v}) \\ + ω (α_{1} (z), {y, x, u}, α_{2} (v)) - ω ({x, y, z}, α_{1} (u), α_{2} (v)) \\ + L^{*} (α_{1} (x), α_{2} (y)) ω (z, u, v) - L^{*} (α_{1} (z), α_{2} (u)) ω (x, y, v) \\ - L^{*} (α_{2} (v), α_{1} (u)) ω (x, y, z) + M^{*} (α_{1} (z), α_{2} (v)) ω (y, x, u) = 0, \end{array}$ (2.6)

其中 $\forall x, y, z, u, v \in J$ ，此时称 $J \oplus J^{*}$ 为J的 $T^{*}$ -扩张。

证明：显然，在 $J \oplus J^{*}$ 上定义的新的运算对三个变量都是线性的。 $\forall x, y, z, u, v \in J$ ， $a^{*}, b^{*}, c^{*}, d^{*}, e^{*} \in J^{*}$ ， $(J \oplus J^{*}, {\cdot, \cdot, \cdot}, α_{1} + α_{1}^{*}, α_{2} + α_{2}^{*})$ 为Hom-约当三系当且仅当在 $J \oplus J^{*}$ 上等式(1.1)、(1.2)成立。

由于

$\begin{array}{l} {x + a^{*}, y + b^{*}, z + c^{*}} - {z + c^{*}, y + b^{*}, x + a^{*}} \\ = {x, y, z} - {z, y, x} + ω (x, y, z) - ω (z, y, x) + L^{*} (x, y) c^{*} \\ - L^{*} (x, y) c^{*} + L^{*} (z, y) a^{*} - L^{*} (z, y) a^{*} + M^{*} (x, z) b^{*} - M^{*} (z, x) b^{*}, \end{array}$

由于 $M^{*} (x, z) = M^{*} (z, x)$ ，因此在 $J \oplus J^{*}$ 上等式(1.1)成立等价于等式(2.5)成立。

在 $J \oplus J^{*}$ 上等式(1.2)左边直接计算得

$\begin{array}{l} {(α_{1} + α_{1}^{*}) (x + a^{*}), (α_{2} + α_{2}^{*}) (y + b^{*}), {z + c^{*}, u + d^{*}, v + e^{*}}} \\ - {(α_{1} + α_{1}^{*}) (z + c^{*}), (α_{2} + α_{2}^{*}) (u + d^{*}), {x + a^{*}, y + b^{*}, v + e^{*}}} \\ - {{x + a^{*}, y + b^{*}, z + c^{*}}, (α_{1} + α_{1}^{*}) (u + d^{*}), (α_{2} + α_{2}^{*}) (v + e^{*})} \\ + {(α_{1} + α_{1}^{*}) (z + c^{*}), {y + b^{*}, x + a^{*}, u + d^{*}}, (α_{2} + α_{2}^{*}) (v + e^{*})} \\ = {α_{1} (x), α_{2} (y), {z, u, v}} - {α_{1} (z), α_{2} (u), {x, y, v}} - {{x, y, z}, α_{1} (u), α_{2} (v)} \end{array}$

$\begin{array}{l} + {α_{1} (z), {y, x, u}, α_{2} (v)} + ω (α_{1} (x), α_{2} (y), {z, u, v}) - ω (α_{1} (z), α_{2} (u), {x, y, v}) \\ - ω ({x, y, z}, α_{1} (u), α_{2} (v)) + ω (α_{1} (z), {y, x, u}, α_{2} (v)) + L^{*} (α_{1} (x), α_{2} (y)) ω (z, u, v) \\ - L^{*} (α_{1} (z), α_{2} (u)) ω (x, y, v) - L^{*} (α_{2} (v), α_{1} (u)) ω (x, y, z) + M^{*} (α_{1} (z), α_{2} (v)) ω (y, x, u) \\ + (L^{*} ({z, u, v}, α_{2} (y)) α_{1}^{*} - L^{*} (α_{1} (z), α_{2} (u)) L^{*} (v, y) \\ - L^{*} (α_{2} (v), α_{1} (u)) L^{*} (z, y) + M^{*} (α_{1} (z), α_{2} (v)) M^{*} (y, u)) ( a * ) \end{array}$

$\begin{array}{l} + (M^{*} (α_{1} (x), {z, u, v}) α_{2}^{*} - L^{*} (α_{1} (z), α_{2} (u)) M^{*} (x, v) \\ - L^{*} (α_{2} (v), α_{1} (u)) M^{*} (x, z) + M^{*} (α_{1} (z), α_{2} (v)) L^{*} (u, x)) (b^{*}) \\ + (L^{*} (α_{1} (x), α_{2} (y)) L^{*} (v, u) - L^{*} ({x, y, v}, α_{2} (u)) α_{1}^{*} \\ - L^{*} (α_{2} (v), α_{1} (u)) L^{*} (x, y) + L^{*} (α_{2} (v), {y, x, u}) α_{1}^{*}) ( c * ) \end{array}$

$\begin{array}{l} + (L^{*} (α_{1} (x), α_{2} (y)) M^{*} (z, v) - M^{*} (α_{1} (z), {x, y, v}) α_{2}^{*} \\ - M^{*} ({x, y, z}, α_{2} (v)) α_{1}^{*} + M^{*} (α_{1} (z), α_{2} (v)) L^{*} (y, x)) (d^{*}) \\ + (L^{*} (α_{1} (x), α_{2} (y)) L^{*} (z, u) - L^{*} (α_{1} (z), α_{2} (u)) L^{*} (x, y) \\ - L^{*} ({x, y, z}, α_{1} (u)) α_{2}^{*} + L^{*} (α_{1} (z), {y, x, u}) α_{2}^{*}) ( e * ) \end{array}$

由 $(J, {\cdot, \cdot, \cdot}, α_{1}, α_{2})$ 为Hom-约当三系，等式(1.2)成立。又由 $(J^{*}, L^{*}, M^{*}, α_{1}^{*}, α_{2}^{*})$ 为 $(J, {\cdot, \cdot, \cdot}, α_{1}, α_{2})$ 的表示知对于 $(J^{*}, L^{*}, M^{*}, α_{1}^{*}, α_{2}^{*})$ 有(1.4)~(1.8)成立，则在 $J \oplus J^{*}$ 上等式(1.2)成立等价于等式(2.6)成立。因此结论成立。

定义2.1设 $(J, {\cdot, \cdot, \cdot}, α_{1}, α_{2})$ 为Hom-约当三系，f为J上的双线性函数，若满足

$f ({x, y, z}, u) = f (z, {x, y, u}) (\forall x, y, z, u \in J),$

$f ({x, y, z}, u) = f (y, {x, u, z}) (\forall x, y, z, u \in J),$

则称f为J上的不变函数。

定义2.2设 $(J, {\cdot, \cdot, \cdot}, α_{1}, α_{2})$ 为Hom-约当三系，若J上存在非退化的、不变的、对称的双线性函数f，则称 $(J, f)$ 为度量Hom-约当三系。

定理2.3设 $(J, {\cdot, \cdot, \cdot}, α_{1}, α_{2})$ 为Hom-约当三系且满足

${x, y, z} = {x, z, y} (\forall x, y, z \in J)$ , (2.7)

三线性映射 $ω : J \times J \times J \to J^{*}$ 满足(2.5)、(2.6)并且

$ω (z, u, x) y = ω (x, y, u) z = ω (x, y, z) u (\forall x, y, z, u \in J),$ (2.8)

在 $J \oplus J^{*}$ 上定义

$q (x + a^{*}, y + b^{*}) = a^{*} (y) + b^{*} (x) (\forall x, y \in J, a^{*}, b^{*} \in J^{*}),$

则 $(J \oplus J^{*}, q)$ 是度量Hom-约当三系。

证明：由 $ω$ 满足等式(2.5)、(2.6)知 $(J \oplus J^{*}, {\cdot, \cdot, \cdot}, α_{1} + α_{1}^{*}, α_{2} + α_{2}^{*})$ 为Hom-约当三系。

对于 $y + b^{*}$ ，若对于 $\forall x + a^{*} \in J \oplus J^{*}$ 满足 $q (x + a^{*}, y + b^{*}) = a^{*} (y) + b^{*} (x) = 0$ ，取 $a^{*} = 0$ ，则 $q (x, y + b^{*}) = b^{*} (x) = 0$ ，因此 $b^{*} = 0$ 。取 $x = 0$ ，则 $q (a^{*}, y + b^{*}) = a^{*} (y) = 0$ ，因此 $y = 0$ ，从而 $y + b^{*} = 0$ ，q是非退化的。

因为 $q (x + a^{*}, y + b^{*}) = a^{*} (y) + b^{*} (x) = q (y + b^{*}, x + a^{*})$ ，所以q是对称的。

$\forall x, y, z, u \in J, a^{*}, b^{*}, c^{*}, d^{*} \in J^{*}$ ，直接计算得

$\begin{array}{l} q ({x + a^{*}, y + b^{*}, z + c^{*}}, u + d^{*}) \\ = q ({x, y, z} + ω (x, y, z) + L^{*} (x, y) c^{*} + L^{*} (z, y) a^{*} + M^{*} (x, z) b^{*}, u + d^{*}) \\ = ω (x, y, z) u + L^{*} (x, y) c^{*} (u) + L^{*} (z, y) a^{*} (u) + M^{*} (x, z) b^{*} (u) + d^{*} ({x, y, z}), \end{array}$

$\begin{array}{l} q (z + c^{*}, {x + a^{*}, y + b^{*}, u + d^{*}}) \\ = q (z + c^{*}, {x, y, u} + ω (x, y, u) + L^{*} (x, y) d^{*} + L^{*} (u, y) a^{*} + M^{*} (x, u) b^{*}) \\ = ω (x, y, u) z + L^{*} (x, y) d^{*} (z) + L^{*} (u, y) a^{*} (z) + M^{*} (x, u) b^{*} (z) + c^{*} ({x, y, u}), \end{array}$

由(2.7)知

$M^{*} (x, u) b^{*} (z) = 〈 b^{*}, {x, z, u} 〉 = M^{*} (x, z) b^{*} (u),$

此外，

$L^{*} (x, y) a^{*} (u) = L^{*} (u, y) a^{*} (z), L^{*} (x, y) c^{*} (u) = c^{*} ({x, y, u}), L^{*} (x, y) d^{*} (z) = d^{*} ({x, y, z}),$

因此，

$q ({x + a^{*}, y + b^{*}, z + c^{*}}, u + d^{*}) = q (z + c^{*}, {x + a^{*}, y + b^{*}, u + d^{*}}) .$

$\begin{array}{l} q (y + b^{*}, {z + c^{*} ， u + d^{*}, x + a^{*}}) \\ = q (y + b^{*}, {z, u, x} + ω (z, u, x) + L^{*} (z, u) a^{*} + L^{*} (x, u) c^{*} + M^{*} (z, x) d^{*}) \\ = ω (z, u, x) (y) + L^{*} (z, u) a^{*} (y) + L^{*} (x, u) c^{*} (y) + M^{*} (z, x) d^{*} (y) + b^{*} ({z, u, x}), \end{array}$

由(2.7)知

$L^{*} (z, u) a^{*} (y) = 〈 a^{*}, {z, u, y} 〉 = L^{*} (z, y) a^{*} (u),$

$L^{*} (x, u) c^{*} (y) = 〈 c^{*}, {x, u, y} 〉 = L^{*} (x, y) c^{*} (u),$

此外，

$b^{*} ({z, u, x}) = M^{*} (x, z) b^{*} (u), M^{*} (z, x) d^{*} (y) = d^{*} ({x, y, z}),$

因此，

$q ({x + a^{*}, y + b^{*}, z + c^{*}}, u + d^{*}) = q (y + b^{*}, {z + c^{*}, u + d^{*}, x + a^{*}}),$

则q是不变的，结论成立。

4. 结论

本文给出了Hom-约当三系的表示的定义及扩张的两种方法，作为约当三系的推广，还可以对Hom-约当三系进入深入研究，比如还可以研究它的上同调，并应用到其他领域。

文章引用

郭乐乐. Hom-约当三系的表示与扩张
Representation and Extension of Hom-Jordan Triple System[J]. 应用数学进展, 2020, 09(07): 1092-1102. https://doi.org/10.12677/AAM.2020.97129

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