Poisson代数源于对Poisson几何的研究, 自然出现于哈密顿力学, 且是量子群研究的中心.简单来说, 一个Poisson代数$ (A, \ast, [-, -]) $, 就是同时带有代数和李代数两种结构的代数, 且这两种代数结构之间要满足Leibniz法则.

鉴于Poisson几何与数学的紧密联系, 使它受到了许多研究者的关注.其中以Lichnerowicz^[1]和Weinstein^[2]最为具有代表性.近年来, 随着Poisson代数的发展, 其结构问题得到了一些新成果, 例如, 文献[3]研究了扭Heisengerg-Virasoro代数上的Poisson结构, 文献[4]研究了Witt代数和Virasoro代数上的Poisson结构, 文献[5]研究了Toroidal李代数上的Poisson结构, 文献[6]研究了扩张仿射李代数上的Poisson结构, 文献[7]确定了李代数W(2, 2)上的Poisson结构, 并改进了文献[4]中的结果.另外, 文献[8]得到Lie理论中一类Poisson结构的构造, 文献[9]研究了Hall代数上的Poisson结构, 文献[10]研究了张量代数上一种带辫子的Poisson结构.

近$ 20 $年, Poisson代数引起了越来越多的研究人员的关注, 并在结构理论和表示理论方面取得了很多结果. Schrödinger-Virasoro代数是一类重要的无限维李代数, 在数学和物理学的许多分支中都有着广泛的应用.许多研究者研究了(广义)Schrödinger-Virasoro代数的结构与表示理论等, 见文献[11]和[12]等, 但其Poisson结构的研究尚未实现.本文将研究Schrödinger-Virasoro代数上的Poisson结构.

本节主要介绍相关李代数及Poisson代数的基本概念.

定义2.1^[11]  Schrödinger-Virasoro代数$ \mathcal{SV} $是一个无限维李代数, 具有$ \mathbb{C} $-基$ \{L_{n}, Y_{n}, M_{n}|n\in\mathbb{Z}\}, \nonumber $且满足如下李关系式

$ \begin{eqnarray*} && [L_{n}, L_{m}] = (m-n)L_{m+n}, \quad [L_{n}, Y_{m}] = (m-\frac{1}{2}n)Y_{m+n}, \quad [Y_{n}, Y_{m}] = (m-n)M_{m+n}, \nonumber \\ && [L_{n}, M_{m}] = mM_{m+n}, \quad [Y_{n}, M_{m}] = [M_{n}, M_{m}] = 0. \end{eqnarray*} $

记$ \mathcal{SV}_{i} $为由$ \{L_{i}, Y_{i}, M_{i}\} $所张成的三维向量空间. $ \mathcal{SV} $关于Cartan子代数$ \mathbb{C}L_0 $有分解$ \mathcal{SV} = \oplus_{i\in \mathbb{Z}} \mathcal{SV}_{i}. $

定义2.2^[3]   Poisson代数$ (A, *, [-, -]) $是域$ \mathbb{C} $上的一个向量空间$ A $, 同时具有代数乘法$ * $和李代数乘法$ [-, -] $, 且满足Leibniz法则: $ [z, x*y] = [z, x]*y+x*[z, y], \forall x, y, z\in A. $若乘法$ * $满足结合律, 则称Poisson代数是结合的; 若乘法$ * $满足交换律, 则称Poisson代数是交换的.

本节将给出Schrödinger-Virasoro代数上的Poisson结构.注意到本文讨论的是非结合Poisson结构, 推广了结合的情形.

引理3.1  若在李代数$ \mathcal{SV} $上存在一个代数乘积$ * $, 使得$ (\mathcal{SV}, *, [-, -]) $成为一个Poisson代数, 则$ \mathcal{SV}_{i}*\mathcal{SV}_{j}\subseteq \mathcal{SV}_{i+j}, \quad \forall i, j\in \mathbb{Z}. $

证  对任意的$ x\in\mathcal{SV}_{i} $, $ y\in\mathcal{SV}_{j} $, 有

$ \begin{eqnarray} [L_{0}, x*y] = [L_{0}, x]*y+x*[L_{0}, y] = ix*y+jx*y = (i+j)x*y, \end{eqnarray} $

即$ x*y\in \mathcal{SV}_{i+j} $.因此, 对任意的$ i, j\in\mathbb{Z} $, 都有$ \mathcal{SV}_{i}*\mathcal{SV}_{j}\subseteq \mathcal{SV}_{i+j} $.

定理3.2   $ \mathcal{SV} $上的任何Poisson结构都具有如下形式

$ \begin{eqnarray} & & L_{m}*L_{n} = c_{1}(m-n)L_{m+n}, \quad L_{m}*Y_{n} = (\frac{m}{2}-n)c_{1}Y_{m+n}, \\ & & L_{m}*M_{n} = -c_{1}nM_{m+n}, \quad Y_{m}*Y_{n} = c_{1}(m-n)M_{m+n}, \\ & & Y_{m}*L_{n} = (m-\frac{n}{2})c_{1}Y_{m+n}, \quad M_{m}*L_{n} = c_{1}mM_{m+n}, \\ & & Y_{m}*M_{n} = M_{m}*M_{n} = M_{m}*Y_{n} = 0, \end{eqnarray} $

$ \forall m, n\in \mathbb{Z} $, 其中$ c_{1}\in \mathbb{C} $.

证  根据引理3.1易知$ M_{m}*M_{n}\in\mathbb{C}M_{m+n} $, 由此可假设

$ \begin{eqnarray} & & L_{m}*L_{n} = a_{m, n}L_{m+n}+b_{m, n}Y_{m+n}+c_{m, n}M_{m+n}, \\ & & L_{m}*Y_{n} = d_{m, n}L_{m+n}+e_{m, n}Y_{m+n}+f_{m, n}M_{m+n}, \\ & & L_{m}*M_{n} = g_{m, n}L_{m+n}+h_{m, n}Y_{m+n}+i_{m, n}M_{m+n}, \\ & & Y_{m}*Y_{n} = j_{m, n}L_{m+n}+k_{m, n}Y_{m+n}+l_{m, n}M_{m+n}, \\ & & Y_{m}*L_{n} = o_{m, n}L_{m+n}+p_{m, n}Y_{m+n}+q_{m, n}M_{m+n}, \\ & & Y_{m}*M_{n} = r_{m, n}L_{m+n}+s_{m, n}Y_{m+n}+t_{m, n}M_{m+n}, \\ & & M_{m}*M_{n} = u_{m, n}M_{m+n}, \\ & & M_{m}*L_{n} = v_{m, n}L_{m+n}+w_{m, n}Y_{m+n}+x_{m, n}M_{m+n}, \\ & & M_{m}*Y_{n} = y_{m, n}L_{m+n}+z_{m, n}Y_{m+n}+\delta_{m, n}M_{m+n}. \end{eqnarray} $

下面分10步来证明定理.

步骤  1   确定$ a_{m, n}, b_{m, n}, c_{m, n} $.由于$ [L_{k}, L_{m}*L_{n}] = [L_{k}, L_{m}]*L_{n}+L_{m}*[L_{k}, L_{n}] $,

$ \begin{eqnarray} & & (m+n-k)a_{m, n} = (m-k)a_{m+k, n}+(n-k)a_{m, n+k}, \end{eqnarray} $

(3.1)

$ \begin{eqnarray} & & (m+n-\frac{k}{2})b_{m, n} = (m-k)b_{m+k, n}+(n-k)b_{m, n+k}, \end{eqnarray} $

(3.2)

$ \begin{eqnarray} & & (m+n)c_{m, n} = (m-k)c_{m+k, n}+(n-k)c_{m, n+k}. \end{eqnarray} $

(3.3)

在(3.1)式中取$ k = n $, 得$ ma_{m, n} = (m-n)a_{m+n, n} $.于是$ \frac{a_{m, n}}{m-n} = \frac{a_{m+n, n}}{m+n-n} = c_{1} $.故$ a_{m, n} = c_{1}(m-n), c_{1}\in\mathbb{C} $.在(3.2)式中取$ m = n = k $, 得$ \frac{3}{2}kb_{k, k} = 0 $.因此$ b_{k, k} = 0, \quad k\neq 0 $; 取$ m = k, n = 0 $, 得$ \frac{k}{2}b_{m, 0} = -kb_{m, m} $.因此有$ b_{m, 0} = 0, \quad k\neq 0 $; 取$ m = n = -k $, 得$ -\frac{5}{2}kb_{-k, -k} = -2kb_{0, -k}-2kb_{-k, 0} $.则$ b_{0, -k} = 0, \quad k\neq 0 $; 取$ m = n = 0, k\neq0 $, 得$ -\frac{k}{2}b_{0, 0} = -kb_{k, 0}-kb_{0, k} $.于是$ b_{0, 0} = 0, \quad k\neq 0 $; 取$ n = k = -m $, 得$ -\frac{k}{2}b_{-k, k} = -2kb_{0, k} $.则有$ b_{-k, k} = 0, \quad k\neq 0 $; 最后在(3.2)式中取$ m = 0, n\neq0, n+k\neq0 $, 得$ (n-\frac{k}{2})b_{0, n} = -kb_{k, n}+(n-k)b_{0, n+k} $.于是有$ -kb_{k, n} = 0 $.因此，对任意的$ m, n\in \mathbb{Z} $, 都有$ b_{m, n} = 0 $.

在(3.3)式中取$ k = n = -m\neq0 $, 得$ 0 = -2nc_{0, n} $.因此$ c_{0, n} = 0, \quad n\neq0 $; 取$ m = 0, n\neq0, n+k\neq0 $, 得$ 0 = -kc_{k, n} $.则$ c_{k, n} = 0, \quad n\neq0, \quad k\neq0, \quad n+k\neq0 $; 取$ k = m\neq0, n = 0 $, 得$ mc_{m, 0} = 0 $.于是有$ c_{m, 0} = 0, \quad m\neq0 $; 取$ k = -m\neq0, n = 0 $, 得$ 2c_{0, 0}+c_{m, -m} = 0 $.于是$ c_{m, -m} = -2c_{0, 0}, \quad m\neq0 $.在(3.3)式中取$ k = -3, m = 2, n = 1 $得$ 3c_{2, 1} = 5c_{-1, 1}+4c_{-2, 2} = -18c_{0, 0} $, 而$ c_{2, 1} = 0 $, 因此$ c_{0, 0} = 0 $.于是, 对任意的$ m, n\in \mathbb{Z} $, 都有$ c_{m, n} = 0 $.

步骤  2   确定$ d_{m, n}, e_{m, n}, f_{m, n} $.由于$ [L_{k}, L_{m}*Y_{n}] = [L_{k}, L_{m}]*Y_{n}+L_{m}*[L_{k}, Y_{n}] $,

$ \begin{eqnarray} & & (m+n-k)d_{m, n} = (m-k)d_{m+k, n}+(n-\frac{k}{2})d_{m, n+k}, \end{eqnarray} $

(3.4)

$ \begin{eqnarray} & & (m+n-\frac{k}{2})e_{m, n} = (m-k)e_{m+k, n}+(n-\frac{k}{2})e_{m, n+k}, \end{eqnarray} $

(3.5)

$ \begin{eqnarray} & & (m+n)f_{m, n} = (m-k)f_{m+k, n}+(n-\frac{k}{2})f_{m, n+k}. \end{eqnarray} $

(3.6)

在(3.4)式中取$ m = k\neq 0, n = 0 $, 得$ -\frac{m}{2}d_{m, m} = 0 $.有$ d_{m, m} = 0, \quad m\neq 0 $; 取$ m = -n\neq 0, k = 2n $, 得$ -2nd_{-n, n} = 0 $.有$ d_{-n, n} = 0, \quad n\neq 0 $; 取$ m = k = -n\neq 0 $, 得$ -\frac{3}{2}md_{m, 0} = 0 $.则有$ d_{m, 0} = 0, \quad m\neq 0 $; 再取$ m = n = -k\neq 0 $, 得$ 2md_{0, m} = 0 $.因此

$ \begin{eqnarray} d_{0, m} = 0, \quad m\neq 0; \end{eqnarray} $

(3.7)

取$ m = 0, n = k\neq 0 $, 得$ \frac{n}{2}d_{0, 2n} = 0 $.于是$ d_{0, 2n} = 0, \quad n\neq 0 $; 取$ m = 0, k = -n\neq 0 $, 得$ 2nd_{0, n} = \frac{3}{2}nd_{0, 0} $.于是结合(3.7)式, 有$ d_{0, n} = 0, \quad n\neq 0 $.最后在(3.4)式中取$ n = 0, m\neq k\neq 0 $, 得$ -\frac{k}{2}d_{m, k} = 0 $.则有$ d_{m, k} = 0, \quad n = 0, \quad m\neq k\neq 0 $.于是, 针对上述对任意的$ m, n\in \mathbb{Z} $, 都有$ d_{m, n} = 0 $.在(3.5)式中取$ k = 2n $, 得$ me_{m, n} = (m-2n)e_{m+2n, n} $.于是$ \frac{e_{m, n}}{m-2n} = \frac{e_{m+2n, n}}{m+2n-2n} = c_{2} $.故$ e_{m, n} = c_{2}(m-2n), c_{2}\in\mathbb{C} $.在(3.6)式中取$ m = -n\neq 0, k = 2n $, 得$ -3nf_{n, n} = 0 $.有$ f_{n, n} = 0, \quad n\neq 0 $; 取$ m = k\neq 0, n = 0 $, 得$ mf_{m, 0} = -\frac{m}{2}f_{m, m} $.于是$ f_{m, 0} = 0, \quad m\neq 0 $; 取$ n = 0, m\neq 0, m+k\neq 0 $, 得$ -\frac{k}{2}f_{m, k} = 0 $.则有$ f_{m, k} = 0, \quad n = 0, \quad m\neq 0, \quad m+k\neq 0 $; 取$ m = n = 0, k\neq 0 $, 得$ 0 = -kf_{k, 0}-\frac{k}{2}f_{0, k} $.就有$ f_{0, k} = 0, \quad k\neq 0 $.在(3.6)式中取$ k = -m\neq 0, n = 0 $, 得$ mf_{m, 0} = 2mf_{0, 0}+\frac{m}{2}f_{m, -m} $.因此

$ \begin{eqnarray} f_{m, -m} = -4f_{0, 0}, \quad k = -m\neq 0, \quad n = 0. \end{eqnarray} $

最后在(3.6)式中取$ k = -3, m = 2, n = 1 $得$ 3f_{2, 1} = 5f_{-1, 1}+\frac{5}{2}f_{2, -2} = -30f_{0, 0} $, 而$ f_{2, 1} = 0 $, 因此$ f_{0, 0} = 0 $.于是, 对任意的$ m, n\in \mathbb{Z} $, 都有$ f_{m, n} = 0 $.

步骤  3   确定$ g_{m, n}, h_{m, n}, i_{m, n} $.由于$ [L_{k}, L_{m}*M_{n}] = [L_{k}, L_{m}]*M_{n}+L_{m}*[L_{k}, M_{n}] $,

$ \begin{eqnarray} & & (m+n-k)g_{m, n} = (m-k)g_{m+k, n}+ng_{m, n+k}, \end{eqnarray} $

(3.8)

$ \begin{eqnarray} & & (m+n-\frac{k}{2})h_{m, n} = (m-k)h_{m+k, n}+nh_{m, n+k}, \end{eqnarray} $

(3.9)

$ \begin{eqnarray} & & (m+n)i_{m, n} = (m-k)i_{m+k, n}+ni_{m, n+k}. \end{eqnarray} $

(3.10)

在(3.8)式中取$ n = 0 $, 且$ m\neq k $得$ g_{m, 0} = g_{m+k, 0} $.于是有

$ \begin{eqnarray} g_{m, 0} = g_{0, 0}, \quad m\in\mathbb{Z}. \end{eqnarray} $

(3.11)

在(3.8)式中取$ m = 0 $, 得

$ \begin{eqnarray} kg_{k, n} = (k-n)g_{0, n}+ng_{0, n+k}. \end{eqnarray} $

(3.12)

令$ k = -n $, 则

$ \begin{eqnarray} -ng_{-n, n} = -2ng_{0, n}+ng_{0, 0}. \end{eqnarray} $

(3.13)

在(3.8)式中取$ k = m $, 得$ g_{m, n} = g_{m, m+n}, n\neq 0. $取$ n = -m\neq0 $, 利用(3.11)式有$ g_{m, -m} = g_{0, 0}, m\in\mathbb{Z}. $结合(3.13)式有$ g_{0, n} = g_{0, 0}, \quad n\in\mathbb{Z} $.再由(3.12)即可得, 对任意的$ k, n\in\mathbb{Z} $, 有$ g_{k, n} = g_{0, 0} $.令$ c_{3} = g_{0, 0} $, 则有$ g_{m, n} = c_{3}, m, n\in\mathbb{Z} $.在(3.9)式中取$ k = m\neq 0, n = 0 $, 得$ \frac{m}{2}h_{m, 0} = 0 $.则$ h_{m, 0} = 0, m\neq 0 $; 取$ k = m = -n $, 得$ -\frac{m}{2}h_{m, -m} = -mh_{m, 0} $.于是有$ h_{m, -m} = 0, m\neq 0 $; 取$ m = -k\neq 0, n = 0 $, 得$ \frac{3}{2}mh_{m, 0} = 2mh_{0, 0} $.因此, $ h_{0, 0} = 0, m\neq 0 $; 取$ n = -k\neq 0, m = 0 $, 得$ \frac{3}{2}nh_{0, n} = 0 $.有$ h_{0, n} = 0, n\neq 0 $.再在(3.9)式中取$ m = n = -k $, 得$ \frac{5}{2}mh_{m, m} = 0 $.则有$ h_{m, m} = 0, m\neq 0 $.最后在(3.9)式中取$ m = 0, n\neq 0, n+k\neq 0 $, 得$ -kh_{k, n} = 0 $.那么$ h_{k, n} = 0, n\neq 0, n+k\neq 0 $.因此综上, 对任意的$ m, n\in \mathbb{Z} $, 都有$ h_{m, n} = 0 $.对于(3.10)式, 取$ k = m $, 得$ (m+n)i_{m, n} = ni_{m, m+n} $.由此可推出$ \frac{i_{m, n}}{n} = \frac{i_{m, m+n}}{m+n} = c_{4} $.故$ i_{m, n} = c_{4}n, c_{4}\in\mathbb{C} $.

步骤  4   确定$ j_{m, n}, k_{m, n}, l_{m, n} $.由于$ [L_{k}, Y_{m}*Y_{n}] = [L_{k}, Y_{m}]*Y_{n}+Y_{m}*[L_{k}, Y_{n}] $,

$ \begin{eqnarray} & & (m+n-k)j_{m, n} = (m-\frac{k}{2})j_{m+k, n}+(n-\frac{k}{2})j_{m, n+k}, \end{eqnarray} $

(3.14)

$ \begin{eqnarray} & & (m+n-\frac{k}{2})k_{m, n} = (m-\frac{k}{2})k_{m+k, n}+(n-\frac{k}{2})k_{m, n+k}, \end{eqnarray} $

(3.15)

$ \begin{eqnarray} & & (m+n)l_{m, n} = (m-\frac{k}{2})l_{m+k, n}+(n-\frac{k}{2})l_{m, n+k}. \end{eqnarray} $

(3.16)

在(3.14)式中取$ n = 0, m = 0, k\neq 0 $得$ -kj_{0, 0} = -\frac{k}{2}j_{k, 0}-\frac{k}{2}j_{0, k} $.于是有

$ \begin{eqnarray} j_{k, 0}+j_{0, k} = 2j_{0, 0}, \quad k\neq 0. \end{eqnarray} $

(3.17)

在(3.14)式中取$ m = n, k = -m $得$ 3mj_{m, m} = \frac{3}{2}mj_{0, m}+\frac{3}{2}mj_{m, 0} $.则根据(3.17)式, 我们可以得到

$ \begin{eqnarray} j_{m, m} = j_{0, 0}, \quad m\in\mathbb{Z}. \end{eqnarray} $

(3.18)

在(3.14)式中取$ k = 2n, n = -m $得$ 2mj_{m, -m} = 2mj_{-m, -m} $.由(3.18)式, 可以得到

$ \begin{eqnarray} j_{m, -m} = j_{0, 0}, \quad m\in\mathbb{Z}. \end{eqnarray} $

(3.19)

在(3.14)式中取$ k = -n, m = 0 $得$ 2nj_{0, n} = \frac{n}{2}j_{-n, n}+\frac{3}{2}nj_{0, 0} $, 即$ 4nj_{0, n} = nj_{-n, n}+3nj_{0, 0} $.再由(3.19)式, 可得到

$ \begin{eqnarray} j_{0, n} = j_{0, 0}, \quad n\in\mathbb{Z}. \end{eqnarray} $

(3.20)

由$ j_{0, n} = j_{0, 0} $, 综合(3.17)式, 可知

$ \begin{eqnarray} j_{n, 0} = j_{0, 0}. \end{eqnarray} $

(3.21)

在(3.14)式中取$ k = 2m, n = 0 $得$ -mj_{m, 0} = -mj_{m, 2m} $.结合(3.21)式, 可得到$ j_{m, 2m} = j_{0, 0}, m\in\mathbb{Z} $; 取$ k = 2n, m = 0 $得$ -nj_{0, n} = -nj_{2n, n} $.结合(3.20)式, 可得到$ j_{2n, n} = j_{0, 0}, n\in\mathbb{Z} $.最后，在(3.14)式中取$ n = 0, m\neq 0, m+k\neq 0 $得到$ (m-k)j_{m, 0} = (m-\frac{k}{2})j_{m+k, 0}-\frac{k}{2}j_{m, k} $.再综合(3.21)式, 可得到$ -\frac{k}{2}j_{0, 0} = -\frac{k}{2}j_{m, k} $, 即$ j_{m, k} = j_{0, 0}, k\neq 0 $.综上所述, 对任意的$ k, m\in\mathbb{Z} $, 有$ j_{k, m} = j_{0, 0} $.令$ c_{5} = j_{0, 0} $, 则有$ j_{m, n} = c_{5}, m, n\in\mathbb{Z} $.在(3.15)式中取$ m = n, k = 2n $得$ mk_{m, m} = 0 $.于是有$ k_{m, m} = 0, m\neq 0 $.再取$ m = n = -k $得$ 0 = \frac{3}{2}mk_{0, m}+\frac{3}{2}mk_{m, 0} $.则

$ \begin{eqnarray} k_{0, m}+k_{m, 0} = 0, \quad m\neq 0. \end{eqnarray} $

(3.22)

取$ m = n = -k $得$ 0 = \frac{3}{2}mk_{0, m}+\frac{3}{2}mk_{m, 0} $.有$ k_{0, m}+k_{m, 0} = 0, m\neq 0 $; 取$ m = n = 0 $得$ -\frac{k}{2}k_{0, 0} = -\frac{k}{2}k_{k, 0}-\frac{k}{2}k_{0, k} $.因此$ k_{0, 0} = 0, \quad k\neq 0 $; 取$ n = 0, k = 2m\neq 0 $得$ -mk_{m, 2m} = 0 $.于是有$ k_{m, 2m} = 0, m\neq 0 $; 取$ m = 0, k = 2n\neq 0 $得$ -nk_{2n, n} = 0 $.则$ k_{2n, n} = 0, n\neq 0 $; 取$ n = 2m, k = -m $得$ 0 = \frac{3}{2}mk_{0, 2m} = 0 $.则有$ k_{0, 2m} = 0, m\neq 0 $; 取$ m = 0, n = k $得$ \frac{n}{2}k_{0, n} = 0 $.有$ k_{0, n} = 0, n\neq 0 $.再结合(3.22)式可得$ k_{n, 0} = 0 $.再在(3.15)式中取$ m = -k, n = 0 $得$ 0 = \frac{m}{2}k_{m, -m} $.于是$ k_{m, -m} = 0, m\neq 0 $.最后在(3.15)式中取$ m = 0, n\neq 0, n+k\neq 0 $得$ -\frac{k}{2}k_{k, n} = 0 $.于是$ k_{k, n} = 0, k\neq 0 $.因此, 对任意的$ m, n\in \mathbb{Z} $, 都有$ k_{m, n} = 0 $.对于(3.16)式, 取$ k = 2m $, 得$ (m+n)l_{m, n} = (n-m)l_{m, n+2m} $.由此可推出$ \frac{l_{m, n}}{n-m} = \frac{l_{m, n+2m}}{n+2m-m} = c_{6} $.故$ l_{m, n} = c_{6}(m-n), c_{6}\in\mathbb{C} $.

步骤  5   确定$ o_{m, n}, p_{m, n}, q_{m, n} $.类似于步骤2中的讨论, 有

$ \begin{eqnarray*} o_{m, n} = 0, \quad p_{m, n} = c_{7}(n-2m), \quad q_{m, n} = 0, \quad c_{7}\in\mathbb{C}. \end{eqnarray*} $

步骤  6   确定$ r_{m, n}, s_{m, n}, t_{m, n} $.由于$ [L_{k}, Y_{m}*M_{n}] = [L_{k}, Y_{m}]*M_{n}+Y_{m}*[L_{k}, M_{n}] $,

$ \begin{eqnarray} & & (m+n-k)r_{m, n} = (m-\frac{k}{2})r_{m+k, n}+nr_{m, n+k}, \end{eqnarray} $

(3.23)

$ \begin{eqnarray} & & (m+n-\frac{k}{2})s_{m, n} = (m-\frac{k}{2})s_{m+k, n}+ns_{m, n+k}, \end{eqnarray} $

(3.24)

$ \begin{eqnarray} & & (m+n)t_{m, n} = (m-\frac{k}{2})t_{m+k, n}+nt_{m, n+k}. \end{eqnarray} $

(3.25)

在(3.23)式中取$ n = 0, k = 2m $得$ -mr_{m, 0} = 0 $.则

$ \begin{eqnarray} r_{m, 0} = 0, \quad m\neq 0. \end{eqnarray} $

(3.26)

在(3.23)式中取$ m = n = 0 $得$ -kr_{0, 0} = -\frac{k}{2}r_{k, 0} $.则有$ r_{0, 0} = 0, \quad k\neq 0 $.在(3.23)式中取$ m = n\neq 0, k = -m $得$ 3mr_{m, m} = \frac{3}{2}mr_{0, m}+mr_{m, 0} $.于是

$ \begin{eqnarray} 3r_{m, m} = \frac{3}{2}r_{0, m}+r_{m, 0}, \quad m\neq 0. \end{eqnarray} $

(3.27)

在(3.23)式中取$ m = 0, n = -k $得$ 2nr_{0, n} = \frac{n}{2}r_{-n, n} $.则

$ \begin{eqnarray} 4r_{0, n} = r_{-n, n}, \quad n\neq 0. \end{eqnarray} $

(3.28)

在(3.23)式中取$ m = -n\neq 0, k = 2m $得$ -2mr_{m, -m} = -mr_{m, m} $.则有

$ \begin{eqnarray} 2r_{m, -m} = r_{m, m}, \quad m\neq 0. \end{eqnarray} $

(3.29)

在(3.23)式中取$ m = n\neq 0, k = 2m $得$ 4mr_{m, m} = 2mr_{-m, m}+mr_{m, -m} $.所以

$ \begin{eqnarray} 4r_{m, m} = 2r_{-m, m}+r_{m, -m}, \quad m\neq 0. \end{eqnarray} $

(3.30)

综合(3.28), (3.29)和(3.30)式可得$ 7r_{m, m} = 16r_{0, m} $.将其代入(3.27)式再结合(3.26)式得到$ r_{m, m} = 0. $因此有$ r_{0, m} = r_{-m, m} = r_{m, -m} = 0 $.再在(3.23)式中取$ m = 0, n\neq 0, n+k\neq 0 $得$ (n-k)r_{0, n} = -\frac{k}{2}r_{k, n}+nr_{0, n+k} $.再由$ r_{0, n} = 0 $可知，对任意的$ k, n\in\mathbb{Z} $, 有$ r_{k, n} = 0 $.因此有$ r_{m, n} = 0, m, n\in\mathbb{Z} $.在(3.24)式中取$ n = -m\neq 0, k = 2m $得$ -ms_{m, -m} = -ms_{m, m} $.则

$ \begin{eqnarray} s_{m, -m} = s_{m, m}, \quad m\neq 0. \end{eqnarray} $

(3.31)

在(3.24)式中取$ n = 0, k = -m $得$ \frac{3}{2}ms_{m, 0} = \frac{3}{2}ms_{0, 0} $.有$ s_{m, 0} = s_{0, 0}, \quad m\in\mathbb{Z} $.在(3.24)式中取$ m = n, k = -2m $得$ 3ms_{m, m} = 2ms_{-m, m}+ms_{m, -m} $.结合(3.31)式可得

$ \begin{eqnarray} s_{m, m} = s_{-m, m} = s_{m, -m}, \quad m\neq 0. \end{eqnarray} $

(3.32)

在(3.24)式中取$ m = -k, n = m $得$ \frac{5}{2}ns_{n, n} = \frac{3}{2}ns_{0, n}+ns_{n, 0} $.因$ s_{m, 0} = s_{0, 0} $, 于是

$ \begin{eqnarray} \frac{5}{2}ns_{n, n} = \frac{3}{2}ns_{0, n}+ns_{0, 0}, \quad n\neq 0. \end{eqnarray} $

(3.33)

在(3.24)式中取$ n = -k, m = 0 $得$ \frac{3}{2}ns_{0, n} = \frac{n}{2}s_{-n, n}+ns_{0, 0} $.将此式代入(3.33)式可得

$ \begin{eqnarray*} \frac{5}{2}ns_{n, n} = \frac{n}{2}s_{-n, n}+2ns_{0, 0}, \quad n\neq 0. \end{eqnarray*} $

再综合(3.32)式有$ s_{m, m} = s_{-m, m} = s_{m, -m} = s_{0, 0}, \quad m\in\mathbb{Z} $.因此在(3.33)式中可得

$ \begin{eqnarray} s_{0, m} = s_{0, 0}, \quad m\in\mathbb{Z}. \end{eqnarray} $

(3.34)

在(3.24)式中取$ m = 0, n\neq 0, n+k\neq 0 $得$ (n-\frac{k}{2})s_{0, n} = -\frac{k}{2}s_{k, n}+ns_{0, n+k} $.再由(3.34)式即可得, 对任意的$ k, n\in\mathbb{Z} $, 有$ s_{k, n} = s_{0, 0} $.令$ c_{8} = s_{0, 0} $, 则有$ s_{m, n} = c_{8}, m, n\in\mathbb{Z} $.在(3.25)式中取$ n = 0, k = 2m $得$ mt_{m, 0} = 0 $.则$ t_{m, 0} = 0, \quad m\neq 0 $; 取$ n = -m, k = 2m $得$ -mt_{m, m} = 0 $.则有$ t_{m, m} = 0, \quad m\neq 0 $; 取$ m = n\neq 0, k = -m $得$ \frac{3}{2}mt_{0, m}+mt_{m, 0} = 0 $.有$ t_{0, m} = 0, \quad m\neq 0 $; 取$ m = -k\neq 0, n = 0 $得$ \frac{3}{2}mt_{0, 0} = 0 $.于是$ t_{0, 0} = 0, \quad m\neq 0 $; 取$ m = 0, n = -k\neq 0 $得$ \frac{n}{2}t_{-n, n} = 0 $.于是有$ t_{-n, n} = 0, \quad n\neq 0 $; 最后在(3.25)式中取$ m = 0, n\neq 0, n+k\neq 0 $得$ -\frac{k}{2}t_{k, n} = 0 $.因此$ t_{k, n} = 0, \quad k\neq 0 $.于是, 对任意的$ m, n\in \mathbb{Z} $, 都有$ t_{m, n} = 0 $.

步骤  7   确定$ u_{m, n} $.根据等式$ [M_{k}, L_{m}*M_{n}] = [M_{k}, L_{m}]*M_{n}+L_{m}*[M_{k}, M_{n}] $, 有

$ \begin{eqnarray} & & c_{3}M_{m+n+k} = M_{m+k}\ast M_{n}, \quad k\neq 0. \end{eqnarray} $

显然, 对任意的$ m, n\in \mathbb{Z} $, 有$ u_{m, n} = c_{3} $.

步骤  8   确定$ v_{m, n}, w_{m, n}, x_{m, n} $.类似于步骤3中的讨论, 有

$ \begin{eqnarray} v_{m, n} = c_{9}, \quad w_{m, n} = 0, \quad x_{m, n} = c_{10}m, \quad c_{9}, c_{10}\in\mathbb{C}. \end{eqnarray} $

步骤  9   确定$ y_{m, n}, z_{m, n}, \delta_{m, n} $.类似于步骤6中的讨论, 则有

$ \begin{eqnarray} y_{m, n} = 0, \quad z_{m, n} = c_{11}, \quad \delta_{m, n} = 0, \quad c_{11}\in\mathbb{C}. \end{eqnarray} $

步骤10   最后, 由以下等式

$ \begin{eqnarray} & & [M_{k}, L_{m}*L_{n}] = [M_{k}, L_{m}]*L_{n}+L_{m}*[M_{k}, L_{n}], \\ & & [M_{k}, Y_{m}*Y_{n}] = [M_{k}, Y_{m}]*Y_{n}+Y_{m}*[M_{k}, Y_{n}], \\ & & [Y_{k}, L_{m}*L_{n}] = [Y_{k}, L_{m}]*L_{n}+L_{m}*[Y_{k}, L_{n}], \\ & & [Y_{k}, L_{m}*Y_{n}] = [Y_{k}, L_{m}]*Y_{n}+L_{m}*[Y_{k}, Y_{n}], \\ & & [Y_{k}, L_{m}*M_{n}] = [Y_{k}, L_{m}]*M_{n}+L_{m}*[Y_{k}, M_{n}], \\ & & [Y_{k}, M_{m}*Y_{n}] = [Y_{k}, M_{m}]*Y_{n}+M_{m}*[Y_{k}, Y_{n}], \\ & & [Y_{k}, M_{m}*L_{n}] = [Y_{k}, M_{m}]*L_{n}+M_{m}*[Y_{k}, L_{n}], \end{eqnarray} $

可以得到

$ \begin{eqnarray} & & c_{1}(m-n) = c_{10}(m+k)+c_{4}(n+k), \quad c_{3}k+c_{9}k = 0, \quad c_{5}k = 0, \\ & & c_{1}(mk-nk-\frac{m^{2}-n^{2}}{2}) = c_{2}(mk-nk-\frac{mn}{2}+n^{2}-2k^{2})+c_{7}(nk-mk-\frac{mn}{2}+m^{2}-2k^{2}), \\ & & c_{3}(n-k)+c_{5}(\frac{m}{2}-k) = 0, \quad c_{2}(m-2n)(m+n-k) = c_{4}(n^{2}-k^{2})+c_{6}(\frac{m}{2}-k)(m+k-n), \\ & & c_{3}(\frac{m+n}{2}-k) = c_{8}(\frac{m}{2}-k), \quad c_{11}m(m+n-k) = c_{3}(n-k), \\ & & c_{9}(\frac{m+n}{2}-k) = c_{11}m(\frac{n}{2}-k). \end{eqnarray} $

由此易得$ c_{2} = \frac{1}{2}c_{1}, \quad c_{4} = -c_{1}, \quad c_{6} = c_{1}, \quad c_{7} = -\frac{1}{2}c_{1}, \quad c_{10} = c_{1}, \quad c_{3} = c_{5} = c_{8} = c_{9} = c_{11} = 0 $, 故此定理成立, 证毕.