众所周知, Schrödinger代数和Virasoro代数与非平衡统计物理密切相关, 它们在数学和物理学(如统计物理学)的许多领域中都有着重要的作用. Schrödinger-Virasoro代数$ \mathfrak{sb} $最初是由Henkel在研究自由Schrödinger方程的不变性时被引入文献[1], 其结构和表示理论被Roger和Unterberger在文献[2]深入研究.比如, Henkel在文献[1]中给出了$ \mathfrak{sb} $只有一维中心扩张. Roger和Unterberger在研究$ \mathfrak{sb} $的同调理论时得出了有三维外导子^[2]. $ \mathfrak{sb} $上的有限维不可约权模在文献[3]中被分类.广义的Schrödinger-Virasoro代数的自同构群及Verma模被完全确定^[4].最近几年Schrödinger-Virasoro代数及其变形的结构和表示理论被许多学者广泛研究^[5-8].为了研究$ \mathfrak{sb} $的顶点表示, Unterberger介绍了一类新的无限维李代数^[9], 称之为扩张的Schrödinger-Virasoro代数$ \widetilde{\mathfrak{sb}} $, 该李代数是复数域$ \mathbb{C} $上的向量空间, 带有一组基$ \{L_{n}, M_{n}, N_{n}, Y_{n+\frac{1}{2}}\mid n\in \mathbb{Z}\} $, 满足李积关系

$ \begin{eqnarray*} && [L_{m}, L_{n}] = (n-m)L_{m+n}, [M_{m}, M_{n}] = 0, [N_{m}, N_{n}] = 0, [Y_{m+\frac{1}{2}}, Y_{n+\frac{1}{2}}] = (m-n)M_{m+n+1}, \\ &&[L_{m}, M_{n}] = nM_{m+n}, [L_{m}, N_{n}] = nN_{m+n}, [L_{m}, Y_{n+\frac{1}{2}}] = (n+\frac{1-m}{2})Y_{m+n+\frac{1}{2}}, \\ && [N_{m}, M_{n}] = 2M_{m+n}, [N_{m}, Y_{n+\frac{1}{2}}] = Y_{m+n+\frac{1}{2}}, [M_{m}, Y_{n+\frac{1}{2}}] = 0, \quad \forall\, m, n\in\mathbb{Z}. \end{eqnarray*} $

该无限维李代数的导子、自同构群及中心扩张等结构理论在文献[10]中被完全刻画.

本文将考虑一类与扩张的Schrödinger-Virasoro代数$ \widetilde{\mathfrak{sb}} $相关的无限维李代数.通过$ \widetilde{\mathfrak{sb}} $与洛朗多项式代数$ \mathbb{C}[t, t^{-1}] $张量成一个新的无限维李代数$ \widetilde{\mathfrak{sb}}\otimes\mathbb{C}[t, t^{-1}] $, 记作$ \widetilde{\mathscr{W}} $, 称之为扩张的圈Schrödinger-Virasoro代数, 满足以下李积关系

$ \begin{eqnarray*} && [L_{m, i}, L_{n, j}] = (n-m)L_{m+n, i+j}, [M_{m, i}, M_{n, j}] = 0, [N_{m, i}, N_{n, j}] = 0, [N_{m, i}, M_{n, j}] = 2M_{m+n, i+j}, \\ &&[L_{m, i}, M_{n, j}] = nM_{m+n, i+j}, [L_{m, i}, N_{n, j}] = nN_{m+n, i+j}, [L_{m, i}, Y_{n+\frac{1}{2}, j}] = (n+\frac{1-m}{2})Y_{m+n+\frac{1}{2}, i+j}, \\ &&[M_{m, i}, Y_{n+\frac{1}{2}, j}] = 0 , [N_{m, i}, Y_{n+\frac{1}{2}, j}] = Y_{m+n+\frac{1}{2}, i+j}, [Y_{m+\frac{1}{2}, i}, Y_{n+\frac{1}{2}, j}] = (m-n)M_{m+n+1, i+j}. \end{eqnarray*} $

对任意的$ m, n, i, j\in\mathbb{Z} $, 其中$ L_{m, i} $表示$ L_{m}\otimes t^{i} $, 其他定义类似.

由于李代数的二上循环在其中心扩张方面起着关键作用, 可以借助它构造许多无限维李代数, 并且可以进一步刻画所得李代数的结构及表示.同时上同调群和李代数的结构密切相关, 比如一阶同调群和李代数的导子代数及李双代数的联系, 从而上同调群的计算就显得比较重要.本文主要确定了扩张的圈Schrödinger-Virasoro代数$ \widetilde{\mathscr{W}} $的所有二上同调群, 并且给出了它的泛中心扩张.我们希望借助于中心扩张能够进一步深刻理解$ \widetilde{\mathscr{W}} $的结构及其表示.

李代数$ \widetilde{\mathscr{W}} $上的双线性型$ \psi:\widetilde{\mathscr{W}}\times \widetilde{\mathscr{W}} \rightarrow \mathbb{C} $若满足以下关系

$ \begin{eqnarray*} &&\psi(x, y) = -\psi(y, x), \\ &&\psi(x, [y, z])+\psi(y, [z, x])+\psi(z, [x, y]) = 0, \end{eqnarray*} $

对任意的$ x, y\in \widetilde{\mathscr{W}} $, 称$ \psi $为$ \widetilde{\mathscr{W}} $上的二上循环.记$ C^{2}(\widetilde{\mathscr{W}}, \ \mathbb{C}) $为$ \widetilde{\mathscr{W}} $上的所有二上循环构成的向量空间.对于任意的线性函数$ f: \widetilde{\mathscr{W}} \rightarrow \mathbb{C} $, 可以定义一个二上循环$ \psi_{f} $,

$ \begin{equation} \psi_{f}(x, y) = f([x, y]), \quad\quad \forall \, x, \, y \in \widetilde{\mathscr{W}}, \end{equation} $

(2.1)

称这样的二上循环为$ \widetilde{\mathscr{W}} $的二上边缘或平凡的二上循环.记$ B^{2}(\widetilde{\mathscr{W}}, \ \mathbb{C}) $为$ \widetilde{\mathscr{W}} $上的所有二上边缘构成的向量空间.如果$ \phi-\psi $是一个平凡的二上循环, 则称二上循环$ \phi $和$ \psi $等价.记$ \bar{\phi} $为所有与二上循环$ \phi $等价的等价类.由所有这样的等价类构成的商空间

$ \begin{eqnarray*} &&H^{2}(\widetilde{\mathscr{W}}, \ \mathbb{C}) = C^{2}(\widetilde{\mathscr{W}}, \ \mathbb{C})/B^{2}(\widetilde{\mathscr{W}}, \ \mathbb{C}) \end{eqnarray*} $

称为$ \widetilde{\mathscr{W}} $上的二上同调群.

设$ \psi $是$ \widetilde{\mathscr{W}} $上的任意一个二上循环, 可以利用$ \psi $定义一个$ \mathbb{C} $上的线性映射$ f:\widetilde{\mathscr{W}}\rightarrow \mathbb{C} $满足

$ \begin{eqnarray*} && f(L_{m, i}) = \frac{1}{m}\psi(L_{0, 0}, L_{m, i}), \quad \forall \ m\neq 0, \\&& f(L_{0, i}) = -\frac{1}{2}\psi(L_{1, 0}, L_{-1, i}), \\&& f(M_{m, i}) = \frac{1}{m}\psi(L_{0, 0}, M_{m, i}), \quad \forall \ m\neq 0, \\&& f(M_{0, i}) = -\psi(L_{1, 0}, M_{-1, i}), \\&& f(N_{m, i}) = \frac{1}{m}\psi(L_{0, 0}, N_{m, i}), \quad \forall \ m\neq 0, \\&& f(N_{0, i}) = -\psi(L_{1, 0}, N_{-1, i}), \\&& f(Y_{n+\frac{1}{2}, i}) = \frac{1}{n+\frac{1}{2}}\psi(L_{0, 0}, Y_{n+\frac{1}{2}, i}), \quad \forall \ n, i\in \mathbb{Z}. \end{eqnarray*} $

令$ \phi = \psi-\psi_{f} $, 其中$ \psi_{f} $即为(2.1)式中的定义, 显然很容易验证

$ \begin{equation} \phi(L_{0, 0}, L_{m, i}) = 0, \quad \forall \ m\neq 0, \end{equation} $

(2.2)

$ \begin{equation} \phi(L_{1, 0}, L_{-1, i}) = 0, \end{equation} $

(2.3)

$ \begin{equation} \phi(L_{0, 0}, M_{m, i}) = 0, \quad \forall \ m\neq 0, \end{equation} $

(2.4)

$ \begin{equation} \phi(L_{1, 0}, M_{-1, i}) = 0, \end{equation} $

(2.5)

$ \begin{equation} \phi(L_{0, 0}, N_{m, i}) = 0, \quad \forall \ m\neq 0, \end{equation} $

(2.6)

$ \begin{equation} \phi(L_{1, 0}, N_{-1, i}) = 0, \end{equation} $

(2.7)

$ \begin{equation} \phi(L_{0, 0}, Y_{n+\frac{1}{2}, i}) = 0, \quad \forall \ n, i\in \mathbb{Z}. \end{equation} $

(2.8)

下面通过几个引理给出主要结果.

引理2.1    $ \phi(L_{m, i}, Y_{n+\frac{1}{2}, j}) = 0, \quad \forall \ m, n, i, j\in \mathbb{Z} $.

证  由二上循环$ \phi $的关系知

$ \begin{eqnarray*} (n+\frac{1}{2})\phi(L_{m, i}, Y_{n+\frac{1}{2}, j}) & = & \phi(L_{m, i}, [L_{0, 0}, Y_{n+\frac{1}{2}, j}])\\ & = & \phi([L_{m, i}, L_{0, 0}], Y_{n+\frac{1}{2}, j})+ \phi(L_{0, 0}, [L_{m, i}, Y_{n+\frac{1}{2}, j}])\\ & = & -m\phi(L_{m, i}, Y_{n+\frac{1}{2}, j})+(n+\frac{1-m}{2}) \phi(L_{0, 0}, Y_{n+m+\frac{1}{2}, j}). \end{eqnarray*} $

整理得

$ \begin{equation*} (n+m+\frac{1}{2})\phi(L_{m, i}, Y_{n+\frac{1}{2}, j}) = (n+\frac{1-m}{2}) \phi(L_{0, 0}, Y_{n+m+\frac{1}{2}, j}). \end{equation*} $

又由(2.8)式, 则上式可得$ (n+m+\frac{1}{2})\phi(L_{m, i}, Y_{n+\frac{1}{2}, j}) = 0 $, 从而有

$ \begin{equation*} \phi(L_{m, i}, Y_{n+\frac{1}{2}, j}) = 0\quad \forall\, m, n, i, j\in\mathbb{Z}. \end{equation*} $

引理2.2    $ \phi(N_{m, i}, Y_{n+\frac{1}{2}, j}) = 0, \quad \forall \ m, n, i, j\in \mathbb{Z} $.

证  由二上循环$ \phi $的关系知

$ \begin{eqnarray*} (n+\frac{1}{2})\phi(N_{m, i}, Y_{n+\frac{1}{2}, j}) & = & \phi(N_{m, i}, [L_{0, 0}, Y_{n+\frac{1}{2}, j}]) \\ & = & \phi([N_{m, i}, L_{0, 0}], Y_{n+\frac{1}{2}, j})+\phi(L_{0, 0}, [N_{m, i}, Y_{n+\frac{1}{2}, j}])\\ & = & -m\phi(N_{m, i}, Y_{n+\frac{1}{2}, j})+\phi(L_{0, 0}, Y_{m+n+\frac{1}{2}, j}). \end{eqnarray*} $

整理得

$ \begin{equation*} (n+m+\frac{1}{2})\phi(N_{m, i}, Y_{n+\frac{1}{2}, j}) = \phi(L_{0, 0}, Y_{n+m+\frac{1}{2}, j}). \end{equation*} $

又由(2.8)式, 则上式可得$ (n+m+\frac{1}{2})\phi(N_{m, i}, Y_{n+\frac{1}{2}, j}) = 0 $, 即有

$ \begin{equation*} \phi(N_{m, i}, Y_{n+\frac{1}{2}, j}) = 0, \quad \forall\, m, n, i, j\in\mathbb{Z}. \end{equation*} $

引理2.3    $ \phi(M_{m, i}, Y_{n+\frac{1}{2}, j}) = 0, \quad \forall \ m, n, i, j\in \mathbb{Z} $.

证  由二上循环$ \phi $的关系知

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

整理得

$ \begin{equation*} (n+m+\frac{1}{2})\phi(M_{m, i}, Y_{n+\frac{1}{2}, j}) = 0. \end{equation*} $

则有

$ \begin{equation*} \phi(M_{m, i}, Y_{n+\frac{1}{2}, j}) = 0, \quad \forall\, m, n, i, j\in\mathbb{Z}. \end{equation*} $

引理2.4    $ \phi(L_{m, i}, M_{n, j}) = \phi(Y_{m+\frac{1}{2}, i}, Y_{n+\frac{1}{2}, j}) = 0, \quad \forall \ m, n, i, j\in \mathbb{Z} $.

证  由关系

$ \begin{eqnarray*} n\phi(L_{m, i}, M_{n, j}) & = & \phi(L_{m, i}, [L_{0, 0}, M_{n, j}]) \\ & = & \phi([L_{m, i}, L_{0, 0}], M_{n, j})+\phi(L_{0, 0}, [L_{m, i}, M_{n, j}]) \\ & = & -m\phi(L_{m, i}, M_{n, j})+n\phi(L_{0, 0}, M_{m+n, i+j}). \end{eqnarray*} $

整理得

$ \begin{equation*} (n+m)\phi(L_{m, i}, M_{n, j}) = n\phi(L_{0, 0}, M_{m+n, i+j}). \end{equation*} $

若$ m+n\neq 0 $, 又由(2.4)式, 则有

$ \begin{equation} \phi(L_{m, i}, M_{n, j}) = 0. \end{equation} $

(2.9)

又由关系

$ \begin{equation*} \phi([L_{m, i}, L_{n, j}], M_{-(m+n), k})+\phi([L_{n, j}, M_{-(m+n), k}], L_{m, i})+\phi([M_{-(m+n), k}, L_{m, i}], L_{n, j}) = 0, \end{equation*} $

整理得

$ \begin{equation} (n-m)\phi(L_{m+n, i+j}, M_{-(m+n), k})+(m+n)\phi(L_{m, i}, M_{-m, j+k})-(m+n)\phi(L_{n, j}, M_{-n, k+i}) = 0. \end{equation} $

(2.10)

取$ n = m $, 则有

$ \begin{equation*} m\phi(L_{m, i}, M_{-m, j+k}) = m\phi(L_{m, j}, M_{-m, k+i}). \end{equation*} $

从而当$ m\neq0 $时, 有

$ \begin{equation} \phi(L_{m, i}, M_{-m, j+k}) = \phi(L_{m, j}, M_{-m, k+i}). \end{equation} $

(2.11)

在(2.10)式中令$ m = -n $, 则有

$ \begin{equation*} 2n\phi(L_{0, i+j}, M_{0, k}) = 0, \end{equation*} $

从而

$ \begin{equation} \phi(L_{0, i+j}, M_{0, k}) = 0. \end{equation} $

(2.12)

由(2.11)和(2.12)式, 得到

$ \begin{equation} \phi(L_{m, i}, M_{-m, j+k}) = \phi(L_{m, j}, M_{-m, k+i}), \quad \forall \, m, i, j, k\in \mathbb{Z}. \end{equation} $

(2.13)

上式说明仅与第二个指标的和$ i+j+k $有关, 而与位置无关, 从而不妨设$ A_{m, i+j} = \phi(L_{m, i+j}, M_{-m, 0}) $.在(2.10)式中取$ m = 1 $, 又由(2.5)式有

$ \begin{equation} (n-1)A_{n+1, i+j+k}-(n+1)A_{n, i+j+k} = 0. \end{equation} $

(2.14)

在(2.14)式中用$ n-1 $替换$ n $, 则有

$ \begin{equation} (n-2)A_{n, i+j+k}-nA_{n-1, i+j+k} = 0. \end{equation} $

(2.15)

在(2.10)式中取$ n = n-1, m = 2 $, 可得

$ \begin{equation} (n-3)A_{n+1, i+j+k}-(n+1)A_{n-1, i+j+k} = -(n+1)A_{2, i+j+k}. \end{equation} $

(2.16)

将(2.14), (2.15)和(2.16)式联立方程, 解得

$ \begin{equation} A_{n, i+j+k} = \frac{n(n-1)}{2}A_{2, i+j+k}, \quad n\neq -1. \end{equation} $

(2.17)

在(2.10)式中令$ m = 2, n = -1 $, 从而

$ \begin{equation} A_{2, i+j+k} = A_{-1, i+j+k}. \end{equation} $

(2.18)

由(2.17)和(2.18)式, 得到

$ \begin{equation} A_{n, i+j} = \frac{n(n-1)}{2}A_{2, i+j}, \quad \forall \, n, i, j\in \mathbb{Z}. \end{equation} $

(2.19)

考虑关系

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

若$ m+n+1\neq 0 $, 由(2.4)式, 则上式可得

$ \begin{equation} \phi(Y_{m+\frac{1}{2}, i}, Y_{n+\frac{1}{2}, j}) = 0. \end{equation} $

(2.20)

又由关系

$ \begin{eqnarray*} \phi([Y_{m+\frac{1}{2}, i}, Y_{n+\frac{1}{2}, j}], L_{-(m+n+1), k})&+&\phi([Y_{n+\frac{1}{2}, j}, L_{-(m+n+1), k}], Y_{m+\frac{1}{2}, i})\\ &+&\phi([L_{-(m+n+1), k}, Y_{m+\frac{1}{2}, i}], Y_{n+\frac{1}{2}, j}) = 0, \end{eqnarray*} $

整理得到

$ (n-m)\phi(L_{-(m+n+1), k}, M_{m+n+1, i+j})+\frac{3n+m+2}{2}\phi(Y_{m+\frac{1}{2}, i}, Y_{-(m+\frac{1}{2}), j+k}) $

$ \begin{equation} -\frac{3m+n+2}{2}\phi(Y_{n+\frac{1}{2}, j}, Y_{-(n+\frac{1}{2}), i+k}) = 0. \end{equation} $

(2.21)

在(2.21)式中取$ m = 0 $, 则有

$ \begin{equation} \frac{n+2}{2}\phi(Y_{n+\frac{1}{2}, j}, Y_{-(n+\frac{1}{2}), i+k}) = n\phi(L_{-(n+1), k}, M_{n+1, i+j})+\frac{3n+2}{2}\phi(Y_{\frac{1}{2}, i}, Y_{-\frac{1}{2}, j+k}). \end{equation} $

(2.22)

在上式取$ n = -2 $, 结合(2.5)式, 从而有

$ \begin{equation} \phi(Y_{\frac{1}{2}, i}, Y_{-\frac{1}{2}, j+k}) = 0. \end{equation} $

(2.23)

由(2.19), (2.22)及(2.23)式, 可得

$ \begin{equation} \phi(Y_{n+\frac{1}{2}, j}, Y_{-(n+\frac{1}{2}), i+k}) = n(n+1)A_{2, i+j+k}, \quad n\neq-2. \end{equation} $

(2.24)

上式也说明仅与第二个指标的和$ i+j+k $有关, 而与位置无关.当$ n = -2 $时, 由$ \phi $的反对称性及上式知

$ \begin{equation} \phi(Y_{-2+\frac{1}{2}, j}, Y_{-(-2+\frac{1}{2}), i+k}) = -\phi(Y_{\frac{3}{2}, i+k}, Y_{-\frac{3}{2}, j}) = -2A_{2, i+j+k}. \end{equation} $

(2.25)

在(2.24)式中分别取$ n = 2, -3 $, 则有

$ \begin{equation*} \phi(Y_{\frac{5}{2}, j}, Y_{-\frac{5}{2}, i+k}) = 6A_{2, i+j+k}, \quad \phi(Y_{-\frac{5}{2}, j}, Y_{\frac{5}{2}, i+k}) = 6A_{2, i+j+k}. \end{equation*} $

由$ \phi $的反对称性知$ A_{2, i+j+k} = 0 $.从而由(2.19), (2.24)和(2.25)式知

$ \begin{equation*} \phi(L_{m, i}, M_{n, j}) = \phi(Y_{m+\frac{1}{2}, i}, Y_{n+\frac{1}{2}, j}) = 0, \quad \forall \ m, n, i, j\in \mathbb{Z}. \end{equation*} $

引理2.5    $ \phi(M_{m, i}, M_{n, j}) = 0, \quad \forall \ m, n, i, j\in \mathbb{Z} $.

证  由于

$ \begin{eqnarray*} n\phi(M_{m, i}, M_{n, j}) & = & \phi(M_{m, i}, [L_{0, 0}, M_{n, j}]) \\ & = & \phi([M_{m, i}, L_{0, 0}], M_{n, j}])+ \phi(L_{0, 0}, [M_{m, i}, M_{n, j}])\\ & = & -m\phi(M_{m, i}, M_{n, j}), \end{eqnarray*} $

从而有

$ \begin{equation} \phi(M_{m, i}, M_{n, j}) = 0, \quad m+n\neq0. \end{equation} $

(2.26)

又由

$ \begin{equation*} \phi([L_{m, i}, M_{n, j}], M_{-(m+n), k})+\phi([M_{n, j}, M_{-(m+n), k}], L_{m, i})+\phi([M_{-(m+n), k}, L_{m, i}], M_{n, j}) = 0, \end{equation*} $

则有

$ \begin{equation} n\phi(M_{m+n, i+j}, M_{-(m+n), k})-(m+n)\phi(M_{n, j}, M_{-n, i+k}) = 0. \end{equation} $

(2.27)

在(2.27)式中令$ m = 0 $, 则有

$ \begin{equation} \phi(M_{n, i+j}, M_{-n, j}) = \phi(M_{n, j}, M_{-n, i+k}), \quad n\neq 0. \end{equation} $

(2.28)

在(2.27)式中令$ n = 0 $, 可得

$ \begin{equation} \phi(M_{0, j}, M_{0, i+k}) = 0. \end{equation} $

(2.29)

综合(2.28)和(2.29)式, 有

$ \begin{equation*} \phi(M_{n, i+j}, M_{-n, k}) = \phi(M_{n, j}, M_{-n, i+k}), \quad \forall\, n, i, j, k\in \mathbb{Z}. \end{equation*} $

上式说明仅与第二个指标的和$ i+j+k $有关, 而与位置无关.从而不妨令$ B_{n, i+j} = \phi(M_{n, i+j}, M_{-n, 0}) $, 在式(2.27)式中取$ n = 1 $, 则有

$ \begin{equation} B_{m+1, i+j+k} = (m+1)B_{1, i+j+k}, \end{equation} $

(2.30)

又由

$ \begin{eqnarray*} \phi([Y_{m+\frac{1}{2}, i}, Y_{n+\frac{1}{2}, j}], M_{-(m+n+1), k})&+&\phi([Y_{n+\frac{1}{2}, j}, M_{-(m+n+1), k}], Y_{m+\frac{1}{2}, i})\\ &+&\phi([M_{-(m+n+1), k}, Y_{m+\frac{1}{2}, i}], Y_{n+\frac{1}{2}, j}) = 0, \end{eqnarray*} $

从而

$ \begin{equation*} (m-n)B_{m+n+1, i+j+k} = 0. \end{equation*} $

在上式中取$ m = 2, n = -2 $, 可得$ B_{1, i+j+k} = 0 $, 由(2.30)式知

$ \begin{equation} B_{m+1, i+j+k} = 0, \quad \forall\, m, i, j, k\in \mathbb{Z}. \end{equation} $

(2.31)

综合(2.26)和(2.31)式, 得到

$ \begin{equation*} \phi(M_{m, i}, M_{n, j}) = 0, \quad \forall \ m, n, i, j\in \mathbb{Z}. \end{equation*} $

引理2.6    $ \phi(M_{m, i}, N_{n, j}) = 0, \; \; \forall \ m, n, i, j\in \mathbb{Z} $.

证  由关系

$ \begin{eqnarray*} n\phi(M_{m, i}, N_{n, j}) & = & \phi(M_{m, i}, [L_{0, 0}, N_{n, j}]) \\ & = & \phi([M_{m, i}, L_{0, 0}], N_{n, j})+\phi(L_{0, 0}, [M_{m, i}, N_{n, j}]) \\ & = & -m\phi(M_{m, i}, N_{n, j})-2\phi(L_{0, 0}, M_{m+n, i+j}), \end{eqnarray*} $

从而由(2.4)式得

$ \begin{equation} \phi(M_{m, i}, N_{n, j}) = 0, \quad m+n\neq0. \end{equation} $

(2.32)

又由关系

$ \begin{equation*} \phi([L_{m, i}, M_{n, j}], N_{-(m+n), k})+\phi([M_{n, j}, N_{-(m+n), k}], L_{m, i})+\phi([N_{-(m+n), k}, L_{m, i}], M_{n, j}) = 0, \end{equation*} $

结合引理2.4, 整理得

$ \begin{equation} n\phi(M_{m+n, i+j}, N_{-(m+n), k})-(m+n)\phi(M_{n, j}, N_{-n, k+i}) = 0. \end{equation} $

(2.33)

若$ m = 0 $, 则有

$ \begin{equation} \phi(M_{n, i+j}, N_{-n, k}) = \phi(M_{n, j}, M_{-n, k+i}), \quad n\neq 0. \end{equation} $

(2.34)

在(2.33)式中令$ n = 0 $, 可得

$ \begin{equation} \phi(M_{0, j}, N_{0, k+i}) = 0. \end{equation} $

(2.35)

由(2.33)和(2.35)式, 有

$ \begin{equation} \phi(M_{n, i+j}, N_{-n, k}) = \phi(M_{n, j}, M_{-n, k+i}), \quad \forall\, n, i, j, k\in\mathbb{Z}. \end{equation} $

(2.36)

上式说明仅与第二个指标的和$ i+j+k $有关, 而与位置无关.不妨设$ C_{n, i+j} = \phi(M_{n, i+j}, N_{-n, 0}) $, 在(2.33)中令$ n = 1 $, 则有

$ \begin{equation} C_{m+1, i+j+k} = (m+1)C_{1, i+j+k}. \end{equation} $

(2.37)

又由关系

$ \begin{eqnarray*} \phi([Y_{n+\frac{1}{2}, i}, Y_{m+\frac{1}{2}, j}], N_{-(m+n+1), k})&+& \phi([Y_{m+\frac{1}{2}, j}, N_{-(m+n+1), k}], Y_{n+\frac{1}{2}, i})\\ &+& \phi([N_{-(m+n+1), k}, Y_{n+\frac{1}{2}, i}], Y_{m+\frac{1}{2}, j}) = 0, \end{eqnarray*} $

又由引理2.4, 整理得

$ \begin{equation*} (n-m)C_{m+n+1, i+j+k} = 0. \end{equation*} $

在上式令$ n = 1, m = -1 $, 则有$ C_{1, i+j+k} = 0 $, 从而由(2.37)式, 知

$ \begin{equation} C_{m+1, i+j+k} = 0, \quad \forall \, m, i, j, k\in \mathbb{Z}. \end{equation} $

(2.38)

由上式结合(2.32)式, 就可得到

$ \begin{equation*} \phi(M_{m, i}, N_{n, j}) = 0, \quad \forall \ m, n, i, j\in \mathbb{Z}. \end{equation*} $

引理2.7    $ \phi(L_{m, i}, N_{n, j}) = \delta_{m+n, 0}\frac{m(m-1)}{2}\phi(L_{2, i+j}, N_{-2, 0}), \quad \forall \ m, n, i, j\in \mathbb{Z} $.

证  由关系

$ \begin{eqnarray*} n\phi(L_{m, i}, N_{n, j}) & = & \phi(L_{m, i}, [L_{0, 0}, N_{n, j}]) \\ & = & \phi([L_{m, i}, L_{0, 0}], N_{n, j})+\phi(L_{0, 0}, [L_{m, i}, N_{n, j}]) \\ & = & -m\phi(L_{m, i}, N_{n, j})+n\phi(L_{0, 0}, N_{m+n, i+j}). \end{eqnarray*} $

由(2.6)式, 则有

$ \begin{equation} \phi(L_{m, i}, N_{n, j}) = 0, \quad \, m+n\neq0. \end{equation} $

(2.39)

又由关系

$ \begin{equation*} \phi([L_{m, i}, L_{n, j}], N_{-(m+n), k})+\phi([L_{n, j}, N_{-(m+n), k}], L_{m, i})+\phi([N_{-(m+n), k}, L_{m, i}], L_{n, j}) = 0, \end{equation*} $

整理得

$ \begin{equation} (n-m)\phi(L_{m+n, i+j}, N_{-(m+n), k})+(m+n)\phi(L_{m, i}, N_{-m, j+k})-(m+n)\phi(L_{n, j}, N_{-n, k+i}) = 0. \end{equation} $

(2.40)

在上式中取$ m = n $, 则有

$ \begin{equation} \phi(L_{n, i}, N_{-n, j+k}) = \phi(L_{n, j}, N_{-n, k+i}), \quad n\neq0. \end{equation} $

(2.41)

在(2.40)式中取$ m = -n $, 可得

$ \begin{equation} \phi(L_{0, i+j}, N_{0, k}) = 0. \end{equation} $

(2.42)

由(2.41)和(2.42)式, 则有

$ \begin{equation} \phi(L_{n, i}, N_{-n, j+k}) = \phi(L_{n, j}, N_{-n, k+i})\quad \forall\, n, i, j, k\in\mathbb{Z}. \end{equation} $

(2.43)

上式说明仅与第二个指标的和$ i+j+k $有关, 而与位置无关.不妨设$ D_{n, i+j} = \phi(L_{n, i+j}, N_{-n, 0}) $, 在(2.40)式中取$ n = 1 $, 结合(2.7)式, 则有

$ \begin{equation} (1-m)D_{m+1}+(m+1)D_{m} = 0. \end{equation} $

(2.44)

在上式用$ m-1 $替换$ m $, 有

$ \begin{equation} (2-m)D_{m}+mD_{m-1} = 0. \end{equation} $

(2.45)

在(2.40)式中令$ n = 2, m = m-1 $, 得到

$ \begin{equation} (3-m)D_{m+1}+(m+1)D_{m-1} = (m+1)D_{2}. \end{equation} $

(2.46)

将式(2.44), (2.45)和(2.46)式联立方程, 解得

$ \begin{equation} D_{m} = \frac{m(m-1)}{2}D_{2}, \quad m\neq-1. \end{equation} $

(2.47)

在(2.40)式中取$ m = 2, n = -1 $, 结合(2.7)式, 易得$ D_{2} = D_{-1} $, 从而由(2.47)式就有

$ \begin{equation*} D_{m} = \frac{m(m-1)}{2}D_{2}, \quad\forall\, m\in \mathbb{Z}. \end{equation*} $

由上式和(2.39)式, 有

$ \begin{equation*} \phi(L_{m, i}, N_{n, j}) = \delta_{m+n, 0}\frac{m(m-1)}{2}\phi(L_{2, i+j}, N_{-2, 0}), \quad \forall \ m, n, i, j\in \mathbb{Z}. \end{equation*} $

引理2.8    $ \phi(N_{m, i}, N_{n, j}) = \delta_{m+n, 0}m\phi(N_{1, i+j}, N_{-1, 0}), \quad \forall \ m, n, i, j\in \mathbb{Z} $.

证  由关系

$ \begin{eqnarray*} n\phi(N_{m, i}, N_{n, j}) & = & \phi(N_{m, i}, [L_{0, 0}, N_{n, j}]) \\ & = & \phi([N_{m, i}, L_{0, 0}], N_{n, j})+\phi(L_{0, 0}, [N_{n, j}, N_{m, i}]) \\ & = & -m\phi(N_{m, i}, N_{n, j}), \end{eqnarray*} $

整理得

$ \begin{equation} \phi(N_{m, i}, N_{n, j}) = 0, \quad m+n\neq0. \end{equation} $

(2.48)

又由关系

$ \begin{equation*} \phi([L_{m, i}, N_{n, j}], N_{-(m+n), k})+\phi([N_{n, j}, N_{-(m+n), k}], L_{m, i})+\phi([N_{-(m+n), k}, L_{m, i}], N_{n, j}) = 0, \end{equation*} $

从而有

$ \begin{equation} n\phi(N_{m+n, i+j}, N_{-(m+n), k})-(m+n)\phi(N_{n, j}, N_{-n, k+i}) = 0. \end{equation} $

(2.49)

在上式取$ m = 0 $, 则有

$ \begin{equation} \phi(N_{n, i+j}, N_{-n, k}) = \phi(N_{n, j}, N_{-n, k+i}), \quad n\neq0, \end{equation} $

(2.50)

在(2.49)式中取$ m = -n $, 可得

$ \begin{equation} \phi(N_{0, i+j}, N_{0, k}) = 0, \end{equation} $

(2.51)

由(2.50)和(2.51)式, 有

$ \begin{equation} \phi(N_{n, i+j}, N_{-n, k}) = \phi(N_{n, j}, N_{-n, k+i}), \quad \forall\, n, i, j, k\in \mathbb{Z}. \end{equation} $

(2.52)

上式说明仅与第二个指标的和$ i+j+k $有关, 而与位置无关.不妨设$ E_{n, i+j} = \phi(N_{n, i+j}, N_{-n, 0}) $, 在(2.49)式中令$ n = 1 $, 就有

$ \begin{equation} E_{m+1, i+j+k} = (m+1)E_{1, i+j+k}, \quad \forall\, m, i, j, k\in \mathbb{Z}. \end{equation} $

(2.53)

由(2.48)和(2.53)式, 则有$ \phi(N_{m, i}, N_{n, j}) = \delta_{m+n, 0}m\phi(N_{1, i+j}, N_{-1, 0}), \quad \forall \ m, n, i, j\in \mathbb{Z}. $

引理2.9    $ \phi(L_{m, i}, L_{n, j}) = \delta_{m+n, 0}\frac{m^{3}-m}{6}\phi(L_{2, i+j}, L_{-2, 0}), \quad \forall \ m, n, i, j\in \mathbb{Z} $.

证  由关系

$ \begin{eqnarray*} n\phi(L_{m, i}, L_{n, j}) & = & \phi(L_{m, i}, [L_{0, 0}, L_{n, j}]) \\ & = & \phi([L_{m, i}, L_{0, 0}], L_{n, j})+\phi(L_{0, 0}, [L_{m, i}, L_{n, j}]) \\ & = & -m\phi(L_{m, i}, L_{n, j})+(n-m)\phi(L_{0, 0}, L_{m+n, i+j}). \end{eqnarray*} $

由上式及(2.2)式可得

$ \begin{equation} \phi(L_{m, i}, L_{n, j}) = 0, \quad m+n\neq0. \end{equation} $

(2.54)

考虑关系

$ \begin{equation*} \phi([L_{m, i}, L_{n, j}], L_{-(m+n), k})+\phi([L_{n, j}, L_{-(m+n), k}], L_{m, i})+\phi([L_{-(m+n), k}, L_{m, i}], L_{n, j}) = 0, \end{equation*} $

整理得

$ \begin{equation} (n-m)\phi(L_{m+n, i+j}, L_{-(m+n), k})+(m+2n)\phi(L_{m, i}, L_{-m, j+k})-(2m+n)\phi(L_{n, j}, L_{-n, k+i}) = 0. \end{equation} $

(2.55)

在上式中令$ n = m $, 则有

$ \begin{equation} \phi(L_{m, i}, L_{-m, j+k}) = 0, \quad m\neq0. \end{equation} $

(2.56)

在(2.55)式中取$ m = -n = 1 $, 结合上式可得

$ \begin{equation} \phi(L_{0, i+j}, L_{0, k}) = 0, \end{equation} $

(2.57)

由(2.56)和(2.57)式, 有

$ \begin{equation} \phi(L_{m, i}, L_{-m, j+k}) = 0, \quad \forall\, m, j, k\in \mathbb{Z}. \end{equation} $

(2.58)

上式说明仅与第二个指标的和$ i+j+k $有关, 而与位置无关.不妨设$ F_{n, i+j} = \phi(L_{m, i+j}, L_{-m, 0}) $, 在(2.55)式中令$ m = 1 $, 结合(2.3)式, 则有

$ \begin{equation} (n-1)F_{n+1, i+j+k}-(n+2)F_{n, i+j+k} = 0. \end{equation} $

(2.59)

将上式$ n $由$ n-1 $替代, 从而

$ \begin{equation} (n-2)F_{n, i+j+k}-(n+1)F_{n-1, i+j+k} = 0. \end{equation} $

(2.60)

在(2.55)式中取$ m = 2, n = n-1 $, 可得

$ \begin{equation} (n-3)F_{n+1, i+j+k}-(n+3)F_{n-1, i+j+k} = -2nF_{2, i+j+k}. \end{equation} $

(2.61)

将(2.59), (2.60)和(2.61)式联立方程组, 解得

$ \begin{equation*} F_{n, i+j+k} = \frac{n^{3}-n}{6}F_{2, i+j+k}. \end{equation*} $

由上式及(2.54)式, 就得到$ \phi(L_{m, i}, L_{n, j}) = \delta_{m+n, 0}\frac{m^{3}-m}{6}\phi(L_{2, i+j}, L_{-2, 0}), \; \forall \ m, n, i, j\in \mathbb{Z}. $

由引理2.1–2.9, 我们就可以得到本文的主要定理.

定理2.1   二上同调群

$ \begin{equation*} H^{2}(\widetilde{\mathscr{W}}, \ \mathbb{C}) = \prod\limits_{k\in \mathbb{Z}, \ i\in I} \mathbb{C}\bar{\phi}_{k, i}, \end{equation*} $

其中$ I = \{1, 2, 3\} $, $ \mathbb{C}\bar{\phi}_{k, i} $表示所有的直积, $ \bar{\phi}_{k, i} $是$ \phi_{k, i} $的同调类, 满足

$ \begin{eqnarray*} \phi_{k, 1}(L_{m, i}, L_{n, j}) & = & \delta_{m+n, 0}\delta_{i+j, k}\frac{m^{3}-m}{12}, \\ \phi_{k, 2}(L_{m, i}, N_{n, j}) & = & \delta_{m+n, 0}\delta_{i+j, k}(m^{2}-m), \\ \phi_{k, 3}(N_{m, i}, N_{n, j})& = & \delta_{m+n, 0}\delta_{i+j, k}m, \end{eqnarray*} $

其他关系均为$ 0 $, 对于$ m, n, i, j, k\in \mathbb{Z} $.

注  考虑中心扩张$ \widehat{\mathscr{W}} = \widetilde{\mathscr{W}}\bigoplus C^{1}\otimes\mathbb{C}[t, t^{-1}]\bigoplus C^{2}\otimes\mathbb{C}[t, t^{-1}]\bigoplus C^{3}\otimes\mathbb{C}[t, t^{-1}] $, 满足李积关系:对于任意给定的$ k\in \mathbb{Z} $, 有

$ \begin{eqnarray*} && [L_{m, i}, L_{n, j}] = (n-m)L_{m+n, i+j}+\delta_{m+n, 0}\delta_{i+j, k}\frac{m^{3}-m}{12}C^{1}_{k}, \\ && [L_{m, i}, N_{n, j}] = nN_{m+n, i+j}+\delta_{m+n, 0}\delta_{i+j, k}(m^{2}-m)C^{2}_{k}, \\ && [N_{m, i}, N_{n, j}] = 2M_{m+n, i+j}+ \delta_{m+n, 0}\delta_{i+j, k}mC^{3}_{k}, \\ && [\widehat{\mathscr{W}}, C^{t}_{k}] = 0, \quad \forall \, m, n, i, j\in \mathbb{Z}, \end{eqnarray*} $

其中$ C^{t}_{k} = C^{t}\otimes t^{k}, t = 1, 2, 3 $, 其他李积关系与$ \widetilde{\mathscr{W}} $一致.定理2.1表明$ \widehat{\mathscr{W}} $是$ \widetilde{\mathscr{W}} $的泛中心扩张.