de Sitter时空上的对称二阶张量场可以构成so(1,4)-模. 爱因斯坦场方程在de Sitter度规附近的线性化方程的解也会构成一个so(1,4)-模, 这个模一定是一些不可约so(1,4)-模的直和, 而有限维的不可约so(1,4)-模可以通过Verma模来构造. 所以, 找出全部二阶对称张量构成的so(1,4)-Verma模对于构造爱因斯坦场方程的线性化方程的通解是很关键的. 文中通过类似量子力学中处理角动量本征值问题的方法, 求出了两个so(1,4)-Verma模的最高权向量.

Symmetric (0,2)-tensor fields on the 4-dimensional de Sitter spacetime form so(1,4)-modules. Also, the solutions of linearized Einstein’s field eqution near de Sitter metric form an so(1,4) module, which should be the direct sum of some irreducible so(1,4)-modules. Finite dimensional irreducible so(1,4)-modules can be structed out of Verma modules. Therefore, finding all the so(1,4)-Verma modules, consisting of symmetric (0,2)-tensor fields on the 4-dimensional de Sitter spacetime, is a key step to solve the linearized Einstein’s field equation about the de Sitter metric. In this paper, we derived two maximal vectors of Verma so(1,4)-modules, using a method similar to which of solving the eigenvalue problem of angular momentum in quantum mechanics.