Show that the inner product $$ex sef{x_1vee cdots vee x_k,y_1vee cdotsvee y_k} eex$$ is equal to the permanent of the $k imes k$ matrix $sex{sef{x_i,y_j}}$.
Solution. $$eex ea &quad sef{x_1vee cdots vee x_k,y_1vee cdots vee y_k}\ &=frac{1}{k!} sum_{sigma, au} sef{x_{sigma(1)},y_{ au(1)}} cdots sef{x_{sigma(k)},y_{ au(k)}}\ &=frac{1}{k!} sum_{sigma, au} sef{x_1,y_{ au(sigma^{-1}(1))}} cdots sef{x_k,y_{ au(sigma^{-1}(k))}} \ &=frac{1}{k!} sum_{sigma}sez{ sum_{ au} sef{x_1,y_{ au(sigma^{-1}(1))}} cdots sef{x_k,y_{ au(sigma^{-1}(k))}}} \ &=frac{1}{k!} sum_{sigma}per sex{sef{x_i,y_j}}\ &=per sex{sef{x_i,y_j}}. eea eeex$$