关于对称排序的注释

Zoran vSkoda
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引用次数: 2

摘要

设A^n是第n个Weyl代数的微分算子的度补全,其生成器为x1,…,xn,∂1,…,∂n。考虑A^n中的n个元素X1,…,Xn,形式为xi =xi+∑K=1∞∑l=1n∑j=1nxlpijK−1,l(∂)∂j,其中pijK−1,l(∂)是∂1,…,∂n中的一个次(K−1)齐次多项式,下标i,j反对称。xi1⋯xik,其中Σ(k)是k个字母上的对称群,并且表示A^n在(交换)多项式空间上的Fock作用。
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A note on symmetric orderings
Let A^n be the completion by the degree of a differential operator of the n-th Weyl algebra with generators x1,…,xn,∂1,…,∂n. Consider n elements X1,…,Xn in A^n of the formXi=xi+∑K=1∞∑l=1n∑j=1nxlpijK−1,l(∂)∂j,where pijK−1,l(∂) is a degree (K−1) homogeneous polynomial in ∂1,…,∂n, antisymmetric in subscripts i,j. Then for any natural k and any function i:{1,…,k}→{1,…,n} we prove∑σ∈Σ(k)Xiσ(1)⋯Xiσ(k)▹1=k!xi1⋯xik,where Σ(k) is the symmetric group on k letters and ▹ denotes the Fock action of the A^n on the space of (commutative) polynomials.
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