Reductive subalgebras of semisimple Lie algebras and Poisson commutativity

Pub Date : 2020-12-07 DOI:10.4310/jsg.2022.v20.n4.a4
D. Panyushev, O. Yakimova
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引用次数: 1

Abstract

Let $\mathfrak g$ be a semisimple Lie algebra, $\mathfrak h\subset\mathfrak g$ a reductive subalgebra such that $\mathfrak h^\perp$ is a complementary $\mathfrak h$-submodule of $\mathfrak g$. In 1983, Bogoyavlenski claimed that one obtains a Poisson commutative subalgebra of the symmetric algebra ${\mathcal S}(\mathfrak g)$ by taking the subalgebra ${\mathcal Z}$ generated by the bi-homogeneous components of all $H\in{\mathcal S}(\mathfrak g)^{\mathfrak g}$. But this is false, and we present a counterexample. We also provide a criterion for the Poisson commutativity of such subalgebras ${\mathcal Z}$. As a by-product, we prove that ${\mathcal Z}$ is Poisson commutative if $\mathfrak h$ is abelian and describe ${\mathcal Z}$ in the special case when $\mathfrak h$ is a Cartan subalgebra. In this case, ${\mathcal Z}$ appears to be polynomial and has the maximal transcendence degree $(\mathrm{dim}\,\mathfrak g+\mathrm{rk}\,\mathfrak g)/2$.
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半单李代数的约化子代数与泊松交换性
设$\mathfrak g$是一个半简单李代数,$\mathfrak h\子集$ mathfrak g$是一个约化子代数,使得$\mathfrak h^\perp$是$\mathfrak g$的补$\mathfrak h$-子模块。1983年,Bogoyavlenski声称,通过取所有$H\ In {\mathcal S}(\mathfrak g)^{\mathfrak g}$的双齐次分量所生成的子代数${\mathcal Z}$,可以得到对称代数${\ mathfrak S}(\mathfrak g)$的一个Poisson交换子代数。但这是错误的,我们提出一个反例。我们也给出了这类子代数的泊松交换性的一个判据。作为副产物,我们证明了如果$\mathfrak h$是阿贝尔的,则${\ mathfrak Z}$是泊松交换的,并且在$\mathfrak h$是Cartan子代数的特殊情况下描述了${\ mathfrak Z}$。在这种情况下,${\mathcal Z}$似乎是一个多项式,并且具有最大超越度$(\ mathm {dim}\,\mathfrak g+\ mathm {rk}\,\mathfrak g)/2$。
本文章由计算机程序翻译,如有差异,请以英文原文为准。
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