{"title":"Reductive subalgebras of semisimple Lie algebras and Poisson commutativity","authors":"D. Panyushev, O. Yakimova","doi":"10.4310/jsg.2022.v20.n4.a4","DOIUrl":null,"url":null,"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 $(\\mathrm{dim}\\,\\mathfrak g+\\mathrm{rk}\\,\\mathfrak g)/2$.","PeriodicalId":0,"journal":{"name":"","volume":null,"pages":null},"PeriodicalIF":0.0,"publicationDate":"2020-12-07","publicationTypes":"Journal Article","fieldsOfStudy":null,"isOpenAccess":false,"openAccessPdf":"","citationCount":"1","resultStr":null,"platform":"Semanticscholar","paperid":null,"PeriodicalName":"","FirstCategoryId":"100","ListUrlMain":"https://doi.org/10.4310/jsg.2022.v20.n4.a4","RegionNum":0,"RegionCategory":null,"ArticlePicture":[],"TitleCN":null,"AbstractTextCN":null,"PMCID":null,"EPubDate":"","PubModel":"","JCR":"","JCRName":"","Score":null,"Total":0}
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 $(\mathrm{dim}\,\mathfrak g+\mathrm{rk}\,\mathfrak g)/2$.