A. Carderi, Matthieu Joseph, F. L. Maitre, R. Tessera
{"title":"Belinskaya定理是最优的","authors":"A. Carderi, Matthieu Joseph, F. L. Maitre, R. Tessera","doi":"10.4064/fm266-4-2023","DOIUrl":null,"url":null,"abstract":"Belinskaya's theorem states that given an ergodic measure-preserving transformation, any other transformation with the same orbits and an $\\mathrm{L}^1$ cocycle must be flip-conjugate to it. Our main result shows that this theorem is optimal: for all $p<1$ the integrability condition on the cocycle cannot be relaxed to being in $\\mathrm{L}^p$. This also allows us to answer a question of Kerr and Li: for ergodic measure-preserving transformations, Shannon orbit equivalence doesn't boil down to flip-conjugacy.","PeriodicalId":55138,"journal":{"name":"Fundamenta Mathematicae","volume":" ","pages":""},"PeriodicalIF":0.5000,"publicationDate":"2022-01-17","publicationTypes":"Journal Article","fieldsOfStudy":null,"isOpenAccess":false,"openAccessPdf":"","citationCount":"1","resultStr":"{\"title\":\"Belinskaya’s theorem is optimal\",\"authors\":\"A. Carderi, Matthieu Joseph, F. L. Maitre, R. Tessera\",\"doi\":\"10.4064/fm266-4-2023\",\"DOIUrl\":null,\"url\":null,\"abstract\":\"Belinskaya's theorem states that given an ergodic measure-preserving transformation, any other transformation with the same orbits and an $\\\\mathrm{L}^1$ cocycle must be flip-conjugate to it. Our main result shows that this theorem is optimal: for all $p<1$ the integrability condition on the cocycle cannot be relaxed to being in $\\\\mathrm{L}^p$. This also allows us to answer a question of Kerr and Li: for ergodic measure-preserving transformations, Shannon orbit equivalence doesn't boil down to flip-conjugacy.\",\"PeriodicalId\":55138,\"journal\":{\"name\":\"Fundamenta Mathematicae\",\"volume\":\" \",\"pages\":\"\"},\"PeriodicalIF\":0.5000,\"publicationDate\":\"2022-01-17\",\"publicationTypes\":\"Journal Article\",\"fieldsOfStudy\":null,\"isOpenAccess\":false,\"openAccessPdf\":\"\",\"citationCount\":\"1\",\"resultStr\":null,\"platform\":\"Semanticscholar\",\"paperid\":null,\"PeriodicalName\":\"Fundamenta Mathematicae\",\"FirstCategoryId\":\"100\",\"ListUrlMain\":\"https://doi.org/10.4064/fm266-4-2023\",\"RegionNum\":3,\"RegionCategory\":\"数学\",\"ArticlePicture\":[],\"TitleCN\":null,\"AbstractTextCN\":null,\"PMCID\":null,\"EPubDate\":\"\",\"PubModel\":\"\",\"JCR\":\"Q3\",\"JCRName\":\"MATHEMATICS\",\"Score\":null,\"Total\":0}","platform":"Semanticscholar","paperid":null,"PeriodicalName":"Fundamenta Mathematicae","FirstCategoryId":"100","ListUrlMain":"https://doi.org/10.4064/fm266-4-2023","RegionNum":3,"RegionCategory":"数学","ArticlePicture":[],"TitleCN":null,"AbstractTextCN":null,"PMCID":null,"EPubDate":"","PubModel":"","JCR":"Q3","JCRName":"MATHEMATICS","Score":null,"Total":0}
Belinskaya's theorem states that given an ergodic measure-preserving transformation, any other transformation with the same orbits and an $\mathrm{L}^1$ cocycle must be flip-conjugate to it. Our main result shows that this theorem is optimal: for all $p<1$ the integrability condition on the cocycle cannot be relaxed to being in $\mathrm{L}^p$. This also allows us to answer a question of Kerr and Li: for ergodic measure-preserving transformations, Shannon orbit equivalence doesn't boil down to flip-conjugacy.
期刊介绍:
FUNDAMENTA MATHEMATICAE concentrates on papers devoted to
Set Theory,
Mathematical Logic and Foundations of Mathematics,
Topology and its Interactions with Algebra,
Dynamical Systems.