{"title":"Tate cohomology of connected k-theory for elementary abelian groups revisited","authors":"Po Hu, Igor Kriz, Petr Somberg","doi":"10.1007/s40062-018-00229-6","DOIUrl":null,"url":null,"abstract":"<p>Tate cohomology (as well as Borel homology and cohomology) of connective K-theory for <span>\\(G=({\\mathbb {Z}}/2)^n\\)</span> was completely calculated by Bruner and Greenlees (The connective K-theory of finite groups, 2003). In this note, we essentially redo the calculation by a different, more elementary method, and we extend it to <span>\\(p>2\\)</span> prime. We also identify the resulting spectra, which are products of Eilenberg–Mac Lane spectra, and finitely many finite Postnikov towers. For <span>\\(p=2\\)</span>, we also reconcile our answer completely with the result of [2], which is in a different form, and hence the comparison involves some non-trivial combinatorics.</p>","PeriodicalId":0,"journal":{"name":"","volume":null,"pages":null},"PeriodicalIF":0.0,"publicationDate":"2019-01-10","publicationTypes":"Journal Article","fieldsOfStudy":null,"isOpenAccess":false,"openAccessPdf":"https://sci-hub-pdf.com/10.1007/s40062-018-00229-6","citationCount":"0","resultStr":null,"platform":"Semanticscholar","paperid":null,"PeriodicalName":"","FirstCategoryId":"100","ListUrlMain":"https://link.springer.com/article/10.1007/s40062-018-00229-6","RegionNum":0,"RegionCategory":null,"ArticlePicture":[],"TitleCN":null,"AbstractTextCN":null,"PMCID":null,"EPubDate":"","PubModel":"","JCR":"","JCRName":"","Score":null,"Total":0}
引用次数: 0
Abstract
Tate cohomology (as well as Borel homology and cohomology) of connective K-theory for \(G=({\mathbb {Z}}/2)^n\) was completely calculated by Bruner and Greenlees (The connective K-theory of finite groups, 2003). In this note, we essentially redo the calculation by a different, more elementary method, and we extend it to \(p>2\) prime. We also identify the resulting spectra, which are products of Eilenberg–Mac Lane spectra, and finitely many finite Postnikov towers. For \(p=2\), we also reconcile our answer completely with the result of [2], which is in a different form, and hence the comparison involves some non-trivial combinatorics.