What is the grading of x(x−1)R[x]? Loopspace for Karoubi-Villamayor K-theory. - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-20T01:13:20Z http://mathoverflow.net/feeds/question/97210 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/97210/what-is-the-grading-of-xx1rx-loopspace-for-karoubi-villamayor-k-theory What is the grading of x(x−1)R[x]? Loopspace for Karoubi-Villamayor K-theory. Simon Markett 2012-05-17T10:42:11Z 2012-05-18T05:15:32Z <p>I am reading the chapter on Karoubi-Villamayor K-theory in Weibel's K-book. In particular he defines $\Omega R=(x^2-x)R[x]$ for a ring. This will eventually lead to a model for the loopspace</p> <p>Corollary 11.13.1 For $n\geq 2$ we have, $$KV_n(R)\cong KV_{n-1}(\Omega R).$$</p> <p>The proof uses the long exact sequence of the GL-fibration $$(x^2-x)R[x]\to xR[x]\to R$$</p> <p>and uses the following fact to show that $KV_n(xR[x])$ vanishes:</p> <p>Exercise 11.5 Let $R=R_0\oplus R_1\oplus\cdots$ be a graded ring. Then for every homotopy invariant functor $F$ on rings, possibly without unit, (i.e. the map $R\to R[x]$ induces homotopy equivalences; true for $KV$) then we have $F(R)\simeq F(R_0)$.</p> <p>Now the question: Can't we just apply the exercise to $\Omega R$ as well? Or is the grading different? I am really confused here.</p> <p>Edit: If anybody is put off by the word Karoubi-Villamayor, just assume $R$ regular, as we then have $KV=K$.</p>