Timeline for Algebraic K-theory of odd-dimensional spheres
Current License: CC BY-SA 3.0
12 events
| when toggle format | what | by | license | comment | |
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| Oct 20, 2013 at 19:59 | vote | accept | Igor Belegradek | ||
| Oct 18, 2013 at 22:55 | comment | added | Igor Belegradek | Tracing back the references I found all the rational homotopy groups of $A(S^n)$ listed on pp.228-229 of 1982 paper ["A Model for Computing Rational Algebraic K-Theory of Simply Connected Spaces", Hsiang-Staffeldt]. | |
| Oct 18, 2013 at 9:31 | history | edited | Ricardo Andrade | CC BY-SA 3.0 |
added authors of article
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| Oct 18, 2013 at 4:23 | comment | added | Tom Goodwillie | Also, the rational answer is older than that paper. Rationally $A(X)$ is the same as the algebraic $K$-theory of the simplicial ring $\mathbb Z[G]$ where $G$ is a Kan loop group for the connected space $X$, and when $X$ is simply connected then the reduced part of this is rationally the same as cyclic homology of $\mathbb Q[G]$ by work of mine, and the latter is the same as rational homology of $S^1$ homotopy orbits of the free loopspace of $X$. | |
| Oct 18, 2013 at 4:17 | comment | added | Tom Goodwillie | John, the description of $\Omega \tilde A(\Sigma Y)$ for connected $Y$ is valid on the nose, just rationally. I've taken the liberty of editing your answer accordingly. | |
| Oct 18, 2013 at 3:15 | comment | added | John Klein | Oh, I did not see the word "even." | |
| Oct 18, 2013 at 2:39 | comment | added | Igor Belegradek | I did not claim that $A(S^n)$ is rationally $(n-1)$-connected, just that the even-dimensional homotopy groups vanish rationally up to dimension $n-1$. | |
| Oct 18, 2013 at 2:34 | history | edited | John Klein | CC BY-SA 3.0 |
added 5 characters in body
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| Oct 18, 2013 at 2:31 | comment | added | John Klein | No, $A(S^n)$ is not rationally $(n-1)$-connected. But the reduced space $\tilde A(S^n)$ is, by the displayed formula in my answer. The reason that $A(S^n)$ is not rationally $(n-1)$-connected is that $A(*)$ splits off of it. | |
| Oct 18, 2013 at 2:30 | comment | added | John Klein | No, it don't think it's hard it seems to me that the rational homology of $S^{jn}_{h\Bbb Z_n}$ is determined completely by the action of $Z_n$ on $H^{nj}(S^{nj};\Bbb Q) = \Bbb Q$. | |
| Oct 18, 2013 at 1:53 | comment | added | Igor Belegradek | Thank you! I gather from your "not hard to compute" that there is no easy to state answer. Am I correct that $\pi_{k}(A(S^n))$ is rationally zero for even $k$ with $k<n$? I think this follows from the formula on page 1 of the linked Berglund-Madsen's paper, could you confirm this (if true)? | |
| Oct 18, 2013 at 1:21 | history | answered | John Klein | CC BY-SA 3.0 |