Timeline for Infinite loop space maps into or out of BAut(F_n)
Current License: CC BY-SA 2.5
5 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Oct 22, 2010 at 15:48 | comment | added | John Rognes | Yes, the homotopy fiber of $Q(S^0) \to A(\ast)$, that you call "Unknown Space", is $\Omega Wh^{Diff}(\ast)$, the stable smooth $h$-cobordism space of a point. There is map to it from the space $H^{Diff}(D^n)$ of $h$-cobordisms on $D^n$, and the connectivity of the map grows to infinity with $n$. The map $G/O \to \Omega Wh^{Diff}(\ast)$ is a rational equivalence, and is precisely $8$-connected after $2$-completion. Refs: M. Bökstedt, The rational homotopy type of $\Omega Wh^{Diff}(\ast)$, MR0764574, and J. Rognes, Two-primary algebraic $K$K-theory of pointed spaces, MR1923990. | |
| Oct 22, 2010 at 15:11 | comment | added | Romeo | Will read those papers, looks like they may be just what I want. Sorry, I wrote $Aut(S^{\infty})$ for some reason; I just meant lim $n \to \infty deg +/- 1 self maps of S^n$, which I I'll call $G$. The map between homotopy fibers G/O $\to$ Unknown Space must be interesting. But I'll hold off any speculations until I read the references you gave. | |
| Oct 22, 2010 at 14:36 | comment | added | John Rognes | I don't quite understand your notation, but, yes, there is a known map $BO \to Q(S^0)$ so that the composite $BO \to Q(S^0) \to A(\ast)$ is homotopic to the composite $BO \to BGL_1(S) \to A(\ast)$, where $BO \to BGL_1(S)$ is the $j$-map. With multiplicative infinite loop structures on the unit components of $Q(S^0)$ and $A(\ast)$ these are even infinite loop maps. Refs: F. Waldhausen, Algebraic $K$K-theory of spaces, a manifold approach, MR0686115 J. Rognes, The Hatcher-Waldhausen map is a spectrum map, MR1282230. | |
| Oct 22, 2010 at 4:42 | comment | added | Romeo | Ah, this is great, thanks. $Q(S^0) = Z \times BS_{\infty}^+ \to A(\ast) = Z \times BGL(S)^+$ is basically (finite set with n elements) $\to$ (n-frame) on the second factor. There's a clear map $BAut(S^{\infty}) \to A(\ast)$, basically inclusion into $BGL_{\infty}$ if I'm getting this right (followed by +-construction and crossign with Z). Looking at the map $BO \to BAut(S^{\infty})$, is there a map $BO \to Q(S^0)$ making a nice square? | |
| Oct 21, 2010 at 9:33 | history | answered | John Rognes | CC BY-SA 2.5 |