Timeline for How commutative is Quillen's Plus-Construction?
Current License: CC BY-SA 2.5
13 events
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| Apr 13, 2017 at 12:58 | history | edited | CommunityBot |
replaced http://mathoverflow.net/ with https://mathoverflow.net/
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| Nov 25, 2010 at 7:04 | vote | accept | Andreas Thom | ||
| Nov 25, 2010 at 0:05 | answer | added | Tyler Lawson | timeline score: 5 | |
| Nov 19, 2010 at 18:44 | comment | added | Tyler Lawson | (Though it factors through plus-constructions applied to higher homotopy groups.) | |
| Nov 19, 2010 at 18:42 | comment | added | Tyler Lawson |
@Johannes: The Kan-Thurston map as stated in their paper is in fact a plus construction on $\pi_1$, though indeed it's not always true that a map of spaces inducing a homology isomorphism and a surjection on fundamental groups is a plus construction.
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| Nov 19, 2010 at 18:10 | comment | added | Andreas Thom | @Johannes: I edited the question to incorporate your comment and I am now also asking whether such a space $X$ exists. I suggest that you post your comment as an answer, it is really nice. | |
| Nov 19, 2010 at 18:09 | history | edited | Andreas Thom | CC BY-SA 2.5 |
incorporated a comment
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| Nov 19, 2010 at 16:29 | comment | added | Johannes Ebert | In fact, I would use Kan-Thurston to construct a counterexample to question 2. 1. Find a space with abelian $\pi_1$ that does not admit a self-map inducing the inversion on $\pi_1$. I do not have a counterexample at hand, but would be \emph{really} surprised if such a space does not exist. Step 2. Apply Kan-Thurston to this space and check that the Kan-Thurston map is a plus construction (I believe that this is not obvious if there is some $H_1$ around). | |
| Nov 19, 2010 at 16:23 | comment | added | Johannes Ebert | Of course not every plus-construction is an $H$-space, in fact, it seems to me that an $H$-space structure on $BG^+$ is a rather rare event. The Kan-Thurston theorem says that for any space $X$, there is a group $G$ and a homology isomorphism $BG \to X$, which will be a plus-construction. Take a simply connected $X$ that is manifestly not an $H$-space ($S^2$, for example) and you find a perfect group that does not have an $H$-space structure on its plus-construction. | |
| Nov 19, 2010 at 12:38 | comment | added | Andreas Thom | I see, every connected H-space has an inverse up to homotopy. This solves the problem for $BGL_{\infty}(R)^+$. Thanks for this comment. However, the $H$-space structure is not directly induced by the multiplication of $GL_{\infty}(R)$, so that there is probably no reason to assume that $BG^+_H$ will be an $H$-space in general. | |
| Nov 19, 2010 at 12:11 | comment | added | Torsten Ekedahl | I think that the second question for $GL_\infty(R)$ follows directly from the fact that we have a connected $H$-space. | |
| Nov 19, 2010 at 7:36 | history | edited | Andreas Thom | CC BY-SA 2.5 |
edited title
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| Nov 19, 2010 at 7:20 | history | asked | Andreas Thom | CC BY-SA 2.5 |