Timeline for Topologizing free abelian groups
Current License: CC BY-SA 2.5
13 events
| when toggle format | what | by | license | comment | |
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| Jun 1, 2010 at 10:42 | comment | added | BS. | Adding to Dev's comment : Z[S] can be thought as a configuration space of particles in $S$ carrying integral charges which add when particles merge. You can replace $Z$ by other groups $Z/2$ for instance. Then Dold-Thom gives nice models of Eilenberg-Maclane spaces K(Z,n) and K(Z/2,n) as configuration spaces of particles on $S^n$ ($n=1$ and $2$ are interesting to compare to the usual models). | |
| Mar 31, 2010 at 21:47 | vote | accept | HenrikRüping | ||
| Mar 30, 2010 at 21:35 | answer | added | Martin Brandenburg | timeline score: 9 | |
| Mar 30, 2010 at 21:19 | comment | added | Dev Sinha | Yes, algebraic topologists do use this construction regularly, but when S is "nice" (a CW complex or something). The remarkable Dold-Thom theorem is essentially that the homotopy groups of Z[S] are the homology groups of S! | |
| Mar 30, 2010 at 14:24 | answer | added | Keenan Kidwell | timeline score: 1 | |
| Mar 30, 2010 at 14:06 | comment | added | Fabrizio Polo | Chris Schomer-Preis' comment made me think a little more. I was under the impression that my construction worked at least when $S$ was something nice like compact metric. If it goes wrong with a stranger space $S$ (as suggested by Dominguez below) then it would be really interesting to see some examples. | |
| Mar 30, 2010 at 14:02 | comment | added | Keenan Kidwell | Your comment reminds me of the complete group algebra of a profinite group G. This object is frequently used in algebraic number theory. If $\mathcal{O}$ is a, say, coefficient ring, meaning a complete, Noetherian local ring with finite residue field, and $G$ a profinite group (so, as you say, a very special space), then there exists a compact topological ring $\mathcal{O}[[G]]$ and a continuous map $G\rightarrow\mathcal{O}[[G]]$ that parametrizes continuous homomorphisms from $G$ into the group of units of any coefficient ring. | |
| Mar 30, 2010 at 13:58 | answer | added | Xabier Domínguez | timeline score: 10 | |
| Mar 30, 2010 at 13:55 | comment | added | Chris Schommer-Pries | I added the algebraic topology tag because I think aspects of this question come up in that context. Some algebraic topologists might know an answer to this. | |
| Mar 30, 2010 at 13:53 | history | edited | Chris Schommer-Pries |
Added Tag: Algebraic Topology
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| Mar 30, 2010 at 13:48 | comment | added | Chris Schommer-Pries | My recollection is that such a topology on $\mathbb{Z}[S]$ does exist (and hence is unique by Fabrizio Polo's argument below), but that it is rather subtle and difficult to construct. I seem to remember that it depends on what exact category of topological spaces you are working in, e.g. compactly generated weakly Hausdorff spaces? vanilla topological spaces? etc. I think that it is easier to show that there exists some "free abelian group" generated by a topological space in the sense that it has the universal property. It's harder to show that it's underlying set is $\mathbb{Z}[S]$. | |
| Mar 30, 2010 at 13:36 | answer | added | Fabrizio Polo | timeline score: 1 | |
| Mar 30, 2010 at 13:20 | history | asked | HenrikRüping | CC BY-SA 2.5 |