Timeline for Motivation of Lawson Homology
Current License: CC BY-SA 4.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| May 8, 2018 at 21:25 | comment | added | Denis Nardin | @VincenzoZaccaro It depends if you're interested in the homotopy type or the weak homotopy type. Look for example at a profinite group: its weak homotopy type is just that of a discrete set (and so in particular an E-ML space). Of course its homotopy type is more complicated. | |
| May 8, 2018 at 20:25 | comment | added | Vincenzo Zaccaro | @DenisNardin It seems quite reasonable! :) Is the local path connctedness necessary? | |
| May 8, 2018 at 20:19 | comment | added | Vincenzo Zaccaro | @JasonStarr You are right. What I dont' understand is why the homotopy type of $\mathcal{Z}_p(X)$ is completely determined by the homotopy groups $\pi_\ast\mathcal{Z}_p(X)$ by using the theorem 3.4 (in the paper). | |
| May 8, 2018 at 19:32 | comment | added | Denis Nardin | I might be wrong, but if G is a (locally path connected) topological group and G_0 is the connected component of the identity, isn't there a (noncanonical) homeomorphism $G\cong G_0 \times \pi_0G$ given by choosing a point in each component, hence $G$ is a product of E-ML spaces? More homotopically, $\mathrm[{Sing}G$ is a simplicial abelian group, hence it's the homotopy type of a chain complex and so a product of E-ML spaces. | |
| May 8, 2018 at 19:06 | comment | added | Jason Starr | Did you read Corollary 8.2 of that article? Lawson does not claim that $\mathcal{Z}_p(X)$ is connected. Quite the contrary, he explicity describes the group of connected components. | |
| May 8, 2018 at 18:36 | comment | added | Vincenzo Zaccaro | I'm using the definition of group completion given by Lawson at pag. 148 of the paper Spaces of Algebraic Cycles. In the same paper the author says (pag. 157, beginnig of par.8) that, by the theorem cited above, the homotopy type of $\mathcal{Z}_p(X)$ is that of a product of E-M spaces. Paper: intlpress.com/site/pub/files/_fulltext/journals/sdg/1993/0002/… | |
| May 8, 2018 at 18:22 | comment | added | Jason Starr | Did you intend to add the hypothesis that the degree of $a$ equals the degree of $b$? If not, then $\text{degree}(a)-\text{degree}(b)$ defines a locally constant function on $\mathcal{Z}_p(X)$ that is not constant, hence $\mathcal{Z}_p(X)$ is not connected. | |
| May 8, 2018 at 17:49 | history | asked | Vincenzo Zaccaro | CC BY-SA 4.0 |