Timeline for Is there a Lie group which is made $S^n \cup_f ~e^m$?
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
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| Oct 1, 2011 at 9:50 | comment | added | Marco Golla | An even easier argument shows that any cell decomposition of a Lie group has to have an even number of cells, since it's Euler characteristic is zero (it's parallelisable, and in particular has a nowhere vanishing vector field). I'm not sure whether the argument can be adapted to H-spaces as well. | |
| Sep 26, 2011 at 9:54 | comment | added | Mark Grant | It's worth mentioning that the same argument rules out H-spaces with three cells, by Hopf's Theorem which says that the rational cohomology ring is an exterior algebra on odd-dimensional generators. | |
| Sep 26, 2011 at 5:51 | comment | added | Mariano Suárez-Álvarez | I know, Ben; but when they are sufficiently apart (that is, not contiguous, in fact :) ) you do not get a bound but an actual count. | |
| Sep 26, 2011 at 3:29 | comment | added | Ben Webster♦ | Mariano- The cells being far apart has nothing to do with it; the total dimension of cohomology over any field gives a lower bound on the number of cells you need. Of course, sometimes it's too low because of cancellation; the rational cohomology of $\mathbb{RP}^3$ looks like it might allow you to get away with 2 cells, but the mod 2 cohomology shows you need 4. | |
| Sep 26, 2011 at 2:11 | comment | added | Jino | @Webster: Can you recommend a reference which contain your explanation dim > 3? | |
| Sep 26, 2011 at 2:09 | comment | added | Jino | @Alvarez: Thank you. I missed that. | |
| Sep 26, 2011 at 2:03 | comment | added | Mariano Suárez-Álvarez | @Jino: if the dimension of your cells are far apart, then computing (co)homology using the cellular complex shows there is a relation. | |
| Sep 26, 2011 at 1:48 | comment | added | Jino | rational cohomology dimension and number of cells are related? | |
| Sep 26, 2011 at 1:36 | history | answered | Ben Webster♦ | CC BY-SA 3.0 |