Timeline for Why is the string group not a Lie group?
Current License: CC BY-SA 2.5
15 events
| when toggle format | what | by | license | comment | |
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| Oct 30, 2016 at 2:09 | answer | added | David Roberts♦ | timeline score: 10 | |
| Sep 24, 2010 at 12:50 | vote | accept | Konrad Waldorf | ||
| Sep 24, 2010 at 12:28 | history | edited | Konrad Waldorf | CC BY-SA 2.5 |
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| Sep 24, 2010 at 8:55 | answer | added | Daniel Pomerleano | timeline score: 7 | |
| Sep 24, 2010 at 8:36 | comment | added | Daniel Pomerleano | Actually this method could never work because rationally the map from $SU(2)=Spin(3) \mapsto K(Z,3)$ is a rational homotopy equivalence. But I think if you do it integrally you get torsion cohomology groups in infinitely many dimensions so it can't be a finite dimensional Lie group. @Mariano Well I interpreted the question as why isn't this group a finite dimensional non-compact Lie group. Of course it could be some kind of infinite dimensional Lie group but then the facts about $pi_3$ don't seem relevant. | |
| Sep 24, 2010 at 8:31 | comment | added | Mariano Suárez-Álvarez | In what way is your cohomology calculation related to the question, Daniel? | |
| Sep 24, 2010 at 8:14 | comment | added | Daniel Pomerleano | G should be Spin there...Hopefully this argument shouldn't be totally wrong but I can't work it out in my head...The idea is to look at rational cohomology. Spin(n), being a compact Lie group should have rational cohomology isomorphic to the exterior algebra on odd spheres, $\Lambda(x_3,...x_{2i+1})$. We have a fibration $String(n) \mapsto Spin(n) \mapsto K(Z,3)$ which should give rise to a fibration $K(Z,2) \mapsto String(n) \mapsto Spin(n)$. Now apply the rational Serre spectral sequence and use the multiplicative structure on K(Z,2) this should be calculable, what happens? | |
| Sep 24, 2010 at 8:03 | comment | added | S. Carnahan♦ | I think your definition needs to be revised to make the string group uniquely defined. In particular, String(n) should be the universal 3-connected term in the Moore-Postnikov tower of $O(n)$, and the definition given above does not exclude higher terms. Some of the commenters may be confused because it is not a covering group in the usual sense of covering spaces. | |
| Sep 24, 2010 at 7:58 | comment | added | Daniel Pomerleano | The string group is given by killing pi_0, pi_1,pi_2,pi_3. So for O(n) it ends up being the homotopy fiber of the map G ---> K(Z,3). Certainly what you guys are saying means it can't be represented by a compact Lie group. What does it's rational cohomology look like? Well if I am thinking about this the right way it doesn't anymore satisfy Poincare duality? | |
| Sep 24, 2010 at 7:47 | answer | added | David Roberts♦ | timeline score: 11 | |
| Sep 24, 2010 at 7:45 | comment | added | algori | ... or $\pi_3\neq 0$. Argh! | |
| Sep 24, 2010 at 7:37 | comment | added | algori | Michele -- a covering group of a Lie group is a Lie group. So either the connected component of the unit is a torus of $\pi_3\neq 0$. I presume the author meant something else but I'm not sure what exactly. | |
| Sep 24, 2010 at 7:32 | comment | added | user47274 | @algori could you give a reference for this (or a sketch of an argument)? | |
| Sep 24, 2010 at 7:29 | comment | added | algori | Konrad -- unless $n\leq2$ there is no group at all satisfying the conditions of the posting. | |
| Sep 24, 2010 at 7:24 | history | asked | Konrad Waldorf | CC BY-SA 2.5 |