Timeline for For which $b$ it is possible that $S^n$ can have a Lorentz metric? Why?
Current License: CC BY-SA 2.5
15 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jan 20, 2012 at 8:31 | vote | accept | Yan Zou | ||
| Jan 20, 2012 at 8:31 | vote | accept | Yan Zou | ||
| Jan 20, 2012 at 8:31 | |||||
| Jan 20, 2012 at 8:31 | vote | accept | Yan Zou | ||
| Jan 20, 2012 at 8:31 | |||||
| Jan 6, 2012 at 2:10 | answer | added | Renato G. Bettiol | timeline score: 21 | |
| Dec 7, 2010 at 16:55 | vote | accept | Yan Zou | ||
| Jan 20, 2012 at 8:31 | |||||
| Nov 30, 2010 at 11:50 | comment | added | Yan Zou | @Andrei, Ryan was right, he pointed out the key point of the problem in just take the proof of Riemannian metric to Lorentz metric. I noticed that but didn't get an idea of solving it. But still I wonder why it must take a section of the projective unit tangent bundle.... | |
| Nov 26, 2010 at 18:54 | comment | added | Ryan Budney | I don't believe it is. My statement was that the proof is essentially identical to the Riemann metric case, once you adapt to the requirement of having a section of the projective unit tangent bundle. Both are convexity arguments. | |
| Nov 26, 2010 at 18:47 | comment | added | Andrei Moroianu | Ryan, yes, I did not read the first comment, sorry. Still, the last sentence of your second comment was misleading. | |
| Nov 26, 2010 at 18:44 | comment | added | Deane Yang | Andrei, I think Ryan is suggesting that when you try to adapt the proof for a Riemannian metric to a Lorentzian one, you run into anecessary condition for the Lorentzian metric to exist. Then you can check that the necessary condition is also sufficient. | |
| Nov 26, 2010 at 18:43 | comment | added | Ryan Budney | @Andrei: it appears you did not read what I said in my first comment "Manifolds have Lorentz metrics if and only if they have a section of the corresponding projective unit tangent bundle". | |
| Nov 26, 2010 at 18:36 | answer | added | Andrei Moroianu | timeline score: 31 | |
| Nov 26, 2010 at 18:00 | comment | added | Andrei Moroianu | @Ryan: should one understand that with a partition of unity argument you can show the existence of Lorenz metrics on every (paracompact) manifold? I have serious doubts about that... | |
| Nov 4, 2010 at 21:16 | comment | added | Ryan Budney | If you don't want to post your question on math.stackexchange, take a look in the nearest manifolds textbook at the proof that abstract manifolds admit Riemann metrics (i.e. avoid proofs that use embeddings in euclidean space). This is a paracompactness/bump function/partition of unity argument. Now consider how you would perturb that proof for Lorentz metrics. | |
| Nov 4, 2010 at 18:52 | comment | added | Ryan Budney | Manifolds have Lorentz metrics if and only if they have a section of the corresponding projective unit tangent bundle. If you want details, please post to the math.stackexchange website listed in the FAQ as this is more of an upper-level undergraduate mathematics issue. | |
| Nov 4, 2010 at 18:48 | history | asked | Yan Zou | CC BY-SA 2.5 |