Timeline for Holonomy of compact manifolds
Current License: CC BY-SA 3.0
16 events
| when toggle format | what | by | license | comment | |
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| Apr 16, 2012 at 6:11 | vote | accept | Earthliŋ | ||
| Apr 11, 2012 at 16:46 | answer | added | Claudio Gorodski | timeline score: 3 | |
| Apr 11, 2012 at 16:32 | comment | added | Claudio Gorodski | A very simple example is a cone of revolution in $\mathbb R^3$ minus the vertex. The restricted holonomy (that is, that generated by parallel transport along loops homotopic to a point) is trivial, because the metric is flat, but the full holonomy can be a countable dense subgroup of $SO(2)$ if you choose correctly the angle of opening of the cone. | |
| Apr 11, 2012 at 14:14 | comment | added | Misha | @s.barmeier: No, in the flat affine case holonomy need not be discrete (it is countable though). Furthermore, if it is discrete, it need not be finite. The simplest example is the Hopf torus: Divide ${\mathbb R}^n\setminus 0$ by a cyclic group of dilations. More interesting example: Take a discrete cocompact torsion-free subgroup $\Gamma<O(n,1)$ and divide the positive light-cone by the product action of $\Gamma$ and an infinite cyclic group acting by dilations. The quotient is homeomorphic to the product of a hyperbolic manifold and $S^1$, its holonomy is $\Gamma\times {\mathbb Z}$. | |
| Apr 11, 2012 at 14:03 | comment | added | Misha | @Mariano: Yes, there are few exceptional cases when pseudo-Riemannian manifold will have compact holonomy, but locally they are products of flat pseudo-Riemannian manifolds, Riemannian manifolds and "anti-Riemannian" manifolds where you take negative of a Riemannian metric. | |
| Apr 11, 2012 at 13:27 | answer | added | Robert Bryant | timeline score: 12 | |
| Apr 11, 2012 at 7:49 | answer | added | Vladimir S Matveev | timeline score: 3 | |
| Apr 11, 2012 at 4:59 | answer | added | Ryan Budney | timeline score: 5 | |
| Apr 11, 2012 at 4:23 | comment | added | Earthliŋ | I'm interested in flat compact manifolds with any kind of affine connection in the tangent bundle. By the Ambrose-Singer holonomy theorem we get that the holonomy of flat manifolds must be discrete... I was hoping it would be finite for compact manifolds. Maybe that was just wishful thinking...? | |
| Apr 11, 2012 at 4:23 | comment | added | Mariano Suárez-Álvarez | @Misha, if we put on $\mathbb R^2$ a "constant" metric of signature $(1,1)$ and pass to the quotient under the usual action of $\mathbb Z^2$, we get a pseudo-Riemannian torus which is still flat, no? The holonomy there is trivial, so compact. | |
| Apr 11, 2012 at 4:02 | comment | added | Misha | Maybe you mean a pseudo-Riemannian manifold. Then it is non-compact, of couse. | |
| Apr 11, 2012 at 3:54 | comment | added | Ryan Budney | Assuming you have a connection and the holonomy is linear, I think it's compact if and only if its Levi-Cevita. | |
| Apr 11, 2012 at 3:52 | comment | added | Somnath Basu | Sorry if I'm being dense here but how would you define the holonomy group for a non-Riemannian (compact) manifold? In my limited understanding, the holonomy groups are associated to a bundle with a connection. For Riemannian manifolds, it is the holonomy group associated with the Levi-Civita connection on the tangent bundle which is of interest and classified by Berger. | |
| Apr 11, 2012 at 3:50 | comment | added | Ryan Budney | What do you want holonomy to mean on a non-Riemannian manifold? Do you want it to be a manifold with some kind of non-Levi-Cevita connection? Any restrictions? Ehresmann minimally? | |
| Apr 11, 2012 at 3:42 | history | edited | Yemon Choi |
added DG tag
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| Apr 11, 2012 at 3:23 | history | asked | Earthliŋ | CC BY-SA 3.0 |