Timeline for Examples and non-examples of Riemannian foliations
Current License: CC BY-SA 3.0
7 events
| when toggle format | what | by | license | comment | |
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| Feb 8, 2012 at 16:21 | comment | added | Robert Bryant | @David: I just realized what may have confused you. I shouldn't have talked about the holonomy being compact, rather, I should have said that a necessary condition for the existence of a transverse metric is that the holonomy of each leaf have compact closure, which is what I was thinking, even though I didn't write it. I'm sorry if this confused you. | |
| Feb 7, 2012 at 15:55 | comment | added | Robert Bryant | @David: You have misread what I wrote or aren't taking into account the fact that I'm talking about the particular example that I gave, in which the leaves are $1$-dimensional and thus are either circles or copies of $\mathbb{R}$. In the former case, the holonomy is generated by a single element (corresponding to the generator of $\pi_1$), in the later case, the holonomy is trivial, in this particular example. The claim about a generator with real eigenvalue greater than $1$ obstructing a transverse metric seems obvious to me, so I'm not sure what you are not getting. | |
| Feb 7, 2012 at 14:15 | comment | added | David Carchedi | Also, are you claiming that if there is such a non-compact leaf such that the generator of the holonomy group induces an endomorphism of the tangent space with a real eigenvalue with modulus greater than 1, then there cannot be a tranverse metric on the foliated manifold? Any references, or clarifications would be most helpful. Thank you! | |
| Feb 7, 2012 at 14:14 | comment | added | David Carchedi | @Robert: I am a little confused. The holonomy group of any leaf of any (regular) foliated manifold is discrete, by definition, since it is a quotient of the fundamental group. Are you speaking of something else? Are you claiming that if I have a foliated manifold $\left(M,F\right)$ and a leaf which is not compact (and I am working here without corners or boundaries) then its holonomy group has to have a single generator? If so, why is this true and where can I read this? | |
| Feb 7, 2012 at 12:58 | comment | added | Robert Bryant | @David: Yes, the holonomy of each leaf is discrete in this case. If the leaf isn't closed, it's trivial and, if the leaf is closed, it's generated by a single element. However, that single element will not preserve any metric if it has a real eigenvalue of modulus greater than $1$, which is the case for closed geodesics on a surface of negative curvature. This is is thoroughly discussed under the topic of Anosov flows, which you should Google for a full explanation. The Wikipedia page on this is quite explicit. | |
| Feb 7, 2012 at 11:59 | comment | added | David Carchedi | The holonomy group of any leaf is a discrete group, so what do you mean by compact? Did you mean finite? By the way, do you have a reference for this non-existence claim? Thanks! | |
| Feb 7, 2012 at 1:21 | history | answered | Robert Bryant | CC BY-SA 3.0 |