Timeline for Can every curve be made transversal to a foliation by applying a pseudo-Anosov?
Current License: CC BY-SA 3.0
12 events
| when toggle format | what | by | license | comment | |
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| Jan 22, 2022 at 21:51 | history | edited | Sam Nead |
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| Mar 15, 2018 at 1:48 | vote | accept | Adam | ||
| Mar 7, 2018 at 16:10 | answer | added | Lee Mosher | timeline score: 6 | |
| Mar 7, 2018 at 16:03 | comment | added | Adam | @SamNead Indeed, $n$ depends on $\gamma.$ If you see a way to proceed, I would be curious to know. | |
| Mar 7, 2018 at 13:52 | comment | added | Lee Mosher | The best you are going to get about support is that every $\mathcal F$ has an invariant measure. It might be that the support is a finite union of simple closed curves each contained in a leaf. Even if there is no simple closed curve contained in any leaf, the support still need not be full, because of the Denjoy construction. | |
| Mar 7, 2018 at 8:36 | comment | added | Sam Nead | I assume that the power $n$ is allowed to vary with the loop or arc $\gamma$? If we have that, and if the foliation $\mathcal{F}$ has a measure of full support, then we should win. | |
| Mar 7, 2018 at 2:31 | history | edited | Adam | CC BY-SA 3.0 |
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| Mar 7, 2018 at 2:30 | comment | added | Adam | Speaking of that, is there some result asserting that every $\cal F$ has an invariant measure with partial, but "reasonably" large support? | |
| Mar 7, 2018 at 0:17 | comment | added | Lee Mosher | As you say, the support of the measure is disjoint from the interior of the Reeb component. | |
| Mar 6, 2018 at 21:35 | comment | added | Adam | @LeeMosher: When you say $\cal F$ has an invariant transverse measure even with a Reeb component, do you mean a measure of full support? Wouldn't an arc connecting the two boundaries of the Reeb annulus intersect each leaf infinitely many times and, hence, have an infinite transverse length? | |
| Mar 6, 2018 at 14:55 | comment | added | Lee Mosher | Its false for every $\cal F$ that has a Reeb component. en.wikipedia.org/wiki/Reeb_foliation#/media/… And even when $\cal F$ has a Reeb component it will always have an invariant transverse measure. | |
| Mar 3, 2018 at 7:49 | history | asked | Adam | CC BY-SA 3.0 |
