Timeline for Can every curve be made transversal to a foliation by applying a pseudo-Anosov?
Current License: CC BY-SA 3.0
5 events
| when toggle format | what | by | license | comment | |
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| Mar 15, 2018 at 1:48 | vote | accept | Adam | ||
| Mar 14, 2018 at 14:12 | comment | added | Lee Mosher | Because if $\mathcal{F}^u_\phi$ were transverse to $\mathcal{F}$ then a long leaf segment of $\mathcal{F}^u_\phi$ which intersects every leaf of $\mathcal{F}$ could be approximated by a closed curve transverse to and intersecting every leaf of $\mathcal F$, which directly violates the property of transverse recurrence. | |
| Mar 14, 2018 at 14:03 | comment | added | Adam | Many thanks for this very comprehensive answer! A quick follow-up: Why $\cal F$ being not trasversely recurrent implies that no $\cal F_\phi^u$ is transversal to $\cal F$? | |
| Mar 7, 2018 at 16:41 | history | edited | Lee Mosher | CC BY-SA 3.0 |
added 1504 characters in body
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| Mar 7, 2018 at 16:10 | history | answered | Lee Mosher | CC BY-SA 3.0 |