Timeline for Is there a bound on the length of the longest Morse trajectory?
Current License: CC BY-SA 3.0
12 events
| when toggle format | what | by | license | comment | |
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| Jul 27, 2018 at 10:08 | comment | added | Pietro Majer | Some compactness is needed, even if not really in crucial way; the assumption of compactness of M may be replaced by the Palais-Smale condition, or even some weaker form of it. What is certainly crucial is the Morse-Smale assumption, whose first consequence is that any flow line may only connect critical points with Morse indices in strict decreasing order. | |
| Jul 26, 2018 at 21:40 | comment | added | Alessio Pellegrini | Does the "monotonicity of Morse indices"-argument use the compactness of $M$ in some crucial way? | |
| Mar 7, 2012 at 11:09 | history | edited | Pietro Majer | CC BY-SA 3.0 |
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| Nov 29, 2010 at 17:50 | comment | added | Orbicular | Excuse me, just a small question: how does one obtain the neighborhood $U_x$ and the bound $c=c(f,g)$? | |
| Oct 4, 2010 at 19:17 | comment | added | Pietro Majer | Yes, that's correct. | |
| Oct 4, 2010 at 19:08 | history | edited | Pietro Majer | CC BY-SA 2.5 |
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| Oct 4, 2010 at 19:02 | comment | added | Orbicular | Thanks, Pietro. The proof will generalize to the case of a fixed Morse function f on a Hilbert manifold (with fixed complete metric) assuming the PS condition if one fixes the endpoints, right? Because the f-values on the Morse trajectory stay in a finite windows. Hence the number of possible critical points it comes close to is finite. Furthermore dim(M) in your equation should be replaced by the relative Morse index. | |
| Oct 4, 2010 at 18:51 | vote | accept | Orbicular | ||
| Oct 4, 2010 at 17:50 | history | undeleted | Pietro Majer | ||
| Oct 4, 2010 at 17:49 | history | edited | Pietro Majer | CC BY-SA 2.5 |
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| Oct 4, 2010 at 16:40 | history | deleted | Pietro Majer | ||
| Oct 4, 2010 at 16:36 | history | answered | Pietro Majer | CC BY-SA 2.5 |