Timeline for Are the higher homotopy groups of the Hawaiian earring trivial?
Current License: CC BY-SA 3.0
6 events
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| Apr 22, 2014 at 23:23 | history | edited | Paul Fabel | CC BY-SA 3.0 |
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| Apr 21, 2014 at 14:29 | comment | added | Paul Fabel | If A and C are disjoint, K =1 suffices, and this is provable by induction on the number of segments comprising A and B. If A and C are not disjoint, let D denote the intersection. Let A=A1*DA2 and C=C1*DC2. The diameter of the planar projection of Ai is at most that of Ci. Hence K=3 is an upper bound on the ratio of diameters of the planar projections of A and C. | |
| Apr 21, 2014 at 14:19 | comment | added | Paul Fabel | It is more straightforward to see K=3 is a universal upper bound. Let $Q$ denote the universal cover of $P$. Note $Q$ is CAT(0). Take any two geodesics segments A and B in $Q$. In $Q$, naturally project A into B. Call the image C. Now map A and C naturally into the plane and compare the ratios of the their diameters, using absolute value. | |
| Apr 21, 2014 at 4:37 | comment | added | Paul Fabel | There is a genuine gap since K=1.5 can actually happen. Place the letter V in the upper half plane with vertex at (0,0) and angles 60 degrees with w.r.t. x axis. Now place a different V with origin vertex, with angles 30 degrees and very long edges. Now project orthogonally the first V into the 2nd V while fixing the origin. This is purportedly the worst case, yielding a kind of poor man's Gehring-Hayman Theorem. | |
| Apr 21, 2014 at 0:18 | comment | added | Anton Petrunin | Nice argument; so you claim that in general planar continua (i.e., not for Hawaiian earring) there is a gap; can you fix it? | |
| Apr 20, 2014 at 11:39 | history | answered | Paul Fabel | CC BY-SA 3.0 |