Timeline for Unknotting knots in 4D
Current License: CC BY-SA 3.0
11 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jan 25, 2012 at 18:42 | vote | accept | Joseph O'Rourke | ||
| Jan 25, 2012 at 13:11 | comment | added | Joseph O'Rourke | This is a remarkable answer, Ian! I think it could become a publishable note. | |
| Jan 25, 2012 at 6:22 | comment | added | Ian Agol | Then we see that the convex hull at some finite stage must be this convex limit polygon, which gives a contradiction. | |
| Jan 25, 2012 at 6:21 | comment | added | Ian Agol | As for a proof of Erdos-Nagy, I didn't realize that this was a subtle problem. Suppose that the reflection process continues forever. Since each polygon contains the previous one, the space of the polygons is compact, so there is a subsequence converging to a convex polygon with the same side lengths. If this polygon is strictly convex, then the sequence must have been finite, since any nearby polygon is also strictly convex. If it has some corners with angle $\pi$, first we see that all of the corners with angle $<\pi$ must eventually be stable, and therefore eventually constant. | |
| Jan 25, 2012 at 3:47 | comment | added | Ian Agol | Sorry, I didn't really address the genericity in my previous comments. One way to view it is like this: take a sequence of chain moves in 3D which send the curve to a planar curve. There will be finitely many times at which there are double points of this sequence of rotations. At each of these times, push a bit into 4D to avoid the crossing, then back into 3D on the other side of the obstruction. | |
| Jan 25, 2012 at 2:19 | comment | added | Ian Agol | One gets a rigid rotation of 3-dim. half-spaces containing $P$, and therefore a rotation of the arc $a$ sending it to its mirror through $P$. | |
| Jan 25, 2012 at 2:18 | comment | added | Ian Agol | Kevin is correct, but I should probably say a bit more since the rotation is supposed to be rigid. Consider the supporting plane $P$ containing the edge $e$. The unit normal bundle to the segment $e$ is a 2-sphere. Any unit vector together with $P$ spans a 3-dimensional subspace and the unit vector defines a half-space in this subspace; for example when this is an orthogonal vector to $P$ in $R^3$, one obtains $R^3$. So take the unit vector pointing to the side of $P$ containing $K$, and rotate it to the unit vector in the side not containing $K$. | |
| Jan 25, 2012 at 2:03 | comment | added | Kevin Walker | Joseph: The track of the isotopy corresponding to the move is 2-dimensional. The other part of the knot that might interfere is 1-dimensional. 1+2 < 4, so generically these two submanifolds 4-space will not intersect. | |
| Jan 25, 2012 at 1:43 | comment | added | Joseph O'Rourke | "making sure that they don't interfere with the other side": How does genericity ensure this, if I may ask? | |
| Jan 25, 2012 at 1:41 | comment | added | Joseph O'Rourke | Just a remark: your 2D "warmup" is called the Erdős–Nagy theorem: en.wikipedia.org/wiki/Erd%C5%91s%E2%80%93Nagy_theorem . | |
| Jan 25, 2012 at 1:22 | history | answered | Ian Agol | CC BY-SA 3.0 |