Timeline for Proof of the Reidemeister theorem
Current License: CC BY-SA 2.5
6 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Oct 6, 2014 at 10:40 | comment | added | Reza Rezazadegan | General position argument (for polygon links) can be found in the book by Crowell and Fox and is pretty simple. | |
| Feb 14, 2010 at 23:07 | comment | added | Ryan Budney | Ah, okay. I take it everyone here is thinking about proofs in the PL category. I prefer the proof in the smooth category as there's less details to fuss over. Of course you have to use transversality so you have a bigger machine at your disposal. | |
| Feb 14, 2010 at 12:37 | comment | added | Dan Margalit | Hi Ryan. As for lemma (2): You start with two diagrams of the equivalent PL knots. Equivalent means they differ by a finite sequence of triangle moves. So now you need to either rotate the projection plane (or modify the triangle moves) so that each move gives a knot with a regular projection. By regular position (lemma 1), you can do this. However, if you rotate too much, the new diagrams of your original two knots are not equivalent to the original diagrams. Therefore, you are just showing that there exist two diagrams of the original knots that differ by R moves. Does that make sense? | |
| Feb 14, 2010 at 9:31 | comment | added | Ryan Budney | Hi Dan, the general position argument (1) has a nice proof using the transversality technique in the proof of the weak Whitney embedding theorem given in Guillemin and Pollack. I'm not quite understanding your lemma (2) statement. | |
| Feb 14, 2010 at 5:22 | comment | added | algori | Dan, thanks! I've seen the proof Prasolov and Sossinsky give (my impression is that they give more details than most sources but not all of them). It would be very interesting to see your notes, if you decide to scan them. I can in turn scan mine and post them somewhere. | |
| Feb 14, 2010 at 0:02 | history | answered | Dan Margalit | CC BY-SA 2.5 |