While preparing for my introduction to topology course, I've realized that I don't know where to find a detailed proof of the Reidemeister theorem (two link diagrams give isotopic links, iff they can be connected by a sequence of Reidemeister moves). The students in my class are not really very advanced and I'm not sure all of them would be able to reconstruct all the details of the proof given in Burde and Zieschang's Knot theory or in other sources I know of. And, being a lazy man, I would like to avoid having to type a detailed proof with pictures when I can refer the students to a book. So does anybody know of a source which would have a proof with all details spelled out?
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$\begingroup$ Did you ever find a satisfactory resource? I find myself in a positions similar to yours seven years ago. $\endgroup$– David SteinbergCommented Apr 20, 2017 at 18:37
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$\begingroup$ I'll post again here, just in case progress has been made in the last 5+ years :-) $\endgroup$– David SteinbergCommented Aug 18, 2022 at 1:00
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5 Answers
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I taught knot theory last semester and ran into the same problem. I looked in every book I could get my hands on, and could not find an undergraduate level proof. In the end, I wrote up my own notes (which I would be happy to scan when I get back into the office). The key ideas for the case-by-case analysis are in the book "Knots, links, braids, and 3-manifolds" by Prasolov and Sosinsky. I also found Louis Kauffman's book "On knots" to be helpful. There are two lemmas I could not find anywhere: (1) the general position argument, which says that there is a nice projection and (2) the argument which says that you can find a general projection so that the associated diagram is equivalent to the original diagram (most books skip this issue). The point of the second lemma is that it is not enough to show that there exist two projections that differ by Reidemeister moves, rather, you want to show that the two given diagrams differ by Reidemeister moves.
answered Feb 14, 2010 at 0:02
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$\begingroup$ 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. $\endgroup$– algoriCommented Feb 14, 2010 at 5:22
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$\begingroup$ 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. $\endgroup$ Commented Feb 14, 2010 at 9:31
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$\begingroup$ 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? $\endgroup$ Commented Feb 14, 2010 at 12:37
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$\begingroup$ 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. $\endgroup$ Commented Feb 14, 2010 at 23:07
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$\begingroup$ General position argument (for polygon links) can be found in the book by Crowell and Fox and is pretty simple. $\endgroup$ Commented Oct 6, 2014 at 10:40
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Reidemeister's proof involves a single move: replacing 2 (or 1) edges of a triangle with the other edge (edges). It is in the English translation of his book.
I don't know if you can ferret out the details of our first movie move theorem in JKTR, or even the CRS version, but when we did those, I felt that we had addressed the general position issues that Dan mentions.
answered Feb 14, 2010 at 3:09
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$\begingroup$ Scott, thanks! Your second paragraph is a bit cryptic. For the record: I know how to prove Reidemeister's theorem, I'm looking for a detailed proof accessible to undergraduate students with a limited experience of mathematics, let alone topology. $\endgroup$– algoriCommented Feb 14, 2010 at 5:30
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1$\begingroup$ I did not mean to be cryptic at all, but it was late when I wrote it. Masahico and I proved a "movie move theorem" that was published in Journ. of Knot Thy. and its Ram. There were problems with the interpretations that people could make of it. So we proved a more precise result with Joachim Rieger (CRS). A lot of the transversality arguments are worked out in both places. In particular, when your knot diagram has a height function, you have to deal with critical interchange, zig-zag, and $\psi$ moves. I just don't remember if we gave a proof of the R thm, or not. Probably not. $\endgroup$ Commented Feb 14, 2010 at 13:48
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Kunio Murasugi's Knot Theory and its Applications contains what you are looking for, I believe. I taught the proof of Reidemeister's theorem as part of a short course on knot theory, from that book. Definitely suitable for undergraduates.
answered Feb 14, 2010 at 13:24
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$\begingroup$ Thanks, Daniel! Will try to borrow this tomorrow from the library. $\endgroup$– algoriCommented Feb 14, 2010 at 16:57
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Messer and Straffin's book "Topology Now!" provides most of the steps starting from their definition of a knot (able to decomposed into a finite number of linear segments), building up the ideas of general position and triangular moves, and then proving the Reidemeister moves. I say "most of" because some of the steps are left as exercises. The material is very accessible to undergraduates.
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$\begingroup$ I think it gives an acceptable outline of how Reidemeister's theorem could be proven, but I don't think it provides a detailed proof. Eg the first half of their 1-page proof is "We begin with diagrams of two equivalent knots. Rotate the knots so that the diagrams of the two knots are based on projections into the same plane. Now adjust the projection so that both knots, as well as the triangles involved in all triangular moves, project in general position. Make this adjustment so that it does not disturb the combinatorics of the knot diagrams." $\endgroup$ Commented Feb 8, 2022 at 19:22
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If you are willing to take a difficult theorem of Moise for granted (which proves equivalence of link isotopy in the smooth (tame) and PL category, in his "Affine Structures in 3 Manifolds VIII"), there is a concise proof in the book "Knot Theory" by Manturov.
