## Proof of the Reidemeister theorem

### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)

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?

-

## 3 Answers

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.

-

Thanks, Daniel! Will try to borrow this tomorrow from the library. – algori Feb 14 2010 at 16:57

### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you.

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.

-
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. – algori Feb 14 2010 at 5:30

1
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. – Scott Carter Feb 14 2010 at 13:48

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.

-

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. – algori Feb 14 2010 at 5:22
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. – Ryan Budney Feb 14 2010 at 9:31
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? – Dan Margalit Feb 14 2010 at 12:37
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.   – Ryan Budney Feb 14 2010 at 23:07
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9559940099716187, "perplexity_flag": "head"}