Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Suppose we call a knot an equivalence class of embeddings of S1 --> R3 under ambient isotopy, a knot representative a particular such embedding, and a knot diagram the "2 1/2 dimensional" shadow of such a knot representative on S2 from a particular vantage point P, i.e. the light source for the shadow is at P, and the "2 1/2" means that the over and under information is shown.

My question involves how the set of knot diagrams of a particular fixed knot representative vary as P, the viewpoint varies. We know about the Reidemeister rules, which generate a group of transformations of knot diagrams in a way such that the underlying knot remains invariant under changes of both ambient isotopy and projection. Intuitively there is also a smaller group of pure "projective" transformations of knot diagrams under which any fixed knot representative remains invariant. Assuming that, what are the relations between these two groups of transformations of knot diagrams?

share|improve this question
    
I'm not sure what you mean by the "pure projective transformations"? Do you mean the changes to a knot diagram (in R^2) resulting from an affine change to the projection R^3 -> R^2? –  Kevin Walker Jul 8 '10 at 13:13
    
Yes, (15 character min). –  sigoldberg1 Jul 8 '10 at 14:04
    
A little motivation. It seemed to me that going from the idea of a knot's equivalence class under ambient isotopy to transformations of its knot diagrams was too much in one step for best understanding. I am looking for a 'natural way' to split the notions via knot equivalence class --> a representative embedding --> knot diagrams of that embedding. If it was a normal subgroup of the group of Reidemeister transformations, we could just factor it out, but I don't know whether that's true. If not, I wonder what the relations actually are. –  sigoldberg1 Jul 8 '10 at 14:59
    
I'm confused by this question. You get all Reidemeister moves by changes of the projection R^3 -> R^2, as (for instance) in the proof of Reidemeister's theorem in Burde-Zieschang, page 9-10, or (in more detail) pages 52-56 of Murasugi's "Knot Theory and its Applications". Try it physically with shadows of pencils or something. Is this what you were asking? –  Daniel Moskovich Jul 8 '10 at 15:16
2  
There is no such collections of moves. The problem is your knot diagram does not faithfully encode the embedding in $\mathbb R^3$. So depending on which way you reconstruct the embedding, rotate tand then project, you will need different sets of moves. In particular you can make the moves you need arbitrarily complicated by choosing "bad" realizations of the knot where the vertical coordinates have a very high-freqency component (in the Fourier decomposition sense). –  Ryan Budney Jul 15 '10 at 0:00

3 Answers 3

up vote 2 down vote accepted

I remember attending a talk by Barbara Jablonska at Knots in Washington (2009) in which she studied a knot in a geometric, rigid fashion. As the direction of projection varies over $S^2$, she obtained interesting surfaces by looking at the locus of a particular crossing (if memory serves). Here's an abstract of the talk, but I cannot find anything published.

http://atlas-conferences.com/c/a/x/q/18.htm

share|improve this answer
    
Thanks, I'll follow up and report back –  sigoldberg1 Jul 15 '10 at 18:35

As Daniel Moskovich points out, you don't get a different set of moves on (isotopy classes of) planar diagrams.

More specifically, given diagrams $D_1$ and $D_2$ which differ by a (any) single Reidemeister move, one can find isotopic diagrams $D'_1$ and $D'_2$ such that $D'_1$ and $D'_2$ are both projections (shadows) of the same fixed knot $K$ in $R^3$.

The only way I see to rescue your question is to not take isotopy classes of knot diagrams in $R^2$ -- just consider "moves" form one rigid diagram to another. But this seems to be a difficult question even for diagrams without any crossings.

share|improve this answer
    
Thank you for clarifying above. I agree with you and Daniel Moskovich, that I have definitely not been asking about isotopy classes of knot diagrams in R<sup>2</sup>, which only manipulate one crossing at a time, but rather about affine changes to the projection of the entire rigid knot (representative) pi(K,P) from R<sup>3</sup>-->S<sup>2</sup>(-->R<sup>2</sup>). But what makes you think this is such a difficult question? Weird, maybe, but why difficult? –  sigoldberg1 Jul 9 '10 at 6:38
    
It's an ill-posed question. If you want to change the projection you need an actual embedding in $\mathbb R^3$ to begin with and your diagram does not give you that (it gives you a continuum of embeddings with one common projection). –  Ryan Budney Jul 15 '10 at 0:02

If you consider knots made of piecewise linear segments, then one can say a couple of things. First, crossings will be between various segments, and for any pair of segments, one could compute the possible projections for any given direction. This will give a quadrilateral of projection directions where the segments cross. If you're interested in generic projections, then one could compute the non-generic phenomena. A Reidemeister 3 move occurs when there's a triple of segments which project to a triple crossing. The set of lines through 3 skew segments forms part of a hyperbolic paraboloid or hyperboloid. So the Gauss image will be part of a quadratic curve in $S^2$. There are some other possible non-generic phenomena, such as the planes containing adjacent segments, or vertices and segments (the boundaries of the quadrilaterals). In any case, once you've computed these, $S^2$ gets partitioned up into regions in which a generic projection occurs, and in principle one could compute each projection for each region. In fact, this is one way to prove that the Reidemeister moves suffice for PL equivalence of knots.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.