# Question about definition of regular projection of a knot

One can define a knot in two ways:

(1) A knot is a closed polygonal curve in $\mathbb R^3$

(2) A knot is an equivalence class of embeddings $S^1 \hookrightarrow \mathbb R^3$

And perhaps also:

(3) A knot is an equivalence class of smooth $1$-dimensional submanifolds of $S^3$

Question 1: Can I replace $S^3$ in (3) with $\mathbb R^3$?

Question 2: I would like to define what a regular projection of a knot is. Unfortunately, it depends on whether I use (1), (2) or (3). I would like to use (2) and I have the definition using (1), which goes as follows:

(Definition) A knot projection is called a regular if no three points on the knot project to the same point, and no vertex projects to the same point as any other point on the knot.

How can I define regular projection without (1), that is, how can I define it for embeddings instead of polygonal curves?

Thanks a lot!

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None of these constitutes a definition of a knot. The content of the definition is in the equivalence relation on these objects whose equivalence classes actually define knots (namely ambient isotopy). –  Qiaochu Yuan Oct 9 '12 at 8:38
@QiaochuYuan Sorry, I'm getting confused. For example, in Kauffman's "On knots" on the first page (page 3) he writes "... Classical knot theory studies embeddings of $S^1$ in $\mathbb R^3$ ...". –  Rudy the Reindeer Oct 9 '12 at 8:51
Yes and yes. A knot is an equivalence class of embeddings of $S^1$ into $\mathbb{R}^3$ up to ambient isotopy (en.wikipedia.org/wiki/Ambient_isotopy). These equivalence classes can also be represented by piecewise linear curves, by smooth curves, or by any other reasonable kind of curve, which is why it doesn't matter which kind of curves you take. –  Qiaochu Yuan Oct 9 '12 at 9:12
Another way of addressing the concern I'm raising here is to define the category of knots to be the category whose objects are embeddings of $S^1$ into $\mathbb{R}^3$ and whose morphisms are ambient isotopies. Then a knot should refer to an object in this category, but up to isomorphism as is usual in category theory. But if you haven't said something equivalent to "ambient isotopy" then you haven't captured arguably the most important part of what people mean when they say "knot." –  Qiaochu Yuan Oct 9 '12 at 9:14
While Qiaochu Yuan raises a good point, your post also seems to contain some other concerns: (a) How to define a knot diagram precisely, (b) how to prove that the notion of knot diagrams makes sense in the first place and (c) how to recognize whether two knots are ambient isotopic using the diagrams. The connection is provided by Reidemeister's theorem: two knots (or links) are ambient isotopic iff they have diagrams that can be transformed into each other by a sequence of Reidemeister moves. This MO thread contains a few references, maybe they help. –  commenter Oct 9 '12 at 18:53

1. Yes, you can replace $\mathbf R^3$ with $S^3$ everywhere and it will essentially not make a difference. Your assumptions imply that the knot misses a point which you can add/remove to go back and forth between either one. This MO question might be stimulating: http://mathoverflow.net/questions/63158/
2. A definition if the embedding is smooth (or just differentiable) is that a projection is regular if the composite map $\phi \colon S^1 \to \mathbf R^2$ has the property that no three points are mapped to the same point, and that if $\phi(p)=\phi(q)$, then nonzero tangent vectors at $p$ and $q$ are mapped to linearly independent tangent vectors under $d\phi$. In other words, the image of the knot crosses itself transversely.