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We can either define a knot to be

(1) a smooth embedding $S^1 \hookrightarrow \mathbb R^3$

or

(2) a piecewise linear, simple closed curve in $\mathbb R^3$

Then these two definitions are equivalent which I assume means that if $K$ is a knot in the sense of $(1)$ then we can find an isotopy to make it into a knot in the sense of $(2)$. (correct me if that's wrong)

Question 1: Could we drop the smoothness requirement in $(1)$ and still get an equivalent definition? I assume not but I don't see why. Say, if we require the embedding to be differentiable it seems to me that it's also piecewise linear.

Question 2: What do smooth maps look like? What should I think of when thinking of smooth maps? I can think of $e^x$ but that seems useless.

Thanks for your help.

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2 Answers 2

up vote 3 down vote accepted

You are correct that one can isotope a piecewise-linear map to an arbitrarily close smooth map.

Answer to 1: No, you cannot. The smoothness of the embedding excludes "wild knots", which are not locally flat:

image

For instance, the complement of such a wild knot has infinitely generated fundamental group. Such pathologies fall outside the intuitive notion of a "knot" and so the definition excludes them.

As well as excluding nasties, smoothness also permits one to bring tools of differential topology to bear. Knots are quite intimately connected with the topology of three-manifolds. For instance, regard a knot as embedded in the three-dimensional sphere isntead of $\mathbb{R}^3$. Remove a tubular neighborhood of the knot and glue in a solid torus by some mapping class of the automorphisms of the torus. The result is a closed three-manifold. (In fact, it's a theorem by, I think, Lickorish that all closed oriented three-manifolds can be obtained by such an operation on links in the three-sphere.)

We have used that the knot has a tubular neighborhood in this construction, which is an object associated to a smooth submanifold in a smooth manifold.

Answer to 2: Smooth maps are infinitely differentiable functions. The standard exponential, polynomial, trigonometric, and hyperbolic functions are all smooth. In this case, think of a curve $\gamma:\mathbb{R}\to \mathbb{R}^3$ which is periodic and has coordinate functions that are all infinitely differentiable like the functions you've learned in calculus. For example, the simplest knot can be written like this: $u(t) = (\cos{t},\sin{t},0)$.

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Thank you! Would one time continuously differentiable not exclude wild knots? It seems to me that it would but I'm not sure how to prove or even verify it. What do you think of this? –  Matt N. Oct 22 '12 at 10:12
    
Or put differently: do we have "smooth embedding" <=> "locally flat" <=> "tame"? –  Matt N. Oct 22 '12 at 12:32
1  
@MattN. I'm asking around - it seems that $C^1$ embeddings would not exclude wild knots. –  Neal Oct 23 '12 at 12:55
1  
(The thinking is that even if you have $C^1$, you don't have control over second derivatives, so the knot can wind more and more tightly on itself.) –  Neal Oct 24 '12 at 12:15
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Maybe. I'm still thinking about the question of $C^1$ versus $C^\infty$, and becoming skeptical of my previous pseudo-answer. –  Neal Oct 25 '12 at 3:31

For your first question, you can't drop the smoothness requirement. This is to prevent the standard definition of knot from including so-called "wild knots", as seen here:

wild_knot

As for the second question, you can think of (at least I do, in some sense) smooth maps as being the higher-dimensional analog of the smooth curves you've dealt with since your first calculus course. They admit derivatives of all orders and so avoid singularities like cusps and self-intersection. Someone with more knowledge than myself on this topic will likely give better intuition into this idea.

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