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I really want to like knot theory but the PL condition seems sort of ugly. I was hoping someone could give me a justification for secretly thinking about smooth knots as I read through a book like Rolfsen which works with PL knots. I'm really hoping for some grand unifying or almost-unifying theorem. Thank you for reading my question.

Edit: I think what I may be hoping for is some sort of "isomorphism of categories" if that makes any sense at all. Where the categories might be knots and functors are (continuous or smooth) ambient isotopies. Is this even a sensible formulation?

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1 Answer 1

Smooth knots are equivalent to PL knots if they are not wild. You can have differentiable curves which are wild, but every $C^1$ closed curve is not wild.

To prove that $C^1$ curves are equivalent to $PL$ I would do the following steps:

  1. find $\epsilon > 0$ such that the in the $\epsilon$-neighborhood of the curve there is a unique projection on the curve.

  2. find a finite number of points on the curve and use them to divide the $\epsilon$-neighbourhood in many cylinders with bases orthogonal to the curve in the given points. If the points are choosen close enough the segment joining two points will stay inside the $\epsilon$-neighbourhood.

  3. construct an homotopy in each small cylinder so that the boundary of the cylinder is kept fixed and the curve goes to the segment. Use the fact that the couple (cylinder-like neighbourhood,curve) is homeomorphic to (real cylinder,axis).

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I see, thank you. Here you have in mind equivalence in the PL sense of a continuous ambient isotopy or an ambient homeomorphism taking one curve to the other? –  user61460 Feb 8 '13 at 10:37
    
a continuous ambient isotopy is possible. In fact in every point of the curve you can find a small neighbourhood where it is possible to make a deformation to let the curve become flat. –  Emanuele Paolini Feb 8 '13 at 10:41
    
@manu-fatto I'm assuming this equivalence is proved using transversality theory... (should be the same as proving the sufficiency of the 3 Reidemeister moves, right?) is there a standard reference? Thanks... –  gmoss Feb 9 '13 at 21:20
    
I don't think that transversality is an issue here. I added something in the answer... –  Emanuele Paolini Feb 10 '13 at 9:42

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