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It's intuitively clear what it means that two knots $K,K'$ are essentially the same, but it can be termed and defined more precisely in different ways. Are all of them equivalent?

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The knots themselves are always homotopy equivalent - indeed, homeomorphic - because they are all homeomorphic to $\mathbb{S}^1$. Perhaps you want to ask if the knot complements are homotopy equivalent? – Zev Chonoles Feb 28 '13 at 17:50
@Zev: You see, I did not understand the concept of homotopy (equivalence) fully, yet. I believed that two knots are h-equivalent if they can be continuously deformed into each other. – Hans Stricker Feb 28 '13 at 17:55
It depends on how you interpret "continuously deformed into each other". For example, the circle $\mathbb{S}^1$, the cylinder $\mathbb{S}^1\times\mathbb{R}^1$, and the solid torus $\mathbb{S}^1\times\mathbb{D}^2$ are all homotopy equivalent to each other; homotopy equivalence does not detect "contractible thickness", in other words. I think the intuitive idea you are referring to with the phrase "continuously deformed into each other" is the one captured by the notion of ambient isotopy, because it requires that the homotopy "keep track of" how the spaces are embedded in the ambient space. – Zev Chonoles Feb 28 '13 at 18:00
@HansStricker As an aside - I think "being continuously deformed into each other" is a perfectly good way of thinking about homotopy, providing you remember that in this continuous deformation the object is allowed to pass through itself, so all knots can be untied. When you don't permit the object to self-intersect, you get an intuitive notion of isotopy. – Matthew Pressland Feb 28 '13 at 18:10
Knots are ambient isotopy classes of embeddings of $S^1$ into things. – Qiaochu Yuan Feb 28 '13 at 18:10
up vote 4 down vote accepted

Isotopy (as opposed to ambient isotopy) is not a good equivalence relation on knots, because any knot is isotopic to an unknot. The idea is to pull the knot tighter and tighter until it shrinks to a point. More precisely (but still not completely precisely, because that would be a mess to write down), the isotopy would start with the given knot at time $t=0$. At a later time $t$ (in $(0,1)$), it would look like a $t$ fraction of a circle (unknotted) with a copy of the original knot compressed into a $(1-t)$-sized part. Finally, at time $t=0$, it's the whole unknotted circle.

This is why people like to define equivalence of knots to mean ambient isotopy rather than mere isotopy.

To the best of my knowledge, there is an equivalent (to ambient isotopy) definition that says essentially that there is a self-homeomorphism of the ambient space sending the one knot to the other. Here, "essentially" refers to the fact that one has to be careful about orientations; it seems the self-homeomorphism of the ambient space should preserve orientation, but maybe there's a subtler point here that I'm overlooking.

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Thank you very much. Maybe I should have asked my question in a different way. What I really aim at is a name and a generalizable definition of what two equivalent (tame) knots have in common. Just like two equivalent (= homeomorphic) (closed connected orientable) surfaces have in common their number of holes (called genus). When we say that genus is the name of the homeomorphism type of (specific) surfaces, what is the name X of the Y-type of those surfaces? (Two unknowns.) Consider - among other things - arbitrarily knotted 2-tori. – Hans Stricker Feb 28 '13 at 19:57
The name is "knot." – Qiaochu Yuan Mar 1 '13 at 21:51
That's obviously consistent with your other comment: X = knot, Y = ambient isotopy. But that's for embeddings of $S^1$. In my comment to Andreas' answer, I was drifting away to embeddings of surfaces. – Hans Stricker Mar 1 '13 at 22:23

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