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I'm trying to understand the equivalence of unoriented knots in oriented 3-manifolds for my thesis, and getting confused. I have not found a satisfactory definition of this equivalence.

My understanding of the equivalence relations is the following:

Let $M$ be an oriented 3-manifold. An unoriented knot is an embedding of $S^1 \to M$. Two unoriented knots are considered equivalent if there is an ambient isotopy between them.

But then, consider the following two knots in the solid torus $[0, 1]^2 \times S^1$.

$\phi: S^1 \to [0, 1]^2 \times S^1$ is given by $e^{2\pi i t} \to ((0, 0), e^{2\pi i t})$

$\psi: S^1 \to [0, 1]^2 \times S^1$ be given by $e^{2\pi i} \to ((0, 0), e^{-2\pi i t})$.

Their images are exactly the same, namely the (1, 0)-curve in the solid torus. I feel like they are equivalent as unoriented knots, however I do not think there is an ambient isotopy between them.

My questions:

  1. Is my definition of an unoriented knot right? I thought of considering only the image of an embedding, but then I don't understand when they are equivalent, as we can no longer use the notion of ambient isotopy.
  2. Is it true that there is no ambient isotopy between $\phi$ and $\psi$?
  3. What is the reason we do not consider $\phi$ and $\psi$ to be equivalent as unoriented knots (if this is the case)?

Hopefully someone can unknot this mess for me!

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So, to resolve this, I think you need to be a little more careful with your definitions. The definition of equivalent knots you give is what I would say out loud if I was talking about knots, but not exactly what you are thinking of. I would say a knot is the image of an embedding, not the map itself. This is more along the lines of thinking of the knots as rope you can hold. So, your two knots are the image of $\phi$ and $\psi$, which is exactly the same set, as you pointed out. So the ambient isotopy is the identity map for all $t\in I$.

If you want to use the maps, you are forcing an orientation upon your knots, even though you are talking about unoriented knots. In this case, your two knots are not equivalent because you can't make $\phi(t)=\psi(t)$ for all $t$. (Hueristically, you cant turn the knot around in the solid torus.) But you can resolve this by noticing that if you ignore orientation, like you want to anyway, you can reverse the orientation on $S^1$ before you apply $\psi$. Let $f:S^1\to S^1$ be $f(t)=-t$ then let $\bar{\psi}=\psi \circ f$ and this gives you that $\phi=\bar{\psi}$ and then these two maps are ambient isotopic.

Hope this helps.

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    $\begingroup$ Thanks for your answer. If we define a knot to be the image of an embedding, then what does an ambient isotopy between two knots mean? Two embeddings can be ambient isotopic, but two sets cannot. Should we then define the equivalence to be (on unoriented knots as images of embeddings): Two unoriented knots $\phi, \psi$ are equivalent if there is a homeomorphism $\alpha$ of $M$ that is ambient isotopic to the identity of $M$ such that $\alpha(\phi) = \beta$. Would this work or does this definition introduce other subtleties? $\endgroup$ – G.L. Dec 11 '14 at 20:04
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    $\begingroup$ I believe that might have the same problem as before. Really, if you want to only consider unoriented embeddings, you could just allow the map $f$ to be applied first. $\endgroup$ – N. Owad Dec 11 '14 at 21:54

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