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Is there any special study of knots in this particular 3-manifold?

A more targeted / simple question: What are some nontrivial examples of knots $S^1\subset S^1\times S^2$, and is there convenient way to view them?

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If $A$ is translation and $C$ is a rational twist, then $\mathbb{S}^2\times \mathbb{R}/(C\times A)$ is homeomorphic to $\mathbb{S}^2\times\mathbb{S}^1$ and the image of $*\times\mathbb{R}$ under the quotient should be a nontrivial knot. – Neal Dec 3 '12 at 20:57
I recently saw a very cool talk which was about "knot theory from a non-knot theorist's perspective" -- it gives a framework in which to study all embeddings of manifolds, of which knots or links (in any given ambient manifold) are a special case. You can see my notes here: – Aaron Mazel-Gee Dec 4 '12 at 13:07
up vote 2 down vote accepted

Typically knot theory in most small 3-manifolds reduces in various ways to knot theory in $S^3$.

For example, if the complement of a knot in $S^1 \times S^2$ isn't irreducible, there's a $2$-sphere which when you do surgery on it turns $S^1 \times S^2$ into $S^3$. So the study of these knots is simply the study of knots in $S^3$, or knots in a ball (which just happens to be in $S^1 \times S^2$).

If the complement is irreducible then the knot theory is a little different, but not all that different. For example, one way to link knot theory in $S^1 \times S^2$ to knot theory in $S^3$ is to observe that $S^1 \times S^2$ is zero surgery on the unknot. So a knot in $S^1 \times S^2$ is a Kirby diagram consisting of a 2-component link, one component is unknotted and labelled with a zero (for zero-surgery) and the other component is un-labelled. From the perspective of living inside $S^1 \times S^2$, what this amounts to doing is choosing a knot in the complement of your original knot, such that projection $S^1 \times S^2 \to S^1$ restricts to a diffeomorphism on the new knot. So there will be "Kirby moves" in addition to link isotopy needed to keep track of how knot theory in $S^1 \times S^2$ reduces to the study of these two-component links in $S^3$.

Those are two things that come to mind, anyhow.

So a standard non-trivial example in this "Kirby notation" would be the Whithead link with $0$-surgery done to one component. This gives you a non-trivial knot in $S^1 \times S^2$, the fundamental group of the complement being $\langle a, b | ab^{-2}aba^{-2}b \rangle$, which is non-abelian.

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Quick question: in Mayazaki's paper Conjugation and the Prime Decomposition of Knots in Closed Orientable 3-Manifolds, he requires the exclusion of a particular knot $\mathcal{R}$ in $S^1\times S^2$ (drawn on the second page). How does this relate to your response? – Chris Gerig Dec 4 '12 at 4:12
I don't understand the $S^1$-projection statement, could you elaborate? And then the standard Kirby-moves should fully track all information? (Is there a reference?) – Chris Gerig Dec 4 '12 at 4:24
Do you have a direct link to the paper? I'm at home and can't access mathscinet. The $S^1$ projection statement means the knot is the graph of a function $S^1 \to S^2$. – Ryan Budney Dec 4 '12 at 4:25
This might not work for you then:… (typo: Miyazaki)... But viewing $S^1\times S^2$ as $[0,1]\times S^2$ (visualized as concentric spheres with a medium between them), then view $\mathcal{R}$ as a braid with two strands (ends at the spheres) and one crossing. He defines a relation between knots called conjugation, and the nonuniqueness statement is that if $K_1$ and $K_2$ are conjugates then $K_1\#\mathcal{R}=K_2\#\mathcal{R}$. – Chris Gerig Dec 4 '12 at 4:38
The connection is the complement of the graph of a function $S^1 \to S^2$ is diffeomorphic to $S^1 \times \mathbb R^2$, which is the complement of an unknot in $S^3$. And I could download the paper. Looking over it now... – Ryan Budney Dec 4 '12 at 4:47

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