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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.2012-12-03
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    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: http://math.berkeley.edu/~aaron/saft/francis.pdf2012-12-04

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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?2012-12-04
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    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?)2012-12-04
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    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 $S^1 \to S^2$.2012-12-04
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    This might not work for you then: http://www.ams.org/journals/tran/1989-313-02/S0002-9947-1989-0997679-2/S0002-9947-1989-0997679-2.pdf (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 $K_1\#\mathcal{R}=K_2\#\mathcal{R}$.2012-12-04
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    Sorry for still misunderstanding, but I don't see the connection between the knot as a function's graph and as a Kirby diagram through zero-surgery.2012-12-04
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    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...2012-12-04
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    Well, Mayazaki's paper doesn't really have anything to do with what I'm saying. Mayazaki's paper is concerned with the connect-sum decomposition of pairs $(M,K)$ where $K$ is an embedded circle in the $3$-manifold $M$. His results seem to be in line with what I'm saying, in that the connect-sum decomposition of pairs $(M,K)$ is similar to the connect-sum decomposition of $3$-manifolds and knots respectively. The not-quite-uniqueness he finds in the decomposition stems from the fact that in the prime decomposition of 3-manifolds the separating spheres are not unique up to isotopy.2012-12-04
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    OK thanks, this has all been very enlightening. I am trying to do this explicitly now for $\mathcal{R}$, but how can I check whether there is a sphere in the complement which doesn't bound a ball (for reducibility)? If there is no such sphere (i.e. the knot complement is irreducible), then I don't see how to build/choose this function-graph and unknot (and its linkage with $\mathcal{R}$). For instance, what stops me from choosing $S^1\times\lbrace(0,0,-1)\rbrace$ away from the knot near the north pole?2012-12-04
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    By the solution to the Poincare conjecture and the sphere theorem, existence of a sphere that does not bound a ball is equivalent to $\pi_2$ being non-trivial. So there's a process you can use. But more simply, any such sphere would have to lift to a sphere in a covering space -- but if you take the cover of your knot complement corresponding to $\mathbb R \times S^2 \to S^1 \times S^2$ you get a submanifold of Euclidean space $\mathbb R^3$, so you can apply the Schoenflies theorem.2012-12-04