Let's have a finite collection of $N$ circles $\mathbb{S}^1$ in $\mathbb{R}^3$. (These circles could not intersect.) Every circle could be "hooked on" other circle and it could be "hooked" for simplicity only once. My question, which I need to solve, is how many combinations of non-homeomorphic structures will I obtain? For example $N=2$: I have $2$ combinations, two unhooked circles and two hooked circles; I know already that this question could be translated to the language of graph theory. Do You know, if anyone has solved such a problem? Thank You!
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1$\begingroup$ The fact that you don't seem to care about the actual embeddings in the way a knot theorist would seems to indicate that your question amounts to finding the number of simple graphs on $N$ vertices. Is it not so? $\endgroup$– Arnaud MortierCommented Jan 11, 2018 at 14:43
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5$\begingroup$ @ArnaudMortier Not exactly - it depends on the notion of equivalence. For example, are borromean rings the same as three unliked circles or not? $\endgroup$– Igor RivinCommented Jan 11, 2018 at 16:01
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2$\begingroup$ What exactly do you mean by "hooked"? Do you mean that the linking number between two distinct circles is at most one? Also please say more about how you know that your problem can be translated into graph theory. $\endgroup$– j.c.Commented Jan 11, 2018 at 17:07
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10$\begingroup$ This seems to be an interesting question when "circle" is interpreted in the narrow sense of a Euclidean circle lying in a plane. This rules out the Borromean rings for example, by an observation of Freedman and Skora. Then the question is, how many ambient homeomorphism classes of links of $N$ components have representatives that are collections of (disjoint) Euclidean circles? Intuition suggests this number should be finite. $\endgroup$– Allen HatcherCommented Jan 11, 2018 at 18:11
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2$\begingroup$ I'm still totally unclear what the actual question is. Here's a guess. "How many $N$-component links are there in $S^3$ such that each component is an unknot and the absolute value of the linking number of each pair of components is at most 1?" $\endgroup$– HJRWCommented Jan 11, 2018 at 22:16
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I don't think that there's much of a literature on this topic. If the circles are all pairwise linked, then this is equivalent to the classification of great circle links in $S^3$. These were studied a bit by Thurston and Viro.
To see this equivalence, use stereographic projection to project the link to a circle link in $S^3$. Each circle is the intersection of a plane in $\mathbb{R}^4$ with $S^3$. Since they pairwise link, each pair of planes intersects once in the interior of $S^3$. Now, project radially to infinity (think of letting the radius of $S^3$ approach infinity), one gets an isotopy to a great circle link.
Great circle links correspond to configurations of skew lines in $\mathbb{R}^3$, which have been studied by the Viros. One natural question: if two circle links are equivalent, are they equivalent by an isotopy that preserves the circularity of each component? The rigid classification of great circle links is the same as the classification of configurations of planes in $\mathbb{R}^4$, of which there is some literature (see e.g. the papers citing the Viros' paper on Google Scholar). Matei and Suciu show that the number of homotopy types of plane arrangements grows like the partition function, so like $C^{\sqrt{n}}$.
As hinted at by Hatcher, if all components of a circle link are pairwise unlinked, then it is the unlink. The proof is the opposite of the previous proof: shrink $S^3$, until each circle disappears when its plane becomes tangent to the sphere. This argument is due to Freedman-Skora.
