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edited Jan 12, 2018 at 5:04

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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.

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).

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.

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.

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created Jan 12, 2018 at 0:09

68.8k
3
194
358

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).

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.