Inspired by the Smith conjecture, is there a finite group action on $S^3$ (by smooth or analytic diffeomorphisms) which possesses an invariant knotted circle?
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asked Sep 10, 2020 at 22:29
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8$\begingroup$ Certainly. To get a simple example, take any toric knot, i.e. something given by equation $z_1^p=z_2^q$ in the unit sphere $|z_1|^2+|z_2|^2=1$. For any such knot there is a (linear) circle action on $S^3$ that preserves the knot, and you can take any finite subgroup of this circle. $\endgroup$– Dmitri PanovCommented Sep 10, 2020 at 22:45
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1$\begingroup$ Since invariant knot include knot Group, Other conventions consider knots to be embedded in the 3-sphere, in which case the knot group is the fundamental group of its complement in $S^3$ $\endgroup$– zeraoulia rafikCommented Sep 10, 2020 at 22:54
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1$\begingroup$ Probably you want a nontrivial action, and then the question reduces to faithful actions of cyclic groups of prime order. Any knot in Euclidean $\mathbf{R}^3$ or in the round sphere, with nontrivial isometry group, yields an example. For example, the usual representation of the trefoil knot has a isometry group of order $6$. $\endgroup$– YCorCommented Sep 11, 2020 at 9:15
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$\begingroup$ Turning your question around a little, you are indirectly asking about the symmetry groups of knots in $S^3$. These are readily computed, by the techniques of Sakuma as mentioned. But you can also compute them for hyperbolic knots fairly rapidly with SnapPea. This was originally done by Jeff Weeks, at essentially the same time as Sakuma. I believe Sakuma's table of symmetry groups for "small" knots is in the back of Kawauchi's reference book on knot theory. A nice feature of the Snappea approach is you can algorithmically describe the fixed point sets. $\endgroup$– Ryan BudneyCommented Feb 1, 2021 at 22:51
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1 Answer
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I think the comments answered the question, but it seems like some pictures would be helpful as well. There are many examples, but one of my favorites is this paper by Sakuma which concludes with a table of knots preserved by orientation-preserving involutions on $S^3$ which reverse the orientation on $K$.
I also want to point out that in general studying prime order group actions is not sufficient. For example, you may ask if the figure eight knot $4_1$ is preserved by a symmetry which reverses the orientation on $S^3$, but preserves the orientation on $4_1$. This is the case, but the symmetry has order 4 (and there is no such order 2 symmetry). Visually, this can be seen in the following image as a $\pi/2$ rotation within the plane of the diagram followed by a reflection across an $S^2$ intersecting the diagram in the shown dotted green circle.
answered Feb 1, 2021 at 22:37
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$\begingroup$ +1, but i suggest you to use inkscape for drawings. I discovered it this year and changed my drawing quality (in my thesis I had very poor graphics XD) $\endgroup$ Commented Feb 1, 2021 at 23:31
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1$\begingroup$ Fair enough - edited to include a better diagram. $\endgroup$ Commented Feb 2, 2021 at 0:30