Uniqueness of rotational symmetry of a link diagram

Question

How can I prove that a connected link diagram can only admit up to one axis perpendicular to the plane through which rotational symmetry lies, i.e there aren't rotational symmetries through different axes in the same diagram?

Answer 1

An idea: There is presumably a finite set $A$ of places where the crossings occur. If the knot is rotation invariant about a point $P$, then so is the finite set $A.$ I would also guess you want to say the least angle $\theta$ of the rotation is of the form $2\pi/n$ so that the knot has symmetry under rotation through $\theta$ [other choices for $\theta$ would give infinitely many points in $A$, I think...]

Now each point $a\in A$ is rotated by $\theta$ around $P$ to another point in $A$, and after doing the rotation $n$ times one has for that $a$ a set of points on a regular $n$ gon centered at $P$. I think this means that $P$ must be the centroid of the collection of points in $A$. Then since a finite set has a unique centroid you could conclude the uniqueness of the rotation center from this.

Note some things need filling in here, this is just an idea of how it might be approached.

Answer 2

Consider a link of four (unlinked) components: two small circles in the $yz$-planes perpendicular to the $x$-axis, with centers at each of $(\pm 1, 0, 0)$, and similarly, small circles in the $xz$-plane, with centers at $(0, \pm 1, 0)$.

There's an axis of rotational symmetry (the $z$-axis, with rotation by $\pi/2$). But rotation about either the $x$- or $y$-axis by $\pi$ is also a symmetry of this knot.

You can presumably enlarge the circles into ellipses and tilt them a little so that they link up into a chain and still have the same $x$- and $y$- symmetries, but it was easiest to describe in the unlinked case.

But maybe I'm missing what you mean by a symmetry in the link diagram. When you say "only one axis perpendicular to the plane through which rotational ...", there's surely only one direction perpendicular to some plane of symmetry. So is the only question whether there's only one point on the plane through which the symmetry line can pass? If so, then someone else's answer about the center of gravity seems spot-on.