(Edit: added definition of strong inversion.)

(2nd edit: added motivation for question at the end.)

Definition: A knot $K$ in $S^3$ is strongly invertible if there is an involution of $(S^3,K)$ which preserves the orientation of $S^3$ and reverses the orientation of $K$.

A strong inversion is an element of Sym$(S^3,K)$ taken up to conjugacy in Sym$^+(S^3,K)$, where Sym$(S^3,K)$ is the the symmetry group of $K$, i.e. the group of diffeomorphisms of the pair $(S^3,K)$ modulo isotopies, and Sym$^+(S^3,K)$ is the group of diffeomorphisms preserving the orientation of $S^3$.

Equivalently, a strong inversion on $K$ is a $π$-rotation of $S^3$ which leaves $K$ invariant and has axis meeting $K$ at exactly two points.

My question:

Let $K$ be a knot (possibly non-alternating) that we know admits a strong inversion. A priori, we do not know anything about the strong inversion. The fixed points of the strong inversion consist of a circle in $S^3$ intersecting $K$ in two points (e.g. by the positive solution to the Smith conjecture). Is it possible to explicitly find these two fixed points of the strong inversion on some projection of $K$? As an additional question: is there any way to find these two fixed points on a minimal crossing projection?

  • For example, the trefoil is a strongly invertible knot. The two fixed points of $K$ under the strong inversion are described by the image below (taken from https://arxiv.org/pdf/1909.08556.pdf).

Fixed points of the strong inversion on the trefoil

  • Another example: take the non-alternating knot $11$n$92$. From KnotInfo (https://knotinfo.math.indiana.edu/), we can see that this hyperbolic knot is reversible with symmetry group D1 ($ = \mathbb{Z}/2\mathbb{Z}$). Thus, it admits exactly one strong inversion. The diagram below includes minimal crossing diagrams of $11$n$92$ taken from KnotInfo. Can we find the two fixed points of the strong inversion in any of the diagrams below? Or, can we find the two fixed points in a a different projection of $11$n$92$?

Projections of 11n92

Motivation for question:

Let $h$ be a strong inversion of a knot $K$. Let $p$ be the projection $S^3 \to S^3/h$. We define $G(K,h) := p(\text{Fix}(h)) \cup p(K)$, which is the $\theta$-curve induced by $h$ and consists of three knots.

In the paper "Identifying tunnel number one knots" by Morimoto, Sakuma, and Yokota (https://projecteuclid.org/euclid.jmsj/1226498920), the authors state the following fact:

Corollary 1.3. Let $K$ be a knot with tunnel number one. Then $K$ admits a a strong inversion such that the set of constituent knots of $G(K,h)$ consists of two trivial knots and a knot with a 2-bridge decomposition.

The authors also mentioned that they use Corollary 1.3 to obtain the tunnel number of almost all knots with $\leq$ 10 crossings (they were missing two), all of which have tunnel number one or two. Many of the knots with $\leq$ 10 crossings have minimal crossing diagrams which make it difficult to explicitly find the fixed points of the strong inversion. I was thus wondering, how can one implement Corollary 1.3?

  • $\begingroup$ Your question is somewhat confusing, and it would help if you included a definition of "strong inversion", with all quantifiers precisely stated. For example, I would guess that the definition of strong invertibility of a knot type says that there exists a knot representing that knot type having such-and-such a property. That being the case, it is quite possible that neither of the two knots that you have depicted in the knot type 11n92 possess the desired property, and hence it makes no sense to ask about which two points are the desired points. $\endgroup$
    – Lee Mosher
    Aug 24, 2020 at 3:25
  • $\begingroup$ @LeeMosher Thank you for the suggestion. I have updated my question to include the definition of strong inversion. Also, I should note that being strongly invertible is a property of the knot, not of the diagram. My question is mainly about whether we can "visualize" the strong inversion in any diagram of the knot. $\endgroup$
    – fcm
    Aug 26, 2020 at 14:59
  • $\begingroup$ Regarding whether we can visualize strong inversion in any diagram of a strongly invertible knot, the cheap answer is "no": starting from a diagram in which strong invertibility is visualizable, take some strands in one half of the diagram and scramble them all up, adding a zillion new crossings. The result is a new diagram of the knot in which strong invertibility has been rather effectively hidden. $\endgroup$
    – Lee Mosher
    Aug 26, 2020 at 15:59

1 Answer 1


For alternating knots there is always a way to see a strong inversion from any alternating diagram as a sequence of flypes plus an elementary symmetry of the diagram, see https://arxiv.org/abs/2002.08191.

Even in this case, the symmetry is not always 'easily' visible in a minimal crossing diagram. For example, there is a strong inversion on $5_2$ for which there is no minimal crossing diagram in which the strong inversion is visible as rotation around an axis contained in the plane of the diagram, or perpendicular to the plane of the diagram (see the strong inversion on $5_2$ shown in the right column of the table at the end of https://arxiv.org/abs/1908.00082 where they show a 6-crossing diagram). This can be checked, for example, by enumerating all minimal crossing diagrams for $5_2$.

For non-alternating knots I don't know about any general visiblity results for strong inversions, although in practice they are also often visible on a minimal-crossing diagram by performing some flypes. In your example 11n92, you can see the strong inversion from your diagram by performing the $\alpha$ flype, then the $\beta$ flype, then $\pi$ rotation around the dotted black axis shown in the image below. In particular, the fixed points are the two intersection points of the black axis with the knot outside of the flypes. (The two intersection points inside the flypes are exchanged by the involution.)



Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.