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The title says it all:

How can I show that a knot $K$ is distinct from its mirror image?

May be I have to try different knot invariants. Not sure, I am new in this area.

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3 Answers 3

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There is a complete solution in theory, which works well in practice generally (for sufficiently small knots).

A hyperbolic knot is amphichiral iff its complement admits an orientation reversing isometry. Using the programs Snappea or SnapPy, if one inputs a knot and it has a hyperbolic complement, then it can usually tell you if it is chiral. Here is a sample:

enter image description here

For non-hyperbolic knots, there is in principle an algorithm to detect if it is chiral, but I don't know if it has been implemented. Using geometric methods, one may detect if each piece of the JSJ decomposition is amphichiral, then see if the mirror symmetries are globally compatible. For a thorough explanation of JSJ decompositions of knot complements, see the paper of Budney.

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  • $\begingroup$ Is this solution mathematically rigorous? I'm not talking about the use of computer. Is it the case that the Snappea kernel use numerical methods to solve for approximate solutions of gluing equations in order to determine hyperbolic structures? Highly precise, but still not exact solutions? $\endgroup$
    – hsp
    Commented Jun 6, 2015 at 4:24
  • $\begingroup$ @hsp: correct. Snappea is not rigorous, but can be made so in principle. There is an underlying rigorous algorithm, but I'm not sure that it has been implemented. $\endgroup$
    – Ian Agol
    Commented Jun 6, 2015 at 5:22
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There are several ways to try and tell apart a knot from its mirror

  • Probably the most "classic" way to do this is using the signature;

  • You can compute Jones polynomial and check that it's not symmetric;

  • You can use Rasmussen's $s$-invariant coming from Khovanov homology;

  • You can use Heegaard Floer homology, and the $\tau$-invariant of Ozsváth and Szabó.

Each of the invariants above is often sensitive to orientation reversal; for example, all four invariants tell apart the left-handed trefoil from the right-handed trefoil.

If you want to show that a knot is amphicheiral (i.e. isotopic to its mirror image), you have to fiddle around with diagrams and Reidemeister moves.

As far as I know, there is no general criterion to answer your question, though.

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  • $\begingroup$ I know that Dowker-Thistlethwaite notation doesn't distinguish mirror image. Does Conway Notation does? $\endgroup$
    – Ahmed Abdullah
    Commented May 23, 2015 at 17:26
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Rigorous answer to this question is to use the Hemion's algorithm. See http://ukcatalogue.oup.com/product/9780198596974.do But it is complicated.

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  • $\begingroup$ The solution Ian outlines is essentially the modernized version of Hemion's algorithm, using all the shortcuts available via geometrization. $\endgroup$
    – Ryan Budney
    Commented Jul 15, 2021 at 20:36

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