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I was recently thinking about links where each component plays the same role: for every permutation of components, there is an isotopy permuting these components in the prescribed way. In the vein of knot/link invariants, we might ask how to tell when this is not the case. The obvious way to do this is by removing components and comparing the isotopy classes of the resulting links. However, this may be insufficient, particularly for Brunnian links. Are there any link invariants that treat components differently, so that they could detect if a Brunnian link does not have this permutability property?

Disclaimer: while I am very interested in topology, my knowledge of knots and links is very limited. This is also my first post to MO, so I’m happy to hear feedback like “you should’ve phrased your question this way” or “this is better suited to math SE.”

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The multivariable Alexander polynomial has the potential to do this. It has one variable $t_i$ for each link component, and if there is an isotopy interchanging the $i^{th}$ and $j^{th}$ component then the polynomial is symmetric with respect to $t_i$ and $t_j$. (Maybe you have to take account of orientations and some indeterminacy in the Alexander polynomial to say this properly.) You can find examples by looking at the tables in Rolfsen, Knots and Links or on Cha-Livingston's Linkinfo. On the latter I found $L9a1\{0\}$ with polynomial $2-2t_1-5t_2+5t_1t_2+5t_2^2-5t_1t_2^2-2t_2^3+2t_1t_2^3$. Both individual components are unknotted but it looks like you can't interchange them.

Other polynomial-type invariants could do something similar if the Alexander polynomial fails.

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  • $\begingroup$ Thank you! These are great resources. After looking around for a while, I found the knot that I was considering, which ended up being L9a35. It turns out that the multivariable Alexander polynomial of this link is symmetric, so I will have to turn to other methods. If you could elaborate on "other polynomial-type invariants," that would be great. Obviously, no amount of playing around with isotopies will give a negative result, but it does make me more confident that the components can't be interchanged. But who knows, maybe I'm missing something obvious. $\endgroup$
    – Nikhil Sahoo
    Commented Oct 21, 2019 at 4:39
  • $\begingroup$ Another choice here is to use Snappy, which reports that your link L9a35 is hyperbolic, with isometry group $D4$, the dihedral group of order 8. It also reports isometries interchanging the link components. So probably you missed something there. $\endgroup$
    – Danny Ruberman
    Commented Oct 21, 2019 at 12:53
  • $\begingroup$ Another choice (what I had in mind at the end of the answer above) is to use the colored Jones polynomial, with different colorings on the two components. $\endgroup$
    – Danny Ruberman
    Commented Oct 21, 2019 at 12:56
  • $\begingroup$ Cool! Thank you so much. $\endgroup$
    – Nikhil Sahoo
    Commented Oct 21, 2019 at 15:15

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