I saw the equation on Wikipedia where it relates Homfly polynomial and it's mirror image, and the equation is as follows


And I can't seem to find a proof of this relation anywhere. So does anyone have any idea how to prove this?

  • $\begingroup$ Welcome to the site. Have you tried looking at introductory texts on knot invariants? If not, have you tried proving this? To be clear, do you want a proof of a place where to find it? $\endgroup$
    – Pedro
    May 18, 2021 at 22:57
  • 1
    $\begingroup$ This follows directly from the skein relation. One presentation of the mirror image is from changing all the crossing types, which swaps $L_+$ an $L_-$ in the skein relation. The coefficients in front of those terms are $\ell$ and $\ell^{-1}$. $\endgroup$ May 19, 2021 at 7:35
  • $\begingroup$ Thank you for the feedback! I've looked up various papers about this, and most of them only mention this relation but none of them prove it. I just want to be able to prove this relation but I'm struggling because I'm not even sure how to start the proof $\endgroup$
    – P.love
    May 19, 2021 at 10:41

1 Answer 1


Consider the HOMFLY polynomial parameterized with variables $\ell$ and $m$ (it's worth specifying, because it turns out there are multiple conventions!) It can be defined using the following skein relations: $$P(\text{unknot})=1\quad\text{and}\quad \ell P(L_+)+\ell^{-1}P(L_-)+m P(L_0)=0$$ where the second relation applies to any case where $L_+$, $L_-$, and $L_0$ are three links the same outside a sphere intersecting the link in four points, but inside the sphere the links have one of three tangles. If you want to think about this in terms of diagrams, then it's that there is a circle on the knot diagram intersecting the link in four points, and $L_+$, $L_-$, and $L_0$ come from replacing the interior of the circle with a right-handed crossing, a left-handed crossing, and a "smoothed" crossing, respectively (or in other words: a +1 twist, a -1 twist, and a 0 twist).

A way the HOMFLY polynomial is defined is that you take the set $\mathcal{L}$ of oriented links in $S^3$, construct the set $\mathbb{Z}[\ell^{\pm1},m^{\pm 1}][\mathcal{L}]$ of formal $\mathbb{Z}[\ell^{\pm1},m^{\pm 1}]$-linear combinations of oriented links, then take the quotient by isotopy and the skein relations: $$\mathcal{H} = \mathbb{Z}[\ell^{\pm1},m^{\pm 1}][\mathcal{L}]/(\text{isotopy},\text{unknot} = 1,\ell L_++\ell^{-1}L_-+mL_0=0).$$ The main theorem is that $\mathcal{H} = \mathbb{Z}[\ell^{\pm1},m^{\pm 1}]$, and we can define the HOMFLY polynomial to be the image of a link in $\mathcal{H}$. (This is all exactly the same as saying there exists a function $P:\mathcal{L}\to \mathbb{Z}[\ell^{\pm1},m^{\pm 1}]$ that is invariant under isotopy and that satisfies the skein relations.)

Let's consider taking the mirror image now. Thinking about knot diagrams, the best mirror image for our purposes is to reflect through the plane of the diagram. This has the effect of turning right-handed crossings into left-handed crossings and vice versa, without changing anything else about the diagram. Denote $\overline{L}$ for the mirror image of $L$. Let's also define something closely related to $\mathcal{H}$: $$\overline{\mathcal{H}} = \mathbb{Z}[\ell^{\pm1},m^{\pm 1}][\mathcal{L}]/(\text{isotopy},\text{unknot} = 1,\ell L_-+\ell^{-1}L_++mL_0=0).$$ By substituting $\ell^{-1}$ for $\ell$, the main theorem implies that $\overline{\mathcal{H}} = \mathbb{Z}[\ell^{\pm1},m^{\pm 1}]$ as well.

Here's the idea: we can either take the image of $\overline{L}$ in $\mathcal{H}$ or the image of $L$ in $\overline{\mathcal{H}}$, and in either case we get the same polynomial. While $\overline{L}$ in $\mathcal{H}$ is the HOMFLY polynomial of the mirror image, $L$ in $\overline{\mathcal{H}}$ is the mirror-image HOMFLY polynomial of $L$. Lastly, the image of $L$ in $\overline{\mathcal{H}}$ is related to the image of $L$ in $\mathcal{H}$ by $\ell \leftrightarrow \ell^{-1}$.

This long account is elaborating a simple fact: in the skein relation, $L_+\leftrightarrow L_-$ has the same effect as $\ell \leftrightarrow \ell^{-1}$.

  • $\begingroup$ Wow thank you so much! This really helps a lot and answers my question! $\endgroup$
    – P.love
    May 24, 2021 at 12:00

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