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Basically my question has to do with the HOMFLY polynomial.
In its wikipedia page, http://en.wikipedia.org/wiki/HOMFLY_polynomial, I see that it says
$$P(L_1 \cup L_2) = \left(\frac{-l-l^{-1}}{m}\right)P(L_1)P(L_2) $$ where $L_1 \cup L_2$ is the split link of the two links
Can anyone show a proof of this fact from the skein relation of the HOMFLY polynomial? I get stuck when I have to prove that $$P(L_1 \# L_2 ') = P(L_1)P(L_2) $$ where # is the knot sum operation and L' is the mirror image of L.

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1 Answer 1

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Note, that is is actually just $P(L_1 \# L_2) = P(L_1) P(L_2)$.

Given a link $L$, we can change crossings using the skein relation to write $L$ as a linear combination (where the coefficients are polynomials in $m$ and $l$) of unlinks with various numbers of components. So, if $U_k$ represents the unlink with $k$ components, and $L_1,L_2$ are diagrams each with $\le n$ crossings, we can write $$L_1 = \sum_{k=1}^n a_k U_k$$ $$L_2 = \sum_{k=1}^n b_k U_k$$ where each $a_k$ is a polynomial in $l,m$. It is not hard to show by induction that $$P(U_k) = \left[\frac{-(l + l^{-1})}{m}\right]^{k-1}$$ so that $P(U_i)P(U_j) = P(U_{i+j-1})$. It follows immediately that $$P(L_1)P(L_2) = \sum_{k=1}^{2n} \sum_{i < k} a_i b_{k-i}P(U_{k-1})$$

Now, it remains to look at the connected sum of the diagrams $L_1 \# L_2$. If we remove all the crossings in $L_1$ by the skein relations, then we get $$L_1 \# L_2 = \sum_{k=1}^n a_k (U_{k-1} \cup L_2)$$ because one of the unknotted components at the end will be stuck together with the diagram for $L_2$. Expanding $L_2$ gives $$L_1 \# L_2 = \sum_{k=1}^n a_k \left(U_{k-1} \cup \sum_{k=1}^n b_k U_k\right) = \sum_{k=1}^{2n} \sum_{i < k} a_i b_{k-i} U_{i-1} \cup U_{k-i} = \sum_{k=1}^{2n} \sum_{i < k} a_i b_{k-i} U_{k-1}$$ so $P(L_1 \# L_2) = P(L_1) P(L_2)$

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  • $\begingroup$ Good answer. Thank you. $\endgroup$
    – wilsonw
    Jun 10, 2013 at 18:08

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