I've been working on some knot invaririants and specialy the Jones Polynomial. I was able to prove that $ V_{K_1 \# K_2} = V_{K_1} V_{K_2} $ for two knots $ K_1 $ and $ K_2 $ . So I found my self asking if the same raltion holds in the case of links. All books states that indeed $ V_{L_1 \# L_2} = V_{L_1} V_{L_2} $ for two knots $ L_1 $ and $ L_2 $ . But I coudn't no proof of that. So my question is how to prove that the jones polynomial of the connected sum of two links is the product of their jones polunomial.

Thanks in advance for the help.


1 Answer 1


Your equation holds for links, but there's a subtlety: The connected sum operation isn't actually well-defined on links; see this question. But it is well-defined if you specify the components that will be connected. Similarly, connected sums of link diagrams $D_1$ and $D_2$ can be made well-defined once you specify the two components that are to be connected (and an arc connecting them, technically).

If you take a careful look at your proof of the equation for knots, you'll probably see something curious and relevant: The Jones polynomial of a connected sum of link diagrams is well-defined, even if the actual diagram isn't. That is, regardless of which link components we connect (and the specific way in which we connect them), $$V_{D_1 \# D_2} = V_{D_1} V_{D_2}.$$ So if $L_1$ and $L_2$ are two links and $L$ and $L'$ are distinct connected sums of $L_1$ and $L_2$, then $$V_{L} = V_{L_1} V_{L_2} = V_{L'}.$$


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