Is this rule for the bracket polynomial redundant?

Question

The rules for the bracket polynomial are generally given as:

[https://tex.stackexchange.com/questions/306004/drawing-the-rules-for-the-bracket-polynomial]

However, in my university lecture notes, it essentially asks for a proof of rule (2) from the other rules. Is this possible?

Answer 1

The second rule is not redundant, that is, you cannot conclude the second rule from the first and third. However, if you rewrote the second rule as $$\langle L \sqcup \bigcirc \rangle = \delta \langle L \rangle$$ where $\delta$ is some function of $A$, then you can use rules one, three, and invariance of the second Reidemeister move to deduce that $\delta=-A^2-A^{-2}$.

Edit. The three rules above are in fact equivalent to your rule 3 and the rule

• $\langle \underbrace{\bigcirc\sqcup\cdots\sqcup\bigcirc}_{n} \rangle = (-A^2 - A^{-2})^{n-1}$.

If $n=1$, then the above rule is your rule 1. The above rule can also replace your rule 2. If one uses your rule 2, then crossingless components can be removed at any stage of the computation. If one uses the above rule instead, then one must remove all crossings using rule 3 before removing any crossingless components. The outcome will be the same.

My comment about the second Reidemeister move forcing $\delta = -A^2-A^{-2}$ can be seen below. If we want invariance under Reidemeister 2, then the coefficient $A^2 + A^{-2} + \delta$ must be zero. Hence $\delta = -A^2 - A^{-2}$.