Compute the Normalized Bracket Polynomial of a Knot, $K$

Question

Compute the normalized bracket polynomial $$(-A^3)^{-w(D)}\langle D \rangle$$ of the following knot diagram $D$ where $w(D)$ is the writhe of $D$.

Thank you!

Answer 1

I will give three methods of computing the normalized Kauffman bracket polynomial.

Method 1: Look it up. The pictured diagram is the mirror of the $5_2$ knot in Rolfsen's tables. The pictured diagram has writhe $w(D)=-5$, and its Jones polynomial (found from KnotInfo by replacing $t$ with $t^{-1}$ in $V_{5_2}(t)$) is $$V_{\overline{5_2}}(t) = t^{-1}-t^{-2}+ 2t^{-3}-t^{-4}+ t^{-5}-t^{-6}.$$ Using the equality $$V_L(t) = (-A^3)^{-w(D)}\langle D \rangle |_{A=t^{-1/4}},$$ we obtain $$(-A^3)^{-w(D)}\langle D \rangle = A^{4} -A^{8} + 2A^{12} - A^{16} +A^{20} - A^{24}.$$

Method 2: Evaluate the Kauffman bracket. Since there are five crossings, there are $2^5=32$ Kauffman states. For each state $s$, let $\alpha(s), \beta(s)$, and $|s|$ denote the number of $A$-resolutions in $s$, the number of $B$-resolutions in $s$ and the number of components of $s$ respectively. Then $$\langle D \rangle = \sum_{s} A^{\alpha(s)-\beta(s)}(-A^2-A^{-2})^{|s|-1}.$$ It is probably best to do this computation with a computer (as in this previous question you asked). The end result of such a computation is $$\langle D \rangle = -A^{-11}+A^{-7}-2A^{-3}+A-A^{5}+A^{9}.$$ Normalizing by $(-A^3)^{-w(D)}=-A^{15}$, we obtain the same answer as in Method 1: $$(-A^3)^{-w(D)}\langle D \rangle = A^{4} -A^{8} + 2A^{12} - A^{16} +A^{20} - A^{24}.$$

Method 3. Resolve to twisted unknots. The idea here is to only resolve crossings until the diagram becomes a twisted unknot, that is, a diagram of the unknot that can be transformed into the crossing-less diagram using only Reidemeister one moves. Reidemeister one moves change the Kauffman bracket as shown below:

Thus we can use the above rules to compute the Kauffman bracket of any twisted unknot. Specifically, for a twisted unknot $T$ of writhe $w(T)$, we have $$\langle T \rangle =(-A^3)^{w(D)}.$$ Start by taking a resolution tree of the diagram $D$, stopping at a leaf whenever one has a twisted unknot, as below. It remains to compute $\langle T_i \rangle$ for each twisted unknot, and then to put everything together. We have \begin{align*} \langle T_1 \rangle = & \; A^6,\\ \langle T_2 \rangle = &\; 1,\\ \langle T_3 \rangle = &\; -A^{-3},\\ \langle T_4 \rangle = &\; -A^3,\\ \langle T_5 \rangle = &\; -A^{-3},\\ \langle T_6 \rangle = &\; A^{-6},~\text{and}\\ \langle T_7 \rangle = &\; -A^{-9}. \end{align*}

The path from the original diagram $D$ to each leaf $T_i$ is labeled with $A$'s and $B$'s. Multiply by $A$ for each $A$-label and $A^{-1}$ for each $B$-label. Then add everything together to get $\langle D \rangle$. We have \begin{align*} \langle D \rangle = & \; A^3\langle T_1 \rangle + A\langle T_2 \rangle + \langle T_3 \rangle + A^2\langle T_4 \rangle + \langle T_5 \rangle + A^{-1}\langle T_6 \rangle + A^{-2}\langle T_7\rangle\\ = &\; -A^{-11}+A^{-7}-2A^{-3}+A-A^{5}+A^{9}. \end{align*} Again we normalize by multiplying by $(-A^3)^{-w(D)}=-A^{15}$ to obtain $$(-A^3)^{-w(D)}\langle D \rangle = A^{4} -A^{8} + 2A^{12} - A^{16} +A^{20} - A^{24}.$$