Colored Jones and the volume conjecture — Examples and Counterexamples

Define $a_{N_0}=\exp (V{N}_0\mathrm{/}(2\pi ))$ and $a_N=1$ for every $N\ge 1$ with $N\mathrm{\ne }{N}_0$. Each term is positive, and $(2\pi \mathrm{/}{N}_0)\log (a_{N_0})=(2\pi \mathrm{/}{N}_0)(V{N}_0\mathrm{/}(2\pi ))=V$.

For every $N>N_0$, $a_N=1$, so $(2\pi \mathrm{/}N)\log (a_N)=0$. The sequence of normalized logarithms is therefore eventually zero and has limit $0\mathrm{\ne }V$. This contradicts the asserted conclusion.

For $N=1$, the basis has one vector, $E=F=0$, $K_L=1$, $c_1=G_1=1$, $d_1={\lambda }_1=1$. Every braid generator therefore acts as $1$, and the normalized value of every such knot closure is $1$. For the unknot at any $N$, the one-strand identity braid has $w=0$ and trace $\mathrm{Tr}(G_N)=d_N$, so its normalized value is also $1$.

The strand permutation of ${\sigma }_1^3$ is $(12{)}^3=(12)$, so its two endpoints are interchanged; pairing each bottom endpoint with the top endpoint in the same position therefore produces one component. At $N=2$, the only nonzero terms of ${\mathcal{R}}_2$ are $r=0,1$. On the ordered basis $(v_{+}{v}_{+},v_{+}{v}_{-},v_{-}{v}_{+},v_{-}{v}_{-})$ they give $c_2=\left(\begin{array}{cccc}A& 0& 0& 0\\ 0& 0& A^{-1}& 0\\ 0& A^{-1}& A-A^{-3}& 0\\ 0& 0& 0& A\end{array}\right)$ and $G_2\otimes G_2=\mathrm{diag}(A^4,1,1,A^{-4})$. The middle block $B=\left(\begin{array}{cc}0& A^{-1}\\ A^{-1}& A-A^{-3}\end{array}\right)$ satisfies $\mathrm{tr}(B^3)=A^3-A^{-9}$ by direct multiplication; the other two diagonal entries of $c_2^3$ are $A^3$, hence $\mathrm{Tr}((G_2\otimes G_2)c_2^3)=A^7+A^3+A^{-1}-A^{-9}$. Here $w({\sigma }_1^3)=3$, ${\lambda }_2=-A^3$, and $d_2=-A^2-A^{-2}$.

Write a braid word as $\beta ={\prod }_{j=1}^m{\sigma }_{i_j}^{{\epsilon }_j}$ with ${\epsilon }_j\in \{1,-1\}$, using [F4] under [A1]. Each letter gives one crossing in the braid box. The closure pairs endpoints as prescribed by [F5]; in the diagram convention fixed in the Example, the joining arcs are routed outside the rectangular braid box without added crossings. A positive letter contributes $+1$ to writhe by [F6]. For a negative letter the opposite half twist [F7] exchanges the over- and under-strands; the determinant in the crossing-sign definition changes sign when its two vectors are exchanged, so that crossing contributes $-1$. Thus $\mathrm{wr}(\widehat{\beta })={\sum }_j{\epsilon }_j=w(\beta )$ by [F8].

Substitution into [F1] using step 1.2 yields $J_2^{\mathrm{g}\mathrm{e}\mathrm{n}}(T;q)=\frac{A^{-9}(A^7+A^3+A^{-1}-A^{-9})}{A^2+A^{-2}}=A^{-4}+A^{-12}-A^{-16}=t+t^3-t^4$, where $t=A^{-4}=q^{-1}$.

Given an oriented knot $K\subset {\mathbb{R}}^3\subset S^3$, choose a braid closure $\widehat{\beta }$ representing it by [F9]. By [F2], the fundamental-color ribbon trace is $\delta$ times the raw bracket and the twist is $-A^3$. Since $d_2=\delta$, formula [F1] gives $J_2^{\mathrm{g}\mathrm{e}\mathrm{n}}(K;q)=(-A^3{)}^{-w(\beta )}⟨\widehat{\beta }⟩$. By step 1.3 this is the writhe-normalized bracket $f_K(A)$. The use of [F9] requires [A1], and the framed trace in [F2] requires [A2].

The bracket/Jones theorem [F3] now gives $f_K(A)=V_K(t)$ at $t=A^{-4}=q^{-1}$. Its hypothesis uses AC [A1], while the ribbon-trace identification in [F2] uses ACω [A2], already supplied in step 2.2; the proof keeps those axiom branches separate. Therefore $J_2^{\mathrm{g}\mathrm{e}\mathrm{n}}(K;q)=V_K(q^{-1})$.