The normalized generic colored Jones quotient is an integral Laurent polynomial

On the displayed color basis, $E^r\mathrm{/}[r]!$ sends $v_i$ to ${\left[\begin{array}{c}a-i+r\\ r\end{array}\right]}_{q_K}{v}_{i-r}$, with zero outside the index range, while $F^r{v}_j={\prod }_{s=1}^r[j+s]v_{j+r}$. By [F2] each coefficient in the finite $R$ series therefore belongs to $R_0$, including the weight factor $A^{hk}$. Its inverse is $({\sum }_{r\ge 0}(-1{)}^r{q}_K^{-r(r-1)\mathrm{/}2}{z}^r{E}^r\otimes F^r\mathrm{/}[r]!)D^{-1}$, again a finite integral matrix by the same argument. To check the inverse formula, the coefficient of $(E\otimes F{)}^m$ in the product of the two nilpotent series is, after multiplication by $[m]!\mathrm{/}{z}^m$, ${\sum }_{r=0}^m(-1{)}^{m-r}{q}_K^{r(r-1)\mathrm{/}2-(m-r)(m-r-1)\mathrm{/}2}{\left[\begin{array}{c}m\\ r\end{array}\right]}_{q_K}$. Its exponent equals $(m-1)r-m(m-1)\mathrm{/}2$, so for $m>0$ it vanishes by [F2]'s alternating Gauss sum; at $m=0$ it is $1$. Thus both $c$ and $c^{-1}$ are integral matrices.

Ordinary dual evaluation and coevaluation have entries $0,1$. The pivotal matrices $G_N$ and $G_N^{-1}$ are diagonal Laurent monomials with signs. Accordingly both orientations of cup and cap are integral: one uses ordinary evaluation/coevaluation, the other their composites with the pivotal comparison or its inverse. Crossings involving a dual strand are obtained by bending one or both strands of $c$ or $c^{-1}$ with those maps, as follows from braiding naturality at evaluation and the snakes; concretely this is a partial transpose followed by the just-listed integral contractions. It introduces no matrix inverse other than $c^{-1}$ and $G_N^{-1}$. Every elementary matrix of a tangle colored by $V_N$ and its dual is thus over $R_0$.

It remains to prove the stronger exponent assertion. Work temporarily over $\mathbb{Q}(i)(A)$, and let a prime denote substitution $A\mapsto iA$. On the basis $v_0,\dots ,v_a$ put $Q{v}_j=(-1{)}^{j(j-1)\mathrm{/}2}{v}_j$ and $D{v}_j=(-1{)}^j{v}_j$. The $r$th term of $R$ sends input $(u,v)$ to $(u-r,v+r)$, before the flip. Every nonzero Laurent monomial in its coefficient has degree, modulo $4$, $(a-2u)(a-2v)+2r(a-u+v+r)\equiv a^2-2a(u+v)+2r(a-u+v+r)(\mathrm{m}\mathrm{o}\mathrm{d}4)\mathrm{.}$ Indeed $[m]$ has all degrees congruent to $2(m-1)$, $z$ has degree $2$, $q_K^{r(r-1)\mathrm{/}2}$ has degree $r(r-1)$, and the weight factor is evaluated at the output weights. The quotient of the output and input $Q\otimes Q$ signs is $(-1{)}^{r(v-u+r)}$. Comparing these two explicit expressions yields $c^{\mathrm{\prime }}=(Q\otimes Q)c(Q\otimes Q{)}^{-1}(a\text{ even}),$ $c^{\mathrm{\prime }}=i(Q\otimes Q)(D\otimes 1)c(D\otimes 1)(Q\otimes Q{)}^{-1}(a\text{ odd})\mathrm{.}$ For the second formula, the remaining sign is $(-1{)}^{u+v+r}$, exactly the product of the left input and left output $D$ signs after the flip. No asymptotic or numerical specialization is used in these matrix equalities.

Cut a small oriented arc of the framed knot closure of $\beta$. The remaining framed $(1,1)$-tangle is a morphism $f:V_N\to V_N$ (choose its endpoint orientation accordingly). A generic planar projection decomposes it into finitely many elementary crossings, cups and caps, so step 2.1 makes every entry of $f$ an element of $R_0$. By [F1] and [F3], $f=sI$ for a scalar $s\in F$; its $v_0$ diagonal entry gives $s\in R_0$. Closing that cut arc is the pivotal trace of $f$, namely $d_{N}s$, and is the same framed knot value as $\mathrm{Tr}(G_N^{\otimes b}{\rho }_N(\beta ))$ by [F3]. Since $d_N\mathrm{\ne }0$ generically, the quotient in the Statement is ${\lambda }_N^{-w(\beta )}s\in R_0$. This proves integral Laurentness in $A$ without dividing in an integral ring or after specialization.

If $a$ is even, let $T=Q^{\otimes b}$. Step 2.2 gives ${\rho }^{\mathrm{\prime }}=T\rho T^{-1}$ for every braid word, including inverses. Direct substitution gives $G_N^{\mathrm{\prime }}=G_N$, $d_N^{\mathrm{\prime }}=d_N$ and ${\lambda }_N^{\mathrm{\prime }}={\lambda }_N$, because $a$ and $a(a+2)$ are even and divisible by $4$, respectively. Since $T$ commutes with $G_N^{\otimes b}$, cyclicity of the ordinary trace shows that the normalized quotient is unchanged by $A\mapsto iA$.

If $a$ is odd, put $H=D^0\otimes D^1\otimes \cdots \otimes D^{b-1}$ and $T=Q^{\otimes b}H$. The operator $c$ preserves the sum of basis indices, so it commutes with $D\otimes D$. Consequently conjugating an adjacent $c$ by $H$ has the effect $(D\otimes 1)c(D\otimes 1)$, regardless of which adjacent position has odd exponent. Step 2.2 thus gives ${\rho }^{\mathrm{\prime }}=i^{w(\beta )}T\rho T^{-1}$. Here $G_N^{\mathrm{\prime }}=-G_N$, $d_N^{\mathrm{\prime }}=-d_N$, and ${\lambda }_N^{\mathrm{\prime }}=-i{\lambda }_N$, since $a(a+2)\equiv 3(\mathrm{m}\mathrm{o}\mathrm{d}4)$. Taking the trace and normalizing therefore multiplies the quotient by $(-1{)}^{b-1+w(\beta )}$. The strand permutation of a knot-closing braid is a single $b$-cycle, so its sign is $(-1{)}^{b-1}$. Each positive or negative generator is a transposition, so that same sign is $(-1{)}^{w(\beta )}$. The multiplier is $1$. This proves invariance under $A\mapsto iA$ also for odd $a$.

Write the integral Laurent polynomial from step 3.1 as ${\sum }_k{n}_k{A}^k$, where $n_k\in \mathbb{Z}$ and only finitely many are nonzero. Steps 3.2 and 3.3 imply ${\sum }_k{n}_k(i^k-1)A^k=0$. Linear independence of Laurent monomials over $\mathbb{Q}(i)$ gives $n_k=0$ unless $4\mid k$. Replacing $A^{4m}$ by $q^m$ gives the required unique integral Laurent polynomial. Its representative independence and unknot value follow from [F1]. Evaluation of a finite Laurent sum at a nonzero complex number is defined without any denominator other than a power of that number, even when the generic quantum dimension specializes to zero. This proves every assertion, including the root-evaluation interpretation.

To check the conventional variable explicitly, write $a=N-1$ and $\{m\}=q^{m\mathrm{/}2}-q^{-m\mathrm{/}2}$, $\{m\}!={\prod }_{s=1}^m\{s\}$. Murakami's defining positive matrix on basis $e_0,\dots ,e_a$ has coefficient, for input $(k,l)$ and output $(I,J)=(l-r,k+r)$, $\frac{\{l\}!\{a-k\}!}{\{l-r\}!\{r\}!\{a-k-r\}!}{q}^{(I-a\mathrm{/}2)(J-a\mathrm{/}2)-r(I-J)\mathrm{/}2-r(r+1)\mathrm{/}4},$ with $0\le r\le \min (l,a-k)$, and pivotal diagonal $\mu e_j=q^{(2j-a)\mathrm{/}2}{e}_j$; these are the explicit defining formulas in §2.4, not an invoked invariance theorem. At $q=A^4$, change basis by $v_j=A^{j^2}([j]![a-j]!{)}^{-1}{e}_{a-j}$. For input $(u,v)$, reversal makes the output $(v+r,u-r)$. The factorial coefficient changes to $z^r[a-u+r]![v+r]!\mathrm{/}([a-u]![v]![r]!)$, exactly the one in step 1.1. Before scaling, the exponent of $A$ differs from the declared $R$ coefficient by $2r(v-u+r)$; the change-of-basis factor $A^{u^2+v^2-(u-r{)}^2-(v+r{)}^2}$ is its inverse. Thus the two positive crossing matrices are conjugate, as are their inverses, and $\mu$ becomes $K=q_K^H$. Murakami's twist-normalizing scalar is $A^{a(a+2)}$. Replacing this unsigned pivotal and twist by the signed ones of our definition multiplies the normalized closed-braid quotient by $(-1{)}^{a(b-1-w(\beta ))}=1$, by the knot-permutation parity proved in step 3.3. Consequently the Laurent polynomial just established is exactly that source's normalized colored Jones polynomial with $q=A^4$.