Colored Jones and the volume conjecture

The diagonal weights give $K_{L}E{K}_L^{-1}{v}_j=q_L^{2}E{v}_j$ and $K_{L}F{K}_L^{-1}{v}_j=q_L^{-2}F{v}_j$. For the commutator, $EF{v}_j-FE{v}_j=([j+1][N-j-1]-[N-j][j])v_j=[N-1-2j]v_j$; the last equality follows by expanding the two quantum-number products over $q_K-q_K^{-1}$. Since $[r{]}_{q_K}=[r{]}_{q_L}$, this is $(K_L-K_L^{-1})(q_L-q_L^{-1}{)}^{-1}{v}_j$. The generators therefore satisfy the defining ${\mathfrak{s}\mathfrak{l}}_2$ relations.

The eigenvalues $q_L^{N-1-2j}$ of $K_L$ are pairwise distinct in $F$. Any module endomorphism is therefore diagonal in the given basis; commuting with $E$, whose successive coefficients $[N-j]$ are nonzero in $F$, forces all diagonal entries to be equal. Thus ${\mathrm{End}}_{U_{q_L}({\mathfrak{s}\mathfrak{l}}_2)}(V_N)=FI$ and the color is absolutely simple by [F6].

If $N=1$, $V_N$ is one-dimensional and irreducible. For $N>1$, let $W$ be a nonzero submodule and choose $x={\sum }_j{c}_j{v}_j\in W$ nonzero. The eigenvalues of $K_L$ are distinct, so the polynomial spectral projection for any $j$ with $c_j\mathrm{\ne }0$ maps $x$ to $c_j{v}_j$; hence $v_j\in W$. Repeated application of $E$ gives $v_0\in W$, since every coefficient $[N-k]$ encountered is nonzero. Repeated application of $F$ to $v_0$ gives each $v_k$ because the coefficients $[k]$ are nonzero. Thus $W=V_N$, proving irreducibility.

The displayed sum for $d_N$ is a geometric sum, so $d_N=(-1{)}^{N-1}[N]$. Its numerator $A^{2N}-A^{-2N}$ is a nonzero Laurent polynomial and its denominator $A^2-A^{-2}$ is nonzero in $F$; hence the generic normalization is defined. The Hopf coproduct [F2] supplies the tensor-power action used in ${\rho }_N(\beta )$. By the braiding, dual, and ribbon axioms [F3] and the all-color ribbon identities [F7], the comparison on $V_N^{\otimes n}$ is $G_N^{\otimes n}$. The basis coevaluation and evaluation in the left categorical trace [F5] then give the ordinary matrix trace $\mathrm{Tr}(G_N^{\otimes n}{\rho }_N(\beta ))$, which is the scalar of the framed closure.

By [F4], [F7], and [A2], ${\lambda }_N^{-w(\beta )}{t}_N(\beta )$ is unchanged by every conjugation and Markov stabilization, since $V_N$ is absolutely simple and its specified twist scalar is nonzero. Countable choice [A3] supplies the evaluator's hypothesis. Dividing by the fixed nonzero $d_N$ preserves that equality, so the displayed generic value is independent of the braid representative.

The unknot is represented by the one-strand identity braid, whose exponent sum is zero and whose pivotal trace is $\mathrm{Tr}(G_N)=d_N$. Hence $J_N^{\mathrm{g}\mathrm{e}\mathrm{n}}(◯;q)=1$. For $N\ge 2$, at $q_N=e^{2\pi i\mathrm{/}N}$ choose $A_N=e^{\pi i\mathrm{/}(2N)}$; then $A_N^4=q_N$ and $A_N^2-A_N^{-2}\mathrm{\ne }0$, while $A_N^{2N}-A_N^{-2N}=e^{\pi i}-e^{-\pi i}=0$. Thus $d_N(q_N)=0$: the generic quotient cannot be evaluated there by dividing by $d_N$.

Apply [F8] to the given knot-closing braid and color under [A2] and [A3]. The generic quotient therefore has a unique integral Laurent representative in $q=A^4$. Substitution into that finite Laurent sum is defined for every nonzero complex $q$, including $q_N$; it does not divide by the specialized quantum dimension. Together with steps 2.1 and 2.2, this proves the stated representative independence, unknot normalization and root-evaluation convention.

Put $K=K_L^{-1}$. Then the generator relations and coproduct in [F1] become exactly $KE{K}^{-1}=q_K^{2}E$, $KF{K}^{-1}=q_K^{-2}F$, $[E,F]=(K-K^{-1})\mathrm{/}(q_K-q_K^{-1})$, $\mathrm{\Delta }(E)=E\otimes K+1\otimes E$, $\mathrm{\Delta }(F)=F\otimes 1+K^{-1}\otimes F$. The antipodes also agree: $S(E)=-E{K}^{-1}$, $S(F)=-KF$. Thus this is a change of parameter and Cartan generator, not an assumption equating the two parameters. Every weight is an integer, and $E,F$ are nilpotent on every fixed generated object by its finite weight range. The finite $R$ series and all dual matrices are therefore rational matrices in $A$.

Fix finitely many objects, module maps, and a single proposed ribbon identity involving them. Realize every image of an idempotent inside its ambient finite tensor sum. There are finitely many rational entries, denominators, weights and matrix sizes in this identity. Specialize $A=s_{\mathrm{\ell }}=e^{\pi i\mathrm{/}(2\mathrm{\ell })}$ for integers $\mathrm{\ell }$ greater than every color, every weight-range width and every divided-power index which can act nontrivially. Exclude the finitely many zeros of the finitely many nonzero denominator polynomials. Every color specializes to $\mathrm{\Delta }(N-1)$ in [F3]; ordinary powers divided by $[r]!$ specialize to its divided powers for all possibly nonzero indices, since those indices are below $\mathrm{\ell }$. Larger divided powers are zero by weight range. A generic module map preserves weights, because the generic $K$ eigenvalues distinguish them, and commutes with these divided powers and the scalar toral characters. It is consequently a full divided-power module map after specialization, not merely a small-quantum-group map. An idempotent specializes to an idempotent module map, whose image is a tilting summand.

The contragredient action and ordinary evaluation and coevaluation in the color definition are module maps by the antipode identities: the action on their composites uses $m(S\otimes 1)\mathrm{\Delta }=m(1\otimes S)\mathrm{\Delta }=\eta \epsilon$. Their vector-space contractions send each basis vector or dual basis vector to itself, proving both snakes. Tensor, sum and idempotent duals use the corresponding transpose maps. Therefore the whole generated category is rigid. The unsigned pivotal comparison is $j_X^{+}(x)(f)=f(Kx)$: $S^2=\mathrm{Ad}K$, and $K$ is group-like, so it is a natural monoidal module isomorphism to the double dual.

At each such specialization the finite generic series becomes exactly the one in [F2]: replacing ordinary powers by divided powers changes its coefficient to $q_K^{r(r-1)\mathrm{/}2}(q_K-q_K^{-1}{)}^r[r]!=q_K^{r(r+1)\mathrm{/}2}(1-q_K^{-2}{)}^r[r]!$. The weight factor is $A^{hk}=s_{\mathrm{\ell }}^{hk}$. Thus the specialized series is a full natural braiding on all the chosen objects and maps, with both hexagons. Each entry of its generic intertwining, naturality, or hexagon equation is a rational function vanishing at infinitely many distinct $s_{\mathrm{\ell }}$. Its numerator polynomial has infinitely many roots and hence is zero. These identities therefore hold generically. Invertibility follows as well: the nilpotent series has constant term $1$, hence a finite geometric-series inverse on each pair, and $D$ is an invertible diagonal matrix. Idempotent restrictions and their inverses are compatible by naturality. This proves braiding on the full generated category.

Let $u$ be [F4]'s Drinfeld isomorphism, formed with the just-proved generic braiding and duality, and put ${\theta }^{+}=u^{-1}{j}^{+}$. Its tensor defect gives the balancing identity. Every matrix entry of its dual-compatibility equation is rational in $A$. Upon each sufficiently large specialization in step 2.1 it is precisely [F2]'s ribbon twist, so the equation holds by the same infinite-root argument as step 3.1. On $V_N$, its specialized matrix is $s_{\mathrm{\ell }}^{(N-1)(N+1)}I$ by [F2]; infinite-root interpolation therefore gives ${\theta }_{V_N}^{+}=A^{N^2-1}I$ generically. This establishes the unsigned ribbon structure without importing an unproved generic universal-$R$ assertion.

On a weight-$h$ vector define $p_X=(-1{)}^h$. This is a module map because $E,F$ change weight by $2$; it is natural because module maps preserve weights, monoidal because weights add, and involutive. On dual weights $-h$ it satisfies $p_{X^{\ast }}=(p_X{)}^{\ast }$, and $p$ commutes with the braiding. Set $j=p{j}^{+}$ and $\theta =p{\theta }^{+}$. Substitution into the unsigned balancing and dual-compatibility equations proves the corresponding equations for $\theta$; monoidality of $p$ proves that $j$ is a pivotal comparison. Since all weights of $V_N$ have parity $N-1$, its pivotal matrix is $G_N=(-1{)}^{N-1}\mathrm{diag}(A^{2(N-1-2j)})$ and its twist is exactly ${\lambda }_N=(-1{)}^{N-1}{A}^{N^2-1}$. The trace of the identity is $(-1{)}^{N-1}{\sum }_{j=0}^{N-1}{A}^{2(N-1-2j)}=d_N$. Every generic ribbon assertion required in the Statement has now been proved, and $N$ was arbitrary.

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 $q=A^4$.

The universal-$R$ sum used for $c_N$ has only the terms $0\le r<N$, since $E^N=F^N=0$ on the displayed $N$-dimensional module. The invertible crossing matrix has finite size, so its inverse is computable by finite Gaussian elimination over $\mathbb{Q}(A)$; forming $J_W=G_N^{\otimes n}$ also takes finitely many field operations. The braid word has finitely many letters, and [F2] performs exactly one matrix update per letter followed by one finite trace contraction. The remaining normalization uses finite arithmetic and integer powers of the nonzero generic scalars $d_N$ and ${\lambda }_N$ in the effective field $\mathbb{Q}(A)$, so every operation terminates.

By [F2], the evaluator returns $t_N(\beta )$, the pivotal trace of the framed colored closure. The last step in [F1] multiplies this value by ${\lambda }_N^{-w(\beta )}$ and divides by $d_N$, exactly the formula in [F3]. Hence the returned value is $J_N^{\mathrm{g}\mathrm{e}\mathrm{n}}(\widehat{\beta };q)$.

Countable Choice, with the meaning in [A2], follows from [A1] by [F6], so the framed evaluator's hypothesis [F5] is satisfied. By [F3], the prescribed category and color satisfy every hypothesis of [F4], including absolute simplicity and the scalar twist. Hence if $\beta$ and ${\beta }^{\mathrm{\prime }}$ close to the same oriented unframed knot, [F4] directly gives equality of their corrected framed traces ${\lambda }_N^{-w}{t}_N$. Dividing by the same nonzero $d_N$ and applying step 2.1 proves equality of the algorithm outputs. Finally [F7] gives their unique common representative in $\mathbb{Z}[q,q^{-1}]$. Substituting any nonzero complex $q$ into its finite Laurent sum is defined, including at $q_N$, without division by a specialized quantum dimension.