Ribbon categories and Reshetikhin–Turaev knot invariants

Under [A1], the evaluator in [F1] exists. The colored framed link is a ribbon graph with no boundary endpoints, so it assigns a morphism from the tensor unit to itself. This is an element of ${\mathrm{End}}_{\mathcal{C}}(\mathbf{1})$, with no field-valued conclusion unless that endomorphism ring has been identified with a field.

If two representatives are framed colored isotopic, [F1] gives equal evaluator morphisms; the empty graph has value $1_{\mathbf{1}}$ by the same fact. Therefore the displayed assignment is well-defined on framed colored isotopy classes and has the stated empty-link value.

The displayed $E,F,K_K$ satisfy the ${\mathfrak{s}\mathfrak{l}}_2$ relations: conjugating $E$ by $K_K$ multiplies its only nonzero entry by $q_K^2$, conjugating $F$ multiplies its only nonzero entry by $q_K^{-2}$, and $EF-FE=\mathrm{diag}(1,-1)=(K_K-K_K^{-1})\mathrm{/}(q_K-q_K^{-1})$, with nonzero denominator since $q_K-q_K^{-1}={\omega }^{-4}-{\omega }^4\mathrm{\ne }0$ in $k^{\mathrm{\prime }}$. Replacing $(q_K,K_K)$ by $(q_L,K_L)=(q_K^{-1},K_K^{-1})$ leaves the commutator quotient unchanged and gives the stated library toral relations. The coproduct conversion in the Definition is exactly [F2] after this substitution.

Any $U_{q_K}({\mathfrak{s}\mathfrak{l}}_2)$-endomorphism of $V$ commutes with $K_K$, whose distinct eigenvalues force it to be diagonal in $(v_{+},v_{-})$. Commuting with $E(v_{-})=v_{+}$ forces the two diagonal entries to agree. Thus $\mathrm{End}(V)=k^{\mathrm{\prime }}{\mathrm{id}}_V$, which is [F7]'s absolute-simplicity condition.

In Korinman's convention the positive braiding is $PD(1+(q_K-q_K^{-1})E\otimes F)$, where $P$ is the flip and $D=\mathrm{diag}(A,A^{-1},A^{-1},A)$ in the displayed tensor basis. Since $E^2=F^2=0$, $(E\otimes F{)}^2=0$ and the source $q$-exponential truncates after its linear term. Applying this operator to the four basis vectors gives, in order, $A{v}_{+}\otimes v_{+}$, $A^{-1}{v}_{-}\otimes v_{+}$, $A^{-1}{v}_{+}\otimes v_{-}+(A-A^{-3})v_{-}\otimes v_{+}$, and $A{v}_{-}\otimes v_{-}$. These are exactly the columns of the displayed matrix; [F6] supplies the half-ribbon completion in which this finite action is a braiding.

The bilinear map $B$ and vector $U(1)$ are $U_{q_K}({\mathfrak{s}\mathfrak{l}}_2)$ intertwiners. The $K_K$-weights in each nonzero pairing and coevaluation term multiply to $1$. For the only nonzero $E$-pairing, $\mathrm{\Delta }(E)(v_{-}\otimes v_{-})=q_K^{-1}{v}_{+}\otimes v_{-}+v_{-}\otimes v_{+}$, whose $B$-value is $q_K^{-1}-q_K^{-1}=0$; the other basis inputs give zero or a multiple of $v_{+}\otimes v_{+}$, also killed by $B$. For $F$, the only nonzero check is $\mathrm{\Delta }(F)(v_{+}\otimes v_{+})=v_{-}\otimes v_{+}+q_K^{-1}{v}_{+}\otimes v_{-}$, whose $B$-value is $-q_K^{-1}+q_K^{-1}=0$. On coevaluation, $\mathrm{\Delta }(E)$ sends $-q_K{v}_{+}\otimes v_{-}+v_{-}\otimes v_{+}$ to $-q_K{v}_{+}\otimes v_{+}+q_K{v}_{+}\otimes v_{+}=0$, and $\mathrm{\Delta }(F)$ sends it to $-q_K{v}_{-}\otimes v_{-}+q_K{v}_{-}\otimes v_{-}=0$; the $K_K$-weight is zero. Since these checks hold on the generators and $\mathrm{\Delta },\epsilon$ are algebra maps, they establish invariance of $B$ and $U$ for every algebra element. The Hopf identity $\sum x_{(1)}\otimes x_{(2)}S(x_{(3)})=x\otimes 1$ then gives $B(xv,w)=\sum B(x_{(1)}v,x_{(2)}S(x_{(3)})w)=\sum \epsilon (x_{(1)})B(v,S(x_{(2)})w)=B(v,S(x)w)$. Thus $\varphi (v)=B(v,-)$ intertwines the stated contragredient action. Its matrix from $V$ to $V^{\ast }$ has nonzero determinant $A^{-2}$, so it is an isomorphism. Transporting the standard coevaluation $1\mapsto v_{+}\otimes v_{+}^{\ast }+v_{-}\otimes v_{-}^{\ast }$ by ${\varphi }^{-1}$ gives exactly the displayed $U$. The snake composites are identities: $(B\otimes 1)(1\otimes U)$ returns $v_{+}$ and $v_{-}$ with coefficients $B(v_{+},v_{-})=1$ and $(-A^2)B(v_{-},v_{+})=1$, respectively, while $(1\otimes B)(U\otimes 1)$ returns them with coefficients $(-A^2)B(v_{-},v_{+})=1$ and $B(v_{+},v_{-})=1$. By [F8], these are the defining left-dual identities. The closed loop is $B(U(1))=-A^2-A^{-2}=\delta$. Thus these maps give the stated self-duality and quantum dimension.

For ${\mathfrak{s}\mathfrak{l}}_2$, the fundamental highest weight $\lambda$ satisfies $(\lambda ,\lambda )=1\mathrm{/}2$, $(\lambda ,\rho )=1\mathrm{/}2$, and $⟨2\lambda ,{\rho }^{\vee }⟩=1$. Snyder–Tingley’s formula [F6] therefore gives $X{v}_{+}=-q_{\mathrm{S}\mathrm{T}}^{3\mathrm{/}4}{v}_{-}$ and $X{v}_{-}=q_{\mathrm{S}\mathrm{T}}^{3\mathrm{/}4}{v}_{+}$. With $q_{\mathrm{S}\mathrm{T}}^{1\mathrm{/}4}={\omega }^{-1}$, this yields $\mathrm{Mat}(\mathrm{\Omega })=\left(\begin{array}{cc}0& {\omega }^{-3}\\ -{\omega }^{-3}& 0\end{array}\right)$ and $\mathrm{Mat}({\mathrm{\Omega }}^{-1})=\left(\begin{array}{cc}0& -{\omega }^3\\ {\omega }^3& 0\end{array}\right)$, whose square is ${\mathrm{\Omega }}^2=-A^{3}I$. The Hopf ribbon identity in [F6] gives $\mathrm{\Delta }(v^{-1})=(v^{-1}\otimes v^{-1})R_{21}R$; since the monodromy $R_{21}R$ is $c_{Y,X}{c}_{X,Y}$, this inverse acts as the categorical twist for the balancing convention in [F4]. Thus the positive categorical twist is $v^{-1}=X^2={\mathrm{\Omega }}^2$.

Under [A1], each braid closure is a framed colored link with empty boundary. By [F1], its value is the endomorphism of the tensor unit assigned by the multicolored framed-tangle evaluator, and its component labels are exactly the colors carried around the permutation cycles.

The isotopy hypothesis is a framed graph isotopy that preserves the labels. By [F2], the two closed tangles have equal evaluation morphisms. Applying [F1] identifies those morphisms with the two braid-closure values, proving equality in ${\mathrm{End}}_{\mathcal{C}}(\mathbf{1})$. This statement concerns framed isotopy; an ordinary Markov stabilization that changes framing is not an additional move here.

By [F6], the assumed Axiom of Choice supplies countable choice, so [F1]'s closure comparison identifies every $t_n$ in the Statement with the algebraic trace. Fix $\gamma \in B_n$ and put $j=j_{X^{\otimes n}}$, $\rho ={\rho }_n$. Apply [F2] with $a=j\rho (\gamma )\rho (\beta )$ and $c=\rho (\gamma {)}^{-1}$ to obtain $t_n(\gamma \beta {\gamma }^{-1})={\mathrm{Tr}}_L((\rho (\gamma {)}^{-1}{)}^{\vee \vee }j\rho (\gamma )\rho (\beta ))$. Naturality of $j$ gives $(\rho (\gamma {)}^{-1}{)}^{\vee \vee }j=j\rho (\gamma {)}^{-1}$, so this is $t_n(\beta )$. The exponent sum is unchanged by [F4], hence $J_X$ is invariant under conjugation.

For a positive or negative stabilization, [F3] multiplies $t_n$ by ${\lambda }^{\pm 1}$ and [F4] changes $w$ by $\pm 1$. Hence ${\lambda }^{-w({\iota }_n\beta {\sigma }_n^{\pm 1})}{t}_{n+1}({\iota }_n\beta {\sigma }_n^{\pm 1})={\lambda }^{-w(\beta )\mp 1}{\lambda }^{\pm 1}{t}_n(\beta )={\lambda }^{-w(\beta )}{t}_n(\beta )$. The scalar is available because [A1] makes the invertible twist a unique $\lambda \in k^{\times }$; the positive sign is fixed by [F3]. Thus $J_X$ is invariant under both stabilizations and their inverses.

By [F5], any two braids with equivalent oriented unframed closures are connected by finitely many conjugations, stabilizations, and destabilizations. Steps 1.1 and 2.1 preserve $J_X$ under each such move, so the corrected value depends only on the oriented unframed link type. The Axiom of Choice [A2] is used in [F5] and supplies the countable-choice hypothesis of [F3] by [F6]; no choice is used in the finite trace or scalar calculations.

After $j$ updates, induction on $j$ shows that the running matrix is the composite of the first $j$ typed crossing maps in composition order. The base case $j=0$ is the identity of $W$; the update rule in [F1] composes exactly the next generator or its inverse, proving the induction step. There are precisely $m$ letters, so there are precisely $m$ updates.

At $j=m$, [F1] identifies the matrix with $M_{\beta }$. Its final-color check makes it an endomorphism of $W$, and the output contraction is the categorical pivotal trace of that endomorphism, hence the defined closure evaluation. Every update is one finite matrix multiplication and the last contraction is finite; the effective-field procedures terminate, so the algorithm terminates and has the asserted value.

In the ordered basis of [F1], direct subtraction gives $e:=A(c_{V,V}-AI)=\left(\begin{array}{cccc}0& 0& 0& 0\\ 0& -A^2& 1& 0\\ 0& 1& -A^{-2}& 0\\ 0& 0& 0& 0\end{array}\right)$. Writing $U(1)=-A^2{v}_{+}\otimes v_{-}+v_{-}\otimes v_{+}$ and $B(v_{+}\otimes v_{-})=1$, $B(v_{-}\otimes v_{+})=-A^{-2}$ shows entrywise that $e=UB$. Hence $B(U(1))=-A^2-A^{-2}=\delta$ and $e^2=\delta e$. The explicit snake equalities in [F2] give the adjacent tensor relations: $e_{12}{e}_{23}{e}_{12}=UB\otimes 1$ because the middle composites $(1\otimes B)(U\otimes 1)$ and then $(B\otimes 1)(1\otimes U)$ are identities; the reversed calculation gives $e_{23}{e}_{12}{e}_{23}=e_{23}$. Distant placements commute because they act on disjoint tensor factors.

Use source coordinates $(u,v)$ with their standard positive orientation and positive projection coordinates $(t,x)$ on the braid box. For $h>0$, the map $(t,x)=(v,-2hu)$ has determinant $2h>0$. It sends the source over direction $(1,1)$ to $(1,-2h)$ and the source under direction $(-1,1)$ to $(1,2h)$, whose determinant is $4h>0$. The source left/right pairs have constant $u$, hence constant braid position $x$ and give the through pairing $I$; the source top/bottom pairs have constant $v$, hence constant braid time $t$ and give the cup-cap $e_i$. The displayed page coordinates $(x,t)$ reverse the positive projection orientation; their visual quarter-turn is not the orientation comparison. Thus the A/B crossing expansion is $C_i=AI+A^{-1}{e}_i$. Its inverse is $C_i^{-1}=A^{-1}I+A{e}_i$, since their product is $I+(A^2+A^{-2}+\delta )e_i=I$. By [A1] and [F8]–[F9], the categorical ribbon trace is the framed tangle evaluation of this braid closure, and that evaluator is monoidal with the stated crossing and duality maps. Expanding a braid word into the two smoothing choices therefore gives the bracket weights in [F4]. Because the closure has $n\ge 1$ strands, each complete smoothing is a nonempty closed 1-manifold, hence a union of at least one circle. Its categorical value is ${\delta }^{\mathrm{\mid }S\mathrm{\mid }}$ by monoidality in [F8] and the loop value [F2], whereas the bracket state sum uses ${\delta }^{\mathrm{\mid }S\mathrm{\mid }-1}$; the extra single factor is independent of the state. Thus $t_n(\beta )=\delta ⟨\widehat{\beta }⟩$.

The matrix trace $t_n$ in [F3] closes the tensor-word braid with the duality maps $B,U$ of [F2]. Substituting each crossing into the trace expansion from step 2.1 therefore gives the first identity in the Statement. The Hecke formula [F6] then yields $t_n(\beta )={\delta }^n(-A{)}^{e(\beta )}{\mathrm{tr}}_n({\pi }_n(\beta )){\mid }_{v=t,z=z_0}$. By [F7], $e(\beta )=w(\beta )$. The twist in [F1] is $\lambda =-A^3$, so multiplying $t_n=\delta ⟨\widehat{\beta }⟩$ by ${\lambda }^{-w(\beta )}$ gives the second displayed identity. The only choice assumption is [A1], used in step 2.1 to invoke the published trace-to-closure identification; no stronger choice principle or division by $\delta$ is used. Since $A$ is indeterminate, the specialization is taken in the rational-function field where $t+1\mathrm{\ne }0$.