Holonomy braidings, biquandles and quantum invariants of links with $\mathsf {SL} _2(\mathbb {C} )$ flat connections

Abstract

R. Kashaev and N. Reshetikhin introduced the notion of holonomy braiding extending V. Turaev’s homotopy braiding to describe the behavior of cyclic representations of the unrestricted quantum group ${U_q{\mathfrak {sl}(2)}}$ at root of unity. In this paper, using quandles and biquandles we develop a general theory for Reshetikhin–Turaev ribbon type functor for tangles with quandle representations. This theory applies to the unrestricted quantum group ${U_q{\mathfrak {sl}(2)}}$ and produces an invariant of links with a gauge class of quandle representations.

Appendices

Appendix A: Proof related with generically define functor

In this appendix we prove Proposition 5.3 and Theorem 5.5.

Proof of Proposition 5.3

Assume the underlying diagrams of D and $D'$ are related by a Reidemeister move. By functoriality (and injectivity) of ${{\mathcal {Q}}}$ it suffices to assume D and $D'$ are diagrams of the form ${\text {Id}}_{w_1}\otimes E\otimes {\text {Id}}_{w_2}:w_1w_3w_2\rightarrow w_1w_4w_2$ and ${\text {Id}}_{w_1'}\otimes E'\otimes {\text {Id}}_{w_2'}:w_1'w_3'w_2'\rightarrow w_1'w_4'w_2'$ where E and $E'$ are the diagrams of the Reidemeister move and $w_i, w_i'$ are words in Y. Since ${{\mathcal {Q}}}$ is injective on Y-colored diagrams, it is clear that the source and target of D and $D'$ are equal if and only if the source and target of ${{\mathcal {Q}}}(D)$ and ${{\mathcal {Q}}}(D')$ are equal. Since ${{{\mathsf {Q} }}}$ is a quandle then in particular it is a biquandle (where $B_1(a,b)=a \rhd b$ and $B_2(a,b)=a$ for $a,b\in {{{\mathsf {Q} }}}$). It follows that the coloring of the incoming boundary of a braid extends uniquely to a ${{{\mathsf {Q} }}}$-coloring of the complete braid. Thus, the coloring ${{\mathcal {Q}}}(D')$ exists, it has the same shape as $D'$ and the same boundary colors as ${{\mathcal {Q}}}(D)$. But there is an unique such ${{{\mathsf {Q} }}}$-colored diagram and it is also obtained by doing a ${{{\mathsf {Q} }}}$-colored Reidemeister move to ${{{\mathsf {Q} }}}(D)$. Therefore, we obtain the if and only if stated in the proposition. $\square$

To prove Theorem 5.5 we first state and prove a key lemma. Recall the (m, n)-cable of the positive and negative crossings given in Fig. 8. Given words, $w,w'\in W_Y$ one can try to use the generic biquandle structure to color one of these diagrams (coloring may not exist because the biquandle maps are not defined everywhere). If such a coloring exists we denote the corresponding Y-colored diagram by $\chi ^+_{w,w'}$ and $\chi ^-_{w,w'}$, respectively. Suppose $\chi ^-_{w,w'}$ and $\chi ^+_{w'',w'''}$ exists and are “inverses,” i.e. $\chi ^+_{w'',w'''}\circ \chi ^-_{w,w'}$ and ${\text {Id}}_w \otimes {\text {Id}}_{w'}$ are related by a sequence of Y-colored Reidemeister moves. In this situation we use a slight abuse of notation and denote $\chi ^+_{w'',w'''}$ by $\left( \chi ^-_{w,w'}\right) ^{-1}$.

Lemma A.1

Let $D:w\rightarrow w'\in {{{\mathcal {D}}}}_Y$, then for generic $x\in Y$ the morphism

$$\begin{aligned} \left( \chi ^-_{x,w}\right) ^{-1}\circ \left[ {{\mathcal {Q}}}^{-1}\left( x{\rightharpoondown }{{\mathcal {Q}}}(D) \right) \otimes {\text {Id}}\right] \circ \chi _{x,w}^{-} \end{aligned}$$

(42)

exists and is related to ${\text {Id}}_{(x,+)}\otimes D$ by a sequence of Y-colored Reidemeister moves.

Proof

The composition of generic bijections of Y is a generic bijection of Y. Thus for any $w\in W_Y$, Axiom 3 of Definition 5.1 implies that for generic $x\in Y$, there exists a (unique) Y-coloring of the diagrams $\chi ^-_{x,w}$ and $(\chi ^-_{x,w})^{-1}$. We conclude that the morphism in Eq.  (42) exist for generic $x\in Y$.

We prove the second statement in the lemma holds in two steps as follows.

Step 1: We will show that statement holds for $D={\text {Id}}\otimes E\otimes {\text {Id}}:w\rightarrow w'$ be where E is an elementary diagram made by a crossing or one cup or cap. From Axiom 6 of Definition  5.1, for generic $x\in Y$ there exist $D'\in {{{\mathcal {D}}}}_Y$ such that ${{\mathcal {Q}}}(D')=x{\rightharpoondown }{{\mathcal {Q}}}(D)$. As explained in the previous paragraph, for generic $x\in Y$ there exists

$$\begin{aligned} \chi ^-_{x,w}:x\otimes w\rightarrow w_x\otimes y {\quad \text {and}\quad }\left( \chi ^-_{x,w}\right) ^{-1} :w_x\otimes y \rightarrow x\otimes w \end{aligned}$$

where $w_x\in W_Y$ and $y\in Y$. The image of the morphism $\chi ^-_{x,w}$ under ${{\mathcal {Q}}}$ is a map with domain ${{\mathcal {Q}}}(x\otimes w)={{\mathcal {Q}}}(x)\otimes (x{\rightharpoondown }{{\mathcal {Q}}}(w))$ and range ${{\mathcal {Q}}}(w_x\otimes y)={{\mathcal {Q}}}(w_x)\otimes q_y$ for some $q_y\in {{{\mathsf {Q} }}}$. Since $\chi ^-_{x,w}$ comes from a negative crossing the quandle color of the over-strand does not change and so we have ${{\mathcal {Q}}}(w_x)=x{\rightharpoondown }{{\mathcal {Q}}}(w)$. The last two sentences imply that the two ${{{\mathsf {Q} }}}$-diagrams

$$\begin{aligned} {{\mathcal {Q}}}(D'\otimes {\text {Id}}_y)\; \text { and } \; \left( x{\rightharpoondown }{{\mathcal {Q}}}(D)\right) \otimes {\text {Id}}_{q_y} \end{aligned}$$

have the same incoming ${{{\mathsf {Q} }}}$-colored boundaries. Now, since both of these ${{{\mathsf {Q} }}}$-diagrams have the same underlying diagram which is elementary and so determined by its incoming ${{{\mathsf {Q} }}}$-colored boundary, we conclude the diagrams are equal in ${{{\mathcal {D}}}}_{{{\mathsf {Q} }}}$. In particular, taking the preimage under ${{\mathcal {Q}}}$ we have

$$\begin{aligned} D'\otimes {\text {Id}}_y ={{\mathcal {Q}}}^{-1}\left( x{\rightharpoondown }{{\mathcal {Q}}}(D)\right) \otimes {\text {Id}}_{y}. \end{aligned}$$

Finally, in ${{{\mathcal {D}}}}_{{{\mathsf {Q} }}}$ we have

$$\begin{aligned} {{\mathcal {Q}}}{\left( \left( \chi ^-_{x,w}\right) ^{-1}\right) } {{\mathcal {Q}}}(D'\otimes {\text {Id}}_y) {{\mathcal {Q}}}{\left( \chi _{x,w}^{-}\right) }&{\mathop {\equiv }\limits ^{1}}{{\mathcal {Q}}}{\left( \left( \chi ^-_{x,w}\right) ^{-1}\right) } {{\mathcal {Q}}}{\left( \chi _{x,w'}^-\right) }{{\mathcal {Q}}}{\left( {\text {Id}}_x\otimes D\right) } \end{aligned}$$

where the Reidemeister equivalence ${\mathop {\equiv }\limits ^{1}}$ comes from the fact that underlying diagrams differ by a single Reidemeister move which is ${{{\mathsf {Q} }}}$-colored. Since both the diagrams in the above equation are in the image of ${{\mathcal {Q}}}$ then Proposition 5.3 implies their preimage are related by a Y-colored Reidemeister move. Thus, the last two equation combined with the facts that ${{\mathcal {Q}}}$ is a functor and ${\left( \chi _{x,w'}^-\right) }^{-1}\circ {\left( \chi _{x,w'}^-\right) }$ is related by Y-colored Reidemeister moves to the identity we can conclude the lemma is true when $D={\text {Id}}\otimes E\otimes {\text {Id}}$.

Step 2: We will show, if the lemma is true for $D_1:w_1\rightarrow w_2$ and $D_2:w_2\rightarrow w_3$ then it is true for $D=D_2\circ D_1$. To simplify notation, for $x\in Y$, we let $H_x: {{{\mathcal {D}}}}_Y\rightarrow {{{\mathcal {D}}}}_Y$ be the partial map given by

$$\begin{aligned} H_x(D)={{\mathcal {Q}}}^{-1}(x{\rightharpoondown }{{\mathcal {Q}}}(D)). \end{aligned}$$

Definition 5.1 implies, for generic $x\in Y$ all six diagrams $\chi ^-_{x,w_1}$, $\chi ^-_{x,w_2}$, $(\chi ^-_{x,w_1})^{-1}$, $(\chi ^-_{x,w_2})^{-1}$, $H_x(D_1)$ and $H_x(D_2)$ exist. Also, for such x, we have the range of $(\chi ^-_{x,w_1})^{-1}$ is $x\otimes w_1$ and the diagram $\chi ^-_{x,w_2}\circ (\chi ^-_{x,w_1})^{-1}$ is related to the identity by a sequence of Y-colored Reidemeister moves (since colorings of Reidemeister II moves reducing the number of crossing always exist). Therefore, we have

$$\begin{aligned} {\text {Id}}_{(x,+)}\otimes D&\equiv {\left( {\text {Id}}_{(x,+)}\otimes D_2\right) }{\left( {\text {Id}}_{(x,+)}\otimes D_1\right) }\\&\equiv {\left( (\chi ^-_{x,w_2})^{-1} {\left( H_x(D_2)\otimes {\text {Id}}\right) } \chi _{x,w_2}^{-}\right) } {\left( (\chi ^-_{x,w_1})^{-1} {\left( H_x(D_1)\otimes {\text {Id}}\right) }\chi _{x,w_1}^{-}\right) }\\&\equiv (\chi ^-_{x,w_2})^{-1} {\left( H_x(D_2)\otimes {\text {Id}}\right) } {\left( \chi _{x,w_2}^{-}(\chi ^-_{x,w_2})^{-1}\right) } {\left( H_x(D_1)\otimes {\text {Id}}\right) }\chi _{x,w_1}^{-}\\&\equiv (\chi ^-_{x,w_2})^{-1} {\left( {\left( H_x(D_2)\circ H_x(D_1)\right) }\otimes {\text {Id}}\right) } \chi _{x,w_1}^{-}\\&\equiv (\chi ^-_{x,w_2})^{-1} {\left( H_x(D)\otimes {\text {Id}}\right) } \chi _{x,w_1}^{-}. \end{aligned}$$

$\square$

Proof of Theorem 5.5

Since ${{\mathcal {Q}}}(D)$ and ${{\mathcal {Q}}}(D')$ represent isotopic ${{{\mathsf {Q} }}}$-tangles there exists a sequence of ${{{\mathsf {Q} }}}$-colored Reidemeister moves

$$\begin{aligned} {{\mathcal {Q}}}(D)= D_0^{{{\mathsf {Q} }}}{\mathop {\equiv }\limits ^{1}}D_1^{{{\mathsf {Q} }}}{\mathop {\equiv }\limits ^{1}}\cdots {\mathop {\equiv }\limits ^{1}}D_n^{{{\mathsf {Q} }}}={{\mathcal {Q}}}(D'). \end{aligned}$$

It is possible that ${{\mathcal {Q}}}^{-1}(D_i^{{{\mathsf {Q} }}})$ does not exist for some $i\in \{1,...,n-1\}$. To address this we use a gauge transformation by a generic $x\in Y$. Axiom 6 of Definition 5.1, implies that for generic $x\in Y$, the diagrams

$$\begin{aligned} D_i={{\mathcal {Q}}}^{-1}{\left( x{\rightharpoondown }D_i^{{{\mathsf {Q} }}}\right) }\in {{{\mathcal {D}}}}_Y \end{aligned}$$

exists for all $i\in \{0,\ldots , n\}$. Since the diagrams underlying ${{\mathcal {Q}}}(D_i)=x{\rightharpoondown }D_i^{{{\mathsf {Q} }}}$ and ${{\mathcal {Q}}}(D_{i+1})=x{\rightharpoondown }D_{i+1}^{{{\mathsf {Q} }}}$ are related by a single Reidemeister move, it extends to a ${{{\mathsf {Q} }}}$-colored move. Then Proposition 5.3 ensures that $D_0{\mathop {\equiv }\limits ^{1}}D_1{\mathop {\equiv }\limits ^{1}}\cdots {\mathop {\equiv }\limits ^{1}}D_n$ is a sequence of Y-colored Reidemeister moves, so the same is true for

$$\begin{aligned} (\chi _{x,{\diamond }}^-)^{-1} {\left( D_0\otimes {\text {Id}}_{\diamond }\right) }\chi _{x,{\diamond }}^{-}{\mathop {\equiv }\limits ^{1}}\cdots {\mathop {\equiv }\limits ^{1}}(\chi _{x,{\diamond }}^-)^{-1} {\left( D_n\otimes {\text {Id}}_{\diamond }\right) }\chi _{x,{\diamond }}^{-} \end{aligned}$$

where ${\diamond }$ represents the appropriate word in $W_Y$. Lemma A.1 implies that

$$\begin{aligned} {\text {Id}}_{(x,+)}\otimes D \equiv (\chi _{x,{\diamond }}^-)^{-1} {\left( D_0\otimes {\text {Id}}_{\diamond }\right) }\chi _{x,{\diamond }}^{-}{\,\text {,}\quad }{\text {Id}}_{(x,+)}\otimes D' \equiv (\chi _{x,{\diamond }}^-)^{-1} {\left( D_n\otimes {\text {Id}}_{\diamond }\right) }\chi _{x,{\diamond }}^{-}. \end{aligned}$$

Combining the above equivalence we have

$$\begin{aligned} {\text {Id}}_{(x,+)}\otimes D\equiv {\text {Id}}_{(x,+)}\otimes D' \end{aligned}$$

for generic $x\in Y$. $\square$

Appendix B: Proof of Proposition 5.6

Here we prove Proposition 5.6. We need to show $(Y,\mathcal {G}, B, {{\mathcal {Q}}}, {\rightharpoondown })$ as defined in Sect. 5.2 satisfy the axioms of Definition 5.1. The maps $B, B^{- 1}, S$ and $S^{- 1}$ defined in Sect.  5.2 by construction satisfy Axiom 2.

For $x\in Y$, the maps

$$\begin{aligned} B_1^\pm (x,{\diamond }),\; S_1^\pm (x,{\diamond }),\; B_2^\pm ({\diamond },x),\; S_2^\pm ({\diamond },x) \end{aligned}$$

are all rational maps (as $\psi ^{\pm 1},\varphi _\pm$ and matrix inversion are rational maps). In fact these maps are pairwise inverse and so they send Zariski open dense sets to Zariski open dense sets. Thus, Axiom 3 of Definition 5.1 is satisfied.

Next we recall the definition of the functor ${{\mathcal {Q}}}$, see [20] and [14]. Let T be a standard tangle in $(0,+\infty )\times \mathbb {R} \times [0,1]$. Let D be a Y-colored diagram in $(0,+\infty )\times \{0\}\times [0,1]$ which we assume is a projection of T in the y-coordinate direction. We assume T is close to D. To define the functor we consider two kinds of paths: (1) positive paths which are in $(0,+\infty )\times [0, +\infty )\times [0,1]$ and to the right of T and (2) negative paths in $(0,+\infty )\times (-\infty ,0]\times [0,1]$ and to the left of T. Let $M_+$ (resp. $M_-$) be the set of points traced out by all positive (resp. negative) paths. Then $M_+\cap M_-$ is a subset of $(0,+\infty )\times \{0\}\times [0,1]$ whose connected components $r_i$ are delimited by the diagram D. The closure of one of these components contains $\{0\}\times \{0\}\times [0,1]$, we denote this component by $r_0$. We fix a point $P_i$ in each region $r_i$. Recall that a quandle $\mathsf {G}$-coloring of D is equivalent to the data of a representation $\rho :\pi _1(M_T,P_0)\rightarrow \mathsf {G}$, see Remark 2.5 and Theorem 2.8.

Let $(\overrightarrow{P_iP_j})_\pm$ be a path in $M_\pm$ from $P_i$ to $P_j$. The space $M_\pm$ is contractible so a path $(\overrightarrow{P_iP_j})_\pm$ in the groupoid $\pi _1(M_\pm ,\{P_i\})$ is uniquely determined by its end points. The positive and negative paths to and from adjacent regions generate the groupoid $\pi _1(M,\{P_i\})$ (for relations see Lemma 3.4 of [14]) so a representation of the groupoid in $\mathsf {G}$ is determined by their image. If an edge e of D with Y-color x has region $r_i$ on the left and $r_j$ on the right, we assign to the path $(\overrightarrow{P_iP_j})_\pm$ the element $\varphi _\pm (x)$. The properties of the Y-coloring imply that this assignment satisfies the defining relations of the groupoid $\pi _1(M_T,\{P_i\})$ given in Lemma 3.4 of [14]. Thus, this assignment extends (uniquely) to a representation $\overline{\rho }$ of $\pi _1(M_T,\{P_i\})$ in $\mathsf {G}$. The meridian of the edge e, as above, is the loop $(\overrightarrow{P_iP_j})_+.(\overrightarrow{P_iP_j})_-$ and its image by $\overline{\rho }$ is $\psi (x)=\varphi _+(x)\varphi _-(x)^{-1}\in \mathsf {G}$. As the fundamental group $\pi _1(M_T,P_0)$ is generated by the meridians of the edges we have the restriction of $\overline{\rho }$ to $\pi _1(M_T,P_0)$ takes values in $\mathsf {G}$ and thus is a $\mathsf {G}$-tangle structure $\rho$ on T.

The partially defined inverse map is constructed as follows: we want to extend a representation

$$\begin{aligned} \rho \in {\text {Hom}}_{\tiny {\text {quandle}}}(Q(T,P_0),{\text {Conj}}(\mathsf {G} )) \cong {\text {Hom}}_{\tiny {\text {group}}}(\pi _1(M_T,P_0),\mathsf {G} ) \end{aligned}$$

to a representation $\overline{\rho }\in {\text {Hom}}_{\tiny {\text {groupoid}}}(\pi _1(M_T,\{P_i\}),\mathsf {G} )$ that will send $(\overrightarrow{P_iP_j})_\pm$ to $\varphi _\pm (x)$. This can be done if and only if

$$\begin{aligned} \rho \left( (\overrightarrow{P_0 P_i})_+.(\overrightarrow{P_iP_0})_-\right) \in \psi ({\mathsf {G} ^*}) \end{aligned}$$

(43)

for all points $P_i$. Assuming that this is true, let

$$\begin{aligned} g_i=\psi ^{-1}\left( \rho \left( (\overrightarrow{P_0 P_i})_+.(\overrightarrow{P_iP_0})_-\right) \right) \end{aligned}$$

then by defining $\overline{\rho }((\overrightarrow{P_iP_j})_\pm )=\varphi _\pm (g_i^{-1}g_j)$ we obtain the desired representation. If an edge e of D has region $r_i$ on the left and $r_j$ on the right, the Y-coloring associated to $\overline{\rho }$ assign to e the color $g_i^{-1}g_j$.

Hence we obtain an injective functor ${{\mathcal {Q}}}$ whose image is formed by $\mathsf {G}$-colorings whose underlying quandle map $\rho$ satisfy Eq.  (43). Next we check that this functor satisfies Eq. (18). Let $x\in Y$ and $w\in W_Y$. As above let $T_1$ and $T_2$ be two standard tangles close the trivial Y-colored diagrams ${\text {Id}}_w$ and ${\text {Id}}_{x\otimes w}$, respectively. Let $\rho _i: \pi _1(M_{T_i},P_0)\rightarrow \mathsf {G}$ be their representation constructed above. Let e be a strand corresponding to a letter in the word w (we think of e in both $T_1$ and $T_2$). Since $T_2$ is just $T_1$ with and extra strand on the left we see that to $\rho _2(e)=\rho _2((\overrightarrow{P_0P_1})_+)\rho _1(e)\rho _2((\overrightarrow{P_0P_1})_+)^{-1}$ where the conjugation of $\rho _2((\overrightarrow{P_0P_1})_+)$ come from passing over first strand of $T_2$ and back. But by definition $\rho _2((\overrightarrow{P_0P_1})_+)=\varphi _+(x)$. Thus, $\rho _2(e)=x{\rightharpoondown }\rho _1(e)$ and the functor satisfies Eq.  (18).

Finally, for fixed $g\in \mathsf {G}$, the set of h in $\varphi _+({\mathsf {G} ^*})$ such that $hgh^{-1}\in \psi ({\mathsf {G} ^*})$ is the complement of an algebraic hypersurface. The preimage by $\varphi _+$ of this set is an open set $Z_g\in \mathcal {G}$. Then the set of $x\in {\mathsf {G} ^*}$ such that ${{\mathcal {Q}}}^{-1}(x{\rightharpoondown }D)$ exists is $\bigcap \left\{ Z_g:g\in \{\rho {\left( \overrightarrow{P_\infty P_i}_+.\overrightarrow{P_iP_\infty }_-\right) }\}\right\} \in \mathcal {G}$. This proves Axiom (6) and completes the proof of Proposition 5.6.

Appendix C: Proof of Theorem 5.8

The proof of first part of the theorem is the same as the proof of Theorem 4.2. To prove the second part of the theorem, recall Theorem 5.5 implies ${\text {Id}}_x\otimes D\equiv {\text {Id}}_x\otimes D'\in {{{\mathcal {D}}}}_Y$ for generic $x\in Y$. As F is invariant by Y-colored Reidemeister moves, we have that

$$\begin{aligned} {\text {Id}}_{V_x}\otimes F(D)=F({\text {Id}}_x\otimes D)=F({\text {Id}}_x\otimes D')={\text {Id}}_{V_x}\otimes F(D'). \end{aligned}$$

Now Part (2) of the theorem follows from the fact that $V_x$ is regular.

To prove Part (3) of the theorem it is convenient to prove a more general proposition about the disjoint union of diagrams of links and 1-1 tangles. With this in mind, consider the disjoint union of ${{{\mathsf {Q} }}}$-colored diagram $\sqcup _i D_i$ where $D_i$ is a diagram representing a ${{{\mathsf {Q} }}}$-link or a 1-1 ${{{\mathsf {Q} }}}$-tangle; let ${{{\mathscr {D}}}_{\Bbbk }}$ be the set of all such unions. Also, let ${{{\mathscr {D}}}_{{\mathcal {Q}}}}$ be the set of all ${{{\mathsf {Q} }}}$-colored diagram that are in the image of ${{\mathcal {Q}}}$. For $D\in {{{\mathscr {D}}}_{{\mathcal {Q}}}}\cap {{{\mathscr {D}}}_{\Bbbk }}$, let

$$\begin{aligned} F_{{{{\mathscr {D}}}_{\Bbbk }}}(D)={\left\langle {F{\left( {{\mathcal {Q}}}^{-1}(D)\right) }}\right\rangle }\in \Bbbk , \end{aligned}$$

where the bracket is defined as the scalar corresponding to the scalar endomorphism $F{\left( {{\mathcal {Q}}}^{-1}(D)\right) }$.

Remark that tensoring on the right by any identity does not affect $F_{{{{\mathscr {D}}}_{\Bbbk }}}$, i.e. $F_{{{{\mathscr {D}}}_{\Bbbk }}}(D\otimes {\text {Id}})=F_{{{{\mathscr {D}}}_{\Bbbk }}}(D)$. Remark also that for a Y-colored diagram D with ${{\mathcal {Q}}}(D)\in {{{\mathscr {D}}}_{\Bbbk }}$ then $F_{{{{\mathscr {D}}}_{\Bbbk }}}({{\mathcal {Q}}}(D))={\left\langle {F(D)}\right\rangle }$. In particular, if D is a Y-colored diagram of a link then $F_{{{{\mathscr {D}}}_{\Bbbk }}}({{\mathcal {Q}}}(D))=F(D)$.

Thus, Theorem 5.8. is a corollary of the following proposition:

Proposition C.1

The partial map $F_{{{{\mathscr {D}}}_{\Bbbk }}}: {{{\mathscr {D}}}_{\Bbbk }}\rightarrow \Bbbk$ is B-gauge invariant, i.e. invariant under the equivalence generated by $(x{\rightharpoondown }{\diamond })_{x\in Y}$ and $(a\rhd {\diamond })_{a\in {{{\mathsf {Q} }}}}$.

Proof

We prove the following five claims:

• (1) If $D\in {{{\mathscr {D}}}_{{\mathcal {Q}}}}\cap {{{\mathscr {D}}}_{\Bbbk }}$ then for generic $x\in Y$, $F_{{{{\mathscr {D}}}_{\Bbbk }}}(D)=F_{{{{\mathscr {D}}}_{\Bbbk }}}(x{\rightharpoondown }D)$.

• (2) Let $D_1, D_2\in {{{\mathscr {D}}}_{{\mathcal {Q}}}}\cap {{{\mathscr {D}}}_{\Bbbk }}$ and $D'_1,D'_2$ be two ${{{\mathsf {Q} }}}$-colored diagrams of the form $D'_i={{\mathcal {Q}}}({\text {Id}}\otimes {{\mathcal {Q}}}^{-1}(D_i))\otimes {\text {Id}}$. Suppose that there exists a ${{{\mathsf {Q} }}}$-colored braid diagram $\sigma$ such that $D'_2\circ \sigma \equiv \sigma \circ D'_1$, then $F_{{{{\mathscr {D}}}_{\Bbbk }}}(D_1)=F_{{{{\mathscr {D}}}_{\Bbbk }}}(D_2)$.

• (3) For any $x,y\in Y$ and any $D\in {{{\mathscr {D}}}_{\Bbbk }}$, if $x{\rightharpoondown }D,y{\rightharpoondown }D\in {{{\mathscr {D}}}_{{\mathcal {Q}}}}$ then $F_{{{{\mathscr {D}}}_{\Bbbk }}}(x{\rightharpoondown }D)=F_{{{{\mathscr {D}}}_{\Bbbk }}}(y{\rightharpoondown }D)$.

• (4) The partial map $F_{{{{\mathscr {D}}}_{\Bbbk }}}$ on ${{{\mathscr {D}}}_{\Bbbk }}$ is invariant for the equivalence relation generated by $(x{\rightharpoondown }{\diamond })_{x\in Y}$.

• (5) The partial map $F_{{{{\mathscr {D}}}_{\Bbbk }}}$ on ${{{\mathscr {D}}}_{\Bbbk }}$ is invariant for the equivalence relation generated by $(a\rhd {\diamond })_{a\in {{{\mathsf {Q} }}}}$.

$\bullet$ To prove Claim (1) we see that Lemma A.1 implies: If $D={{\mathcal {Q}}}(D')$ with $D':w\rightarrow w$ then

$$\begin{aligned} {\text {Id}}_{V_x}\otimes F_{{{{\mathscr {D}}}_{\Bbbk }}}(D).{\text {Id}}_{F(w)}=F{\left( \chi ^-_{x,w}\right) }^{-1}\circ {\left( F_{{{{\mathscr {D}}}_{\Bbbk }}}(x{\rightharpoondown }D).{\text {Id}}_{V_{x'}}\otimes {\text {Id}}_{F(w')}\right) }\circ F{\left( \chi ^-_{x,w})\right) }. \end{aligned}$$

The claim then follows because $V_x\otimes F(w)$ is regular.

$\bullet$ For Claim (2), let $D^Y_i= {{\mathcal {Q}}}^{-1}(D_i)$ for i=1,2. We have that for generic $x\in Y$, the diagrams $x{\rightharpoondown }(D'_2\circ \sigma )$ and $x{\rightharpoondown }(\sigma \circ D'_1)$ are in ${{{\mathscr {D}}}_{{\mathcal {Q}}}}$ and they represent isotopic tangles so $F\circ {{\mathcal {Q}}}^{-1}(x{\rightharpoondown }(D'_2\circ \sigma ))=F\circ {{\mathcal {Q}}}^{-1}(x{\rightharpoondown }(\sigma \circ D'_1))$. Thus,

$$\begin{aligned} F_{{{{\mathscr {D}}}_{\Bbbk }}}(x{\rightharpoondown }D'_2).F\circ {{\mathcal {Q}}}^{-1}(x{\rightharpoondown }\sigma )=F_{{{{\mathscr {D}}}_{\Bbbk }}}(x{\rightharpoondown }D'_1).F\circ {{\mathcal {Q}}}^{-1}(x{\rightharpoondown }\sigma ) \end{aligned}$$

and as $F\circ {{\mathcal {Q}}}^{-1}(x{\rightharpoondown }\sigma )$ is invertible and objects are regular, we get $F_{{{{\mathscr {D}}}_{\Bbbk }}}(x{\rightharpoondown }D'_1)=F_{{{{\mathscr {D}}}_{\Bbbk }}}(x{\rightharpoondown }D'_2)$. Next

$$\begin{aligned} F_{{{{\mathscr {D}}}_{\Bbbk }}}(x{\rightharpoondown }D'_i)=F_{{{{\mathscr {D}}}_{\Bbbk }}}(x{\rightharpoondown }[{{\mathcal {Q}}}({\text {Id}}\otimes D^Y_i)\otimes {\text {Id}}])=F_{{{{\mathscr {D}}}_{\Bbbk }}}(x{\rightharpoondown }[{{\mathcal {Q}}}({\text {Id}}\otimes D^Y_i)]), \end{aligned}$$

then as ${{\mathcal {Q}}}({\text {Id}}\otimes D^Y_i)\in {{{\mathscr {D}}}_{{\mathcal {Q}}}}\cap {{{\mathscr {D}}}_{\Bbbk }}$ we can apply Claim (1) twice and we get that for generic x,

$$\begin{aligned} F_{{{{\mathscr {D}}}_{\Bbbk }}}({{\mathcal {Q}}}({\text {Id}}\otimes D^Y_1))= & {} F_{{{{\mathscr {D}}}_{\Bbbk }}}(x{\rightharpoondown }[{{\mathcal {Q}}}({\text {Id}}\otimes D^Y_1)])=F_{{{{\mathscr {D}}}_{\Bbbk }}}(x{\rightharpoondown }[{{\mathcal {Q}}}({\text {Id}}\otimes D^Y_2)])\\= & {} F_{{{{\mathscr {D}}}_{\Bbbk }}}({{\mathcal {Q}}}({\text {Id}}\otimes D^Y_2)). \end{aligned}$$

Finally, we see Claim (2) follows from $F_{{{{\mathscr {D}}}_{\Bbbk }}}({{\mathcal {Q}}}({\text {Id}}\otimes D^Y_i))=F_{{{{\mathscr {D}}}_{\Bbbk }}}(D_i)$.

$\bullet$ We are now ready to prove Claim (3): Let $D:w\rightarrow w$ with $D\in {{{\mathscr {D}}}_{\Bbbk }}$. Define $q_x,q_y\in {{{\mathsf {Q} }}}$ by

$$\begin{aligned} {{\mathcal {Q}}}({\text {Id}}_{(x,-)}\otimes {{\mathcal {Q}}}^{-1}(x{\rightharpoondown }D))={\text {Id}}_{(q_x,-)}\otimes D \end{aligned}$$

and

$$\begin{aligned} {{\mathcal {Q}}}({\text {Id}}_{(y,-)}\otimes {{\mathcal {Q}}}^{-1}(y{\rightharpoondown }D))={\text {Id}}_{(q_y,-)}\otimes D. \end{aligned}$$

Let $\sigma$ be the ${{{\mathsf {Q} }}}$-colored braid

then $({\text {Id}}\otimes D\otimes {\text {Id}})\circ \sigma \equiv \sigma \circ ({\text {Id}}\otimes D\otimes {\text {Id}})$ and Claim (2) implies that $F_{{{{\mathscr {D}}}_{\Bbbk }}}(x{\rightharpoondown }D)=F_{{{{\mathscr {D}}}_{\Bbbk }}}(y{\rightharpoondown }D)$.

$\bullet$ To prove Claim (4) let us first extend the partial map $F_{{{{\mathscr {D}}}_{\Bbbk }}}$ to a total map $\widehat{F}$ on ${{{\mathscr {D}}}_{\Bbbk }}$ by $\widehat{F}(D):=F_{{{{\mathscr {D}}}_{\Bbbk }}}(x{\rightharpoondown }D)$ for any $x\in Y$ such that $x{\rightharpoondown }D\in {{{\mathscr {D}}}_{{\mathcal {Q}}}}$. Claim (3) implies that this extension is well defined and Claim (1) implies that $F_{{{{\mathscr {D}}}_{\Bbbk }}}$ is indeed the restriction of $\widehat{F}$ on ${{{\mathscr {D}}}_{{\mathcal {Q}}}}\cap {{{\mathscr {D}}}_{\Bbbk }}$. Now Claim (4) is equivalent to

$$\begin{aligned} \widehat{F}(x{\rightharpoondown }D)=\widehat{F}(D), \quad \forall (x,D)\in Y\times {{{\mathscr {D}}}_{\Bbbk }}. \end{aligned}$$

(44)

If $x{\rightharpoondown }D\in {{{\mathscr {D}}}_{{\mathcal {Q}}}}$, this follows from the definition of $\widehat{F}$. Else, we have

$$\begin{aligned} \widehat{F}(x{\rightharpoondown }D)=F_{{{{\mathscr {D}}}_{\Bbbk }}}(z{\rightharpoondown }(x{\rightharpoondown }D)) \end{aligned}$$

for generic $z\in Y$. But for generic z, $(z{\rightharpoondown }{\diamond })\circ (x{\rightharpoondown }{\diamond })=(x'{\rightharpoondown }{\diamond })\circ (z'{\rightharpoondown }{\diamond })$ where $(x',z')=B^{-1}(z,x)$ and in particular $z'$ is the image of z by the generic bijection $B^{-1}_2({\diamond },x)$. Thus for generic z, $z'{\rightharpoondown }D\in {{{\mathscr {D}}}_{{\mathcal {Q}}}}$ and then $F_{{{{\mathscr {D}}}_{\Bbbk }}}(z{\rightharpoondown }(x{\rightharpoondown }D))=F_{{{{\mathscr {D}}}_{\Bbbk }}}(x'{\rightharpoondown }(z'{\rightharpoondown }D))=F_{{{{\mathscr {D}}}_{\Bbbk }}}(z'{\rightharpoondown }D)=\widehat{F}(D)$.

$\bullet$ Finally we prove

$$\begin{aligned} \widehat{F}(q\rhd D)=\widehat{F}(D), \quad \forall (q,D)\in {{{\mathsf {Q} }}}\times {{{\mathscr {D}}}_{\Bbbk }}, \end{aligned}$$

which is equivalent to Claim (5). By (44), we can replace q with $x{\rightharpoondown }q$ and D with $x{\rightharpoondown }D$ because $x{\rightharpoondown }(q\rhd D)=(x{\rightharpoondown }q)\rhd (x{\rightharpoondown }D)$. Doing this for generic x we are left with the case where $D,q\rhd D\in {{{\mathscr {D}}}_{{\mathcal {Q}}}}$. Then, if $D:x\rightarrow w$, the result follows by applying Claim (2) for $D_1=D$, $D_2=q\rhd D$ and

This completes the proof of the proposition. $\square$

Appendix D: Proof of colored braid relation

In this appendix we prove Lemma 6.4. Let ${\bar{R}}$ be the matrix chosen when defining $c_{{\chi _1},{\chi _2}}$. Since $\mathscr {R}$ satisfies the YB equation then ${\bar{R}}$ satisfies a holonomy YB equation up to a scalar. Again, as all determinants are 1 then the holonomy YB equation is true up to a $r^3$-root of unity. The following lemma shows that the this equation is actually true up to a $r^2$-root of unity (and thus implies Lemma 6.4).

Lemma D.1

The holonomy YB equation is true up to a ${r^2}$-root of unity.

Proof

For $n\in \mathbb {N}$, let $\Theta _n$ be the set of complex n^th roots of unity which acts by multiplication on complex vector spaces. If f is a bijection between two n-dimension $\mathbb {C}$-vector spaces equipped with volume forms then, up to $\Theta _n$, there is a unique normalization $\lambda .f$ which send one volume form to the other. The R-matrix is such a renormalization where the volume forms are induced from the maps $\phi _\chi$. Hence, if two representations $V,V'$ equipped with volume forms

have the same character then they are isomorphic by an unique volume preserving isomorphism up to $\Theta _{r}$. Furthermore, as the tensor product of volume preserving isomorphisms is a volume preserving isomorphism, the R-matrices constructed from V and those constructed from $V'$ correspond through these isomorphisms.

$\bullet$ Consider now the subset $A_3$ of $Y_{\ell }^3$ formed by triplet (x, y, z) where the set YB equation is defined

$$\begin{aligned} \begin{array}{rl} ({\text {Id}}_{V_x}\otimes c_{y,z})^{-1}\circ (c_{{\diamond },{\diamond }}\otimes {\text {Id}}_{\diamond })^{-1} \circ ({\text {Id}}_{\diamond }\otimes c_{{\diamond },{\diamond }})^{-1}\circ &{}(c_{{\diamond },{\diamond }}\otimes {\text {Id}}_{\diamond })\\ \circ ({\text {Id}}_{\diamond }\otimes c_{{\diamond },{\diamond }})\circ (c_{x,y}\otimes {\text {Id}}_{V_z}) =&{}\lambda (x,y,z){\text {Id}}_{V_x\otimes V_y\otimes V_z} \end{array} \end{aligned}$$

(45)

where the ${\diamond }$ objects are completed with the biquandle structure B and where $\lambda :A_3\rightarrow \mathbb {C}$ is a function. The set $A_3$ is a 9 dimensional complex variety obtain from $\mathbb {C} ^9$ by removing complex hyper-surfaces and taking finite coverings. In particular, it is path connected. Moreover, the R-matrix has determinant 1, so we have that $\lambda ^{r^3}=1$ and $\lambda \in \Theta _{r^3}$ has discrete values. Remark that the previous paragraph implies that $\lambda$ induces a partial map $\overline{\lambda }:A_3\rightarrow \Theta _{r^3}/\Theta _{r^2}$ which is independent of the representation $(V_\chi )_{\chi \in Y_\ell }$.

$\bullet$ We now construct a bundle of r-dimensional $U_\xi$-module ${{\mathcal {V}}}\rightarrow P'$ where $P'\rightarrow Y_{\ell }$ is onto and is a local homeomorphism.

Let $P=\{(y,k)\in Y_{\ell }\times \mathbb {C} ^*: k^r=(-1)^{r-1}y(K^r)\}$ be the r fold cover of $Y_{\ell }$. If an element $(g,\omega )\in Y_{\ell }$ satisfies $g(F^r)=0$, then ${\text {Cb}}_r{\left( \omega \right) }=-(-1)^\ell g(K^r+K^{-r})$ and there exists $k\in \mathbb {C}$ such that $k^r=(-1)^{r-1}g(K^r)$ and $\omega =(-1)^{\ell -1}{\left( k+k^{-1}\right) }$. We denote by $P^0$ the set of such $(g,\omega ,k)$. Remark that if $(g,\omega ,k)\in P^0$, then $k\ne \pm \xi ^i$ for $i=1\cdots r-1$ else one would have ${\text {Cb}}_r{\left( \omega \right) }=\pm 2$ and $\omega =2(-1)^{\ell -1}$.

Let $P'=P{\setminus }\{(g,\omega ,k):g(F^r)=0\}\cup P^0$ and let $\varepsilon :P'\rightarrow \mathbb {C}$ be defined on $p=(g,\omega ,k)$ by

$$\begin{aligned} \varepsilon (p)=\left\{ \begin{array}{l} \dfrac{\omega +(-1)^\ell {\left( k+k^{-1}\right) }}{{\left\{ 1\right\} }^2g(F^r)}\text { if }p\notin P^0\\ \dfrac{(-1)^{r-1}{\left\{ 1\right\} }^{2r}g(E^r)}{\prod _{i=1}^{r-1}{\left\{ i\right\} }(k\xi ^{-i}-k^{-1}\xi ^i)} \text { if }p\in P^0 \end{array}\right. . \end{aligned}$$

One easily checks that $\varepsilon$ is continuous on $P'$. Let $Z_\ell \subset {\mathcal {C}}^0(P',\mathbb {C} )$ be the polynomial algebra $Z_\ell =\mathbb {C} [\varepsilon ,k^{\pm 1},\varphi ]$ where $\varphi :(g,\omega ,k)\mapsto g(F^r)$. Following [22, Section VI.5] we define a bundle of $U_\xi$-module ${{\mathcal {V}}}=P'\times \mathbb {C} ^r$ by giving an algebra map $\rho _{{\mathcal {V}}}:U_\xi \rightarrow \mathsf {Mat} _r(Z_\ell )$ as follows:

$$\begin{aligned} \rho _{{\mathcal {V}}}(K)_{i,i}= & {} k\xi ^{r+1-2i}, \quad \rho _{{\mathcal {V}}}(F)_{i+1,i}=1,\quad \rho _{{\mathcal {V}}}(F)_{1,r}=\varphi \\ \rho _{{\mathcal {V}}}(E)_{i,i+1}= & {} \varphi \varepsilon -(-1)^\ell \dfrac{(k\xi ^{-i}-k^{-1}\xi ^i){\left\{ i\right\} }}{{\left\{ 1\right\} }^2},\quad \rho _{{\mathcal {V}}}(E)_{r,1}=\varepsilon \end{aligned}$$

and other coefficients are 0.

The obvious map $\pi :P'\rightarrow Y_{\ell }$ is surjective and it is locally an homeomorphism. For any $(p_1,p_2,p_3,p_4)\in (P')^4$, if $B(\pi (p_1),\pi (p_2))=(\pi (p_4),\pi (p_3))$ then the R-matrix is up to a scalar the unique solution in

$$\begin{aligned} {\text {Hom}}_\mathbb {C} {\left( {{\mathcal {V}}}_{p_1}\otimes {{\mathcal {V}}}_{p_2},{{\mathcal {V}}}_{p_4}\otimes {{\mathcal {V}}}_{p_3}\right) } \end{aligned}$$

of the six linear equations

$$\begin{aligned} \left\{ \begin{array}{l} \rho _{{{\mathcal {V}}}_{p_3}\otimes {{\mathcal {V}}}_{p_4}}{\left( \mathscr {R}(X\otimes 1)\right) }R=R\rho _{{{\mathcal {V}}}_{p_1}\otimes {{\mathcal {V}}}_{p_2}}{\left( 1\otimes X\right) }\\ \rho _{{{\mathcal {V}}}_{p_3}\otimes {{\mathcal {V}}}_{p_4}}{\left( \mathscr {R}(1\otimes X)\right) }R=R\rho _{{{\mathcal {V}}}_{p_1}\otimes {{\mathcal {V}}}_{p_2}}{\left( X\otimes 1\right) } \end{array},\,X\in {\left\{ K,E,F\right\} }.\right. \end{aligned}$$

Hence the R-matrix is continuous in the following way: there exists open neighbors $N_i$ of $p_i$ such that B is a bijection $\pi (N_1)\times \pi (N_2)\rightarrow \pi (N_4)\times \pi (N_3)$ which lift to a continuous isomorphism of bundle ${{\mathcal {V}}}_{|N_1}\otimes {{\mathcal {V}}}_{|N_2} \rightarrow {{\mathcal {V}}}_{|N_4}\otimes {{\mathcal {V}}}_{|N_3}$. This implies that the map $\overline{\lambda }$ defined above is continuous thus locally constant.

$\bullet$ Finally as $\overline{\lambda }$ is locally constant and $A_3$ is connected, $\overline{\lambda }$ is constant and this constant is $\overline{\lambda }(\chi _0,\chi _0,\chi _0)=1$ where $\chi _0$ is the character of the Steinberg module.

This finishes the proof of Lemma D.1. $\square$

Appendix E: Proof of Lemma 6.7

We need the following lemma which appears in [10].

Lemma D.2

(Cutting Coupon Lemma) Let H be a pivotal Hopf algebra over a field $\Bbbk$. Let $\mathscr {C}$ be the pivotal $\Bbbk$-category of finite dimensional H-modules. If $f:V_1\otimes V_2 \rightarrow V_3\otimes V_4$ is a morphism in $\mathscr {C}$ such that

$$\begin{aligned} f (hx \otimes y) =(h \otimes 1) f (x\otimes y) \;\; \text { or } \;\; f (x \otimes hy) =(1 \otimes h) f (x\otimes y) \end{aligned}$$

for all $h\in H$ and $x\otimes y\in V_1\otimes V_2$ then there exists $a_i:V_1\rightarrow V_3$ and $b_i:V_2\rightarrow V_4$ such that $f=\sum _i a_i \otimes b_i$.

Let V and W be generic simple $U_\xi$-modules. Recall the morphism R given in Eq. (35). Let $V'$ be a simple $U_\xi$-modules such that there exists an $\mathscr {C}$-isomorphism $\beta : V'\rightarrow V^*$. Consider the following morphisms

$$\begin{aligned} f_1:=\tau _{12}R_{12}\tau _{23}R_{23}=\tau _{12}\tau _{23}R_{13}R_{23}: V'\otimes V\otimes W\rightarrow W\otimes U'_1\otimes U_1 \end{aligned}$$

and

$$\begin{aligned} f_2:=\tau _{23}R_{23}\tau _{12}R_{12}=\tau _{23}\tau _{12}R_{13}R_{12}: W\otimes V'\otimes V\rightarrow U'_2\otimes U_2 \otimes W \end{aligned}$$

where the colors $U_i$ and $U'_i$ are determined by the morphisms. There exists isomorphisms $\alpha _i:U'_i \rightarrow U_i^*$ for $i=1,2$. Consider the following morphisms:

$$\begin{aligned} g_1=({\text {Id}}\otimes {\mathop {{\text {ev}}}\limits ^{\longleftarrow }}_{U_1})({\text {Id}}\otimes \alpha _1 \otimes {\text {Id}})f_1: V'\otimes V\otimes W\rightarrow W \end{aligned}$$

and

$$\begin{aligned} g_2=({\mathop {{\text {ev}}}\limits ^{\longleftarrow }}_{U_2} \otimes {\text {Id}})( \alpha _2\otimes {\text {Id}}\otimes {\text {Id}})f_2:W\otimes V'\otimes V\rightarrow W. \end{aligned}$$

Lemma D.3

There exist a scalar $\lambda _1, \lambda _2 \in \mathbb {C}$ such that

$$\begin{aligned} g_1=\lambda _1 ({\mathop {{\text {ev}}}\limits ^{\longleftarrow }}_V \otimes {\text {Id}})( \beta \otimes {\text {Id}}\otimes {\text {Id}}) \;\; \text { and } \;\; g_2=\lambda _2 ({\text {Id}}\otimes {\mathop {{\text {ev}}}\limits ^{\longleftarrow }}_{V})( {\text {Id}}\otimes \beta \otimes {\text {Id}}). \end{aligned}$$

Proof

We prove the theorem for $g_1$ the case for $g_2$ is similar. To do this we need to derive several equations. Let $v\in V'\otimes V\otimes W$ and $x,y\in U_\xi$. Then

$$\begin{aligned} f_1 \big ((\Delta (x)\otimes y) . v\big )&= \tau _{12}\tau _{23}R_{13}R_{23} \big ((\Delta (x)\otimes y) . v\big ) \nonumber \\&= \tau _{12}\tau _{23}\mathscr {R}_{13}\mathscr {R}_{23}\big (\Delta (x)\otimes y\big ) R_{13}R_{23} (v) \nonumber \\&= \tau _{12}\tau _{23} \Delta _1 \big ( \mathscr {R}(x\otimes y)\big ) R_{13}R_{23} (v) \nonumber \\&= \Delta _2 \big (\tau \big ( \mathscr {R}(x\otimes y)\big )\big ). f_1 ( v) \end{aligned}$$

(46)

where the second equality comes Eq. (36) and the third equality follows from Eq.  (30). Consider the morphism $\alpha '= {\mathop {{\text {ev}}}\limits ^{\longleftarrow }}_{U_1}(\alpha _1 \otimes {\text {Id}})$. We have

$$\begin{aligned} g_1\big ((\Delta (x)\otimes y) . v\big )&=({\text {Id}}\otimes \alpha ') \Big ( \Delta _2 \big (\tau \big ( \mathscr {R}(x\otimes y)\big )\big ) . f_1 (v) \Big )\\&= \epsilon _2 \big ( \tau \big ( \mathscr {R}(x\otimes y)\big ) \big ) .\big (({\text {Id}}\otimes \alpha ') f_1 \big )(v)\\&= \epsilon _1 \big ( \mathscr {R}(x\otimes y)\big ). g_1(v)\\&= \epsilon (x)y. g_1(v) \end{aligned}$$

where the first equality comes from Eq. (46), the second since $\alpha '$ is an invariant morphism and the fourth from Eq. (31).

Thus, we have proved that $\mathscr {C}$-morphism $g:V'\otimes V\otimes W\rightarrow \mathbb {C} \otimes W$ satisfies the hypothesis of Lemma D.2 and so there exists $\mathscr {C}$-morphisms $a_i:V'\otimes V\rightarrow \mathbb {C}$ and $b_i :W\rightarrow W$ so that $g=\sum _i a_i\otimes b_i$. But

$$\begin{aligned} {\text {Hom}}_{\mathscr {C}}(V'\otimes V, \mathbb {C} )\cong {\text {End}}_{\mathscr {C}}(V)\cong \mathbb {C} \end{aligned}$$

and ${\text {End}}_{\mathscr {C}}(W)\cong \mathbb {C}$ since V and W are simple. Thus, for each i,

$a_i$ and $b_i$ are proportional to ${\mathop {{\text {ev}}}\limits ^{\longleftarrow }}_V(\beta \otimes {\text {Id}}_V)$ and ${\text {Id}}_W$, respectively. $\square$

Lemma D.3 is an algebraic version of Lemma 6.7.

Appendix F: Proof semi cyclic module give a representation

Here we discuss why the biquandle give in Sect. 4.6 has a representation coming from the semi cyclic modules. Let $U_\xi {\mathfrak {sl}(2)}$ be quantum quantum ${\mathfrak {sl}(2)}$ where $\xi$ is the $2r^{th}$-root of unity. Let $\mathscr {C}$ be the category of $U_\xi {\mathfrak {sl}(2)}$ weight modules has defined in Sect. 6.1.3 of [14]. Recall the set X given in Sect. 4.6. For $x\in X$ there exists a semi cyclic module $V_x$ in $\mathscr {C}$, see Theorem 6.6 of [14]. Moreover, Eq.  (23) of [14] defines a Yang-Baxter model for $\{V_x\}_{x\in X}$: for all $x,y\in X$

$$\begin{aligned} c_{x,y}:V_x\otimes V_y\rightarrow B_1(V_x,V_y)\otimes B_2(V_x,V_y) \end{aligned}$$

(47)

which satisfies the colored braid relations. Remark that the semi cyclic modules are cyclic modules as in Sect. 6.1. The braiding given in Eq. (47) intertwines the actions of the generators K, E and F of $U_\xi {\mathfrak {sl}(2)}$ and their image by $\mathscr {R}$. Thus, up to a scalar, this braiding is equal to the braiding given in Lemma 6.4 when the modules are semi cyclic. Thus, the braiding in Eq. (47) is sideways invertible, up to a root of unity.

In Sects. 6.4 and 6.5 we showed the Yang-Baxter model associated to the cyclic modules was sideways invertible and induced a twist. Since the Yang-Baxter model of the cyclic modules was only defined up to a root of unity we were only able to show the sideways invertibility and twist held up to a root of unity. However, the proof works in the context of semi cyclic modules and in this situation we have equalities independent of any root of unity. We explain this now.

As mentioned above the braiding is sideways invertible, up to a root of unity. In other words, Yang-Baxter model satisfies the two negative Reidemeister moves $RII_{-+}$ and $RII_{+-}$ up to a scalar. Then the proof of Theorem 39 holds in for the semi cyclic. The only difference is that our braiding is defined on the nose and not just up to a scalar. Thus, we get that $\{V_x, c_{x,y}\}$ satisfies the $RII_{-+}$ and $RII_{+-}$. Also, the proof of Lemma 6.10 show that $\{V_x, c_{x,y}\}$ induces a twist.