Yang–Baxter Operators and Quantum Braid Representations — Examples

The cubic equation holds for $R=0$. Every factor in both composites $(R\otimes 1_X)(1_X\otimes R)(R\otimes 1_X)$ and $(1_X\otimes R)(R\otimes 1_X)(1_X\otimes R)$ is a tensor product containing the zero morphism $R$; a composite with a zero factor is zero, so both sides are the zero endomorphism of $X\otimes X\otimes X$ and the cubic relation holds.

The morphism $R$ is not invertible. By [F1] the object $X\otimes X$ is nonzero and $0$ has no inverse as an endomorphism of it. The local operator $R_i=1^{\otimes (i-1)}\otimes R\otimes 1^{\otimes (n-i-1)}$ is the tensor product of $R=0$ with identities, hence is the zero endomorphism of the nonzero object $X^{\otimes n}$, so it too has no inverse.

No braid-group representation arises. The assignment ${\sigma }_i\mapsto R_i$ does not even land in $\mathrm{Aut}(X^{\otimes n})$, because $R_i=0$ is not invertible by step 1.2. Moreover there is no homomorphism $\rho \text{\hspace{0.17em}⁣}:B_n\to \mathrm{End}(X^{\otimes n})$ with $\rho ({\sigma }_i)=R_i$: from ${\sigma }_i{\sigma }_i^{-1}=1$ one would get $R_i\rho ({\sigma }_i^{-1})=\mathrm{id}$, contradicting $R_i\rho ({\sigma }_i^{-1})=0$. So the non-invertible solution $R=0$ of the cubic equation gives no representation of the braid group.

Conclusion. The zero solution satisfies the Yang–Baxter equation but is not a Yang–Baxter operator in the sense of [L1], and it produces no braid-group action; the claim that the cubic equation alone suffices is therefore refuted, and invertibility is a genuine part of the definition.

Invertibility. Define $R^{-1}$ on the graded basis by $R^{-1}(e_h\otimes e_g):=\chi (g,h{)}^{-1}{e}_g\otimes e_h$ and extend linearly. Then $R(R^{-1}(e_h\otimes e_g))=\chi (g,h{)}^{-1}R(e_g\otimes e_h)=e_h\otimes e_g$ and $R^{-1}(R(e_g\otimes e_h))=\chi (g,h)R^{-1}(e_h\otimes e_g)=e_g\otimes e_h$, so $R$ and $R^{-1}$ are mutually inverse linear bijections.

The cubic relation. Applying the left-hand composite $(R\otimes 1_X)(1_X\otimes R)(R\otimes 1_X)$ to $e_g\otimes e_h\otimes e_l$, from right to left, produces first $\chi (g,h)e_h\otimes e_g\otimes e_l$, then $\chi (g,l)e_h\otimes e_l\otimes e_g$, then $\chi (h,l)e_l\otimes e_h\otimes e_g$. Applying the right-hand composite $(1_X\otimes R)(R\otimes 1_X)(1_X\otimes R)$ produces first $\chi (h,l)e_g\otimes e_l\otimes e_h$, then $\chi (g,l)e_l\otimes e_g\otimes e_h$, then $\chi (g,h)e_l\otimes e_h\otimes e_g$. Both sides therefore act on $e_g\otimes e_h\otimes e_l$ by multiplication by $\chi (g,h)\chi (g,l)\chi (h,l)$, and these scalars are equal in the commutative group $k^{\times }$; since the pure tensors of graded basis vectors span $X^{\otimes 3}$, the cubic equation holds.

The braid-group actions. By steps 1.1 and 1.2 the map $R$ is an invertible solution of the cubic equation, hence a Yang–Baxter operator on $X$ in the sense of [L1], and [L2] gives the homomorphisms ${\rho }_n$ with ${\rho }_n({\sigma }_i)$ acting on the graded basis by exchanging the $i$-th and $(i+1)$-st entries with the scalar $\chi$ attached to the two exchanged degrees.

Involutivity criterion. On a pure tensor, $R^2(e_g\otimes e_h)=R(\chi (g,h)e_h\otimes e_g)=\chi (g,h)\chi (h,g)e_g\otimes e_h$, so $R^2=1_{X\otimes X}$ if and only if $\chi (g,h)\chi (h,g)=1$ for every pair $(g,h)$ with $k{e}_g,k{e}_h\mathrm{\ne }0$. If this holds, then each local operator satisfies $R_i^2=1$ (it is a tensor product of identities with $R^2$) and the $R_i$ satisfy the braid relations, so the assignment $s_i\mapsto R_i$ respects the Coxeter relators of $S_n$ and [F1] gives homomorphisms ${\psi }_n\text{\hspace{0.17em}⁣}:S_n\to \mathrm{Aut}(X^{\otimes n})$ with ${\rho }_n={\psi }_n\circ {\pi }_n$; the actions factor through the symmetric groups. If instead $\chi (h,h{)}^2\mathrm{\ne }1$ for some $h$ with $k{e}_h\mathrm{\ne }0$, then $R^2(e_h\otimes e_h)=\chi (h,h{)}^2{e}_h\otimes e_h\mathrm{\ne }{e}_h\otimes e_h$, so $R^2\mathrm{\ne }{1}_{X\otimes X}$ and $R$ is not involutive.

Conclusion. The diagonal map $R(e_g\otimes e_h)=\chi (g,h)e_h\otimes e_g$ is always an invertible Yang–Baxter operator on the free graded vector space, its square is the diagonal map with coefficients $\chi (g,h)\chi (h,g)$, and it is involutive exactly when those coefficients are $1$. All computations are on a spanning set of pure tensors and use no choice principle.

Invertibility and the cubic equation. Multiplication by $2$ on the one-dimensional space $X\otimes X\cong \mathbb{Q}$ is invertible with inverse multiplication by $\frac{1}{2}$. Both the left- and the right-hand side of the cubic equation are, on the single basis vector $e\otimes e\otimes e$ of the one-dimensional space $X^{\otimes 3}$, multiplication by $2\cdot 2\cdot 2=8$; hence they agree, and $R$ is a Yang–Baxter operator on $X$ in the sense of [L1].

The one-dimensional braid character. Since $X^{\otimes n}$ is one-dimensional, each local operator $R_i$ is multiplication by $2$, so [L1] gives ${\rho }_n({\sigma }_i)=2$ for every $i$, and additivity of the exponent sum [F1] on Artin words gives ${\rho }_n(\beta )=2^{w(\beta )}$ for every $\beta \in B_n$.

Non-involutivity and failure of factorization. From $R^2(e\otimes e)=4e\otimes e$ we get $R^2=4\cdot 1_{X\otimes X}\mathrm{\ne }{1}_{X\otimes X}$. By [L2] the two-strand action ${\rho }_2$ does not factor through ${\pi }_2$. For any $n\ge 2$, if ${\rho }_n$ factored through ${\pi }_n$ then [L2] would give $R_i^2=1_{X^{\otimes n}}$; but $R_i^2$ is multiplication by $4$ on the one-dimensional space $X^{\otimes n}$, so $R_i^2\mathrm{\ne }1$. Hence no ${\rho }_n$ with $n\ge 2$ factors through $S_n$; for $n=1$ the braid and symmetric groups are trivial and the action factors through them, and the character of step 2.1 is a genuine non-involutive braid character. Finally, since $w({\iota }_n(\beta ){\sigma }_n^{\pm 1})=w(\beta )\pm 1$, the values $2^w$ on the two stabilizations of a braid differ by a factor $4$, so the character separates them.

Conclusion. The one-dimensional operator $R=2\cdot \mathrm{id}$ is an invertible, non-involutive solution of the Yang–Baxter equation, and its braid actions are the one-dimensional characters $2^w$ that, for $n\ge 2$, do not factor through the symmetric groups. All computations are finite and use no choice principle.

The flip is an involutive Yang–Baxter operator. On the basis, $R^2(e_i\otimes e_j)=R(e_j\otimes e_i)=e_i\otimes e_j$, so $R^2=1_{X\otimes X}$. For the cubic relation, the left-hand composite applied to $e_i\otimes e_j\otimes e_l$ reverses the order of the three factors: $R\otimes 1$ exchanges the first two, then $1\otimes R$ exchanges the last two, then $R\otimes 1$ the first two, giving $e_l\otimes e_j\otimes e_i$; the right-hand composite produces the same by the mirror computation. Since the pure tensors span $X^{\otimes 3}$, the cubic equation holds and $R$ is a Yang–Baxter operator on $X$.

The action on pure tensors. For $m\ge 2$, by [L1] the local operator at position $j$ is $1^{\otimes (j-1)}\otimes R\otimes 1^{\otimes (m-j-1)}$, which on the basis vector $e_{i_1}\otimes \cdots \otimes e_{i_m}$ exchanges the entries in positions $j$ and $j+1$; thus each ${\rho }_m({\sigma }_j)$ is the corresponding place permutation.

Factorization and identification of the representation. By [L2] and $R^2=1$ the action ${\rho }_m$ factors as ${\psi }_m\circ {\pi }_m$ with ${\psi }_m(s_j)={\rho }_m({\sigma }_j)$; by step 2.1 the value ${\psi }_m(s_j)$ is the place permutation exchanging positions $j$ and $j+1$. Since the $s_j$ generate $S_m$, ${\psi }_m$ is the place-permutation representation of $S_m$ on $X^{\otimes m}$, and ${\rho }_m$ is that representation composed with ${\pi }_m$.

The two-strand case. For $m=2$ and $n\ge 2$ the operator $R$ swaps $e_1\otimes e_2$ with $e_2\otimes e_1$ and fixes $e_i\otimes e_i$ for $i=1,2$; this is the place-permutation representation of $S_2$, in agreement with step 3.1.

Conclusion and small strand counts. For $m=0,1$, the braid and symmetric groups are trivial and their actions send the sole element to the identity, the place permutation on $X^{\otimes m}$. If $n=0$ and $m\ge 1$, the tensor power is zero and its unique endomorphism is its identity. The flip operator is an involutive Yang–Baxter operator, and its braid actions are exactly the place-permutation representations of the symmetric groups, pulled back along the canonical surjections $B_m\to S_m$. All computations are finite and linear and use no choice principle.

The braiding already gives the actions. In ${\mathbf{V}\mathbf{e}\mathbf{c}\mathbf{t}}_k$ with the transposition braiding the canonical actions ${\rho }_n$ of [F1] exist for every $n$, even though the category is not rigid. Hence whatever is missing from a link trace is not the braid action.

The closure of the one-braid has no value in the non-rigid model. For the object $X$ coloring the band, the left trace of $j_X:X\to X^{\vee \vee }$ begins with ${\mathrm{coev}}_X$, applies $j_X\otimes 1_{X^{\vee }}$, and ends with ${\mathrm{ev}}_{X^{\vee }}:X^{\vee \vee }\otimes X^{\vee }\to \mathbf{1}$ [F2, F4]. If $X=k[x]$ had a left dual $Y$, write $\mathrm{coev}(1)={\sum }_{i=1}^N{v}_i\otimes y_i$, a finite sum by [F3]. The zig-zag identity would give $v={\sum }_i{v}_i\mathrm{ev}(y_i\otimes v)$ for every $v\in X$, putting all of $X$ in the finite-dimensional span of the $v_i$. The monomials $1,x,x^2,\dots$ are linearly independent, so this is impossible. The right-dual version is the same mirror argument. Therefore the closure of $e$ has no categorical trace value for $X=k[x]$, and a braiding alone does not evaluate it.

A rigid model with two different ribbon evaluations. Take finite-dimensional $\mathbb{Z}\mathrm{/}2$-graded vector spaces over a field of characteristic $\mathrm{\ne }2$, with even linear maps and $c(v\otimes w)=(-1{)}^{\mathrm{\mid }v\mathrm{\mid }\mathrm{\mid }w\mathrm{\mid }}w\otimes v$. Graded duals, ordinary evaluation and basis coevaluation obey the zig-zags. The sign rule is natural, and both hexagons follow from $(-1{)}^{p(q+r)}=(-1{)}^{pq}(-1{)}^{pr}$; the double braiding is the identity. Both ${\theta }^0=1$ and ${\theta }^1(v)=(-1{)}^{\mathrm{\mid }v\mathrm{\mid }}v$ are natural monoidal automorphisms preserving duals, hence ribbon twists by [F4]. On the odd line $L=ke$ with odd dual basis $e^{\vee }$, the Drinfeld composite of [F2] sends $e$ to $-e^{\vee \vee }$: its sole crossing is $c_{L,L^{\vee }}(e\otimes e^{\vee })=-e^{\vee }\otimes e$. Thus $j_L^0=-1$ and $j_L^1=1$ under the usual double-dual identification; coevaluation $1\mapsto e\otimes e^{\vee }$ and evaluation $e^{\vee \vee }\otimes e^{\vee }\mapsto 1$ give left traces $-1$ and $1$. The same braiding and duals therefore give different ribbon unknot values.

Conclusion. The non-rigid witness lacks the dual pair needed for closure, and the rigid witness has two different ribbon evaluations for the same braiding and chosen duals. Hence it does not suffice to define a link trace, and the claim is refuted.

The model is a ribbon category. The usual graded duals of finite-dimensional spaces give rigidity. The sign braiding is a symmetric braiding, so it is in particular braided; the parity twist is a natural automorphism of the identity, and on homogeneous vectors ${\theta }_{V\otimes W}=(-1{)}^{\mathrm{\mid }v\mathrm{\mid }+\mathrm{\mid }w\mathrm{\mid }}$ equals $({\theta }_V\otimes {\theta }_W)c_{W,V}{c}_{V,W}$ because the double braiding is the identity; the dual of a homogeneous space has the same parity, so ${\theta }_{V^{\vee }}=({\theta }_V{)}^{\vee }$. Hence [L1] makes $\mathcal{C}$ a ribbon category.

The odd line and its twist eigenvalue. The odd one-dimensional space $X$ has $\mathrm{End}(X)=k$, so it is absolutely simple, and ${\theta }_X=-{\mathrm{id}}_X$, so $\lambda =-1\in k^{\times }$ because $\mathrm{char}k\mathrm{\ne }2$. Its categorical dimension is $d_X=t_1(e)=1$: the defining composite of [F1] is ${\mathrm{Tr}}_L(j_X)$ with $j_X=u_X{\theta }_X$, and on the odd line the Drinfeld morphism is $u_X=-{\mathrm{id}}_X$ while ${\theta }_X=-{\mathrm{id}}_X$, so $j_X={\mathrm{id}}_X$ and $d_X={\mathrm{Tr}}_L({\mathrm{id}}_X)=1$.

The two values differ. By [L2] applied to $\beta =e\in B_1$ and its positive stabilization ${\sigma }_1\in B_2$, $t_2({\sigma }_1)=\lambda t_1(e)=(-1)\cdot 1=-1$, while $t_1(e)=1$ by step 1.2. The two values differ.

The closures are the same oriented link. By [L3] the closures of $e$ and of its stabilization ${\sigma }_1$ are equivalent oriented links; both are the unknot, the closure of ${\sigma }_1$ being a one-component diagram with a single positive kink. If the unnormalized evaluation were an invariant of oriented unframed links, it would take equal values on the closures of $e$ and ${\sigma }_1$; step 2.1 shows that it does not.

Conclusion. The unnormalized ribbon trace takes the values $-1$ and $1$ on two braids whose closures are equivalent oriented links, so it is not invariant under positive stabilization and not an unframed link invariant; the writhe factor is necessary, as the companion example shows. All computations are finite; the only choice principle used is ${\mathrm{A}\mathrm{C}}_{\omega }$, consumed through the stabilization lemma [L2].

Verify the model and dimension. Even linear maps and graded duals give the rigid $k$-linear category, with the ordinary evaluation and basis coevaluation satisfying the zig-zags. The sign braiding is natural; its hexagons are $(-1{)}^{p(q+r)}=(-1{)}^{pq}(-1{)}^{pr}$ and the analogous identity in the first variable, and its square is the identity. Parity is natural, multiplicative on tensor products and unchanged on duals, so it is a ribbon twist by [F1]. The odd line has only scalar endomorphisms and twist $-1$. For a dual basis $e,e^{\vee }$, the Drinfeld composite gives $u_X(e)=-e^{\vee \vee }$, hence $j_X(e)=e^{\vee \vee }$; coevaluation $1\mapsto e\otimes e^{\vee }$ and evaluation $e^{\vee \vee }\otimes e^{\vee }\mapsto 1$ give $d_X=1$.

The writhe exponents. The trivial one-braid has exponent sum $w(e)=0$, and its positive stabilization ${\sigma }_1$ has $w({\sigma }_1)=1$ by [L2].

The unnormalized values and the kink. By [L1], $t_1(e)=1$ and $t_2({\sigma }_1)=\lambda t_1(e)=-1$. By [F2], the closures both represent the unknot, so the unnormalized evaluation changes by the factor $-1$ under the positive stabilization, the kink contribution modelled by the scalar twist $\lambda =-1$.

The normalized values agree. Substituting steps 2.1 and 1.2, $J_X(\widehat{e})={\lambda }^{-w(e)}{t}_1(e)=(-1{)}^0\cdot 1=1,J_X(\widehat{{\sigma }_1})={\lambda }^{-w({\sigma }_1)}{t}_2({\sigma }_1)=(-1{)}^{-1}\cdot (-1)=1.$ The two normalized values are equal, in accordance with [L3].

Conclusion. The writhe factor ${\lambda }^{-w}$ cancels exactly the kink contribution $\lambda$ contributed by the stabilization, so the normalized evaluation is the same on the closure of $e$ and on its positive stabilization, as the invariant theorem predicts; the common value is $1$, and the dimension normalization by $d_X=1$ is trivial here. All computations are finite; the Axiom of Choice is assumed through [L3] and supplies the countable-choice input to [L1], as recorded there.