Quasi-Hopf, Weak Hopf, and Relative Tensor Products: Examples and Counterexamples

The monoid laws make multiplication associative and unital. On basis elements, $\mathrm{\Delta }(mn)=mn\otimes mn=(m\otimes m)(n\otimes n)$, $\mathrm{\Delta }(1)=1\otimes 1$, $\epsilon (mn)=1=\epsilon (m)\epsilon (n)$, and coassociativity and the counit equations hold. Thus $H$ is a bialgebra and, with $\mathrm{\Phi }=1$, a quasi-bialgebra.

Suppose $(S,\alpha ,\beta )$ is a quasi-antipode. With $\mathrm{\Phi }=1$, its normalization equations give $\beta \alpha =1=\alpha \beta$, so ${\alpha }^{-1}=\beta$. Define $S^{\mathrm{\prime }}(h)={\alpha }^{-1}S(h)\alpha$. The quasi-antipode equations imply $\sum S^{\mathrm{\prime }}(h_{(1)})h_{(2)}=\epsilon (h)1$ and $\sum h_{(1)}{S}^{\mathrm{\prime }}(h_{(2)})=\epsilon (h)1$; thus $S^{\mathrm{\prime }}$ is an ordinary Hopf antipode of the bialgebra in step 1.1.

The basis element $e$ is group-like: $\mathrm{\Delta }(e)=e\otimes e$ and $\epsilon (e)=1$. The antipode equations in [F3] would make $e$ invertible, since $S^{\mathrm{\prime }}(e)e=1=e{S}^{\mathrm{\prime }}(e)$. But an invertible idempotent is $1$ (multiply $e^2=e$ by $e^{-1}$), contradicting $e\mathrm{\ne }1$. Hence no quasi-antipode exists, refuting the statement.

Multiplication is associative and unital. The coproduct is coassociative, and $(\epsilon \otimes \mathrm{i}\mathrm{d})\mathrm{\Delta }(e_i)=e_i=(\mathrm{i}\mathrm{d}\otimes \epsilon )\mathrm{\Delta }(e_i)$. Also $\mathrm{\Delta }(e_i)\mathrm{\Delta }(e_j)={\delta }_{ij}{e}_i\otimes e_i=\mathrm{\Delta }(e_i{e}_j)$, so $\mathrm{\Delta }$ is multiplicative.

Put $E=\mathrm{\Delta }(1)={\sum }_i{e}_i\otimes e_i$. Both iterated coproducts of $E$ equal ${\sum }_i{e}_i\otimes e_i\otimes e_i$, and $(E\otimes 1)(1\otimes E)=(1\otimes E)(E\otimes 1)={\sum }_i{e}_i\otimes e_i\otimes e_i$ by orthogonality. For basis triples $e_i,e_j,e_k$, each weak-counit product equals $1$ exactly when $i=j=k$ and is $0$ otherwise; trilinearity proves both weak-counit identities. The counital-map formulas give ${\epsilon }_t(e_i)=e_i={\epsilon }_s(e_i)$. Finally, $S(e_i)=e_i$ gives $e_{i}S(e_i)={\epsilon }_t(e_i)$, $S(e_i)e_i={\epsilon }_s(e_i)$, and $S(e_i)e_{i}S(e_i)=S(e_i)$. Thus all weak Hopf axioms hold by linearity.

The counital formulas also give ${\epsilon }_s^{\mathrm{\prime }}(e_i)={\sum }_j{e}_j\epsilon (e_j{e}_i)=e_i$, hence $H_t=H_s=H$. Since $E$ has no off-diagonal terms whereas $1\otimes 1$ has them, $\mathrm{\Delta }(1)\mathrm{\ne }1\otimes 1$. Moreover $\epsilon (e_0{e}_1)=0$ but $\epsilon (e_0)\epsilon (e_1)=1$, so the structure is not an ordinary bialgebra.

Orthogonal idempotents summing to $1$ give direct sums $V=e_{0}V\oplus e_{1}V$ and $W=e_{0}W\oplus e_{1}W$ by the left module axioms. On each elementary summand $e_{i}V\otimes e_{j}W$, $E$ acts as multiplication by ${\delta }_{ij}$, so its image is $(e_{0}V\otimes e_{0}W)\oplus (e_{1}V\otimes e_{1}W)$. Also $v◃e_i=e_{i}v$. The balancing relation for $x=e_i$ kills $e_{i}V\otimes e_{j}W$ when $i\mathrm{\ne }j$, while for $i=j$ the left and right actions agree. For $v_i=e_{i}v$ and $w_j=e_{j}w$, the relation for $x=a_0{e}_0+a_1{e}_1$ is $(a_i-a_j)v_i\otimes w_j$. It vanishes when $i=j$; when $i\mathrm{\ne }j$, choosing $x=e_i$ generates the whole summand $e_{i}V\otimes e_{j}W$. Thus the relation subspace is exactly the sum of the two off-diagonal summands.

The quotient and image are both the direct sum of the diagonal summands by step 4.1, and the quotient map and its inverse are the identity on each such summand. On each diagonal summand $e_{j}V\otimes e_{j}W$, both the image and quotient actions of $e_i$ are multiplication by ${\delta }_{ij}$; hence the isomorphism is $H$-linear. If $f:V\to V^{\mathrm{\prime }}$ and $g:W\to W^{\mathrm{\prime }}$ are $H$-linear, then [F3] gives $f(e_{i}v)=e_{i}f(v)$ and $g(e_{i}w)=e_{i}g(w)$, so $f\otimes g$ preserves each diagonal summand and the isomorphism is natural in $V,W$.

The pointwise idempotents are orthogonal and sum to $1$. The formula for $\mathrm{\Delta }$ is multiplicative because $\mathrm{\Delta }(p_g)\mathrm{\Delta }(p_h)=0$ for $g\mathrm{\ne }h$ and equals $\mathrm{\Delta }(p_g)$ for $g=h$; it is unital since $\mathrm{\Delta }(p_0+p_1)=(p_0+p_1)\otimes (p_0+p_1)$. Both iterated coproducts send $p_g$ to ${\sum }_{a+b+c=g}{p}_a\otimes p_b\otimes p_c$, and the two counit equations hold on each $p_g$.

Each coefficient of $\mathrm{\Phi }$ is $1$ or $-1$ on an orthogonal idempotent, so ${\mathrm{\Phi }}^{-1}=\mathrm{\Phi }$. The algebra $H^{\otimes 3}$ is commutative and $\mathrm{\Delta }$ is coassociative, hence $\mathrm{\Phi }$ commutes with $(\mathrm{\Delta }\otimes \mathrm{i}\mathrm{d})\mathrm{\Delta }(h)$ and quasi-coassociativity follows. The normalization holds because $\omega (a,0,c)=1$. The pentagon, evaluated at $(a,b,c,d)$, is the scalar identity $\omega (b,c,d)\omega (a,b+c,d)\omega (a,b,c)=\omega (a+b,c,d)\omega (a,b,c+d)\mathrm{.}$ Writing all sums modulo $2$, both exponents are $bcd+abd+acd+abc$, so the identity holds for all sixteen quadruples.

For $g\in C_2$, $\mathrm{\Delta }(p_g)={\sum }_{x+y=g}{p}_x\otimes p_y$. In the first quasi-antipode equation, a product $p_x\alpha p_y$ vanishes unless $x=y$; this happens for both pairs when $g=0$ and for neither pair when $g=1$. The sum is therefore $\alpha$ for $g=0$ and $0$ for $g=1$, exactly $\epsilon (p_g)\alpha$. The second equation has the same two cases with $\beta =1$, and yields $\epsilon (p_g)\beta$.

In $\mathrm{\Phi }$ and ${\mathrm{\Phi }}^{-1}$, a product of three basis idempotents is nonzero only when $a=b=c$. For $a=0$ its coefficient times ${\alpha }_0$ is $1$; for $a=1$ it is $(-1)(-1)=1$. Hence both normalization sums are $p_0+p_1=1$. The map $S$ is the identity and is bijective, so all quasi-Hopf axioms hold.

The one-dimensional module $V_1$ on which $p_1$ acts as $1$ and $p_0$ as $0$ has reassociator scalar $\omega (1,1,1)=-1$ by [F2]. Since that coefficient of $\mathrm{\Phi }$ is not the coefficient of $1^{\otimes 3}$, $\mathrm{\Phi }\mathrm{\ne }1$. The calculation establishes only the stated associator and makes no claim about twist equivalence.