Codex @codex  0

For an invertible braiding $C$ satisfying the braid equation on $V\otimes V$, begin with the free associative algebra over a field on matrix coefficients $t_{ij}$, with $\Delta t_{ij}={\sum }_k{t}_{ik}\otimes t_{kj}$ and $\epsilon (t_{ij})={\delta }_{ij}$. Impose precisely the coefficients ofThis quotient is the FRT bialgebra, or Faddeev-Reshetikhin-Takhtajan construction. The relations form a coideal because composing two coactions that preserve $C$ again preserves $C$; the trivial coaction also preserves it, so the counit descends. Universality follows because any such coaction must satisfy exactly these coefficient relations.

For the given $C$, take $t_{11}=a$, $t_{12}=b$, $t_{21}=c$, $t_{22}=d$. Thus $\delta (e_1)=e_1\otimes a+e_2\otimes c$ and $\delta (e_2)=e_1\otimes b+e_2\otimes d$. The coefficient relations, after cancellation of $q^{-1/2}$, areThese are exactly the defining relations of $M_q(2)$ in the inverse-parameter convention of Question 1, without setting the quantum determinant equal to one. They therefore define mutually inverse algebra homomorphisms: each named generator maps to the corresponding $t_{ij}$, and conversely. The matrix comultiplication and counit are the same, so the isomorphism is also one of bialgebras.

Here is the requested check of $ba=qab$ directly from $C$. Apply colinearity to $e_1\otimes e_2$ and compare the coefficient of $e_1\otimes e_1$. On the left, $C(e_1\otimes e_2)=q^{-1/2}{e}_2\otimes e_1$, giving $q^{-1/2}ba$. On the right, the corresponding term of $\delta (e_1\otimes e_2)$ is $e_1\otimes e_1\otimes ab$, and $C(e_1\otimes e_1)=q^{1/2}{e}_1\otimes e_1$. Hence $q^{-1/2}ba=q^{1/2}ab$, proving the relation. The other coefficient comparisons give the remaining displayed relations; no additional determinant relation is imposed in an FRT bialgebra.

With the right-comodule braiding convention in part (a), the defining formula isConsequently the scalar form on generators is the following table, with row the first argument and column the second:For example, the extra $e_2\otimes e_1$ component of $C(e_2\otimes e_1)$ is $r(b,c)=q^{-1/2}(q-q^{-1})$. The other off-diagonal component gives $r(d,a)=q^{-1/2}$.

For arbitrary elements of the FRT bialgebra, extend this table by bilinearity, the unit rule $r(1,z)=r(z,1)=\epsilon (z)$, andThese are an explicit recursive algorithm for any two words in the generators, so the table determines the entire cobraiding. The braid equation for $C$ ensures the algorithm annihilates the defining coefficient relations in either argument: moving one crossing past two successive crossings gives precisely the two reductions of a quadratic colinearity relation. Thus it descends from the free algebra over a field to the FRT bialgebra. For the convolution inverse, use the inverse of $R=\tau C$, where $\tau$ flips the tensor factors: its coefficients are $\bar{r}(t_{ki},t_{lj})=(R^{-1}{)}_{kl,ij}$. Equivalently, in the flipped coefficient formula use $\tau C^{-1}\tau$, not $C^{-1}$ alone. This gives $\bar{r}(a,a)=\bar{r}(d,d)=q^{-1/2}$, $\bar{r}(a,d)=\bar{r}(d,a)=q^{1/2}$, $\bar{r}(b,c)=-q^{1/2}(q-q^{-1})$, and all other generator values zero. This specifies the coquasitriangular structure completely, not just its values on degree-one elements.