Let $u^i\_j$, $i,j = 1, . . . N$, and det$\_q^{-1}$ be the standard generators of the quantum group $U\_q(N,C)$, and define the matrices $U$ and $U^{\ast}$ by setting $U\_{ij} := u^i\_j$ and $U^{\ast}\_{ij}:=(u^j\_i)^{\ast}$. It is "well known" that $U^{\ast}U=UU^{\ast}=1$. How does one prove this?


Moreover, how does this imply that $u^i\_j(u^r\_s)^{\ast} = (u^r\_s)^{\ast}u^i\_j$, for $r\neq i,s\neq j$?



Prove that $U^*U=UU^*=1$ for $U_q(N,C)$
I assume that $U^\ast\_{ij}=S(u\_j^i)$, where $S$ is the antipode. (See for example the book by Klimyk & Schmüdgen, Section 9.2.4.) If so, then $UU^\ast=1=U^\ast U$ follows from the antipode axiom


$\mu\circ(\mathrm{Id}\otimes S)\circ \Delta =\varepsilon =\mu\circ( S\otimes\mathrm{Id})\circ \Delta$


after applying both sides to an arbitrary generator $u\_j^i$.


To prove $u\_j^iS(u\_s^r)=S(u\_s^r)u\_j^i$ for $r\neq i, s\neq j$, one can multiply both sides by the quantum determinant and use that it is central to get the equivalent statement $u\_j^iM\_s^r=M\_s^r u\_j^i$, where $M\_s^r$ are quantum minors of size $(N-1)\times(N-1)$. 
Now one can observe that the subalgebra generated by $u\_b^a$ with $a\neq r, b\neq s$ is isomorphic to $M\_q(N-1,C)$, the quantized algebra of regular functions on all $(N-1)\times(N-1)$ complex matrices. $u\_j^i$ and $M\_s^r$ belong to this subalgebra and moreover $M\_s^r$ is the quantum determinant in $M\_q(N-1,C)$, hence it commutes with $u\_j^i$. 


