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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 $r\neq i,s\neq j$?

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Is something missing after "for"? – darij grinberg Nov 6 '10 at 22:56
It's fixed now. – Dyke Acland Nov 7 '10 at 1:50
up vote 2 down vote accepted

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 $u_j^i$.

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Nice answer, but I think there's a slight error. I find in K & S $(u^i_j)^* = S(u^j_i)$, and not, $(u^i_j)^* = S(u^i_j)$. Reworking your proof then gives $u^i_jS(u^r_s) = S(u^r_s)u^i_j$, for $r \neq j,s \neq j$. The result you have stated is not true. No worries though, you still gave me the answer I was looking for. Do you think you could vary your proof to find a formula for $u^i_jS(u^j_s) - S(u^j_s)u^i_j$, which is certainly non-zero? – Dyke Acland Nov 11 '10 at 20:16
Glad you found it useful! Actually I think it should be $(U^\ast)_{ij}=(u_i^j)^\ast$ and not $(u_j^i)^\ast$ as you write, because the "hermitian conjugate" of $U$ should be the transpose matrix with all entries "conjugated" i.e. starred. This is also consistent with KS. If I'm not mistaken. If you would like to prove that something is nonzero, sometimes a good way is to find a natural representation where it is nonzero. However, I think one can prove that a factor q enters there, and this is discussed somewhere on mathoverflow. – Jonas Hartwig Nov 11 '10 at 23:13
Check this out:… – Jonas Hartwig Nov 11 '10 at 23:20

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