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As is well known, the set

$\{a^ib^jc^k | i,j,k \in \mathbb{Z}_{\geq 0}\0},k>0\} \cup \{b^lc^md^n | l,m,n \in \mathbb{N}_{\geq 0}, k > 0\}$mathbb{Z}_{\geq 0}\}$forms a basis for quantum$SU(2)$. Does anyone know of a basis for quantum$SU(n)$? My guess would be that a similar result holds. Namely that the set made up of all products of powers of the matrix entries ordered with respect to the canonical ordering such that the first entry in the q-det(n) does not appear would forms a basis. How to prove this, however, I do not know. 8 added 2 characters in body; added 4 characters in body As is well known, the set${a^ib^jc^k \{a^ib^jc^k | i,j,k \in \mathbb{Z}{\geq 0}mathbb{Z}_{\geq 0}\} \cup {b^lc^md^n \{b^lc^md^n | l,m,n \in \mathbb{N}{\geq mathbb{N}_{\geq 0}, k > 0}$0\}$

forms a basis for quantum $SU(2)$. Does anyone know of a basis for quantum $SU(n)$?

My guess would be that a similar result holds. Namely that the set made up of all products of powers of the matrix entries ordered with respect to the canonical ordering such that the first entry in the q-det(n) does not appear would forms a basis. How to prove this, however, I do not know.

7 deleted 6 characters in body

As is well known, the set

$\{a^ib^jc^k {a^ib^jc^k | i,j,k \in \mathbb{Z}_{\geq 0}\mathbb{Z}{\geq 0}} \cup \{b^lc^md^n {b^lc^md^n | l,m,n \in \mathbb{N}_{\geq mathbb{N}{\geq 0}, k > 0\}$0}$forms a basis for quantum$SU(2)$. Does anyone know of a basis for quantum$SU(n)\$?

My guess would be that a similar result holds. Namely that the set made up of all products of powers of the matrix entries ordered with respect to the canonical ordering such that the first entry in the q-det(n) does not appear would forms a basis. How to prove this, however, I do not know.

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