As is well known, the set 


$\{a^ib^jc^k | i,j,k \in \mathbb{Z}\\_{\geq 0},k>0\} \cup \{b^lc^md^n | l,m,n \in \mathbb{Z}\\_{\geq 0}\}$


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.



Basis of quantum SU(n)
[edit: Following John's helpful comments below, I made this answer much more complete.]


Yes, this is the statement that $O\_q(G)$ is a flat deformation of $O(G)$ for any semi-simple group G. See the book by Klimyk and Schmuedgen, "Quantum Groups and Their Representations" for a proof of this: on page 311 they state the relevant theorem for $Mat\_q(n)$ (although the proof is just a reference to the original source). In the following section, they prove that det\_q is central, which allows us to identify $O\_q(SL\_N)$ with $Mat\_q(n)/(det\_q-1)$. The OP asked about $SU(N)$, but in the context of algebraic groups one studies SL\_N, which has a compact real form $SU(N)$, and morally the same representation theory.


In general we have to be careful when either inverting or specializing to a scalar any element in a noncommutative algebra, because this can in general drastically change the size of the algebra relative to what you'd expect from the commutative situation (it is bigger in the former case and smaller in the latter than expected). For inverting, you need the element to lie in a "denominator set", which assures that you don't have to add too many more things to invert it (imagine inverting $y$ in the free algebra $k((x,y))$ on two generators x and y: it would be a lot bigger than the vector space $k((x,y))[1/y]$). [edit: I can't get carot's or braces to work, hence the awkward symbol for free algebra; I hope it's clear.] For specializing, your element should honestly lie in the center of the noncommutative algebra, since it's image in the quotient will be a scalar (thus central). For instance, if you take $A^2\_{q}=k((x,y))/(yx=qxy)$, this has the same basis as $A^2=k[x,y]$. However, quotienting A\_q by y-1 forces x=0, which doesn't happen in A.


So far as I remember, the standard proof of the PBW theorem in this example (and many examples) relies on a technical lemma called the diamond lemma, Lemma 4.8 from KS, which gives an ordering on the monomials of O\_q(G) compatible with the defining relations, allowing one to prove the existence of PBW basis.


