Basis of quantum SU(n)

Question

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.

Answer 1

[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.

Answer 2

This answer is not really all that different from David Jordan's answer, but is a somewhat different take. The coordinate ring $O_q(M_n)$, for all matrices, is an example of a multivariate "skew polynomial ring", which means a graded or filtered ring with certain axioms that imply that any map from a vector of exponents to monomials is a basis. (I am getting this from "Noncommutative Gröbner bases and filtered-graded transfer", by Huishi Li.) In any ring like this, there is a theory of Gröbner bases for any ideal, in particular for the determinant ideal. If you eliminate monomials that contain the leading term of a Gröbner generator, then remaining ones are a basis for the quotient ring. It is an equivalent answer, but more general, because it boils down to the same diamond lemma.

Answer 3

The question is slightly mistated, from the example for $n=2$ it is seen that the question is about $SL_q(n)$ and not $SU_q(n)$; the answer is of course those standard normally ordered monomials

$(t^1_1)^{a_{11}}(t^1_2)^{a_{12}}...(t^n_n)^{a_{nn}}$

satisfying the condition that at least one of the diagonal exponents $a_{ii}$ is zero. Unlike in $O_q(M_n)$ literal application of the Bergman's diamond lemma does not produce the algorithm, because the diagonal enetries are not one next to another so if one wants to exclude the diagonal extra occurences one needs to go against the semigroup law. This is possible to do with great effort, I have checked this in 1999 with lots of algorithmic combinatorics; namely the set of reductions used is infinite and given algorithmically rather than by explicit formulas. Unlike the general rule advised by Bergman, it is not wise in the straight diamond lemma approach to exclude the nested ambiguities. Some other Grober arguments not relying on standard diamond lemma can give easy answer though.

For generic $q$ it is of course enough to use the classical commutative case and deformation arguments (Edit: alluded in David's answer).

It is not true, what is stated above in the accepted answer that the simple technique for $O_q(M_n)$ via diamond lemma and with the relations taken as reductions works when setting $det_q =1$. Imagine you have expression $(x^1_1)^2 (x^2_2)^2 (x^3_3)^2$ in $SL_q(3)$. How will you use centrality of the quantum determinant to translate this into something what does not have all three diagonal generators ? You need first to rearrange thing to be able to complete to a quantum determinant to exclude a bad diagonal generator, but this is not very compatible with the ordering. It can be done systematically but by now means is trivial or implied by Klimyk-Schmuedgen book.