MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
I tried to fix the LaTeX, but I think there are still some mathematical typos which I don't want to guess how to correct. – Reid Barton Nov 25 '09 at 20:10
The latex compiles fine now on my computer. – Abtan Massini Nov 25 '09 at 20:21
Please add a backslash before each underscore. The first compiler, which italicizes, etc., will remove each backslash but leave the underscore, and then JSMath will do its thing. – Theo Johnson-Freyd Nov 25 '09 at 20:32
The LaTeX is fine, but the mathematical formula is still poorly quantified. Why is there a k in the second set, any what is the difference between nonnegative elements of Z and nonnegative elements of N? – S. Carnahan Nov 25 '09 at 22:12
The accepted answer by David is, while true for formal deformations, not correct for specializations: the diamond lemma proof stalls if done the same way as for $O_q(Mat_n)$. Klimyk-Schmuedgen certainly do not prove that result for $SL_q(n)$. In fact I talked to Schmuedgen about it in about 2000 at MSRI. Greg's answer however works as I learned from Brown around 2002 via email. – Zoran Skoda Jul 25 '11 at 15:08
up vote 8 down vote accepted

[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.

share|cite|improve this answer
Where exactly in Klimyk and Schmuedgen is the proof? I'm having trouble finding it. – John McCarthy Feb 20 '10 at 18:03
Hi John, on page 311, they prove the claim for O_q(Mat_n) (well I should have been more careful; this book rarely proves things, but rather points to precise locations in the original literature. Their proof in this case consists of 3 citations). From that claim about O_q(Mat_N), together with the proof in the section following that the quantum determinant is central, one can conclude the PBW property for O_q(SL_N). I also find on page 103 a brief overview of the diamond lemma, also with references. – David Jordan Feb 21 '10 at 18:31
I also recently found while trying to understand the diamond lemma more completely. It has a really nice exposition, and references to a good original source. – David Jordan Feb 21 '10 at 18:32
Sorry, but I don't understand what you mean by "which allows us to identify O_q(SL_N) with Mat_q(n)/." – John McCarthy Feb 22 '10 at 21:53
sorry, tex error =[. The js tex compiler doesn't like underscores or slashes. I'll fix it now. – David Jordan Feb 22 '10 at 22:17

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.

share|cite|improve this answer
I'm very glad you added this reference, which I didn't know about. The one I provided gives a brief explanation in this special case, but I didn't know there was a whole book full of tricks. Thanks! – David Jordan Nov 26 '09 at 17:21

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


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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.