MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

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

Let $(\Omega,d)$ be a differential calculus over an algebra $A$. It is easy to show that $\Omega$ is always equal to a quotient of $\Omega_u(A)$, the universal calculus over $A$, by some ideal $N$ of $\Omega_u(A)$. Now in practical terms, given a presentation of $\Omega$ in terms of generators and relations, how does one find the ideal $N$? I have some confused memory of there being a concrete recipe (due to Woronowicz?) for doing this when $A$ is a Hopf algebra and the calculus is left-covariant (or bi-covariant).

share|cite|improve this question
Ideal of what algebra? – Mariano Suárez-Alvarez Feb 11 '10 at 20:44
Edited to make clearer. Does it make sense now? – Abtan Massini Feb 11 '10 at 21:29

I don't know if you still care, but I think I found the answer to your question.

Look at Proposition 1 in Chapter 14 of Quantum Groups and Their Representations by Klimyk and Schmudgen. It shows that there is a bijection between left-covariant first-order differential calculi over a Hopf algebra $H$ and right ideals of the kernel of the counit of $H$, and it shows how the relations in a first-order calculus are obtained from the ideal. I have never worked with these things, so I don't know how tractable the computations are, but as far as I can tell from quickly scanning it seems that all the maps are at least explicitly defined.

As for higher order calculi, I am not sure whether/how these results extend. But at least maybe this is a good start?

share|cite|improve this answer

I don't much about differential calculi, but back in my head I also remember somewhat like as a HINT (?): Could N be the kernel of something like a "quantum shuffle map" or "quantum symmetrizer"? Then chances are you are talking about a Nichols algebra....and then one had a good description of N but NOT easily explicit relations....

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.