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).
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asked Feb 11, 2010 at 16:48
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1$\begingroup$ Ideal of what algebra? $\endgroup$– Mariano Suárez-ÁlvarezCommented Feb 11, 2010 at 20:44
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$\begingroup$ Edited to make clearer. Does it make sense now? $\endgroup$– Abtan MassiniCommented Feb 11, 2010 at 21:29
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2 Answers
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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?
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I don't much about differential calculi, but back in my head I also remember somewhat like this....so 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....
