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As is well known (see Kassel), when $q$ is not a root of unity, the centre or the quantum enveloping algebra $U_q({\mathfrak sl}_2)$ of ${\mathfrak sl}_2$ is generated by the element $$ C_q = EF + \frac{q^{-1}K+qK^{-1}}{(q-q^{-1})^2}. $$ The element is called the quantum Casimir. My questions are as follows:

(i) Does this situation extend to the general setting of $U_q({\mathfrak sl}_N)$?

(ii) If it does, is there a general formula for $C_q$?

(iii) How would this formula relate to the usual formula for the classical Casimir? (The uasual formula I refer to is $\sum X^iX_i$, for some basis $X_i$ and its dual $X^i$, see wikipedia for details.)

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for the third part, set K=q^h and take the limit as q goes to 1. – Peter McNamara Nov 19 '10 at 18:42
@Peter McNamara: here you mean "$q^H$", where $H$ is the third element of usual $\mathfrak{sl}(2)$ (as opposed to $h = \log q$). – Theo Johnson-Freyd Nov 20 '10 at 2:02
up vote 7 down vote accepted

The centers of the Drinfeld-Jimbo quantum groups $U_q(\mathfrak{g})$ are well-understood and quite analogous to the classical case. See the book by Klimyk and Schmüdgen, Section 6.3, where in particular the quantum Casimirs are constructed.

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Does the object they construct there reduce to the ordinary Casimir in the $q=1$ case? – Dyke Acland Nov 19 '10 at 19:07
What do you mean by "reduce to"? They have properties analogous to the usual Casimirs. However, looking at their eigenvalues on finite-dimensional modules, it might be necessary to add some constant (rational function of $q$) before you will get the exact $q$-numbers of the usual eigenvalues. I guess it depends on what you need them for. If you would like to take a formal limit you need to look at the $h$-adic versions of these quantum groups. There I am sure there exist proper direct analogues. – Jonas Hartwig Nov 19 '10 at 19:30
I'm just trying to understand these objects. The way I do this with a new quantum object is to see exactly what it is when $q=1$. – Dyke Acland Nov 19 '10 at 19:45
Sure, morally $q=1$ should correspond to classical things. But for example in the quantum casimir you mention one cannot just put $q=1$... – Jonas Hartwig Nov 20 '10 at 3:49
.... of course. – Dyke Acland Nov 20 '10 at 15:25

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