What is $\left[ \begin{array}{c} K_i;0\\ \ell\\ \end{array} \right] _{\varepsilon _i}$ in the restricted specialization in QUE algebras?

Question

I have a question about the book A Guide to Quantum Groups written by Vyjayanthi Chari and Andrew Pressley. It comes from section $9.3$ on page $300$ of this book

In Section 9.1, the authors define the algebra $U_q(\mathfrak{g})$ over $\mathbb{Q}(q)$ where $\mathfrak{g}$ is a finite-dimensional complex Lie algebra. Let $\mathbb{A} =\mathbb{Z}[q,q^{-1}]$, they define $U^{res}_{\mathbb{A} }$ and $U^{res}_{\varepsilon}=U^{res}_{\mathbb{A} }\otimes_{\mathbb{A} }\mathbb{Q}(\varepsilon)$. Here $\varepsilon $ is a primitive $\ell$th root of unity, where $\ell$ is odd and greater than $d_i$ for all $i$.

Let $$\left[ \begin{array}{c} K_i;0\\ s\\ \end{array} \right] _{q_i}=\prod_{r=1}^{s}{\frac{K_iq^{1-r}-{K_i}^{-1}q^{r-1}}{q_{i}^{r}-q_{i}^{-r}}}.$$ Then in page $300$ remark $[1]$, the author write an element $\left[ \begin{array}{c} K_i;0\\ \ell\\ \end{array} \right] _{\varepsilon _i}\in U_\varepsilon^{res}$ .

Here is my question:

If I replace $q_i$ with $\varepsilon_i$, because $\varepsilon^l=\varepsilon^{-l}=1$, then there will have $0$ in the denominator.

So I am very confused about it. How should I deal with $\left[ \begin{array}{c} K_i;0\\ \ell\\ \end{array} \right] _{\varepsilon _i}$?

Any help and references are greatly appreciated.

Thanks!

Answer 1

You are correct and your observation is precisely the point. It is all about choices. You define the K-symbol as an element in the quantum group over $\mathbb{Q}(q)$. At a later point you more or less define $U_\mathbb{A}^{res}$ as being generated as $\mathbb{Z}[q,q^{-1}]$-module by by a specified set of elements (such that this definition is compatible with product and coproduct etc). If you then specialize to $\varepsilon$, by definition the K-symbol stays and is nonzero, while for example $$(\varepsilon^\ell-1)\left[ \begin{array}{c} K_i;0\\ \ell\\ \end{array} \right] _{\varepsilon _i}\in U_\varepsilon^{res}$$ specializes to zero, even though as a polynomial in $q$ it looks more innocent.

More prominent and maybe more transparent is this for another generator: the divided power $E_i^{(n)}$. Because we choose $E_i^{(n)}$ as a generator of $U_\mathbb{A}^{res}$ as $\mathbb{Z}[q,q^{-1}]$-module, it specializes to something nonzero, while $E^n=[n]_q! E_i^{(n)}$ specializes to zero. Had we chosen $E^n$ as generator, then it would specialize to something nonzero and $E_i^{(n)}$ would be undefined (i.e. not contained in $U_\mathbb{A}^{res}$ altogether). This choices are called restriced versus unrestricted specialization. Something similar has been previousley done for Lie algebras in finite characteristic

Concretely, you make all your computations over $q$, then you rewrite everything in terms of K-symbols and divided powers ("grouping singular terms") and then you specialize and see what the remaining polynomials in $q$ do. See G. Lusztig: Quantum groups at unit of 1.