2
$\begingroup$

I have a question about a lemma $9.3.6$ in the book A Guide to Quantum Groups written by Vyjayanthi Chari and Andrew Pressley. This question comes from page 301, "The restricted specialization".

Here is the lemma:

Let $\epsilon$ be a primitive $\ell$th root of unity, where $\ell$ is odd and greater than one. Let $r,s\in \mathbb{N}$ and write $r=r_0+\ell r_1,s=s_0+\ell s_1$, where $0\le r_0,s_0<\ell$, $r_1,s_1\ge 0$, $r\ge s$. Then $$\left[ \begin{array}{c} r\\ s\\ \end{array} \right] _{\varepsilon}=\left[ \begin{array}{c} r_0\\ s_0\\ \end{array} \right] _{\varepsilon}\left( \begin{array}{c} r_1\\ s_1\\ \end{array} \right).$$ where $\left( \begin{array}{c} r_1\\ s_1\\ \end{array} \right)$ is an ordinary binomial coefficient

Here are some notations: \begin{align} [n]_q& =\frac{q^n-q^{-n}}{q-q^{-1}}, \\ [n]_q!& =[n]_q.[n-1]_q\cdots [2]_q.[1]_q, \\ \left[\begin{array}{c}m\\n\end{array}\right]_q& =\frac{[m]_q!}{[n]_q![m-n]_q!}. \end{align}

Here is my question:

Since $\epsilon$ is a primitive $\ell$th root of unity, then I think $[l]_\varepsilon$ equals to $0$.

So the denominator will have $0$. Does this lemma make sense?

I am confused about it.

Any help and references are greatly appreciated.

Thanks!

Improve this question
$\endgroup$
9
  • 1
    $\begingroup$ In my copy of the book is says $0\le r_0,s_0<\ell$. $\endgroup$
    – Dave Benson
    Commented Sep 7 at 11:07
  • 2
    $\begingroup$ There is an important fact that $q$-binomials are Laurent polynomials in $q$, so you may substitute every non-zero $q$ $\endgroup$
    – Fedor Petrov
    Commented Sep 7 at 11:59
  • $\begingroup$ @DaveBenson Thank you. I have edited it. $\endgroup$
    – fusheng
    Commented Sep 7 at 12:09
  • $\begingroup$ @FedorPetrov Should I replace $q$ with $\varepsilon$ in the final Laurent polynomials in q? $\endgroup$
    – fusheng
    Commented Sep 7 at 12:21
  • 4
    $\begingroup$ This is often called the $q$-Lucas theorem (see e.g. symmetricfunctions.com/q-analogues.htm#prelimQanalogsQLucas). As Fedor said these are polynomials (or Laurent polynomials because of the quantum groups convention) so plugging in anything makes sense. But also, even if you want to leave it as a rational expression, yes you get some $0$'s in the denominator, but you get more $0$'s in the numerator which cancel. Maybe even another elementary way to think about it is: use L'Hôpital's rule. $\endgroup$
    – Sam Hopkins
    Commented Sep 7 at 12:32

0

Reset to default

You must log in to answer this question.