The case $ q^l=1$

Let $ q^l=1$ and $ q^s\neq 1$ for $ s=1,\dots,l-1$. Then we have $ [n]_q=[n+l]_q$. And there are only $ l$ different quantum numbers, in particular $ [l]_q=0$. Note that the quantum numbers are still polynomials in $ q$. The only difference is that we can now reduce high powers of $ q$ to the smaller ones using the relation $ q^l=1$.

The situation is somewhat similar to the reduction of all integers to reminders modulo $ l$. In fact this similarity is very deep. Here we show some examples.


Exercise:Let $ l$ be a prime number. Let $ m_1,m_2$, be natural numbers. Devide $ m_1,m_2$ by $ l$ with a reminder: $ m_1=k_1l+r_1,m_2=k_2l+r_2$, where $ 0\leqslant r_1,r_2< l$. Prove that

$\displaystyle (x+y)^l=x^l+y^l \;\;({\rm mod}\; l), $

$\displaystyle {m_2 \choose m_1}= {k_2 \choose k_1}{r_2\choose r_1} \;\;({\rm mod}\;l). $


We have the following $ q$-analog of this identities.


Exercise:Let $ q^l=1$ and $ q^s\neq 1$ for $ s=1,\dots,l-1$. Let $ m_1,m_2$, be natural numbers. Devide $ m_1,m_2$ by $ l$ with a reminder: $ m_1=k_1l+r_1,m_2=k_2l+r_2$, where $ 0\leqslant r_1,r_2< l$. Prove that (we recall the relations $ YX=qXY$).

$\displaystyle (X+Y)^l=X^l+Y^l, $

$\displaystyle {m_2 \choose m_1}_q= {k_2 \choose k_1}{r_2\choose r_1}_q. $