DAHA of rank 1 is defined by the relation $$ (T - t^{1/2})(T + t^{-1/2})=0~, \quad TXT=X^{-1}~, \quad TY^{-1}T=Y~, \quad Y^{-1}X^{-1}YXT^2q^{1/2}=1 .$$ To understand its representations, it is useful to introduce non-symmetric polynomials $e_n(x)$ for $n\in\mathbb{Z}$ $$e_n(x)\equiv X^n \quad \textrm{mod} \ X^m \prec X^n ~, \quad Ye_n=q^{-n_\sharp}e_n~.$$ See Eqn. (2.6.1) of Cherednik's book. In addition, there are operations $$ 1=e_0\xrightarrow{A_0}e_1\xrightarrow{B_1}e_{-1}\xrightarrow{A_{-1}}e_2\xrightarrow{B_2}e_{-2}\xrightarrow{A_{-2}}\cdots $$ See Corollary 2.6.1 of Cherednik's book for more detail.
In sec 2.8.2 of Cherednik's book, he mentions that there are finite-dimensional representations of DAHA of rank 1 when $t = - q^{-n/2}$ for $n\in\mathbb{Z}_{>0}$ (Eqn. (2.8.11)). As far as I understand, the convention throughout the book is $t=q^k$. Hence, my first question is:
In this case, can one consider $k=-n/2+\log_q(-1)$?
Then, after Eqn. (2.8.11), it says that \begin{equation} T(e_{-n})=t^{1/2} e_{-n} \quad \textrm{and} \quad s(e_{-n})=e_{-n} \quad (**) \end{equation} My second question is how one can obtain this formula? It seems to me that if I use the definition of B_n (Corollary 2.6.1) $$ B_n = t^{1/2} (T + \frac{t^{1/2}-t^{-1/2}}{t q^n-1}) \quad \textrm{for}\quad n>0 $$ $$ B_{-n} = t^{1/2} (T + \frac{t^{1/2}-t^{-1/2}}{t^{-1} q^{-n}-1}) \quad \textrm{for}\quad n>0~, $$ I got $$ e_{-n} = t^{1/2} (T - t^{1/2}) e_n \quad \textrm{and} \quad (T + t^{-1/2}) e_{-n} =0 $$ as in sec 2.8.1. I cannot see how the condition of Eqn. (2.8.11) change the situation from sec 2.8.1. I wonder if somebody could possibly tell me how to derive $(**)$ by using Eqn. (2.8.11).
