PBW basis for quantum groups and braid group action

Question

I'm trying to check some details from this paper. Let $U_q(\mathfrak{sl}_3)$ be the usual Drinfeld-Jimbo quantum group, with generators $E_1,E_2,F_1,F_2,K_1^{\pm},K_2^{\pm}$ and well known relations, for example $$K_1E_1K_1^{-1} = q^2E_1$$ $$E_1F_1 - F_1E_1 = \frac{K_1 - K_1^{-1}}{q-q^{-1}}$$ etc... For notational ease let us also define $$[K_i;a] := \frac{q^aK_i - q^{-a}K_i^{-1}}{q-q^{-1}}$$

Define operators $T_1, T_2$ as follows : $T_i(F_j) = - K_i^{-1}E_i$ if $i=j$ and $F_iF_j - qF_jF_i$ else, $T_i(E_j) = -F_iK_i$ if $i=j$, and $E_iE_j - q^{-1}E_jE_i$ else. Finally $T_i(K_j) = K_i^{-1}$ if $i=j$ and $K_iK_j$ else.

Question : how to check the equalities $T_1T_2(F_1) = F_2$ ?

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Details of my computation :

\begin{equation} T_1T_2F_1= T_1(F_2F_1 - qF_1F_2) = \\ (F_2F_1 - qF_1F_2)(-K_1^{-1}E_1) -q(-K_1^{-1}E_1)(F_2F_1-qF_1F_2) = \\ -q^{-1}K_1^{-1}(F_2F_1-qF_1F_2)E_1 +q(K_1^{-1}E_1)(F_1F_2 - qF_1F_2) \end{equation}

Where we used $K_1F_2 = q F_2K_1$ so $F_2K_1^{-1} = qK_1^{-1}F_2$ and $F_1K_1^{-1} = q^{-2}K_1^{-1}F_1$.

Now since $E_1F_2 = F_2E_1$ and $F_1E_1 = E_1F_1 - [K_1;0]$ we get $$(F_2F_1 - qF_1F_2)E_1 = (F_2F_1 - qF_1F_2) +q[K_1;0]F_2 - F_2[K_1;0]$$

Using this we obtain

$\begin{equation} -q^{-1}K_1^{-1}(F_2F_1-qF_1F_2)E_1 -q(K_1^{-1}E_1)(F_1F_2 - qF_1F_2) = \\ -q^{-1}K_1^{-1}E_1(F_2F_1 - qF_1F_2) - q^{-1}K_1^{-1}(q[K_1;0]F_2-F_2[K;0]) + q K_1^{-1}E_1(F_2F_1 - q F_1F_2) = \\ (q-q^{-1})K_1^{-1}E_1(F_2F_1-qF_1F_2) - q^{-1}K_1^{-1}F_2(q[K_1;1] + [K_1;0]) \end{equation}$

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Edit : the first version asked why $T_1T_2T_1 = T_2T_1T_2$ but it is pretty easy once you know that $T_iT_j(F_i) = F_i$ and similarly with $E_i$. Indeed we get $T_2T_1T_2(F_1) = T_2(F_2) = -K_2^{-1}E_2$ and $T_1T_2T_1(F_1) = T_1T_2(-K_1^{-1}E_1) = -T_1(K_1^{-1}K_2^{-1}T_2(E_1))) = -K_2^{-1}T_1T_2(E_1) = -K_2^{-1}E_2$.

Answer 1

Since you didn't give any details, I can't tell what went wrong with your calculation, but I will note that one definition you have is incorrect. You should define $T_i(F_i)=-K_i^{-1}E_i$ (as opposed to $-K_i^{+1}E_i$, which you have written multiple times in your post).

I double checked and if you apply commutation relations correctly the result follows quite easily.

Answer 2

The definition of $T_i$ appearing in the linked paper is $$T_i(F_j)=\begin{cases}F_j&\mbox{if }a_{ij}=0\\F_jF_i-qF_iF_j&\mbox{if }a_{ij}=-1\\-K_i^{-1}E_i&\mbox{if }i=j\end{cases} $$ Therefore, \begin{align} T_1T_2(F_1)=&T_1(F_1F_2-qF_2F_1)\\ =&(-K_1^{-1}E_1)(F_2F_1-qF_1F_2)-q(F_2F_1-qF_1F_2)(-K_1^{-1}E_1)\\ =&qK_1^{-1}E_1F_1F_2-K_1^{-1}E_1F_2F_1+qF_2F_1K_1^{-1}E_1-q^2F_1F_2K_1^{-1}E_1\\ =&qK_1^{-1}(F_1E_1+[K_1,0])F_2-K_1^{-1}F_2(F_1E_1+[K_1,0])+qF_2F_1K_1^{-1}E_1-q^2F_1F_2K_1^{-1}E_1\\ =&qK_1^{-1}F_1F_2E_1+qK_1^{-1}[K_1,0]F_2-K_1^{-1}F_2F_1E_1-K_1^{-1}F_2[K_1,0]\\ &+qF_2F_1K_1^{-1}E_1-q^2F_1F_2K_1^{-1}E_1\\ =&qK_1^{-1}[K_1,0]F_2-K_1^{-1}F_2[K_1,0],\\ \end{align} where the last equality follows from the relation $K_1^{-1}F_1=q^2F_1K_1^{-1}$ and $K_1^{-1}F_2=q^{-1}F_2K_1^{-1}$ and cancellation. Finally, \begin{align} qK_1^{-1}[K_1,0]F_2-K_1^{-1}F_2[K_1,0]=&(q-q^{-1})^{-1}(qK_1^{-1}(K_1-K_1^{-1})F_2-K_1^{-1}F_2(K_1-K_1^{-1}))\\ =&(q-q^{-1})^{-1}(qK_1^{-1}(K_1-K_1^{-1})-K_1^{-1}(q^{-1}K_1-qK_1^{-1}))F_2\\ =&F_2 \end{align}