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I am reading the paper: ELEMENTARY CONSTRUCTION OF LUSZTIG’S CANONICAL BASIS and want to compute $F_{\beta_3}$ in Example 3 on page 3. Maybe there is some mistake in my computations but I could not find it. The following are my computations.

Let $U_q(\mathfrak{g})$ be the quantum group associated to a simple Lie algebra $\mathfrak{g}$. Let $w_0=s_{i_1} \cdots s_{i_N}$ be a reduced expression of the longest word $w_0$ in the Weyl group of $\mathfrak{g}$. The $F_{\beta_j}$'s are defined by \begin{align} F_{\beta_j} = T_{i_1}T_{i_2} \cdots T_{i_{j-1}}(F_{i_j}), \end{align} where $T_i$'s are Lusztig's symmetry. For each $i$, there is an algebra automorphism $T_i$ of $U_q(\mathfrak{g})$ defined by \begin{align} T_i(F_j) = \begin{cases} F_j, & |i-j|\geq 2, \\ F_j F_i - q F_i F_j, & |i-j|=1, \\ -K^{-1}_j E_j, & i=j, \end{cases} \end{align} \begin{align} T_i(E_j) = \begin{cases} E_j, & |i-j|\geq 2, \\ E_i E_j - q^{-1} E_j E_i, & |i-j|=1, \\ -F_j K_j, & i=j, \end{cases} \end{align} \begin{align} T_i(K_j) = \begin{cases} K_j, & |i-j|\geq 2, \\ K_i K_j, & |i-j|=1, \\ -K^{-1}_j, & i=j. \end{cases} \end{align}

The commutation relations in $U_q(\mathfrak{g})$ are: for $i \neq j$, \begin{align} & K_i K_i^{-1} = K_i^{-1} K_i = 1, \quad K_i K_j = K_j K_i, \\ & K_i E_i K_i^{-1} = q^2 E_i, \quad K_i F_i K_i^{-1} = q^{-2} F_i, \\ & [E_i, F_j]=0, \quad [E_i, F_i] = \frac{K_i-K_i^{-1}}{q-q^{-1}}, \end{align} for $|i-j|=1$, \begin{align} & E_i^2 E_j + E_j E_i^2 = (q+q^{-1})E_i E_j E_i, \\ & F_i^2 F_j + F_j F_i^2 = (q+q^{-1}) F_i F_j F_i, \\ & K_i E_j K_i^{-1} = q^{-1} E_j, \quad K_i F_j K_i^{-1} = qF_j, \end{align} for $|i-j|>1$, \begin{align} & E_i E_j = E_j E_i, \quad F_i F_j = F_j F_i, \\ & K_i E_j K_i^{-1} = E_j, \quad K_iF_jK_i^{-1}=F_j. \end{align}

Let $w_0 = s_1 s_2 s_1$. Then \begin{align} F_{\beta_3} & = T_1 T_2 (F_1) \\ & = T_1( F_1 F_2 - q F_2 F_1 ) \\ & = T_1(F_1) T_1(F_2) - q T_1(F_2) T_1(F_1) \\ & = ( -K_1^{-1} E_1 )( F_2 F_1 - q F_1 F_2 ) - q ( F_2 F_1 - q F_1 F_2 )( -K_1^{-1} E_1 ) \\ & = -K_1^{-1} E_1 F_2 F_1 + q K_1^{-1} E_1 F_1 F_2 + q F_2 F_1 K_1^{-1} E_1 - q^2 F_1 F_2 K_1^{-1} E_1 \\ & = -K_1^{-1} E_1 F_2 F_1 + q K_1^{-1} E_1 F_1 F_2 \\ & \quad + qK_1^{-1}E_1 F_2 F_1 - \frac{K_1-K_1^{-1}}{q-q^{-1}}F_2 - q^2 K_1^{-1} E_1 F_1 F_2 + q \frac{K_1-K_1^{-1}}{q-q^{-1}}F_2. \end{align} Therefore $F_{\beta_3} \neq F_2$. But in Example 3, it is said that $F_{\beta_3} = F_2$. Where did I make a mistake? Thank you very much.

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The mistake happens in the last step.

Note that

\begin{align*}qF_2F_1K_1^{-1}E_1 &=q^{1-2+1}K_1^{-1}F_2F_1E_1\\ &=K_1^{-1}F_2\left(E_1F_1-\frac{K_1-K_1^{-1}}{q-q^{-1}}\right)\\ &=K_1^{-1}E_1F_2F_1-\left(\frac{q^{-1}-qK_1^{-2}}{q-q^{-1}}\right)F_2, \end{align*}

and similarly

$$-q^{2}F_1F_2K_1^{-1}E_1=-qK_1^{-1}E_1F_1F_2+\left(\frac{q-qK_1^{-2}}{q-q^{-1}}\right)F_2.$$

Now everything cancels except for $$\frac{q-q^{-1}}{q-q^{-1}}F_2=F_2.$$

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