I am reading the paper: [ELEMENTARY CONSTRUCTION OF LUSZTIG’S CANONICAL BASIS](https://arxiv.org/pdf/1602.04895.pdf) 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.



Question about a computation of $F_{\beta_3}$ related to canonical basis of $U_q(sl_3)$.
**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 $$\frac{q-q^{-1}}{q-q^{-1}}F\_2=F\_2.$$


