Let us proceed to the unrestricted T-systems. Choose $h\in {\mathbb{C}\backslash 2\pi \sqrt{-1} \mathbb{Q}}$ arbitrarily. The unrestricted T-system for $U_{q}(X_{N}^{(\mathfrak{k})})$ is the following relations. For $X_{N}^{(\mathfrak{k})}=A_{2r}^{(2)}$, \begin{align} T_{m}^{(a)}(u-1)T_{m}^{(a)}(u+1)&=T_{m-1}^{(a)}(u)T_{m+1}^{(a)}(u)+T_{m}^{(a-1)}(u)T_{m}^{(a+1)}(u), (1\leq a\leq r-1)\\ T_{m}^{(r)}(u-1)T_{m}^{(r)}(u+1)&=T_{m-1}^{(r)}(u)T_{m+1}^{(r)}(u)+T_{m}^{(r-1)}(u)T_{m}^{(r)}(u+\Omega), \end{align} where $\Omega=2\pi \sqrt{-1}/{\mathfrak{k}h}$.
My question is how to value $\Omega$ in T-system for twisted quantum affine algebras?
For example in type $A_{r}$, the untwisted case. \begin{align} T_{m}^{(a)}(u-1)T_{m}^{(a)}(u+1)&=T_{m-1}^{(a)}(u)T_{m+1}^{(a)}(u)+T_{m}^{(a-1)}(u)T_{m}^{(a+1)}(u). \end{align} $A_{3}$, if $a=m=2$, then \begin{align} 2_{-4}2_{-2}\ast2_{-2}2_{0}=2_{-2}*2_{-4}2_{-2}2_{0}+1_{-3}1_{-1}*3_{-3}3_{-1} \end{align} then we have \begin{align} T_{2}^{(2)}(-3)T_{2}^{(2)}(-1)&=T_{1}^{(2)}(-2)T_{3}^{(2)}(-2)+T_{2}^{(1)}(-2)T_{2}^{(3)}(-2), \end{align} which satisfies the above equation. I want to get the similar equation like $A_{3}$
