and $v_l(2)=0$. Since $g\cdot u_l-\xi^{i+j}u_l=0$, $v_l+g\cdot u'_{(l)}-\xi^{i+j}u'_{(l)}=0$. Hence $(g\cdot u'_{(l)}-\xi^{i+j}u'_{(l)})(1)=l-1$ and $(g\cdot u'_{(l)}-\xi^{i+j}u'_{(l)})(2)=0$. By Lemma \[3.2\], we know that $l-1=(g\cdot u'_{(l)}-\xi^{i+j}u'_{(l)})(1)\leqslant u'_{(l)}(1)
<l$, which forces that $u'_{(l)}(1)=l-1$. Since $u'_{(l)}(1)+u'_{(l)}(2)\leqslant h(u'_{(l)})\leqslant l$, $u'_{(l)}(2)\leqslant 1$. If $u'_{(l)}(2)=0$, then it follows from Lemma \[3.2\] that $l-1=(g\cdot u'_{(l)}-\xi^{i+j}u'_{(l)})(1)< u'_{(l)}(1)=l-1$, a contradiction. Therefore, $u'_{(l)}(2)=1$, and so $h(u'_{(l)})=l$. Thus we have $$u'_{(l)}=\alpha b^{l-1}\cdot v\otimes b\cdot w+\beta b^{l-1}\cdot v\otimes w
+\sum\limits_{i_1<l-1}\alpha_{i_1,j_1}b^{i_1}\cdot v\otimes b^{j_1}\cdot w.$$ Again by Lemma \[3.2\], one gets $$g\cdot u'_{(l)}-\xi^{i+j}u'_{(l)}=-\alpha\xi^{i+j}(1-\xi^j) b^{l-1}\cdot v\otimes w
+\sum\limits_{i_1<l-1}\beta_{i_1,j_1}b^{i_1}\cdot v\otimes b^{j_1}\cdot w.$$ Since $v_l+g\cdot u'_{(l)}-\xi^{i+j}u'_{(l)}=0$, $\alpha=-l(1-\xi^i)(1-\xi^j)^{-1}$, and hence $$u'_{(l)}=-l(1-\xi^i)(1-\xi^j)^{-1}b^{l-1}\cdot v\otimes b\cdot w+\beta b^{l-1}\cdot v\otimes w
+\sum\limits_{i_1<l-1}\alpha_{i_1,j_1}b^{i_1}\cdot v\otimes b^{j_1}\cdot w.$$ Since $ga=ag$, $a\cdot V_{\xi^{i+j}}\subseteq V_{\xi^{i+j}}$. Consider the action of $a$ on $V_{\xi^{i+j}}$. Then $a\cdot u_0=(1-\xi^{i+j})u_0$. For $1\leqslant l\leqslant p-1$, let $u=u_l+\alpha_1u_{l-1}+\ldots+\alpha_lu_0$ be an element in $V_{\xi^{i+j}}$. If $a\cdot u=\alpha u$ for some $\alpha\in k$, then by comparing their coefficients of the item $b^l\cdot v\otimes w$, we find that $\alpha=1-\xi^{i+j}$. It follows that $1-\xi^{i+j}$ is the unique eigenvalue for the action of $a$ on $V_{\xi^{i+j}}$. Using Lemma \[3.2\], one finds that the coefficient of the item $b^{l-1}\cdot v\otimes w$ in $a\cdot u-(1-\xi^{i+j})u$ is $-\frac{l}{2}(1-\xi^i)(1-\xi^{i+j})$. We divide the discussion into the following two cases.

For case 1: $i+j\neq t$. In this case, $a\cdot 