l$ and $u'(1)=$ $v_1(1)$ such that $g\cdot u''-\xi^{i+j}u''=0$, or $(g\cdot u''-\xi^{i+j}u'')(1)<v_1(1)$, or $(g\cdot u''-$ $\xi^{i+j}u'')(1)=v_1(1)$ and $(g\cdot u''-\xi^{i+j}u'')(2)<v_1(2)$, where $u''=u+u'$.

Let $v_1(1)=m$ and $v_1(2)=s$.

\(1) It follows from Lemma \[3.2\](1).

\(2) Assume $s=0$. By Part (1), we have $0\leqslant m<l$. Hence $v_1=\alpha b^m\cdot v\otimes w+\sum\limits_{i_1<m}\alpha_{i_1, j_1}b^{i_1}\cdot v\otimes b^{j_1}\cdot w$ for some $\alpha$, $\alpha_{i_1, j_1}\in k$ with $\alpha\neq 0$. Take $u'=\alpha\xi^{-(i+j)}(1-\xi^j)^{-1}b^m\cdot v\otimes b\cdot w$ and let $u''=u+u'$. Then $h(u')=m+1\leqslant l$, $u'(1)=m$ and $$g\cdot u'-\xi^{i+j}u'=-\alpha b^m\cdot v\otimes w+\sum\limits_{i_1<m, j_1\leqslant 1}
\beta_{i_1, j_1}b^{i_1}\cdot v\otimes b^{j_1}\cdot w.$$ Since $g\cdot u''-\xi^{i+j}u''=v_1+g\cdot u'-\xi^{i+j}u'$, we know that $g\cdot u''-\xi^{i+j}u''=0$, or $(g\cdot u''-\xi^{i+j}u'')(1)<m$.

\(3) Assume $s>0$. Then $$v_1=\sum\limits_{0\leqslant j_1\leqslant s}\alpha_{j_1}b^m\cdot v
\otimes b^{j_1}\cdot w+\sum\limits_{i_1<m}\alpha_{i_1, j_1}b^{i_1}\cdot v\otimes b^{j_1}\cdot w$$ for some $\alpha_{j_1}$, $\alpha_{i_1, j_1}\in k$ with $\alpha_s\neq 0$. Note that $m+s\leqslant h(v_1)<l\leqslant p-1$. Hence $s<p-1$ and so $1<s+1<p$. Let $u'=\alpha_s(s+1)^{-1}\xi^{-(i+j)}(1-\xi^j)^{-1}b^m\cdot v\otimes b^{s+1}\cdot w$ and $u''=u+u'$. Then $h(u')=m+s+1\leqslant l$, $u'(1)=m$ and $$g\cdot u'-\xi^{i+j}u'=-\alpha_s b^m\cdot v\otimes b^s\cdot w+\sum\limits_{j_1<s}
\beta_{j_1}b^m\cdot v\otimes b^{j_1}\cdot w+\sum\limits_{i_1<m, j_1\leqslant s+1}
\beta_{i_1, j_1}b^{i_1}\cdot v\otimes b^{j_1}\cdot w.$$ Since $g\cdot u''-\xi^{i+j}u''=v_1+g\cdot u'-\xi^{i+j}u'$, we know that $g\cdot u''-\xi^{i+j}u''=0$, or $(g\cdot u''-\xi^{i+j}u'')(1)<m$, or $(g\cdot u''-\xi^{i+j}u'')(1)=m$ and $(g\cdot u''-\xi^{i+j}u'')(2)<s$.

\[3.6\] Let $0\neq u\in S_i\otimes S_j$ with $h(u)=u(1)=l>0$. If $g\cdot u\neq\xi^{i+j}u$, then there is an element $\overline{u}\in S_i\otimes S_j$ with $h(\overline{u})\leqslant l$ and $\overline{u}(1)<l$ 