All modules here are finite dimensional. The field is over complex number. Let $U$ be an irreducible $\mathfrak{sl}_n$-module, and $V$ is a highest weight $\mathfrak{sl}_n$-module. Suppose $U\otimes V$ is not a highest weight representation. Then $U\otimes V$ has an irreducible quotient $W$. My question is: what is the maximal vector of $W$? or if $u^{-}\otimes v^{+}$ can be a maximal vector of $W$? It is known that $u^{-}\otimes v^{+}$ generates $U\otimes V$.
Thank you in advance.
Your notation suggests that that $u^-$ is a lowest weight vector, so I will asssume that this is the case. Then $u^-\otimes v^+$ generates $U\otimes V$. Indeed, let $W$ be the $\mathfrak{sl}_n$-submodule of $U\otimes V$ generated by $u^-\otimes v^+$ and write $\mathfrak{b}_+$ and $\mathfrak{b}_ -$ for the positive and the negative Borel subalgebras of $\mathfrak{sl}_n$. As $U$ is generated by $u^-$ over $U(\mathfrak{b}_+)$, we have that $W$ contains $u\otimes v^+$ for all $u\in U$. If $x\in \mathfrak{b}_-$ then $x(u\otimes v^+)=x(u)\otimes v^++u\otimes x(v^+)$, which implies that $u\otimes x(v_+)\in W$ for all $u\in U$ and $x\in\mathfrak{b}_-$. Since $V=U(\mathfrak{b}_{-})v_+$ we obtain $W=U\otimes V$ as wanted.
