# Irreducible quotient of $U\otimes V$

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 $U\otimes V$.

• Why will that tensor product have a unique irreducible quotient? Aug 7, 2013 at 7:04
• The formulation is a bit loose and symbols undefined, as comments indicate. In any case, you should specify what kind of field you are working over and whether any of the modules is assumed to be finite dimensional. (And some motivation for the question might help.) Aug 7, 2013 at 16:08
• P.S. Given the new formulation, complete reducibility makes it easy to control the highest weight occurring in such a tensor product. So there seems to be no real problem here. Aug 7, 2013 at 19:23
• Given the new formulation, I would say this is not a research level question. The tensor product is a direct sum of simple modules, and the highest weights of those are given by the Littlewood-Richardson rule. Also, the formulation is a bit strange, as a finite dimensional highest weight module is the same as a finite dimensional simple module. Aug 8, 2013 at 6:54

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.
• $W$ is irreducible, the maximal vector is unique up to scalar. Aug 7, 2013 at 13:22