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.

  • $\begingroup$ Why will that tensor product have a unique irreducible quotient? $\endgroup$ Aug 7, 2013 at 7:04
  • $\begingroup$ 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.) $\endgroup$ Aug 7, 2013 at 16:08
  • $\begingroup$ 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. $\endgroup$ Aug 7, 2013 at 19:23
  • 2
    $\begingroup$ 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. $\endgroup$ Aug 8, 2013 at 6:54

1 Answer 1


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.

  • 1
    $\begingroup$ But he wrote that he knew that this was the case. The question was about the highest weight of an irreducible quotient of the tensor product (I am still not sure why there should be a unique such). $\endgroup$ Aug 7, 2013 at 11:59
  • $\begingroup$ $W$ is irreducible, the maximal vector is unique up to scalar. $\endgroup$
    – Yilan Tan
    Aug 7, 2013 at 13:22
  • 1
    $\begingroup$ @YilanTan Yes, but why will the irreducible quotient be unique? $\endgroup$ Aug 7, 2013 at 13:27
  • $\begingroup$ any cyclic module will have a unique irreducible quotient: take the maximal submodule not containing the cyclic vector (Zorn's lemma) and go modulo that maximal one. $\endgroup$ Aug 7, 2013 at 14:04
  • 1
    $\begingroup$ @Aakumadula: But why should the maximal submodule be unique? $\endgroup$ Aug 7, 2013 at 16:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.