Let $\tilde{\frak g}=\frak g \otimes \mathbb C[t,t^{-1}]$ be the loop algebra associated to a finite-dimensional simple Lie algebra $\frak g$. Consider $U,V$ and $W$ universal finite-dimensional highest-weight $U(\tilde{\frak g})$-modules of different highest weights and suppose that they generated by $u,v$ and $w$, respectively. Suppose these modules have unique irreducible quotient $U'$, $V'$ and $W'$ respectively such that

1) $W$ is generated by $u\otimes v$ as $U(\tilde{\frak g})$-module.

2) There exists a surjective homomorphism $W \to U\otimes V$ defined by $w\mapsto u\otimes v$.

3) $W'\cong U'\otimes V'$ as $U(\tilde{\frak g})$-modules.

For this scenery, could it be shown that $W\cong U\otimes V$ as $U(\tilde{\frak g})$-module?

Motivation: These modules are the well known Weyl modules and its irreducible quotients. So, it is very interesting to work with tensor products and so on.

  • 2
    I'm curious about condition 3: How often is a tensor product of finite dimensional irreducibles irreducible? – S. Carnahan Dec 2 '11 at 4:42
  • @Chris: Your formulation is confusing to me, for instance the use of tilde in the first line. And what is meant here by "universal finite dimensional highest weight" modules (presumably for the loop algebra)? What is the source of "well known Weyl modules" here? Etc. – Jim Humphreys Dec 2 '11 at 14:38
  • @Scott: Certainly in characteristic 0 representation theory, it's very uncommon for a tensor product of finite dimensional irreducibles to be irreducible when both factors are nontrivial (though somewhat less so in prime characteristic). I can't immediately decipher the question. – Jim Humphreys Dec 2 '11 at 14:41
  • The modules U, V, and W are not finite dimensional.... – Andy B Dec 2 '11 at 16:12
  • 1
    @Chris: You made it sound like they are finite dimensional was all. – Andy B Dec 2 '11 at 16:42

Your Answer


By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Browse other questions tagged or ask your own question.