Let $\mathfrak{g}$ be a semisimple Lie algebra and let $(\pi,V)$ be an irreducible $n \geq 1$ dimensional representation. Then how does the universal enveloping algebra act on the tensor product $V \otimes V$?

For instance, say $\{x_1,x_2,\ldots,x_m\}$ is a vector space basis of $\mathfrak{g}$ and $\omega = x_1 x_2 \in U(\mathfrak{g})$. Further suppose $\{e_1,\ldots,e_n\}$ is a basis of $V$. What is the action $$\omega(e_1 \otimes e_2)?$$

I know the action by the Lie algebra is given by $$x(v \otimes w) = (x.v) \otimes w + v \otimes (x.w).$$

  • $\begingroup$ If you know the action of the Lie algebra, the action of the universal enveloping algebra is determined by multiplication in the obvious way, e.g. $x_1 x_2 (v \otimes w) = x_1 (x_2 (v \otimes w))$ as one expects. $\endgroup$ – Qiaochu Yuan Sep 10 '15 at 20:35
  • $\begingroup$ Argh. I was applying it like $(x_1 x_2 v) \otimes w + v \otimes (x_1 x_2 w)$ and I was confused as to why things were going wrong. Thanks! $\endgroup$ – nigel Sep 10 '15 at 20:45
  • 2
    $\begingroup$ To help you remember in the future, you can think of $x(v \otimes w) = xv \otimes w + v \otimes xw$ as being a variant of the product rule for differentiation. But $x_1 x_2$ doesn't act like differentiation; it acts like differentiating twice, which satisfies a more complicated product rule. $\endgroup$ – Qiaochu Yuan Sep 10 '15 at 20:50
  • 3
    $\begingroup$ And of course this has nothing to do with $\mathfrak{g}$ being semisimple or $V$ being irreducible. $\endgroup$ – Qiaochu Yuan Sep 10 '15 at 21:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.