$ch(M\otimes N)=chMchN$

Question

Let L be a semi-simple Lie algebra and M be a finite-dimensional L-module. $M=\oplus _{\lambda \in H^*}M_{\lambda}$ be the weight space decomposition of M.

Define $chM=\sum_{\lambda \in H^*} dimM_{\lambda}e^{\lambda}$.

I am trying to prove $ch(M\otimes N)=chMchN.$

I intended to write the weight space decomposition of $M\otimes N$, then calculate $ch(M\otimes N)$.

But there are two problems.

1. I don't know what the weight space decomposition of $M\otimes N$ is like.

2. I think it would be a complicated solution. In Serre's book, it says the conclusion is obvious. Is there a better solution?

Answer 1

Let $$M=\oplus_{\lambda \in H^*} M_{\lambda},\ N=\oplus_{\mu \in H^*} N_\mu.$$

We have $$M\otimes N= \oplus_{\lambda ,\mu \in H^*}(M_{\lambda} \otimes N_{\mu}) \subset \oplus_{\lambda , \mu \in H^*}(M\otimes N)_{\lambda + \mu}$$

Hence, $$ch(M\otimes N)=\sum\limits_{\lambda , \mu \in H^*} dim M_\lambda dim N_\mu e^{\lambda + \mu}=(\sum\limits_{\lambda \in H^*} dim M_\lambda e^\lambda)(\sum\limits_{\mu \in H^*} dim N_\mu e^\mu)=chMchN.$$