1
$\begingroup$

I am taking a course in representation theory of Lie algebras. We are now dealing with semisimple algebras, their nice decomposition into direct sum of simple ideals, the structure of their derivations and we have talked a bit about toral and cartan subalgebras. In the notes we follow there exists the following claim:

Let $\mathfrak{g}$ be a semisimple Lie algebra and $(\rho,V)$ a (finite dimensional) representation of $\mathfrak{g}$. Then $\rho$($\mathfrak{g})\subset$ $\mathfrak{sl}$(V).

There is also a theorem for semisimple algebras and their representations, which we did not prove:

If $(\rho,V)$ is a representation of a semisimple Lie algebra, then the Jordan decomposition of an element $\rho(x)$ is $\rho(x)=\rho(x_{s})+\rho(x_{n})$, where $x_{s}$ and $x_{n}$ are the semisimple and nilpotent part respectively of the decomposition of $x$ in $\mathfrak{g}$.

$\textbf{Note}$:We are working over $\mathbb{C}$.

I was thinking that the claim above might be false or cannot be proven with the tools developed so far in the course.Taking the theorem in mind, the condition that $\rho(x) \in \mathfrak{sl}(V)$, meaning $\operatorname{tr} (\rho(x))=0$, reduces to $\operatorname{tr}(\rho(x_{s}))=0$ since $\rho(x_{n}) $ is a nilpotent operator and hence has zero trace.But how can this be true for all $x\in \mathfrak{g}$? $\rho(x_{s})$ is simply a diagonizable operator on $V$...

Any ideas or enlightments are more than welcome, thanks!!

$\endgroup$
3
  • 2
    $\begingroup$ For the 1st claim all you need is that commutators of matrices are traceless. Therefore $\rho([\mathfrak{g},\mathfrak{g}])\subseteq \mathfrak{sl}(V)$. But $\mathfrak{g}$ is spanned by its commutators, so... The same way Diertich Burde handled the linked question. $\endgroup$ May 17, 2020 at 10:44
  • $\begingroup$ @Jyrki yeaah, I completely forogt about the fact that $[\mathfrak{g},\mathfrak{g}]=\mathfrak{g}$ in the semisimple case..With that in mind, the rest are pretty trivial.. $\endgroup$
    – mits314
    May 17, 2020 at 10:48
  • $\begingroup$ Dietrich Burde ^ Sorry, about lack of proof-reading. $\endgroup$ May 17, 2020 at 10:54

1 Answer 1

3
$\begingroup$

For the first point have a look here.

The second one is the Corollary in Section 6.4 page 30 of Humphreys Introduction to Lie algebras and representation theory 3ed

$\endgroup$
1
  • $\begingroup$ thanks, i ll have a look at it ! $\endgroup$
    – mits314
    May 17, 2020 at 10:39

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .