0
$\begingroup$

Theorem: Let $L\subseteq \mathfrak{gl}(V)$ be finite dimensional semisimple Lie algebra. Then every $x\in L$ has Jordan decomposition within $L$ (semi-simple + nilpotent).

Part of proof:(Humphreys, p.29) Let $W\subseteq V$ be an $L$-submodule. Define $$L_W :=\{ y\in \mathfrak{gl}(V): y(W)\subseteq W \mbox{ and } Tr(y|_W)=0\}.$$ Since $L$ is semi-simple, $L=[L,L]$.

Thus, $L$ is contained in $L_W$. How?

Here $L=[L,L]$ implies that $L$ is sitting inside $\mathfrak{sl}(V)$, so every element of $L$ has trace $0$ on $V$; but how can it be zero also on any subspace of $V$ which is invariant under it?

  • 0
    But trace zero on $V$, how it will imply that trace zero on $W$? (This may be trivial, but I am confusing.) – 2017-01-10
  • 1
    The point is that not only the commutator leaves $W$ invariant, but so do its constituents - hence the restriction of the commutator to $W$ is a commutator of endomorphisms of $W$, hence has trace $0$ on $W$, too. – 2017-01-10
  • 0
    Sorry, I was not thinking quite right when I commented at first. Hanno has the right hint. – 2017-01-10
  • 0
    @Tobias, Hanno: I will think on these hints (could be trivial, but not clicking immediately to me!) – 2017-01-10
  • 0
    OK; got it. One point I was missing is that **$W$ is $L$-invariant**; so if $x\in L$ and $x=\sum [x_i,y_i]$ for $x_i,y_i\in L$ then $x_i(W)\subset W$ and $y_i(W)\subset W$ and comments of Hanno imply the expected claim. – 2017-01-10

1 Answers 1

1

$L\subset gl(W)=\{M\in gl(V): M(W)\subset W\}$ since $W$ is a submodule. Now use the fact that $[gl(V),gl(V)]=sl(V)=\{M: tr(M)=0\}$, this implies that $[L,L]\subset [gl(V),gl(V)]=sl(V)$, we deduce $L=[L,L]\subset sl(V)$ and $L=\subset sl(V)\cap gl(W)=L_W$.

  • 0
    sorry; $L\subset gl(W)$ is not clear to me a little; some elements of $L$ can act as zero on $W$; sorry, a little more explaination? – 2017-01-10
  • 1
    I define $gl(W)=\{M\in gl(V): M(W)\subset W\}$, since $W$ is a submodule, for every $l\in L, w\in W, l(w)\in W$. Thus $l(W)\subset W$ and $l\in gl(W)$ and this means according to the definition $L\subset gl(W)$. – 2017-01-10