# Lie algebra homomorphism preserves Jordan form

Fact : $\phi : L_1\rightarrow L_2$ is $surjective$ Lie algebra homomorphism. If $h\in L_1$ and ${\rm ad}_h$ is diagonalizable then ${\rm ad}_{\phi(h)}$ is diagonalizable

Defn $x\in {\rm gl}\ ({\bf C}^n)$ has Jordan decomposition if $$x= d+n$$ where $[d,n]=0$, $d$ is diagonalizable and $n$ is nilpotent.

EXE (cf. 90page in Erdmann and Wildon's book) If under the same assumption in the above fact $L_i$ are complex semisimples and if $$x=d+n$$ is Jordan, then we have Jordan $$\varphi x= \varphi d + \varphi n$$

Proof : Commutativeness : $$[\phi d, \phi n] = \phi [dn]=0$$

Question : I cannot prove "Diagonalizability of $\varphi d$ and nilpotency of $\varphi n$"

My try : ${\rm ad}_x$ has Jordan That is $${\rm ad}_x = {\rm ad}_{d'} + {\rm ad}_{n'},\ d',\ n'\in L_1$$ Note that we have Lie algebra homomorphism $$\Phi : {\rm ad} \ L_1 \rightarrow {\rm ad}\ L_2,\ \Phi\ ({\rm ad}_y) = {\rm ad}_{\phi y}$$ and $${\rm ad}_{\phi x} = {\rm ad}_{\phi d'} + {\rm ad}_{\phi n'}$$ where $[{\rm ad}_{\phi d'}, {\rm ad}_{\phi n'}]=0$, and ${\rm ad}_{\phi d'}$ is diagonalizable.

And note that ${\rm ad}_{\varphi n'}$ is nilpotent : Since $$0= \varphi (0) = \varphi [ ({\rm ad}_{n'})^k(y) ] = \varphi [ n',\ \cdots, [n',y]\cdots]=[\varphi n',\cdots, [ \varphi n', \varphi y]\cdots ] = ({\rm ad}_{\varphi n'})^k (\varphi y)$$ ${\rm ad}_{\varphi n'}$ is nilpotent.

My try is right ? I need helpful comment Thank you !

• You have also to show that $\phi (d)$ and $\phi(n)$ are again in $L_2$. Here you need that $L_2$ is semisimple, over an algebraically closed field of characteristic zero. See also math.stackexchange.com/questions/487384/…. – Dietrich Burde Dec 20 '13 at 9:46