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 !

  • 1
    $\begingroup$ 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/…. $\endgroup$ – Dietrich Burde Dec 20 '13 at 9:46

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