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Good evening everybody,

I have a question about the red tagged argument in Prop. 8.5.1: Let $g$ be a semisimple Lie subalgebra of $gl(V )$.

For $x \in g$ we have J-C decomposion $ x= x_s +x_n$ but that's not clear to me why $ x_s ,x_n \in n_{gl(V)}(g) $ where $ n_{gl(V)}(g) = \{y \in gl(V) | [y,x] \in g $ for every $ x \in g \} $.

My idea: We know that $ x_s = P(x) ,x_n = S(x) $ for appropriate polynomials without constant terms. Thats clear that $x \in n_{gl(V)} $ so because $ g$ is a subspace it would be enough to show that $[z,x^n] \in g $ for every $z \in g$. But how?

user267839
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1 Answers1

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You know that $\mathrm{ad}(x)(\mathfrak{g})\subset\mathfrak{g}$, because after all you have a Lie algebra. Now here is a general fact: if $\psi$ is a linear endomorphism of a vector space $V$ with subspaces $A\subset B\subset V$ and $\psi(A)\subset B$, then each of $A_s$ and $A_n$ map $A$ into $B$. This follows from what you said: $A_s$ and $A_n$ are polynomials in $\psi$ without constant term. For example, if $A_n=\psi^2+2\psi$, then $A_n$ maps $A$ into $B$ because $\psi$ does and $B$ is a subspace.

This argument is the proof of proposition 4.2 (c) which is given in Humphreys' book.

Stefan Dawydiak
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