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I'm reviewing a paper about Lie algebras for class and I'm finding the following sentence hard to grasp: "It is known and easy to see that if $L = L'$, then $Z(Der(L)) = 0$." where $\mathrm{Der}(L)$ is the subalgebra of $gl(L)$ containing all the derivations on $L$, $Z$ denotes the centre as usual, and $L'$ is the derived algebra $[L, L] \subseteq L$. I'm certainly not finding it easy to see! Here's what I have so far:

Clearly if $T \in Z(\mathrm{Der}(L))$ then $\mathrm{ad}_T D = 0 \,\forall\, D \in \mathrm{Der}(L)$, i.e. $\mathrm{ad}_T D (x) = 0$ for any $x \in L$. Also, if $L = L'$ then there exist $y,z \in L$ s.t. $\forall \, x \in L, x = [y,z]$. Therefore

$0 = \mathrm{ad}_T D (x) = \mathrm{ad}_T D ([y,z]) = [\mathrm{ad}_T D (y), z] + [y, \mathrm{ad}_T D (z)]$

hence $[(\mathrm{ad}_T D) (y), z] = [(\mathrm{ad}_T D) (z), y]$

However this gets me no closer to the conclusion that Z(Der(L)) is the zero map, i.e. that in the above working we must have that $T = 0$.

A Hughes
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1 Answers1

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This is well known for perfect Lie algebras with trivial center. Suppose that $D\in Z(Der(L))$. Then $D$ commutes with all $ad(x)$, i.e., we have $$ 0=[D,ad(x)](y)=D([x,y])-[x,D(y)]=[y,D(x)] $$ for all $x,y \in L$. This implies that $D(x)\in Z(L)$ for all $x\in L$. If $Z(L)=0$ we can conclude that $D(x)=0$ for all $x\in L$, so that $D=0$.

However, a perfect Lie algebra need not have trivial center in general. In that case, I am no longer sure that $Z(Der(L))$ is really trivial - do you have a link of the paper which claims this ?

Remark: $L=L'$ does not imply that every $x\in L$ can be represented as $x=[y,z]$, but only that $x$ is a linear combination of elements $[y_i,z_i]$.

Dietrich Burde
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