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To say more explicitly if we state that L is a Lie algebra:

Show that if IDerL is the set of inner derivations of L, then IDer L is an ideal of Der L

To give some of my thinking, the adjoint map "ad x" maps to the inner derivations, and since an ideal is a subalgebra that maps components from the broader algebra back into that subalgebra, I could take the adjoint mapping and apply it to derivations to show they map back to the inner derivations.

I'm still fairly new to Lie algebras, so any suggestions would be appreciated.

Rusk
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1 Answers1

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The set $\operatorname{Der}(L)$ consisting of derivations of $L$, i.e. linear maps $d: L \to L$ satisfying $d([x,y]) = [d(x),y] + [x,d(y)]$, is a Lie algebra under the commutator $[d_1,d_2] = d_1 \circ d_2 - d_2 \circ d_1$.

The subset $\operatorname{IDer}(L)$ of inner derivations, i.e. derivations of the form $ad_x$ for $x \in L$, where $ad_x(y) = [x,y]$, is an ideal in $\operatorname{Der}(L)$.

Proof. We have to show $[\operatorname{Der}(L),\operatorname{IDer}(L)] \subseteq \operatorname{IDer}(L)$. So let $d \in \operatorname{Der}(L)$ and $ad_x \in \operatorname{IDer}(L)$ be two derivations, and let $y \in L$ be an element to apply their commutator to. Then we have: $$\begin{align*}[d,ad_x](y) &= d(ad_x(y)) - ad_x(d(y)) \\ &= d([x,y]) - [x,d(y)]\\&=([d(x),y] + [x,d(y)]) - [x,d(y)]\\&=[d(x),y]\\&=ad_{d(x)}(y),\end{align*}$$ so $[d,ad_x] = ad_{d(x)}$ is inner, as desired.

Ricardo Buring
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