About derivations of lie algebra

Please, I am looking for a simple example of derivation on ##sl_2##, if possible, I try to use identity map, but not work with me,

A derivation of the Lie algebra ##\mathfrak{g}## is a linear map ##\delta: \mathfrak{g} \rightarrow \mathfrak{g}## such that ##\delta([x, y])=[\delta(x), y]+[x, \delta(y)]##, for all ##x## and ##y## in ##\mathfrak{g}##.Thanks in advance,

dear @fresh_42 , if you could help, I would appreciate it,

HDB1 said:

TL;DR Summary: Derivation on lie algebra ##sl_2##

Please, I am looking for a simple example of derivation on ##sl_2##, if possible, I try to use identity map, but not work with me,

A derivation of the Lie algebra ##\mathfrak{g}## is a linear map ##\delta: \mathfrak{g} \rightarrow \mathfrak{g}## such that ##\delta([x, y])=[\delta(x), y]+[x, \delta(y)]##, for all ##x## and ##y## in ##\mathfrak{g}##.Thanks in advance,

Well, how should I say it?

For every semisimple Lie algebra ##L##, and ##\mathfrak{sl}(2)## is simple and therefore semisimple, all derivations are inner derivations. Inner derivations are all transformations ##\operatorname{ad}X## with ##X\in L.## This is the second Whitehead lemma IIRC and it should be somewhere in Humphreys, too.

So for semisimple Lie algebras, ##\delta =\operatorname{ad}(X)## for some ##X\in L.## Hence ##X \longmapsto (Y\longmapsto \operatorname{ad}(X)(Y)=[X,Y]## are already all derivations you can get for ##sl(2).## And since the center of ##L## which is the kernel of ##\operatorname{ad}## is zero, we even have
$$
L\cong \operatorname{ad}L =\operatorname{Der}(L)
$$
Derivations are basically the Leibniz rule in calculus:
$$
(f\cdot g)' = f' \cdot g+f\cdot g' \Leftrightarrow D(f\cdot g)=D(f)\cdot g+f\cdot D(g) \Leftrightarrow
\delta([f, g])=[\delta(f), g]+[f, \delta(g)]
$$
and the similarity is not by chance!

So if you need a derivation of ##\mathfrak{sl}(2)## then how about
$$
\operatorname{ad}H=\begin{pmatrix}2&0&0\\0&0&0\\0&0&-2\end{pmatrix}
$$

fresh_42 said:

$$
L\cong \operatorname{ad}L =\operatorname{Der}(L)
$$

Thank you, please, could you explain here why we have isomorphism?

Thanks in advance,

HDB1 said:

Thank you, please, could you explain here why we have isomorphism?

Thanks in advance,

We have a surjective transformation ##L\longrightarrow \operatorname{ad}L=\{\operatorname{ad}X\,|\,X\in L\}## per definition. ##\operatorname{ad}## is linear and for the sake of the Jacobi identity also a Lie algebra homomorphism. So all that remains to show is that it is injective, or that ##\ker\operatorname{ad}(L)=\{0\}.## However,
\begin{align*}
\ker\operatorname{ad}(L)&=\{X\in L\,|\,\operatorname{ad}(X)\equiv 0\}=
\{X\in L\,|\,\operatorname{ad}(X)(Y)=0\text{ for all }Y\in L\}\\
&=\{X\in L\,|\,[X,Y]=0\text{ for all }Y\in L\}=Z(L)
\end{align*}

I used to write ##\mathfrak{Z}(L)## as the center of ##L##. I'm used to fraktura that leaves more letters for elements and ordinary sets and is a better distinction if also Lie groups are involved. I think Humphreys writes it ##\mathfrak{Z}(L)=Z(L)=C(L),## one of these.

Anyway, the center is an abelian ideal, hence a solvable ideal, hence contained in the maximal solvable ideal, which is the radical, but the radical of a semisimple Lie algebra is zero by definition of semisimplicity. The center of semisimple ideals is thus the zero ideal.

Thank you so much for your help, @fresh_42 , words can't help me to express how grateful I am to you,