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Let $L$ be a central extension of a simple Lie algebra $\mathfrak{g}$ such that $L=[L,L]$. It is not difficult to see that if $H^1(\mathfrak{g}, \mathfrak{g})=0$ then $H^1(L,L)=0$. In other words, if all derivations of $\mathfrak{g}$ are inner, then all derivations of $L$ are inner.

Is the converse true?

For instance, this is indeed the case when $L$ is the universal central extension of $\mathfrak{g}$. (See Theorem 2.2 of [G.M. Benkart - R.V. Moody: Derivations, central extensions, and affine Lie algebras. Algebras Groups Geom. ${\bf 3}$ (1986), no. 4, 456--492.])

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  • $\begingroup$ Maybe useful to recall that $H^1(\mathfrak{h},\mathfrak{h})=0$ means that every self-derivation of $\mathfrak{h}$ is inner. $\endgroup$
    – YCor
    Commented May 27, 2019 at 13:39
  • $\begingroup$ @Yves, I agree. The problem is just that a non-inner derivation of $\mathfrak{g}$ is not necessarily induced by a derivation of $L$. I would expect that the answer to my question is negative, however I have not a counterexample. $\endgroup$
    – Salvatore Siciliano
    Commented May 29, 2019 at 9:01
  • $\begingroup$ Yes, I got this. My idea was to have a simple Lie algebra with a sufficiently big $H_2$, and a small but nontrivial Out, and mod out its universal central extension by a "random enough" subspace. Unfortunately my knowledge of simple Lie algebras of infinite dimension or in finite characteristic is too narrow to think of such examples. $\endgroup$
    – YCor
    Commented May 29, 2019 at 9:06

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I don't known if I interprete correctly your question. First, if $g$ is simple, then always $H^1(g,g)=0$. ALSO, $H^1(L,L)=0$ do not implies that $L$ is a central extensión of a simple algebra.For example, take $L$=the non abelian 2-dimensional Lie algebra, you can compute by hand Der(L,L) and check that they are all inner. (Same happens for borel subalgebras of semisimple ones).

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    $\begingroup$ Ciao Marco, in fact you are interpreting uncorrectly my question...First of all, I am not assuming that the simple Lie algebra $\mathfrak{g}$ is finite-dimensional and defined over a field of characteristic zero, so it is not true in general that $H^1(\mathfrak{g},\mathfrak{g} )=0$. Second, $L$ is by hypothesis a central extension of a simple Lie algebra because this is just the specific situation I am interested in. $\endgroup$
    – Salvatore Siciliano
    Commented May 30, 2019 at 12:59

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