In its most general form, I look at the following. Let $g$ be a dg Lie algebra and $Z$ be its center. The sequence $Z \to g \to g/Z$ gives me a short exact

$0\to C(g,Z) \to C(g,g) \to C(g,g/Z) \to 0$

of Chevalley-Eilenberg cohomology complexes. Are there any known standard tricks associated to computations with the corresponding long exact in cohomology? Any conditions that ensure the induced maps $H(g,g)\to H(g,g/Z)$ in the long exact sequence are isomorphisms (apart from $H(g,Z)=0$)? In the particular case I look at $g$ has a filtration

$g=F^0g \supset F^1g \supset \dots$,

the bracket maps $F^rg\otimes F^sg$ into $F^{r+s-1}g$ and $Z=F^0g/F^1g$ is one-dimensional. Apart from that my $g$ does not have any nice properties that I know of; in particular, it is not even finite-dimensional.

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Johan, the homomorphisms you indicate are isos if and only if the remaining cohomology is zero. – Fernando Muro Oct 8 '13 at 9:54
I meant to emphasize that I was looking for a "trick", rather than having to compute, e.g., $H(g,Z)$ directly. – Johan Alm Oct 9 '13 at 8:02