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Is there any comprehensive list of examples for computed

1) de-Rham cohomology-groups

2) Lie-algebra-cohomology groups $H^i(\mathfrak{g},\mathbb{R})$

3) equivariant de-Rham cohomology groups ?

Especially I am interested in a comprehensive list of examples (or classes of examples) of

a) manifolds with vanishing first and second de-Rham-cohomology

b) lie-algebras with vanishing first and second lie-algebra-cohomology

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I believe that all semi-simple Lie algebras (in char. 0) have vanishing first and second lie-algebra cohomology. (The first cohomology measures exts. of the trivial rep'n by itself, which must split because of semi-simplicity. The second cohomology measures central extensions of the Lie algebra, which must split because of the existence of Levi decompositions.) – Emerton Jul 25 '10 at 20:13
@Emerton, indeed: those are the two Whitehead lemmas (the purely cohomological arguments can be found in Hilton-Stammbach, for example) – Mariano Suárez-Alvarez Jul 25 '10 at 21:30
I do not understand what is the point of the question? On one hand, the first part is rather different to the second one (which might deserve its own question) On the other hand, the first part is rather too wide: you probably are interested in something more specific that "tell me which cohomologies of which objects have been computed?" – Mariano Suárez-Alvarez Jul 25 '10 at 21:35
Seconding Mariano's comment. For a start, by taking Cartesian products of manifolds one gets (by Kunneth) lots of different possible cohomology groups, but this is somehow an uninteristing "list". As it stands, the first part of the question creates an unfortunate impression of "I want to know about X. Tell me all about X." – Yemon Choi Jul 25 '10 at 23:20
this is also an cool resource – Sean Tilson Jul 26 '10 at 6:07

Chevalley and Borel calculated the cohomology of many Lie groups and Landweber collected those results here in just two pages: it describes the cohomology of $U(n)$, $SU(n)$, $SO(n)$, $G_2$, $F_4$, $E_6$, $E_7$ and $E_8$.

You might also enjoy Neil Strickland's bestiary of topological spaces and spectra, which lists the (co)homology of many topological spaces.

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You might be interested in the Equivariant cohomology wiki, which has a table of which cohomology theories have been computed for which symmetric spaces.

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Consolidating the previous comments, one can start by understanding the topology of a compact simple group. Here is an example: the Lie group Sp(3) of dimension 21. It has rank 3 and the same Betti numbers as the product S^3 x S^7 x S^11 of spheres of dimension 2i+1; the numbers i (here 1,3,5) are the exponents of Sp(3). This theory dates back to Hopf, and was nicely explained by Bott using Morse theory. The (trivial) Lie algebra cohomology ring is generated by the associated forms in dimension 3,7,11. All this theory can be extended to compact irreducible symmetric spaces G/H, which (if not Hermitian) will also have b_1=b_2=0.

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You have to be a little bit more specific in your question (b), since Lie algebra cohomology groups are defined with respect to a module.

Indeed, there is a cohomological criterion for semisimplicity of (real) Lie algebras, which says that a (real) Lie algebra $\mathfrak{g}$ is semisimple if and only if for any finite-dimensional $\mathfrak{g}$-module $\mathfrak{M}$, $H^1(\mathfrak{g},\mathfrak{M})=0$.

This would seem to answer your question (b) if you mean vanishing in this strong sense. If, on the other hand, you mean vanishing of $H^i(\mathfrak{g},\mathbb{R})$ for $i=1,2$, with $\mathbb{R}$ the trivial one-dimensional module, then I am not aware of any very general results. The vanishing of $H^1(\mathfrak{g},\mathbb{R})$ says that $[\mathfrak{g},\mathfrak{g}] = \mathfrak{g}$ and such Lie algebras are called perfect. Such algebras cannot be solvable, but there are non-semisimple examples.

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I took question (b) to mean Lie alg. cohom. with trivial coeffs. – Emerton Jul 25 '10 at 20:16
The criterion in your 2nd paragraph works for all fin.dim. Lie algebras over all fields: the condition that $H^1(\mathfrak g,\mathord-)=0$ identically means that the trivial $\mathfrak g$-module is projective in the category of finite dimensional modules; now pick a $\mathfrak g$-module $D$ and a short exact sequence $0\to A\to B\to C\to 0$ of $\mathfrak g$-modules: appling the functor $\hom_k(D,\mathord-)$ (this is the $\hom$ as vector spaces!), you get another short exact sequence $0\to\hom_k(D,A)\to\hom_k(D,B)\to\hom_k(D,C)\to0$ of $\mathfrak g$-modules, – Mariano Suárez-Alvarez Jul 25 '10 at 22:49
(continued) and if you apply now the functor $\hom_{\mathfrak g}(k,\mathord-)$ and do the identification $\hom_{\mathfrak g}(k,\hom_k(\mathord -,\mathord-))=\hom_{\mathfrak g}(\mathord-,\mathord-)$, you see that $0\to\hom_{\mathfrak g}(D,A)\to\hom_{\mathfrak g}(D,B)\to\hom_{\mathfrak g}(D,C)\to0$ is exact: it follows that $D$ is projective, and, since $D$ was arbitrary, that $\mathfrak g$ is semisimple. – Mariano Suárez-Alvarez Jul 25 '10 at 22:49
Thanks, Mariano. I have only ever worked over $\mathbb{R}$ or $\mathbb{C}$ and hence I've learnt to be careful here in MO, to be careful to state the ground field before I make a general statement. I'll remember that this particular criterion works in general! – José Figueroa-O'Farrill Jul 25 '10 at 23:45

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