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Let $\mathfrak g$ be a Lie algebra over $\mathbb R$. I need to understand the following statement I found in a book (Discrete Subgroups of Lie Groups, Raghunathan p.123): the cohomology $H^p(\mathfrak g,\mathbb R)$ is defined by considering $\mathbb R$ as a module over $\mathfrak g$ via the trivial representation.

So what does that mean? And how do the $\mathfrak g$-submodules of $\mathbb R$ look like?

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  • $\begingroup$ That is not a definition. You can find the definition in textbooks on homological algebra; it involves Ext groups, which you may or may not be familiar with. There are no nontrivial submodules. $\endgroup$ – Qiaochu Yuan Nov 17 '17 at 1:14
  • $\begingroup$ Ok, why there are no nontrivial submodules? $\endgroup$ – Ronald Nov 17 '17 at 1:17
  • $\begingroup$ I mean in this article encyclopediaofmath.org/index.php/Cohomology_of_Lie_algebras for nilpotent Lie algebras case he considered two cases with and without nontrivial submodules!! what is the difference? $\endgroup$ – Ronald Nov 17 '17 at 1:23
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    $\begingroup$ (1) $\mathbb R$ has no nontrivial submodules because it doesn't even have any nontrivial vector subspaces, being $1$-dimensional. (2) The distinction, in the article you linked to, between the cases where $V$ (not necessarily $\mathbb R$) has or doesn't have nontrivial submodules is that one case has lots of nontrivial cohomology and the other case has all cohomology equal to zero. $\endgroup$ – Andreas Blass Nov 17 '17 at 1:59
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    $\begingroup$ No, nilpotent Lie algebras have $\dim H^p(\mathfrak{g},\mathbb{R})\ge 1$. In fact, according to Dixmier all Betti numbers for nilpotent Lie algebras, with the exception of the zeroeth and the highest, are at least two. $\endgroup$ – Dietrich Burde Nov 17 '17 at 20:09

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