The way I learned Lie algebra cohomology in the context of Lie groups was a direct construction: one defines the Chevalley-Eilenberg complex with coefficients in a vector space $V$ (we assume the real case) as $$C^p(\mathfrak g; V) := \operatorname{Hom}(\bigwedge^p \mathfrak g, V)$$ and explicitly defines the boundary map $\delta^p: C^p(\mathfrak g; V) \to C^{p+1}(\mathfrak g; V)$ by a formula similar to the coordinate-free definition of the de Rham differential. Finally, one takes the homology of this cochain complex. With this approach, the connection between the Chevalley-Eilenberg cohomology and the left-invariant de Rham cohomology is obvious. This was roughly the approach used by Chevalley and Eilenberg themselves

However, Wikipedia and some other sources rather define $$H^n(\mathfrak g; V) = \operatorname{Ext}^n_{U(\mathfrak g)}(\mathbb R, V)$$ where one constructs so-called universal enveloping algebra $U(\mathfrak{g})$, whose motivation isn't clear to me, even though I understand the formal definition. Even when one knows what a derived functor is (which I do), this definition still requires a lot of work, such as the introduction of the universal enveloping algebra, finding the projective resolutions, etc.

At first I thought that the $\operatorname{Ext}$ approach might just be abstract restatement of the same procedure that we carry out while defining the cohomology through the Chevalley-Eilenberg complex, but I don't really see why it should be that way. Well, we take homology of the $\operatorname{Hom}$ complex, but it's where the analogy seems to end because of this universal enveloping algebra, which doesn't have a clear counterpart in the explicit construction.

Is there any advantage to use the second definition of the Lie algebra cohomology? The only reason I could see is the derived functor LES, but it would probably be much easier to show it directly. There's also a clear analogy with the group cohomology defined that way - one just takes the group ring $\mathbb Z[G]$ instead of the universal enveloping algebra $U(\mathfrak{g})$, but group cohomology isn't hard to construct explicitly and the derived functor definition seems so abstract that it's useless.

Maybe my confusion stems from the fact that I learned homological algebra separately in a very abstract setting, roughly following Weibel's book, and while I understood the definitions I don't think I understood the motivations and the big ideas. I asked a similar question on Math.SE, but now I realized that MO is a better place to ask.

  • 7
    $\begingroup$ Essentially all homology and cohomology groups of all kinds of objects can be expressed as derived functors; that is the key unifying principle of (co)homology theory. So we want to express Lie algebra cohomology in that way, to clarify the relationship with other kinds of cohomology. It also explains why Lie algebra cohomology can be computed using any projective resolution, not just the one corresponding to the Chevalley-Eilenberg complex. In many examples there will be ad-hoc choices of resolutions that are much smaller an more convenient than the canonical one. $\endgroup$ Mar 31, 2021 at 20:07
  • 3
    $\begingroup$ If the introduction of the universal enveloping algebra is problematic, can't you just do the derived functor cohomology in the category of representations of $\mathfrak g$? As far as I know, the only reason to introduce $U(\mathfrak g)$ is to connect with the Ext functors for $R$-modules if one is already familiar with those. $\endgroup$
    – Will Sawin
    Mar 31, 2021 at 20:47
  • 3
    $\begingroup$ @WillSawin: Introducing the universal enveloping algebra also makes it clear that projective resolutions exist. It also give you access to things like the Casimir which are important for cohomology stuff but whose nature is kind of mysterious if all you know is the Lie algebra. $\endgroup$ Mar 31, 2021 at 20:57
  • 2
    $\begingroup$ @AndyPutman I think this resolutionless definition of Ext was the original definition and the reason it's called Ext. $\endgroup$ Mar 31, 2021 at 21:16
  • 1
    $\begingroup$ Regarding the motivation for the universal enveloping algebra, you might be interested in some of the comments to my MO question, Motivating the Casimir element. $\endgroup$ Mar 31, 2021 at 22:15

1 Answer 1


What is your motivation for thinking about Lie algebra homology? If all you want to do is compute it for a single fairly explicit Lie algebra, then the Chevalley-Eilenberg complex will do the job. But for actually proving non-homological theorems using Lie algebra homology, it's much better to have the more flexible derived functor definition.

For instance, one of my favorite easy applications is to give very efficient proofs of things like semi-simplicity of finite-dimensional representations of semisimple Lie algebras over fields $k$ of characteristic $0$ (like $\mathbb{R}$). For this, the key fact is that extensions of $U$ by $V$ are classified by $\text{Ext}^1_{U(\mathfrak{g})}(U,V)$, and this can be identified with Lie algebra cohomology via the sequence of isomorphisms $$\text{Ext}^1_{U(\mathfrak{g})}(U,V) = \text{Ext}^1_{U(\mathfrak{g})}(k,\text{Hom}_k(U,V)) \cong H^1(\mathfrak{g};\text{Hom}_k(U,V)).$$ So what we want to do is show that $H^1(\mathfrak{g};W)=0$ for all finite-dimensional $W$ (assuming semisimplicity for $\mathfrak{g}$). You do this in two steps. First, you use the quadratic Casimir element in the universal enveloping algebra to kill $\text{Ext}^1_{U(\mathfrak{g})}(k,W)$ for nontrivial simple $W$ (this is the key use of semisimplicity!), and then you use the fact that $\mathfrak{g}$ has trivial abelianization to see that $H^1(\mathfrak{g};k)=0$, and then finally you deal with arbitrary finite-dimensional $W$ by filtering them so that the associated graded pieces are simple.

If you're comfortable with homological algebra, then the above is a pretty natural proof outline, and it carefully isolates the non-formal part (the existence of the Casimir) from the formal bookkeeping. But if you insisted on defining cohomology using the Chevalley-Eilenberg complex, I would have no idea how to come up with it. Why would cohomology have something to do with extensions?

  • 1
    $\begingroup$ (by the way, I'm not saying that you couldn't write out all of the above using the Chevalley-Eilenberg complex, though it sounds pretty awkward. But certainly the identifications I list are not obvious from that complex, but are trivial if you understand the bigger picture.) $\endgroup$ Mar 31, 2021 at 20:29
  • $\begingroup$ In a nutshell, my motivation for studying Lie algebra cohomology is its applications in mathematical physics, such as arxiv.org/pdf/1705.05854.pdf. I've actually seen the connection between between $H^2({\mathfrak g; \mathfrak a})$ and central extensions proved directly. $\endgroup$
    – marmistrz
    Apr 1, 2021 at 9:39
  • 1
    $\begingroup$ Why is it the case that $\text{Ext}^1_{U(\mathfrak{g})}(U,V) = \text{Ext}^1_{U(\mathfrak{g})}(k,\text{Hom}_k(U,V))$? $\endgroup$
    – marmistrz
    Apr 1, 2021 at 9:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.