Suppose $\mathfrak{g}$ is a real Lie algebra integrating to the connected Lie group $G$. One may consider the $G$-equivariant cohomology of $\mathfrak{g}$ ($\mathfrak{g}^*$) where the $G$-action is the adjoint (coadjoint) representation. That is, the cohomology induced by the cochain complex of $G$-invariant differential forms on $\mathfrak{g}$. (Or, more generally, $G$-invariant $V$-valued forms where $V$ is the target of a linear representation of $G$). When $G$ is compact, it follows from a theorem of Chevalley and Eilenberg that this cohomology vanishes in all degrees. I am however interested in the semisimple case. What is known in general about this cohomology and its vanishing? What is known for specific semisimple Lie algebras? Can this cohomology be tied to the standard Lie algebra cohomology?


1 Answer 1


It vanishes as well. The complex of differential forms on $\mathfrak{g}$ is null homotopic through the standard null-homotopy coming from the linear homotopy to the origin. Since the $G$-action preserve this null-homotopy, it is $G$-equivariantly null and so its $G$-invariants subcomplex is null as well.

  • $\begingroup$ Could you please also explain where is semisimplicity used? $\endgroup$ Commented Jul 5, 2020 at 11:55
  • $\begingroup$ it is not used anywhere. Its just the fact that as a $G$-space the representation is a contractible space. $\endgroup$
    – S. carmeli
    Commented Jul 5, 2020 at 12:02
  • $\begingroup$ So it is also true when $G$ is, say, solvable?? $\endgroup$ Commented Jul 5, 2020 at 12:10
  • $\begingroup$ I might be confusing with something else of course, but it seem unconditional to me, its just a contractible complex. Maybe you have another complex then I in mind? $\endgroup$
    – S. carmeli
    Commented Jul 5, 2020 at 12:18
  • 1
    $\begingroup$ Maybe it would help to note that the terms in this complex are not acyclic objects in the category of representations of the group? $\endgroup$
    – S. carmeli
    Commented Jul 5, 2020 at 13:43

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.