MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $\mathfrak{L}_n(\mathbf{C})$ be the set of all $n$-dimensional complex Lie algebras. One know that in $\mathfrak{L}_n(\mathbf{C})$ there exist only a finite number of Lie algebras with trivial second cohomology group with coefficients in the adjoint representation (by using a rigidity argument [Nijenhuis, Richardson]).

Question: Fixing $n$, are Lie algebras $\mathfrak{g} \in \mathfrak{L}_n(\mathbf{C})$ such that $H^2(\mathfrak{g},\mathfrak{g})=0$ classified?

[Nijenhuis, Richardson] Nijenhuis, Richardson - Deformations of Lie algebra structures

[Nijenhuis, Richardson 1] Nijenhuis, Richardson - Cohomology and deformations in Graded Lie algebras

[Nijenhuis, Richardson 2] Nijenhuis, Richardson - Cohomology and deformations of algebraic structure

share|cite|improve this question
I edited your latex - I guessed \mathds{C} meant to indicate the complex numbers, so substituted \mathbf{C}. – David Roberts Jun 14 '11 at 3:02
Could one equivalently ask "Of the finitely many components of the scheme of Lie brackets on ${\bf C}^n$, how many have an open set of Lie algebras with $H^2=0$, and how many of those are not pairwise isomorphic?" – Allen Knutson Jun 14 '11 at 4:00
No, such algebras were not classified and I agree with Nicola Ciccoli that this seems to be a hopeless task. There are lot of works, though, devoted to such algebras (often coupled with some other conditions), see, for example, references in arXiv:math/0211416. – Pasha Zusmanovich Jun 19 '11 at 5:35

Classifying them for any $n$ seems to me an hopeless task, as every classification problem in Lie algebras that extends out of semisemplicity. However some classification results for rigid Lie algebras in dimension up to 8 (also something in dim. 9) were obtained, mainly by Ancochea-Bermudez and Goze. Lie algebras such that $H^2(\mathfrak g,\mathfrak g)=0$ are also called cohomologically (or sometimes algebraically) rigid and form a subclass of rigid Lie algebras; in the papers by the authors mentioned above it is usually specified which rigid Lie algebras are also cohomologically rigid.


Ancochea-Bermudez, Goze On the classification of rigid Lie algebras Journal of Algebra 245, 68-91 (2001).

Results are based on some structural properties of rigid Lie algebras that may be found in works by Roger Carles published in the 80's (I can provide the exact reference if needed).

share|cite|improve this answer

This is merely an additional comment, too long for a comment field. To see that classification of finite-dimensional Lie algebras $L$ with $H^2(L,L) = 0$ is unfeasible, consider, for example, Lie algebras of the form $g \otimes A$, for a classical simple $g$ and finite-dimensional associative commutative algebra $A$. By (cohomological interpretation of) the result of J.-L. Cathelineau, Homologie de degre trois d'algebres de Lie simple deployees etendues a une algebre commutative, Enseign. Math. 33 (1987), 159-173 , $H^2(g\otimes A, g\otimes A)$ is isomorphic either to Hochschild of dihedral cohomology of $A$, depending on the type of $g$. So, the purported classification would include, as a very particular case, classification of associative commutative algebras with the vanishing second Hochschild cohomology. By modification of this tensor product construction, one possible to get other variants of this classification problem which seem to be out of reach.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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