10 votes
Accepted

Poincaré duality for (co)homology of Lie algebras?

First, let me expand on the reply of Dietrich Burde: I got hold of the paper of Hazewinkel, and can now be more precise about what is and what is not there (last time I saw it was some years ago). ...
10 votes

What's the advantage of defining Lie algebra cohomology using derived functors?

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. ...
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9 votes

Invariants of exterior powers

To offer a slightly more geometric viewpoint on the same, the space $\bigoplus_q \mathop{\mathrm{Hom}}_K(\Lambda^q(\mathfrak{p}),\mathbb{C})$, which is the direct sum of all spaces you are considering,...
8 votes

Invariants of exterior powers

You are basically looking for trivial $O(\mathbb{C}^n)$ representations contained in $\Lambda^k \mathfrak{p}^*.$ Since your representation $\mathfrak{p}$ is the space of symmetric matrices, you can ...
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8 votes

Invariants of exterior powers

Edit: I misread the question and understood $\mathfrak{p}$ as corresponding to antisymetric instead of symmetric matrices. So what immediately follows answers a different question. See the end of this ...
8 votes

mod p cohomology ring of alternating groups

Regarding your first question, you have a group extension $$1\rightarrow (\mathbb{Z}/2)^2\rightarrow A_4\rightarrow \mathbb{Z}/3\rightarrow 1.$$ Then we can use the Lyndon-Hochschild-Serre spectral ...
  • 9,520
8 votes

Lyndon–Hochschild–Serre spectral sequence for a non-normal subgroup

Sorry for reviving an old question, but it seems that the Kropholler spectral sequence exactly answers the first 3 questions: Kropholler, P.H., A generalization of the Lyndon-Hochschild-Serre ...
  • 33.5k
8 votes
Accepted

Chevalley complex and $\text{BG}$

It’s the ring of functions on the formal completion at the base point: CE is self ext of the trivial rep of the Lie algebra, equivalently of the formal group of G, ie global functions on B of the ...
8 votes
Accepted

Infinite dimensional Lie algebras with trivial homology

There exist acyclic infinite-dimensional Lie algebras, i.e., for which the trivial homology vanishes in all nonzero degree (recall that $H_0$ of every Lie algebra is always 1-dimensional over the ...
  • 54k
6 votes

Poincaré duality for (co)homology of Lie algebras?

As far as I know, a generalisation of Poincare duality for Lie algebra cohomology over rings is given in M. Hazewinkel, "A duality theorem for the cohomology of Lie algebras" Math. USSR-Sb. , 12 (...
6 votes
Accepted

What is known about the morphism $H^*_{Lie}(L,L)\to H^*_{Lie}(L,UL)$ induced by $L\hookrightarrow UL$

The symmetrization mapping $\sigma$ from the symmetric algebra $S(L)$ to the universal enveloping algebra $U(L)$ is an isomorphism of $L$-modules. Since $L$ is a direct summand of $S(L)$, its ...
6 votes

Homology of solvable (nilpotent) Lie algebras

Let $K$ be a field. Let $\mathfrak{g}$ be a Lie algebra over $K$. Let $\lambda:\mathfrak{g}\to K$ be a homomorphism. Let $V_\lambda$ be $K$ endowed with the structure of right $\mathfrak{g}$-module ...
  • 54k
6 votes

Is the Chevalley-Eilenberg cohomology the only interesting cohomology for Lie algebra?

No, it's not the only one. For instance, Leibniz cohomology is interesting for Lie algebras themselves. See this answer of mine for references.
  • 54k
5 votes
Accepted

Is this sequence of Lie algebra cohomology a part of spectral sequence?

It's part of the Pirashvili exact sequence, relating Lie algebra and Leibniz cohomologies of Lie algebras. This is discussed (in homology terms) in the end of p2 of this paper of mine on Koszul's ...
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5 votes

cohomological representations of GL(N)

Yes. (That's assuming your definition of “representation” gives at least a $(\mathfrak g, K_\infty)$-module.)
5 votes

Poincaré duality for (co)homology of Lie algebras?

Let $g$ be the $3$-dimensional complex Lie algebra with basis $\{x,y,z\}$ and $[x,y]=y$, $[x,z]=y+z$ and $[y,z]=0$. Then $H_0(g,\mathbb C)=\mathbb C$ and $H^3(g,\mathbb C)=0$, if I computed correctly. ...
5 votes

Derivations of universal enveloping algebra of Lie algebras

Associated to any associative algebra $A$ is the Hochschild cochain complex \begin{align*} HH^n(A) &= \operatorname{Hom}(A^{\otimes n},A),\\ \mathrm d f(a_0,\dots,a_n) &= a_0f(a_1,\dots,a_n) +...
4 votes

Lyndon–Hochschild–Serre spectral sequence for a non-normal subgroup

I don't think so. The LHS spectral sequence can be thought of as the Serre spectral sequence associated to the fiber sequence $$BN \to BG \to B(G/N)$$ where $G$ is a group and $N$ is a normal ...
4 votes
Accepted

Matsushima-Murakami Isomorphism for $L^2$-cohomology

Morally yes, by Franke, Jens Harmonic analysis in weighted $L_{2}$-spaces. Ann. Sci. École Norm. Sup. (4) 31 (1998), no. 2, 181--279 See especially theorem 4 (and note that if you want an ...
  • 9,640
4 votes

Cohomology of Infinite Dimensional Lie Algebra

A very useful source to learn about cohomology of infinite-dimensional Lie algebras is the book by D.B.Fuks "Cohomology of Infinite-Dimensional Lie Algebras" (shocking, I know). This source discusses ...
4 votes
Accepted

Lie algebras : Deformations and Rigidity

Let me consider the finite-dimensional case (as YCOR was commenting it is basically impossible to say something in the infinite-dimensional one unless you clarify in a more detailed manner your ...
4 votes
Accepted

Homology of solvable (nilpotent) Lie algebras

Let me provide an elementary proof that for $\mathfrak{g}$ nilpotent finite-dimensional and $\lambda\neq 0$ we have $H_*(\mathfrak{g},V_\lambda)=0$. (In general it seems to be particular case of ...
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4 votes

Universal central extension of Lie algebras

In my opinion, it's probably not so interesting to consider all extensions. If you consider the Lie subalgebra $\mathfrak{n}_n\subset \mathfrak{gl}_n$ of all strictly upper triangular matrices, then $...
4 votes
Accepted

Spectral sequence from standard/Verma filtration/flag to compute Lie algebra cohomology of tensor product with respect to $\mathfrak{n}$

I think the issue here is that the subquotients in the standard filtration have weights (your $\lambda_i$'s) which are ordered the other way. To be clear, if the weights of $\nu_0$, ..., $\nu_n$ of $L$...
3 votes
Accepted

Simple identity on Lie algebras in a note of Koszul

Are you sure about the factor of $3$ in Koszul's formula? The computation below does not give such a factor, and it gives a stronger result. Let $X_i$ be any basis of the left-invariant vector ...
3 votes

Nilpotency of Lie Algebra from Structure Constants

The book of Willem de Graaf "Lie Algebras: Theory and Algorithms" contains some algorithms for determining whether or not a Lie algebra $L$ is nilpotent. Of course, Engel's theorem is one of the main ...
3 votes

What's the most simple proof of Kostant's version of Borel-Weil-Bott for Lie Algebra cohomology?

I'm not sure exactly what your header means, but maybe I can suggest partial answers to your questions. First of all, there are by now many ways to approach the original Borel-Weil theorem, ...
3 votes

What's the most simple proof of Kostant's version of Borel-Weil-Bott for Lie Algebra cohomology?

See page 351 of Andreas Cap, Jan Slovak: Parabolic Geometries I: Background and General Theory Mathematical Surveys and Monographs 154, Amer. Math. Soc. 2009, 628 pp. This includes the parabolic ...
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3 votes

What are cohomology of Lie algebra with coefficients geometrically?

Examining the complex used to calculate $H^*(\mathfrak{g};\mathbb{R})$, one sees that it is precisely the complex of $G$-invariant differential forms on $G$. (To see this, take the left-trivialization ...
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3 votes

Homology of solvable (nilpotent) Lie algebras

To fill out my comment with a partial answer to the question. Note that given any (left) $\mathfrak{g}$-module $V$ one can compute $H_i(\mathfrak{g},V)$ as $\mathrm{Tor}_i^{U(\mathfrak{g})}(\mathbb{C}...

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