23

Although the germ of the idea might've appeared in Koszul's earlier work on the cohomology of Lie algebras and homogeneous spaces, it seems that the first full-fledged appearance of the Koszul complex/resolution is in Koszul, Sur un type d'algèbres différentielles en rapport avec la transgression, Colloque de topologie (espaces fibrés), Bruxelles (1950), 73–...


10

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). Hazewinkel's most general result (for not necessarily trivial coefficients) is over a field. A result that he proves over a ring is as follows: Let $R$ be a ...


9

For the Hochschild cohomology, the celebrated Hochschild-Kostant-Rosenberg theorem gives a canonical algebra isomorphism, for a smooth $k$-algebra $A$, $\ HH^*(A,A)\cong \wedge^*T_A$, where $T_A$ is the $A$-module of $k$-derivations of $A$. If $A=\mathbb{C}[t_1,\ldots ,t_p]$ $\ T_A$ is a free $A$-module of rank $p$, so $HH^*(A,A)\cong \wedge^*A^p$.


9

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, is the cohomology of the Lie algebra $\mathfrak{gl}_n$ relative to its subalgebra $\mathfrak{o}_n$, which, by a standard argument, is the same as $H^*(U(n)/O(...


8

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 sequence $$E_2^{p,q}=H^p(\mathbb{Z}/3,H^q((\mathbb{Z}/2)^2,\mathbb{Z}/3))\implies H^{p+q}(A_4,\mathbb{Z}/3)$$ a priori the $E_2$-page involves local coefficient ...


8

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 post for a complete solution of the OP's case. I don't know what you mean exactly by "compute the invariants": perhaps finding an explicit linear basis? ...


8

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 find the whole decomposition into irreducibles of $\Lambda^k S^2 \mathbb{C}^n$ as a special case of plethysm for the Lie group $O(\mathbb{C}^n).$ In fact, this ...


7

The map $\langle\cdot,\cdot\rangle\mapsto\langle\cdot,[\cdot,\cdot]\rangle$ is usually called Koszul homomorphism. Indeed Koszul showed that for a semisimple Lie algebra over a field of characteristic zero it is an isomorphism. For an arbitrary Lie algebra (say over a field of characteristic zero), it always maps into the space of closed 3-forms $Z^3(\...


6

This is rather late, and not a full answer to your question, but may be interesting nonetheless. Concerning point 1), I would like to propose the following natural refinement which avoids the problems mentioned in Pavel Etingof's answer. If you want to know the relation between cohomology of the group and the Lie algebra over $\mathbb{Q}$, you should work ...


6

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 spectral sequence with applications to group cohomology and decompositions of groups., J. Group Theory 9, No. 1, 1-25 (2006). ZBL1115.20042.


6

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 (1970) pp. 638–644. There is a large literature on cohomology of Lie algebras. Here is a very short list of articles and books: C. Chevalley, S. Eilenberg, ...


6

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 given by $\lambda$. For a representation $V$ of $\mathfrak{g}$, let us say that $\lambda$ is a weight of $V$ is $V_\lambda$ is isomorphic to a subquotient of $V$ ...


6

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


5

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. This Lie algebra is not unimodular; if you add that hypothesis, then it should be true over fields at least. Indeed, in that case the enveloping algebra is $n$...


5

No, there are no counter-examples. Note that a generic coadjoint orbit is $G$-equivariantly diffeomorphic to $G/T$, for a maximal torus $T\subseteq G$. However, $G/T$ (also known as the full flag variety of $G_{\mathbb{C}}$) has a cell decomposition into even-dimensional cells (the Bruhat decomposition). Therefore, $G/T$ has vanishing integral cohomology in ...


5

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 isomorphic image $\sigma(L)$ is a direct summand of $U(L)$. Thus the same direct summand property holds for their Lie algebra cohomology. In particular, the induced map ...


5

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


5

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) + (-1)^{n+1} f(a_0,\dots,a_{n-1})a_n + \sum_{i=1}^n (-1)^i f(a_0,\dots,a_{i-1}a_i,\dots,a_n) \end{align*} In particular, $\operatorname{ker} d^1\subset \...


4

The theorems of Borel-Weil and then Bott, along with Kostant's translation of the ideas into the language of Lie algebra cohomology, do much to illuminate classical representation theory (Cartan-Weyl) but probably can't be viewed as a computational tool. As in other situations, cohomological language provides a natural setting for concrete older ideas and ...


4

That was a large part of the subject of my 1964 PhD thesis. While the motivation came from algebraic topology, the relevant algebra was published separately in the paper http://www.math.uchicago.edu/~may/PAPERS/3.pdf. It is very obvious from the case of abelian restricted Lie algebras with zero restriction what the minimal size of a $V(L)$-projective ...


4

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 subgroup of it. If $N$ is not required to be normal then the third term in this fiber sequence no longer exists, so it's unclear to me in what sense we can have a ...


4

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 equivariant isomorphism, you need to restrict to the twisted action by some character).


4

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 homomorphism (arxiv link). Pirashvili's paper is freely accessible here on Numdam; it's also written in terms of homology but the cohomological statement follows.


4

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 definitions). From a geometrical point of view a ``rigid Lie algebra'' is an open subset in the variety of all Lie algebras such that all Lie algebras in this set ...


4

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 results of Delorme in the 70s, and possibly known earlier.) Recall that the homology of $V_\lambda$ is the homology of the Chevalley-Eilenberg complex, with $C_p=\...


4

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 $\mathfrak{n}_n$ is an extension of $\mathbf{C}$ because it has a quotient Lie algebra of dimension 1. It follows that even for nilpotent extensions the ...


3

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 of the tangent bundle of $G$, and re-write all exterior power bundles in terms of this trivialization.) I suppose the geometric content is that the cohomology ...


3

1) Any choice of path (or cylinder) objects will give the same homotopy classes, using the model category structure. Unless M is abelian, these path objects are harder to construct than path objects for chain complexes (exercise). You can't just ignore the coalgebra structure. 2) Hinich's definition of weak equivalences of dg coalgebras was reflected from ...


3

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, depending on what machinery you are inclined to use. While it can be formulated in several related settings, this theorem basically provides a model for the finite ...


3

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 tools. We can very efficiently see that a given Lie algebra $L$ is not nilpotent, by checking first the trace condition, i.e., $tr(ad(x))=0$ for all $x\in L$, ...


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