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Results tagged with lie-algebras
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user 1306
Lie algebras are algebraic structures which were introduced to study the concept of infinitesimal transformations. The term "Lie algebra" (after Sophus Lie) was introduced by Hermann Weyl in the 1930s. In older texts, the name "infinitesimal group" is used. Related mathematical concepts include Lie groups and differentiable manifolds.
14
votes
Accepted
Uncle of Witt algebra
Interesting/uninteresting is a very subjective thing, so let me try to just say several things that I see immediately.
0) This algebra, unlike the Witt algebra, does not have any [obvious] grading, …
- 16.9k
12
votes
Accepted
Counting degrees of freedom in Lie algebra structure constants (aka why are there any nontri...
Linear independence does not really say much.
This algebraic variety is discussed in some detail in an old paper of Kirillov and Neretin: The variety $A_n$ of $n$-dimensional Lie algebra structures.
T …
- 16.9k
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).
…
- 16.9k
10
votes
Accepted
Hopf algebra structure on the universal enveloping algebra of a Leibniz algebra?
Do you know the paper of Loday and Pirashvili? They discuss what, in their opinion, should replace the notion of a Hopf algebra in Leibniz setting, "Hopf algebras in the category of linear maps".
- 16.9k
10
votes
Accepted
Breaking up the free Lie algebra into GL irreps
The Whitehouse module referred to in one of the other answers is not necessary, since it is related to the cyclic operad Lie, that is to the representation of $S_{n+1}$ in $Lie(n)$.
The decomposition …
- 16.9k
10
votes
Accepted
The dimension of the Grassmannian cohomology ring $H^*(\mathrm{Gr}_{n,d})$ and the fundament...
This is very standard. For a compact complex variety admitting a cell decomposition, the (co)homology is the free Abelian group generated by the cells (over $\mathbb{C}$ there is no room for the diffe …
- 16.9k
10
votes
Accepted
Reference for an old result of P. M. Cohn
This is in P. M. Cohn, "On the Embedding of Rings in Skew Fields", Proceedings of the London Mathematical Society, Volume s3-11, Issue 1 (1961), Pages 511-530. I do not think that the zero characteris …
- 16.9k
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 …
- 16.9k
7
votes
factorization of the regular representation of the symmetric group
The fact that the Lie module (as proposed by Darij Grinberg) works, as well, as an explicit isomorphism of modules, follows from the theory of cyclic operads: see Corollary 6.9 in http://sites.math.no …
- 16.9k
7
votes
Three-dimensional simple Lie algebras over the rationals
Disclaimer: This answer is mostly an extended comment coming from my attempt to understand the answer of BR. However, the time I invested in it made me think that someone else would find it useful. Es …
- 16.9k
7
votes
Accepted
surjectivity of irreducible representation
What's the ground field? Of course if it's $\mathbb{R}$ and $A=\mathbb{R}[t]/(t^2+1)$, then the regular module is irreducible, but the corresponding $p$ is not surjective. Over an algebraically closed …
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6
votes
Accepted
Nilpotent Lie algebras of vector fields
I am afraid not: the subalgebra of $W_2$ spanned by $\partial_x, \partial_y, x\partial_y, \ldots, x^m\partial_y$ has the nilpotency index $m+1$.
- 16.9k
6
votes
Accepted
Lie algebra admitting some hyperbolic automorphism is nilpotent
First, if $x$ and $y$ are generalised eigenvectors of $\phi$ with generalised eigenvalues $\alpha$ and $\beta$, that is $(\phi-\alpha\mathrm{Id}_\mathfrak{g})^N(x)=(\phi-\beta\mathrm{Id}_\mathfrak{g}) …
- 16.9k
6
votes
Accepted
CE(g) for g infinite dimensional
A definition that always works and does agree with that one in the finite-dimensional case is the following: put
$$
C^k(\mathfrak{g})=({\Lambda}^k\mathfrak{g})^*=\operatorname{Hom}({\Lambda}^k\mathfra …
- 16.9k
5
votes
Whitehead lemmas in Lie algebra cohomology for non-algebraically closed fields
Cohomology does not change under field extensions, so just extend everything to the algebraic closure to prove your result in char 0 in general. Of course, field extensions do not preserve irreducibil …
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