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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.
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 $m+1$.
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1
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A positive formula for the dimensions of homogeneous components of free Lie algebras
Now, to the matter of "positive formulas".
Classical result of Kraskiewicz and Weyman (preprint W. Kraskiewicz, J. Weyman. Algebra of Invariants and the Action of a Coxeter Element. Math. Inst. Copern …
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2
votes
A positive formula for the dimensions of homogeneous components of free Lie algebras
For your question about Lie triple systems:
In my article "Veronese powers of operads and pure homotopy algebras" joint with Martin Markl and Elisabeth Remm, Eur. J. Math. 6 (2020), 829-863 (https://l …
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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 …
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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}) …
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2
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Are S(g) and U(g) isomorphic as g-modules for g Lie algebra over F_p ? Are S(g)^g and U(g)^g...
It seems that in some particular cases it is known, see e.g. this recent result.
I was looking at that a while ago, and at least it became clear to me that one has to be very careful with lifting the …
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6
votes
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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 …
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12
votes
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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 …
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14
votes
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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, …
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Whitehead's second Lemma and invariants of exterior square
The way the question is formulated, it is trivial.
If $V\ne\mathfrak{g}$ ($\mathfrak{g}$ here being the adjoint module, in which case you already know everything), the corresponding sum is clearly d …
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2
votes
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Linear independence in (graded) Lie algebras
Let me say that since you are interested in square-free elements where the weight is equal to the number of generators, you actually are asking questions about multilinear elements, that is elements o …
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2
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Is Nijenhuis–Richardson bracket a BV bracket?
There is one silly answer to your question, and you are probably aware of it: if you use as coefficients $S(g)$, the symmetric algebra of $g$, and not just $g$, everything will work wonderfully. Alas …
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0
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Representations of reductive Lie group
Edit: I was too fast with my answer, but I am going to keep it here to possibly prevent others from the misunderstanding that I had. The problem is that you ask about Lie groups in the title of your q …
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Applications of the PBW theorem on enveloping algebras
One immediate corollary is that you get to know the "sizes" of induced representations. If $\mathfrak{h}\subset\mathfrak{g}$ is a Lie subalgebra, and $M$ is an $\mathfrak{h}$-module, then the underlyi …
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7
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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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