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Results tagged with lie-algebras
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user 703
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.
4
votes
How about the Lie algebra over commutative ring?
One interesting fact about Lie algebras over commutative rings is that the PBW theorem can fail. I highly recommend the paper A remark on the Birkhoff-Witt theorem by P.M. Cohn. The fact that PBW ca …
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3
votes
Accepted
Almost-Lie Algebras?
See Section 2.3 of the lecture notes called Geometric Models for Noncommutative Algebras by Ana Cannas da Silva and Alan Weinstein. There they define an "almost Lie algebra" to be something with an a …
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8
votes
2
answers
2k
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BGG resolution and representations of parabolic subalgebras
Everything here is over $\mathbb{C}$.
Let $\mathfrak{g}$ be a finite-dimensional simple Lie algebra and let $\mathfrak{p}$ be a parabolic subalgebra (relative to some fixed Borel subalgebra that is un …
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10
votes
Lie algebras over non-algebraically closed fields
Well, certainly things get more complicated when the field is not algebraically closed, as you can see from the classification of finite-dimensional simple Lie algebras over $\mathbb{R}$. But there a …
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7
votes
Accepted
Compute formal character of semisimple Lie algebras.
The software package LiE is good for this. It is no longer maintained, but it has a lot of functionality, and the documentation is good. There is an online demonstration here. There is a bit of a l …
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5
votes
Clifford Lie algebras
A little bit of what you want can be found in Chapter 5 of Gracia-Bondia, Varilly, and Figueroa's book Elements of Noncommutative Geometry. They don't say much about subalgebras, I think, but they do …
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11
votes
Is a retract of a free object free?
The answer is yes in the category of groups. Suppose that $f: G \to H$ is a retraction with $G$ a free group. Then there is a homomorphism $g: H \to G$ such that $fg = \mathrm{id}_H$.
Thus $g$ is in …
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3
votes
Accepted
How can one find generators of basic differential forms on homogeneous spaces?
This actually can be done in much greater generality.
Let $G$ be a compact group and $K \subseteq G$ a closed subgroup.
Then for any finite-dimensional representation $(V,\pi)$ of $K$ you can form the …
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14
votes
Accepted
Relationship between "different" quantum deformations
There is certainly a way to quantize the algebra of functions on a Lie group in a way that is compatible with the $q$-deformation of the universal enveloping algebra of its Lie algebra. The standard …
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8
votes
Accepted
Criterion for nilradical of a maximal parabolic subalgebra to be abelian?
Denote by $\mathfrak{l}$ the Levi factor of the parabolic, so that $\mathfrak{p} = \mathfrak{l} \oplus \mathfrak{n}$, and note that this is a splitting as $\mathfrak{l}$-modules. Also denote by $\mat …
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