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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
0
answers
180
views
On an interesting subalgebra of the functions on the cotangent bundle of the flag variety
Setup
Let $G$ be a semisimple algebraic group over a field $k$ of characteristic $p$ where $p = 0$ or $p > 0$ is a good prime for $G$. Fix a Borel subgroup $B \subseteq G$ corresponding to the positi …
1
vote
quasi-minuscule representations
There is a list here.
11
votes
What would you want on a Lie theory cheat poster?
Handy identities for rings associated to Lie groups and algebraic groups, such as the enveloping algebra, hyperalgebra, or quantum groups of various kinds. For example, I often find myself needing to …
3
votes
0
answers
223
views
Generators and relations for the enveloping algebra of a unipotent radical
Let $G$ be a semisimple algebraic group over an algebraically closed field $k$ of characteristic 0 and let $B \subseteq G$ be a Borel subgroup with unipotent radical $U$. Let $P \supseteq B$ be a para …
3
votes
1
answer
557
views
Springer isomorphisms and parabolics
Let $G$ be a semisimple, simply-connected algebraic group over an algebraically closed field $k$ of positive characteristic. Fix a Borel subgroup $B \subseteq G$ with unipotent radical $U$. Also let $ …
24
votes
Accepted
Which is the correct universal enveloping algebra in positive characteristic?
The notions do indeed diverge in positive characteristic: there is the enveloping algebra, and then (in the case that $\mathfrak g$ is the Lie algebra of an algebraic group G) there is also the hypera …
2
votes
Accepted
Does there exist a canonical "degree" filtration on quantum groups?
Nobody has answered this yet, so maybe I'll expand on my comment above, with the caveat that I'm no expert in this area. I believe the answer to your question is yes; the reference for all of this is …
8
votes
Accepted
About $G$-modules versus $Lie(G)$-modules for algebraic groups
I'm sure that there are a number of ways of answering this in the affine case; here's one. Let's assume $G$ affine and let's say that $k$ is our algebraically closed field. The following argument may …
4
votes
subgroups with the same number of roots that the group.
A good modern introduction to the subject is given in Martin Liebeck's survey article "Introduction to the subgroup structure of algebraic groups." It's a chapter in the book "Representations of Reduc …
3
votes
1
answer
121
views
Lower bound on the degree of a product of elements in a hyperalgebra/enveloping algebra
Background:
Fix a linear algebraic group $G$ over an algebraically closed field $k$ of arbitrary characteristic and let $B \subseteq G$ be a Borel subgroup with unipotent radical $N$. Let $\Delta^+$ d …
9
votes
1
answer
879
views
On q-Demazure operators
Setup
Let $G$ be a semisimple algebraic group over an algebraically closed field of arbitrary characteristic with Borel subgroup $B$. Let $\Lambda$ denote the weight lattice of $G$; we write elements …
4
votes
0
answers
203
views
The Killing form on quantized enveloping algebras and reduction to the classical case
Let $U_q$ be the quantized enveloping algebra associated to a semisimple Lie algebra $\mathfrak g$. It is a result due to Tanisaki (see here; also see Chapter 6 of Jantzen's book Lectures on Quantum G …
9
votes
2
answers
1k
views
Relative Lie Algebra cohomology and sheaf cohomology
(I apologize in advance for the vagueness of my question). Let $G$ be a reductive algebraic group over $\mathbb C$ with Lie algebra $\frak g$ and Borel $B$. I have seen casual references to the fact t …
7
votes
2
answers
986
views
Kostant's theorem on invariant polynomials in positive characteristic
Let $k$ be an algebraically closed field and let $G$ be a reductive linear algebraic group over $k$ with Lie algebra $\mathfrak g$. If the characteristic of $k$ is $0$ then, by a classical result of K …