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Results tagged with lie-algebras
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user 391
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.
1
vote
character formula for demazure modules
Littelmann's, which gives a positive formula (counting Littelmann paths). His ICM address is here: http://www.mathunion.org/ICM/ICM1994.1/Main/icm1994.1.0298.0308.ocr.pdf
He proves its validity using …
- 27.9k
7
votes
Outer automorphisms of simple Lie Algebras
In characteristic 2, $B_2$ and $F_4$ each have an outer automorphism, and in characteristic 3, so does $G_2$. This is relevant when you want to construct the twisted Chevalley groups, which use a Chev …
- 27.9k
3
votes
Matrices of Lie algebra of Dynkin diagram B2
While I'm not sure this question is appropriate for this site, here goes.
First, you need a maximal torus. Inside SO(5) we have SO(4) and then SO(2) x SO(2).
Write your antisymmetric matrices as
$$\be …
- 27.9k
35
votes
Accepted
Figure out the roots from the Dynkin diagram
Here's an answer in the simply-laced case. Its proof, and generalization to non-simply-laced, are left to the reader.
1) Start with a simple root, and think of it as a labeling of the Dynkin diagram …
- 27.9k
7
votes
Accepted
How to write down the map $T(V)_n \to S(Lie(V))_n$ explicitly?
The natural map is rather from $TV \to U(FreeLie(V))$: consider the forgetful functors $Assoc \to Lie \to Vec$ and compose their left adjoints to get the left adjoint $T$ of the composite. Then, as Al …
- 27.9k
17
votes
What is significant about the half-sum of positive roots?
While I appreciate Dave Ben-Zvi's half-densities answer, I'm going to
put forth the contrary opinion that it's largely a bookkeeping artifact.
The most familiar place that $\rho$ shows up is in the W …
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22
votes
1
answer
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Modern reference for maximal connected subgroups of compact Lie groups
What's the nicest place to see a list of the maximal connected subgroups of compact Lie groups? Is there anything on-line?
I looked at Tits' Bourbaki talk on Dynkin's and others' work, but he admits …
- 27.9k
9
votes
Accepted
Which linear combinations of simple roots are roots
My favorite answer to #2 and #3 is Kostant's "Find the highest root game", which is written up in detail in section 5.4 of Balázs Elek's notes on reflection groups. It is not hard to show that all pla …
- 27.9k
6
votes
Accepted
Motivating the existence of Canonical Bases for Representations
I don't really see how to get there from just compact groups, so in that sense this is not an answer. My take on the question is something like: how might one have guessed the existence of canonical b …
- 27.9k
5
votes
Convention about "long" roots for simple Lie algebras of types ADE?
I like being able to say "the highest root is always long".
- 27.9k
1
vote
Question on irreducible representation of tensor products
Your question is about the vectors in $V_1\otimes V_2$ that provide $1$-dimensional $B_\Delta$-subrepresentations of weight $\mu$, where $B_\Delta$ is the diagonal in $B\times B \leq G\times G$. Let's …
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8
votes
Accepted
How to find faces of polytope defined by a Weyl orbit
The faces are all of the following form: $w W_P / Stab_W(\xi)$, where $W_P$ varies over the subgroups generated by subsets of the simple reflections. In particular, for $\xi$ regular, the number of th …
- 27.9k
3
votes
What would you want on a Lie theory cheat poster?
For each real form, the poset K_C\G/B, as calculable by the Atlas here.
4
votes
very very basic question on semi-simple Lie algebras
I think this is more stackexchange-worthy, but here goes.
I'm a little afraid you're mixing up general weight diagrams with
the root system, which is the weight diagram of the adjoint representation. …
- 27.9k
5
votes
Accepted
An innocent looking subgroup of $U(n)$
Yup. Your condition is $M \vec v = 0$ for $\vec v = [1 1 1\ldots 1]$. Exponentiating, that's $\exp(M) \vec v = \vec v$.
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