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Lie Groups are Groups that are additionally smooth manifolds such that the multiplication and the inverse maps are smooth.
14
votes
Why study Lie algebras?
Although the title is about Lie algebras, the question body mentions Lie groups, and my answer will deal more with these. As mentioned in other answers, Lie groups show up frequently in geometry as gr …
14
votes
Accepted
Iwasawa Decomposition & Polar Decomposition related how ?
You can obtain the $G=KAK$ decomposition from a decomposition of the type $F=UR$. To avoid unnecessary complications, let's assume that our reductive group $G$ is a selfadjoint subgroup of $\operatorn …
14
votes
Accepted
Complex structure on flag manifolds
This is essentially a more "condensed" version of Johannes Ebert's answer.
From the root space decomposition
$$ \mathfrak g /\mathfrak t \otimes \mathbb C = \oplus_{\alpha \in \Phi} R_\alpha, $$
one …
14
votes
3
answers
3k
views
Are complex semisimple Lie groups matrix groups?
Actually, my question is a bit more specific: Does every complex semisimple Lie group $G$ admit a faithful finite-dimensional holomorphic representation? [As remarked by Brian Conrad, this is enough t …
13
votes
connected compact semisimple lie group finite fundamental group
There is a quick proof via Lie algebra cohomology: Let $G$ denote your compact, connected, semisimple Lie group, and let $\mathfrak g$ denote its Lie algebra. Then $$ H^1(G;\mathbb R) = H^1(\mathfrak …
5
votes
Parallel forms and cohomology of symmetric spaces
I think there's some confusion in the question. For example by "Levi-Civita connection" you must really mean some kind of Laplacian. Anyway, your end result about the cohomology of $G/H$ is essentiall …
5
votes
real orbits of highest weight vectors
There seems to be some confusion in the question, so let me try to recap the basic setup. Thus let $G$ be a complex simple Lie group, $V^\lambda$ the irrep of $G$ of highest weight $\lambda$, and $G/P …
5
votes
Accepted
why are all characters of the maximal torus in a Lie group weights?
"Easiest" depends on how you set things up: everything really hinges on how you want to identify $X^\ast(T)$ with $\mathbb Z^n$. It's probably cleanest if you don't work explicitly with $\mathbb Z^n$, …
4
votes
weyl group representations
I'm not sure I know what you mean by Mackey theory or how it relates to the representation theory of Weyl groups. I guess you could mean Mackey's approach to the representation theory of semidirect pr …
4
votes
Which groups have only real and quaternionic irreducible representations?
This was a comment on Torsten's answer, but it got too long.
Suppose $G$ is connected and semisimple. Fixing a choice $\Phi^+$ of positive roots for $G$, we can describe $w_0$ as the unique element o …
4
votes
2
answers
500
views
Equivariance of vector bundles over G/B
Let $G$ be a complex semisimple group, $B$ a Borel subgroup of $G$ and $X=G/B$ the flag variety of $G$. If $G$ is simply connected, then every line bundle $L$ on $X$ can be made $G$-equivariant (see t …
3
votes
Accepted
Topological properties of $K$ orbits in $G/B$
If $\theta$ is an involution of a complex linear algebraic group $G$ and if $K=G^\theta$ is its fixed-point set, then $K/K^0$ will always have exponent 2. This follows from a generalized "Cartan deco …
3
votes
Accepted
Geometric structure of flag manifolds, Borel -Weil-Bott theorem
Correct. You can be fairly explicit here. For each root $\alpha$, let $\omega_\alpha \in \mathfrak g^\ast$ be a left-invariant form on $G$ that is dual to $\mathfrak g_\alpha$. Then for $\lambda \in …
3
votes
Kostant's theorem on principal 3-dimensional subalgebras
Is it true that the centralizer $Z_{\frak{g}}(\frak{a})=\{\xi\in\frak{g}:[\xi,\eta]= $0$ \text{ }\forall\eta\in\frak{a}\}$ is trivial (or equivalently, that the trivial one-dimensional representati …
3
votes
Accepted
maximal tori cover compact Lie group
There are proofs that avoid algebraic topology: see for example Chapter 16 in Bump's Lie Groups or IV.5 in Knapp's Lie Groups Beyond an Introduction.