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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 …

"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 $\mathbb Z^n$, …

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 …

They're the ones perpendicular to $\lambda$. To see this, note, first off, that $\mathfrak g_\alpha v_\lambda \in V_{\lambda+\alpha}$ so if $g_\alpha v_\lambda \neq 0$ then $\lambda+\alpha$ must be a …

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 …

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 …

In general you can't expect to get an action of $G$ on $V^\omega$. Instead, what one has is that if $U$ is a $U(\mathfrak g_{\mathbb C})$-invariant subspace of $V^\omega$ then its closure $\overline{U …

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 …

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 …

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 …

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.

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 …

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 …

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 …

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 …