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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.

"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 $\mathbb Z^n$, …

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 …

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 …

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 …

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 …

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 …

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 …

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 …

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 …

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 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 …

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 …

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 …

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 …