I would encourage you to consider "Representation Theory and Complex Geometry" by Chriss and Ginzburg. In particular, I think you might enjoy the realization of irreducible representations of the Weyl ...

Let $\mathfrak{g}$ be the Lie algebra of the group $G$. You might consider reading about the Springer resolution $$\mu:T^*(G/B)\rightarrow\mathcal{N},$$ where $\mathcal{N}$ is the nilpotent cone of $\... View answer Accepted answer 10 votes Take a ring$A$that is not an integral domain, and let$d\in A$be a zero-divisor. Consider the polynomial$dx-1$. Since$d$is a zero-divisor in$A$, it is a zero-divisor in every ring$B$... View answer Accepted answer 9 votes There are holomorphic line bundles over a compact Riemann surface$X$that are topologically trivial, yet not holomorphically trivial. To see this, note that smooth complex line bundles are classified ... View answer 8 votes I believe the answer is yes. Since$X$is an affine$G$-variety,$G$acts on$\mathbb{C}[X]$by$\mathbb{C}$-algebra automorphisms. This yields an action of$G$on the fraction field of$\mathbb{C}[X]$... View answer 8 votes Let$G$be a Lie group and$H\subseteq G$a closed subgroup. The quotient map$G\rightarrow G/H$is a principal$H$-bundle. In particular, it is an example of a fibration. We then have an associated ... View answer Accepted answer 7 votes The answer is that$\pi_1(G/G_x)=0$. Use the long-exact sequence of homotopy groups one obtains from a fibration. In this case, the fibration is$G_x\rightarrow G\rightarrow G/G_x$. We have $$\ldots\... View answer Accepted answer 7 votes Suppose that G is compact, connected, and semisimple. Let T\subseteq G be a maximal torus. Take the complexification G_{\mathbb{C}} of G, and choose a Borel subgroup B\subseteq G_{\mathbb{C}}... View answer 5 votes No, I think this need not be the case. Consider the usual action of S^1 on \mathbb{C}^2. The symplectic quotient is \mathbb{P}^1, which is not hyper-Kahler for dimension reasons. View answer Accepted answer 5 votes No, there are no counter-examples. Note that a generic coadjoint orbit is G-equivariantly diffeomorphic to G/T, for a maximal torus T\subseteq G. However, G/T (also known as the full flag ... View answer 5 votes The action of GL_k on Mat_{k\times n} is linear. Therefore, the scaling retraction of Mat_{k\times n} to \{0\} is GL_k-equivariant. It follows that the restriction map$$H^*_{GL_k}(Mat_{k\... View answer 4 votes Working over$\mathbb{C}$, the simplicity of$S$implies that the adjoint representation is irreducible. Picking a Cartan subalgebra of$S$and a collection of positive roots, it follows that the ... View answer Accepted answer 4 votes Suppose one relaxes the condition that the action is free, replacing it with the condition that every point in$M$has a finite$G$-stabilizer. In this case, the topological quotient$M/G$carries a ... View answer 4 votes I'm not sure you will find this answer to be satisfactory, as it addresses only a special case. Nevertheless, a unipotent conjugacy class in$SL_n(\mathbb{C})$is the same as a conjugacy class of a ... View answer Accepted answer 4 votes You can prove that$G/H$is quasi-projective, and a reference is Theorem 4.4.1 of Algebraic Quotients, Torus Actions, and Cohomology by A. Bialynicki-Birula, ‎J. Carrell, ‎and W.M. McGovern. View answer Accepted answer 4 votes Let$\frak{b}\subseteq\frak{g}$denote the Lie algebra of your Borel$B$. There is a natural$G$-equivariant isomorphism$\tilde{\frak{g}}\cong G\times_B\frak{b}$of vector bundles over$G/B$, where$...

Examining the complex used to calculate $H^*(\mathfrak{g};\mathbb{R})$, one sees that it is precisely the complex of $G$-invariant differential forms on $G$. (To see this, take the left-trivialization ...

@Alex Degtyarev's comment gives the right idea. The tangent bundle of $SU(2)$ can be given a left-trivialization, which will identify it with $SU(2)\times\mathfrak{su}_2$. In this way, you can think ...

Consider $PSU(2)$, the three-dimensional projective special unitary group (or just $\mathbb{R}\mathbb{P}^3$). It is a Lie group, and therefore orientable. Yet, it is the quotient of the simply-...

I think the idea is to write $P$ as a semidirect product of its unipotent radical $N$ and its (maximal reductive) Levi subgroup $M$. (If $P$ is a Borel, then $M$ is a maximal torus.) You can then ...

I'm not sure I have a direct answer to your question, but here are some thoughts. I would suggest you investigate some of the literature on "flat" connections. It is precisely the flatness condition ...

No, the bundle is not trivial in general. If we consider the case $n=1$, then the associated Lagrangian Grassmannian is actually the ordinary Grassmannian, namely $\mathbb{R}\mathbb{P}^1$. Also, the ...

I believe the answer is yes. There are perhaps several ways to see this, but a decent reference is the book "Affine Flag Manifolds and Principal Bundles" by Schmitt. It contains an article called "...

I am not an expert, but the affine Grassmannian is intimately related to the representation theory of the Langlands dual group $G^{\vee}$. Assume that $G$ is complex semisimple and simply-connected (...

I believe the answer is yes. For a $G$-manifold $X$, let $X_G$ denote its Borel mixing space. Recall that $$H_G^*(X)=H^*(X_G).$$ Now, note that $E_G$, $F_G$, and $(E\oplus F)_G$ are the total spaces ...
You might consider your problem in a generalized setting. Let $H$ be a Lie group and $X$ an $H$-manifold on which $H$ acts freely and properly. There is a unique manifold structure on $X/H$ for which $... View answer 2 votes As peoples' comments suggest, it is perhaps somewhat ambitious to hope for a classification of all Lie subalgebras of a given complex simple Lie algebra$\mathfrak{g}$. However, Jacobson and Morozov ... View answer 2 votes I believe that the answer is yes. First note that the complexification$G_{\mathbb{C}}$of$G$is reductive and contains$G$as a maximal compact subgroup. Secondly,$G_{\mathbb{C}}$acts ... View answer 2 votes Consider the conjugation representation of$GL_n$on$M_n$. Suppose that$A$is an$n\times n$matrix. Then$C(A)$is the$\mathfrak{gl}_n$-stabilizer of$A$, so that$\dim(C(A))$is equal to the ... View answer 2 votes This is not true as stated. Let$n=1$and consider$V(x_1^2)=\{0\}$. This subvariety is irreducible, but$x_1^2\$ is reducible.