Peter Crooks
  • Member for 9 years, 6 months
  • Last seen more than 3 years ago
Introductory References for Geometric Representation Theory
Accepted answer
14 votes

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 ...

View answer
Cohomology ring of a flag variety and representation theory
Accepted answer
14 votes

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
Polynomial without roots in a ring
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
Is there a toplogically trivial line bundle over a compact Riemann surfaces that isn't holomorphically trivial?
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
Equivariant normalization?
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
How does one calculate homotopy classes for group coset spaces?
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
Computing the fundamental group of a flag variety
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
Is $G/T$ a projective variety?
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
If $M$ has hyper-kaehler structure then $M//G$ has hyper-kaehler structure?
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
computing second cohomology $H^2(O_a,\mathbb{Z})\cong \mathbb{Z}^n$ of a generic coadjoint orbit
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
$GL_k$-equivariant cohomology of $k\times n$ matrices
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
An algorithm to compare two representations of a simple Lie algebra?
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
Local structure of the quotient of a Lie group action
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
Unipotent conjugacy classes
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
Quotient of an algebraic group by a closed algebraic subgroup
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
Computing the Grothendieck-Springer resolution for $G = SL_2$
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 $...

View answer
What are cohomology of Lie algebra with coefficients geometrically?
3 votes

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 ...

View answer
SU(2) and differential forms
3 votes

@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 ...

View answer
Example of torsion in orientable manifolds?
Accepted answer
3 votes

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-...

View answer
Representations of parabolic subgroups of the general linear group over the complex numbers
Accepted answer
3 votes

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 ...

View answer
A cohomology group which depends on the connection
3 votes

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 ...

View answer
Canonical n plane bundle over Lagrangian Grassmanian
3 votes

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 ...

View answer
Whitney stratification and affine grassmanian
3 votes

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 "...

View answer
Relations between affine Grassmannian and Grassmannian
Accepted answer
3 votes

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 (...

View answer
equivariant whitney product formula
3 votes

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 ...

View answer
Cross section for closed Lie subgroup in a Lie group
Accepted answer
3 votes

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
Equivalence of Lie subalgebras, within a (irreducible) representation
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
Symplectic quotient of projective variety is projective?
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
Dimension of Commutator Space
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
if V(f) is irreducible, then how to show that the polynomial f itself is irreducible?
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.

View answer