1
vote
1answer
145 views

The set of leaves of the distribution $D$ on coadjoint orbit $O_{\mu}$

Let $G$ be a compact connected Lie group and $O_{\mu}$ be a coadjoint orbit where $\mu\in \mathfrak{g}^*$ and $\mathfrak{g}^*$ is the dual of the Lie algebra of $\mathfrak{g}=\mathrm{Lie}(G)$. Let ...
1
vote
1answer
120 views

Each coadjoint orbit of a compact connected Lie group $G$ admits a $G$-invariant generalized complex structure

I am looking for a proof or counterexample for following assertion Each coadjoint orbit of a compact connected Lie group $G$ admits a $G$-invariant generalized complex structure (In sense of ...
6
votes
0answers
226 views

Injectivity of Lie group exponential function

If $G$ is a (finite-dimensional) Lie group, then the exponential function $\exp\colon\mathfrak{g}\to G$ is injective on some identity neighbourhood. If, moreover, $\mathfrak{g}$ is semi-simple and ...
0
votes
2answers
97 views

Derivations of algebra of smooth $g$-valued function?

Let $M$ is a smooth n-manifold and $g$ is a $Z_2$-graded Lie algebra, we denote the algebra of smooth $g$-valued function on $M$ by $C^{\infty} (M,g)$. I wanna find all graded derivation of ...
3
votes
1answer
400 views

How was this Lie algebra found?

In a paper the author lists, without justification, generators for a Lie algebra. I would be grateful if someone could justify these choices and perhaps suggest how I might have found them for myself. ...
1
vote
1answer
129 views

Right Invariant Randers metrics

I'm hoping to determine the geodesic equation for a right invariant Randers metric $F(x) = \sqrt{a(x,x)} + b(x)$ on $SU(N)$. In my special case the navigation data $(h,W)$ for the Randers metric are ...
5
votes
1answer
191 views

Determining the Lie algebra elements exponentiating to the center of a Lie group

For a semi-simple compact Lie group $G$ with center $Z(G)$, one can characterize the preimage of $Z(G)$ in the Cartan subalgebra under the exponential map as the nodes of the Stiefel diagram (see for ...
1
vote
0answers
119 views

About the Lie algebra of polyvector fields

I would like to know if someone already did some computations of the group of Lie algebra automorphisms of the algebra of polyvector fields on $\mathbb{R}^n$ equiped with the Schouten bracket (or ...
2
votes
1answer
121 views

Split real form of $SL(2,\mathbb{C})$ description of the two sphere?

If we denote the parabolic subgroup of $SL(2,\mathbb{C})$ by $P$, then we have the well known isomorphism $SL(2,\mathbb{C})/P \simeq S^2$, where $S^2$ is the two sphere. Now the compact real form of ...
1
vote
1answer
414 views

Para-Complexification of Lie Groups

Let $G$ be a real Lie group. Then the complexification $G_\mathbb{C}$ of $G$ is the unique complex Lie group equipped with a map $φ:G\to G_\mathbb{C}$ such that any map $G\to H$ where $H$ is a ...
2
votes
1answer
109 views

Casimir of a three dimensional solvable lie algebra

Good morning everyone. I've encountered recently during my computations the following lie algebra $$\mathfrak g=\text{span}(f_0,f_1,f_2),$$ with $$\begin{eqnarray}[f_2,f_1]&=&f_0+a f_2,\\ ...
0
votes
0answers
55 views

compute the determinant of a conjugacy map

Let $k$ be an algebraically closed field. Let $F=k((\pi))$ and $\mathcal{O}$ the ring of integers, let $\gamma\in T(F)$ regular semisimple for a connected reductive group $G$. We consider the map ...
2
votes
1answer
455 views

Kirillov-Kostant-Souriau Theorem on $\mathfrak{g}\oplus \mathfrak{g^*} $

My question is about the extention of kirillov's symplectic structure on coadjoint orbits. The most remarkable feature of the coadjoint representation is the fact that all coadjoint orbits possess a ...
0
votes
1answer
255 views

fiber bundle on an orbit of $\mathfrak{g}\oplus\mathfrak{g^*}$

Let $G$, be a Lie Group and $\mathfrak{g}$ be its Lie algebra ,i.e, $Lie(G)=\mathfrak{g}$. Let $\zeta=(\ X,F)\ \in \mathfrak{g}\oplus\mathfrak{g^*}$. Here $X\in \mathfrak{g} $ and $F\in ...
0
votes
1answer
127 views

A question about G-Manifolds

I am looking for a clear reason for following fact:Is there any reference ? Why a $G$-invariant differential form $\omega$ on a homogeneous $G$-manifold $M=G/H$ is uniquely determined by its value at ...
2
votes
1answer
260 views

Can this Lie group written as a direct product?

Let $G=G_1.G_2$ be a Lie Subgroup of $SO(k) \times SO(2) \subset SO(k,2)$, where $G_1=SU(k/2) \subset SO(k)$ and $G_2$ is a Lie subgroup of $SO(k) \times SO(2)$ isomorphic to $SO(2)$. Let $G_1 \cap ...
2
votes
1answer
236 views

symmetry of generationg function of PDE

We know that for finding the solutions of PDE equations, one of methods is "reduction of PDE", . For nonlinear equation $v_t=(v^{-4/3}v_x)_x+\lambda v$ how can we compute the generators of Lie ...
3
votes
2answers
291 views

degrees of the invariants for the action of $SL(V)$ on $\wedge^4V$

How can we find the degrees of the invariants for the action of $SL(V)$ on $\wedge^4V$ , $dimV=8$ by the model in the Lie algebra $E_7$.
3
votes
0answers
229 views

The normalizer a maximal compact subgroup of a semi-simple Lie group

Let $G$ be a semi-simple real Lie group such that $|\pi_0(G)|<\infty$ and let $K$ be a maximal compact subgroup of $G$. Q1: How does one prove that $N_G(K)=K$? So I know a nice (and low-tech) ...
4
votes
2answers
248 views

Are maximal connected semisimple subgroups automatically closed?

(Yet another question in a series demonstrating my rather embarrassing ignorance of standard Lie theory... I hope this is not too basic for MO!) To be a little more precise: let $G$ be a real ...
1
vote
2answers
175 views

Copies of ax+b inside the AN part of an Iwasawa decomposition?

As a relative novice to the structure theory of Lie algebras and Lie groups, the following is what I can gather from reading parts of Helgason's book DG, Lie groups and symmetric spaces and Knapp's ...
2
votes
2answers
540 views

Maurer-Cartan structure equation derivation

Dear all. I'm a theoretical physicist trying to understand the structure equations and their geometrical significance, this for their gravitational applications. I know the relation between the Lie ...
3
votes
1answer
244 views

How can one find generators of basic differential forms on homogeneous spaces?

Dear all, In short, my problem is that I would like to have a better control of the 1-forms on a homogeneous space. Contrary to the group case, the module of differential form is not trivialisable. ...
29
votes
5answers
2k views

Beautiful descriptions of exceptional groups

I'm curious about the beautiful descriptions of exceptional simple complex Lie groups and algebras (and maybe their compact forms). By beautiful I mean: simple (not complicated - it means that we need ...
23
votes
8answers
3k views

“Modern” proof for the Baker-Campbell-Hausdorff formula

Does someone has a reference to a modern proof of the Baker-Campbell-Hausdorff formula? All proofs I have ever seen are related only to matrix Lie groups / Lie algebras and are not at all geometric ...
2
votes
1answer
178 views

Topologic or geometric mean of the structure constants of a semi simple lie algebra

Let $G$ be a semi simple Lie group (or real reductive), $\mathfrak{g}$ its lie algebra and $B$ its killing form. We can defined the 3-form $k$ by $$k(X,Y,Z)=B([X,Y],Z).$$ with $X,Y,Z\in \mathfrak{g}$. ...
1
vote
0answers
202 views

volume form in a symmetric space of real rank one

I want to compare the two canonical volume forms on a noncompact symmetric space fo real rank one. The first one is the volume form induced by the Riemannian structure given by the Killing form ...
12
votes
1answer
1k views

Isometry group of a homogeneous space

Background Let $(M,g)$ be a finite-dimensional riemannian (or more generally pseudoriemannian) manifold. Suppose that I know that a certain Lie group $G$ acts transitively and isometrically on $M$ ...
0
votes
1answer
341 views

Higher order Approximation of Lie groups [closed]

Maybe the following is trivial or folklore, but I can't find any concrete proof of the theorem, that higher order derivatives of Lie groups don't give any new information above what is coded in its ...
4
votes
2answers
385 views

Semi-Simple Kahler Groups?

We say that a Kahler manifold is a Kahler group if it is also a Lie group. I would like to know which semi-simple Lie groups are also Kahler groups?
3
votes
3answers
1k views

Is the Lie algebra-valued curvature two-form on a principal bundle P the curvature of a vector bundle over P?

I am an analyst struggling through some geometry used in physics. Some background: For some Lie group $G$, let $P$ be a principal $G$-bundle over the smooth manifold $M$. Let $\omega$ be a connection ...
4
votes
5answers
759 views

Lie group operation and tangent vectors

Suppose we have two differentiable paths $\alpha$ and $\beta$ thru the identity of a Lie group $G$, $\alpha(0)=\beta(0)=e$ the identity element. Denote $\alpha\beta$ the path given by ...
2
votes
3answers
674 views

Is there any relation between deformation and extension of Lie algebras?

In a paper of A. Weinstein on the geometry of Poisson manifolds, he relates the formal linearization around a zero, p, of the Poisson bivector to extensions of the Lie algebra induced by the bivector ...
1
vote
2answers
271 views

Explanation of $y = x \exp(\triangle)$ for a Lie Group

Let $M$ be a non-compact matrix Lie group and $T_e M$ its lie algebra. Consider a point $x \in M $ and $ \triangle \in T_e M$. To move from $x$ to a point $y \in M$ along $\triangle$, below group ...
38
votes
10answers
10k views

why study Lie algebras?

I don't mean to be rude asking this question, I know that the theory of Lie groups and Lie algebras is a very deep one, very aesthetic and that has broad applications in various areas of mathematics ...
4
votes
1answer
380 views

Convexity radius of a Lie Group

Is there a nice formula/method to find the convexity radius of a matrix Lie group (the manifold can be noncompact) ? Edited based on comments: Definition : Convexity Radius (Berger - Panoramic View ...
13
votes
4answers
971 views

homotopy type of connected Lie groups

Is there a simple proof (short and low-tech) of the following fact: (E. Cartan) A connected real Lie group $G$ is diffeomorphic (as a manifold) to $K\times\mathbb{R}^n$ where $K$ is a maximal ...
6
votes
4answers
1k views

Quotient space of $\mathbb{C}^5$ under the action of $SL(2,\mathbb{C})$

One sees that given the $SL(2,\mathbb{C})$ action on $\mathbb{C}^5$, thought of as the space of polynomials of the form, $$a_0 x^4 + 4a_1 x^3 y + 6a_2x^2y^2 + 4a_3xy^3 + a_4 y^4$$ the ring of ...
10
votes
1answer
296 views

Is there much theory of superalgebras acting on manifolds by alternating polyvector fields?

Usual story: vector fields on $M$, with their Lie bracket, form a Lie algebra. We can consider "actions" of some other Lie algebra ${\mathfrak g}$ on $M$ by looking at Lie homomorphisms ${\mathfrak ...
4
votes
2answers
3k views

Maurer-Cartan form

I suppose given a Lie Group (G) and its corresponding Lie Algebra (g) every element in its dual defines a Maurer-Cartan form on the whole Lie Group? Let $\omega \in g^*$ be a Maurer-Cartan form and ...
2
votes
1answer
328 views

How do you exponentiate a section of the adjoint bundle to get a gauge transformation?

Suppose $E$ is a vector bundle with structure group $G$ and let $P = F(E)$ be the frame bundle. Let $\mathfrak{g}_P$ denote the associated bundle to the adjoint representation of $G$ on its Lie ...
8
votes
4answers
728 views

lists of computed cohomologies?

Is there any comprehensive list of examples for computed 1) de-Rham cohomology-groups 2) Lie-algebra-cohomology groups $H^i(\mathfrak{g},\mathbb{R})$ 3) equivariant de-Rham cohomology groups ? ...
12
votes
4answers
3k views

Formal Geometry

[edit: I posted an answer to this which summarizes one that I received verbally a few weeks after posting this question. I hope it is useful to someone.] I am presently seeking references which ...
27
votes
8answers
5k views

What is the symbol of a differential operator?

I find Wikipedia's discussion of symbols of differential operators a bit impenetrable, and Google doesn't seem to turn up useful links, so I'm hoping someone can point me to a more pedantic ...