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4 votes
0 answers
236 views

Jacobian of exponential map

I am playing around with the coarea formula and came across the problem of finding the Jacobian of the exponential map. Let $G$ be a compact, semisimple Lie group with associated Lie algebra $\...
DarkViole7's user avatar
3 votes
1 answer
100 views

Representation of Lie groups inducing a quasi-isometric embedding of their symmetric spaces

Let $G_{1}$ and $G_{2}$ be connected semisimple real Lie groups with no compact factors and finite center and let $K_{1}$ and $K_{2}$ denote some fixed choice of their maximal compact subgroups, ...
Aleksander Skenderi's user avatar
4 votes
1 answer
254 views

Isometry group of a left-invariant Riemannian metric on $\mathrm{SU}(2)$

Recall that \begin{equation} \mathbb{S}^3=\operatorname{SU}(2)=\left\{ \begin{pmatrix} z&w\\ -\bar{w}&\bar{z} \end{pmatrix} ,|z|^2+|w|^2=1 \right\} \end{...
Adterram's user avatar
  • 1,441
4 votes
0 answers
114 views

Mean-value type property for eigenfunctions of Casimir operator on $\operatorname{SL}(2,\mathbb{R})$

The purpose of this post is to ask for the community's opinion on a conjecture about a certain mean-value property for functions on $\operatorname{SL}(2,\mathbb{R})$. This conjecture appears at the ...
burtonpeterj's user avatar
  • 1,769
5 votes
0 answers
203 views

Differential equation involving Casimir operator on $\operatorname{SL}(2,\mathbb{R})$

Throughout this post, we let $G$ denote the Lie group $\mathrm{SL}(2,\mathbb{R})$. For $t,\theta \in \mathbb{R}$ we define the following elements of $G$: \begin{align*} A(t) &= \begin{bmatrix}e^t &...
burtonpeterj's user avatar
  • 1,769
1 vote
0 answers
62 views

Expression of the Riemannian metric on the Siegel domain?

I'm looking for proof that, for the complex Siegel domain in $\mathbb C^{n}$ defined by: $$\mathcal H_{n} = \{ z=(z_1,\dots,z_n) \in \mathbb C^{n} \mid \operatorname{Im}(z_{n}) > \sum_{j=1}^{n-1} |...
Z. Alfata's user avatar
  • 650
1 vote
1 answer
181 views

For proper group action on closed Riemannian manifold, must the union of orbits with non-unique closest points to a given point be of 0 volume measure

Let $(M,g)$ be a closed (compact without boundary) Riemannian manifold of finite dimension, with the volume measure $\mu:= \mu(E):=\int_{E}d\operatorname{vol}_g \forall E \in \mathcal{B}(M),$ the ...
Learning math's user avatar
1 vote
0 answers
109 views

A homogeneous manifold that does not admit an equivariant Riemannian metric?

Let $M = G/H$ be a homogeneous space, where $G$ is a Lie group and $H$ is a closed Lie subgroup. Can it happen that $M$ does not admit an invariant Riemannian metric?
Jake Wetlock's user avatar
  • 1,144
0 votes
0 answers
126 views

Expressing the union of principal orbits as a disjoint union of global slices for proper group actions

Setup: I was reading about slices and principal orbit theorems (Theorem 3.4.6) from these notes. Let the Lie group $G$ act on a complete Riemannian manifold $(M,g)$ isometrically on $M$, i.e. $\phi^{*}...
Learning math's user avatar
2 votes
0 answers
165 views

A property of the Riemannian metric on the symmetric space $X_\textsf{d}=\textsf{SL}(d,\Bbb R)/\textsf{SO}(d)$

Consider the symmetric space $X_\textsf{d}=\textsf{SL}(d,\Bbb R)/\textsf{SO}(d)$. For any $M\in X_d$, by transporting the Killing form on $T_I(X_d)$ (the space of symmetric matrices with trace zero) ...
Random's user avatar
  • 1,097
0 votes
0 answers
128 views

How to build a representation of the diffeomorphism group of $U(n)$?

Given that $U(n)$ is a smooth manifold I would like to know if there is a way of building a representation of $\text{Diff}(U(n))$ once you pick a particular (finite dimensional) representation of $U(n)...
Nicolas Medina Sanchez's user avatar
0 votes
1 answer
183 views

Finite subgroups of $\mathrm{O}_n(\mathbb R)$ from finite subgroups of $\mathrm{SO}_n(\mathbb R)$

Let $G$ be a finite subgroup of $\mathrm{SO}_n(\mathbb R)$. We also assume $G$ to be "maximal" in the sense that for every $g\in\mathrm{SO}_n(\mathbb R)\setminus G$, we have that $\overline{\...
Andrea Aveni's user avatar
1 vote
1 answer
181 views

Gradient descent under the presence of symmetries

Let $M$ be a Riemannian manifold (I'm happy to assume it is Euclidean space) with a function $f: M \to \mathbb R$ and a group of isometries $G$ acting on $M$ and preserving $f$, i.e., $f(gm) = f(m)$ ...
Asvin's user avatar
  • 7,746
1 vote
0 answers
110 views

Questions on the differential of the Lie logarithm

Let $G$ be a Lie group. Recall the Lie logarithm is well-defined about a neighborhood $U \subset G$ of the identity: $\log:U\to \mathfrak{g}$. I am dealing with a research problem that concerns the ...
Spencer Kraisler's user avatar
0 votes
1 answer
304 views

A Lie group whose Lie algebra is equal to (the Lie algebra? of )all functions with fibrewise polynomial growth

Let $M$ be a Riemannian manifold. We denote by $\mathfrak{g}$ the space of all smooth function $f:TM\to \mathbb{R}$ with fibre wise polynomial growth. Is it a Lie algebra wrt the Poisson bracket ...
Ali Taghavi's user avatar
5 votes
1 answer
530 views

Geodesic distance on $\mathrm{SO}(n)$

$\DeclareMathOperator\SO{SO}$Recently I came across this old MSE post or this paper (w.o. proof) discussing the geodesic distance on $\SO(n)$ when it is equipped with the left-invariant Riemannian ...
Math_Newbie's user avatar
4 votes
1 answer
263 views

Geodesics on orthogonal matrix

Let $ O(n) $ be the manifold of orthornormal matrix, i.e. $$ O(n)=\{A\in\mathbb{R}^{n\times n}:A^TA=I\}. $$ Then $ O(n) $ is a submanifold of $ \mathbb{R}^{n\times n} $. On $ O(n) $, there is a ...
Luis Yanka Annalisc's user avatar
1 vote
0 answers
70 views

Orbit projection geometry

Background: As shown in [1] and [2], for a closed smooth submanifold $M$ of $\mathbb R^d$, the domain $D_M$ of the projection map $P_M:D_M\rightarrow M$ has a dense interior $\Omega_M$ over which $P_M|...
miniii's user avatar
  • 71
8 votes
2 answers
440 views

Show $\langle \log(R), \log(R^{-1}S) \rangle \geq \langle \log(R), \log(S) - \log(R) \rangle $ for all $R,S \in \mathrm{SO}(3)$

$\DeclareMathOperator\SO{SO}$I have a similar question to one I asked a few days ago. Lately, I've been researching Lie groups equipped with bi-invariant Riemannian metrics. One common object is $\SO(...
Spencer Kraisler's user avatar
5 votes
0 answers
146 views

Is every linear Lie group of bounded geometry?

$\newcommand\norm[1]{\lVert#1\rVert}$Given any point $p$ of a smooth Riemannian manifold $M$ there exists $r\in (0,\infty]$ such that the Riemannian exponential is a diffeomorphism in the geodesic ...
Marco's user avatar
  • 293
8 votes
2 answers
642 views

Is it true $\left\|\log(RS)\right\|≤\left\|\log(R)+\log(S)\right\|$ for all $R,S \in \mathrm{SO}(3)$, where $\|\cdot\|$ is the Frobenius norm?

$\DeclareMathOperator\SO{SO}$I asked this initially in math stack exchange, but thought to ask it here since it is more advanced and related to my research topic. I study optimization on Lie groups ...
Spencer Kraisler's user avatar
6 votes
1 answer
323 views

Deformations of the 4-sphere with 8-dimensional isometry groups

I am looking for deformations of the 4-sphere with 8-dimensional isometry group, like a 4-dimensional Berger sphere.
Thomas Schucker's user avatar
5 votes
2 answers
732 views

Can the exponential map be used to define geodesics (and hence, generalisations of geodesics)?

Let $(M,g)$ be a (connected, paracompact, $C^{\infty}$-smooth) Riemannian manifold with Riemannian metric $g$. The exponential map is defined for each point $p \in M$ to be the map $\exp_p : T_p M \to ...
AmorFati's user avatar
  • 1,379
13 votes
2 answers
2k views

Why is $\operatorname{SO}(4)$ not a simple Lie group?

$\DeclareMathOperator\SO{SO}$I'm not asking for a proof but for some hint that might be helpful to understand this "anomaly" in 4 dimensions. I'm aware of the parallelism with the $A_4$ ...
bonif's user avatar
  • 301
3 votes
1 answer
203 views

Free $S^1$-action on compact homogeneous spaces

Let $M = G/K$ be a compact homogeneous space, i.e. $G$ a connected compact Lie group, and $K$ a closed subgroup inside $G$ that contains no nontrivial normal subgroup of $G$. If $r(G) > r(K)$ (...
abracadabra12345's user avatar
6 votes
0 answers
147 views

Maximum symmetry metric on irreducible compact symmetric space

Let $ M $ be a compact connected manifold. The degree of symmetry of $ M $, denoted $ N(M) $, is the maximum of the dimensions of the isometry groups of all possible Riemannian structures on $ M $. ...
Ian Gershon Teixeira's user avatar
5 votes
2 answers
480 views

Maximum symmetry metric on $ \mathbb{C}P^n $

Let $ M $ be a compact connected manifold. The degree of symmetry of $ M $, denoted $ N(M) $, is the maximum of the dimensions of the isometry groups of all possible Riemannian structures on $ M $. ...
Ian Gershon Teixeira's user avatar
12 votes
0 answers
247 views

Symmetric spaces are quandles. Is this important?

For concreteness, let's consider a connected reductive Lie group $G$, and an involution $\theta$ on it. Then the associated symmetric space $X=G/G^\theta$ has the structure of an involutive quandle: ...
Spencer Leslie's user avatar
9 votes
1 answer
646 views

Explicit construction of a (the?) dual symmetric space

I am looking for a reference, proof or disproof of the fact that every Riemannian globally symmetric space of compact (non-compact) type has a "dual", which is of non-compact (compact) type. ...
S.T.'s user avatar
  • 113
1 vote
0 answers
83 views

Right-invariant metrics on the unitary groups and embeddings in the complexification

Let $G = SU(n)$ and $G_c = SL(n, \mathbb{C})$. Let $g$ be a right-invariant metric on $G$ and let $g_k$ be the Killing metric on $G_c$. Define the map $p$ from $G_c$ to $G$ which maps $h \in G_c$ to $$...
Malkoun's user avatar
  • 5,215
2 votes
1 answer
137 views

What does the boundary of convex hulls look like in matrix Lie groups?

Let $G$ be a compact matrix Lie group under the Killing form metric $\langle \xi, \eta \rangle_g = -\frac{1}{2}\text{tr}((g^{-1}\xi)^T(g^{-1}\eta))$ for $g \in G$ and $\xi,\eta \in T_gG$. Let $C \...
Spencer Kraisler's user avatar
3 votes
0 answers
53 views

Decomposition about splitting of symmetric spaces of compact type

I get stuck in the following question: Why does a locally symmetric space of compact type $M$ split locally irreducible components of dimension $\geq 2$ which are Einstein? In particular, why are all ...
Radeha Longa's user avatar
2 votes
0 answers
108 views

Questions about symmetric spaces

I'm a little confused with the following questions: (1) Why does a symmetric space $M=G/K$ of compact type have $\mathcal{R}^M\geq 0$? (2) Moreover, why does $\chi(M)\neq 0$ if and only if ${\rm rk}\ ...
Radeha Longa's user avatar
1 vote
0 answers
196 views

Homogeneous metrics on compact Lie groups

Given a differentiable manifold $M$, a Riemannian metric $g$ on $M$ is called homogeneous if its isometry group acts transitively on $M$. In that case, given any group $G$ acting transtitively on $(M,...
emiliocba's user avatar
  • 2,446
10 votes
4 answers
710 views

Palais's and Kobayashi's theorems on automorphism groups of geometric structures

My question concerns two results in the neighborhood of the standard theorem of Myers-Steenrod that isometry groups of Riemannian manifolds are Lie groups. Both appear in the first chapter of ...
Chris Wendl's user avatar
5 votes
0 answers
132 views

geometry and connected sum of aspherical closed manifolds

Let $ G $ be a Lie group with finitely many connected components, $ K $ a maximal compact subgroup, and $ \Gamma $ a torsion free cocompact lattice. Then $$ \Gamma \backslash G/K $$ is an aspherical ...
Ian Gershon Teixeira's user avatar
9 votes
1 answer
444 views

Compact flat orientable 3 manifolds and mapping tori

There are 10 compact flat 3 manifolds up to diffeomorphism, 6 orientable and 4 non orientable. I am looking to better understand how to construct the orientable ones. The six orientable ones are ...
Ian Gershon Teixeira's user avatar
3 votes
1 answer
425 views

3 dimensional solvmanifolds and Thurston geometries

Does every three dimensional compact solvmanifold admit either Euclidean, nil, or sol geometry? definitions/motivation/background: A solvmanifold is a manifold $ M $ admitting a transitive action by a ...
Ian Gershon Teixeira's user avatar
5 votes
0 answers
276 views

Fundamental group of compact globally symmetric spaces

The fundamental group of a globally symmetric space $M$ of compact type is known (see Loos [1], Borel [2]). The result can be formulated as follows: it is isomorphic to the quotient $$(*) \quad \pi_1(...
Lucas Seco's user avatar
  • 1,123
2 votes
1 answer
484 views

Mapping torus of orientation reversing isometry of the sphere

$\DeclareMathOperator\SL{SL}\DeclareMathOperator\SO{SO}\DeclareMathOperator\SU{SU}\DeclareMathOperator\O{O}\DeclareMathOperator\Iso{Iso}$ Let $ f_n $ be an orientation reversing isometry of the round ...
Ian Gershon Teixeira's user avatar
6 votes
0 answers
341 views

When exponential map is 1-1 from vector fields to diffeomorphisms

Let $M$ be a connected and complete Riemannian manifold of positive dimension, $k$ be a positive integer, and let $\mathfrak{X}^k_c$ be the set of class $C^k$-vector fields on $M$ of compact support. ...
ABIM's user avatar
  • 5,405
7 votes
1 answer
367 views

Does complexified isometry group act transitively on tangent bundle of compact Riemannian manifold?

$\DeclareMathOperator\SL{SL}\DeclareMathOperator\SO{SO}\DeclareMathOperator\SU{SU}\DeclareMathOperator\O{O}\DeclareMathOperator\Iso{Iso}$Let $ g $ be the round metric on the sphere $ S^n $. Since $ S^...
Ian Gershon Teixeira's user avatar
4 votes
1 answer
230 views

Generalizing a result about hyperbolic 2-folds to hyperbolic 3-folds

$\DeclareMathOperator\SL{SL}\DeclareMathOperator\SO{SO}\DeclareMathOperator\SU{SU}$Let $ \Sigma_g $ be a compact orientable surface of genus $ g $. Let the subgroup $ \pi_1(\Sigma) $ of $ \SL_2(\...
Ian Gershon Teixeira's user avatar
8 votes
1 answer
599 views

Exact condition for smooth homogeneous to imply Riemannian homogeneous for compact manifolds

Let $ (M,g) $ be a homogeneous Riemannian manifold. That is, the isometry group $ Iso(M,g) $ acts transitively on $ M $. Let $ \pi_1(M) $ be the fundamental group of $ M $. Then $ \pi_1(M) $ has ...
Ian Gershon Teixeira's user avatar
2 votes
1 answer
137 views

noncompact Riemannian homogeneous is trivial vector bundle over compact homogeneous

Is it true that a manifold $ E $ admits a metric with respect to which the isometry group is transitive ($ E $ is Riemannian homogeneous) if and only if $ E $ is the total space of a $ K $ equivariant ...
Ian Gershon Teixeira's user avatar
3 votes
1 answer
470 views

Is the Moebius strip Riemannian homogeneous?

Let $ M $ be the Moebius band. In other words, the total space of the nontrivial line bundle over the circle. Can we equip $ M $ with a metric such the the isometry group acts transitively? My ...
Ian Gershon Teixeira's user avatar
2 votes
2 answers
213 views

Riemannian homogeneous equivalent to linear group orbit

Let $ M $ be a smooth manifold. Recall that a manifold $ M $ is smooth homogeneous if there exists a Lie group acting transitively on $ M $. Recall that a manifold $ M $ is Riemannian homogeneous if ...
Ian Gershon Teixeira's user avatar
2 votes
0 answers
78 views

Examples of curvature-adapted subgroups of semi-Riemannian groups

Let $G$ be a semi-Riemannian group, i.e., a Lie group equipped with a bi-invariant semi-Riemannian metric. I am looking for examples of Lie subgroups that are curvature adapted to $G$. First, allow me ...
Matteo Raffaelli's user avatar
2 votes
0 answers
192 views

Submanifold of Lie group whose tangent bundle is "almost" left-invariant

Let $G$ be a Lie group equipped with a left-invariant Riemannian metric, and let $M$ be a submanifold of $G$ containing the identity $e\in G$. It is not difficult to show that, if the tangent bundle $...
Matteo Raffaelli's user avatar
7 votes
2 answers
499 views

Submanifolds of Lie groups with abelian normal bundle

Let $M$ be a submanifold of a symmetric space $Q$. The normal bundle $NM$ is called abelian if $\exp(N_{p}M)$ is contained in some totally geodesic and flat submanifold of $Q$ for all $p \in M$; see ...
Matteo Raffaelli's user avatar