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4 votes
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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
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
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
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
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
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
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
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
7 votes
1 answer
368 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
138 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
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
2 votes
0 answers
54 views

Want a minimal subgroup, whose orbits cover a submanifold, that is contained in a maximal subgroup which leaves the submanifold invariant

Say we have a Lie group $G$ acting transitively on smooth manifold $M$ and take a submanifold $S\subseteq M$. It seems to me that there should be some minimal subgroup $G'\subseteq G$ such that $S\...
Brock A.'s user avatar
1 vote
0 answers
134 views

Hausdorff dimension of a compact Lie group [closed]

Let $G$ be a compact Lie group (for simplicity assume $G=SO(3)$). Equip $G$ with a left-invariant Riemannian metric and let $m$ be left-invariant Haar measure on $G$. Now that $G$ is a metric space ...
Adam's user avatar
  • 323
3 votes
0 answers
142 views

Homogeneous Carnot group, its Lie algebra and Carnot-Carathéodory ball

Background: Let the smooth vector fields $X=(X_1,\cdots,X_m)$ define on $\mathbb{R}^n$ and they satisfy the following assumption: (H1): There is a dilation structure $$\delta_{t}:\mathbb{R}^n\to \...
Houa's user avatar
  • 561
9 votes
0 answers
326 views

Maximal geodesic spheres in the "octooctonic projective plane"

Boris Rosenfeld claimed that the 128-dimensional compact Riemannian symmetric space on which $\mathrm{E}_8$ acts as isometries could be seen as the "octooctonionic projective plane", $(\...
John Baez's user avatar
  • 22.3k
-1 votes
1 answer
230 views

Orthonormal frame on $\mathbb{S}^3$ orthogonal to foliations [duplicate]

Does there exist a smooth orthonormal frame $X_1,X_2,X_3$ on $\mathbb{S}^3$ such that the distribution spanned by $X_i$ and $X_j$ is integrable for all $1\leq i,j\leq 3$?
SubGeo's user avatar
  • 89
8 votes
1 answer
610 views

Are invariant forms on homogeneous spaces necessarily closed?

Take a compact homogeneous space $G/K$, and a left $G$-invariant differential $k$-form $\omega \in \Omega^k(G/K)$. Will $\omega$ necessarily be closed? Might it even be harmonic when $G/K$ is endowed ...
Quin Appleby's user avatar
1 vote
0 answers
517 views

Horizontal lift of fundamental vector field

Suppose $\theta\colon G\times M\to M$ is a transitive smooth left action of a compact Lie group $G$ on a manifold $M$ and $\pi\colon G\to M\cong G/K$ the corresponding smooth submersion for some ...
Sven Pistre's user avatar
-1 votes
1 answer
97 views

A manifold or Riemannian structure on the space of all conjugacy classes of a compact Lie group [closed]

Let $G$ be a compact Lie group. Is each conjugacy class a closed subset of $G$? Define the conjugacy equivalent relation $g\sim h$ if $g$ is conjugate to $h$.Is $G/\sim$ a Haussdoef space with ...
Ali Taghavi's user avatar
3 votes
1 answer
347 views

On functional equation $f\circ \exp=\exp \circ Df$ on a Riemannian manifold or a Lie Group

Let $M$ be a Riemannian manifold or a Lie group whose corresponding exp map (in corresponding context) is denoted by "exp" which is a map $\exp:TM\to M$ We search for the set $\mathcal{H}...
Ali Taghavi's user avatar
7 votes
1 answer
1k views

Help with definition of Liouville measure

$\require{AMScd}$For a Riemannian manifold $M$, I have read authors talking about a 'Liouville measure' on the unit tangent bundle $\operatorname{T}^1(M)$ and then proceed to claim/prove that it is ...
Calamardo's user avatar
  • 675
2 votes
1 answer
328 views

Totally geodesic submanifolds of bi-invariant Lie groups

Let $G$ be a Lie group equipped with a bi-invariant metric. I have some questions concerning the totally geodesic and the flat (all sectional curvatures zero) submanifolds of $G$. I known that every ...
Matteo Raffaelli's user avatar
4 votes
3 answers
862 views

Is every homogeneous space Riemannian homogeneous?

A manifold $M$ together with a transitive $G$-action is always diffeomorphic a quotient $G/H$ for $H < G$ Lie groups. On the other hand, there might be a proper subgroup of $G$ that also acts ...
Lezkus's user avatar
  • 177
2 votes
0 answers
260 views

A geometric property of certain Lie groups

I call Poincaré $n$-half-space group the semidirect product of $\mathbb{R}^{n-1}$ and $\mathbb{R}^+$, where the action is by homotheties; equivalently as the group of translations and positive ...
Ali Taghavi's user avatar
2 votes
1 answer
86 views

Compact group actions with uniformly bounded derivatives

Suppose we have a smooth action of a compact Lie group $G$ on a non-compact smooth manifold $M$, denoted by $$\phi:G\times M\rightarrow M.$$ Differentiating $\phi$ at a point $x\in M$ gives a map that ...
geometricK's user avatar
  • 1,903
3 votes
0 answers
60 views

Transformation between nearby tangent planes [closed]

This question is kinda long, but the picture is quite clear. Question: Let $(M,g)$ be a Riemannian manifold, $p$ a point on $M$, $U$ an open neighborhood of $0\in T_pM$ such that $exp_p|_U$ is a ...
Student's user avatar
  • 5,230
5 votes
2 answers
377 views

Existence of an isotopy in Riemannian manifold

Let $(M,g)$ be a Riemannian manifold, and $p,q\in M$ be two fixed points. We assume $p,q$ are close enough. Say, we assume $p$ and $q$ are in the same normal coordinate chart. It is clear that there ...
Hang's user avatar
  • 2,789
4 votes
0 answers
244 views

Gram-Schmidt map as a Riemannian submersion

We equip $\mathrm{GL}(n,\mathbb{R})$ and $\mathrm{O}(n)$ with their left-invariant metrics, whose restrictions to the corresponding neutral elements is the standard inner product $\mathrm{Trace}(AB^{\...
Ali Taghavi's user avatar