All Questions
15 questions
20
votes
3
answers
9k
views
Curvature of a Lie group
Since a lie group is a manifold with the structure of a continuous group, then each point of the manifold [Edit: provided we fix a metric, for example an invariant or bi-invariant one] has some scalar ...
- 251
12
votes
1
answer
2k
views
Isometry group of pseudo Riemannian manifold always a Lie group? (Myers-Steenrod)
Myers-Steenrod states that the isometry group of a Riemannian manifold is a Lie group. Is that also true for pseudo Riemannian manifolds? I didn't find anything related to that.
Cheers
- 123
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 $. ...
- 4,081
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 ...
- 486
8
votes
1
answer
673
views
Classification of compact globally symmetric spaces
It is known that any connected compact Lie group $G$ is a finite quotient of the product of a compact simply connected semisimple Lie group $\tilde{G}$ and a torus $\mathbb{T}^n$ (see for example ...
- 193
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 ...
- 4,081
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^...
- 4,081
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 ...
- 985
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 $. ...
- 4,081
6
votes
2
answers
903
views
Ricci curvature of the symplectic group
Is the Ricci curvature of the compact symplectic group $Sp(n)$ bounded below by $cn$ for some constant $c > 0$ independent of $n$?
For $O(n)$ and $U(n)$ I know many references which state such a ...
- 11.4k
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 &...
- 1,769
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 ...
- 4,081
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 ...
- 4,081
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 ...
- 4,081
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 ...
- 356