All Questions
Tagged with riemannian-geometry gt.geometric-topology
14 questions
7
votes
2
answers
1k
views
G-spaces and manifolds
In his book "The geometry of geodesics" H. Busemann defines the notion of a G-space to be a space which satisfies the following axioms:
The space is metric
The space is finitely compact, i.e., a ...
- 579
11
votes
2
answers
2k
views
Retraction of a Riemannian manifold with boundary to its cut locus
This question is edited following the comment of Joseph. He pointed out that the main object of the first version of this question is the cut locus.
Recall that the cut locus of a set $S$ in a ...
- 28.9k
20
votes
0
answers
540
views
Homeomorphisms of the sphere mapping a geodesic triangulation to another one
Consider the standard Riemannian 2-sphere $S$, equipped with a geodesic triangulation $T$. Let $L(S,T)$ be the space of homeomorphisms of $S$ which map
$T$ to a geodesic triangulation. What is the ...
16
votes
3
answers
1k
views
SO(3) action on (simply connected) 6 manifold with discrete fixed point
If a 6-dimensional orientable smooth manifold $M$ admits a smooth effective $SO(3)$ action with discrete fixed point set, what can we say about the topology of $M$? What if we assume that in addition ...
- 591
14
votes
1
answer
3k
views
How metric is Riemannian geometry
Let $(M, g)$ be a finite-dimensional Riemannian manifold. It is well-known, that the Riemannian metric induce a metric on the manifold by
$$d(x, y) = \text{inf} \int_a^b \| \dot\gamma(t) \| \, dt\,,$$...
- 5,824
12
votes
3
answers
930
views
Voronoi cells and the dual complexes in Riemannian manifolds
I would like to use some "intuitively clear" properties of Voronoi cells in general Riemannian manifolds, but I have trouble finding references.
Let $(X,d)$ be a connected Riemannian ...
- 3,897
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
5
votes
1
answer
1k
views
Does every smooth manifold admit a metric with bounded geometry and uniform growth?
Let $M$ be a smooth manifold, $g_M$ a Riemannian metric, and consider for $x\in M$ the volume growth function, $gr_x$ that maps $r>0$ to the volume $vol_{g_M}(B(x,r))$. My interest is to see ...
- 541
4
votes
0
answers
182
views
Symmetric line spaces are homeomorphic to Euclidean spaces
For points $x,y,z$ of a metric space $(X,d)$ we write $\mathbf Mxyz$ and say that $y$ is a midpoint between $x$ and $z$ if $d(x,z)=d(x,y)+d(y,z)$ and $d(x,y)=d(y,z)$.
Definition: A metric space $(X,d)$...
- 41.8k
4
votes
0
answers
74
views
Determinant twist and $Pin _{\pm}$ structure on $4k$-dimensional bundles [Reference request]
Consider the automorphism $\varphi$ of $O(2n)$ given by $\varphi (A)=det(A)\cdot A$. The induced map in cohomology $H^*(BO(2n))$ sends $w_2$ to $w_2+(n+1)w_1^2$ (a proof is given at the end of the ...
- 3,051
3
votes
2
answers
265
views
For a closed Riemannian manifold $M$, must the set of points with non-unique closest points to a closed submanifold $S$ of $M$ be of 0 volume measure?
Let $(M,g)$ be a closed (compact without boundary) Riemannian manifold of finite dimension, with the volume measure $\mu:= \mu(E):=\int_{E}dvol_g \forall E \in \mathcal{B}(M),$ the Borel sigma algebra ...
- 1,467
3
votes
1
answer
123
views
Involution on the set of isomorphism classes of 2n-dimensional vector bundles induced by an involution of $O(2n)$
Denote by $\varphi$ the automorphism of $O(2n)$ given by $\varphi (A)=det(A)\cdot A$.This induces a self-map $B\varphi$ of $BO(n)$, so it induces a self-map (actually an involution)
$B\varphi ^*$ on $[...
- 3,051
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
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