All Questions
9 questions
17
votes
1
answer
1k
views
Hyperbolic manifolds which fiber over the circle
If $N^2$ is a closed, orientable surface of genus at least $2$, and if $\phi$ is an (orientation-preserving) pseudo-Anosov mapping on $N$, then one can form the closed orientable 3-manifold $M^3$ by ...
- 4,713
14
votes
1
answer
860
views
Mapping torus of Klein bottle
This got 5 upvotes but no answers on MSE (Mapping torus of Klein bottle), so I'm cross-posting to MO:
The mapping torus of a Klein bottle $ K $ is a compact flat 3 manifold.
The mapping class group of ...
- 12.9k
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
395
views
Embedding round manifolds into low dimensional spheres
Robert Bryant's answer to Isometric embedding of SO(3) into an euclidean space mentions that there is an isometric embedding of the round tetrahedral space $ SO_3/A_4 $ into the round sphere $ S^6 $.
...
- 4,081
9
votes
0
answers
336
views
Nash embedding for 3 manifolds
The Nash embedding theorem tells us that every smooth Riemannian m-manifold can be embedded in $R^n$ for, say, $n = m^2 + 5m + 3$ (edit: 14 is a better bound for compact 3 manifolds thanks @mme). What ...
- 4,081
2
votes
0
answers
205
views
Can a non-compact manifold become compact by cutting it?
I'm trying to understand a step in a proof, where one starts with a non-compact manifold $V$ containing a trapped (2-sided, closed) surface $\Sigma$ that's non-separating. In order to complete the ...
- 121
12
votes
0
answers
381
views
Two ways a manifold can have little symmetry
Let $M$ be a closed connected smooth oriented manifold. The following two properties - that $M$ can either enjoy or not - intuitively both mean that $M$ has very little symmetry:
(a) Every self-map $...
- 11.9k
25
votes
2
answers
1k
views
Is there a smooth manifold which admits only rigid metrics?
Does there exist a (finite dimensional) smooth manifold $M$, such that every Riemannian metric on $M$ has no isometries except the identity?
Of course, such a manifold must not admit a diffeomorphism ...
CommunityBot
- 1
8
votes
1
answer
696
views
Geodesics on manifolds with boundary
Let $(M,g)$ be a Riemannian manifold with non-empty boundary. Is there any notion of injectivity radius on $(M,g)$ in points away from the boundary? By this I mean points lying in $M- \partial M$. ...
- 108k