All Questions
Tagged with riemannian-geometry curves-and-surfaces
7 questions with no upvoted or accepted answers
8
votes
0
answers
295
views
Intuition for the volume form - combinatorial definition?
I apologize that this is short of research level but I have realized that I am not happy with my understanding of the volume form on an oriented Riemannian manifold and I was hoping to find some ...
8
votes
0
answers
125
views
Blaschke points
A Blaschke point of a metric space is a point so that every geodesic (i.e. locally shortest path) starting at that point and of length less than the diameter of the metric space is the unique shortest ...
6
votes
1
answer
604
views
When is the cut locus a finite tree?
Let $\Omega \subset \mathbf{R}^2$ be a bounded, simply connected domain, with a regular boundary, say of class $C^2$ at least. Let the cut locus $C$ of $\Omega$ be the set of points $x \in \Omega$ for ...
3
votes
0
answers
165
views
A higher-dimensional "line of curvature"?
Let $M$ be a submanifold of a Riemannian manifold $Q$, and let $\Gamma$ be a $d$-dimensional submanifold of $M$, i.e., $\Gamma^{d} \subset M \subset Q$.
Suppose that, for all (unit) normal vectors of $...
3
votes
0
answers
189
views
The classification of developable surfaces: Are these statements equivalent?
This is a cross-post from MSE (https://math.stackexchange.com/q/4330772/242708).
I thought to know very well the answer to the classification problem for developable surfaces, so I sought for some ...
2
votes
0
answers
134
views
Hypersurfaces whose unit normal $N$ satisfies $[N,X] =0$ for every tangent vector field $X$
Let $M$ be a hypersurface of a Riemannian manifold, and assume that $M$ satisfies the following property:
For each $p \in M$, given a unit normal vector field $N$ defined in a neighborhood $U$ of $...
0
votes
0
answers
121
views
Positive curvature of the boundary away from a point implies regularity?
In a paper I'm refereeing, the authors make use of the following geometric fact:
Let $U$ be an open subset of $\mathbb{R}^2$. If there is a point $p\in \partial U$ so that $\partial U \backslash p$ ...