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Results tagged with riemannian-geometry
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user 172802
Riemannian Geometry is a subfield of Differential Geometry, which specifically studies "Riemannian Manifolds", manifolds with "Riemannian Metrics", which means that they are equipped with continuous inner products.
12
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Are there $n$ points dividing a compact Riemannian manifold into equal regions?
Let $M$ be a compact, connected $m$-dimensional Riemannian manifold, and let $n\in\mathbb{N}$. Can we always find distinct points $p_1,\dotsc,p_n\in M$ such that for $i=1,\dotsc,n$ the regions $A_i=\{ …
- 10.6k
2
votes
Accepted
Closed almost geodesics in a Riemannian manifold
Any curve $\gamma:[a,b]\to M$ parametrized by arc length is an $\varepsilon$-geodesic for any $\varepsilon>0$.
The inequality $(1-\varepsilon)dist_M(\gamma(x),\gamma(y))\leq length[\gamma(x),\gamma(y) …
- 10.6k
5
votes
Accepted
Pythagorean theorem in Riemann metrics of non constant curvature
I think (see (??) below) there is no connected complete Riemannian manifold $(M,g)$ which is not flat and such that there is an isometry $f:(M,g)\to(M,\lambda^2g)$, where $\lambda\neq0,1$. I will assu …
- 10.6k
3
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On diffeomorphisms that preserve the metric
This is true if $\Omega\neq\mathbb{R}^2$ (so that every path component of $\Omega$ has nonempty boundary). Firstly, $F^*e=e$ means that $F$ is a local isometry. We know that two local isometries $f,g: …
- 10.6k
1
vote
Packing a Riemannian manifold with disjoints balls
For any smooth Riemannian manifold $(M,g)$ there is a countable disjoint union of balls with complement of measure $0$.
Let $\mu$ be Riemannian measure and for each $p\in M$ let $B_p$ be a small preco …
- 10.6k
1
vote
Accepted
Bisector of two points in a Riemannian manifold has measure $0$
Here is a short proof supposing that $M$ is complete (if not the statement is false, see the last paragraph) using the idea from Leo Moos' answer of using Rademacher's theorem.
Suppose $\mathcal{B}(p, …
- 10.6k
7
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2
answers
177
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Bisector of two points in a Riemannian manifold has measure $0$
Let $p,q\in M$, $p\neq q$, where $M$ is a Riemannian manifold. We will let the bisector of $p,q$ be $\mathcal{B}(p,q)=\{x\in M;d(p,x)=d(q,x)\}$. Does $\mathcal{B}(p,q)$ have measure $0$?
I was thinkin …
- 10.6k
23
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2
answers
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Can we make distances in a finite subset of a manifold whatever we want?
Given a connected smooth manifold $M$ of dimension $m>1$, points $p_1,\dots,p_n\in M$ and positive values $\{d_{i,j};1\leq i<j\leq n\}$ satisfying the strict triangle inequalities $d_{i,j}<d_{i,k}+d_{ …
- 10.6k
7
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Accepted
For a closed Riemannian manifold $M$, must the set of points with non-unique closest points ...
More generally, for any closed subset $S$ of a complete manifold $M$, the set of points $x$ at whose minimal distance to $S$ is attained at more than point has measure $0$.
Indeed, consider the distan …
- 10.6k
3
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Equidistant points on a compact Riemannian manifold
$K(M,g)$ depends on the metric, as shown by this question, which implies that we can change the metric of $\mathbb{R}^3$ so it has as many points pairwise at distance $1$ as we want.
- 10.6k
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For proper group action on closed Riemannian manifold, must the union of orbits with non-uni...
Perhaps this is not the strongest result one can get, but it is true that, if $M$ is a complete Riemannian manifold, then for almost all $p^*\in M$ the set $F_{p^*}$ you define in the question has mea …
- 10.6k