Search Results

Not necessarily. Consider the sphere $(\mathbb{S}^2,g)$ with its usual metric and give it a new metric $hg$, where $h\leq1$, $h$ has three local minima $h(p_1)=0.7,h(p_2)=0.8,h(p_3)=0.9$ (where $p,q,r …

If $V$ is closed in $\mathcal{O}$, then $\tau^V$ is continuous at $x_0$. If $V$ is not closed, then it is not hard to find counterexamples (e.g. imagine $\mathcal{O}=\mathbb{R}^3$, $V$ is an open disk …

I was thinking of a way to prove this and I realised that for my approach the lemma from the title would be useful, and it´s an interesting question on its own. Obviously it is true if the manifold is …

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: …

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_{ …

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 …

$K(M,g)$ 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.

Those constants don't exist for any $d\geq4$, here is an idea of why. For each $\varepsilon>0$ let $A_\varepsilon=\{(x_1,\dots,x_d)\in[-1,1]^d;\lvert (d-1)x_d-\sum_{i=1}^{d-1} x_i\rvert\leq\varepsilon …

It seems $\gamma:\mathbb{R}\to\mathbb{R}^2$ (I assume $\gamma$ is injective and continuous) is indeed a line. My argument is very similar to the one by Ilkka Törmä (I thought I could write a shorter o …

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 …

(This answer was posted before the convexity condition on the curve $\gamma$ was added to the question) Suppose you have any finite set of points in $\mathbb{R}^2$, and rotate $\mathbb{R}^2$ so that t …

As Matt F. says, his answer is not optimal, but the optimal solution, for most polygons (see (!!) below) comes from a similar construction using just arcs of circumference and segments. This answer gi …

My comment turned answer: Any smooth $m$-manifold $M$ admits a complete Riemannian metric (for example, as this answer says, any manifold embeds into some Euclidean space as a closed subset by Whitney …