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This may not reallly be an answer that you like, but I think that, maybe you misunderstood what Ben McKay was trying to describe. Here is a more explicit, extrinsic description that may help:
Suppose that $M^m\subset\mathbb{E}^n$ is an isometrically embedded submanifold of Euclidean $n$-space. Let $\gamma:(a,b)\to M^m$ be a smooth curve in $M$ and let $v:(...

13

No. If $G$ is $2$-transitive $-$ or even transitive on pairs $-$ then
$X$ must be the set of vertices of a regular simplex,
which has isometry group $S_n$. But there are plenty of
examples of $2$-transitive groups $G$ properly contained in $S_n$
(such as the $ax+b$ group if $n$ is a prime power and $n>3$).

answered Mar 30 at 23:30

Noam D. Elkies

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12

I agree with Ben McKay and Robert Bryant that the best way to introduce parallel transport to students, or to provide some motivation and intuition for it, is via an extrinsic approach, i.e., by the example of a tangent vector field to a surface whose derivative along a curve is orthogonal to the surface. Then I explicitly compute for them the parallel ...

10

If any officer can move more than $\delta$, then that officer can simply chase down the fugitive. Thus, I propose modifying the question to allow each officer to move at most distance $\delta$, but they all pull from a given pool of movement of size $c\delta$. Then we can ask what values of $c$ guarantee the capture of the fugitive. We can squeeze $c$ from ...

9

I prefer to introduce the Levi-Civita connection before the concept of parallel transport. On the surface, parallel transport seems like it should be more intuitive and easier to visualize geometrically. On the other hand, as you yourself indicate, it's not so obvious.
Although I have a preference for introducing manifolds abstractly and not as a submanifold ...

9

I like the bike wheel interpretation introduced by Mark Levi (A “bicycle wheel” proof of the Gauss–Bonnet theorem Exposition. Math. 12.2 (1994), 145–164).
It takes no time in class and helps to build right intuition.

9

Indeed that paper I cited in the comments describes how to determine all
symmetries of a polygon $P$ in $O(n)$ time.
The polygon is first translated so that its centroid is at the origin.
Then a "representation" $L(P)$ of $P$ is constructed.
$L$ is an $n$-tuple of pairs $(d_i,\alpha_i)$, where $d_i$ is the length
of polygon edge $i$, and $\alpha_i =...

answered Mar 25 at 13:36

Joseph O'Rourke

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8

CLAIM. The only fair cutting with at least one quadrilateral is the square grid (2). It was already shown in the original answer (below) that there cannot be $n$-agons for $n\ge 5$.
LEMMA. No fair cutting can contain a non-simple quadrilateral (i.e. one with a non-simple vertex, intersection of more than 2 lines).
PROOF of Lemma. The proof in my original ...

7

The answer to question 4 is no - every fair cutting is affinely equivalent to one of your four. This implies the answers to questions 1, 2, and 3 are no, yes, and yes, respectively. (Question 1 was already answered by Yaakov Baruch.)
My argument will use Yaakov Baruch's partial answer. Beyond what he does, there is one tricky algebraic calculation, and the ...

5

This is the "reflection property" of the hyperbola. In the context of "academic references", it is good practice to cite the original source, which is
• Apollonius of Perga, 200 BCE, Treatise on Conic Sections, book III, proposition 48; English translation by T.L. Heath, page 116 (Cambridge University Press, 1896).
Alternatively, you may ...

answered Apr 18 at 15:06

Carlo Beenakker

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5

These are called Minkowski functionals, see Wikipedia article.
They come up in the study of locally convex vector spaces.

5

Here is, I think, a proof that the only fair cutting with no triangles is the one you call (2). Hopefully some ideas might help answer other questions.
I call “block” a finite union of cells.
Lemma.
If, in a triangle-free fair cutting, there exists a triangular block such that one its (open) sides is not cut by any other line, then there exists a smaller ...

3

Fix small $\epsilon>0$. We shall show that the union of balls with center in a given ball of radius $\epsilon$ must have bounded perimeter. Note each ball $D_i$ has boundary expressible as a polar graph $$r=f_i(\theta)$$ where $f_i$ are uniformly bounded and uniformly Lipschitz. Then the boundary of $\cup_iD_i$ is given by the polar graph $$r=\sup_if_i(\...

3

### Does convergence of convex sets in Hausdorff distance implies convergence of the complementary sets?

Yes. For each unit vector $v$ in $\mathbb{R}^d$ and for any set $B \subseteq \mathbb{R}^d$, let $$r(B,v) = \inf_{b \in B} \langle b,v\rangle,$$ where $\langle x, y \rangle$ is the inner product of $x$ and $y$. It is not too hard to show that for any $B,C \subseteq \mathbb{R}^d$, $$|r(B,v) - r(C,v)| \leq d_H(B,C),$$ with the appropriate conventions regarding $...

3

I did not see the comment section under Yaakov Baruch's answer to this question. I think these ideas are interesting enough to be posted as an answer; I'm making it comminuty wiki. This provides a negative answer to your first question: all cells have to be triangles or quadrilaterals.
I call “block” a finite union of cells.
Consider a cell with at least 5 ...

3

This is only a partial answer. Though it hints that the situation is already complicated enough even asymptotically for $m$ large.
In the following we define $\lambda(n, m):=\min D/d$, where minimum is taken over all possible sets of $m$ points in $\mathbb{R}^n$.
Turns out, asymptotically, for $m$ large we have $\lambda(2,m) = (12/\pi^2)^{1/4}m^{1/2}$, ...

2

$\DeclareMathOperator{\area}{area}$ The region $X$ is a non-convex polygon and for simplicity I will asume it is simply connected. Its vertices are of two types: convex vertices and non-convex vertices. Denote by $V_+$ the set of convex vertices and by $V_-$ the set of non-convex vertices. For a vertex $v$ denote by $\alpha_v$ the exterior angle at $...

2

I think the answer is no. There is a metrizable version of the Tangent Disc Topology, namely where instead of extending the Euclidean topology in the upper half-plane to all of $\mathbb{R}$ you extend it to a countable subset of $\mathbb{R}$. This will be locally path-connected and connected (see Counterexamples in Topology by Steen/Seebach), but not ...

2

This paper (and its references) may help:
O’Rourke, Joseph. "An extension of Cauchy’s arm lemma with application to curve development." In Japanese Conference on Discrete and Computational Geometry, pp. 280-291. Springer, Berlin, Heidelberg, 2000.
Springer link.

answered Apr 2 at 14:29

Joseph O'Rourke

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2

First of all, we should assume that the original knot $K(t)$ is twice differentiable and has non vanishing curvature, so that its principal normal $N(t)$ exists and the curve $c(t)$ you want is well defined. Then we will have $c(t)=K(t)+r N(t)$.
By the tubular neighborhood theorem, for small $r$ the perturbation $c(t)$ will be isotopic to $K(t)$; however, ...

1

I am not sure if this is known. Two remarks.
Consider a slightly different problem, where your quantity is changed into the one that measures the surface area instead of the volume.
Take a body of constant width in $R^3$, say, K. Then $K-K=B$ (Euclidean ball), i.e., the numerator for the new quantity is constant independently of the direction (all ...

1

Well, I still havent found a solution for a hemisphere, but for a full sphere the solution was normalizing my cube hilbert curve around the zero and then mapping it with the equations from "Mapping cube to a sphere"
The results:

1

It seems that one can in many cases construct an extended metric space from what I'll call a merometric space $X$ (i.e. the kind of partially defined metric space considered in the question) by proceeding as follows:
Call a finite sequence $\gamma$ of points $\gamma_i\in X$ a polygonal path if the distances $d(\gamma_i,\gamma_{i+1})$ are all defined.
Assign ...

1

It turns out this statement is not true without additional conditions on the space. See this counterexample for a closed convex set in as strictly convex Banach space which does not admit a metric projection from the origin (every strictly convex Banach space is Busemann).

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