24

Yes, there is such a body. Actually there is one very close to the standard unit ball and containing disjoint representatives of each combinatorial type (but these representatives are very small).
Indeed, every combinatorial type of a $d$-polytope has a realization which looks as follows: there is a "large" $(d-1)$-dimensional facet and the remaining ...

20

It is known (and follows from an easy calculation) that solutions of the isoperimetric problem on a surface have constant geodesic curvature. In his 1887 classic Leçons Sur La Théorie Générale Des Surfaces Et Les Applications Géométriques Du Calcul Infinitésimal, G. Darboux states, without indicating a proof, that a surface for which all of the curves of ...

answered Jun 26 '12 at 11:59

Robert Bryant

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17

The following paper shows that if chronological order on $\mathbb R^n$ is defined by cone
(i.e., $x\in \mathbb R^n$ chronologically precedes $y\in \mathbb R^n$ iff $y − x$ belongs to some fixed cone)
then any bijection which preserve the chronological order has to be linear.
Alexandrov A.D. Contribution to chronogeometry Canad. J. Math. - 1967.- V.19, N.6. ...

17

There are several related and very interesting problems and theorems:
Schinzel's theorem - solves the problem in $\mathbb{R}^2$ using so-called Schinzel circles. It seems intuitively clear that it generalizes to higher dimensions by slightly adjusting radius of a hypersphere so that it contains exactly the same lattice points as its section in lower ...

16

Guy, Unsolved Problems In Number Theory, problem D22: Simplexes with rational content. "Are there simplexes in any number of dimensions, all of whose contents (lengths, areas, volumes, hypewrvolumes) are rational?"
Guy notes the answer is "yes" in 2 dimensions, by Heron triangles. Also "yes" in three dimensions: "John Leech notes that four copies of an ...

answered Aug 3 '11 at 6:00

Gerry Myerson

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15

Yes, such $A$ exists, and the radius can be much smaller than $10$; indeed radius $2^{1/2}$ suffices regardless of the area of $E$.
Write the ellipse $A(E)$ as $ax^2+bxy+cy^2 \leq 1$. By reduction theory of binary quadratic forms, we can choose the ${\rm SL}_2({\bf Z})$ transformation $A$ to obtain coefficients satisfying $|b| \leq a \leq c$. But $a \geq ...

answered Feb 9 '12 at 5:30

Noam D. Elkies

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15

Assume $\gamma$ is smooth and it has two points $p_1$ and $p_2$ with different curvatures, say $\kappa_1>\kappa_2$.
Then one can touch $\gamma$ at $p_2$ from inside by a $g(\gamma)$ such that $g(p_1)=p_2$.
Since $\gamma$ and $g(\gamma)$ bound the same area, they intersect at some other points.
(In fact at least 2, so together with $p_2$ it will be ...

14

You are asking for conditions on a metric space $X$ for which any distance preserving map $X\to X$ is bijective. (Usually isometry is defined as bijective distance preserving map).
Well, there are arbitrarily bad spaces $X$ such that the only distance preserving map $X\to X$ is identity.
In this case distance preserving maps (well, the only one) form the ...

14

Dear Marco,
If you take unparameterized geodesics (the quotient of the unit co-sphere bundle by the action of the geodesic flow), here is an answer:
The space of geodesics of a Riemannian or even Finsler manifold of dimension $n$ is itself a manifold of dimension $2n - 2$ in a variety of special, albeit important, cases. For example, the space of geodesics ...

14

Another good place to start is Donaldson's 8 page paper "A new proof of a theorem of Narasimhan and Seshadri" (available here) - this is a little less bulky than Atiyah-Bott but has the same flavour and concerns the same circle of ideas. Following Liviu Nicolaescu's advice (reading this paper and looking up what you need when you need it, e.g. stability of ...

13

I've got a letter from Idjad Sabitov which answer the question completely. Here is a short extract from it:
half-torus has rigidity of second order (Rembs' theorem, see Е. Rembs. Verbiegungen hoeherer Ordnung und ebene Flaechenrinnen. Math. Zeitschrift 36 (1932) or Ефимов, УМН, 1948, т.3, вып.2, стр. 135)
Any second order rigid surface does not admit ...

13

No. Let $G = \mathbb{Z}/p$ and $Y = S^1$. The group $G$ acts on $Y$ in the usual way (the generator acts as rotation by $2\pi/p$), and we have $Y/G \cong S^1$. Thus all the homology groups of $Y/G$ above degree $1$ vanish, but the homology groups of $G$ are nonzero in infinitely many degrees.

13

This is not really an answer, but I think it might help you look at things a little bit differently. It is not at all clear that Cartan chose the word 'torsion' to describe the 'translation' component of the curvature because it was related to the torsion of a curve in flat space or had anything to do with developing maps associated to what are now called "...

answered Feb 18 '13 at 22:28

Robert Bryant

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12

To 2, the answer is yes. This is easy. Re-enumerate points according to their cyclic order. For every fixed $k$ ($1\le k<n$), the product
$$
D_k:= \prod_i |P_i P_{i+k}|
$$
(where the indices are taken modulo $n$), equals
$$
D_k=\prod_i 2\sin (t_i/2) = 2^n\exp \left(\sum_i \log\sin(t_i/2) \right)
$$
where $t_i$ is the (oriented) angle from $P_i$ to $P_{i+...

12

1. This is the answer under the assumption that the condition 2. means exactly what it says.
Consider a regular triangle with side $2+\varepsilon$ and three diameters in the middles of its sides. If you construct the half-circles towards the triangle on these diameters, you obtain the desired example.
Now about the four copies.
Lemma. Assume that the two ...

12

Take a look at a notion of rigid geometric structure defined by Gromov, see for instance here for the detailed definition and examples. Reproducing the definition here would take a bit too much room, but, I think, it matches the notion of rigidity you have in mind.

11

There cannot be an asymptotic answer, because $\Delta r_i$ can be as small as $c/r_i$ (with $c \rightarrow 1/2$ as $i \rightarrow \infty$), but $\Delta r_i$ is of order $(\log r_i)^{1/2} / r_i$ on average. Equivalently, $\Delta(r_i^2)$ can be as small as $1$ but is of order $\sqrt{\log r_i}$ on average. One can also construct gaps of length $\gg \log r_i / ...

answered Aug 17 '11 at 3:01

Noam D. Elkies

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11

It turns out that these polyhedra that have congruent vertices and faces have a name. They are the Noble Polyhedra. If one insists that they also be convex the Noble polyhedra are the regular polyhedra plus the disphenoids mentioned in Douglas Zare's answer.
When one allows intersecting faces, however, new collections turn up, such as the stephanoids, ...

11

This problem is one of the easiest applications of Frenet formulas for planar curves and can be found in differential geometry textbooks.
Some minor corrections: First, $q$ is usually called "turning number" rather than "winding number". (The winding number is how many times a curve goes around a marked point; the turning number is how many times its ...

11

Assuming that "geodesic" in this question means "simple closed geodesic", then every complete hyperbolic surface $S$ of finite area is "bad": You cannot even separate an arbitrary pair of points. The reason is that the union of simple closed geodesics on $S$ is nowhere dense (even more, its closure has Hausdorff dimension 1) by the result of Birman and ...

11

It seems easier to answer the (current) title than the body of the question, so here goes: The notion of "space" has been vastly enlarged in the 20th century, even including some notions that Riemann would probably not recognize as sufficiently like classical spaces to deserve the name. Topological spaces (and their relatives like uniform spaces and ...

answered Nov 17 '12 at 3:51

Andreas Blass

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11

Yes, it is. Take a point P, and let us check if it belongs to Nth Brillouin zone. The Bragg planes (rather, lines, as we are in $R^2$) that we have to cross while going from the origin $O$ to $P$, correspond to the points $L$ of the lattice $\Gamma$ such that $P$ is closer to $L$ than to $O$.
In other words, we are looking for the number of the points $L$ of ...

10

Besides our paper that Alessandro mentions, Sageev, Whyte and I also have a paper "Maximally Symmetric Trees" which comes pretty close to doing what you ask. For certain trees it gives nice descriptions of $Aut(T)$ in terms of the representation into the quasi-isometry group $QI(T)$ (which, by the paper "Un groupe hyperbolique est determine par son bord" of ...

10

Well, you'll never need more than $k-1$ extra dimensions. Let $\hat{v}_1 = v_1 \oplus (1,0,\ldots, 0)$, $\hat{v}_2 = v_2 \oplus (a,1,0,\ldots,0)$, $\hat{v}_3 = v_3 \oplus (b,c,1,0,\ldots, 0)$, etc., and choose $a, b, c, \ldots$ to ensure orthogonality. But this does not tell us under what conditions we can get by with fewer extra dimensions.
If we write $\...

10

The spherical and hyperbolic versions may be proved in a uniform way.
Consider the cross product $\times$ on $\mathbb{R^3}$ or on $\mathbb{R}^{2,1}$. If the vertices of the triangle are $a,b,c$ thought of as vectors in the unit sphere or hyperboloid, then the line through $a,b$ is perpendicular to $a\times b$, etc. The altitude of $c$ to $\overline{ab}$ is ...

10

One of the standard generalizations is Knaster's conjecture: for every function $f: \mathbb{S}^{n-1}\rightarrow \mathbb{R}^m, m\lt n,$ and $k=n-m+1$ points $p_1, \dots, p_k \in \mathbb{S}^{n-1}$ does there always exista rotation $\rho \in SO(n),$ such that $f(\rho(p_1) = \dots = f(\rho(p_k)).$ That this is true for $k=2$ is a theorem of H. Hopf (which ...

10

Dick Lashof once told me that "differential topology is about first derivatives; differential geometry is about second derivatives".

answered Dec 12 '12 at 4:09

Steven Landsburg

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9

The finite case is pretty cheap. Take the free group $G$ with countably many generators $R_1,R_2,\dots$. Consider any injection $S:F(G)\to \mathbb N$ where $F(G)$ is the set of finite subsets of $G$ such that $S(W)$ is different from any index of a generator contained in a word $w\in W$ and take $E$ to be all (irreducible) words that do not start with $R_{S(...

9

I'm not exactly sure what he has in mind. If the smooth Poincare conjecture is false in 4 dimensions, then one could imagine taking a connect sum with a fake 4-sphere (homeomorphic but not diffeomorphic to $S^4$), and getting a manifold which is homeomorphic but not diffeomorphic to a hyperbolic 4-manifold.
However, if the smooth Poincare conjecture is true,...

9

Well, all simplicial polytopes can be made rational (i.e. to have rational coordinates) by perturbing vertices. Similarly, all simple polytopes can be made rational by perturbing hyperplanes. This covers all regular polytopes except for the 24-cell which already has a natural rational realization (see WP page). This resolves Q1. I am not sure about Q2.

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