User yoav kallus - MathOverflow

Answer by Yoav Kallus for Is there an "accepted" jamming limit for hard spheres placed in the unit cube by random sequential adsorption?

2013-05-19T16:59:32Z

The process you describe is usually called Random Sequential Addition (RSA). In this paper, Torquato, Uche, and Stillinger compute the saturation density up to $d=6$ (see Table I).

For $d=3$ they have $\phi_s\approx 0.38278$ as the saturation density.

Answer by Yoav Kallus for When does the rigidity matrix of a graph have full row rank?

2013-05-08T20:41:54Z

The term used by Connelly and Whiteley is self-stressed, and they discuss this notion extensively in their paper "Second-Order Rigidity and Prestress Stability for Tensegrity Frameworks."

Answer by Yoav Kallus for Constraint optimization problem for any dimensionality $n>1$.

2013-04-23T19:15:41Z

Since you asked so nicely, here we go: Let $a_k = \sum_{q=0}^{n-1}b_q \exp(2\pi i k q/n)$ (where $b_q=b_{n-q}^*$ by realness). Up to normalization factors I have dropped, $F=\sum_{q=0}^{n-1}|b_q|^2\exp(2\pi i q/n)$ and your constraints give $\sum_{q=0}^{n-1} |b_q|^2 = 1$ and $b_0=0$. So it is easy now to see that $F$ is maximum when $b_1=b_{n-1}=\tfrac{1}{\sqrt{2}}$. In real space, $a_k = C \cos(2\pi k/n+\phi)$.

Answer by Yoav Kallus for Strong notions of general position

2013-04-11T15:30:47Z

As I indicated in my comment, if you're looking for an intuitive and easy to state strong notion of general position, you might ask that the coordinates of the points are algebraically independent over the rationals (that is, there is no non-zero rational-coefficient polynomial in the coordinates that vanishes). This notion is used, for example in this paper on generic rigidity. This is a pretty arbitrary condition and may be unnecessarily strong, but it is also very powerful and could effectively capture the intuitive idea of a "generic" set of points.

Answer by Yoav Kallus for Packing and isoperimetrics

2013-04-09T17:55:34Z

László Fejes Tóth considered the planar version in his book Regular Figures. On page 175, he gives the following theorem, which verifies your guess (discs here does not necessarily mean circular discs):

If P denotes the total perimeter of n convex discs, each of area a, lying in a convex hexagon of area H without mutual overlapping, then $$\frac{P}{H}\ge \sqrt{\frac{n}{H}} p(na/H)\text.$$

The function $p(a')$ is defined as follows: it is equal to the perimeter of a circle of area $a'$ for $a'\le\pi/\sqrt{12}$ and equal to the perimeter of a rounded hexagon of area $a'$. A rounded hexagon being a regular hexagon of unit area, whose corners have been rounded to arcs of circles.

If there was no restriction to convex discs, this would easily imply the honeycomb conjecture. Fejes Tóth writes "It may be assumed that this proposition remains valid without the restriction to convex faces. In the case of isoperimetric faces this conjecture turns out to be true, but for faces of equal area its proof seems to involve considerable difficulties". These difficulties were eventually overcome by Thomas Hales, as pointed out by jc.

Regarding your last question "how to compute them", you can use Surface Evolver to simulate a bubble confined to the interior of a rhombic dodecahedron. Here, for example, is the result of a simulation of a bubble confined to a cube:

If you're interested, I can provide the input file I used to do the simulation.

Maximal cross sections of the Cartesian product of two planar domains

2013-04-06T19:39:24Z

Let $K$ and $L$ be two two-dimensional convex bodies, and consider their Cartesian product $K\times L\subseteq \mathbb R^4$. Now let $U_\theta\subseteq \mathbb R^4$ be the two-dimensional subspace spanned by $(1,0,\cos\theta,-\sin\theta)$ and $(0,1,\sin\theta,\cos\theta)$. Let $f(\theta) = \max_{\mathbf{x}\in\mathbb{R}^4} |(K\times L)\cap(U_\theta+\mathbf{x})|$, that is, the maximal cross-sectional area of $K\times L$ parallel to $U_\theta$. I have reason to believe, and would like to prove, that $f(\theta)$ cannot be unimodal. I call $f(\theta)$ unimodal if it is strictly increasing for $a\le\theta\le b$ and then strictly decreasing for $b\le\theta\le a+2\pi$. I will not say for now what my reasons are to believe that $f(\theta)$ cannot be unimodal, because I don't want to bias the reader's impression of what the best approach to this problem might be.

If anyone has any suggestions for approaches which might be beneficial to consider I would be very happy. Also, if you have some ideas for solving special cases, such as the case $K=L$, that would also be helpful. In the case $K=L$, we have $f(\theta)=f(-\theta)$, $f(0)$ is automatically a maxmimum, and $f(\pi)$ is either a local maximum or a local minimum. So, unimodality occurs if and only if $f(\pi)$ is a minimum and there are no other local extrema.

Answer by Yoav Kallus for Intersecting hyperplanes.

2013-01-25T14:46:04Z

Is this what you want?

Halperin, Israel The product of projection operators. Acta Sci. Math. (Szeged) 23 1962 96–99.

MathSciNet review: It is proved that if E1,E2,⋯,En are projections in a Hilbert space H and if T=E1E2⋯En, then Tm converges strongly as m→∞ to the projection E whose range is ⋂ni=1(EiH). For n=2 this was discovered by J. von Neumann [Ann. of Math. (2) 50 (1949), 401–485, p. 475; MR0029101 (10,548a)]. For general n, F. E. Browder [J. Math. Mech. 7 (1958), 69–80; MR0092070 (19,1057a)] says that weak convergence of Tm to E was previously proved by S. Kakutani; Browder modified Kakutani's result to allow an infinite sequence (En)n≥1 of projections.

Answer by Yoav Kallus for packing disks tightly in the plane

2012-12-14T21:33:12Z

Hopkins, Stillinger and Torquato give putative minima for $d_n$ for $n\le348$. In many cases these are improvements over a triangular lattice packing.

From the abstract:

The densest local packings of $N$ identical nonoverlapping spheres within a radius $R_{min}(N)$ of a fixed central sphere of the same size are obtained using a nonlinear programming method operating in conjunction with a stochastic search of configuration space. The knowledge of $R_{min}(N)$ in $d$-dimensional Euclidean space $R^d$ allows for the construction both of a realizability condition for pair-correlation functions of sphere packings and an upper bound on the maximal density of infinite sphere packings in $R^d$. In this paper, we focus on the two-dimensional circular disk problem. We find and present the putative densest packings and corresponding $R_{min}(N)$ for selected values of $N$ up to $N=348$ and use this knowledge to construct such a realizability condition and an upper bound. We additionally analyze the properties and characteristics of the maximally dense packings, finding significant variability in their symmetries and contact networks, and that the vast majority differ substantially from the triangular lattice even for large N. Our work has implications for packaging problems, nucleation theory, and surface physics.

Answer by Yoav Kallus for A "known" tangent half-angle formula?

2012-12-13T04:56:00Z

Using the tangent double-angle formula $\tan\gamma=\frac{2\tan\tfrac{\gamma}{2}}{1-\tan^2\tfrac{\gamma}{2}}$ we get \begin{align} \tan\gamma & = \frac{2\tan\tfrac{\beta}{2}\tan\tfrac{\alpha}{2}}{1-\tan^2\tfrac{\beta}{2}\tan^2\tfrac{\alpha}{2}} \\[10pt] & = \frac{2\sin\tfrac{\beta}{2}\sin\tfrac{\alpha}{2}\cos\tfrac{\beta}{2}\cos\tfrac{\alpha}{2}}{\cos^2\tfrac{\beta}{2}\cos^2\tfrac{\alpha}{2}-\sin^2\tfrac{\beta}{2}\sin^2\tfrac{\alpha}{2}} \\[10pt] & =\frac{\sin\beta\sin\alpha}{2(\cos\tfrac{\beta}{2}\cos\tfrac{\alpha}{2}-\sin\tfrac{\beta}{2}\sin\tfrac{\alpha}{2})(\cos\tfrac{\beta}{2}\cos\tfrac{\alpha}{2}+\sin\tfrac{\beta}{2}\sin\tfrac{\alpha}{2})} \\[10pt] & =\frac{\sin\beta\sin\alpha}{2\cos\tfrac{\beta+\alpha}{2}\cos\tfrac{\beta-\alpha}{2}} \\[10pt] & =\frac{\sin\beta\sin\alpha}{\cos\beta+\cos\alpha} \end{align}

A Wirtinger-like inequality involving two functions

2012-12-11T18:25:13Z

Let $f(t)$ and $g(t)$ be periodic functions on $t\in[0,2\pi]$. By using the Fourier series of the two functions, we can easily prove the inequality $$\left|\int_0^{2\pi}f(t)g'(t)dt\right|= \left|\int_0^{2\pi}f'(t)g(t)dt\right|\le \frac{1}{2}\int_0^{2\pi}[f'(t)^2+g'(t)^2]dt\text.$$

I have been trying to find a reference for this inequality because I need to use it to solve some problem. The closest I have been able to find is Pachpatte 1986, which gives $$\frac{1}{2}\int_0^{2\pi}\left[|f(t)||g'(t)|+|f'(t)||g(t)|\right]dt\le \frac{\pi}{2}\int_0^{2\pi}[f'(t)^2+g'(t)^2]dt\text.$$

The extra factor of $\pi$ is highly undesirable and the absolute values inside of the integral unnecessary for me. I can easily provide a short proof in the text, but if anybody can think of where the first inequality might appear, that would be better.

Answer by Yoav Kallus for Convex curves with many inscribed triangles maximizing perimeter

2012-12-06T15:50:05Z

This 1988 paper of Innami gives a construction of a convex curve, all of whose points are vertices of billiard triangles and therefore also maximal-perimeter inscribed triangles. In Innami's construction, all triangles are isosceles.

Answer by Yoav Kallus for Important open problems that have already been reduced to a finite but infeasible amount of computation

2012-11-13T20:17:42Z

Voronoi gave an algorithm to enumerate all perfect quadratic forms in $n$ variables and consequently to identify the densest lattice packing of spheres in $\mathbb{R}^n$.

Answer by Yoav Kallus for An exact counting solution for the number of points within a circle of radius $r$ centered on a lattice point in a $A_2$ hexagonal lattice

2012-10-20T22:09:14Z

This Mathematica formula reproduces the numbers in the OEIS and should be self explanatory:

n[r_] := Sum[ 1 + 2 Floor[Sqrt[r^2 - 3 x^2]], {x, -Floor[r/Sqrt[3]], Floor[r/Sqrt[3]]}] + Sum[2 Floor[ Sqrt[r^2 - 3 x^2] + 1/2], {x, -Floor[(r/Sqrt[3]) + 1/2] + 1/2, Floor[(r/Sqrt[3]) + 1/2] - 1/2}]

http://oeis.org/A053416 has the values corresponding to n[1/2], n[1], n[3/2], n[2] etc.

Assuming the lattice is generated by $(0,1)$ and $(\sqrt{3}/2,1/2)$, then the first sum counts the number of points of the form $(\sqrt{3}x,y)$ where $x,y\in\mathbb{Z}$, $\sqrt{3}x\le r$, and $3x^2+y^2\le r^2$. The second sum counts the number of points of the form $(\sqrt{3}x,y)$ where $x,y\in(\mathbb{Z}+1/2)$, $\sqrt{3}x\le r$, and $3x^2+y^2\le r^2$.

Generalization of the non-existence of a monostatic planar body

2012-08-27T21:09:01Z

Domokos, Papadopulos, and Ruina showed that there does not exist a convex planar rigid body of uniform density which has only one orientation of stable equilibrium and one orientation of unstable equilibrium when placed on a level surface under the influence of gravity. In three dimensions, such a body does exist. Therefore, let us stay in two dimensions, but consider different ways of trying to generalize this result (which I will refer to as Problem 0). Consider the following two problems.

Problem 1: Is there a convex body of uniform density $0<\rho<1/2$ which has only one stable equilibrium when floating in an incompressible liquid of unit density? This problem has been considered by Varkonyi, without any definite results, but with some results that suggest at least that for $\rho\simeq0$ and $\rho\simeq1/2$ no such body exists. Note that the limit $\rho\to0$ corresponds to the original problem and the symmetry of the case $\rho=1/2$ makes the existence of at least two stable orientations trivial.

Problem 2: Consider two congruent convex rigid bodies, one fixed and the other moving under the influence of a potential energy function given by the negative of the overlap area between the bodies. Can the shape of such a body be chosen so that there is only one stable equilibrium (namely the configuration of perfect overlap)? I first heard this problem from Paul Chaikin and then later saw some numerical results, unpublished at the moment, by Etienne Marcotte, a physics student at Princeton University, which suggest that no such shape exists.

It is clear that these problems have a common flavor. What is the best way to unify them? Here is my attempt: Let $K$ be a convex two-dimensional body, and denote by $K_{\mathbf{x},\theta}$ the congruent body obtained from $K$ by rotating about the origin by an angle $\theta$ and then translating by a vector $\mathbf{x}$. Let $\mu$ be a log-concave measure, and define a potential function $U(\mathbf{x},\theta)=-\mu(K_{\mathbf{x},\theta})$. For any fixed $\theta$, it is easy to show that $U(\mathbf{x},\theta)$ is convex as a function of $\mathbf{x}$ and therefore achieves its minimum in a convex region. Let $u(\theta)$ denote this minimum. Conjecture: if there are angles $\theta_1$ and $\theta_2$ such that $u(\theta)$ is monotonic non-decreasing for $\theta_1\le\theta\le\theta_2$ and monotonic non-increasing for $\theta_2\le\theta\le\theta_1+2\pi$, then $u(\theta)$ is constant.

For Problem 1, the density of $\mu$ is given by $C-\rho y$ for $0<y<\operatorname{diam} K$ and $C+(1-\rho)y$ for $-\operatorname{diam} K<y<0$ , where $C$ is large enough to make sure the density is non-negative. For Problem 2, $\mu$ is simply the Lebesgue measure restricted to $K$.

My question is whether anyone can point out a counterexample? I could not think of one, but it is possible I am missing something obvious. Other comments would be welcome as well.

Edit: since I did not receive any suggestions for counterexamples of my conjecture, I would like to ask a different question to go with the same exposition. The solution of Problem 0 is often identified as a special case of the four-vertex theorem (4VT). However, the conventional proof does not use the 4VT in any way. In the proof that does use the 4VT, it is not the boundary of the body that is the curve on which the theorem is invoked, and it is really unclear, at least to me, what the significance is of the curve on which it is invoked (see Appendix B of this paper). Can anybody clear this up? What is the connection, really, of the 4VT to Problem 0? Is there any hope of its extending to Problems 1 and 2 and the conjecture? What is the significance of the curve on which the 4VT is invoked in Problem 0?

Answer by Yoav Kallus for A function that is defined everywhere but has unknown values

2012-09-14T13:16:27Z

Your (1) could be substituted for pedagogical ease of presentation with the Conway's Game of Life "halting problem" (note that they are equivalent). That is, the function which for any starting configuration returns a value indicating whether the system eventually reaches a stationary state (return 0), periodic cycle (return 1), or never becomes stationary or periodic (return 2). You can show cool movies of Game of Life evolutions which would hopefully make it clear how unpredictable the behavior can be. The only tricky part pedagogically is assigning a number to each configuration, but that's not really that hard to understand.

Answer by Yoav Kallus for Building a Physical Model to Solve Sudoku

2012-08-14T21:40:30Z

I believe the so-called "adiabatic algorithm" fits your criteria, but perhaps not in the way you intend. In the adiabatic algorithm you encode the problem in a quantum Hamiltonian $H(s_1,s_2,\ldots,s_n)$ whose degrees of freedom are, say, Ising spins, in such a way that the ground state of $H$ is a simple product $|+\rangle|-\rangle...|+\rangle$, where the $\pm$ are bits encoding the solution. You bring the system to the ground state of $H$ by using a time-dependent Hamiltonian $H_t = (t/\tau) H + (1-t/\tau) H_0$, where $H_0$ is some simple Hamiltonian whose ground is well known is easily prepared at $t=0$. If $\tau$ is large enough, the final state at $t=\tau$ will be mostly in the ground state of $H$.

Answer by Yoav Kallus for How to find the minimum number of hyperplanes to define a convex hull

2012-05-16T16:13:57Z

There are two ways to represent convex polytopes: as the convex hull of its vertices, or as the intersection of the half-spaces whose boundaries contain the faces. If you store both of these representations, checking if a new constraint is redundant is easy: if all current vertices satisfy it, then so do all points in the convex hull. Now, the problem is (a) how large is the vertex representation? -- in general it can be exponential with the number of constraints -- and (b) how to update the vertex representation in case a new constraint is relevant. It might be more efficient in your situations, where constraints are added one by one, to store the vertex representation, but it might not, depending on the situation.

P.S. Also note that by duality, the problem you describe is equivalent to checking whether a new point lies in the convex hull of a set of old points. So as you search the literature, you might have more luck finding the dual version treated.

Answer by Yoav Kallus for Inequality of arithmetic and geometric means for the lattice polytopes?

2012-05-14T19:08:44Z

Undeleted for the sake of a full record

Your inequality seems to follow easily from the Brunn-Minkowski inequality. Namely,

$$ |(K+L)/2|^2 \ge \left[|K/2|^{1/n}+|L/2|^{1/n}\right]^{2n} \ge\left[4|K/2|^{1/n}|L/2|^{1/n}\right]^{n} =|K||L|\text. $$

Identifying lattices

2012-05-14T00:55:15Z

I wrote a program that numerically searches for lattices in $\mathbb{R}^d$ with high sphere packing densities. As I have been running the program, it has been able to find, in addition to well-known lattices such as the laminated lattice $\Lambda_d$ and the Coxeter Todd-related lattices $K_d$, a few interesting looking lattices, which I have been unable to identify. The lattices I have found so far are not as dense as $\Lambda_d$ or $K_d$, but are reasonably dense, and are nice integral lattices. Since I found them through a sort of a local optimization, I suppose they are probably perfect. I looked through the lattices listed on the Sloane-Nebe Catalog of Lattices and did not find any matches, but there do not seem to be many lattices listed there.

Here is an example of one of the lattices I find in $\mathbb{R}^{11}$. The Gram matrix is given by

$$ G = \left(\begin{array}{ccccccccccc} 8&3&3&2&3&4&4&4&4&4&4\\ 3&8&4&4&4&-1&4&-1&4&4&4\\ 3&4&8&0&0&-1&0&-1&0&4&4\\ 2&4&0&8&2&2&2&1&2&4&0\\ 3&4&0&2&8&3&4&-1&4&1&1\\ 4&-1&-1&2&3&8&0&4&0&0&0\\ 4&4&0&2&4&0&8&0&4&2&2\\ 4&-1&-1&1&-1&4&0&8&0&0&0\\ 4&4&0&2&4&0&4&0&8&2&2\\ 4&4&4&4&1&0&2&0&2&8&4\\ 4&4&4&0&1&0&2&0&2&4&8 \end{array}\right)\text. $$

The number of spheres in successive shells (equiv. theta function) are: norm 8, 308; norm 10, 320; norm 12, 680; norm 14, 1472. The packing density is $1/14\sqrt{7}=0.02699\ldots$ (number density for non-overlapping spheres of radius 1, compare to $0.03208\ldots$ for $K_{11}$ and $0.03125$ for $\Lambda_{11}$).

Does anybody know where I might be able to find if these lattices have been studied before?

Answer by Yoav Kallus for What recent discoveries have amateur mathematicians made?

2012-05-12T19:51:30Z

While this is on the front page again, I wanted to make mention of Joan Taylor, who discovered an aperiodic single tile, which she published with Joshua Socolar of Duke University in 2010. This is her bio blurb as it appears on their article in The Mathematical Intelligencer:

JOAN M. TAYLOR took up mathematics in 1991 at age 34 after being inspired by a magazine article on quasicrystals featuring Penrose’s rhombus tiling. She began but did not complete a degree, preferring to conduct her own research. Since then she has pursued tiling and related topics in abstract algebra and number theory including original work on constructible polygons. She likes to unwind with knitting and reading.

Answer by Yoav Kallus for Covering a Cube with a Square

2012-05-03T21:10:16Z

You can cut a $\sqrt{6}\times\sqrt{6}$ square into 24 pieces that then cover the $1\times1\times1$ cube. Two triangles from the figure below plus one parallelogram make up one $1\times1$square. parts of pieces sticking out to the left can obviously fit back in the right, so 18 pieces, plus 6 parts sticking out equals 24. You can improve on this by stitching pieces across the cube edge to make one bent piece and by stitching some of the parallelograms back to the triangles.

[Added by O'Rourke:] Just to make Yoav's construction more explicit, here is how two triangles and a parallelogram fit together to form a $1 \times 1$ square:

[Added by Kallus:] Here's an illustration of a construction similar to Fedja's construction but with only five pieces. The first figure is the $\sqrt{6}\times\sqrt{6}$ square. The second is the $2\times3$ rectangle, which we fold into a cube by taking away the two yellow squares, folding the remainder, and adding the squares as the two missing faces.

[Added by O'Rourke:]

Answer by Yoav Kallus for Examples of conjectures that were widely believed to be true but later proved false

2012-05-03T21:42:43Z

Euler's conjecture about the nonexistence of $n\times n$ Graeco-Latin squares for $n=4k+2$. Disproved for all $k>1$ by the so called Euler's Spoilers Bose, Shrikhande, and Parker.

Answer by Yoav Kallus for Tangled Knot Function

2012-04-29T03:26:46Z

If you parametrize a torus in $\mathbb{R}^3$ as $(x(u,v),y(u,v),z(u,v))$, $0\le u,v<1$, you can easily generate the torus knot $(3,q)$ (with crossing number $2q$) for $q$ large enough and not divisible by three by letting $f(u,v) = (u+3/q^2 \mod 1,v+1/q \mod 1)$ and $n=q^2$. So you just have to embed a continuum of these tori with $q$ varying continuously, and you'll have a function $f$ that generates every knot $(3,q)$. You can probably also improve on how $n$ grows with $q$.

Answer by Yoav Kallus for Metric on the set of Polyhedral Decompositions of a Compact Metric Space

2012-04-26T18:03:00Z

How about we look at the set $S$ of pairs of points $(x,y)$ that are in the same cell of $P_1$ but in different cells in $P_2$ or vice-versa. Then the volume of this set might work as a metric if you have a measure on $X\times X$.

Edit. On second thought, this might be more the kind of idea you're looking for: for any set $A$ let $A_\epsilon = {x:d(x,A)\le\epsilon}$ be its $\epsilon$-parallel body. The Hausdorff distance $d(A,B)$ is minimum value of $\epsilon$ such that $B\subseteq A_\epsilon$ and $A\subseteq B_\epsilon$. Therefore, for decompositions $P$ and $Q$ let $d(P,Q)$ be the minimum value of epsilon such that for each $A\in P$ there is a $B\in Q$ such that $A\subseteq B_\epsilon$ and for each $A\in Q$ there is a $B\in P$ such that $A\subseteq B_\epsilon$.

Parameterizing infinitesimal perturbations of the sphere using signed measures

2012-04-19T15:40:22Z

Let $\mathcal K^3$ be the space of convex bodies, with some metric $\delta$, and let $B$ be the unit ball. Let us define a volume-preserving perturbation of the sphere to be a continuous (substitute Lipschitz or something else if necessary for the sequel) map $K:[0,\epsilon_0]\to \mathcal K^3$ where $\epsilon_0>0$, $\delta(K(\epsilon),B) = \epsilon$, and $V(K(\epsilon))=V(B)$. Let $\delta(K,B) = V(K\setminus B)+V(B\setminus K)$, which is the symmetric difference metric, and let $\rho_K(x) = \max_{\lambda x\in K} \lambda$, which is the radial distance function, then I have functions $f_\epsilon(x) = (\rho_{K(\epsilon)}^3(x)-1)/3\epsilon$ that for $\epsilon>0$ all have $\int_{S^2}|f_\epsilon(x)|d^2x=1$ and $\int_{S^2}f_\epsilon(x)d^2x =0$.

Q1. Under what reasonable conditions can I say that $\lim_{\epsilon\to0}f_\epsilon = f_0$ where $f_0(x)$ is a signed measure? What kind of convergence will I have?

Q2. The signed measure $f_0(x)$ is in a sense the derivative $d\rho_{K(\epsilon)}(x)/d\epsilon$ at $\epsilon=0$. Can this relationship between the limit measure and the point-wise derivative be made exact by saying for example that $\lim\inf (\rho_{K(\epsilon)}(x)-1)/\epsilon = f_0(x)$ (i.e. the point-wise Dini derivative is given by the density of the limit measure)?

Q3. I would much rather have a result in terms of $h_K(x)$ (the support height) rather than $\rho_K(x)$ (the radial distance). Can a statement such as in Q2 also be made for the point-wise derivative of $h_{K(\epsilon)}(x)$? For example if I let $\delta(K,B)=\int_{S^2}|h_K(x)-1|d^2x$ and let $f_\epsilon = (h_{K(\epsilon)}(x)-1)/\epsilon$? Will I still have $\int_{S^2} f_0(x) d^2x = 0$?

Edit: I did a bit of reading and figured out part of my question. As I understand, Prokhorov's theorem guarantees that $f_\epsilon$ have a weak limit $f_0$. Also, some simple examples such as the convex hull of a sphere and a point above the north pole show that no stronger convergence is possible. So if I understood my reading correctly, that settles Q1. Q2 boils down to whether the limit taken in Q1 and the limit involved in taking the Radon-Nikodym derivative can be interchanged. I have not been able to find information and when that is possible.

Answer by Yoav Kallus for German mathematical terms like "Nullstellensatz"

2012-03-30T23:57:33Z

There is also the Quermassintegral (mixed volumes of the form $V(K,K,\ldots,B,B)$ where $B$ is the unit ball, see Wikipedia), which I'm not even sure is German (not a lot of Qs in German usually).

Generalized widths and reverse Urysohn inequalities

2012-01-27T06:33:36Z

This question is inspired by the discussion in MO questions "Local minimum from directional derivatives in the space of convex bodies" and "Bodies of constant width?" about generalized notions of minimum widths and constant widths. Some interesting ideas came up in those discussions which I would like to bring together with some ideas I had and some things I'm not clear on yet.

Let the $\mu$-width of a three-dimensional body $K$ be a function $$w_\mu(\vartheta)=\int_{S^2} h(\vartheta(\mathbf{u})) d\mu(\mathbf{u})\text,$$ where $\vartheta\in SO(3)$ is a rotation, $h(\mathbf{u})=\max_{\mathbf{x}\in K} \mathbf{u}\cdot\mathbf{x}$ is the support height function, and $\mu$ is a (signed?) measure on $S^2$ with $\int \mathbf{u} d\mu(\mathbf{u})=0$ (to ensure translation invariance). The standard width corresponds to a measure concentrated at two opposite points. The mean width corresponds to the uniform measure. The main properties I want to talk about are the minimum $\mu$-width (the minimum of $w_\mu(\vartheta)$) and bodies of constant $\mu$-width. The mean $\mu$-width is not interesting because it reduces to the mean standard width (assuming $\mu(S^2)\neq 0$).

The minimum standard width is the "mailslot width", the smallest mailslot through which the body can pass. If $\mu$ is concentrated uniformly on the equator (this is related to the spherical Radon transform) then the minimum width is the "loop width", giving the smallest length of string loop through which the body can pass (this follows from the fact that the mean width of a planar body is proportional to its perimeter). If $\mu$ is concentrated with equal weight at the vertices of a regular tetrahedron, the minimum "tetrahedral width" gives the linear size of the smallest regular tetrahedron that contains the body. Bodies of constant tetrahdral width are the rotors of a tetrahedral cavity (see "Bodies of constant width?").

Let us define the "harmonic support" (h.s.) of a function or measure on $S^2$ as the set of integers $n>1$ such that the projection of the function to the space of spherical harmonics of degree $n$ does not vanish. Also let $\mathcal{K}_I$ be the space of convex bodies such that the harmonic support of their height function is a subset of $I$. If $I$ and $J$ are disjoint and their union is ${n>1}$, I call $\mathcal K_I$ and $\mathcal K_J$ complementary spaces. Then the space of bodies of constant $\mu$-width is the complementary space to $\mathcal{K}_{h.s.(\mu)}$. Thus, it follows that bodies of constant width and bodies of constant loop width are the same.

Urysohn's inequality says that the ratio $vol/\bar{w}^3$, where $\bar{w}$ is the mean width is maximized by balls. In general, the ball also maximizes the ratio $vol/w_\mu^3$ among bodies of constant $\mu$-width $w_\mu$. I am interested in the complementary space, and whether $vol/w_\mu^3$, where $w_\mu$ is the minimum $\mu$-width, is minimized by balls among bodies in $\mathcal K_{h.s.(\mu)}$. Clearly, this holds for the standard width: among all centrally-symmetric bodies of a given volume, balls maximize the mailslot width (not true if central symmetry is not assumed). However, based on some experiments I made, I find that this is not true in general as a global statement. Still, I believe that balls are local minima. This is because it is pretty easy to show that if $K\in\mathcal K_{h.s.(\mu)}$ is not a ball, then for some $\alpha_0>0$ the body $K_\alpha=(1-\alpha)B+\alpha K$ obtains a greater ratio than that of the ball for all $0<\alpha<\alpha_0$. (See "Local minimum from directional derivatives in the space of convex bodies"). My question is, can you find a counterexample of my claim that $B$ is a local minimum of $vol/w_\mu^3$ among bodies in $\mathcal K_{h.s.(\mu)}$; or can you see a way of proving it?

Answer by Yoav Kallus for General Isoperimetric Inequality via Representation Theory of SO(n)

2012-01-26T22:09:14Z

For the second time today (see http://mathoverflow.net/questions/86742), I give the answer: see "Geometric Applications of Fourier Series and Spherical Harmonics" by Helmut Groemer. That book deals exactly with your question, and the answer is yes.

Answer by Yoav Kallus for Bodies of constant width?

2012-01-26T19:28:58Z

I found the reference I was looking for. The full list of cases under which $K$ is a rotor in a cavity shaped like the polytope $P$ is available on page 27 of the notes title "The use of spherical harmonics in convex geometry" by Rolf Schneider. They are available under "Course Materials" on his website. As I recall, there is one more non-trivial case in $d=3$ if the cavity is allowed to be open (e.g. a cone), and this case appears in the more complete list in Groemer's book.

Local minimum from directional derivatives in the space of convex bodies

2012-01-25T20:42:20Z

I have a function $f(K)$ defined on the space of three-dimensional convex bodies for which I want to show that the unit ball $B$ is a local minimum. I have been able to show if $K$ is not homothetic to $B$, then the function $g(\alpha) = f(K_\alpha) = f((1-\alpha)B+\alpha K)$ is positively sloped at $\alpha=0$. (If it is homothetic then $g(\alpha)$ is constant). This, of course, implies that for any $K$, there is $\alpha>0$ such that $f(K_\alpha)\ge f(B)$. However, what I actually want to show is that there exists $\epsilon$ such that if $d(K,B)<\epsilon$ then $f(K)\ge f(B)$ (where $d$ is the Hausdorff metric). Question: can I use the fact that the space of convex bodies is locally compact (i.e. the Blaschke Selection Theorem) to go from one result to the other?

Here are some more background and details which might be useful: the function is defined as a ratio $f(K)=f_1(K)/f_2(K)$ where $f_1(K) = V(K)^{1/3}$ is the cube-root of the volume, and $f_2(K) = \min_{\vartheta\in SO(3)} f_\vartheta (K)$ is the minimum of a family of functions that are each linear in the support height function of $K$, $h_K(\mathbf{u})$. Namely, $f_\vartheta(K) = \int_{S^2} h_K(\vartheta(\mathbf{u})) d\mu(\mathbf{u})$, where $\mathbf{u}\in S^2$, $\vartheta\in SO(3)$, and $\mu$ is some measure on $S^2$ that has $\mu(S^2)=1$ and $\int_{S^2} \mathbf{u} d\mu(\mathbf{u}) = 0$. Therefore, $f(\lambda K + \mathbf{t}) = f(K)$, and we can limit our attention to bodies $K$ with a mean width of $2$ and Steiner point at the origin. I have that the projection of $\mu$ to the space of spherical harmonics of degree $n$ never vanishes for $n>1$, and therefore $f_2(K) = 1$ if and only if $K=B$; otherwise, $f_2(K)<1$. Since $g_1(\alpha) = f_1(K_\alpha)$ has zero slope at $\alpha=0$ (by the definition of mixed volumes, the slope is given by the difference in mean widths of $K$ and $B$) and $g_2(\alpha) = (1-\alpha) + \alpha f_2(K)$, then $g(\alpha)$ is positively sloped at $\alpha=0$. I have tried to put more definite bounds on $f_1$ and $f_2$ as a function of $h_K(\mathbf{u})$. I think I can obtain $f_1(K)-f_1(B) \ge -c ||\nabla_0 h_K||^2$ (i.e. the $L^2$ norm of the magnitude of the gradient of the height function restricted to the sphere) and $f_2(K)-f_2(B)\le c' (\min_\mathbf{u} h_K(\mathbf{u}) - 1)$ (but not, it seems, $-c'' (\max h -1)$), but I'm not sure those give me anything.

If the answer to my original question is no, can you suggest a way to obtain my desired result ($B$ is a local minimum of $f$) by other means?

Comment by Yoav Kallus

2013-05-20T03:12:10Z

I believe polymer people call the correlation time for this kind of dynamics the "Rouse relaxation time," so that should give a clue to search for how it is calculated.

Comment by Yoav Kallus

2013-05-19T22:03:46Z

@Carlo, since Allen says "I expect this would converge quickly," I take it to be implied that the process is iterated.

Comment by Yoav Kallus

2013-05-19T18:45:12Z

The shape of the container should not make a difference in the limit of a very large container.

Comment by Yoav Kallus

2013-05-19T17:27:18Z

With unit vectors, I think this would be very hard. But if you're willing to go with Gaussians, I think you can calculate the conditional probability distributions without much trouble.

Comment by Yoav Kallus

2013-05-19T17:24:00Z

Also, I don't think anybody would have computed a value for the expected RSA saturation density for any specific size finite box.

Comment by Yoav Kallus

2013-05-16T22:16:11Z

I guess T doesn't have to be bisectible into two convex regions, but, still, I don't see a good reason why the minimum should be achieved.

Comment by Yoav Kallus

2013-05-16T21:08:02Z

Looking again at Sergei's answer to the other question, it seems like what you're asking for here should fail: let T=S∪(−S), then among all S′ such that T=S′∪(−S') (and therefore Δ(S)=Δ(S′)) we minimize the volume by letting S′ be the intersection of T with a half-space through the origin. Since this minimum is never achieved so long as the origin is in the interior, the minimum of vol(S)/Δ(S) is never achieved. Am I missing something?

Comment by Yoav Kallus

2013-05-16T20:50:01Z

Also worth noting, the even easier case: the minimum of $S\mapsto\operatorname{vol}(S)/\Delta(S-S)$ is actually known among two-dimensional convex bodies (the triangle).

Comment by Yoav Kallus

2013-05-16T20:35:20Z

Comment by Yoav Kallus

2013-05-15T22:28:16Z

The Wikipedia page on the bounded inverse theorem (en.wikipedia.org/wiki/Bounded_inverse_theorem), which is equivalent to the open mapping theorem, has some counterexamples if one of the spaces is not complete and if the map is not surjective.

Comment by Yoav Kallus

2013-05-14T01:36:56Z

Since you're looking for a basis for the full space, you're actually looking at the unitary group, not the Stiefel manifold. For topological considerations, look for example at Kuiper's theorem: en.wikipedia.org/wiki/Kuiper's_theorem

Comment by Yoav Kallus

2013-05-09T22:42:24Z

fixed some malformatted tex

Comment by Yoav Kallus

2013-05-08T19:05:28Z

I reiterate Anton's (trollish?!) question. I think this question should be closed unless it is edited. From the FAQ: "MathOverflow is not the right place to ask open problems. You should post questions you're actually seriously thinking about. If you're thinking about a well-known open problem, provide some background and ask about something specific related to the problem."

Comment by Yoav Kallus

2013-05-06T23:08:43Z

Indeed, as you have surmised, this is not the right place to ask for feedback on one's results. Please take a look at the "faq" and "how to ask" links at the top of the page.

Comment by Yoav Kallus

2013-05-05T17:31:25Z

Does the fact that for two different bicycles different points on the front track match the same point on the rear track lead to a problem with this argument?