User anton petrunin - MathOverflow

Answer by Anton Petrunin for Closed geodesic loops around points in compact manifolds

2013-05-20T02:01:35Z

Here is a standard argument, but I do not know a reference.

Choose the smallest $k>0$ such that $\pi_kM\ne 0$. Choose a spheroid which represents a nontricial element of $\pi_kM$. We can assume that the spheroid is swapped by an $\mathbb S^{k-1}$-parameter family of broken geodesics such that the length of each edge is smaller than the injectivity radius of $M$; denote it by $i_M$.

Start a natural curve-shortening process. You can choose one which keep the lengths of the edges below $i_M$ and such that the rate of length (or energy) decay is estimated through rate of change of the broken geodesic.

Note that there is a lower bound for the maximal length of broken geodesics; "maximal" means "maximal in the $\mathbb S^{k-1}$-family after spending arbitrary time in the process". Say, this value can not go below $i_M$.

It follows that after long time in the process, one broken line in the family almost does not change the length. Hence this it almost does not move; the later implies that and all the angles almost $\pi$. Pass to the limit and you get the loop.

Answer by Anton Petrunin for Can one (block) diagonalize the curvature matrix of 2 forms on a Riemannian manifold?

2013-05-17T04:55:37Z

Let us start it from the other end. Assume you can block-diagonalize the curvature tensor. Then in the most of coordinate sectional directions the curvature is zero.

In fact your curvature tensor equals to a curvature of product of few surfaces and maybe the real line --- this is a very special case.

Answer by Anton Petrunin for On a version of gradient descent

2013-05-16T18:45:51Z

I guess $|f'(y)|$ stays for $|d_yf|$. If so, $$|f'(y)| \geq \frac{f(y) - f(x^{\rm min})}{|y - x^{\rm min}|}.$$ Since $|x(t)-x^{\rm min}|$ is decreasing, the statement follows.

Answer by Anton Petrunin for Triangle area on surfaces of constant curvature

2013-05-12T18:16:20Z

All the "elementary derivations of that type" are cheating (it may look nice but it proves nothing).

The only elementary way to introduce area is adding it as an axiom (which is already kind of cheating). You have to say that there is a additive area-functional on the set of all polygons. Then you probably want to prove that this functional is unique (or include it in the same axiom).

It remains to notice that your functional $A$ satisfies the same properties and nothing left to prove.

Is a free alternative to MathSciNet possible?

2010-02-01T03:41:55Z

How could a free (i.e. free content) alternative for MathSciNet and Zentralblatt be created?

Comments

Some mathematicians have stopped writing reviews for MathSciNet because they feel their output should be freely available. (The Pricing for MathSciNet is not high, but it is not the point.)
Two related questions: Are there any good websites... and Errata database?; see also r-forum, nLab and wikademic.

What can be done (based on answers below)

One thing that can be really useful and doable is to create (and maintain) a database of articles (and maybe abstracts), where you can find all the articles that were referring to a given one.
Once it is done we can add lists of errors --- it will add something new and valuable for the project (but this will take a while).
The above two things might be already enough for practical purpose. It will be even better if it will attract enough reviewers to the project.

Answer by Anton Petrunin for Converse to Milnor's theorem on manifolds with nonnegative Ricci curvature.

2013-05-08T15:54:03Z

A better question is:

Given a group $\Pi$, is there a compact manifold $M$ with non-negative Ricci curvature such that $\pi_1(M)=\Pi$?

The answer is given in "On fundamental groups of manifolds of nonnegative curvature" by Wilking. Here is the main result:

Answer by Anton Petrunin for Measuring the distance of a convex set from a ball (Nikodym distance)

2013-05-06T18:04:13Z

For convex closed sets with uniform upper-diameter and lower-volume bounds, the Nikodym distance is equivalent to Hausdorff distance.

Your condition for the Hausdorff distance, implies that $K$ is close to any ball which does not contain $K$ and is not contained in $K$. Hence the statement follows.

Answer by Anton Petrunin for Does a connected manifold with vanishing Euler characteristic admit a nowhere-vanishing vector field?

2013-05-05T18:53:52Z

Yes, if M is closed and connected and χ(M)=0, then M admits a nowhere-vanishing vector field.

Start with generic vector field, it has zero of index $\pm 1$. Two zeros of opposite sign can kill each other (maybe it is called Whitney trick?).

So you get a field with zeros of the same sign. The result follows since the sum of the indexes is the Euler characteristic.

Answer by Anton Petrunin for Polynomial growth of the Betti number of balls of the Cayley graphs

2013-05-04T01:13:09Z

Assume the group is not free. Then its Cayley graph has a nontrivial loop. Applying shifts of the group you get a loop which starts at any element.

If the group grows faster then polynomially, then the number of disjoint cycles and therefore first Betti number grows faster than polynomial.

So, the answer is "yes" --- only free groups and groups of polynomial growth.

Answer by Anton Petrunin for Intersection points of straight line segment with Voronoi diagram

2013-05-03T19:29:29Z

The question seem to be nearly as hard as finding Voronoi diagram.

In other words, you have a collection of functions of the form $$f_i(t)=t^2+a_i{\cdot}t+b_i$$ and you need to find the maximal interval $[c_i,d_i]$ where $$f_i=\min_{j}\{f_j\}.$$

Yo will get the same answer, if instead of $f_i$, you use the functions $$h_i(t)=a_i{\cdot}t+b_i=f_i(t)-t^2$$ which are linear. I.e., your problem is equivalent to the classical problem in linear programming --- choose you favorite method.

Answer by Anton Petrunin for Positively curved manifold with a codimension 1 totally geodesic submanifold.

2013-04-30T21:08:47Z

Compact positively curved manifold with totally geodesic hypersurface has to be sphere, or its double cover has to be a sphere.

To prove this cut along the hypersurface and apply the soul theorem to each part.

Answer by Anton Petrunin for Alexandrov angles in Riemannian manifolds

2013-04-27T05:50:02Z

Your equality is two inequalities.

To show the upper bound you can use the triangle inequality --- come closer to $p$ along the geodesic and apply the local estimates. (This is the "first variation inequality" it holds in any metric space where angles defined.)

The lower bound follows since the injectivity radius at $p$ is positive. Indeed, if the angle is smaller, the geodesic $[\gamma_1(t)\gamma_2(\tau)]$ converges as $\tau\to0$ to an other geodesic distinct from $\gamma_1$.

This question is the baby case of so called "lemma about strong angle" in Alexandrov geometry it also can be called "first variation formula".

(Hope it helps, sorry if I misinterpret your question.)

Answer by Anton Petrunin for When does the finite union of convex sets have a hole in it?

2013-04-27T05:30:39Z

Let $K=K_1\cup\dots\cup K_n$ and $K_i$ are convex.

Consider the nerve $N$ of your covering $K_i$. Note that $N$ is homotopically equivalent to $K$. (To find $N$ you only need an algorithm which decides that given subcollection of $K_i$ has nonempty intersection.)

Calculate the homology groups of $N$ and you may get a "no" answer if you are (un)lucky.

Answer by Anton Petrunin for Riemannian manifolds with small geodesics and bounded curvature

2013-04-26T17:41:29Z

It seems that you want $M$ to be a sphere (although you do not mention it).

The natural example is a tube with curvature $-1$ closed by two spherical cups with curvature $1$. (This is a surface of revolution.) This example should have the minimal area and minimal diameter among the spheres with given length of closed geodesic. But at the moment I do not see a proof.

Some rough estimates will follow from the Klingenberg's result. Take the neighborhood $U$ of your closed geodesic of size near $e^{1/\ell}$ where $\ell$ is the length of the geodesic. Note that $U$ has to be a cylinder otherwise you end in singular point. It implies that diameter has to have order $e^{1/\ell}$.

Rough bounds for the area follow Rbega comment. They are not far from the area of the surface of revolution, somewhere in the range between $4{\cdot}\pi$ to $(4+2{\cdot}\sqrt{2}){\cdot}\pi$.

Answer by Anton Petrunin for What is the doubling dimension of convex functions?

2013-04-26T01:14:39Z

No the space is not doubling.

Take a strongly convex function, say $f(x)=x^2$. It is sufficient to show that there is $N(\varepsilon)$ which goes to $\infty$ as $\varepsilon\to0$, such that $\varepsilon$-neighborhood of $f$ contains $N(\varepsilon)$ points on distance $>\varepsilon$ from each other.

To see this take any finite collection of distinct points $\{x_i\}$ in $[0,1]$ and note that for all small $\varepsilon>0$ there is a function $f_i$ in the $\varepsilon$-neighborhood of $f$ such that $f_i(x_i)=f(x_i)+\varepsilon/2$ and $f_i(x_j)=f(x_j)-\varepsilon/2$ for all $i\ne j$.

Bolyai's construction

2010-11-07T23:46:58Z

Here is Bolyai's construction (in Klein model), which I learned recently from answer of Will Jagy to this question. It is a Compass-and-straightedge construction of asymptotically parallel line in absolute geometry.

Do you know an elementary proof showing that Bolyai's construction really does what it suppose to? "Elementary" means without calculations and without referring to the models.

Comments:

I know a simple proof in Klein model (thanks to A. Akopyan). (It is easy to guess from the picture above. Draw one more vertical line and note that cross ratios for two quadruples of points on two asymptotically parallel lines is the same.)
If there is a simple argument, I would include it in the course on Foundation of geometry which I'm teaching at the moment.
At the moment I do not know a simple argument in the Poincaré disk model.

Answer by Anton Petrunin for Decide inside/outside convex hull using only distances in graph

2013-04-22T23:27:39Z

Assume you want to know if $k_0$ lies in the convex hull of $k_1,\dots,k_n$. Consider the $n\times n$ matrix $A$ with the components $$a_{ij}=(d_{ij}^2-d_{0i}^2-d_{0j}^2)/2$$

(Note that $x^TAx\ge 0$ for any vector $x$; it means that the distances come from Euclidean distances.)

If $x^TAx=0$ for a nonzero vector $x$ with all nonnegative components then the answer is "yes".

Answer by Anton Petrunin for examples of space of direction at a point in an infinite dim Alexandrov space compact

2013-04-22T17:13:18Z

Yes, the space of direction of an infinite dim Alexandrov space can be compact at some point.

Take for example the pyramid with Hilbert cube as the base.

On closed simple curve with curvature at most 1

2013-03-31T19:16:35Z

I am looking for the reference to the following theorem.

I have to apply a similar statement, and it would be nice to trace the source. Please note, I know few proofs in fact it is Problem 3 in my collection of exercises.

Theorem. Let $\gamma$ be a closed simple plane curve with curvature at most 1. Then the disc bounded by $\gamma$ contains a circle of radius $1$.

Here is an illustration for those who are lazy to read.

P.S. So, do you think I can claim it as mine?

Answer by Anton Petrunin for Dose closed Alexandrov space admit a bi-Lipschitz embedding into $\mathbb R^N$?

2013-04-12T15:22:40Z

See Distance embedding (27.5) in our book

Answer by Anton Petrunin for Infimum of a finite number of distances in the plane

2013-04-02T13:25:58Z

It can be zero.

Take the standard metric on $\mathbb R^3$ and the one given in this example by S. Ivanov. Below I give simplification of his example which works in your case.

Simplified example. Choose two points $x$ and $y$ in $\mathbb R^2$ and construct a sequence of arcs $\gamma_n$ between them. For each $n$ choose small $\varepsilon_n>0$, so that $\varepsilon_n\to 0$ very fast. Choose disjoint $\varepsilon_n$-intervals on $\gamma_n$, so that it cover all $\gamma_n$ except set of lenght $\varepsilon_n$. Construct metric $d_1$ so that it is very cheap to go along each such interval, but compensate it by making very expancive to get to such interval, so you can not use it as a shortcut.

Let $d_2$ be the Euclidean metric. Then the $d$-length of $\gamma_n$ has order of $\varepsilon_n$. In particular $d(x,y)=0$.

Answer by Anton Petrunin for Difference between parallel transport and derivative of the exponential map

2013-03-31T17:16:13Z

Let $P_t:T_p\to T_{c(t)}$ be the parallel transport and $Q_t:T_p\to T_{c(t)}$ be your map. Given $J\in T_p$, the field $J(t)=Q_t(J)$ is a Jacoby field and $J'(0)=0$. Set $$W_t=P_t-Q_t.$$ It follows that

$W_t=O(t^2)$;
the value $W_t''$ can be expressed from the curvature tensor.
As you noticed $W_t(\tau)\equiv 0$ where $\tau=c'(0)\in T_p$.

I guess that as much as you can get.

Answer by Anton Petrunin for A question about closed curves

2013-03-28T20:58:06Z

Note that $S$ is a connected 1-dimensional manifold. Since it is not compact we get that $S$ is homeomorphic to $\mathbb R$, a contradiction.

Answer by Anton Petrunin for Diameter estimate of distance sphere of positive curved manifold

2013-03-21T22:49:19Z

I guess you want to ask is it true that $$\mathop{\rm IntrinsicDiameter}[S(r)]\le\mathop{\rm IntrinsicDiameter}[\tilde S(r)],$$ where $\tilde S(r)$ denotes the sphere of radius $r$ in the standard sphere.

This is true if $r\ge \tfrac\pi2$; it follows since $S(r)$ has bigger curvature than $\tilde S(r)$ in the sense of Alexandrov.
Note that if $r<\tfrac\pi2$ then $S(r)$ might be not connected; in this case $$\mathop{\rm IntrinsicDiameter}[S(r)]=\infty.$$ If sectional curvature $\ge 1$, I do not see other counterexamples. It reminds me some questions related to the conjecture that boundary of Alexandrov space is an Alexandrov space.
If you find a way to prove it then likely you will get some nontrivial corollaries of this conjecture say if $\Sigma$ is an Alexandrov space with curvature $\ge 1$ then $\mathop{\rm diam}\partial\Sigma\le \pi$ or perimeter of any triangle in $\partial\Sigma$ is at most $2{\cdot}\pi$.
If $r\le\tfrac\pi2$, it is possible to construct a short map $h_r\colon \tilde S(r)\to M$ so that its image covers $S(r)$. In particular $$\text{area}[S(r)]\le\text{area}[\tilde S(r)]$$ (which is obvious anyway). In general the image of $h_r$ contains creases which stick inside $S(r)$ which in principle might be used as a shortcut.
For Ricci curvature the statement does not hold even if $S(r)$ is connected. You may take a small disc in hyperbolic plane and take a warp product with the sphere to make the Ricci curvature of obtained manifold to be colose to $+\infty$. The sphere $S(r)$ will have intrinsic diameter bigger than $\tilde S(r)$ as far as $S(r)\ne\emptyset$.

Answer by Anton Petrunin for Does a Riemannian manifold with bounded geometry admit an isometric proper embedding into Euclidean space with uniformly thick tubular neighborhood

2013-03-18T17:07:18Z

It does not hold for hyperbolic plane. It follows since the volume growth of the hyperbolic plane is exponential, while volume growth of $\mathbb{R}^N$ is polynomial.

Answer by Anton Petrunin for Does convex set in Alexandrov space has positive reach?

2013-03-08T07:21:06Z

The answer is "no".

Say consider the doubling $M$ of plane region $F=\{y\ge x^2\}$. $M$ is known to be an Alexandrov space.

Let $A\subset M$ is the doubling of the intersection of $F$ with the disc with center $(0,-1)$ passing through $(1,1)$. Note that $A$ is convex but all the points on the parabola have two minimizing geodesics to $A$ (one in each of to copies of $F$ in $M$).

So $A$ has zero reach.

Answer by Anton Petrunin for What are the worst notations, in your opinion ?

2010-03-18T17:06:17Z

I hate the short cut $ab$ for $a\cdot b$. Everyone get used to it, BUT it creates very deep problem with all other notation; say you never can be sure what $f(x+y)$ or $2\!\tfrac23$ might be...

Also in modern mathematics people do not multiply things too often, so it does not have sense to make such a short cut.

Yet the shortcut $x^n$ is really bad one. One can not use upper indexes after this. It would be easy to write $x^{\cdot n}$ instead.

Answer by Anton Petrunin for Positivity of second fundamental form implies global convexity?

2013-02-18T20:11:31Z

From the comment of Willie Wong , the answer is NO.

Yet simpler example is $\mathbb{S}^1\times\mathbb{R}$, where you have locally convex discs which are not globally convex.

If you want "yes" as an answer, you have to assume bit more. For example simply connectedness plus sectional curvature $\le 0$ (It was proved by S. Alexander).

An other related statement: if an immersed hypersurface is locally convex plus curvature is nonnegative then it bounds an immersed locally convex ball. (this was conjectured by S. Alexander and proved by Gromov independently).

Answer by Anton Petrunin for Area of triangles vs. comparison triangles.

2013-02-17T23:59:52Z

No.

Blow a bubble of the same size at each integer point of $\mathbb R^2$. Clearly coarse isoperimetric inequality will hold.

On the other hand, the global metric on the plane can be made arbitrary close to the Manhattan metric, in particular there will be triangles which bound arbitrary large area while its comparison area is near zero.

Answer by Anton Petrunin for Do plane projections determine a convex polytope?

2013-02-13T17:15:34Z

Here is a more direct proof of this statement.

A Short Proof of Klee's Theorem by John J. Zanazzi

Comment by Anton Petrunin

2013-05-20T18:29:00Z

@Igor, the manifold of broken geodesics is not complete. So you have to be bit more careful, to make sure that there is a critical point. In other words, read my answer.

Comment by Anton Petrunin

2013-05-20T03:30:48Z

@Misha, Yes, I do use minimax. I made an update, it should be more clear now.

Comment by Anton Petrunin

2013-05-18T01:36:09Z

@Robert, You are right, but note that I state "curvature tensor equals to a curvature of product"; it does not mean that space is a product.

Comment by Anton Petrunin

2013-05-17T17:45:11Z

Do you know if $\cal{T}$ is finite-imensional?

Comment by Anton Petrunin

2013-05-17T17:41:22Z

P.S. For example, if sectional curvature has definite sign then the curvature tensor can not be block-diagonalized. Further, the rank of generic curvature tensor is $n\cdot(n-1)/2$, but any block-diagonalized has rank at most $\lfloor\tfrac n2\rfloor$

Comment by Anton Petrunin

2013-05-17T05:12:12Z

@Robinson1. I would better leave it as an exercise :) BTW, if it would be wrong then so is the original statement.

Comment by Anton Petrunin

2013-05-17T05:04:22Z

@Steven, this proof is cheating. The Pythagorean theorem is way simpler than existence of the area functional which is used in the proof (and many proofs of existence of the area use Pythagorean theorem).

Comment by Anton Petrunin

2013-05-16T23:37:29Z

@Alexandre, Euclid (and Kiselev) did not prove the existence, essentially they add the existence as an axiom, but they did not say that it is an "axiom". This axiom follows from the rest of axioms, but it takes 20 pages at least. Instead of unit square you have to use other normalization (which essentially defines curvature). A rigourous way to introduce area given in "Elementary Geometry From An Advanced Standpoint" by Moise 35 pages Euclidean plane onlyy,and in "Geometry: A Metric Approach with Models" by Millman and Parker 40 pages neutral plane and contains a gap.

Comment by Anton Petrunin

2013-05-16T18:33:11Z

@BS, check this question mathoverflow.net/questions/119953/… I would be very happy if you know a better answer.

Comment by Anton Petrunin

2013-05-15T04:50:17Z

It use the properties of the area which (if you look carefully) already include the original statement inside.

Comment by Anton Petrunin

2013-05-15T04:13:27Z

@Alexandre, I do not see what exactly you disagree with. A rigorous intro to area from the axioms takes 20-40 pages, and once it is done the formula is already proved. So these sort of "proofs" confuse poorly educated students and they prove nothing to those who know what area is.

Comment by Anton Petrunin

2013-05-13T23:20:44Z

Chapter 5 is "Affine-projective relationship" did you really mean this?

Comment by Anton Petrunin

2013-05-13T17:45:49Z

P.S. surface of constant curvature κ are spheres plane or Lobachevsky plane. All these things are "elementary" for me.

Comment by Anton Petrunin

2013-05-13T17:02:59Z

@Kofi, you ask for an elementary derivation. For me "Riemannian metric" and "integral" are not elementary and the geometry as it was used to be covered in the school (but not any more) is elementary.

Comment by Anton Petrunin

2013-05-13T02:17:12Z

@Sergei, yes sure, all I wanted to say is that if one knows what is area and curvature then there is nothing to prove.