User anton petrunin - MathOverflow

Perpetuum Mobile

2010-01-06T03:41:55Z

In 2 hours after posting this, I realized that preserving Liouville measure solves the problem completely. Sorry for disturbing...

Construction of perpetuum mobile: Consider room with mirror walls formed by two arcs of two ellipses with common foci and two segments of on bisecting perpendicular for the focuses as on the picture:

Place one-point-bodies with the same temperature in each focus --- they radiate and

All rays from blue focus come to the red one.
Big portion of rays from red focus goes back to it-self (that are all rays which reflect in bisector) while the rest goes to the blue focus.

Thus red focus getting hotter than blue one; i.e., we have a perpetual motion machine of the second kind...

Questions:

Why exactly it does not work? My guess is: if instead of one-point-bodies we have bodies with real size (no matter how small) it will no longer work, but I'm too lazy to do calculations, and I also it should be a good explanation (with no calculations).
For those of you who think it is not math, here is math formulation: Assume instead of one-point-bodies we have very small bodies of arbitrary shape. Then physics tells us it should not longer work. BUT I do not see mathematical proof of it...

Comment:

I know this construction from Vladimir Troitsky. A similar (but not as elegant) construction appears in comment on the Brain Teaser in the September 1972 issue of Physics Education (page 414). (maybe earlier?) --- thanks to Scott Carnahan for the ref. On page 446, there is a "solution", it only says that "it would not work because of the finite sizes of [the bodies]". Next year (June 1973, p.292) a letter with a better explanation was pubilshed in the same journal, this "better explanation" roughly says that it does not work by a "well established law".

Answer by Anton Petrunin for Theorems in Euclidean geometry with attractive proofs using more advanced methods

2010-07-11T15:34:29Z

Take a triangle with a circle $\Gamma_0$ tangent to two of three sides (you may also think that sides of the triangle are made out of circle arcs). Construct a chain of circles $\Gamma_1,\Gamma_2,\dots$ on such a way that $\Gamma_{n+1}\not=\Gamma_{n-1}$ is tangent to $\Gamma_n$ and two of the sides of triangle. Prove that $$\Gamma_6=\Gamma_0.$$

I do not know the proof, but I was told that it is hard to do without knowing elliptic functions.

P.S. I do not know the references --- please feel free to add it :)

Answer by Anton Petrunin for Frobenius Theorem

2010-07-07T13:37:40Z

I assume that $M$ is compact. [More generally you may assume that the vector fields are complete i.e. they have infinite integral curves.]

All diffeomorphism obtained by integrating your vector fields give an $S^3$-action on your manifold --- your assumption is just a reformulation in terms of Lie algebra.

For any compact Lie group acting smoothly on a manifold there is a invariant Riemannian metric --- this can be constructed as an average of a given metric by the group action.

Euclid with Birkhoff

2010-07-05T13:27:27Z

I'm looking for an short and elementary book which does Euclidean geomety with Birkhoff's axioms.

It would be best if it would also include some topics in projective and hyperbolic geometry.

Why: I will have to teach a course in "Foundation of geometry"; students should have rough an idea "what is mathematical proof". My plan is to spend 2/3 of time on Euclidean geometry then do a bit of affine and projective geometry and end up at Klein model for hyperbolic geometry. Surprisingly I can not find a reasonable book (or two) which would cover this subjects the way I want. On the other hand there SO MANY books on the subject that I could easely miss one.

Archaeogenetics

2010-01-17T16:54:36Z

This question is meant to be applied to recover historic information from genetic data. The following model, is (probably) the simplest possible which takes recombinations into account.

First, let us introduce some terms:

human is a finite set of numbers in (0,1); further these numbers will be called "scars".
population is a finite set of humans

One can perform one operation with a population:

recombination; i.e. take two humans $A$ and $B$ from the population, choose randomly a number $x\in(0,1)$ and produce a new human $C$ which has scar $x$, all scars in $A$ which are $ < x$ and all scars in $B$ which are $ > x$.

Assume someone starts with a population formed by two empty sets and does these operations for a while then stop. Assume you have all information about a portion of the population, BUT you do not know how this portion had been chousen. What one can say (even in which terms) about the history of population.

Comments:

It is not exactly mathematical problem, an answer might be something like "colored graph".
Clearly one can not say everything; yet there are many questions which can be answered. For example, assume you want to know if it is likely that at some moment your population was divided in two and there were no interbreeding between these groups for a while. You can even estimate "time" when they split. BUT I'm interested in a more abstract way to describe the history --- I want to say something without any assumptions.
Something is possible: First note that with probability 1, there is one-to-one correspondence between humans and scars. Given scar $x$, let us denote by $H_x$ the corresponding human. Assume in the portion of population you have two humans: human $A$ who has scars $x < y$ and nothing in between $x$ and $y$ and human $B$ who has scars $x < z$ and nothing in between. Assume $z < y$ then $H_z$ is a descendant of $H_x$. That gives a partial order on all such humans. Is it all one can do?

Answer by Anton Petrunin for Tetrahedron splitting/subdivision

2010-06-18T11:46:45Z

Answer 1: Look what happens on a face of the big tetrahedron where some edges of small ones come together: you have to make angle 180° out of some dihedral angles of the tetrahedra (which is about 70°) --- that is impossible.

Answer 2: There is the so-called Dehn invariant. If a polyhedron $X$ is split into a number of polyhedra $Y_1,Y_2,\dots,Y_n$ then the Dehn invariant of $X$ is equal to the sum of the Dehn invariants of $Y_i$.

For the regular tetrahedron, the Dehn inveriant is nonzero and proportional to the length of a side. Suppose you could split a regular tetrahedron with side $a$ into a number of tetrahedra with sides $a_1, a_2,\dots, a_n$. Then from the volume you have $$a_1^3+a_2^3+\dots+a_n^3=a^3$$ and from the Dehn invariant you have $$a_1+a_2+\dots+a_n=a.$$ It follows that there is no nontrivial splitting.

Answer by Anton Petrunin for Surfaces all of whose geodesics are both closed and simple

2010-06-18T13:09:16Z

From Guillemin's "The Radon transform on Zoll surfaces", it follows that there are deformations of $S^2$ which keep all geodesics closed AND simple.

Inverse function theorem for DC-functions

2010-06-08T22:33:37Z

I would like to have an inverse (or/and) implicite function theorem for DC-functions. It seems that I have right definitions, but I fail to prove it...

Definitions:

Let $h:\mathbb R^n\to\mathbb R$ be a convex function and $x\in\mathbb R^n$, the set of all linear functionals $\ell:\mathbb R^n\to\mathbb R$ such that $$h(y)\ge h(x)+\ell(y-x)$$ is called subdifferential of $h$ at $x$ --- it will be denoted as $\partial_{x}h$. (In general $\partial_{x}h$ is a nonempty bounded convex set)
$f:\mathbb R^n\to\mathbb R$ is called DC-function if it is a difference between two convex functions.
$F:\mathbb R^n\to\mathbb R^k$ is called DC-map each coordinate function of $F$ is DC.
$x\in\mathbb R^n$ is called regular value of a DC-function $f:\mathbb R^n\to\mathbb R$ if of $f=h_1-h_2$ for some convex functions $h_1$ and $h_2$ and $\partial_x h_1 + (-\partial_x h_2)\not\ni0$. Here $+$ denotes Minkowski sum and $(-\partial_x h_2)$ is reflection of $\partial_x h_2$ in the origin.
$x\in\mathbb R^n$ is called regular value of a DC-map $f:\mathbb R^n\to\mathbb R^k$ if $x$ is a regular value of $\ell\circ F$ for any non-zero linear map $\ell:\mathbb R^k\to\mathbb R$.

A variation of Minkowski sum

2010-06-04T12:19:27Z

I have to work with the following variation of Minkowski sum:

Let $\mathbb E$ be a Euclidean space and $K$ be a convex set in $\mathbb E\times \mathbb E$. Set $$K^+=\{\,x+y\in\mathbb E\mid(x,y)\in K\,\}.$$

Note that if $K=K_x\times K_y$ for some convex sets $K_x$ and $K_y$ in $\mathbb E$ then $K^+$ is the usual Minkowski sum of $K_x$ and $K_y$.

Questions:

Did anyone consider this construction?
Does it have a name?

The set of non-smooth points of a convex function is (m - 1)-rectifiable

2010-05-23T19:53:25Z

I am looking for a reference to the following result.

Let $f:\mathbb R^m\to\mathbb R$ be a convex function. Then $f$ is differentiable at all points of outside of a countable union of $(m-1)$-rectifiable sets.

Comments:

$n$-rectifiable set is an image of Lipschitz map from bounded domain in $\mathbb R^n$
I checked Federer's "Geometric Measure Theory", but I might miss the right place.
Extract from the Greg's answer (for those who are lazy to read the paper): In this paper, it is given a complete characterization of subsets of nondifferatiable points of a convex function. Namely, it is proved that $A$ is a such a set if and only if it can be covered by countably many graphs of DC-functions.

Answer by Anton Petrunin for Maximal area coverable by $k$ disjoint isosceles triangles contained in a triangle of area 1.

2010-04-28T16:42:04Z

Three trivial observations (not an answer).

Let us denote by $a_k(\Delta)$ the maximal covered area in $\Delta$. Then

for any right triangle $a_2=1$,
for any acute triangle $a_3=1$,
for any triangle $a_4=1$.

Two discs with no parallel tangent planes

2010-03-08T16:19:01Z

Is it possible to construct smooth embedded of 2-discs $\Sigma_1$ and $\Sigma_2$ in $\mathbb R^3$ with the same boundary curve such that there is no pair of points $p_1\in \Sigma_1$ and $p_2\in \Sigma_2$ with parallel tangent planes?

Comments:

The question is inspired by this; you will find there a construction three such discs with no triples of points.
More formally, "the same boundary curve" means that $\Sigma_1=f_1(D^2)$ and $\Sigma_2=f_2(D^2)$ for some smooth embedding $f_1$ and $f_2$ of the unit disc $D^2$ such that $f_1|_ {\partial D^2}\equiv f_2|_ {\partial D^2}$.

Ramified cover of 4-sphere

2009-12-12T17:49:58Z

Is it true that any closed oriented $4$-dimensional manifold can be obtained as a result of the following construction:

Take $S^4$ with a finite collection of immersed closed 2-manifolds (with transversal intersections and self-intersections) and construct ramified cover of $S^4$ with a ramification of order at most 2 only at these submanifolds.

Comments:

Two related questions: Ramified covers of 3-torus, Ramified covers of $S^n$
According to Feighn's Branched covers according to J.W. Alexander any closed oriented 4-manifold is a branched cover of $S^4$ with a ramification along 2-skeleton of 4-tetrahedron embedded in $S^4$ (which is not at all a 2-manifold).

Intrinsic metric with no geodesics

2010-02-17T16:29:27Z

It seems that I have the needed example, but I want it to be simple and self-explaining...

Construct a nontrivial complete metric space $X$ with intrinsic metric which has no nontrivial minimizing geodesics.

Definitions:

A metric $d$ is called intrinsic if for any two points $x$, $y$ and any $\epsilon>0$ there is an $\epsilon$-midpoint $z$; i.e. $d(x,z),d(x,y)<\tfrac12 d(x,y)+\epsilon$.
A minimizing geodesic is nontrivial if it connects two distinct points.
A meric space is nontrivial if it contains two distinct points.

Comments:

Clearly, $X$ can not be locally compact.

Minimal surface in a ball

2009-11-03T17:27:18Z

Assume a minimal surface $\Sigma$ has boundary on the unit sphere in the Euclidean space and $r$ is the distance from $\Sigma$ to the center of the ball. Is it true that $$\mbox{Area } \Sigma\ge \pi(1-r^2).$$

Comments:

If $r=0$, the statement follows directly from the monotonicity formula.
If $\Sigma$ is topological disc the answer is YES, see answer of Oleg Eroshkin below.
There is an analog in all dimension and codimension for area minimizing surfaces, see Alexander, H.; Hoffman, D.; Osserman, R. Area estimates for submanifolds of Euclidean space. 1974.
The general question is formulated as a conjecture in 1975 --- see comment of Ian Agol.

Answer by Anton Petrunin for On Alexandrov embedding theorem

2010-04-22T02:54:00Z

Is the metric embeddable as the boundary of a convex subset of 3?

YES, it is a limit case of standard Alexandrov's theorem. Moreover one can choose any embedding of cone at infinity and construct the embedding. This is a theorem of Olovyanishnikov --- one of three students of Alexandrov who died in the war.

Is the embedding unique?

NO, but I suspect it is unique once you fixed the convex embedding of the cone at infinity. It might follow from the proof of Pogorelov's theorem but I was not able to check his proof.

Are there generalizations of 1-2 to complete noncompact surfaces of nonnegative sectional curvature?

I'm not sure what you mean --- if it has strictly positive curvature at one point then it is automatically $\mathbb R^2$. If it is $\mathbb R^2$ then it is all the same.

Answer by Anton Petrunin for Is there Domain Invariance for Alexandrov spaces?

2010-04-16T04:22:37Z

The following lemma from Grove--Petersen, A radius sphere theorem does the trick.

Lemma 1. Let $X$ be a compact Alexandrov space without boundary. Then $X$ has a fundamental class in Alexander-Spanier cohomology with $\mathbb Z_2$ coefficients; i.e. $\bar H^n(X,\mathbb Z_2 ) = \mathbb Z_2$.

Why: First note that it is true for compact spaces --- in this case the map has $\mathbb Z_2$-degree one. Moreover, in this case the same it true for any continuous map which is injective around one point in the target.

Now let $X$ and $Y$ be $m$-dimensional Alexandrov spaces, $\Omega\subset X$ be an open subset and $f:\Omega\to Y$ be an injective continuous map and $y=f(x)$. One can use $f$ to construct a continuous map between sperical suspensions over spaces of directions $\mathbb S\,\Sigma_x\to \mathbb S\,\Sigma_y$ which is injective around one point in the target --- take a small sherical neghborhood $W\ni y$ and collapse everything outside of $W$ to the south pole of $\mathbb S\,\Sigma_y$. (We can do it sinse small spherical neighborhood of a point $x$ in an Alexandrov space is homeomorphic to cone over space of directions at $x$.)

The same is true for the second question --- link of any pseudomanifold is a pseidomanifold, thus it has $\mathbb Z_2$-fundamental class.

Answer by Anton Petrunin for Number of subgroups in a Bieberbach group.

2010-04-15T17:29:13Z

That is not an answer. I want to give an example where the argument of Erdős does not work directly.

Consider an action of group $\Gamma$ on $\mathbb R^3$ generated by the reflections $r_1, r_2 and $r_3$ correspondingly in the lines $x=z=0$ and $x+1=z=0$ and $x-y=z-1=0$.

Each of the reflections $r_i$ generate a maxiamal $\mathbb Z_2$-subgroups, all of them are nonconjugate. These groups corespond to three singular circles, say $\Sigma_i$ in the factor $X=\mathbb R^3/\Gamma$.

Let us try to mimic argument of Erdős. Take subsets $X_i$ of $X$ of midpoints $m$ between $x\in X$ and a closest $x_0\in\Sigma_i$ to $x$. As in the argument of Erdős we have* $\mathrm{vol}\, X_i>\tfrac{1}{2^3}\cdot\mathrm{vol}\, X$. BUT $X_1\cap X_3$ has interior points and here argument brakes into parts.

Comments

Since fixed point sets are 1-dimensional, it would be enough to take $m\in [xx_0]$ such that $\tfrac{|mx_0|}{|xx_0|}=\tfrac1{2\sqrt[3]{2}}$. But even in this case one has interior points in $X_1\cap X_2$ (the borderline in this example seems to be $\tfrac13$).
There is a natural bisecting hyperplane for any two affine subspaces. We may use it to cut a cylinder domain around each fixed point set of a maximal subgroup. The projection of these cylinders in $X$ gives Voronoi-like domains, but they do not cover whole space in general --- that is OK as far as we have lower bound on their volumes...

Answer by Anton Petrunin for Improving a sequence of 1s and -1s

2010-01-15T14:20:23Z

"Quasiperiodic" means that there is a function $L:\mathbb N\to\mathbb N$ such that any sequence of length $\ell$ appears in any sequence of length $L(\ell)$. You can get a quasiperiodic sequence as a limit, but you you will not get more. The later follows from your remark (which is added later).

P.S. If you start with a quasiperiodic sequence, then you can get a new sequence as a limit. But, the same way, you also can get your original sequence from the new one.

P.P.S. It seems to be interesting to consider quasiperiodic sequences with $L(\ell)=const\cdot\ell$. Was it considered anywhere? Is it true that if $L=\tfrac{3}{2}\cdot\ell$ then sequence is periodic?

Diameter of m-fold cover

2009-12-04T00:28:20Z

Let $M$ be a closed Riemannian manifold. Assume $\tilde M$ is a connected Riemannian $m$-fold cover of $M$. Is it true that $$\mathop{diam}\tilde M\le m\cdot \mathop{diam} M\ ?\ \ \ \ \ \ \ (*)$$

Comments:

A complete solution is given by S. Ivanov [it can not be marked as accepted due to software limitations].
This is a modification of a problem of A. Nabutovsky. Here is yet related question about universal covers.
You can reformulate it for compact length metric space --- no difference.
The answer is YES if the cover is regular (but that is not as easy as one might think).
The estimate $\mathop{diam}\tilde M\le 2(m-1)\cdot \mathop{diam} M$ for $m>1$ is trivial.
We have equality in $(*)$ for covers of $S^1$ and for some covers of figure-eight.

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.)
This question is related; 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 article that were reffering 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 Orbifold fundamental group in terms of loops?

2010-04-06T15:49:35Z

If you want to think in terms of loops, then choose a local lift of each loop in every chart of some atlas on such a way that they agree by some of transition maps. Then think about homotopy with such liftings.

In case your orbifold is spin and oriented and dim $\ge 3$, you may also pass to the double cover of frame bundle and think there.

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.

Answer by Anton Petrunin for In a locally CAT(k) space, does uniqueness of geodesics imply the lack of conjugate points?

2010-04-04T00:03:23Z

Consider a surface of revolution with an equator $\ell$ of lenght $2{\cdot}\pi$ such that its Gauss curvature $$K=1/\left(1+\sqrt[5]{\mathrm{dist}_ \ell}\right).$$ Choose $z\in \ell$ and let $\Sigma=B_{\pi/2}(z)$. Clearly $\Sigma$ is a $\mathrm{CAT}(1)$-space it has just one pair of conjugate points (say $p$ and $q$ --- the ends of $\Sigma\cap\ell$) and it has unique geodesics between each pair.

It remains to make geodesics extensible. To do this, we take $\Lambda=(S^1\times [0,\infty), d)$ with flat metric and concave boundary $\partial \Lambda=\partial\Sigma$. Then we glue $\Lambda$ and $\Sigma$ along the boundary.

The metric on $\Lambda$ is completely described by curvature $k(u)$ of its boundary [$u\in \partial \Lambda=\partial \Sigma$]. We only need to choose a function $k$ which is

on one had is large enough so that the glued surface still has unique geodesics between each pair (in particular $k(p)=k(q)=\infty$).
on the other hand is $\int_{S^1} k<\infty$, so that glued space is locally compact.

I believe it is possible...

One-step problems in geometry

2009-12-08T20:56:46Z

I'm collecting advanced exercises in geometry. Ideally, each exercise should be solved by one trick and this trick should be useful elsewhere (say it gives an essential idea in some theory).

If you have a problem like this please post it here.

Remarks:

I'm collecting such problems for many years. The current collection is here, it is about 80 problems.
At the moment, I have just few problems in topology and in geometric group theory and just one in algebraic geometry.
Thank you all for nice problems --- I gave bounty to one, but would give it to 4 problems if I could :). Some of them are in the collection by now --- please post more...

Universal group?

2010-03-30T23:30:31Z

I can construct a finitely presented group $G$ with the following property (which I use to construct something else).

Given a finitely preseted group $\Gamma$, there is a subgroup $G'\le G$ of finite index such that $$\Gamma=G'/\langle\mathrm{Tor}\, G'\rangle ,$$ where $\mathrm{Tor}\, G'\subset G'$ is the set of all elements of finite order.

I think to call such group $G$ universal.

Questions:

Was it already constructed?
Does it already has a name? Is there any closely related terminology?

P.S.

The group which I construct is in fact hyperbolic.
My construction is simple, but it takes 2--3 pages. Let me know if you see a short way to do it.

Answer by Anton Petrunin for The Symmetry of a Soccer Ball

2010-03-29T18:45:22Z

Only soccer ball or dodecahedron.

Clearly 3 hexagons can not meet at one vertex. Thus we have only 3 choices for one vertex:

3 pentagons
2 pentagons + 1 hexagon
1 pentagons + 2 hexagon

Note that if $[pq]$ is an edge then $p$ has the same type as $q$ (the type is determined by angle at $[pq]$). Thus the polyhedron is completely determined by one vertex. Further:

Once you have a vertex of the first type you have a regular dodecahedron.
If you have a vertex of the second type then you will get one hexagon surrounded by pentagons. Then it is easy to see that you can not continue.
For the third type you will get a soccer ball or "truncated icosahedron" as some people call it :)

Answer by Anton Petrunin for Contractible manifold with boundary - is it a disc?

2010-03-27T16:54:10Z

Given a function $\psi:\mathbb R\to \mathbb R$, set $$\Psi=\psi\circ\mathrm{dist}_ {\partial M},\ \ \ \ \ f=\Psi\cdot(R-\mathrm{dist}_ p)$$ for some fixed $R>\mathrm{diam}\, M$.

Further, $$d\,f= (R-\mathrm{dist}_ p)\cdot d\,\Psi-\Psi\cdot d\,\mathrm{dist}_ p$$ Thus, we may choose smooth increasing $\psi$, such that $\psi(0)=0$ and it is constant outside of little nbhd of $0$ so that $\Psi$ is smooth. (It is possible since the function $\mathrm{dist}_ {\partial M}$ is smooth and has no critical points in a small neighborhood of $\partial M$.) Note that $d\,\Psi$ is positive muliple of $d\,\mathrm{dist}_ {\partial M}$. Thus $d_x\,f=0$ means that geodesic from $x$ to $p$ goes directly in the direction of minimizing geodesic from $x$ to $\partial M$, which can not happen.

Now we can apply Morse theory for $f$...

Answer by Anton Petrunin for Is every smooth function Lebesgue-Lebesgue measurable?

2010-03-26T22:12:42Z

It seems that your example of bijection that sends one Cantor set with positive measure to an other Cantor set with zero measure can be made $C^\infty$.

Am I missing something?

Metric on one-point compactification

2010-03-18T02:43:55Z

Is there a standard construction of a metric on one-point compactification of a proper metric space?

Comments:

A metric space is proper if all bounded closed sets are compact.
Standard means found in literature.

From the answers and comments:

Here is a simplification of the construction given here (thanks to Jonas for ref). Let $d$ be the original metric. Fix a point $p$ and set $h(x)=1/(1+d(p,x))$. Then take the metric $$\hat d(x,y)=\min\{d(x,y),\,h(x)+h(y)\},\ \ \ \ \hat d(\infty,x)=h(x).$$

A more complicated construction is given here (thanks to LK for ref), some call it "sphericalization". One takes $$\bar d(x,y)=d(x,y)\cdot h(x)\cdot h(y),\ \ \ \ \hat d(\infty,x)=h(x).$$ The function $\bar d$ does not satisfies triangle inequality, but one can show that there is a metric $\rho$ such that $\tfrac14\cdot \bar d\le \rho\le \bar d$.

Comment by Anton Petrunin

2010-07-10T16:32:43Z

The point is if you have an action of Lie algebra then it gives a (local) action of the Lie group. You can glue a global action if you fields are complete, but this will be in general an action of SIMPLY CONNECTED Lie group (and $SO(3)$ is not s.c. and its cobver is $SU(2)=S^3$).

Comment by Anton Petrunin

2010-07-10T16:27:58Z

@Anirbit, Sorry, I did not read your comment correctly and said something irrelevant... I do NOT need "completeness of the geodesics" --- I need "completeness of the fields". A field is complete if it defines a global flow; i.e. one can start at any point and go along integral curve to the field for arbitrary time. I'm sorry I could not find a nice book --- the problem is I did not ready study the subject --- ask someone else OR if you know how to get from Lie algebra to its Lie group then try to mimic the same argument for your vector fields and you will get what you want...

Comment by Anton Petrunin

2010-07-08T13:02:50Z

@Anirbit (1) instead of compactness you may assume that your vector fields are complete (see my answer) --- that is to rule out case when your manifold is an open set in a big manifold with an $S^3$-action. (2) Yes (3) There is no canonical metric, you may start with any, pass to an average and get an invariant one --- it is possible if the group is compact. (4) Essentially you need a link between Lie algebra and Lie groups... --- I will think of a good book...

Comment by Anton Petrunin

2010-07-06T16:03:42Z

Yes, "it moves on to the fancy stuff a lot sooner"...

Comment by Anton Petrunin

2010-07-06T14:40:51Z

I checked Prenowitz and Jordan --- too long... I could not find Moise's "Elementary Geometry From An Advanced Standpoint" in the library but it is also 500 pages --- that probably means it is wrong choice; the review of Coxeter in MAthSciNet suggests the same.

Comment by Anton Petrunin

2010-07-06T13:17:39Z

I checked this book --- it will definetely scary any student...

Comment by Anton Petrunin

2010-07-06T13:12:29Z

Thank you. I checked the book, it is written for school students, a lot of motivation, but it does not go far enough for me (even in Euclidean geometry).

Comment by Anton Petrunin

2010-06-24T14:23:58Z

Still, could you include the def of Sasaki metric?

Comment by Anton Petrunin

2010-06-23T21:12:19Z

Essentially, you do not understand the definition. I would suggest to ask any geometer around --- it is very easy to explain by "talking", but by "writing" it is hard to add anything to the definition...

Comment by Anton Petrunin

2010-06-21T08:20:51Z

@Hugo, it is corrected now

Comment by Anton Petrunin

2010-06-18T12:04:46Z

The Dehn invariant of cube is $0$.

Comment by Anton Petrunin

2010-06-16T10:08:47Z

Where did you see it? My feeling that "minimal" can not be the right choice; maybe barycentric (?). Also, you probably want to work only with Hadamard spaces (i.e. curvature $\le0$ and s.c.), is it correct?

Comment by Anton Petrunin

2010-06-10T16:57:33Z

This theorem only says that implicit function is DC if it exists... Just in case, here is a link to the paper: math.psu.edu/petrunin/papers/scans/DC.pdf

Comment by Anton Petrunin

2010-06-10T10:09:13Z

@Greg, Yes --- Thurston and Conway.

Comment by Anton Petrunin

2010-06-04T22:03:28Z

Thank you, "sumset" sounds nice. By accident it was the first name which I came up with...