User sergei ivanov - MathOverflow

Answer by Sergei Ivanov for Lower bound on $L^2$ norm of mean curvature in general dimensions

2013-05-16T20:29:04Z

I have no idea about the general case but in the convex case the sphere is indeed optimal. Moreover the $L^1$ norm of $H$ attains its minimum at the sphere (among the convex surfaces with the same area). To deduce the result for the $L^2$ norm, just apply Cauchy-Schwarz,

Let $A$ be the convex body bounded by $\Sigma$ and $B$ the unit ball in $\mathbb R^{n+1}$. Then the area $|\Sigma|$ is proportional to $V_1(B,A)$ and the integral of the mean curvature is proportional to $V_2(B,A)$, where $V_k(B,A)$ is the mixed volume of $k$ copies of $B$ and $n+1-k$ copies of $A$.

By the Alexandrov-Fenchel inequality, $\log V_k(A,B)$ is a concave function of $k$ ( $k\in\{0,1,\dots,n+1\}$ ). This fact yields a lower bound for $V_2(B,A)$ in terms of $V_1(B,A)=C(n)|\Sigma|$ and $V_{n+1}(B,A)=V(B)$, a constant. Namely $$ C(n)\int_\Sigma H = V_2(B,A) \ge V(B)^{1/n} V_1(B,A)^{(n-1)/n} = C_1(n)|\Sigma|^{(n-1)/n} $$ If $A$ is a ball, the inequality turns to equality because so does the Alexandrov-Fenchel inequality.

Answer by Sergei Ivanov for A characterization of Hilbert spaces?

2013-05-12T22:16:49Z

Yes it is true. Let me show that the existence of $\phi$ implies that the norm of $B^*$ is associated to an inner product. Then it follows easily that the spaces are Hilbert.

It suffices to verify that the norm is an inner product one on every two-dimensional subspace. (Indeed, this property is equivalent to the parallelogram law, which involves only two-dimensional configurations.)

Let $V$ be a two-dimensional subspace of $B^*$ (equipped with the restriction of the norm of $B^*$ ) and $V^*$ the dual of $V$ (equipped with the norm dual to this restriction). There is a natural map $\pi:B\to V^*$ dual to the inclusion $V\to B^*$ . Namely $\pi(x)(f)=f(x)$ for $x\in B$, $f\in V$. Note that $\pi$ does not increase the norm: $\|\pi(x)\|\le\|x\|$ for all $x\in B$.

Let $D$ be the unit ball of $V$ and $E\subset V$ the maximum-area ellipse contained in $D$ (i.e., the John ellipsoid of $D$). Let $\Sigma$ be the set of points where the boundaries of $D$ and $E$ meet. It is easy to see that $\Sigma$ contains at least two pairs of opposite points. The ellipse $E$ is a unit ball of a Euclidean norm $\|\cdot\|_E$ on $V$. Since $E\subset D$, we have $\|f\|_E\ge\|f\|$ for all $f\in V$ and equality is attained only if $x$ is proportional to an element of $\Sigma$.

Now consider two linearly independent vectors $f,g\in\Sigma$ and look at the distance between $f$ and $-g$. Since $\phi$ is an isometry, we have $$ \|f+g\|=\|\phi(f)-\phi(-g)\| \ge \|\pi(\phi(f)-\phi(-g))\| . $$ The r.h.s equals $$ \|\pi(\phi(f))-\pi(\phi(-g))\| = \|I(f)-I(-g)\|=\|I(f+g)\| $$ since $\pi$ and $I$ are linear and $f,-g\in\Sigma$.

Thus $\|I(f+g)\|\le \|f+g\|$ and therefore $$ \|I(f+g)\|_E^*\le\|I(f+g)\|\le \|f+g\| \le \|f+g\|_E . $$ But $I$ preserves the Euclidean norm, so the inequalities turn into equalities. In particular, $\|f+g\|= \|f+g\|_E$, hence $f+g$ is proportional to an element of $\Sigma$.

Thus we have proved that, for every $f,g\in\Sigma$, the normalized bisector $\frac{f+g}{\|f+g\|}$ also belongs to $\Sigma$. Since $\Sigma$ is a closed subset of an ellipse and contains more than two points, it follows that $\Sigma$ is the whole ellipse. This means that $D=E$, so the norm on $V$ is Euclidean. Q.E.D.

Remark. The proof would be much easier (in fact, nearly trivial) if we assumed in advance that $\phi$ is a restriction of an isometry between $B^*$ and $B$. Then it would be linear by Mazur-Ulam and one could just define the inner product of $f,g\in B^*$ by $2\langle f,g\rangle=g(\phi(f))+f(\phi(g))$.

Answer by Sergei Ivanov for Measuring the distance of a convex set from a ball (Nikodym distance)

2013-05-06T18:35:51Z

Let $r$ be the radius of the maximal ball contained in $K$ and $R$ the radius of the minimal ball containing $K$. I claim that $\frac rR>1-2\epsilon^{1/n}$. It follows that the Hausdorff distance from $K$ to a ball is bounded by $C_n\epsilon^{1/n}$. The argument does not use symmetry (although the constant can be improved in the symmetric case).

Let $v$ and $v'$ be unit vectors such that $Rv$ and $rv'$ belong to the boundary of $K$. Apply a reflection which sends $v'$ to $v$ and let $K'$ be the resulting body. Let $h$ be the homothety centered at $Rv$ with ratio $\frac{R-r}{2R}$. Then $h(K)$ does not intersect the interior of $K'$ because they are separated by the hyperplane through $rv$ orthogonal to $v$. On the other hand, $h(K)\subset K$ due to convexity. Thus $h(K)\subset K\setminus int(K')$, hence $Vol(h(K))\le Vol(K\Delta K')<\epsilon$. But $Vol(h(K))=\left(\frac{R-r}{2R}\right)^n=\frac1{2^n}\left(1-\frac rR\right)^n$. Hence $1-\frac rR<2\epsilon^{1/n}$ as claimed.

Added later.

Now let us show that the Nikodym distance from $K$ to a ball is bounded by $O(\epsilon)$. Let $f:S^{n-1}\to\mathbb R_+$ be the radial function of $K$. Then $$ Vol(K\Delta R_u(K)) =\frac1n \int_{S^{n-1}} |f(x)^n-f(R_ux)^n| dx \sim c(n) \int_{S^{n-1}} |f(x)-f(R_ux)| dx . $$ The last equivalence holds because we already know that $f$ is uniformly close to a known constant. Thus $f$ lies within $L^1$ distance $O(\epsilon)$ from every its reflection and hence from any rotation. Averaging over all rotations yields that $$ \int_{S^{n-1}}\int_{S^{n-1}} |f(x)-f(y)| dxdy \le C\epsilon $$ (where $C$ depends on $n$). This easily implies that $f$ lies within $L^1$ distance $O(\epsilon)$ from a constant. Indeed there is $r_0$ such that the volumes of both sets $A=\{x:f(x)\le r_0\}$ and $B=\{x:f(x)\ge r_0\}$ are at least half the total volume. Restricting the above integral to $x\in A$ and $y\in B$ shows that the integral mean of $|f(x)-f(y)|$ is at least 1/4 of the integral mean of $|f(x)-r_0|$ over $x\in S^{n-1}$. Thus $$ \int_{S^{n-1}} |f(x)-r_0| dx \le C_1\epsilon $$ for some $C_1=C_1(n)$. This means that $K$ lies within Nikodym distance $C_2\epsilon$ from the ball of radius $r_0$ and hence within Nikodym distance $C_3\epsilon$ from the ball of volume 1.

Isoperimetric inequality on a Riemannian sphere

2013-05-05T20:31:19Z

Consider a two-dimensional sphere with a Riemannian metric of total area $4\pi$. Does there exist a subset whose area equals $2\pi$ and whose boundary has length no greater than $2\pi$?

(To avoid technicalities, let's require that the boundary is a 1-dimensional smooth submanifold.)

If that fails, does there exists a set, say, with area between $\pi$ and $2\pi$ and length of the boundary no greater than that of the spherical cap of the same area? Or at least no greater than $2\pi$?

More generally, I am interested in any results saying that the isoperimetric profile of the round metric on the sphere is maximal in some sense (among all Riemannian metrics of the same area).

Notes.

The answer is affirmative for central symmetric metrics (i.e. if the metric admits an $\mathbb{RP}^2$ quotient). This follows from Pu's isosystolic inequality: in $\mathbb{RP}^2$ with a metric of area $2\pi$ there exists a non-contractible loop of length at most $\pi$. The lift to the sphere is a loop of length at most $2\pi$ dividing the area in two equal parts.
One should not require that the set is bounded by a single loop. A counter-example is the surface of a small neighborhood of a tripod (formed by three long segments starting from one point) in $\mathbb R^3$. Here one can divide the area in half by two short loops, but one loop would be long. (However one can cut off 1/3 of the area by one short loop.)
In $S^3$ the similar assertion is false, moreover the minimal area of the boundary of a half-volume set can be arbitrary large.

Answer by Sergei Ivanov for spectral radius monotonicity

2013-05-05T22:18:56Z

Update. This answer answers completely different question, see comments. Namely "positive" is substituted by "positive definite", norm is used instead of spectral radius, and quantifiers are different.

Not true in general: take $S=T=-I$. Then the inequality boils down to $\frac{b}{b+1}<\frac{a}{a+1}$ which is always false for $b>a>1$ .

For positive symmetric matrices, yes. Fix $a$ and let $b\to+\infty$. The l.h.s equals to $\rho((I-\frac1bS)^{-1}T)$ which goes to $\rho(T)$. And the r.h.s. is greater than $\rho(T)$. Indeed, the matrix $S':=(I-\frac1aS)^{-1}$ satisfies $|S'(v)|>|v|$ for all $v\in\mathbb R^n\setminus 0$ (where $n$ is the size of the matrices). Let $v$ be an eigenvector of $T$ corresponding to the maximal eigenvalue $\lambda=\rho(T)$. Then $|S'T(v)|>|T(v)|=\lambda |v|$, hence $\rho(S'T)>\lambda$ by the minimax principle.

Answer by Sergei Ivanov for Monotonicity of Loewner ellipsoid?

2013-04-10T18:19:31Z

No, the Loewner ellipsoid is not monotone w.r.t. inclusion. Let $K$ be a square, whose Loewner ellipsoid is its circumcircle. Let $L$ be any other ellipse through the four vertices of $K$. The Loewner ellipsoid of $L$ is $L$ itself but it does not contain the circle. (I assume that the Loewner ellipsoid is the minimal circumscibed one. If you meant the maximal inscribed one, just consider the dual bodies.)

The continuity of Loewner ellipsoid follows from its uniqueness via a standard compactness argument. Its volume is continuous for trivial reasons. If the Loewner ellipsoid $E(K)$ were discontinuous at some convex body $K$, then by compactness there would be a sequence $K_i\to K$ such that $E(K_i)$ converge to some ellipsoid other than $E(K)$ but of the same volume, contrary to the uniqueness.

Answer by Sergei Ivanov for The sum of same powers of all matrices modulo p

2013-04-07T20:21:54Z

The sum is zero for all $k<p^2-1$ .

Assume that $k$ is a multiple of $p-1$ and $k<p^2-1$ . Divide all matrices into classes of the form $\{A,A+1,A+2,\dots,A+(p-1)\}$ . Summimg the $k$th powers over such a class yields $$ \sum_{s=0}^k A^{k-s}\binom ks \sum_{j=0}^{p-1} j^s $$ where $0^0=1$. If $s=0$ or $s$ is not a multiple of $p-1$, the inner sum is 0 mod $p$. If $s$ is a multiple of $p-1$ and $s<k$ , then the binomial coefficient is 0 mod $p$. (This is what breaks for $k=p^2-1$.) The only remaining term is the one $s=k$ and it does not depend on $A$. So the sum over a class is the same for all classes. The number of classes is a multiple of $p$, hence the result.

Answer by Sergei Ivanov for Maximal cross sections of the Cartesian product of two planar domains

2013-04-07T18:44:02Z

Here is a non-constant counter-example to the original version (with weak monotonicity).

First observe that $f(\theta) $ (up to a factor $\sqrt2$ or so) equals the maximum area of intersection of $K$ and a translate of $R_\theta(L)$ where $R_\theta$ is the rotation of the plane through the angle $\theta$.

Let $K$ be the triangle with vertices $(1,0)$, $(-1,0)$ and $(0,1)$. Let $L$ be a narrow isosceles triangle $ABC$ with $AB=AC=1$ and $BC=\varepsilon$. I assume that $BC$ is horizontal and $A$ is above $BC$.

For most values of $\theta$ (more precisely, for all but a small neighborhood of $\pi$) $R_\theta(L)$ can be translated so as to fit inside $K$, hence $f(\theta)=area(L)$. But for $\theta\approx\pi$ this is not possible and hence $f(\theta)<area(L)$ . It seems that (although it is cumbersome to verify) $f(\pi)$ is the minimum and $f$ is monotone at both sides of $\pi$.

_{(Image added by J.O'Rourke.)}

Invariance of the l.h.s. of Euler-Lagrange equation

2013-04-02T14:52:33Z

Let $M^n$ be a smooth manifold equipped with a nondegenerate Lagrangian $L:TM\to\mathbb R$, $L=L(x,y)$, $x\in M$, $y\in T_xM$. The stationary points of the corresponding integral functional on curves are the solutions of the Euler-Lagrange equation, which in coordinates reads $$ \frac d{dt} \frac{dL}{dy_i}(x(t),\dot x(t)) - \frac{\partial L}{\partial x_i}(x(t),\dot x(t)) = 0, \qquad i=1,\dots,n . $$ Consider a smooth curve $t\mapsto x(t)$ which is not stationary. Plugging it into the l.h.s. of the equation yields coordinates of a co-vector (from $T^*_{x(t)}M$) which depends on the curve but not on the coordinate system. The invariance of this co-vector can be seen e.g. from the first variation formula for the functional.

Question: Is there a coordinate-free definition of this co-vector?

Actually I am interested only in the case when $L$ is the Lagrangian associated to a Finsler metric (i.e. $L$ is quadratically homogeneous).

Notes

The equation itself (i.e. the property that the co-vector is zero) has an invariant expression e.g. with Hamiltonian formalism.
In the Riemannian case, the co-vector in question (for a unit-speed curve) corresponds to the geodesic curvature vector under the isomorphism between $TM$ and $T^*M$ defined by the metric. This can be defined invariantly via the Levi-Civita connection.

Answer by Sergei Ivanov for Infimum of a finite number of distances in the plane

2013-04-02T15:39:29Z

This is possible even on the real line.

There is a strictly increasing continuous function $f:[0,1]\to\mathbb R$ whose derivative is zero almost everywhere. It is a suitable sum of a series of Cantor functions. See, for example, Gelbaum and Omsted, "Counterexamples in analysis" (1964), Chapter 8, Example 30.

This function has the following property: for every $\varepsilon>0$ there is a collection of disjoint intervals $[a_i,b_i]\subset[0,1]$ with total length greater than $1-\varepsilon$ and total variation of $f$ less than $\varepsilon$: $$ \sum (b_i-a_i)>1-\varepsilon, \qquad \sum (f(b_i)-f(a_i)) < \varepsilon . $$

Now define metrics $d_1$ and $d_2$ on $[0,1]$ as follows: $d_1$ is the standard metric and $d_2$ is the pull-back of the standard metric by $f$, i.e. $d_2(x,y)=|f(x)-f(y)|$. You can go from 0 do 1 through points $a_1,b_2,a_2,b_2,\dots$ using the distance $d_2$ between $a_i$ and $b_i$ and $d_1$ between $b_i$ and $a_{i+1}$. Thus $d(0,1)<\varepsilon$ for every $\varepsilon>0$.

Answer by Sergei Ivanov for Iterates converging to a continuous map

2013-03-31T10:12:54Z

I don't know a reference but maybe the following proof is shorter than yours.

By continuity, $\varphi\circ\varphi_\infty=\varphi_\infty$. Hence $\varphi$ is identity on the set $I:=\varphi_\infty([0,1])$. Hence $I$ is the set of fixed points of $\varphi$. And it is compact and connected. Then there are two cases: either $I$ is a nontrivial interval $[a,b]$ or it is a single point $\{a\}$ .

In the first case, we have $f(x)<x$ for all $x>b$ and $f(x)>x$ for all $x<a$ , otherwise there is a fixed point outside $I$. There is an $\varepsilon>0$ such that $f(x)>a$ for all $x\in(b,b+\varepsilon)$ and $f(x)<b$ for all $x\in(a-\varepsilon,a)$. Let $U=(a-\varepsilon,b+\varepsilon)$. Then $\varphi(U)\subset U$ and for every $x\in U$ we have $$ |\varphi^n(x)-x|<\delta(\varepsilon) $$ where $\delta(\varepsilon)\to $ as $\varepsilon\to 0$ ($\delta(\varepsilon)$ is the maximum of the diameters of the sets $\varphi([a-\varepsilon,a])$ and $\varphi([b,b+\varepsilon])$) For every $x\in[0,1]$, the iterations $\varphi^n(x)$ eventually get into $U$, hence $$ \bigcup_n \varphi^{-n}(U) = [0,1] . $$ By compactness, this implies that $[0,1]$ is covered by finitely many sets $f^{-n}(U)$ and hence there exist $n$ such that $\varphi^n([0,1])\subset U$. After this $n$, all iterations stay within the distance $\delta(\varepsilon)$ from the limit. Since $\varepsilon$ can be arbitrarily small, this implies uniform convergence.

In the second case (when $I$ is a single point $a$), similarly observe that $f(x)<x$ for $x>a$ and $f(x)>x$ for $x<a$ , and the same holds for $\varphi^2$, $\varphi^3$, etc. This implies that if e.g. $x>a$ and $f(x)<a$ then all iterations $\varphi^n(x)$ lie between $x$ and $f(x)$. This easily implies that there is an arbitrarily small neighborhood $U$ of $a$ such that $f(U)\subset U$. Then uniform convergence follows similarly to the first case.

Smoothness of $f(\sqrt x)$

2011-10-04T10:19:35Z

I found that I need to use the following facts in a paper that I am writing.

Let $f\in C^\infty(\mathbb R)$, then

If $f(0)=0$, then $f(x)=x g(x)$ for some $g\in C^\infty(\mathbb R)$.
If $f$ is even, i.e. $f(-x)=f(x)$ for all $x$, then $f(x)=g(x^2)$ for some $g\in C^\infty(\mathbb R)$.

These facts are trivial for analytic functions (just look at the Taylor series). In the smooth case, one can prove them by analyzing the derivatives of $f(x)/x$ and $f(\sqrt x)$, respectively, and showing that they have certain limits at 0. However this is somewhat cumbersome (especially if one wants to analyze how $g$ depends on $f$). This is not a problem for me because the facts fall into the category "you should be able to prove this yourself if you are reading my paper". But I would like to know if there is a nicer proof.

For the first statement, I know the following trick (which can be found in textbooks): define $$ g(x) = \int_0^1 f'(tx) \ dt $$ and observe that $f(x)=xg(x)$, and $g\in C^\infty$ since the function $t\mapsto f'(tx)$ under the integral is smooth in the parameter $x$. As a bonus, this argument also shows easily that $g$ (as a point of $C^\infty$) depends smoothly on $f$. (This is another fact that I need to use.)

Is there a similarly nice proof of the second statement? And, by the way, is there a textbook reference for it?

Added. Here is a more precise mathematical question that more or less formalizes what I mean. The function $g$ such that $f(x)=g(x^2)$ is uniquely defined only on $\mathbb R_+$, and its extension to negative arguments involves some choice.

Is there a canonical way to associate $g$ to $f$? Even more precisely, can one make a mapping $f\mapsto g$ which is linear, preserves the pointwise multiplication, and is continuous as a map from $C^\infty$ to $C^\infty$?

Answer by Sergei Ivanov for A question of compactness in the geometry of numbers

2013-03-25T15:04:16Z

It is not compact unless you allow the origin at the boundary (or maybe impose some kind of uniform strict convexity).

Consider a rectangle $K=[-1,1]\times[-\delta,1]$ in the plane, where $\delta$ is positive and goes to 0. If $K$ intersects some lattice only at the origin then so does $-K$, by symmetry. But $K\cup -K$ is a convex symmetric set of area 4, hence $\Delta(K)=\Delta(K\cup -K) \ge 1$ by Minkowski's theorem.

Therefore the ratio $vol/\Delta$ is bounded for all such $K$, but they have no limit in $\mathcal K^2_0$ as $\delta\to 0$.

Answer by Sergei Ivanov for On Lipschitz embeddability of certain compact metric spaces into $\mathbb{R}^n$

2013-03-22T16:49:03Z

No. There is a length metric counter-example in dimension 3. See Theorem 1 in this paper.

Let me briefly explain the construction here. For a large $n$, consider a unit segment $[p_nq_n]\subset \mathbb R^3$ and let $U_n$ be its neighborhood of radius $2/n$. Inside $U_n$, consider a "chain" formed by $n$ small solid tori. Each solid torus is a $(1/10n)$-neighborhood of a circle of radius $1/n$, and each of these circles is linked to the next one. The first link in the chain is nearby $p_n$ and the last one is nearby $q_n$. In each of these solid tori introduce a conformal Riemannian metric as follows. At the middle circle, the conformal factor is very small, so that the length of this circle equals $1/n^2$. And the conformal factor equals 100 almost everywhere between the middle circle and the boundary of the torus. So despite the fact that the middle circle is very short, the distances between points outside the torus are no shorter than the Euclidean distances. It follows that the distance between $p_n$ and $q_n$ in the resulting Riemannian metric is at least 1.

Repeat this constriction for $n=10,100,1000,\dots$, choosing the segments $[p_nq_n]$ so that they converge to some $[pq]$ (but the neighborhoods $U_n$ are disjoint). This yields a length metric on $\mathbb R^3$ which defines the standard topology (i.e., the identity is a homeomorphism). Restrict it to a compact ball containing all the segments of the construction.

There is no Lipschitz homeomorphism from this ball with this metric to $\mathbb R^3$. Indeed, suppose that $f$ is a 1-Lipschitz homeomorphism. Consider the image of the neighborhood of $[p_nq_n]$. The images of the chained circles have lengths (and hence diameters) at most $1/n^2$, Since they are linked, each next circle is within distance $1/n^2$ from the previous one. So the endpoints of the chain are mapped to points at distance at most $2/n$ or so from each other: $|f(p_n)-f(q_n)|\le 2/n$. It follows that $f(p)=f(q)$, a contradiction.

Answer by Sergei Ivanov for How to define an "anisotropic vector" for a given object?

2013-03-22T11:11:46Z

There are various ways to associate an ellipsoid to a solid body. For example, the inertia ellipsoid. (This is the ellipsoid centered at the center of gravity of the body and having the same second momentum as the body. It is easy to compute if you can integrate over the volume.) In the 2-dimensional case, you get an ellipse and can direct the "anisotropic vector" along the longer axis of the ellipse and define its length as a function of the difference of axes. In dimension 3, I think there is no construction with the same intuitive meaning. For example, think of an ellipsoid $x^2+y^2+10z^2\le 1$ with two equal axes of length 1 and a short third axis.

If your set is represented as a discrete mesh which is not evenly distributed, the inertia ellipse of this discrete set does not approximate the set geometrically (because essentially you are dealing with the momentum of non-uniformly distributed mass). If this is the case you want to handle, there are other ellipses/ellipsoids that can be used. For example, consider the Loewner ellipsoid. This is the the ellipsoid of minimal volume containing the given set. It is uniquely defined and not too hard to compute.

Answer by Sergei Ivanov for Is displacement controled by stable norm?

2013-03-07T17:28:50Z

If you allow an additive term $C(n)diam(g)$ rather than $2diam(g)$, then yes, the statement is true. In the paper D.Burago, "Periodic metrics", Adv. Soviet Math. 9, (1992), 205-210, he proves that for every periodic metric on $\mathbb R^n$ there is a constant $C$ such that $$ | d(x,y)-\|x-y\|_{st} | \le C $$ for all $x,y\in\mathbb R^2$. In this non-invariant formulation, $C$ depends not only on the metric but also from a particular choice of coordinates used to identify the torus with the standard $T^n$. However, if $x$ and $y$ are from one orbit of $\mathbb Z^n$, then going through the proof shows that one can take $C$ of the form $C(n)diam(g)$.

Unfortunately the paper is published in an obscure place and I can no longer find it on Google Books. But the proof is rather short and I think I can write down all details here if needed.

Added later. I'm adding a proof outline upon a request from comments.

Fix $p\in \mathbb R^n$ and $v\in\mathbb Z^n$. Define $f:\mathbb Z_+\to\mathbb R$ by $f(m)=d(p,p+mv)$. Note that $ \|v\|_{st} = \lim_{k\to\infty} f(k)/k$ . The function $f$ satisfies the following properties:

(1) $f(m_1+m_2)\le f(m_1)+f(m_2)$

(2) $f(2m)\ge 2f(m)-C$,

where $C$ depends only on the metric (and in fact equals $C(n)diam(g)$). These properties immediately imply that $k\|v\|_{st}\le f(k)\le k\|v\|_{st}+C$ with the same $C$. For $k=1$, this gives us the desired estimate on $d(p,p+v)=f(1)$.

The property (1) is just the triangle inequality. The property (2) is based on the following topological lemma:

Lemma. Let $s:[0,1]\to\mathbb R^n$ be a continuous path. Then there is a collection of disjoint intervals $[a_i,b_i]\subset[0,1]$, $i=1,2,\dots,k\le (n+1)/2$, such that $$ \sum (s(b_i)-s(a_i)) = \frac{s(1)-s(0)}2 . $$

Let me derive property (2) from the lemma. Apply the lemma to a shortest path $s$ connecting $p$ to $p+2mv$. This gives us a collection of at most $n/2$ vectors $v_i=s(b_i)-s(a_i)$ in $\mathbb R^n$ whose sum equals $mv$. Let $p_0=p,p_1,\dots,p_k$ be points such that $p_i-p_{i-1}=v_i$. Then $p_k=p+mv$, and each pair of points $p_{i-1},p_i$ is a (possibly non-integer) parallel translation of $a(t_i),b(t_i)$. Due to periodicity, this implies that $d(p_{i-1},p_i)\le d(a_i,b_i)+C_1$ for some constant $C_1$ depending on $g$. I will address this dependence later. Therefore $f(m)=d(p,p_k)$ is bounded by a constant $C_1n$ plus the length of the parts of $s$ covered by the intervals $[a_i,b_i]$.

Consider the complement intervals $[0,a_1]$, $[b_i,a_{i+1}]$, $[b_k,1]$. The corresponding vectors $s(a_1)-s(0)$ etc, also add up to $mv$. Hence the above argument applies to these intervals as well and implies that $f(m)$ is no greater than $C_1n$ plus the remaining part of the length of $s$. One of these two parts is no greater that half of the total length of $s$, hence the result.

The problem is how to control the constant $C_1$ by the diameter only. To do so, one moves the division points $s(a_i)$, $s(b_i)$ to nearby lattice points (losing at most the diameter on each move), then the required parallel translations preserve the distance and the dependence on the metric goes away. But the new collection of vectors no longer sums up to $mv$. So we have to choose those "nearby" lattice points wisely, so that the accumulated error is bounded by the diameter times $C(n)$. This (hopefully) can be done as follows: approximate $s$ by a sequence of lattice points (with distances between neighboring ones at most twice the diameter), then apply the lemma to the Euclidean broken line connecting these points, then move each division point to the next vertex of this broken line. The error in the sum of the resulting vectors has bounded stable norm and this should imply a bound on the metric distance. (I have not worked through this detail yet.)

Proof of lemma. It was discussed in this answer but the Google Book link there died since that (probably some copyright maniac killed it, now it looks as if the volume was never digitized). The proof it the following, Consider the standard unit sphere $S^n$ of points $x\in\mathbb R^{n+1}$ with $\sum x_i^2=1$. To each point $x\in S^n$, associate a partition $0=t_0\le t_1\le\dots t_{n+1}=1$ of $[0,1]$ such that $t_i-t_{i-1}=x_i^2$, and define $f(x)\in\mathbb R^n$ by $$ f(x) = \sum sign(x_i) (s(t_i)-s(t_{i+1}) . $$ The resulting function $f:\mathbb S^n\to\mathbb R^n$ is continuous and odd (i.e. $f(-x)=f(x)$). Therefore $f(x)=0$ for some $x$. Now let $[a_j,b_j]$ be the segments of the partition $\{t_i\}$ associated to this $x$, whose corresponding coordinates $x_i$ are positive. If there are too many of them, take the negative ones instead.

Answer by Sergei Ivanov for Fattening of totally convex sets

2013-01-20T20:29:52Z

No. Consider a rotation-symmetric metric on $\mathbb R^2$ resembling a small spherical cap extented by a flat cone. A sufficiently short geodesic segment at the origin is totally convex in your sense. But a small neighborhood of such a segment is not convex due to positive curvature.

Answer by Sergei Ivanov for A generalization of intermediate value theorem on R^k.

2013-01-16T17:39:23Z

The statement is true. It is almost precisely Lemma 2 in the paper D.Burago, "Periodic metrics", Adv. Soviet Math. 9, (1992), 205-210. The proof is short but not easy to invent. The paper can be read on Google Books here.

Notes on the text: the intervals in the formulation of Lemma 2 are in fact disjoint, the term "antipodal map" means "odd map" (i.e. $\varphi(-x)=-\varphi(x)$).

For $k=2$, this fact is classic and one can replace 0.5 by any inverse integer ($1/3, 1/4,\dots$) but for every other value there is a counter-example.

Answer by Sergei Ivanov for Is a manifold with flat ends of bounded geometry?

2013-01-15T23:44:13Z

Here is a self-contained proof not using any classification. With some effort, it can be made to work under weaker assumptions: the curvature is nonnegative outside a compact set and is bounded from above. It is a variation of the proof of the Soul Theorem via Sharafutdinov's retraction.

Fix a reference point $o\in M$ and define a Busemann function $b:M\to\mathbb R$ by $$ b(x) = \limsup_{y\in M,\ d(o,y)\to\infty} ( d(o,y)-d(x,y)) $$ where $d$ is the Riemannian distance (note the non-standard sign convention). The flat ends assumption implies the following key features:

Sublevel sets of $b$ are compact.
$b$ is (locally) convex outside a compact set.

To prove this, first observe some general properties of $b$. First, $b$ is 1-Lipschitz. Second, $b$ goes to $+\infty$ along any geodesic ray (essentially by the triangle inequality). Third, for any $x\in M$ there exists a geodesic ray starting at $x$ such that $b$ grows at unit speed along this ray (take a limit of geodesic segments $[xy_i]$ where a sequence $y_i$ realizes the $\limsup$). Let me call such rays "calibrating". Finally, for every compact set $K$ there is a compact set $K'$ such that no calibrating ray starting outside $K'$ intersects $K$. Indeed, otherwise a sequence of such rays would converge to a geodesic line, and $b$ would going to $-\infty$ in one of the directions along this line.

Now convexity of $b$ outside a compact set follows easily. Let $x_0\in M$ and $\gamma$ be a calibrating ray starting from $x_0$. Then $b(\gamma(t))=b(x_0)+t$ for all $t\ge 0$. Since $b$ is 1-Lipschitz, it follows that $$ b(x) \ge b(x_0) + (t-d(\gamma(t),x)) $$ for every $y\in M$, with equality for $x=x_0$. So $b$ is supported from below at $x_0$ by a function of the form $x\mapsto const - d(\gamma(t),x)$. If $x_0$ is sufficiently far from $o$, the ray $\gamma$ is contained in the flat part. Hence the distance to $\gamma(t)$ is bounded from above by a similar Euclidean distance function which is nearly linear near $x_0$ if $t$ is large. Existence of such lower bounds for $b$ implies that $b$ is convex.

Now let me show that sublevels of $b$ are compact. Suppose the contrary, then there is a sequence $x_i$ with $d(o,x_i)\to\infty$ but $b(x_i)\le const$. Choose a subsequence such that geodesic segments $[ox_i]$ converge to a geodesic ray $\gamma$. On this ray, mark a point $y$ where it leaves a ball containing the non-flat part of $M$. Once a segment $[ox_i]$ contains a point $y_i$ near $y$ and almost the same direction as $\gamma$ there, one observes that the derivative of $b$ along this segment at $y$ is greater than, say, 1/2. This and convexity of $b$ yield a linear lower bound for $b(x_i)$, contrary to the assumption $b(x_i)\le const$.

This proves the two key properties of $b$. Let $B_r$ denote the sublevel set $\{x:b(x)\le r\}$ . Let $r_0$ be such that $b$ is convex outside $B_{r_0}$. Then, for every $R\ge r\ge r_0$, there is a distance non-increasing retraction (homotopic to identity) $f:B_R\to B_r$. Indeed, let $\varepsilon$ be smaller than the minimum injectivity radius on $B_R$. Then one retracts $B_R$ to $B_{R-\varepsilon}$ via a nearest-point projection, Locally it is just a nearest-point projection to a convex set in a Euclidean space, so it is well-defined and does not increases distances. Iterating this construction yields a retraction from $B_R$ to $B_r$. Moreover the complement of $B_r$ is mapped to the boundary of $B_r$.

Now let us turn to the injectivity radius. Let $r$ be such that $B_{r/2}$ contains the not-flat part of $M$ and $\rho_0$ is the minimum of $r/10$ and the injectivity radius on $B_{2r}$. I claim that the injectivity radius is no less than $\rho_0$ everywhere. Suppose the contrary. Then, somewhere outside $B_r$ there is a geodesic loop of length $2\rho<2\rho_0$ . This loop is not contractible in the flat part. Apply the above retraction $f:B_R\to B_r$ where $R$ is so large that the loop is contained in $B_R$. The image is a loop in the boundary of $B_r$ with length $\le 2\rho$ and still non-contractible in the flat part. Hence the injectivity radius at $f(x)$ is at most $\rho$, a contradiction.

Answer by Sergei Ivanov for computational complexity

2013-01-09T08:58:27Z

It is NP-complete. Here is a reduction from the graph 3-coloring problem.

For a given graph, consider a sphere with $n$ holes where $n$ is the number of vertices. Enumerate the holes by the vertices of the graph. Fill each hole by 3 discs (red, green and blue) by identifying each disc boundary with the hole boundary. Glue the centers of the 3 discs into one point (so that nor two of them cannot be included in one surface), Then, for every edge of the graph, pick one new point in each of the 6 discs filling the holes corresponding to the edge endpoints, and glue together each of the 3 pairs of chosen points lying on discs of the same color. The points chosen for different edges must be distinct.

The resulting topological space is homeomorphic to a simplicial complex (which is easy to construct from the graph in polynomial time). It contains a sphere if and only if the original graph is 3-colorable.

Remark. You can alter the formulation by saying that an immersed sphere which self-intersects itself in finitely many points is still a sphere. For this problem, the construction can be modified as follows. To make a pair of discs "forbidden", glue together two pairs of small discs in them rather than one pair of point. This creates a surface of positive genus if you try to pick a pair of discs connected this way.

Answer by Sergei Ivanov for Polyominoes with double contact

2013-01-03T17:40:26Z

Here is a formalization of your argument. Let $A$ and $B$ be two poliominoes in question. Construct the following figure $K$ made of unit square: a square with integer coordinates $(x,y)$ is included in $K$ if and only if the translate of $B$ by the vector $(x,y)$ overlaps with $A$. ($K$ is essentially the Minkowski difference of $A$ and $B$ regarded as subsets of the integer lattice rather than figures with interior.)

Since $A$ and $B$ are connected. so is $K$. First consider the case when $K$ is not convex (i.e. not a rectangle) and find a concave corner of $K$. The cell outside $K$ near this corner is adjacent to at least 2 cells of $K$. This means there is a translate $B'$ of $B$ which does not overlap with $A$ but there are two coordinate directions (such as "up" and "left") such that if one moves $B'$ one step in any of these directions, the resulting figure overlaps with $A$. If you cannot move $B'$ up, then there is a cell of $B'$ right below a cell of $A$, so they share a boundary segment. Two such "forbidden" directions give us two common boundary segments.

Now consider the case when $K$ is a rectangle. Clearly the upper side of $K$ corresponds to the position where the bounding box (=minimal enclosing rectangle) of $B$ touches the bounding box of $A$ from above. Similarly for the other three sides of $K$. This means that an integer cell $(x,y)$ belongs to $K$ if and only if the bounding box of $B$ translated by $(x,y)$ overlaps with the bounding box of $A$. By the definition of $K$ this means the following: translates of $A$ and $B$ overlap if and only if their bounding boxes overlap.

This implies that $A$ and $B$ contain the corners of their bounding boxes. Indeed, if e.g. $A$ does not contain the upper right corner of its bounding box, we can translate $B$ so that the intersection of the two bounding boxes is just the cell at that corner, contrary to the above.

Once we know that $A$ and $B$ contain the corners of their bounding boxes, we can find boundary segments of length 2 near these corners and position our poliominoes so that they contact by these segments (this is already pointed out in the question).

Remark. If we disallow rotations and reflections. There is a counter-example. Let $A$ be the $3\times 6$ rectangle with the following 6 cells removed: one in the middle of each short side and two in the middle of each long side. (Sorry I don't know how to make a picture here.) Let $B$ be the same figure rotated 90 degrees. Then $A$ and $B$ cannot have two common edges without rotation.

Answer by Sergei Ivanov for The kissing number of a square, cube, hypercube?

2012-12-31T15:05:56Z

The square case was posed as a problem at Leningrad (now St. Petersburg) high school math olympiad in 1963. I wrote a solution of this problem for the volume "St. Petersburg mathematical olympiads 1961-1993", D.V.Fomin, K.P.Kokhas eds., Lan' Publ. 2007 (in Russian), it is Problem 63.31 in that book.

Here is the original very detailed draft of the solution from my archive (only a sketch made it into the final book). It is in Russian but the pictures and formulae might be enough to follow the proof. Up to elementary but cumbersome case chasing, the solution is the following.

Consider the boundary of the twice bigger square with the same center and parallel sides. It is a broken line of length 8. It turns out that each of the kissing squares takes away a piece of length at least 1 from this broken line. Hence there are at most 8 kissing squares (and furthermore one analyse the equality case).

_{(Snapshots of the figures added by J.O'Rourke)}

The proof of the fact that the length of the intersection is at least 1 is essentially an exhaustion of cases, each of which is trivial. It is helpful to observe that the length of the intersection is a piecewise linear function of the relative position of the squares (if the orientation is fixed), so one has to consider only "borderline" positions (i.e. those where one of the corners is on one of the lines). This leaves about 10 cases to consider.

Answer by Sergei Ivanov for Discretization of a complete manifold

2012-12-11T20:27:05Z

My knowledge of this subject is obsolete, but anyway here are some partial answers.

If you have a Ricci curvature bound in addition to the injectivity radius, then you can recover the diffeomorphism type, e.g. with harmonic coordinates, see M. Anderson, Convergence and rigidity of manifolds under Ricci curvature bounds, Invent. Math., 102 (1990), no. 2, 429-445. But 100 should be replaced by some radius depending on the dimension and the Ricci curvature bound.

Correction. It is the constant 1 that would depend on the Ricci curvature bound. (I did the rescaling argument wrong.)

Without curvature bounds things get more complicated. It is very easy to recover homotopy type, In fact, you don't need injectivity radius, only a contractibility function: every ball of radius $r\le r_0$ is contractible within a ball of radius $f(r)$. Given this, you can construct homotopy equivalence from an almost isometry defined on a net: extend it step by step to skeletons of a cell decomposition, just make sure that $n$ iterations of $f$ (or maybe $2f$) do not grow bigger than your constant 100.

If the dimension is not 3 (and maybe for 3 as well, since the Poincare conjecture is now solved), you can recover homeomorphism type under a pre-compactness assumption (in your set-up, this pre-compactness boils down to something like that every ball of radius 2 is covered by a bounded number of balls of radius 1, and then your radius 100 depends on this number). This follows from the arguments in Grove-Petersen-Wu, Geometric finiteness theorems via controlled topology. Invent. Math. 99 (1990), no. 1, 205-213.

Correction. Again, the pre-compactness would translate into something more complicated after rescaling the injectivity radius to 1.

They also use only local contractibility function, and in this more general context diffeomorphism stability fails in dimension 4. Of course this does not answer your question since the injectivity radius assumption is so much stronger. I don't know how essential the pre-compactness is. They certainly need it for their result (which is finiteness of topology types) but perhaps it may be relaxed if you only need stability.

Answer by Sergei Ivanov for Nonzero convex combinations of convex hull vertices to yield an inner point

2012-11-25T20:58:57Z

This is indeed easy. Let $p$ be a point that you want to represent, $m$ the barycenter of all vertices and $\varepsilon>0$ so small that the point $q=(1+\varepsilon)p-\varepsilon m=p+\varepsilon(p-m)$ is still in the convex hull. Represent $q$ as a convex combination of some vertices, add $\varepsilon m$ with $m$ represented as the arithmetic mean of all vertices, and finally divide all the coefficients by $1+\varepsilon$.

Answer by Sergei Ivanov for A Question on Exterior Forms

2012-11-22T00:14:29Z

For what is worth, here is a simple coordinate example in dimension 6. Take $$ \omega_0 = dx_1\wedge dx_2 + dx_3\wedge dx_4 + dx_5\wedge dx_6 , $$ $$ \omega_1 = dx_1\wedge dx_3 + dx_2\wedge dx_4 , $$ $$ \omega_2 = dx_3\wedge dx_5 + dx_4\wedge dx_6 , $$ $$ \omega_3 = dx_5\wedge dx_1 + dx_6\wedge dx_2 . $$ The sum of squares of these forms is zero, but the square of any nontrivial linear combination is not.

Answer by Sergei Ivanov for Counterexample to Sard Theorem for a not-C1 map

2012-11-20T23:23:44Z

No, such functions do not exist. More precisely, let $f:\mathbb R\to\mathbb R$ be an arbitrary function, $\Sigma$ is the set of $x\in\mathbb R$ such that $f'(x)$ exists and equals 0. Then $f(\Sigma)$ has measure 0.

By countable subadditivity of measure, we may assume that the domain of $f$ is $[0,1]$ rather that $\mathbb R$. Fix an $\varepsilon>0$. For every $x\in\Sigma$ there exists a subinterval $I_x\ni x$ of $[0,1]$ such that $f(5I_x)$ is contained in an interval $J_x$ with $m(J_x)<\varepsilon m(I_x)$. Here $m$ denotes the Lebesgue measure and $5I_x$ the interval 5 times longer than $I_x$ with the same midpoint. Now by Vitali Covering Lemma there exists a countable collection ${x_i}$ such that the intervals $I_{x_i}$ are disjoint and the intervals $5I_{x_i}$ cover $\Sigma$. Since $I_{x_i}$ are disjoint, we have $\sum m(I_{x_i})\le 1$. Therefore $f(\Sigma)$ is covered by intervals $J_{x_i}$ whose total measure is no greater than $\varepsilon$. Since $\varepsilon$ is arbitrary, it follows that $f(\Sigma)$ has measure 0.

Answer by Sergei Ivanov for Composition of (topologically) connected binary relations

2012-11-20T10:54:32Z

No. There are two (discontinuous) surjective maps $f,g:S^1\to S^1$ whose graphs are connected but the graph of $g\circ f$ (as well as its closure) is not.

The map $f$ is defined as follows, using the standard parametrization of $S^1$ by $\mathbb R/2\pi\mathbb Z$. It is the identity on the complement from the arc (parametrized by) $[\pi/2,\pi]$. The arc $[\pi/2,\pi]$ is slightly stretched from the point $\pi/2$ so that its image is the arc $[\pi/2,\pi+\varepsilon]$. The map $g$ is similar but with a discontinuity at 0 rather than $\pi$.

The graphs of $f$ and $g$ can be parametrized by an interval and hence connected. The graph of $g\circ f$ is not, because the map has two simple discontinuities (at 0 and $\pi$) that divide the circle into two components.

The graphs are not closed in $S^1\times S^1$, but adding points $(\pi,\pi)$ and $(0,0)$ fixes this issue.

Answer by Sergei Ivanov for Characterization of bounded geometry - Reference-request

2012-11-19T14:12:52Z

I assume that by "all derivatives" you mean derivatives of every order.

Suppose that all transitions between normal coordinates have uniformly bounded derivatives within some radius $r$. For every point $p$ we have a unit radial vector field $V=V(p)$ whose derivatives are uniformly bounded at distances between, say $r/10$ and $r$ from $p$. (This holds in normal coordinates centered at $p$ and hence in normal coordinates centered at any nearby point.)

Now fix a point $q\in M$ and arrange points $p_k$, $k=1,\dots,\frac{n(n+1)}2$, where $n=\dim M$, so that the distances from $q$ to $p_k$ equal $r/2$ and the corresponding vector fields $V_k=V(p_k)$ at $q$ are generic in the sense that the symmetric tensors $V_k\otimes V_k$ are linearly independent. Then the same holds at every point in a neighborhood of $q$. Actually the points $p_k$ should have some prescribed coordinates in normal coordinate system centered at $q$, then all subsequent estimates are uniform.

We have a system of equations $g(V_k,V_k)=1$, $k=1,\dots,\frac{n(n+1)}2$. At every point near $q$ this is a nondegenerate linear system on the components of the metric tensor $g$. Hence it uniquely determines $g$ via some explicit formulae in terms of $V_k$. Since the derivatives of $V_k$ are uniformly bounded, so are the derivatives of the metric tensor.

Answer by Sergei Ivanov for Does the metric space of compact metric spaces satisfy the binary intersection property?

2012-11-17T18:10:29Z

No, Let $B_n\in M$ be the $n$-dimensional Euclidean unit ball and $r=\frac12+\varepsilon$ where $\varepsilon=\frac1{100}$. Then the $r$-balls in $M$ centered at $B_n$ intersect pairwise. Indeed, for $m>n$ the $m$-dimensional Euclidean ball of radius 1/2 lies within Gromov-Hausdorff distance 1/2 from both $B_n$ and $B_m$ (as seen from their natural inclusion into $\mathbb R^m$).

However there is no compact metric space $K$ which stays within distance $1/2+\varepsilon$ from every $B_n$. Indeed, suppose the contrary, then there is a map $f_n:B_n\to K$ which distorts distances by at most $1+2\varepsilon$. But $B_n$ contains $2n$ points with pairwise distances $\sqrt 2$, hence the $f_n$-images of these points are separated by distances at least $\sqrt2-1-2\varepsilon>\frac1{10}$. Thus $K$ contains arbitrarily many $\frac1{10}$-separated points, hence it is not compact.

Answer by Sergei Ivanov for Do sufficiently regular distances on manifolds come from riemannian metrics?

2012-11-15T19:46:49Z

As pointed out in other answers, only length metrics can be Riemannian and there are plenty of non-length metrics satisfying your condition.

If your metric is a length metric, then yes, it is Riemannian. Indeed, for every $x\in M$ the function $f_x=d(x,\cdot)^2$ is smooth near $x$. Since it attains its minimum at $x$, its derivative at $x$ is zero. Therefore its second derivative at $x$ is well-defined as a quadratic form on $T_xM$. Define the metric tensor at $x$ as one half of this second derivative. Now from the Taylor expansion it is easy to see that for every $\varepsilon>0$ there is a neighborhood of $x$ where the original distance and the new Riemannian distance are Lipschitz equivalent with Lipschitz constant $1+\varepsilon$. Since both metrics are length metrics, this property implies that they are $(1+\varepsilon)$-Lipschitz equivalent globally. Since $\varepsilon$ is arbitrary, this means that they are equal.

Comment by Sergei Ivanov

2013-05-15T20:56:17Z

The exponent at $|\Sigma|$ should be $(n-2)/n$ for scale invariance.

Comment by Sergei Ivanov

2013-05-14T09:54:52Z

There is something wrong with the inequality on $f$. Take $t=0$ for example.

Comment by Sergei Ivanov

2013-05-13T10:01:47Z

The question came by association from a totally different problem. I am hoping that there might be a kind of rigidity result where one concludes that the metric is round (or not far from round) by looking at some rough measures of isoperimetric profile. If there is one, I could try to apply the technique of the proof in another context.

Comment by Sergei Ivanov

2013-05-13T09:54:36Z

@Bill: this is plausible but I don't know for sure. In finite dimensions, you can differentiate the map at some point and get that a codimension 1 section is within a bounded distance from a Euclidean norm. I don't know how to carry this over to infinite dimensions.

Comment by Sergei Ivanov

2013-05-12T22:38:32Z

You would need some normalization axiom, in order to distinguish between proportional measures. Do you have a specific one in mind?

Comment by Sergei Ivanov

2013-05-12T19:01:07Z

@katz: This is just an example of statement that might be true if dividing in half fails. It might as well be an integral inequality on the isoperimetric profile. In other words, I am more interested in a affirmative answer to a slightly different question than in a counter-example to the precise one. There is nothing magic about the constant $\pi$, I just want to avoid the trivial answer "look at a small neighborhood of a point where curvature is greater than 1".

Comment by Sergei Ivanov

2013-05-10T12:01:34Z

@alex: I think you are right. This simplification did not occur to me.

Comment by Sergei Ivanov

2013-05-06T18:38:43Z

It is not immediately clear that the diameter is bounded.

Comment by Sergei Ivanov

2013-05-06T14:19:47Z

@Hans: yes, sure. I misread the question as "there exist $a$ and $b$ such that...". I am not sure about monotonicity. And actually I now see a flaw in the argument: $S'T$ is not symmetric, so its spectral radius is not equal to the norm. Sorry about this confusion.

Comment by Sergei Ivanov

2013-05-06T10:38:01Z

@Hans: sorry, I meant "positive definite", not "positive elementwise". The inequality follows by diagonalization. The transition from $b\to\infty$ to a large $b$ is basically the definition of limit.

Comment by Sergei Ivanov

2013-05-05T16:29:23Z

How about $w=\pi$ and arbitrarily short $\gamma$? I think in this case $\gamma(w)$ covers the sphere.

Comment by Sergei Ivanov

2013-04-07T08:07:23Z

The function can be a constant and constants seem to be unimodal by your definition.

Comment by Sergei Ivanov

2013-04-05T16:33:00Z

What do you mean by convexity of the boundary? It has no meaning without additional structure (e.g. a Riemannian metric or an embedding into $\mathbb R^n$).

Comment by Sergei Ivanov

2013-04-02T20:02:38Z

Thank you. It seems that this description gets better when translated to $T^*M$ by Legendre transform. If $s(t)\in T^*M$ is the Legendre transform of $\gamma'(t)$, the analog of $\beta$ is a 1-form $\omega(s'(t),\cdot)+dH(\cdot)\in T^*_{s(t)}T*M$, where $\omega$ is the symplectic form and $H$ is the Hamiltonian. And it projects down to $M$ because vanishes on the fiber of $T^*M$.

Comment by Sergei Ivanov

2013-03-31T09:21:12Z

The limit is not continuous.