(1-Lipschitz) + (length-preserving) = isometry - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-18T08:40:22Z http://mathoverflow.net/feeds/question/71479 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/71479/1-lipschitz-length-preserving-isometry (1-Lipschitz) + (length-preserving) = isometry Anton Petrunin 2011-07-28T11:17:34Z 2011-08-09T21:04:37Z <p>I am looking for an elementary way to prove the following theorem.</p> <blockquote> <p><strong>Theorem.</strong> Let $\alpha$ and $\beta$ be two simple convex closed curves in $\mathbb R^2$. Assume $$\mathop{\rm length} \alpha=\mathop{\rm length}\beta$$ and there is a 1-Lipschitz bijecction $f\colon\alpha\to\beta$. Then $f$ is an isometry.</p> </blockquote> <p>It would be better if the same proof would work for Lobachevsky plane and unit sphere (for the sphere one has to assume that the length of the curves is $&lt;2{\cdot}\pi$).</p> <p>The proof I know is simple, but it use Alexandrov geometry quite a bit: <em>If we cut from the plane the region bounded by $\alpha$ and glue instead the region bounded by $\beta$ then the obtained space will have curvature $\ge0$ in the sence of Alexandrov and it is easy to show that it has to be isometric to the Euclidean plane. Hence the result.</em></p> <p><strong>P.S.</strong> This morning I realized that this also follows from the following continuos version of Cauchy's Arm Lemma:</p> <p>Let $\alpha,\beta\colon[0,\ell]\to\mathbb R^2$ be closed convex curves with unit-speed parameter. Assume that for any $t$ in a subinterval $[a,b]\subset [0,\ell]$, the curvature of $\alpha$ at $\alpha(t)$ is at most the curvature $\beta$ at $\beta(t)$. Then $|\alpha(a)-\alpha(b)|\ge|\beta(a)-\beta(b)|$ and equality holds only if the resriction $\alpha|[a,b]$ is isometric to the resriction $\beta|[a,b]$.</p> http://mathoverflow.net/questions/71479/1-lipschitz-length-preserving-isometry/71481#71481 Answer by Benoît Kloeckner for (1-Lipschitz) + (length-preserving) = isometry Benoît Kloeckner 2011-07-28T11:44:52Z 2011-07-31T16:32:01Z <p>It seems there should a proof along the following lines, but I did not took the time to check every detail.</p> <p><strong>Edit:</strong> there is a problem in what follows; the fact that $\beta$ must be contained in the interior of $\alpha$ is not true. It is possible that another normalization makes it hold, but I know feel that the postscriptum of the question is the good point of view.</p> <p>First, the assumptions give that $f$ is an isometry with respect to the length metrics on $\alpha$ and $\beta$. Let $a$, $a'$ be two farthest points on $\alpha$, $b=f(a)$ and $b'=f(a')$. Without lost of generality, apply to $\beta$ an isometry of $\mathbb{R}^2$ so that $b=a$ and $\beta$ is contained in the half-plane delimited by the line orthogonal to $[aa')$ at $a$. Using that $f$ is $1$-Lipschitz and that $[aa']$ is a diameter, we get that $\beta$ must be contained in the interior of $\alpha$. Considering the projection to the domain delimited by $\beta$, we get a $1$-Lipschitz map $\tilde f:\alpha\to\beta$ that contracts strictly distances around any point $x\in\alpha$ such that $\alpha$ at $x$ and $\beta$ at $f(x)$ do not share a common supporting direction. Since $\alpha$ and $\beta$ have the same length, this never happens and $\beta$ must be an homothetic image of $\alpha$. Since they have the same length, the homothety constant must be $1$ and we are (hopefully) done.</p> http://mathoverflow.net/questions/71479/1-lipschitz-length-preserving-isometry/71595#71595 Answer by Gjergji Zaimi for (1-Lipschitz) + (length-preserving) = isometry Gjergji Zaimi 2011-07-29T17:47:06Z 2011-07-29T17:47:06Z <p>The proof for polygonal paths is almost trivial. Suppose $\alpha=P_1P_2\cdots P_n$ and $\beta=Q_1Q_2\cdots Q_m$. Subdivide the edges of $\beta$ to include $f(P_i)$'s as fake vertices and denote this polygon $\beta'$. Do the same with $\alpha$ and $f^{-1}(Q_j)$'s and use the Lipschitz condition to prove that $\alpha'$ and $\beta'$ have corresponding edges of equal length. Now assume that $f(P_i)\in [Q_j,Q_{j+1}]$, then $$|Q_jQ_{j+1}|=|Q_jf(P_i)|+|Q_{j+1}f(P_i)|=|P_if^{-1}(Q_j)|+|P_if^{-1}Q_{j+1}|$$ $$\geq |f^{-1}(Q_j)f^{-1}(Q_{j+1})|$$ so $f(P_i)=Q_j$ for some $j$, for all $i$. A similar argument shows that $\angle Q_j\le \angle P_i$ and we conclude that $f$ is an isometry.</p> <p>Now it seems to me that this argument can be modified to include an approximation argument to imply your statement. (Approximate the curves by polygons, use the fact that $f$ is almost Lipschitz to conclude that it is almost an ismoetry and take the limit.)</p> http://mathoverflow.net/questions/71479/1-lipschitz-length-preserving-isometry/71598#71598 Answer by akopyan for (1-Lipschitz) + (length-preserving) = isometry akopyan 2011-07-29T18:04:27Z 2011-07-29T19:51:15Z <p>If α is a convex shape and f is a 1-lipshitz map then perimeter of convex hull of f(α) ⩽ length of α. (Similar statement for higher dimension has been proven by Alexander)</p>