User dror atariah - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T18:48:43Z http://mathoverflow.net/feeds/user/8047 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/129204/intersection-of-2-visibility-polygons Intersection of 2 visibility polygons Dror Atariah 2013-04-30T10:52:01Z 2013-05-01T11:36:01Z <p>Let $P$ be a <em>simple, closed and bounded</em> polygon and $p_1,p_2 \in \mathrm{int}(P)$ be two points in its interior. <strong>Is it true that the intersection of the visibility polygons of $p_1$ and $p_2$ is connected?</strong></p> <p>The visibility polygon of a point $p\in P$ is the set of all $x \in P$ such that the line segment connecting $p$ and $x$ is contained in $P$.</p> <p>It might be elementary, but we fail either finding a proof or a counterexample. Furthermore, in the literature I could only find details about the computational aspects of the visibility polygon, but not a single word about its properties. Any reference, example, sketch of proof etc. is welcomed!</p> http://mathoverflow.net/questions/48629/algebraic-surfaces-and-their-intrinsic-geometry Algebraic surfaces and their (intrinsic) geometry Dror Atariah 2010-12-08T11:51:40Z 2012-05-21T19:00:33Z <p>Recently I began to consider algebraic surfaces, that is, the zero set of a polynomial in 3 (or more variables). My algebraic geometry background is poor, and I'm more used to differential and Riemannian geometry. Therefore, I'm looking for the relations between the two areas. I should also mention, that I'm interested in the realm of real surfaces, i.e. subsets of $\mathbb{R}^n$.</p> <p>On my desk you could find the following books: <strong>Algebraic Geometry</strong> by <em>Hartshorne</em>, <strong>Ideals, Varieties, and Algorithms</strong> by <em>Cox &amp; Little &amp; O'Shea</em>, <strong>Algorithms in Real Algebraic Geometry</strong> by <em>Basu &amp; Pollack &amp; Roy</em> and <strong>A SINGULAR Introduction to Commutative Algebra</strong> by <em>Greuel &amp; Pfister</em>. Unfortunately, neither of them introduced notions and ideas I'm looking for.</p> <p>If I get it right, please correct me if I'm wrong, locally, around non-singular points, an algebraic surface behaves very nicely, for example, it is smooth. Here's the first question: <em>is it locally (about non-singular point) a smooth manifold? Is it a Riemannian manifold, having, for instance, the metric induced from the Euclidean space?</em></p> <p>Further questions I have are, for example:</p> <ol> <li>Can I define <em>geodesics</em> (either in the sense of length minimizer or straight curves) in the non-singular areas of the surface? Can they pass singularities?</li> <li>How about <em>curvature</em>? Is it defined for these objects?</li> <li>Can we talk about <em>convexity</em> of subsets of the algebraic surface?</li> <li>What other tools and term can be imported from differential/Riemannian geometry?</li> </ol> <p>I will be grateful for any hint, tip and lead in the form of either answers to my questions, or references to books/papers which can be helpful, or any other sort of help.</p> http://mathoverflow.net/questions/54742/gaussian-curvature-radius Gaussian curvature radius Dror Atariah 2011-02-08T09:47:07Z 2011-02-17T14:00:01Z <p>In the paper <em>Surface sampling and the intrinsic Voronoi diagram (2008)</em>, Ramsay Dyer defines the <em>Gaussian curvature radius</em> at a point $x$ of a surface $S$ to be $\rho_K(x) = 1/\sqrt{K(x)}$ where $K(x)=\kappa_1(x) \kappa_2(x)$ is the Gaussian curvature at $x$.</p> <p>Trying to track back the notion in Berger's <em>A panoramic view of Riemannian geometry</em>, and in Lee's <em>Riemannian manifolds</em> and in Chavel's <em>Riemannian Geometry</em> yielded nothing.</p> <p>My question is two-folded:</p> <ol> <li>Where can I find more information about this notion?</li> <li>Is there a reason not to define it as $\rho_K(x) = 1/|K(x)|$? Otherwise, this definition is only valid for non-negatively curved surfaces.</li> </ol> <p><strong>EDIT</strong> As pointed out by Deane Yang, there is no sense in the definition I suggested. Nevertheless, if one wants to relate the Gaussian curvature to a radius (for either negatively or positively curved surfaces) how about this alternative: $\rho_{K}(x)=1/\sqrt{|K(x)|}$?</p> http://mathoverflow.net/questions/49946/principal-curvatures-and-curvature-directions Principal curvatures and curvature directions Dror Atariah 2010-12-20T09:51:05Z 2010-12-20T17:15:25Z <p>Last week I considered again <em>principal curvature (pc)</em> and <em>principal curvature directions (pcd)</em> of a, for the sake of simplicity, 2-manifold embedded in 3-space. In this simple case, the pc and pcd of at a point are the eigenvalues and eigenvectors of the shape operator. The magnitude of the pc's corresponds to the minimal and maximal normal curvature at the point. My question, however, is:</p> <blockquote> <p>What does the principal curvatures direction <strong>magnitude</strong> represents?</p> </blockquote> <p>In the textbooks I looked up in (Kühnel and do Camro) I couldn't find a reference to the principal curvature direction's magnitude. Is there something known about this? Is it something basic (maybe even from linear algebra)?</p> <p>Edit 1: A somewhat more general, but related, question is:</p> <blockquote> <p>What is the geometrical meaning of an eigenvector's magnitude?</p> </blockquote> http://mathoverflow.net/questions/37708/nash-embedding-theorem-for-2d-manifolds/37712#37712 Answer by Dror Atariah for Nash embedding theorem for 2D manifolds Dror Atariah 2010-09-04T11:35:26Z 2010-09-04T11:35:26Z <p>Another counter example, and somehow more exotic, is the Klein bottle.</p> http://mathoverflow.net/questions/36437/convexity-and-strong-convexity-of-subsets-of-surfaces Convexity and Strong convexity of subsets of Surfaces Dror Atariah 2010-08-23T11:45:27Z 2010-08-23T21:37:24Z <p>In the book <em>Riemannian geometry - modern introduction</em> by Isaac Chavel, three different definitions of convexity are introduced. I am looking for an example of a set which is convex but not strongly convex according to the following definitions (cited from the book):</p> <blockquote> <p>Let $M$ be a complete Riemannian manifold, and $A \subset M$. $A$ is:</p> <ol> <li><em>convex</em> if for any $p,q\in A$ there exists a geodesic $\gamma_{pq} \subset A$ such that $\gamma_{pq}$ is the unique minimizer in $M$ connecting $p$ to $q$.</li> <li><em>strongly convex</em> if for any $p,q\in A$ there exists a geodesic $\gamma_{pq}\subset A$ such that $\gamma_{pq}$ is the unique minimizer in $M$ connecting $p$ to $q$, and $\gamma_{pq}$ is the only geodesic contained in $A$ joining $p$ and $q$.</li> </ol> </blockquote> <p>I believe that <a href="http://page.mi.fu-berlin.de/atariah/Files/convex_bu_not_strongly.png" rel="nofollow">this picture</a> gives an example. This is not an accurate figure! The set I'm referring to is the one centered at around a geodesic segment with end points conjugate one to the other along it, and bounded symmetrically by two close non minimal geodesics.</p> <p>Is my example correct? If not, then can someone provide a valid example?</p> <p>Edit: <em>Caution</em> I know of something like 7-8 different definitions of convexity in the case of a Riemannian manifold. Try to refer to the definitions I gave above.</p> http://mathoverflow.net/questions/36082/injectivity-radius-and-the-cut-locus Injectivity radius and the cut locus Dror Atariah 2010-08-19T13:41:55Z 2010-08-20T14:09:56Z <p>Consider a connected, complete and compact Riemannian manifold $M$. Is it correct that the following equality holds: $\text{inj}(x)=\text{dist}\left(x,\text{CuL}(x)\right)$? Or in words that the injectivity radius of a point is the distance from the point to its cut locus.</p> <p>Here is my explanation: As the manifold is compact and complete, then the cut locus $\text{CuL}(x)$ is compact as well[1]. Thus, there exists a point $y\in \text{CuL}(x)$ such that $\text{dist}\left(x,\text{CuL}(x)\right)=\text{dist}(x,y)$. Since $y$ is a cut point of $x$, there exists a tangent vector $\xi_0\in T_x M$ such that $y=\exp_x\left(c(\xi_0)\xi_0\right)$[2], where $c(\xi_0)$ is the distance from $x$ to its cut point in the $\xi_0$ direction. This in turn means that $\text{dist}(x,y) = c(\xi_0)$. </p> <p>Recall that $\text{inj}(x)=\inf_{\xi\in T_x M}(c(\xi))$. This means that $\text{inj}(x) \leq c(\xi_0) = \text{dist}(x,y)=\text{dist}\left(x,\text{CuL}(x)\right)$. If $\text{inj}(x)&lt; c(\xi_0)$, then since $M$ is compact, it means that there exists some other tangent vector $\xi\in T_x M$ with $c(\xi) &lt; c(\xi_0)$. But this means that $\exp_x(c(\xi)\xi)$ is a cut point of $x$ closer to it then $y$, and this is a contradiction.</p> <hr> <p>[1] See <em>Contributions to Riemannian Geometry in the Large</em> by W. Klingenberg</p> <p>[2] Here I'm using the notation of I. Chavel in his book <em>Riemannian Geometry - Modern Introduction</em>.</p> <hr> <p><strong>Update(@dror)</strong> Today I finally found a copy of the book *Riemannian Geometry" by Takashi Sakai, and there the above is stated as proposition 4.13 in chapter 3. Thanks anyway.</p> http://mathoverflow.net/questions/34229/compact-cover-of-a-hausdorff-compact-space Compact cover of a Hausdorff compact space Dror Atariah 2010-08-02T09:13:50Z 2010-08-17T10:01:42Z <p>In his book "Riemannian Geometry" do Carmo cites the Hopf-Rinow theorem in chapter 7. (theorem 2.8). One of the equivalences there deals with the cover of the manifold using nested sequence of compact subsets. This made me wonder whether the following lemma holds:</p> <p><strong>Lemma:</strong> Let $M$ be a compact Hausdorff space, and let $K_i \subset M$ be a sequence of compact subsets such that: $K_i\subset K_{i+1}$ and $\cup_{i=1}^{\infty} K_i = M$. Then there exists an index $i_0$ such that $K_i = M$ for all $i \geq i_0$.</p> <p>Here is my proof to this:</p> <p><strong>Proof</strong>: Assume that $K_i \neq M$ for all $i$, that is all $K_i$'s are proper subsets of $M$. With out loss of generality we can then assume that $K_i\subsetneq K_{i+1}$. This implies that $\forall i$ there exists $x_i$ such that $x_i\notin K_i$ but $x_i\in K_{i+1}$. Since $K_i$ is compact subset of a Hausdorff space, there exist open $U_i$ and $V_i$, such that $K_i \subset U_i$, $x_i\in V_i$ and $U_i \cap V_i = \emptyset$.</p> <p>Now, if $x\in \cup_{i=1}^{\infty} U_i$ then clearly $x\in M$ since $U_i \subset M$. On the otehr hand, if $x\in M$, then $x\in K_{i_0}$ for some $i_0$, and thus it is also in $U_{i_0}$. This yields that $M= \cup_{i=0}^{\infty}U_i$. </p> <p>Let us now assume that $\cup_{j=1}^n U_{i_j}=M$ is a finite cover. If $n_0 = \max_{j=1,\ldots,n}{i_j}$ then $x_{n_0+1}\notin \cup_{j=1}^nU_{i_j}$. This in turn means, that we cannot find a finite sub-cover of $M$ using the open cover $\cup_{i=1}^{\infty}U_i$. But this contradicts the compactness of $M$. This completes the proof. $\square$</p> <p>Finally, here's my question. Is this lemma correct? Is my proof correct?</p> <p>Thanks in advance and all the best!</p> <p>Dror, Edit: As I verified with the author, he meant that the last equivalence is valid when the manifold is <em>not compact</em>. Thus, my false lemma, is irrelevant from the first place.</p> http://mathoverflow.net/questions/34394/g-spaces-and-manifolds G-spaces and manifolds Dror Atariah 2010-08-03T14:42:39Z 2010-08-04T13:55:18Z <p>In his book <em>"The geometry of geodesics"</em> H. Busemann defines the notion of a G-space to be a space which satisfies the following axioms:</p> <ol> <li>The space is metric</li> <li>The space is finitely compact, i.e., a bounded infinite set has at least one accumulation point</li> <li>For every $x\neq z$ there exists a third point $y$ different from $x$ and $z$ such that $d(x,y)+d(y,z)=d(x,z)$</li> <li>To every point $p$ there corresponds $\rho_p>0$ such that for every two point $x,y\in S(p,\rho_p)$ there exists a point $z$ such that $d(x,y)+d(y,z)=d(x,z)$</li> <li>If $d(x,y)+d(y,z_1)=d(x,z_1)$ and $d(x,y)+d(y,z_2)=d(x,z_2)$ and $d(y,z_1)=d(y,z_2)$ then $z_1=z_2$.</li> </ol> <p>Busemann conjectured that every $G$-space is a topological manifold. My question is does every topological/smooth/Riemannian manifold is also a $G$-space?</p> <p>As for connected complete Riemannian manifold, I figured out that <strong>1</strong> holds since by the metric. <strong>3</strong> holds since every two points can be joined by a minimal geodesic, and then we can pick $y$ to be a point on it. <strong>4</strong> holds since it is a manifold and locally it is homeomorphic to some Euclidean space. Unfortunately, even in this case, I couldn't figure out <strong>5</strong> and <strong>2</strong>.</p> http://mathoverflow.net/questions/129204/intersection-of-2-visibility-polygons/129208#129208 Comment by Dror Atariah Dror Atariah 2013-04-30T19:01:18Z 2013-04-30T19:01:18Z As far as I understood this book is mainly about the algorithms related to the art gallery, isn't it? Anyway, I'll have a closer look. http://mathoverflow.net/questions/129204/intersection-of-2-visibility-polygons/129208#129208 Comment by Dror Atariah Dror Atariah 2013-04-30T14:40:59Z 2013-04-30T14:40:59Z Are you familiar with some references where these kind of things are shown/proven? There might be more things of that sort that will be useful for me. Thank any way! http://mathoverflow.net/questions/121052/reference-question-poncelet-theorem Comment by Dror Atariah Dror Atariah 2013-02-08T09:00:39Z 2013-02-08T09:00:39Z Here's an image illustrating the theorem, for the case of two circles <a href="https://dl.dropbox.com/u/9709624/Poncelet%20example.pdf" rel="nofollow">dl.dropbox.com/u/9709624/Poncelet%20example.pdf</a> http://mathoverflow.net/questions/117374/why-is-the-leibniz-rule-a-definition-for-derivations/117376#117376 Comment by Dror Atariah Dror Atariah 2012-12-31T10:00:00Z 2012-12-31T10:00:00Z For the sake of reference: W. K&#252;hnel provides in <i>Differential Geometry - Curves, Surfaces, Manifolds</i> (in 5B) three possible ways to defined the tangent space. The one given here is one of them. http://mathoverflow.net/questions/89198/smooth-a-matrix Comment by Dror Atariah Dror Atariah 2012-02-24T11:02:18Z 2012-02-24T11:02:18Z Another keyword you might want to check is &quot;outliers&quot;. This is some type of noise in this setting. http://mathoverflow.net/questions/81622/probing-a-manifold-with-geodesics Comment by Dror Atariah Dror Atariah 2011-11-23T11:52:40Z 2011-11-23T11:52:40Z By recording the geodesics intersections you can collect information on the cut locus of the surface, which in turn provide you with insight into the topology of the surface. http://mathoverflow.net/questions/54742/gaussian-curvature-radius/54765#54765 Comment by Dror Atariah Dror Atariah 2011-02-21T15:39:05Z 2011-02-21T15:39:05Z @Deane: I don't know what is the hidden message in your example. Here, the Gaussian curvature at $(0,0)$ is nothing but the determinant of the Hessian of $f$ and the mean curvature is the trace. Were you aiming at something more specific? http://mathoverflow.net/questions/54742/gaussian-curvature-radius/55734#55734 Comment by Dror Atariah Dror Atariah 2011-02-18T10:22:29Z 2011-02-18T10:22:29Z @Ramsay: Why is this smaller? Smaller then what? I agree that it is not very helpful definition w.r.t. Morse-Schoenberg lemma. But it is a sizing function which is well defined for non-flat surfaces. At the moment I'm not sure what I want to do with this definition; I found it a natural generalization to a negatively curved surface case and I was wondering if it was investigated in the literature. http://mathoverflow.net/questions/54742/gaussian-curvature-radius/54765#54765 Comment by Dror Atariah Dror Atariah 2011-02-10T09:13:02Z 2011-02-10T09:13:02Z @Deane: Considering eq. (3) mathworld.wolfram.com/MeanCurvature.html what you actually defined is nothing but the quotient $H/K$. Is it correct? Is there some standard geometrical meaning of this quantity? http://mathoverflow.net/questions/54742/gaussian-curvature-radius/54765#54765 Comment by Dror Atariah Dror Atariah 2011-02-08T15:53:00Z 2011-02-08T15:53:00Z @Deane: By <i>mean radius</i>, do you refer to $1/H(p)$ where $H(p)=\kappa_1(p)+\kappa_2(p)/2$ is the mean curvature at the point $p$. http://mathoverflow.net/questions/54742/gaussian-curvature-radius/54755#54755 Comment by Dror Atariah Dror Atariah 2011-02-08T15:15:58Z 2011-02-08T15:15:58Z In short, and hopefully as correct, $\sqrt{\frac{1}{K}}$ is the arithmetic-harmonic mean of the two principal curvature radii $\frac{1}{\kappa_i}$. http://mathoverflow.net/questions/51644/convex-subsets-of-hadamard-manifolds Comment by Dror Atariah Dror Atariah 2011-01-10T14:48:01Z 2011-01-10T14:48:01Z @Henrik: Can you specify what you consider a <i>convex</i> subset of a Riemannian manifold? http://mathoverflow.net/questions/49946/principal-curvatures-and-curvature-directions/49951#49951 Comment by Dror Atariah Dror Atariah 2010-12-20T10:59:05Z 2010-12-20T10:59:05Z That is my understanding as well. However, is it possible that some geometrical meaning of the pcd's magnitude does exist? http://mathoverflow.net/questions/48629/algebraic-surfaces-and-their-intrinsic-geometry/48631#48631 Comment by Dror Atariah Dror Atariah 2010-12-09T10:05:53Z 2010-12-09T10:05:53Z @Ariyan: First, thanks for your answer! Unfortunately, I'm not sufficiently comfortable with the notions you mentioned. If I get the general picture right, then my initial understanding is correct, namely, that wherever the surface is non-singular, then it is also a (Riemannian manifold). At least this is sort of a good start. http://mathoverflow.net/questions/48629/algebraic-surfaces-and-their-intrinsic-geometry Comment by Dror Atariah Dror Atariah 2010-12-09T09:20:58Z 2010-12-09T09:20:58Z @Donu: As my surfaces most likely contain singularities, I'm looking for some theory that will extend the Riemannian geometry tools to the singularities. By removing the singularities the surface's connectivity will change - this is something I would like to avoid.