Determining a surface in $\mathbb{R}^3$ by its Gaussian curvature - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-23T03:56:07Z http://mathoverflow.net/feeds/question/76955 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/76955/determining-a-surface-in-mathbbr3-by-its-gaussian-curvature Determining a surface in $\mathbb{R}^3$ by its Gaussian curvature Joseph O'Rourke 2011-10-02T00:37:54Z 2011-10-28T21:48:33Z <p>A curve in the plane is determined, up to orientation-preserving Euclidean motions, by its curvature function, $\kappa(s)$. Here is one of my favorite examples, from Alfred Gray's book, <a href="http://library.wolfram.com/infocenter/Books/3759" rel="nofollow"> <em>Modern Differential Geometry of Curves and Surfaces with Mathematica</em></a>, p.116: <br /> &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; <img src="http://cs.smith.edu/~orourke/MathOverflow/Curvature.sSin2s.jpg" alt="s Sin2 s" /> <br /></p> <blockquote> <p><b>Q1</b>. Is there an analogous theorem stating that a surface in $\mathbb{R}^3$ is determined (in some sense) by its Gaussian curvature?</p> </blockquote> <p>I know such a reconstruction path (curvature $\rightarrow$ surface) is needed in computer vision, and so there are approximation algorithms, but I don't know what is the precise theorem underlying this work.</p> <blockquote> <p><b>Q2</b>. Are there higher-dimensional generalizations, determining a Riemannian manifold by its curvature tensor?</p> </blockquote> <p>I have no doubt this is all well known to the cognoscenti, in which case a reference would suffice. Thanks!</p> <p><b>Addendum</b> (<em>4Oct11</em>). Permit me to augment this question with a relevant reference which loosens the notion of "determines" and answers my <b>Q1</b> with that notion replaced by "find some." The paper by Gluck, Krigelman, and Singer, entitled "The converse to the Gauss-Bonnet Theorem in PL," <a href="http://projecteuclid.org/DPubS?service=UI&amp;version=1.0&amp;verb=Display&amp;handle=euclid.jdg/1214432558" rel="nofollow"><em>J. Diff. Geom</em>, 9(4): 601-616, 1974</a>, poses this question:</p> <blockquote> <p>Suppose that a closed smooth two-manifold $M$ and a smooth real-valued function $K \;:\; M \rightarrow \mathbb{R}$ are given, and that one is asked to find a Riemannian metric for $M$ having $K$ as its Gaussian curvature. [...] With these restrictions on $K$ [just elided], the problem has been completely solved for all closed smooth two-manifolds by: Melvyn Berger [...], Gluck [...], Moser [...], Kazdan and Warner [...]. Recently Kazdan and Warner have obtained a uniform solution. The problem for compact two-manifolds with boundary, however, seems not to have been addressed in the smooth category.</p> </blockquote> <p>The <a href="http://www.ams.org/mathscinet/search/publdoc.html?arg3=&amp;co4=AND&amp;co5=AND&amp;co6=AND&amp;co7=AND&amp;dr=all&amp;pg4=AUCN&amp;pg5=TI&amp;pg6=PC&amp;pg7=ALLF&amp;pg8=ET&amp;r=1&amp;review_format=html&amp;s4=&amp;s5=the%20converse%20to%20the%20gauss-bonnet%20theorem%20in%20PL&amp;s6=&amp;s7=&amp;s8=All&amp;vfpref=html&amp;yearRangeFirst=&amp;yearRangeSecond=&amp;yrop=eq" rel="nofollow">MathSciNet review of this paper</a> was written by Gromov. </p> http://mathoverflow.net/questions/76955/determining-a-surface-in-mathbbr3-by-its-gaussian-curvature/76957#76957 Answer by José Figueroa-O'Farrill for Determining a surface in $\mathbb{R}^3$ by its Gaussian curvature José Figueroa-O'Farrill 2011-10-02T01:20:23Z 2011-10-02T01:20:23Z <p>I'm not sure what you mean by "determining". One natural notion of equivalence is for two surfaces to be related by an ambient isometry (a euclidean motion).</p> <p>A basic result is that two surfaces in $\mathbb{R}^3$ are related by an isometry of $\mathbb{R}^3$ if and only if their first and second fundamental forms agree.</p> <p>A weaker condition is that of isometry. Two surfaces are said to be <em>isometric</em> if their first fundamental forms agree. Gauss's <em>Theorema Egregium</em> says that isometric surfaces have the same Gaussian curvature, but the converse is not true: there are examples of surfaces with the same Gaussian curvature, but which are not isometric.</p> <p>In dimension $\geq 4$ Kulkarni in his paper <a href="http://www.jstor.org/stable/1970580" rel="nofollow"><em>Curvature and Metric</em></a> showed that a diffeomorphism which preserves the sectional curvature is an isometry, except possibly in the case of constant sectional curvature. In dimension $\leq 3$ there are counterexamples which are mentioned in his paper.</p> http://mathoverflow.net/questions/76955/determining-a-surface-in-mathbbr3-by-its-gaussian-curvature/76958#76958 Answer by Anton Petrunin for Determining a surface in $\mathbb{R}^3$ by its Gaussian curvature Anton Petrunin 2011-10-02T01:23:11Z 2011-10-02T04:40:48Z <p>Insted of Q2, I will answer the following question:</p> <blockquote> <p>Are there higher-dimensional generalizations, determining a submanifold in $\mathbb R^q$.</p> </blockquote> <p>Yes, there are some analogs, but I am sure you do not need them.</p> <p>They work well for $n$-dimensional submanifolds in $\mathbb R^{n{\cdot}(n+3)/2}$. (curves in $\mathbb R^2$, surfaces in $\mathbb R^5$ and so on). Instead of natural parametrization you remember metric tensor $g$; which is a degree 2 homogeneous polynomial on the tangent space. Instead of curvature you remember the following degree 4 homogeneous polynomial $h(X)=|s(X,X)|^2$, where $s\colon T\times T\to N$ is the second fundamental form (for two tangent vectors $X$ and $Y$ the value $s(X,Y)$ is a normal vector).</p> <p>The proofs are the same as Frenet–Serret formulas. You can find it in Spivak's book. </p> http://mathoverflow.net/questions/76955/determining-a-surface-in-mathbbr3-by-its-gaussian-curvature/76968#76968 Answer by Jean-Marc Schlenker for Determining a surface in $\mathbb{R}^3$ by its Gaussian curvature Jean-Marc Schlenker 2011-10-02T06:06:37Z 2011-10-02T06:06:37Z <p>There is a satisfactory answer to Q1 if you restrict to convex surfaces: one way to state the question is then as the Minkowski problem. That is, you choose a positive function $k$ on the sphere and look for a surface $S$ in $R^3$ with Gauss curvature $k(n)$ at the point where the unit normal vector is $n$. This problem was solved in the early 50', see the Math Review of MR0058265 (15,347b) Nirenberg, Louis The Weyl and Minkowski problems in differential geometry in the large. Comm. Pure Appl. Math. 6, (1953). 337–394. In higher dimensions you can still play the same game, for convex hypersurfaces (or sometimes using a weaker form of convexity) and find one with prescribed "curvature", where the curvature can be a symmetric function of the eigenvalues of the shape operator. For instance the determinant of the shape operator, which corresponds to the Minkowski problem in higher dimension, which was also solved in the early 50', but it's not exactly what you're asking for in Q2. </p> http://mathoverflow.net/questions/76955/determining-a-surface-in-mathbbr3-by-its-gaussian-curvature/76990#76990 Answer by Matt Noonan for Determining a surface in $\mathbb{R}^3$ by its Gaussian curvature Matt Noonan 2011-10-02T15:42:39Z 2011-10-04T22:05:12Z <p>The associated family of a minimal surface gives a tangible counterexample. The Weierstrass representation lets you cook up a conformally parameterized minimal surface from a meromorphic pair $f \sqrt{dz}$, $g \sqrt{dz}$.</p> <p>The parameterization is then given by $$F(x,y) = Re\int_0^{x+iy} (f^2 - g^2, i(f^2 + g^2), 2 fg) ~dz$$</p> <p>The normal map of $F$ can be obtained by thinking of $g/f$ as a map to the Riemann sphere, and the metric induced by $F$ is just $4(|f|^2 + |g|^2)^2 |dz|^2$. From this data, you can cook up the Gauss and mean curvatures, and it happens to be true that if $f \sqrt{dz}$ and $g \sqrt{dz}$ are meromorphic, you get a minimal surface.</p> <p>But then consider what happens if you multiply both $f$ and $g$ by $e^{i \theta}$ --- the normal map and the metric are both unchanged, and $e^{i\theta} f \sqrt{dz}, e^{i\theta} g \sqrt{dz}$ are still quite meromorphic, so you get a new minimal surface which is isometric to your old one. This means you have made a new surface whose principal curvatures agree with your old one! </p> <p>I think the moral here is that even knowing the metric and the complete set of principal curvatures isn't enough to reconstruct a surface --- the curvature <i>directions</i> are also vital data.</p> <p>To see all this in action, here is a video with strange music showing the <a href="http://www.youtube.com/watch?v=-Pa6FOK3gpM" rel="nofollow">helicoid transforming into the catenoid</a>, which starts with the Weierstrass data for the catenoid and then multiplies by $e^{i \theta}$, with $\theta$ increasing as the movie progresses. Every one of the surfaces is isometric to the catenoid! But they <i>do</i> have different second fundamental forms. </p> http://mathoverflow.net/questions/76955/determining-a-surface-in-mathbbr3-by-its-gaussian-curvature/77223#77223 Answer by Sebastian for Determining a surface in $\mathbb{R}^3$ by its Gaussian curvature Sebastian 2011-10-05T11:43:04Z 2011-10-05T11:43:04Z <p>This is not really an answer to one of the questions, but it might be interesting: It is a well-known fact that CMC surfaces come along in 1dimensional families, the special case of $H=0$ is described in Matt's answer. Of course, these associated surfaces are generally not closed, but the first and second fundamental form are well-defined globally. By a theorem of Tribuzy and Lawson (On the mean curvature function on compact surfaces, JDG), this cannot happen for non-constant mean curvature $H$ on compact oriented surfaces. In fact, they have shown that there are at most two different isometric immersions from a compact Riemannian surface (surface with a metric and orientation) into $\mathbb R^3$ which have the same non-constant mean curvature.</p> http://mathoverflow.net/questions/76955/determining-a-surface-in-mathbbr3-by-its-gaussian-curvature/78432#78432 Answer by AO for Determining a surface in $\mathbb{R}^3$ by its Gaussian curvature AO 2011-10-18T08:17:37Z 2011-10-28T21:48:33Z <p>In section 4.5 of his <a href="http://www.amazon.com/Panoramic-View-Riemannian-Geometry/dp/3540653171" rel="nofollow">big book on Riemannian geometry</a>, Berger has a discussion on the question of to what degree (and in what sense) the curvature determines the metric. He quotes the following theorem by Cartan on the two-dimensional case.</p> <blockquote> Given two surfaces with Riemannian metrics, so that the functions $K$ and $\|dK\|^2$ have everywhere independent differentials, a map between these surfaces is an isometry precisely when it preserves the four functions \begin{eqnarray} I_1 &=& K\\ I_2 &=& \|dK\|^2\\ I_3 &=& \langle dK,dI_2\rangle\\ I_4 &=& \|dI_2\|^2 \end{eqnarray} [where $K$ is the Gaussian curvature.] </blockquote> http://mathoverflow.net/questions/76955/determining-a-surface-in-mathbbr3-by-its-gaussian-curvature/78964#78964 Answer by taha for Determining a surface in $\mathbb{R}^3$ by its Gaussian curvature taha 2011-10-24T11:12:18Z 2011-10-24T11:12:18Z <p>An important classical result due to Bonnet [53] states that the sextuplet {E, F, G\ e, f, g] determines the surface £ up to its position in space. and there exist a progrme for solving this proplem by mathematica where {E, F, G\ e, f, g] . are the first and seconed fundamental form on a surface . I am taha yousif from egypt my emial is tahaa8080@yahoo.com</p> http://mathoverflow.net/questions/76955/determining-a-surface-in-mathbbr3-by-its-gaussian-curvature/78969#78969 Answer by taha for Determining a surface in $\mathbb{R}^3$ by its Gaussian curvature taha 2011-10-24T11:33:56Z 2011-10-24T11:33:56Z <p>this link <a href="http://eom.springer.de/b/b016910.htm" rel="nofollow">http://eom.springer.de/b/b016910.htm</a> give you the main theorem and I am sure that there exist aprogram for solving this proplem by mathematica </p>