From: jeffcoat@stat.rice.edu (Mark Evans Jeffcoat) Subject: Re: Immersion Theory Date: 13 May 1999 16:12:44 GMT Newsgroups: sci.math.research Keywords: Isometric immersion of H^2 into R^4? Roger Arbogast (roar0000@stud.uni-sb.de) wrote: : Hi everybody! I am studying mathematics at an university in Germany and my : question is the following: Does anybody know if there is an isometric : immersion of the two- dimensional Hyperbolic Space (i.e. the Hyperbolic : Plane) into the four- dimensional (n=4) Euclidean Space (i.e. IR^4)? I know : that the answer is yes if n=5 and that the answer is no if n=3. I would be : very thankful for a hint to any literature for the case n=4. : Thanks Roger (email: roar0000@stud.uni-sb.de)! : According to Do Carmo, _Differential Geometry of Curves and Surfaces_, this is still an open question. (The edition I have was printed in 1976, so any solution that might exist ought to be quite recent.) ============================================================================== From: lrudolph@panix.com (Lee Rudolph) Subject: Re: embedding the hyperbolic plane? Date: 12 Jul 1999 12:30:10 -0400 Newsgroups: sci.math Pertti Lounesto writes: >Dave Rusin wrote: > >> Nick Halloway wrote: >> >Can the hyperbolic plane be embedded isometrically in 4-space? > >I do not understand the question. Please, explain. "The hyperbolic plane" can be taken to mean "the smooth manifold R^2, equipped with a complete Riemannian metric of constant curvature -1" (there is a unique such metric on R^2). "Embedded isometrically in 4-space" must mean, here, "embedded as a smooth, closed submanifold of R^4 in such a way that the Riemannian metric induced from the flat metric of R^4 is the given metric on R^2". >> Unknown (!). There's an embedding into R^5, none into R^3. > >The answer does not give me glue, to anything. Please, expand. John Klein sometimes posts here. He, or someone else who is familiar with the work of Gromov, might comment on whether it's known if there is a *non-smooth* embedding of R^2 as a closed subset of R^4, such that (the non-Riemannian) metric induced on R^2 (in which the distance between two points of R^2 is the infimum of the lengths of curves on the image in R^4 joining the images of the points) is isometric (in the sense of metric spaces) to the hyperbolic metric on R^2. Lee Rudolph ============================================================================== From: rusin@vesuvius.math.niu.edu (Dave Rusin) Subject: Re: embedding the hyperbolic plane? Date: 12 Jul 1999 23:03:32 GMT Newsgroups: sci.math In article <3788CBDD.7F46FBD6@hut.fi>, Pertti Lounesto wrote: > >Dave Rusin wrote: > >> Nick Halloway wrote: >> >Can the hyperbolic plane be embedded isometrically in 4-space? >> >> Unknown (!). There's an embedding into R^5, none into R^3. > >Can you give some details, both of the embedding into 5D and the >unknowledgeability of an embedding into 4D. Sorry, not really. This is recent work which only crossed my desk by reference. Some Math Reviews citations are below. I didn't say an embedding into R^4 was "unknowable", rather, that it simply isn't yet known. If I had to bet, I'd guess there is no embedding into R^4, but that we haven't yet found the tools to show this. Just to clarify the problem here, the hyperbolic plane can be thought of as the topological space usually known as the open unit disk in R^2. Clearly, it can be embedded into R^2 ! But in addition to the topological structure, it can be given the structure of a Riemannian manifold, that is, we can give a method for computing distances on it _distinct from_ the usual metric. (Ultimately this metric comes from an inner product on each tangent plane T_p(M) of the manifold; the hyperbolic plane is a space in which the inner product is of signature (+1, -1). ) An isometric embedding would then be one in which distances between points are preserved. I'm embarassed to say I cannot recall quite why there is no embedding of H^2 into R^3. It isn't quite the constant negative curvature which prohibits this, as there _is_ such a surface in R^3 which is homeomorphic not to the disk but to the punctured disk. It looks rather like a bugle horn with Gabriel infinitely far away... By the way, you can _always_ embed a Riemannian manifold into _some_ Euclidean space; that's the Nash embedding theorem. See e.g. index/57RXX.html dave 96e:53098 53C42 (53A35) Oláh-Gál, Róbert The $n$-dimensional hyperbolic space in ${E}\sp {4n-3}$. (English. English summary) Publ. Math. Debrecen 46 (1995), no. 3-4, 205--213. D. Blanusa constructed an isometric immersion of class $C\sp \infty$ of the $n$-dimensional hyperbolic space $H\sp n$ into the Euclidean space $E\sp {6n-5}$ [Monatsh. Math. 59 (1955), 217--229; MR 17, 188f]. By modifying this construction, the author constructs an isometric immersion of class $C\sp \infty$ of the $n$-dimensional hyperbolic space $H\sp n$ into the Euclidean space $E\sp {4n-3}$. Reviewed by Vladislav Goldberg _________________________________________________________________ 96e:53084 53C42 (53A35) Azov, D. G.(RS-CHLT) Isometric embedding of $n$-dimensional metrics into Euclidean and spherical spaces. (Russian. Russian summary) Vestnik Chelyabinsk. Univ. Ser. 3 Mat. Mekh. 1994, no. 1, 12--16. D. Blanusa constructed an isometric immersion of the $n$-dimensional hyperbolic space $H\sp n$ into the Euclidean space $E\sp {6n-5}$ [Monatsh. Math. 59 (1955), 217--229; MR 17, 188f] and into the spherical space $S\sp {6n-4}$ [see Glasnik Mat.-Fiz. Astronom. Ser. II Drustvo Mat. Fiz. Hrvatske 19 (1964), 53--61; MR 30 #2421]. E. R. Rozendorn [Akad. Nauk Armyan. SSR Dokl. 30 (1960), 197--199; MR 24\#A3568] constructed an embedding of the metric $ds\sp 2 = du\sp 2 + f\sp 2 (u) dv\sp 2$ into the Euclidean space $E\sp 5$. In the paper under review, the author considers two classes of Riemannian metrics: $ds\sp 2 = du\sb 1\sp 2 + f\sp 2 (u\sb 1) \sum\sb {i=2}\sp n du\sb i\sp 2$, $f > 0$, and $ds\sp 2 = g\sp 2 (u\sb 1) \sum\sb {i=2}\sp n du\sb i\sp 2$, $g > 0$, and proves that if $n > 2$, then these two metrics admit an isometric embedding into the Euclidean space $E\sp {4n-4}$ and into the spherical space $S\sp {4n-4}$, and if $n = 2$, then these two metrics admit an isometric embedding into $E\sp 4$ and $S\sp 4$ in the form of surfaces with singular lines. The class $C\sp m$ of all these embeddings coincides with the class of the functions $f$ and $g$. The author indicates that the isometric embedding of the Lobachevskii plane into $E\sp 4$ was obtained earlier by I. Kh. Sabitov [Sibirsk. Mat. Zh. 30 (1989), no. 5, 179--186, 218; MR 90k:53099]. Reviewed by Vladislav Goldberg _________________________________________________________________ 90k:53099 53C42 (53A35) Sabitov, I. Kh. Isometric immersions of the Lobachevski\u\i plane in $E\sp 4$. (Russian) Sibirsk. Mat. Zh. 30 (1989), no. 5, 179--186, 218; translation in Siberian Math. J. 30 (1989), no. 5, 805--811 It is well known that the Lobachevskii plane can be isometrically imbedded in $E\sp 6$ or immersed in $E\sp 5$ in the class of $C\sp \infty$-surfaces. In this paper, the author proves the following result: The Lobachevskii plane can be isometrically immersed in $E\sp 4$ in the form of a piecewise-analytic surface with smoothness in the class $C\sp {0,1}$. Moreover, the Lobachevskii plane cannot be isometrically immersed in $E\sp 4$ as a generalized $C\sp 1$-surface of revolution: $x\sb 1+ix\sb 2=F(u)e\sp {iPv}$, $x\sb 3+ix\sb 4=G(u)e\sp {iQv}$. Reviewed by Yi Bing Shen © Copyright American Mathematical Society 1990, 1999 ============================================================================== From: rusin@vesuvius.math.niu.edu (Dave Rusin) Subject: Re: embedding the hyperbolic plane? Date: 16 Jul 1999 15:58:34 GMT Newsgroups: sci.math "Nick Halloway" wrote: >Can the hyperbolic plane be embedded isometrically in 4-space? Hoping to address a few of the finer points in the recent discussion of the hyperbolic disk, I looked up the article by Olah-Gal whose review I cited in another post. It contains a nice summary of the situation in just a few pages. I won't display the whole article of course, but I will quote sections which refer to our previous discussion. 1. The "best" embedding of the hyperbolic plane into Euclidean space seems to be a construction of Blanusa (Monatshefte Math. 59 (1955) 217-229) which is an explicit, smooth (C-infinity) embedding of the plane onto a subset of R^6 which preserves the hyperbolic metric. I will give the formulas below. 2. Olah-Gal notes in his article that by deleting the last coordinate (i.e. projecting to R^5) we obtain a surface contained in a smaller space, which still has constant negative curvature. However, this projection is not one-to-one (it is only a local embedding) and has a singular point at the origin. A handy illustration is provided in the article showing an analogous situation in which a surface in R^4 [drawn on paper on my 2-dimensional desk...] projects to a surface in R^3. 3. From the introduction: "[Blanusa's map] is of class C^\infty, but it is not analytic. A construction of an analytical embedding of the hyperbolic plane into E^n (with sufficiently large n) is unknown even these days." [1995] I have a reference to Gromov's "Partial Differential Relations" which indicates he proves there that hyperbolic k-space can be analytically isometrically _immersed_ into Euclidean (5k-5)-space. I don't know for certain that this applies to k=2. 4. From p.208: "In 1955 Amsler [Math. Ann. 130, 234-256] proved that each surface of E^3 with constant negative curvature has an edge (i.e. it contains a curve consisting of singularities), and showed the nonexistence in E^3 of surfaces of constant negative curvature with singularity consisting of one point alone." In a previous post I described a surface called "Gabriel's horn" of constant negative curvature, with a singular curve around the bell of the horn. Another poster noted that the non-immersibility of H^2 in R^3 had something to do with singularities. I guess it was Hilbert who first showed H^2 does not embed (or even immerse) in R^3. The situation in R^4 is open as far as I can tell. Any _compact_ domain in the hyperbolic plane can be isometrically immersed into R^4; see Gromov's book. (Indeed, from all reports, one should see Gromov's book for just about everything regarding smoothly embedding Riemannian manifolds.) Olah-Gal's article goes on to describe the constructions in substantial detail and to comment on generalizations. Now I will describe the embedding of R^2 into R^6. This is not exactly what one would call an "obvious" construction, but there are certain patterns to it which suggest how Blanusa might have been led to it. We will send the point (u,v) to a point with six coordinates (x1, ..., x6) which are functions of u and v of the special forms x1 = x1(u) x2 = f1(u) sin( v psi1(u) ) x3 = f1(u) cos( v psi1(u) ) x4 = f2(u) sin( v psi2(u) ) x5 = f2(u) cos( v psi2(u) ) x6 = v I will describe the functions x1, f1, f2, psi1, psi2 (of one variable each) in stages. Let [x] denote the integer part of x. The functions psi1 and psi2 are periodic functions of |u| (period = 2) which on [0,2] are exponentials of linear maps: psi1(u) = exp( 2*[ (|u|+1)/2 ] + 5 ) psi2(u) = exp( 2*[ ( |u| )/2 ] + 6 ) There are discontinuities in the psi_i at certain integers but other parts of the construction will keep the x_i smooth. Define two functions phi_i via certain normalized antiderivatives: writing F(x) = sin( pi x )/exp( sin^{-2}(pi x) ) we have phi1(u) = { (1/A) integral( F(x), x=0 to x= u+1 ) }^(1/2) phi2(u) = { (1/A) integral( F(x), x=0 to x= u ) }^(1/2) where A = integral( F(x), x=0 to x=1 ) = 0.141327... These functions phi_i are non-negative, periodic, and satisfy phi1^2 + phi2^2 = 1 and phi1(u) = phi2(u+1). You can think of the phi_i as being very smooth versions of |sin(pi u)| and |cos(pi u)|. Now set f_i(u) = sinh(u) phi_i(u)/psi_i(u) for i=1,2 and define x1 to be an antiderivative of 1-(f1')^2-(f2')^2 having x1(0)=0. Loosely speaking the mapping x1 sends lines in R^2 far enough away, and the the coordinates x2, x3, x4, x5 allow the points in these line to spin around in four perpendicular directions, with enough spinning to account for the fact that the images of lines are supposed to grow very long. The last coordinate x6 merely adds a motion in another perpendicular direction to separate points so that these curves don't self-intersect. The metric which R^2 inherits from this embedding into R^6 comes out to ds^2 = du^2 + cosh(u)^2 dv^2, from which one finds the curvature to be constant and negative, making R^2 into the hyperbolic plane. dave ============================================================================== [additional related review -- djr] 80c:53071 53C40 Yang, Paul Curvatures of complex submanifolds of ${C}\sp{n}$. J. Differential Geom. 12 (1977), no. 4, 499--511 (1978). The author treats the following question: Does there exist a complete complex submanifold of complex space with holomorphic sectional curvature bounded away from zero? The motivation for the question is due to the fact proved by S. Bochner (Bull. Amer. Math. Soc. 53 (1947), 179 - 195; MR 8, 490) that the Poincare metric of constant negative curvature on the unit disc cannot be holomorphically embedded in complex Euclidean space even locally. The author gives partial results to the above question. The first claim is that a negative answer to this question would imply that there is no bounded complete submanifold of complex space. Secondly, the question is considered for the hypersurfaces of complex space. Namely, the fact that complete complex hypersurfaces of complex space cannot have strongly negative holomorphic sectional curvatures is proved. Thirdly, the author shows that for arbitrary codimension this question can be reduced to the consideration of holomorphic curves. Using the higher order curvature functions, the author shows that two such functions are enough to determine a holomorphic curve uniquely up to a rigid motion in complex space, thus providing a justification for the generalization of the fact for hypersurfaces. Fourthly, the author derives a criterion, involving the curvature behavior at infinity of a simply connected metric Riemannian surface, for it to be confor- mally equivalent to the disc, which is a complement to results of R. E. Greene and H.-H. Wu (ibid. 77 (1971), 1045 - 1049; MR 44 #473). Lastly, a curvature estimate is proved for a piece of a curve in complex 2-space which is a graph over a domain in complex line. Reviewed by Yoshiaki Maeda Cited in reviews: 80j:53063 © Copyright American Mathematical Society 1980, 1999