A Simple 3D Isometric Embedding of the Flat Square Torus
AA Simple 3D Isometric Embedding of the FlatSquare Torus
J. Richard Gott III ∗ and Robert J. Vanderbei Department of Astrophysical Sciences Department of Operations Research and Financial EngineeringPrinceton University, Princeton, NJ, 08544, USA
Abstract
Start with Gott (2019)’s envelope polyhedron (Squares—4 around apoint): a unit cube missing its top and bottom faces. Stretch by a factor of2 in the vertical direction so its sides become (2 × F − E + V = 0,implying a (toroidal) genus = 1. It is isometric to a flat square torus. Likeany polyhedron it has zero intrinsic Gaussian curvature on its faces andedges. Since 4 right angled rectangles meet at each vertex, there is noangle deficit and zero Gaussian curvature there as well. All meridian andlatitudinal circumferences are equal (4 units long). Many video games including
Pac-Man and
Asteroids are played on a square TVscreen with a toroidal geometry. Pac-Man, or a spaceship that disappears offthe top of the screen immediately reappears at the bottom at same horizontallocation. Likewise, if Pac-Man or a spaceship disappears off the right-hand edgeof the screen it immediately reappears on the left-hand edge of the screen at thesame vertical location. These games thus have a square Euclidean flat geometrybut a toroidal topology. It takes seven colors in general to color nations on sucha TV game map, rather than the four required on the plane. It is a map ofa toroidal flatland universe of two dimensions, that obeys Euclidean geometry,has zero Gaussian curvature everywhere (every small circle around any pointhas a circumference of 2 πr , and the sum of angles in any triangle on the screenhas a sum of angles of 180 ◦ ). But the map has two boundary cuts, one acrossthe top of the screen, one across the right of the screen, connected to the samecuts on the opposite sides. This game can be mapped onto a donut (or tire ∗ Corresponding author: [email protected] a r X i v : . [ m a t h . M G ] J un nner tube shape) in 3D space, but with distortion, since the tire has a curvedsurface. It has positive Gaussian curvature on the outer circumference of thetire, and negative Gaussian curvature on the interior of the tire, where thegeometry is saddle shaped. Can the game geometry be embedded in 3D space?Videos illustrating this usually start with a square, flat sheet of paper. Thedemonstrator then bends it into a cylinder and tapes it together. If the sidelength of the square is S , the cylinder has a height of S and a radius of r , where S = 2 πr . The demonstrator bends the paper; this does not change its intrinsicGaussian curvature of zero everywhere. A cylinder has extrinsic curvature, butzero intrinsic curvature everywhere. One of the two boundary cuts has beensown together and healed when the demonstrator tapes the two opposite edgestogether. This leaves at the ends of the cylinder, two circular edges, which nowneed to be taped together. Then the demonstrator tries fruitlessly to bend thecylinder around to bring them together, and illustrates that it would tear andcrumple the paper to try to do so. One is trying to produce a tire inner tubewith curvature from a flat cylinder with no curvature. Doesn’t work we are told.But if we lived in 4D Euclidean space with a metric ds = dx + dy + dz + dw , we could do it by bending the cylinder around smoothly in the w direction.The result is a 2D flat torus living in 4D space (a Clifford torus): r = x + y = 12 = z + w . This is a 2D surface defined by two equations, living in 4D Euclidean space.Here r is a constant r = 1 / √
2. Define θ and φ by x = r sin( θ ), y = r cos( θ ), z = r sin( φ ), w = r cos( φ ) and we can show that the equations of the surface aresatisfied automatically since sin + cos = 1, and that using ( θ, φ ) as coordinateson the surface, the metric on the surface is ds = r dθ + r dφ , where 0 ≤ θ < π, ≤ φ < π. This is a flat metric with zero Gaussian curvature. It is a flat square with side S = 2 πr = √ π , our original TV game with no distortion, and no boundarycuts. The vertical and horizontal coordinates are cyclic, and the boundary cutshave been healed. This is an isometric embedding of the TV game geometry in4D Euclidean space.Nash (1954) famously proved that there must exist a C isometric embed-ding of the flat square torus in 3D Euclidean space – but he provided no con-structive proscription for doing so. (A C isometric embedding has continuousfirst derivatives, in other words it admits tangent planes that very smoothly,being defined at every point. (A C isometric map which is not C has nodefined extrinsic curvature everywhere.) Recently, Borrelli, Jabrane, Lazarus,and Thibert in their paper “Isometric Embedding of the Square Flat Torus inAmbient Space”, Borrelli et al. (2013) have shown how to do it. It looks like atorus, but with ripples on it. And there are ripples on those ripples tipped atan angle, and tinier ripples on those ripples at another angle, ad infinitum. Itis sort of a smooth fractal. The amplitude of the ripples on the inner edge of2he torus is larger, making its circumference exactly as large as the smoothercircumference on the outside. Beautiful pictures of it are shown in their paper.For his original work in this field, John Nash (who by the way often visitedour Astrophysics Library in Princeton) received the Abel Prize in Mathematics.Many mathematicians regard this as a greater work on his part than his workon game theory for which he won the Nobel Prize.While the embedding of Borelli, et al. is beautiful and important mathe-matically, it does not make a very practical object on which to play the game.It is so crinkled that it would be hard to even see the game portrayed on itscorrugated surface. Yet it is isometric. We wish to propose here a simple C isometric mapping of the flat square torusinto 3D Euclidean space. Note that in Figure 1, Gott’s envelope polyhedronSquares—4 around a point has a toroidal topology. It is a cube with its top andbottom missing.Figure 1: Some Finite Envelope Polyhedra from Gott (2019). Envelope poly-hedra are regular polyhedra with regular polygons as faces, the arrangementof polygons around each vertex must be identical, but not all dihedral anglesare equal, and some are 0 ◦ , allowing polygons to appear back to back. Fromleft to right at top: Squares—4 around a point, Triangles—6 around a point,Squares—4 around a point (a unit cube missing its top and bottom—tippedon its side in this picture—a subject of this paper), and Triangles—8 around apoint. At bottom, Squares—2 around a point (a dihedron–see Coxeter (1938)).To see this stereo view, touch your nose to the page and slowly withdraw it.You will see a blurry 3D image in the center which will come into focus.The cube with its top and bottom faces missing has 4 exterior squares linkedby the square holes at the top and bottom to 4 interior squares. It thus has8 faces, 4 outside edges, 4 inside edges, and 8 edges at the top and bottomthat link the exterior and interior faces. It has 8 vertices. Thus F − E + V =8 −
16 + 8 = 0 = 2(1 − g ), giving this a genus of 1, which is a torus. Notice3hat there is no Gaussian curvature on any of the faces, they are flat. Thereis no Gaussian curvature on any of the edges; they are limits of flat surfaces ofpartial cylinders whose radii shrink to zero. As in any polyhedron, the Gaussiancurvature is contained in delta-functions at the vertices. In this case since thereare 4 squares around every vertex, the angle deficit is zero at each vertex andthe integral of the Gaussian curvature at the vertex is zero. This is a geometricsurface which contains zero volume. The circumference around its equator onthe outside is 4 units (if the cube is of one-unit side length). This circumferencecircles the 4 outside faces. It is obviously a geodesic. The inner circumferencecircling the 4 inside faces is also 4 units. It is also a geodesic. What aboutthe upper edge: the four straight edges at the top that form the opening thatconnects the outer and inner faces at the missing top face of the cube. Theseseem to turn 90 ◦ four times as one circles the top of the cube. Yet this circuitis also a geodesic. For any 2D surface imbedded in 3D Euclidean space, theshortest curve connecting points A and B in the surface is a geodesic. Pickany two points A and B in the four top edges. The shortest path from A to B staying in the surface will necessarily travel along those top edges. If the curvedips down into any of the faces during part of its journey it will be longer.Take this cube and lengthen it in the vertical direction by a factor of 2. Nowthere is a missing square at the top and bottom, and there are 4 rectangularsides made up of 2:1 rectangles. There are 4 exterior rectangles of length 2units and width 1 unit, and 4 interior rectangles of length 2 units and width 1unit. The equatorial circumference of any horizontal geodesic traversing the 4exterior rectangular faces is 4 units as before. But now consider the longitudinalcircumference of the torus. Starting at a bottom edge, one first traverses anoutside rectangular face traveling 2 units, one then travels over the top edge andtraverses the inside rectangular face traveling another 2 units and reaches thebottom edge where one started. These are now the vertical lines in the originalgame, 4 units long, while the equatorial geodesics, also 4 units in circumference,are the horizontal lines in the original game. This is a flat square torus. Go back to the original demonstration. Bend the original square piece of pa-per containing an
Asteroids game into a cylinder with the game on the out-side. Tape it together to form a complete cylinder as before: the top of theoriginal square piece of paper is now taped to the bottom. Now instead of givingup, we are solving the problem with origami.Origami has an interesting history in mathematics. In 1980, Hisashi Abesolved the angle trisection problem by using origami: the angle trisection takes7 steps, see for example Alperin (2005) and Fuchs (2011) for discussion, proofsand the axiomatic basis for this. It didn’t occur to the ancient Greeks tothink of a plane as a piece of paper that could be folded. Likewise, the C embeddings of Nash (1954) and Borrelli et al. (2013) by definition do not allowfolds, although it is only the intrinsic geometry of the embedded surface that4e are really interested in.Now take the horizontal cylinder you have just formed lying on a table, wherethe taping line is flush with the table, and squash the cylinder flat: crease itmaking two folds, opposite each other on the cylinder and equidistant from thetaping line. Let the square have side length of 4 units. One now has a flattened,open-ended envelope (4 units long and 2 units wide) with the game printed onthe top and bottom outside sides of this envelope. Flip it over so the tapededges are on the top side. Now pick up the two open ended ends (which are 2units wide) and 4 units apart and fold them inward: fold it in quarters so thatit makes 4 rectangles, each 2 units by 1 unit. Mate the two open-ended slits andtape them together to form a tall, square, open-ended box with 4 rectangularsides, each 2 units tall and 1 unit wide. This is the elongated box-like structurewe have just discussed above. See Figures 2 and 3 below. We have made theopen-ended boxlike surface out of our original flat square and we have tapedthe opposite sides of the square together as required to make the square torus.The spaceship shoots an asteroid with a ray gun.A literature search revealed that people have come close, but failed, to findthis solution earlier. The Zalgaller (2000) paper, “Some Bendings of a LongCylinder” came up with some polyhedron solutions for a long rectangular torus.These enclose volume. Basically, he makes a long triangular prism by origamifirst, makes origami fold crimps in it and ends up with an n-gonal toroidalpolyhedron, where the sum of face angles around each vertex is 360 ◦ . As partof his introduction, he gives a construction where he takes a long rectangle andfolds in in half and tapes it together to make a long, ironed sleeve. Then hetakes the open cuff ends and creates two folds in the sleeve so that he can bringthe two cuff ends back together at the center and tape them together in thecenter. Now he has a four-folded surface ironed flat in the plane. This is reallya long rectangular flat torus. It has a disadvantage that half of its map of thevideo game is buried on inside faces of the ironed flat surface of the plane sothat you can’t see them. He calls this a direct flat torus. He seems to not realizethat this construction could be used to make a flat square torus embedded in a2D Euclidean space as well, instead of the long one he is making.There is a wonderful YouTube video uploaded on June 8, 2015 by Segerman(2015) titled “Hinged Flat Torus” showing him folding a flat polyhedron to forma torus. It is a bending of a long cylinder as well. It has 10 rectangles around itsouter equatorial surface forming a decagonal prism with open top and bottom.Then a zig-zag configuration of 20 long triangles that connect top vertices withrotated vertices on the decagonal edge below. This encloses a volume. This iscalled a piecewise linear flat torus. Unfolded, it shows a row of 10 rectanglesalong the bottom and 20 long, tipped triangles making a parallelogram on thetop. This figure is flat and would tile the plane. There is no Gaussian curvatureat any vertex; the sum of angles around each vertex is 360 ◦ with three trianglesand two rectangles around each point. This is longer than it is wide and sodoes not solve the flat torus embedding. Guy Valette saw a torus like this atOberwolfach over 30 years ago. Thus, we don’t know who invented it.Had they allowed polygons to appear back to back with dihedral angles of 0 ◦ Asteroids game, top and bottomedges are identified as are left and right edges.6igure 3: Folding the square flat torus into a simple boxlike surface isometri-cally embedded in 3D space that is C .7s in Gott’s envelope polyhedra they could have used the 10 rectangles alone tohave formed an open-ended decagonal prism with rectangles meeting 4 around apoint (at the top and bottom decagonal edges). Make the rectangles tall enough(5:1) and one would have an envelope polyhedron that was an embedding of thesquare flat torus.We propose another origami construction which embeds the flat torus in 3Dspace. Start with a flat square 4 units on a side. Fold it in half, and tape thetwo sides together. Now bend this 4-unit long, 2-unit wide rectangular envelopeof zero thickness into a cylinder by taping the two open ends together. One nowhas an open-ended cylinder 2 units tall with a circumference of 2 πr = 4 units.The equatorial circumference is 4 units. The longitudinal circumference is also 4units, since starting at the bottom circular edge traveling upward on the outsideof the cylinder 2 units, crossing over the upper edge and traveling downward2 units one reaches the bottom edge where one started. All the longitudinaland equatorial circumferences are geodesics. The circular edges at the top andbottom are also geodesics, being straight lines in the original flat square. Takethe original square, tile the plane with images of it. Pick any two points A and B in the original square and connect them with a straight line connectingtheir closest pair of images in the tiling and this will be the geodesic connectingthem on the surface. Such a geodesic encountering the top or bottom edge willmake an equal angle with the edge on the other side. Unfolded, the line willbe straight. This is a surface that is the limit of an N -gonal prism withoutits top and bottom N -gons as N goes to infinity. The prisms have N externalrectangular faces, N internal rectangular faces, N external edges, N internaledges, 2 N top and bottom edges linking the internal and external edges, and2 N vertices. Thus, they have F − E + V = 2 N − N + 2 N = 0 = 2(1 − g ), so g = 1 and each is a torus.An interesting side note: starting with the same flat square torus 4 unitsby 4 units, one can make this out of 16 squares. These meet 4 around a point,like in a checkerboard. This can be embedded in 3D space as we have describedabove: 8 squares on the inside of a 2x1x1 hollow box, and 8 squares on theoutside. This is a cube stretched vertically by a factor of 2 missing its top andbottom. Dihedral angles at the top and bottom edges are 0 ◦ . But it is alsopossible to embed this in 4D Euclidean space as 16 square faces which are asubset of the square faces of a tesseract (or hyper-cube). The tesseract has asurface composed of 8 cubic cells. Unfold the tesseract in a cross like shapeas shown in the Salvador Dali 1954 painting Corpus Hypercubus. Four cubesform the vertical post of this cross. Consider the four sides of each of these fourcubes, they form a tube with cylindrical topology whose surface consists of 16squares. When this polyhedral net is folded back up into the tesseract, the topof the cross will meet the bottom, to make this a torus with 16 square faces.This is a two-sided surface. It divides the tesseract surface into two equivalentparts, like the Clifford Torus divides the 3-sphere into two equivalent parts. Onthis torus, the square faces meet 4 around a point.8 eferences Alperin R.C.
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