An Introduction to Topology the Classification Theorem for Surfaces by E
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Note on 6-Regular Graphs on the Klein Bottle Michiko Kasai [email protected]
Theory and Applications of Graphs Volume 4 | Issue 1 Article 5 2017 Note On 6-regular Graphs On The Klein Bottle Michiko Kasai [email protected] Naoki Matsumoto Seikei University, [email protected] Atsuhiro Nakamoto Yokohama National University, [email protected] Takayuki Nozawa [email protected] Hiroki Seno [email protected] See next page for additional authors Follow this and additional works at: https://digitalcommons.georgiasouthern.edu/tag Part of the Discrete Mathematics and Combinatorics Commons Recommended Citation Kasai, Michiko; Matsumoto, Naoki; Nakamoto, Atsuhiro; Nozawa, Takayuki; Seno, Hiroki; and Takiguchi, Yosuke (2017) "Note On 6-regular Graphs On The Klein Bottle," Theory and Applications of Graphs: Vol. 4 : Iss. 1 , Article 5. DOI: 10.20429/tag.2017.040105 Available at: https://digitalcommons.georgiasouthern.edu/tag/vol4/iss1/5 This article is brought to you for free and open access by the Journals at Digital Commons@Georgia Southern. It has been accepted for inclusion in Theory and Applications of Graphs by an authorized administrator of Digital Commons@Georgia Southern. For more information, please contact [email protected]. Note On 6-regular Graphs On The Klein Bottle Authors Michiko Kasai, Naoki Matsumoto, Atsuhiro Nakamoto, Takayuki Nozawa, Hiroki Seno, and Yosuke Takiguchi This article is available in Theory and Applications of Graphs: https://digitalcommons.georgiasouthern.edu/tag/vol4/iss1/5 Kasai et al.: 6-regular graphs on the Klein bottle Abstract Altshuler [1] classified 6-regular graphs on the torus, but Thomassen [11] and Negami [7] gave different classifications for 6-regular graphs on the Klein bottle. In this note, we unify those two classifications, pointing out their difference and similarity. -
Hydraulic Cylinder Replacement for Cylinders with 3/8” Hydraulic Hose Instructions
HYDRAULIC CYLINDER REPLACEMENT FOR CYLINDERS WITH 3/8” HYDRAULIC HOSE INSTRUCTIONS INSTALLATION TOOLS: INCLUDED HARDWARE: • Knife or Scizzors A. (1) Hydraulic Cylinder • 5/8” Wrench B. (1) Zip-Tie • 1/2” Wrench • 1/2” Socket & Ratchet IMPORTANT: Pressure MUST be relieved from hydraulic system before proceeding. PRESSURE RELIEF STEP 1: Lower the Power-Pole Anchor® down so the Everflex™ spike is touching the ground. WARNING: Do not touch spike with bare hands. STEP 2: Manually push the anchor into closed position, relieving pressure. Manually lower anchor back to the ground. REMOVAL STEP 1: Remove the Cylinder Bottom from the Upper U-Channel using a 1/2” socket & wrench. FIG 1 or 2 NOTE: For Blade models, push bottom of cylinder up into U-Channel and slide it forward past the Ram Spacers to remove it. Upper U-Channel Upper U-Channel Cylinder Bottom 1/2” Tools 1/2” Tools Ram 1/2” Tools Spacers Cylinder Bottom If Ram-Spacers fall out Older models have (1) long & (1) during removal, push short Ram Spacer. Ram Spacers 1/2” Tools bushings in to hold MUST be installed on the same side Figure 1 Figure 2 them in place. they were removed. Blade Models Pro/Spn Models Need help? Contact our Customer Service Team at 1 + 813.689.9932 Option 2 HYDRAULIC CYLINDER REPLACEMENT FOR CYLINDERS WITH 3/8” HYDRAULIC HOSE INSTRUCTIONS STEP 2: Remove the Cylinder Top from the Lower U-Channel using a 1/2” socket & wrench. FIG 3 or 4 Lower U-Channel Lower U-Channel 1/2” Tools 1/2” Tools Cylinder Top Cylinder Top Figure 3 Figure 4 Blade Models Pro/SPN Models STEP 3: Disconnect the UP Hose from the Hydraulic Cylinder Fitting by holding the 1/2” Wrench 5/8” Wrench Cylinder Fitting Base with a 1/2” wrench and turning the Hydraulic Hose Fitting Cylinder Fitting counter-clockwise with a 5/8” wrench. -
Chapter 11. Three Dimensional Analytic Geometry and Vectors
Chapter 11. Three dimensional analytic geometry and vectors. Section 11.5 Quadric surfaces. Curves in R2 : x2 y2 ellipse + =1 a2 b2 x2 y2 hyperbola − =1 a2 b2 parabola y = ax2 or x = by2 A quadric surface is the graph of a second degree equation in three variables. The most general such equation is Ax2 + By2 + Cz2 + Dxy + Exz + F yz + Gx + Hy + Iz + J =0, where A, B, C, ..., J are constants. By translation and rotation the equation can be brought into one of two standard forms Ax2 + By2 + Cz2 + J =0 or Ax2 + By2 + Iz =0 In order to sketch the graph of a quadric surface, it is useful to determine the curves of intersection of the surface with planes parallel to the coordinate planes. These curves are called traces of the surface. Ellipsoids The quadric surface with equation x2 y2 z2 + + =1 a2 b2 c2 is called an ellipsoid because all of its traces are ellipses. 2 1 x y 3 2 1 z ±1 ±2 ±3 ±1 ±2 The six intercepts of the ellipsoid are (±a, 0, 0), (0, ±b, 0), and (0, 0, ±c) and the ellipsoid lies in the box |x| ≤ a, |y| ≤ b, |z| ≤ c Since the ellipsoid involves only even powers of x, y, and z, the ellipsoid is symmetric with respect to each coordinate plane. Example 1. Find the traces of the surface 4x2 +9y2 + 36z2 = 36 1 in the planes x = k, y = k, and z = k. Identify the surface and sketch it. Hyperboloids Hyperboloid of one sheet. The quadric surface with equations x2 y2 z2 1. -
Surfaces and Fundamental Groups
HOMEWORK 5 — SURFACES AND FUNDAMENTAL GROUPS DANNY CALEGARI This homework is due Wednesday November 22nd at the start of class. Remember that the notation e1; e2; : : : ; en w1; w2; : : : ; wm h j i denotes the group whose generators are equivalence classes of words in the generators ei 1 and ei− , where two words are equivalent if one can be obtained from the other by inserting 1 1 or deleting eiei− or ei− ei or by inserting or deleting a word wj or its inverse. Problem 1. Let Sn be a surface obtained from a regular 4n–gon by identifying opposite sides by a translation. What is the Euler characteristic of Sn? What surface is it? Find a presentation for its fundamental group. Problem 2. Let S be the surface obtained from a hexagon by identifying opposite faces by translation. Then the fundamental group of S should be given by the generators a; b; c 1 1 1 corresponding to the three pairs of edges, with the relation abca− b− c− coming from the boundary of the polygonal region. What is wrong with this argument? Write down an actual presentation for π1(S; p). What surface is S? Problem 3. The four–color theorem of Appel and Haken says that any map in the plane can be colored with at most four distinct colors so that two regions which share a com- mon boundary segment have distinct colors. Find a cell decomposition of the torus into 7 polygons such that each two polygons share at least one side in common. Remark. -
Examples of Manifolds
Examples of Manifolds Example 1 (Open Subset of IRn) Any open subset, O, of IRn is a manifold of dimension n. One possible atlas is A = (O, ϕid) , where ϕid is the identity map. That is, ϕid(x) = x. n Of course one possible choice of O is IR itself. Example 2 (The Circle) The circle S1 = (x,y) ∈ IR2 x2 + y2 = 1 is a manifold of dimension one. One possible atlas is A = {(U , ϕ ), (U , ϕ )} where 1 1 1 2 1 y U1 = S \{(−1, 0)} ϕ1(x,y) = arctan x with − π < ϕ1(x,y) <π ϕ1 1 y U2 = S \{(1, 0)} ϕ2(x,y) = arctan x with 0 < ϕ2(x,y) < 2π U1 n n n+1 2 2 Example 3 (S ) The n–sphere S = x =(x1, ··· ,xn+1) ∈ IR x1 +···+xn+1 =1 n A U , ϕ , V ,ψ i n is a manifold of dimension . One possible atlas is 1 = ( i i) ( i i) 1 ≤ ≤ +1 where, for each 1 ≤ i ≤ n + 1, n Ui = (x1, ··· ,xn+1) ∈ S xi > 0 ϕi(x1, ··· ,xn+1)=(x1, ··· ,xi−1,xi+1, ··· ,xn+1) n Vi = (x1, ··· ,xn+1) ∈ S xi < 0 ψi(x1, ··· ,xn+1)=(x1, ··· ,xi−1,xi+1, ··· ,xn+1) n So both ϕi and ψi project onto IR , viewed as the hyperplane xi = 0. Another possible atlas is n n A2 = S \{(0, ··· , 0, 1)}, ϕ , S \{(0, ··· , 0, −1)},ψ where 2x1 2xn ϕ(x , ··· ,xn ) = , ··· , 1 +1 1−xn+1 1−xn+1 2x1 2xn ψ(x , ··· ,xn ) = , ··· , 1 +1 1+xn+1 1+xn+1 are the stereographic projections from the north and south poles, respectively. -
Simple Infinite Presentations for the Mapping Class Group of a Compact
SIMPLE INFINITE PRESENTATIONS FOR THE MAPPING CLASS GROUP OF A COMPACT NON-ORIENTABLE SURFACE RYOMA KOBAYASHI Abstract. Omori and the author [6] have given an infinite presentation for the mapping class group of a compact non-orientable surface. In this paper, we give more simple infinite presentations for this group. 1. Introduction For g ≥ 1 and n ≥ 0, we denote by Ng,n the closure of a surface obtained by removing disjoint n disks from a connected sum of g real projective planes, and call this surface a compact non-orientable surface of genus g with n boundary components. We can regard Ng,n as a surface obtained by attaching g M¨obius bands to g boundary components of a sphere with g + n boundary components, as shown in Figure 1. We call these attached M¨obius bands crosscaps. Figure 1. A model of a non-orientable surface Ng,n. The mapping class group M(Ng,n) of Ng,n is defined as the group consisting of isotopy classes of all diffeomorphisms of Ng,n which fix the boundary point- wise. M(N1,0) and M(N1,1) are trivial (see [2]). Finite presentations for M(N2,0), M(N2,1), M(N3,0) and M(N4,0) ware given by [9], [1], [14] and [16] respectively. Paris-Szepietowski [13] gave a finite presentation of M(Ng,n) with Dehn twists and arXiv:2009.02843v1 [math.GT] 7 Sep 2020 crosscap transpositions for g + n > 3 with n ≤ 1. Stukow [15] gave another finite presentation of M(Ng,n) with Dehn twists and one crosscap slide for g + n > 3 with n ≤ 1, applying Tietze transformations for the presentation of M(Ng,n) given in [13]. -
Volumes of Prisms and Cylinders 625
11-4 11-4 Volumes of Prisms and 11-4 Cylinders 1. Plan Objectives What You’ll Learn Check Skills You’ll Need GO for Help Lessons 1-9 and 10-1 1 To find the volume of a prism 2 To find the volume of • To find the volume of a Find the area of each figure. For answers that are not whole numbers, round to prism a cylinder the nearest tenth. • To find the volume of a 2 Examples cylinder 1. a square with side length 7 cm 49 cm 1 Finding Volume of a 2. a circle with diameter 15 in. 176.7 in.2 . And Why Rectangular Prism 3. a circle with radius 10 mm 314.2 mm2 2 Finding Volume of a To estimate the volume of a 4. a rectangle with length 3 ft and width 1 ft 3 ft2 Triangular Prism backpack, as in Example 4 2 3 Finding Volume of a Cylinder 5. a rectangle with base 14 in. and height 11 in. 154 in. 4 Finding Volume of a 6. a triangle with base 11 cm and height 5 cm 27.5 cm2 Composite Figure 7. an equilateral triangle that is 8 in. on each side 27.7 in.2 New Vocabulary • volume • composite space figure Math Background Integral calculus considers the area under a curve, which leads to computation of volumes of 1 Finding Volume of a Prism solids of revolution. Cavalieri’s Principle is a forerunner of ideas formalized by Newton and Leibniz in calculus. Hands-On Activity: Finding Volume Explore the volume of a prism with unit cubes. -
The Sphere Project
;OL :WOLYL 7YVQLJ[ +XPDQLWDULDQ&KDUWHUDQG0LQLPXP 6WDQGDUGVLQ+XPDQLWDULDQ5HVSRQVH ;OL:WOLYL7YVQLJ[ 7KHULJKWWROLIHZLWKGLJQLW\ ;OL:WOLYL7YVQLJ[PZHUPUP[PH[P]L[VKL[LYTPULHUK WYVTV[LZ[HUKHYKZI`^OPJO[OLNSVIHSJVTT\UP[` /\THUP[HYPHU*OHY[LY YLZWVUKZ[V[OLWSPNO[VMWLVWSLHMMLJ[LKI`KPZHZ[LYZ /\THUP[HYPHU >P[O[OPZ/HUKIVVR:WOLYLPZ^VYRPUNMVYH^VYSKPU^OPJO [OLYPNO[VMHSSWLVWSLHMMLJ[LKI`KPZHZ[LYZ[VYLLZ[HISPZO[OLPY *OHY[LYHUK SP]LZHUKSP]LSPOVVKZPZYLJVNUPZLKHUKHJ[LK\WVUPU^H`Z[OH[ YLZWLJ[[OLPY]VPJLHUKWYVTV[L[OLPYKPNUP[`HUKZLJ\YP[` 4PUPT\T:[HUKHYKZ This Handbook contains: PU/\THUP[HYPHU (/\THUP[HYPHU*OHY[LY!SLNHSHUKTVYHSWYPUJPWSLZ^OPJO HUK YLÅLJ[[OLYPNO[ZVMKPZHZ[LYHMMLJ[LKWVW\SH[PVUZ 4PUPT\T:[HUKHYKZPU/\THUP[HYPHU9LZWVUZL 9LZWVUZL 7YV[LJ[PVU7YPUJPWSLZ *VYL:[HUKHYKZHUKTPUPT\TZ[HUKHYKZPUMV\YRL`SPMLZH]PUN O\THUP[HYPHUZLJ[VYZ!>H[LYZ\WWS`ZHUP[H[PVUHUKO`NPLUL WYVTV[PVU"-VVKZLJ\YP[`HUKU\[YP[PVU":OLS[LYZL[[SLTLU[HUK UVUMVVKP[LTZ"/LHS[OHJ[PVU;OL`KLZJYPILwhat needs to be achieved in a humanitarian response in order for disaster- affected populations to survive and recover in stable conditions and with dignity. ;OL:WOLYL/HUKIVVRLUQV`ZIYVHKV^ULYZOPWI`HNLUJPLZHUK PUKP]PK\HSZVMMLYPUN[OLO\THUP[HYPHUZLJ[VYH common language for working together towards quality and accountability in disaster and conflict situations ;OL:WOLYL/HUKIVVROHZHU\TILYVMºJVTWHUPVU Z[HUKHYKZ»L_[LUKPUNP[ZZJVWLPUYLZWVUZL[VULLKZ[OH[ OH]LLTLYNLK^P[OPU[OLO\THUP[HYPHUZLJ[VY ;OL:WOLYL7YVQLJ[^HZPUP[PH[LKPU I`HU\TILY VMO\THUP[HYPHU5.6ZHUK[OL9LK*YVZZHUK 9LK*YLZJLU[4V]LTLU[ ,+0;065 7KHB6SKHUHB3URMHFWBFRYHUBHQBLQGG The Sphere Project Humanitarian Charter and Minimum Standards in Humanitarian Response Published by: The Sphere Project Copyright@The Sphere Project 2011 Email: [email protected] Website : www.sphereproject.org The Sphere Project was initiated in 1997 by a group of NGOs and the Red Cross and Red Crescent Movement to develop a set of universal minimum standards in core areas of humanitarian response: the Sphere Handbook. -
Equivelar Octahedron of Genus 3 in 3-Space
Equivelar octahedron of genus 3 in 3-space Ruslan Mizhaev ([email protected]) Apr. 2020 ABSTRACT . Building up a toroidal polyhedron of genus 3, consisting of 8 nine-sided faces, is given. From the point of view of topology, a polyhedron can be considered as an embedding of a cubic graph with 24 vertices and 36 edges in a surface of genus 3. This polyhedron is a contender for the maximal genus among octahedrons in 3-space. 1. Introduction This solution can be attributed to the problem of determining the maximal genus of polyhedra with the number of faces - . As is known, at least 7 faces are required for a polyhedron of genus = 1 . For cases ≥ 8 , there are currently few examples. If all faces of the toroidal polyhedron are – gons and all vertices are q-valence ( ≥ 3), such polyhedral are called either locally regular or simply equivelar [1]. The characteristics of polyhedra are abbreviated as , ; [1]. 2. Polyhedron {9, 3; 3} V1. The paper considers building up a polyhedron 9, 3; 3 in 3-space. The faces of a polyhedron are non- convex flat 9-gons without self-intersections. The polyhedron is symmetric when rotated through 1804 around the axis (Fig. 3). One of the features of this polyhedron is that any face has two pairs with which it borders two edges. The polyhedron also has a ratio of angles and faces - = − 1. To describe polyhedra with similar characteristics ( = − 1) we use the Euler formula − + = = 2 − 2, where is the Euler characteristic. Since = 3, the equality 3 = 2 holds true. -
Section 2.6 Cylindrical and Spherical Coordinates
Section 2.6 Cylindrical and Spherical Coordinates A) Review on the Polar Coordinates The polar coordinate system consists of the origin O,the rotating ray or half line from O with unit tick. A point P in the plane can be uniquely described by its distance to the origin r = dist (P, O) and the angle µ, 0 µ < 2¼ : · Y P(x,y) r θ O X We call (r, µ) the polar coordinate of P. Suppose that P has Cartesian (stan- dard rectangular) coordinate (x, y) .Then the relation between two coordinate systems is displayed through the following conversion formula: x = r cos µ Polar Coord. to Cartesian Coord.: y = r sin µ ½ r = x2 + y2 Cartesian Coord. to Polar Coord.: y tan µ = ( p x 0 µ < ¼ if y > 0, 2¼ µ < ¼ if y 0. · · · Note that function tan µ has period ¼, and the principal value for inverse tangent function is ¼ y ¼ < arctan < . ¡ 2 x 2 1 So the angle should be determined by y arctan , if x > 0 xy 8 arctan + ¼, if x < 0 µ = > ¼ x > > , if x = 0, y > 0 < 2 ¼ , if x = 0, y < 0 > ¡ 2 > > Example 6.1. Fin:>d (a) Cartesian Coord. of P whose Polar Coord. is ¼ 2, , and (b) Polar Coord. of Q whose Cartesian Coord. is ( 1, 1) . 3 ¡ ¡ ³ So´l. (a) ¼ x = 2 cos = 1, 3 ¼ y = 2 sin = p3. 3 (b) r = p1 + 1 = p2 1 ¼ ¼ 5¼ tan µ = ¡ = 1 = µ = or µ = + ¼ = . 1 ) 4 4 4 ¡ 5¼ Since ( 1, 1) is in the third quadrant, we choose µ = so ¡ ¡ 4 5¼ p2, is Polar Coord. -
Quick Reference Guide - Ansi Z80.1-2015
QUICK REFERENCE GUIDE - ANSI Z80.1-2015 1. Tolerance on Distance Refractive Power (Single Vision & Multifocal Lenses) Cylinder Cylinder Sphere Meridian Power Tolerance on Sphere Cylinder Meridian Power ≥ 0.00 D > - 2.00 D (minus cylinder convention) > -4.50 D (minus cylinder convention) ≤ -2.00 D ≤ -4.50 D From - 6.50 D to + 6.50 D ± 0.13 D ± 0.13 D ± 0.15 D ± 4% Stronger than ± 6.50 D ± 2% ± 0.13 D ± 0.15 D ± 4% 2. Tolerance on Distance Refractive Power (Progressive Addition Lenses) Cylinder Cylinder Sphere Meridian Power Tolerance on Sphere Cylinder Meridian Power ≥ 0.00 D > - 2.00 D (minus cylinder convention) > -3.50 D (minus cylinder convention) ≤ -2.00 D ≤ -3.50 D From -8.00 D to +8.00 D ± 0.16 D ± 0.16 D ± 0.18 D ± 5% Stronger than ±8.00 D ± 2% ± 0.16 D ± 0.18 D ± 5% 3. Tolerance on the direction of cylinder axis Nominal value of the ≥ 0.12 D > 0.25 D > 0.50 D > 0.75 D < 0.12 D > 1.50 D cylinder power (D) ≤ 0.25 D ≤ 0.50 D ≤ 0.75 D ≤ 1.50 D Tolerance of the axis Not Defined ° ° ° ° ° (degrees) ± 14 ± 7 ± 5 ± 3 ± 2 4. Tolerance on addition power for multifocal and progressive addition lenses Nominal value of addition power (D) ≤ 4.00 D > 4.00 D Nominal value of the tolerance on the addition power (D) ± 0.12 D ± 0.18 D 5. Tolerance on Prism Reference Point Location and Prismatic Power • The prismatic power measured at the prism reference point shall not exceed 0.33Δ or the prism reference point shall not be more than 1.0 mm away from its specified position in any direction. -
Surface Physics I and II
Surface physics I and II Lectures: Mon 12-14 ,Wed 12-14 D116 Excercises: TBA Lecturer: Antti Kuronen, [email protected] Exercise assistant: Ane Lasa, [email protected] Course homepage: http://www.physics.helsinki.fi/courses/s/pintafysiikka/ Objectives ● To study properties of surfaces of solid materials. ● The relationship between the composition and morphology of the surface and its mechanical, chemical and electronic properties will be dealt with. ● Technologically important field of surface and thin film growth will also be covered. Surface physics I 2012: 1. Introduction 1 Surface physics I and II ● Course in two parts ● Surface physics I (SPI) (530202) ● Period III, 5 ECTS points ● Basics of surface physics ● Surface physics II (SPII) (530169) ● Period IV, 5 ECTS points ● 'Special' topics in surface science ● Surface and thin film growth ● Nanosystems ● Computational methods in surface science ● You can take only SPI or both SPI and SPII Surface physics I 2012: 1. Introduction 2 How to pass ● Both courses: ● Final exam 50% ● Exercises 50% ● Exercises ● Return by ● email to [email protected] or ● on paper to course box on the 2nd floor of Physicum ● Return by (TBA) Surface physics I 2012: 1. Introduction 3 Table of contents ● Surface physics I ● Introduction: What is a surface? Why is it important? Basic concepts. ● Surface structure: Thermodynamics of surfaces. Atomic and electronic structure. ● Experimental methods for surface characterization: Composition, morphology, electronic properties. ● Surface physics II ● Theoretical and computational methods in surface science: Analytical models, Monte Carlo and molecular dynamics sumilations. ● Surface growth: Adsorption, desorption, surface diffusion. ● Thin film growth: Homoepitaxy, heteroepitaxy, nanostructures.