HOMEWORK 3 - GEOMETRY OF CURVES AND SURFACES. where ν is the unit normal consistent with the orientation of α (right hand rule).

Size: px
Start display at page:

Download "HOMEWORK 3 - GEOMETRY OF CURVES AND SURFACES. where ν is the unit normal consistent with the orientation of α (right hand rule)."

Transcription

1 HOMEWORK 3 - GEOMETRY OF CURVES AND SURFACES ANDRÉ NEVES ) If α : I R 2 is a curve on the plane parametrized by arc length and θ is the angle that α makes with the x-axis, show that α t) = dθ dt ν, where ν is the unit normal consistent with the orientation of α right hand rule). The bridge with the framework of what I have done in class is that we can see R 2 as the plane P = {z = 0} in R 3 and so α is a curve on P where Dα k = = α ). dt But N = 0, 0, ) is the unit normal vector to P and α = x, y, 0), which implies α ) = α. In other words k is just α for a curve on the plane. If E =, 0, 0) and E 2 = 0,, 0), then it is simple to see that DE = 0 dt E.α = cos θ, ν.e = α.e 2 = sin θ. Thus sin θ dθ dt = d cos θ = d dt dt α.e = α.e +α. DE = α.ν)ν.e = sin θα.ν) dt and so dθ dt = α.ν. Chapter 4.2, Exercise 8: The first step is to show that G is a linear map. Fix a point p and consider α t) = p + tx, where X is any vector in R 3. Set F t) = G α t) Gq). By differentiation in the t variable when t = 0 we have F t) 2 = tx + p q 2 = F 0).F 0) = X.p q) = DG p X).Gp) Gq)) = X.p q) = X.DG p ) Gp) Gq)) = X.p q)

2 2 ANDRÉ NEVES for all X and all q. In the last line we use the fact that if A is a 3 3 matrix with transpose A T, then AX).Y = X.A T Y ) for every vectors X, Y. Because the identity we derived is valid for all X it is simple to conclude that DG p ) Gp) Gq)) = p q for every q R 3. In particular the linear map DG p ) T is surjective and thus injective. Denoting by B its inverse we have from the above formula that Gp) = Gq) + Bp q)for every p, q R 3. Using again the hypothesis we obtain that Bp q) = Gp) Gq) = p q. Using Exercise 7 of Chapter 4.2 more precisely, that b) is equivalent to a) which is simple to show) it follows at once that G is a linear isometry of R 3. Chapter 4.2, Exercise : First a). From the previous exercise we know that F is a linear isometry of R 3, which means that DF p = DF 0 for all p in R 3 i.e. F is linear) and DF 0 X).DF 0 Y ) = X.Y for all vectors X, Y in R 3. In particular for every p S we have that DF p = DF 0 and DF p X).DF p Y ) = X.Y for all X, Y in T p S, which means that F is an isometry of S. Technically speaking one should show that if F is an ambient map of R 3, then the restriction of F to S has the property that for every X T p S, then DF p X) as it was defined in class is nothing but DF p X) as it was defined in multivariable calculus. This is just the chain rule and so I will not say anything else to avoid the risk of just ending up making it more confusing. Now b). The orthogonal linear transformations are just those 3 3 matrices which have A T A = Id, i.e., A = A T. In this case we have AX) 2 = AX).Ax) = X.A T AX) = X.X = X 2 and so A send the unit sphere into the unit sphere. Moreover AX) AY ) = AX Y ) = X Y, and so its distance preserving in the sense of Exercise 8. Thus a) can be applied to conclude A is an isometry of the sphere. Finally c). Take F x, y, 0) = cos x, sin x, y) which is an isometry from part of the plane {z = 0} into part of the cylinder {x 2 + y 2 = }. F is not distance preserving because F 0, 0, 0) F θ, 0, 0) = 2 cos θ) and 0, 0, 0) θ, 0, 0) = θ. Chapter 4.2, Exercise 2: Consider F x, y, z) = x, y, z). This is an isometry of the cylinder S = {x 2 + y 2 = } because F send the S into S and F is an orthogonal linear transformation. Finally F x, y, z) = x, y, z) then z = 0 and y = 0.

3 HOMEWORK 3 - GEOMETRY OF CURVES AND SURFACES 3 But x 2 + y 2 =, which means x = ±, i.e., the only fixed points are, 0, 0) and, 0, 0). Chapter 4.2, Exercise 5: To be in accordance with the stuff we have been learning in class, the slight modified set up should be P = {z = 0} and F is a map from P into P which is given by To see that is conformal we have F x, y, 0) = ux, y), vx, y), 0). F x. F y = xu, x v, 0). y u, y v, 0) = x u y u+ x v y v = x v y v+ x v y v = 0 F x. F x = xu) 2 + x v) 2 F y. F y = yu) 2 + y v) 2 = x v) 2 + x u) 2 = F x. F x. Thus E = G and F = 0, which means the map is conformal. Chapter 4.3, Exercise : It becomes slightly simpler if we use E = e 2u and G = e 2v. Note that my u, v are different from the u, v in do Carmo. The do Carmo u, v correspond to my x, x 2. In this case, using the formula 2) in page 232 of Do Carmo we have Γ = x u, Γ 2 = e 2u v) x2 u, Γ 2 = x2 u, Γ 2 2 = x v, Γ 22 = e 2v u) x u, Γ 2 22 = x2 v. Using formula 5) in page 234 of Do Carmo we have K = E x Γ 2 2 x2 Γ 2 + Γ 2Γ 2 + Γ 2 2Γ 2 2 Γ 2 Γ 2 22 Γ Γ 2 ) 2 = e 2u x 2 x v + x2 e 2u v) x2 u) e 2u v) x2 u) 2 + x v) 2 ) +e 2u v) x2 u x2 v x u x v ) = e 2u x 2 x v + e 2u v) x2 u v) x2 u + e 2u v) x 2 2 x 2 u + x v) 2 x u x v ) = e 2u x 2 x v + e 2u v) x2 u v) x2 u + e 2u v) x 2 2 x 2 u + x v u) x v = e 2u x 2 x v e 2v x2 u v) x2 u e 2v x 2 2 x 2 u e 2u x v u) x v.

4 4 ANDRÉ NEVES On the other hand 2EG) /2 x2 ) x2 E EG) /2 + = e u+v) x2 x x2 u e v u) )) x G EG) ) /2 + x x v e u v) ) = e u+v) e u v) x 2 2 x 2 u + e u v) x2 u v) x2 u + e v u) x 2 x v + e v u) x v u) x v = e 2v 2 x 2 x 2 u e 2v x2 u v) x2 u e 2u 2 x x v e 2u x v u) x v. Thus, from the previous exercise, K = 2EG) /2 Chapter 4.3, Exercise 2: x2 x2 E EG) /2 )) ) + )) x G x EG) /2. We follow the previous exercise and write λ = e 2u, i.e.,define u = ln λ. Note that this u is not the u of do Carmo.). Then we have EG = e 2u, Thus x2 E EG) /2 = 2 x 2 u, x G EG) /2 = 2 x u. K = e 2u u = ln λ. 2λ Recall that ln λ = 2 ln λ). To be consistent with the rest of the notation I should use λ = x 2 + x2 2 + c) 2 for the last part. Then and Thus and so xi ln λ = 2 xi lnx 2 + x 2 4x i 2 + c) = x 2 + x2 2 + c x 2 4 i x i ln λ = x 2 + x2 2 + c + 8x 2 i x 2 + x c)2 8 ln λ = x 2 + x2 2 + c + 8x2 + x2 2 ) x 2 + x2 2 + = 8c c)2 x 2 + = 8cλ x2 2 + c)2 K = ln λ = 4c. 2λ Chapter 4.3, Exercise 2: We have x u = cos v, sin v, u ), x = u sin v, u cos v, 0) v

5 and so HOMEWORK 3 - GEOMETRY OF CURVES AND SURFACES 5 E = x u. x u = + u 2, F = x u. x v Because v E = 0 we obtain from Exercise 2u + u 2 We have and so K = 2 + u 2 u = 0, G = x v. x v = u2. ) = + u 2 ) 2 x x = cos v, sin v, 0), = u sin v, u cos v, 0) u v Ē = x. x u u =, F = x. x x = 0, G =. x u v v v = + u2. Using the formula in Exercise we have ) K = 2 2u = + u 2 u + u 2 + u 2 ) 2. Therefore the surfaces have the same curvature. If the map x x were an isometry then by Proposition, page 220, we would have E = Ē, F = F, and G = Ḡ. Because this is false the surfaces are not isometric. Chapter 4.3, Exercise 6: Let s ignore the part about using Bonnet s Theorem because I did not talk about that. If there were a surface with E = G =, F = 0 and e = g =, f = 0, then from the definition of Gaussian curvature we would have K = eg f)eg F ) =. But from Gauss s Theorem we know that K is intrinsic, i.e., depends only on E, F and G. Thus K must be zero because we can surely put coordinates on a plane which have E = G =, F = 0 and a plane has Gaussian curvature zero. Chapter 4.4, Exercise 5 a) and 6: Let s do thing s a bit more generally here because it will be useful. Consider a parametrization of S Then φs, θ) = rs) cos θ, rs) sin θ, zs)). φ s = r cos θ, r sin θ, z ), 2 φ s) 2 = r cos θ, r sin θ, z ), 2 φ θ s = r sin θ, r cos θ, 0). φ = r sin θ, r cos θ, 0), θ 2 φ = r cos θ, r sin θ, 0), θ) 2

6 6 ANDRÉ NEVES and so E = r ) 2 + z ) 2 =, F = 0, G = r 2, where I am assuming, without loss of generality, that rs), zs)) is parametrized by arc-length. If αt) = φst), θt)) is a curve parametrized by arc-length then, denoting differentiation with respect to t with a dot, we have α = d dt ṡ sφ + θ θ φ) = s s φ + θ θ φ + ṡ) 2 2 ssφ + 2ṡ θ 2 sθ φ + θ) 2 2 θθ φ. Consider the orthonormal basis X, X 2 for T S where Then X = φ s, X φ 2 = r = r sin θ, r cos θ, 0). θ α.x = s + ṡ) 2 r r + z z ) θ) 2 rr = s θ) 2 rr, where we use the fact that and Therefore r ) 2 + z ) 2 = = r r + z z = 0, α.x 2 = θ + 2ṡ θrr. D α k = dt = α)t = α.x )X + α.x 2 )X 2 = s θ) 2 rr ) φ s + θ + 2ṡ θrr φ r θ. For the case in point we have rs) = a + r 0 coss/r 0 ), zs) = r 0 sins/r 0 ) so that r ) 2 +z ) 2 = ). Note that my r 0 is the parameter r in Do Carmo s notation for this exercise. The curves are maximum parallel: α t) = φ0, t/r 0 ), minimum parallel: α 2 t) = φr 0 π, t/r 0 ), upper parallel: α 3 t) = φr 0 π/2, t/r 0 ) I use the term t/r 0 instead of t because I want the curves to be parametrized by arc-length. In any case we have θ = r0, ṡ = θ = 0, and r = sins/r 0 ). Using the formula above for k we get kα ) = r 2 0 r0)r 0) φ s = 0 kα ) = r0 2 rr 0π)r r 0 π) φ s = 0 and kα3 ) = r0 2 rr 0π/2)r r 0 π/2) φ s = a φ r0 2 s. Note that kα 3 ) compute the geodesic curvature of the upper parallel in Exercise 6. Chapter 4.4, Exercise 7:

7 HOMEWORK 3 - GEOMETRY OF CURVES AND SURFACES 7 We use the coordinate chart φstd, z) = cos s, sin s, z) which makes it into a local isometry with Euclidean plane, i.e., E = = G, F = 0. The plane which intersects the cylinder is P = {z = y tan θ} and so the curve C is parametrized by Ct) = cos t, sin t, sin t tan θ) = φt, sin t tan θ). If e =, 0, 0), e 2 = 0, cos θ, sin θ) is an orthonormal basis for P, then C = {xe + ye 2 x 2 + cos 2 θy 2 = } and so we see that C is indeed an ellipse in the plane P. To compute its curvature vector we have the little annoying thing that Ct) is not a parametrization by arc-length. Suppose then that t = tu) is a change of variable which makes Cu) parametrized by arc length. Then, if ν denotes a unit vector lying in the tangent plane to to the cylinder which is orthogonal to C t), we know that k has the direction of ν and so k = k.ν)ν. Now and thus k.ν = d du C t) dt ).ν = du = C t) 2 C t).ν. dc du = C t) dt du = dt du = C t) ) dt 2 C t).ν + d2 t du du) 2 C t).ν = ) dt 2 C t).ν du Let s use this formula. C intersects the axis when t = 0, t = π. In this case C t) = Ct) and so C 0) =, 0, 0), C π) =, 0, 0). In both cases C has the direction of the normal vector to the cylinder and so the term C t).ν is zero because ν lies in the tangent plane of the cylinder. Thus k = 0 at those points. Chapter 4.4, Exercise 7: There is some terminology that needs further explanation. Essentially what the hypothesis of the exercise is saying is that there is F : U R 2 : S so that i) the curves αt) = F t, s 0 ) or αs) = F t 0, s) are geodesics and ii) t F t 0, s 0 ). s F t 0, s 0 ) = 0 for every s 0, t 0 fixed. One obvious consequence of i) is that s F and t F are never zero. Hence from ii) we get that s F t, s), t F t, s) are linearly independent which means that for every t 0, s 0 ) U, there is a small neighborhood V so that F restricted to V is a chart. We will show that for this chart we have E and G constant and F = 0. This implies K must be zero for the following reason: If E = a 2 and G = b 2, then we consider the map φx, y) = ax, by, 0) from R 2 into P = {z = 0}. For this map we have E = a 2, G = b 2 and F = 0 as well. Thus S is locally isometric to P and so the intrinsic invariance of K implies that the Gaussian

8 8 ANDRÉ NEVES curvature of S must be the same as the Gaussian curvature of P which is zero. That F = s F. t F is zero is essentially condition ii). We show that E is constant. t E = t t F. t )F = 2 2 ttf. t F = 2 D tf t. t F = 0 because t F t, s) is a geodesic for every s fixed. Likewise s E = s t F. t F ) = 2 2 stf. t F = 2 t s F. t F ) 2 s F. 2 ttf = 2 D tf t. s F = 0. Thus E is constant. The same reasoning shows that G is constant which is what we wanted to show.

HOMEWORK 2 - RIEMANNIAN GEOMETRY. 1. Problems In what follows (M, g) will always denote a Riemannian manifold with a Levi-Civita connection.

HOMEWORK 2 - RIEMANNIAN GEOMETRY. 1. Problems In what follows (M, g) will always denote a Riemannian manifold with a Levi-Civita connection. HOMEWORK 2 - RIEMANNIAN GEOMETRY ANDRÉ NEVES 1. Problems In what follows (M, g will always denote a Riemannian manifold with a Levi-Civita connection. 1 Let X, Y, Z be vector fields on M so that X(p Z(p

More information

HOMEWORK 2 - SOLUTIONS

HOMEWORK 2 - SOLUTIONS HOMEWORK 2 - SOLUTIONS - 2012 ANDRÉ NEVES Exercise 15 of Chapter 2.3 of Do Carmo s book: Okay, I have no idea why I set this one because it is similar to another one from the previous homework. I might

More information

1 The Differential Geometry of Surfaces

1 The Differential Geometry of Surfaces 1 The Differential Geometry of Surfaces Three-dimensional objects are bounded by surfaces. This section reviews some of the basic definitions and concepts relating to the geometry of smooth surfaces. 1.1

More information

9.1 Mean and Gaussian Curvatures of Surfaces

9.1 Mean and Gaussian Curvatures of Surfaces Chapter 9 Gauss Map II 9.1 Mean and Gaussian Curvatures of Surfaces in R 3 We ll assume that the curves are in R 3 unless otherwise noted. We start off by quoting the following useful theorem about self

More information

Pullbacks, Isometries & Conformal Maps

Pullbacks, Isometries & Conformal Maps Pullbacks, Isometries & Conformal Maps Outline 1. Pullbacks Let V and W be vector spaces, and let T : V W be an injective linear transformation. Given an inner product, on W, the pullback of, is the inner

More information

SELECTED SAMPLE FINAL EXAM SOLUTIONS - MATH 5378, SPRING 2013

SELECTED SAMPLE FINAL EXAM SOLUTIONS - MATH 5378, SPRING 2013 SELECTED SAMPLE FINAL EXAM SOLUTIONS - MATH 5378, SPRING 03 Problem (). This problem is perhaps too hard for an actual exam, but very instructional, and simpler problems using these ideas will be on the

More information

Euler Characteristic of Two-Dimensional Manifolds

Euler Characteristic of Two-Dimensional Manifolds Euler Characteristic of Two-Dimensional Manifolds M. Hafiz Khusyairi August 2008 In this work we will discuss an important notion from topology, namely Euler Characteristic and we will discuss several

More information

HOMEWORK 4 - GEOMETRY OF CURVES AND SURFACES

HOMEWORK 4 - GEOMETRY OF CURVES AND SURFACES HOMEWORK 4 - GEOMETRY OF CURVES ND SURFCES NDRÉ NEVES DISCLIMER: This homework was very tough and it involves being comfortable with facts you might have seen in other classes but are not so fresh. The

More information

Math 5378, Differential Geometry Solutions to practice questions for Test 2

Math 5378, Differential Geometry Solutions to practice questions for Test 2 Math 5378, Differential Geometry Solutions to practice questions for Test 2. Find all possible trajectories of the vector field w(x, y) = ( y, x) on 2. Solution: A trajectory would be a curve (x(t), y(t))

More information

MATH 31BH Homework 5 Solutions

MATH 31BH Homework 5 Solutions MATH 3BH Homework 5 Solutions February 4, 204 Problem.8.2 (a) Let x t f y = x 2 + y 2 + 2z 2 and g(t) = t 2. z t 3 Then by the chain rule a a a D(g f) b = Dg f b Df b c c c = [Dg(a 2 + b 2 + 2c 2 )] [

More information

CHAPTER 3. Gauss map. In this chapter we will study the Gauss map of surfaces in R 3.

CHAPTER 3. Gauss map. In this chapter we will study the Gauss map of surfaces in R 3. CHAPTER 3 Gauss map In this chapter we will study the Gauss map of surfaces in R 3. 3.1. Surfaces in R 3 Let S R 3 be a submanifold of dimension 2. Let {U i, ϕ i } be a DS on S. For any p U i we have a

More information

Lecture 13 The Fundamental Forms of a Surface

Lecture 13 The Fundamental Forms of a Surface Lecture 13 The Fundamental Forms of a Surface In the following we denote by F : O R 3 a parametric surface in R 3, F(u, v) = (x(u, v), y(u, v), z(u, v)). We denote partial derivatives with respect to the

More information

Chapter 3. Riemannian Manifolds - I. The subject of this thesis is to extend the combinatorial curve reconstruction approach to curves

Chapter 3. Riemannian Manifolds - I. The subject of this thesis is to extend the combinatorial curve reconstruction approach to curves Chapter 3 Riemannian Manifolds - I The subject of this thesis is to extend the combinatorial curve reconstruction approach to curves embedded in Riemannian manifolds. A Riemannian manifold is an abstraction

More information

Part IB Geometry. Theorems. Based on lectures by A. G. Kovalev Notes taken by Dexter Chua. Lent 2016

Part IB Geometry. Theorems. Based on lectures by A. G. Kovalev Notes taken by Dexter Chua. Lent 2016 Part IB Geometry Theorems Based on lectures by A. G. Kovalev Notes taken by Dexter Chua Lent 2016 These notes are not endorsed by the lecturers, and I have modified them (often significantly) after lectures.

More information

Lectures in Discrete Differential Geometry 2 Surfaces

Lectures in Discrete Differential Geometry 2 Surfaces Lectures in Discrete Differential Geometry 2 Surfaces Etienne Vouga February 4, 24 Smooth Surfaces in R 3 In this section we will review some properties of smooth surfaces R 3. We will assume that is parameterized

More information

Math 426H (Differential Geometry) Final Exam April 24, 2006.

Math 426H (Differential Geometry) Final Exam April 24, 2006. Math 426H Differential Geometry Final Exam April 24, 6. 8 8 8 6 1. Let M be a surface and let : [0, 1] M be a smooth loop. Let φ be a 1-form on M. a Suppose φ is exact i.e. φ = df for some f : M R. Show

More information

DIFFERENTIAL GEOMETRY HW 4. Show that a catenoid and helicoid are locally isometric.

DIFFERENTIAL GEOMETRY HW 4. Show that a catenoid and helicoid are locally isometric. DIFFERENTIAL GEOMETRY HW 4 CLAY SHONKWILER Show that a catenoid and helicoid are locally isometric. 3 Proof. Let X(u, v) = (a cosh v cos u, a cosh v sin u, av) be the parametrization of the catenoid and

More information

Surfaces JWR. February 13, 2014

Surfaces JWR. February 13, 2014 Surfaces JWR February 13, 214 These notes summarize the key points in the second chapter of Differential Geometry of Curves and Surfaces by Manfredo P. do Carmo. I wrote them to assure that the terminology

More information

Differential Geometry III, Solutions 6 (Week 6)

Differential Geometry III, Solutions 6 (Week 6) Durham University Pavel Tumarkin Michaelmas 016 Differential Geometry III, Solutions 6 Week 6 Surfaces - 1 6.1. Let U R be an open set. Show that the set {x, y, z R 3 z = 0 and x, y U} is a regular surface.

More information

Riemannian geometry of surfaces

Riemannian geometry of surfaces Riemannian geometry of surfaces In this note, we will learn how to make sense of the concepts of differential geometry on a surface M, which is not necessarily situated in R 3. This intrinsic approach

More information

Chapter 14. Basics of The Differential Geometry of Surfaces. Introduction. Parameterized Surfaces. The First... Home Page. Title Page.

Chapter 14. Basics of The Differential Geometry of Surfaces. Introduction. Parameterized Surfaces. The First... Home Page. Title Page. Chapter 14 Basics of The Differential Geometry of Surfaces Page 649 of 681 14.1. Almost all of the material presented in this chapter is based on lectures given by Eugenio Calabi in an upper undergraduate

More information

Chapter 5. The Second Fundamental Form

Chapter 5. The Second Fundamental Form Chapter 5. The Second Fundamental Form Directional Derivatives in IR 3. Let f : U IR 3 IR be a smooth function defined on an open subset of IR 3. Fix p U and X T p IR 3. The directional derivative of f

More information

Solutions to old Exam 3 problems

Solutions to old Exam 3 problems Solutions to old Exam 3 problems Hi students! I am putting this version of my review for the Final exam review here on the web site, place and time to be announced. Enjoy!! Best, Bill Meeks PS. There are

More information

d F = (df E 3 ) E 3. (4.1)

d F = (df E 3 ) E 3. (4.1) 4. The Second Fundamental Form In the last section we developed the theory of intrinsic geometry of surfaces by considering the covariant differential d F, that is, the tangential component of df for a

More information

A PROOF OF THE GAUSS-BONNET THEOREM. Contents. 1. Introduction. 2. Regular Surfaces

A PROOF OF THE GAUSS-BONNET THEOREM. Contents. 1. Introduction. 2. Regular Surfaces A PROOF OF THE GAUSS-BONNET THEOREM AARON HALPER Abstract. In this paper I will provide a proof of the Gauss-Bonnet Theorem. I will start by briefly explaining regular surfaces and move on to the first

More information

Contents. MATH 32B-2 (18W) (L) G. Liu / (TA) A. Zhou Calculus of Several Variables. 1 Multiple Integrals 3. 2 Vector Fields 9

Contents. MATH 32B-2 (18W) (L) G. Liu / (TA) A. Zhou Calculus of Several Variables. 1 Multiple Integrals 3. 2 Vector Fields 9 MATH 32B-2 (8W) (L) G. Liu / (TA) A. Zhou Calculus of Several Variables Contents Multiple Integrals 3 2 Vector Fields 9 3 Line and Surface Integrals 5 4 The Classical Integral Theorems 9 MATH 32B-2 (8W)

More information

INTRODUCTION TO GEOMETRY

INTRODUCTION TO GEOMETRY INTRODUCTION TO GEOMETRY ERIKA DUNN-WEISS Abstract. This paper is an introduction to Riemannian and semi-riemannian manifolds of constant sectional curvature. We will introduce the concepts of moving frames,

More information

Stereographic projection and inverse geometry

Stereographic projection and inverse geometry Stereographic projection and inverse geometry The conformal property of stereographic projections can be established fairly efficiently using the concepts and methods of inverse geometry. This topic is

More information

Chapter 16. Manifolds and Geodesics Manifold Theory. Reading: Osserman [7] Pg , 55, 63-65, Do Carmo [2] Pg ,

Chapter 16. Manifolds and Geodesics Manifold Theory. Reading: Osserman [7] Pg , 55, 63-65, Do Carmo [2] Pg , Chapter 16 Manifolds and Geodesics Reading: Osserman [7] Pg. 43-52, 55, 63-65, Do Carmo [2] Pg. 238-247, 325-335. 16.1 Manifold Theory Let us recall the definition of differentiable manifolds Definition

More information

Gauss Theorem Egregium, Gauss-Bonnet etc. We know that for a simple closed curve in the plane. kds = 2π.

Gauss Theorem Egregium, Gauss-Bonnet etc. We know that for a simple closed curve in the plane. kds = 2π. Gauss Theorem Egregium, Gauss-Bonnet etc. We know that for a simple closed curve in the plane kds = 2π. Now we want to consider a simple closed curve C in a surface S R 3. We suppose C is the boundary

More information

7.1 Tangent Planes; Differentials of Maps Between

7.1 Tangent Planes; Differentials of Maps Between Chapter 7 Tangent Planes Reading: Do Carmo sections 2.4 and 3.2 Today I am discussing 1. Differentials of maps between surfaces 2. Geometry of Gauss map 7.1 Tangent Planes; Differentials of Maps Between

More information

Dr. Allen Back. Sep. 10, 2014

Dr. Allen Back. Sep. 10, 2014 Dr. Allen Back Sep. 10, 2014 The chain rule in multivariable calculus is in some ways very simple. But it can lead to extremely intricate sorts of relationships (try thermodynamics in physical chemistry...

More information

3 Applications of partial differentiation

3 Applications of partial differentiation Advanced Calculus Chapter 3 Applications of partial differentiation 37 3 Applications of partial differentiation 3.1 Stationary points Higher derivatives Let U R 2 and f : U R. The partial derivatives

More information

Hyperbolic Geometry on Geometric Surfaces

Hyperbolic Geometry on Geometric Surfaces Mathematics Seminar, 15 September 2010 Outline Introduction Hyperbolic geometry Abstract surfaces The hemisphere model as a geometric surface The Poincaré disk model as a geometric surface Conclusion Introduction

More information

HILBERT S THEOREM ON THE HYPERBOLIC PLANE

HILBERT S THEOREM ON THE HYPERBOLIC PLANE HILBET S THEOEM ON THE HYPEBOLIC PLANE MATTHEW D. BOWN Abstract. We recount a proof of Hilbert s result that a complete geometric surface of constant negative Gaussian curvature cannot be isometrically

More information

CALCULUS ON MANIFOLDS. 1. Riemannian manifolds Recall that for any smooth manifold M, dim M = n, the union T M =

CALCULUS ON MANIFOLDS. 1. Riemannian manifolds Recall that for any smooth manifold M, dim M = n, the union T M = CALCULUS ON MANIFOLDS 1. Riemannian manifolds Recall that for any smooth manifold M, dim M = n, the union T M = a M T am, called the tangent bundle, is itself a smooth manifold, dim T M = 2n. Example 1.

More information

MATH Final Review

MATH Final Review MATH 1592 - Final Review 1 Chapter 7 1.1 Main Topics 1. Integration techniques: Fitting integrands to basic rules on page 485. Integration by parts, Theorem 7.1 on page 488. Guidelines for trigonometric

More information

Arc Length and Surface Area in Parametric Equations

Arc Length and Surface Area in Parametric Equations Arc Length and Surface Area in Parametric Equations MATH 211, Calculus II J. Robert Buchanan Department of Mathematics Spring 2011 Background We have developed definite integral formulas for arc length

More information

Vectors, dot product, and cross product

Vectors, dot product, and cross product MTH 201 Multivariable calculus and differential equations Practice problems Vectors, dot product, and cross product 1. Find the component form and length of vector P Q with the following initial point

More information

Introduction to Geometry

Introduction to Geometry Introduction to Geometry it is a draft of lecture notes of H.M. Khudaverdian. Manchester, 18 May 211 Contents 1 Euclidean space 3 1.1 Vector space............................ 3 1.2 Basic example of n-dimensional

More information

j=1 ωj k E j. (3.1) j=1 θj E j, (3.2)

j=1 ωj k E j. (3.1) j=1 θj E j, (3.2) 3. Cartan s Structural Equations and the Curvature Form Let E,..., E n be a moving (orthonormal) frame in R n and let ωj k its associated connection forms so that: de k = n ωj k E j. (3.) Recall that ωj

More information

ν(u, v) = N(u, v) G(r(u, v)) E r(u,v) 3.

ν(u, v) = N(u, v) G(r(u, v)) E r(u,v) 3. 5. The Gauss Curvature Beyond doubt, the notion of Gauss curvature is of paramount importance in differential geometry. Recall two lessons we have learned so far about this notion: first, the presence

More information

A local characterization for constant curvature metrics in 2-dimensional Lorentz manifolds

A local characterization for constant curvature metrics in 2-dimensional Lorentz manifolds A local characterization for constant curvature metrics in -dimensional Lorentz manifolds Ivo Terek Couto Alexandre Lymberopoulos August 9, 8 arxiv:65.7573v [math.dg] 4 May 6 Abstract In this paper we

More information

1. Geometry of the unit tangent bundle

1. Geometry of the unit tangent bundle 1 1. Geometry of the unit tangent bundle The main reference for this section is [8]. In the following, we consider (M, g) an n-dimensional smooth manifold endowed with a Riemannian metric g. 1.1. Notations

More information

Exercises for Multivariable Differential Calculus XM521

Exercises for Multivariable Differential Calculus XM521 This document lists all the exercises for XM521. The Type I (True/False) exercises will be given, and should be answered, online immediately following each lecture. The Type III exercises are to be done

More information

Mathematics 2203, Test 1 - Solutions

Mathematics 2203, Test 1 - Solutions Mathematics 220, Test 1 - Solutions F, 2010 Philippe B. Laval Name 1. Determine if each statement below is True or False. If it is true, explain why (cite theorem, rule, property). If it is false, explain

More information

Chapter 5: Integrals

Chapter 5: Integrals Chapter 5: Integrals Section 5.3 The Fundamental Theorem of Calculus Sec. 5.3: The Fundamental Theorem of Calculus Fundamental Theorem of Calculus: Sec. 5.3: The Fundamental Theorem of Calculus Fundamental

More information

Directional Derivatives in the Plane

Directional Derivatives in the Plane Directional Derivatives in the Plane P. Sam Johnson April 10, 2017 P. Sam Johnson (NIT Karnataka) Directional Derivatives in the Plane April 10, 2017 1 / 30 Directional Derivatives in the Plane Let z =

More information

Final Exam Topic Outline

Final Exam Topic Outline Math 442 - Differential Geometry of Curves and Surfaces Final Exam Topic Outline 30th November 2010 David Dumas Note: There is no guarantee that this outline is exhaustive, though I have tried to include

More information

DIFFERENTIAL GEOMETRY HW 5. Show that the law of cosines in spherical geometry is. cos c = cos a cos b + sin a sin b cos θ.

DIFFERENTIAL GEOMETRY HW 5. Show that the law of cosines in spherical geometry is. cos c = cos a cos b + sin a sin b cos θ. DIFFEENTIAL GEOMETY HW 5 CLAY SHONKWILE Show that the law of cosines in spherical geometry is 5 cos c cos a cos b + sin a sin b cos θ. Proof. Consider the spherical triangle depicted below: Form radii

More information

MATH 332: Vector Analysis Summer 2005 Homework

MATH 332: Vector Analysis Summer 2005 Homework MATH 332, (Vector Analysis), Summer 2005: Homework 1 Instructor: Ivan Avramidi MATH 332: Vector Analysis Summer 2005 Homework Set 1. (Scalar Product, Equation of a Plane, Vector Product) Sections: 1.9,

More information

DIFFERENTIAL GEOMETRY CLASS NOTES INSTRUCTOR: F. MARQUES. September 25, 2015

DIFFERENTIAL GEOMETRY CLASS NOTES INSTRUCTOR: F. MARQUES. September 25, 2015 DIFFERENTIAL GEOMETRY CLASS NOTES INSTRUCTOR: F. MARQUES MAGGIE MILLER September 25, 2015 1. 09/16/2015 1.1. Textbooks. Textbooks relevant to this class are Riemannian Geometry by do Carmo Riemannian Geometry

More information

Isometries ( ) Dr. Emanuele Rodolà Matthias Vestner Room , Informatik IX

Isometries ( ) Dr. Emanuele Rodolà Matthias Vestner Room , Informatik IX Isometries (22.05.2014) Dr. Emanuele Rodolà Matthias Vestner {rodola,vestner}@in.tum.de Room 02.09.058, Informatik IX Seminar «The metric approach to shape matching» Alfonso Ros Wednesday, May 28th 14:00

More information

Physics 411 Lecture 8. Parametrized Motion. Lecture 8. Physics 411 Classical Mechanics II

Physics 411 Lecture 8. Parametrized Motion. Lecture 8. Physics 411 Classical Mechanics II Physics 411 Lecture 8 Parametrized Motion Lecture 8 Physics 411 Classical Mechanics II September 14th 2007 We have our fancy new derivative, but what to do with it? In particular, how can we interpret

More information

+ dxk. dt 2. dt Γi km dxm. . Its equations of motion are second order differential equations. with intitial conditions

+ dxk. dt 2. dt Γi km dxm. . Its equations of motion are second order differential equations. with intitial conditions Homework 7. Solutions 1 Show that great circles are geodesics on sphere. Do it a) using the fact that for geodesic, acceleration is orthogonal to the surface. b ) using straightforwardl equations for geodesics

More information

Totally quasi-umbilic timelike surfaces in R 1,2

Totally quasi-umbilic timelike surfaces in R 1,2 Totally quasi-umbilic timelike surfaces in R 1,2 Jeanne N. Clelland, University of Colorado AMS Central Section Meeting Macalester College April 11, 2010 Definition: Three-dimensional Minkowski space R

More information

THEODORE VORONOV DIFFERENTIAL GEOMETRY. Spring 2009

THEODORE VORONOV DIFFERENTIAL GEOMETRY. Spring 2009 [under construction] 8 Parallel transport 8.1 Equation of parallel transport Consider a vector bundle E B. We would like to compare vectors belonging to fibers over different points. Recall that this was

More information

MTHE 227 Problem Set 2 Solutions

MTHE 227 Problem Set 2 Solutions MTHE 7 Problem Set Solutions 1 (Great Circles). The intersection of a sphere with a plane passing through its center is called a great circle. Let Γ be the great circle that is the intersection of the

More information

Introduction - Motivation. Many phenomena (physical, chemical, biological, etc.) are model by differential equations. f f(x + h) f(x) (x) = lim

Introduction - Motivation. Many phenomena (physical, chemical, biological, etc.) are model by differential equations. f f(x + h) f(x) (x) = lim Introduction - Motivation Many phenomena (physical, chemical, biological, etc.) are model by differential equations. Recall the definition of the derivative of f(x) f f(x + h) f(x) (x) = lim. h 0 h Its

More information

Summer 2017 MATH Solution to Exercise 5

Summer 2017 MATH Solution to Exercise 5 Summer 07 MATH00 Solution to Exercise 5. Find the partial derivatives of the following functions: (a (xy 5z/( + x, (b x/ x + y, (c arctan y/x, (d log((t + 3 + ts, (e sin(xy z 3, (f x α, x = (x,, x n. (a

More information

Distances, volumes, and integration

Distances, volumes, and integration Distances, volumes, and integration Scribe: Aric Bartle 1 Local Shape of a Surface A question that we may ask ourselves is what significance does the second fundamental form play in the geometric characteristics

More information

Math 6455 Nov 1, Differential Geometry I Fall 2006, Georgia Tech

Math 6455 Nov 1, Differential Geometry I Fall 2006, Georgia Tech Math 6455 Nov 1, 26 1 Differential Geometry I Fall 26, Georgia Tech Lecture Notes 14 Connections Suppose that we have a vector field X on a Riemannian manifold M. How can we measure how much X is changing

More information

4.7 The Levi-Civita connection and parallel transport

4.7 The Levi-Civita connection and parallel transport Classnotes: Geometry & Control of Dynamical Systems, M. Kawski. April 21, 2009 138 4.7 The Levi-Civita connection and parallel transport In the earlier investigation, characterizing the shortest curves

More information

Math 205 Integration and calculus of several variables

Math 205 Integration and calculus of several variables Math 05 Integration and calculus of several variables week 8 - May 8, 009. Geometry We have developed the calculus of differential forms algebraically, focusing on algebraic manipulations which can be

More information

Surface x(u, v) and curve α(t) on it given by u(t) & v(t). Math 4140/5530: Differential Geometry

Surface x(u, v) and curve α(t) on it given by u(t) & v(t). Math 4140/5530: Differential Geometry Surface x(u, v) and curve α(t) on it given by u(t) & v(t). α du dv (t) x u dt + x v dt Surface x(u, v) and curve α(t) on it given by u(t) & v(t). α du dv (t) x u dt + x v dt ( ds dt )2 Surface x(u, v)

More information

4 Partial Differentiation

4 Partial Differentiation 4 Partial Differentiation Many equations in engineering, physics and mathematics tie together more than two variables. For example Ohm s Law (V = IR) and the equation for an ideal gas, PV = nrt, which

More information

MA1131 Lecture 15 (2 & 3/12/2010) 77. dx dx v + udv dx. (uv) = v du dx dx + dx dx dx

MA1131 Lecture 15 (2 & 3/12/2010) 77. dx dx v + udv dx. (uv) = v du dx dx + dx dx dx MA3 Lecture 5 ( & 3//00) 77 0.3. Integration by parts If we integrate both sides of the proct rule we get d (uv) dx = dx or uv = d (uv) = dx dx v + udv dx v dx dx + v dx dx + u dv dx dx u dv dx dx This

More information

Geometric Modelling Summer 2016

Geometric Modelling Summer 2016 Geometric Modelling Summer 2016 Exercises Benjamin Karer M.Sc. http://gfx.uni-kl.de/~gm Benjamin Karer M.Sc. Geometric Modelling Summer 2016 1 Dierential Geometry Benjamin Karer M.Sc. Geometric Modelling

More information

ISOMETRIES AND THE LINEAR ALGEBRA OF QUADRATIC FORMS.

ISOMETRIES AND THE LINEAR ALGEBRA OF QUADRATIC FORMS. ISOMETRIES AND THE LINEAR ALGEBRA OF QUADRATIC FORMS. Please review basic linear algebra, specifically the notion of spanning, of being linearly independent and of forming a basis as applied to a finite

More information

Exercise 1 (Formula for connection 1-forms) Using the first structure equation, show that

Exercise 1 (Formula for connection 1-forms) Using the first structure equation, show that 1 Stokes s Theorem Let D R 2 be a connected compact smooth domain, so that D is a smooth embedded circle. Given a smooth function f : D R, define fdx dy fdxdy, D where the left-hand side is the integral

More information

5.2 The Levi-Civita Connection on Surfaces. 1 Parallel transport of vector fields on a surface M

5.2 The Levi-Civita Connection on Surfaces. 1 Parallel transport of vector fields on a surface M 5.2 The Levi-Civita Connection on Surfaces In this section, we define the parallel transport of vector fields on a surface M, and then we introduce the concept of the Levi-Civita connection, which is also

More information

CS Tutorial 5 - Differential Geometry I - Surfaces

CS Tutorial 5 - Differential Geometry I - Surfaces 236861 Numerical Geometry of Images Tutorial 5 Differential Geometry II Surfaces c 2009 Parameterized surfaces A parameterized surface X : U R 2 R 3 a differentiable map 1 X from an open set U R 2 to R

More information

Complete Surfaces of Constant Gaussian Curvature in Euclidean Space R 3.

Complete Surfaces of Constant Gaussian Curvature in Euclidean Space R 3. Summary of the Thesis in Mathematics by Valentina Monaco Complete Surfaces of Constant Gaussian Curvature in Euclidean Space R 3. Thesis Supervisor Prof. Massimiliano Pontecorvo 19 May, 2011 SUMMARY The

More information

Differential Geometry of Surfaces

Differential Geometry of Surfaces Differential Forms Dr. Gye-Seon Lee Differential Geometry of Surfaces Philipp Arras and Ingolf Bischer January 22, 2015 This article is based on [Car94, pp. 77-96]. 1 The Structure Equations of R n Definition

More information

Introduction to Algebraic and Geometric Topology Week 14

Introduction to Algebraic and Geometric Topology Week 14 Introduction to Algebraic and Geometric Topology Week 14 Domingo Toledo University of Utah Fall 2016 Computations in coordinates I Recall smooth surface S = {f (x, y, z) =0} R 3, I rf 6= 0 on S, I Chart

More information

Math 11 Fall 2018 Practice Final Exam

Math 11 Fall 2018 Practice Final Exam Math 11 Fall 218 Practice Final Exam Disclaimer: This practice exam should give you an idea of the sort of questions we may ask on the actual exam. Since the practice exam (like the real exam) is not long

More information

Math 433 Outline for Final Examination

Math 433 Outline for Final Examination Math 433 Outline for Final Examination Richard Koch May 3, 5 Curves From the chapter on curves, you should know. the formula for arc length of a curve;. the definition of T (s), N(s), B(s), and κ(s) for

More information

II. Unit Speed Curves

II. Unit Speed Curves The Geometry of Curves, Part I Rob Donnelly From Murray State University s Calculus III, Fall 2001 note: This material supplements Sections 13.3 and 13.4 of the text Calculus with Early Transcendentals,

More information

MATH 12 CLASS 5 NOTES, SEP

MATH 12 CLASS 5 NOTES, SEP MATH 12 CLASS 5 NOTES, SEP 30 2011 Contents 1. Vector-valued functions 1 2. Differentiating and integrating vector-valued functions 3 3. Velocity and Acceleration 4 Over the past two weeks we have developed

More information

Calculus III. Math 233 Spring Final exam May 3rd. Suggested solutions

Calculus III. Math 233 Spring Final exam May 3rd. Suggested solutions alculus III Math 33 pring 7 Final exam May 3rd. uggested solutions This exam contains twenty problems numbered 1 through. All problems are multiple choice problems, and each counts 5% of your total score.

More information

HOMEWORK 3 MA1132: ADVANCED CALCULUS, HILARY 2017

HOMEWORK 3 MA1132: ADVANCED CALCULUS, HILARY 2017 HOMEWORK MA112: ADVANCED CALCULUS, HILARY 2017 (1) A particle moves along a curve in R with position function given by r(t) = (e t, t 2 + 1, t). Find the velocity v(t), the acceleration a(t), the speed

More information

Exact Solutions of the Einstein Equations

Exact Solutions of the Einstein Equations Notes from phz 6607, Special and General Relativity University of Florida, Fall 2004, Detweiler Exact Solutions of the Einstein Equations These notes are not a substitute in any manner for class lectures.

More information

HOMEWORK 2 SOLUTIONS

HOMEWORK 2 SOLUTIONS HOMEWORK SOLUTIONS MA11: ADVANCED CALCULUS, HILARY 17 (1) Find parametric equations for the tangent line of the graph of r(t) = (t, t + 1, /t) when t = 1. Solution: A point on this line is r(1) = (1,,

More information

An Introduction to Gaussian Geometry

An Introduction to Gaussian Geometry Lecture Notes in Mathematics An Introduction to Gaussian Geometry Sigmundur Gudmundsson (Lund University) (version 2.068-16th of November 2017) The latest version of this document can be found at http://www.matematik.lu.se/matematiklu/personal/sigma/

More information

MATH 4030 Differential Geometry Lecture Notes Part 4 last revised on December 4, Elementary tensor calculus

MATH 4030 Differential Geometry Lecture Notes Part 4 last revised on December 4, Elementary tensor calculus MATH 4030 Differential Geometry Lecture Notes Part 4 last revised on December 4, 205 Elementary tensor calculus We will study in this section some basic multilinear algebra and operations on tensors. Let

More information

Math 461 Homework 8. Paul Hacking. November 27, 2018

Math 461 Homework 8. Paul Hacking. November 27, 2018 Math 461 Homework 8 Paul Hacking November 27, 2018 (1) Let S 2 = {(x, y, z) x 2 + y 2 + z 2 = 1} R 3 be the sphere with center the origin and radius 1. Let N = (0, 0, 1) S 2 be the north pole. Let F :

More information

Notes on the Riemannian Geometry of Lie Groups

Notes on the Riemannian Geometry of Lie Groups Rose- Hulman Undergraduate Mathematics Journal Notes on the Riemannian Geometry of Lie Groups Michael L. Geis a Volume, Sponsored by Rose-Hulman Institute of Technology Department of Mathematics Terre

More information

Linear Models Review

Linear Models Review Linear Models Review Vectors in IR n will be written as ordered n-tuples which are understood to be column vectors, or n 1 matrices. A vector variable will be indicted with bold face, and the prime sign

More information

DIFFERENTIAL GEOMETRY HW 5

DIFFERENTIAL GEOMETRY HW 5 DIFFERENTIAL GEOMETRY HW 5 CLAY SHONKWILER 1 Check the calculations above that the Gaussian curvature of the upper half-plane and Poincaré disk models of the hyperbolic plane is 1. Proof. The calculations

More information

Summary for Vector Calculus and Complex Calculus (Math 321) By Lei Li

Summary for Vector Calculus and Complex Calculus (Math 321) By Lei Li Summary for Vector alculus and omplex alculus (Math 321) By Lei Li 1 Vector alculus 1.1 Parametrization urves, surfaces, or volumes can be parametrized. Below, I ll talk about 3D case. Suppose we use e

More information

Calculus III: Practice Final

Calculus III: Practice Final Calculus III: Practice Final Name: Circle one: Section 6 Section 7. Read the problems carefully. Show your work unless asked otherwise. Partial credit will be given for incomplete work. The exam contains

More information

Material for review. By Lei. May, 2011

Material for review. By Lei. May, 2011 Material for review. By Lei. May, 20 You shouldn t only use this to do the review. Read your book and do the example problems. Do the problems in Midterms and homework once again to have a review. Some

More information

x + ye z2 + ze y2, y + xe z2 + ze x2, z and where T is the

x + ye z2 + ze y2, y + xe z2 + ze x2, z and where T is the 1.(8pts) Find F ds where F = x + ye z + ze y, y + xe z + ze x, z and where T is the T surface in the pictures. (The two pictures are two views of the same surface.) The boundary of T is the unit circle

More information

MATH The Chain Rule Fall 2016 A vector function of a vector variable is a function F: R n R m. In practice, if x 1, x n is the input,

MATH The Chain Rule Fall 2016 A vector function of a vector variable is a function F: R n R m. In practice, if x 1, x n is the input, MATH 20550 The Chain Rule Fall 2016 A vector function of a vector variable is a function F: R n R m. In practice, if x 1, x n is the input, F(x 1,, x n ) F 1 (x 1,, x n ),, F m (x 1,, x n ) where each

More information

Introduction to Algebraic and Geometric Topology Week 3

Introduction to Algebraic and Geometric Topology Week 3 Introduction to Algebraic and Geometric Topology Week 3 Domingo Toledo University of Utah Fall 2017 Lipschitz Maps I Recall f :(X, d)! (X 0, d 0 ) is Lipschitz iff 9C > 0 such that d 0 (f (x), f (y)) apple

More information

DIFFERENTIAL GEOMETRY OF CURVES AND SURFACES 5. The Second Fundamental Form of a Surface

DIFFERENTIAL GEOMETRY OF CURVES AND SURFACES 5. The Second Fundamental Form of a Surface DIFFERENTIAL GEOMETRY OF CURVES AND SURFACES 5. The Second Fundamental Form of a Surface The main idea of this chapter is to try to measure to which extent a surface S is different from a plane, in other

More information

OHSx XM521 Multivariable Differential Calculus: Homework Solutions 14.1

OHSx XM521 Multivariable Differential Calculus: Homework Solutions 14.1 OHSx XM5 Multivariable Differential Calculus: Homework Solutions 4. (8) Describe the graph of the equation. r = i + tj + (t )k. Solution: Let y(t) = t, so that z(t) = t = y. In the yz-plane, this is just

More information

3.4 Conic sections. Such type of curves are called conics, because they arise from different slices through a cone

3.4 Conic sections. Such type of curves are called conics, because they arise from different slices through a cone 3.4 Conic sections Next we consider the objects resulting from ax 2 + bxy + cy 2 + + ey + f = 0. Such type of curves are called conics, because they arise from different slices through a cone Circles belong

More information

z x = f x (x, y, a, b), z y = f y (x, y, a, b). F(x, y, z, z x, z y ) = 0. This is a PDE for the unknown function of two independent variables.

z x = f x (x, y, a, b), z y = f y (x, y, a, b). F(x, y, z, z x, z y ) = 0. This is a PDE for the unknown function of two independent variables. Chapter 2 First order PDE 2.1 How and Why First order PDE appear? 2.1.1 Physical origins Conservation laws form one of the two fundamental parts of any mathematical model of Continuum Mechanics. These

More information

M435: INTRODUCTION TO DIFFERENTIAL GEOMETRY

M435: INTRODUCTION TO DIFFERENTIAL GEOMETRY M435: INTODUCTION TO DIFFNTIAL GOMTY MAK POWLL Contents 7. The Gauss-Bonnet Theorem 1 7.1. Statement of local Gauss-Bonnet theorem 1 7.2. Area of geodesic triangles 2 7.3. Special case of the plane 2 7.4.

More information