M435: INTRODUCTION TO DIFFERENTIAL GEOMETRY

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1 M435: INTRODUCTION TO DIFFERENTIAL GEOMETRY MARK POWELL Contents 5 Second fundamental form and the curvature 5 The second fundamental form 52 Second fundamental form alternative derivation 2 53 The shape operator 3 54 Principal curvature 5 55 The Theorema Egregium of Gauss References 4 5 Second fundamental form and the curvature 5 The second fundamental form We will give several ways of motivating the definition of the second fundamental form Like the first fundamental form, the second fundamental form is a symmetric bilinear form on each tangent space of a surface Σ Unlike the first, it need not be positive definite The idea of the second fundamental form is to measure, in R 3, how Σ curves away from its tangent plane at a given point The first fundamental form is an intrinsic object whereas the second fundamental form is extrinsic That is, it measures the surface as compared to the tangent plane in R 3 By contrast the first fundamental form can be measured by a denizen of the surface, who does not possess 3 dimensional awareness Given a smooth patch r: U Σ, let n be the unit normal vector as usual Define Ru, v, t : ru, v tnu, v, with t ε, ε This is a -parameter family of smooth surface patches How is the first fundamental form changing in this family? We can compute that: 2 t t0 Edu 2 + 2F dudv + Gdv 2 Ldu 2 + 2Mdudv + Ndv 2 where L : r u n u 2M : r u n v + r v n u N : r v n v The reason for the negative signs will become clear soon

2 M435: INTRODUCTION TO DIFFERENTIAL GEOMETRY 2 For example: 2 t t0 Eu, v, t ru 2 t t0 tn u r u tn u ru 2 t t0 r u 2tr u n u + t 2 n u n u 2 2ru n u + 2tn u n u t0 r u n u Exercise 5 Check the analogous computations for 2M and N The form: Ldu 2 + 2Mdudv + Ndv 2 is called the second fundamental form Like the first fundamental form, it also defines a bilinear form on the tangent space: ar u + br v, cr u + dr v a b L M c d 52 Second fundamental form alternative derivation We can think of the second fundamental form as measuring distance from the tangent plane Consider the Taylor expansion around a point u, v: ru + δu, v + δv ru, v + δur u + δvr v + 2 r uuδu 2 + 2r uv δuδv + r vv δv 2 + hot Therefore ru + δu, v + δv ru, v n where 2 r uu nδu 2 + 2r uv nδuδv + r vv nδv 2 + hot 2 Lδu2 + 2Mδuδv + Nδv 2 + hot L r uu n M r uv n N r vv n We prefer this definition of L, M and N with second derivatives of r, since usually n is tricky to differentiate, as there is a square root in the denominators Note that since r u n 0, we have r uu n + r u n u 0 so Also r v n 0, so r uu n r u n u L r vv n + r v n v 0

3 which implies that and which imply M435: INTRODUCTION TO DIFFERENTIAL GEOMETRY 3 r vv n r v n v N r uv n + r u n v 0 r vu n + r v n u 0 r uv n r u n v r v n u M Example 52 a For the plane L M N 0 b For the sphere of radius a, r an, so that r u n u a r u r u a E Repeating this argument for F and M and for G and N, we have Ldu 2 + 2Mdudv + Ndv 2 a Edu 2 + 2F dudv + Gdv 2 c Consider the graph z fx, y of a smooth function f : R 2 R So the parametrisation is r x, y, fx, y The second fundamental form is a multiple of the Hessian: + fx 2 + fy 2 fxx f xy f xy f yy d Fix a point p Σ Since every surface is locally a graph we didn t prove this but we could have: See Do Carmo Section 2-2, Proposition 3, Page 63, which uses the Inverse Function Theorem We can parametrise by the inverse of projection onto the tangent plane T p Σ Our point is a critical point of fx, y, and we can understand the local behaviour: maximum, minimum or saddle point in terms of the Hessian matrix ie in terms of the second fundamental form This will be described soon in detail using the principal curvatures 53 The shape operator The shape operator, also known as the Weingarten map, is a function S : T p Σ T p Σ which we can use to define the second fundamental form in terms of covariant derivatives of the normal vector; the Gauss map Σ S 2 Define S by v p vp n So Sr u ru n n u and Sr v rv n n v This measures how the normal vector tilts as we move infinitesimally away from our point All 3 versions of the second fundamental form are really the same idea in different guises Note that Sr u r u L Sr u r v M

4 M435: INTRODUCTION TO DIFFERENTIAL GEOMETRY 4 Sr v r v N We can define a bilinear form on each tangent space by Bv, w : Sv w The matrix of this with respect to the basis {r u, r v } is that of the second fundamental form L M and Now, for later, what is the matrix of the shape operator? Let a, b, c, d be such that Sr u ar u + br v Then dotting with r u and r v we have: Therefore L M so that a b Sr v cr u + dr v L Sr u r u ae + bf M Sr u r v af + bg M Sr v r u ce + df N Sr v r v cf + dg c E F L M d This matrix need not be symmetric Compute the second fundamental form of a surface of rev- Exercise 53 olution E F a c b d EG F 2 G F L M F E ru, v fu cos v, fu sin v, u 2 Compute the second fundamental form of the patch of the torus: ru, v a + b cos u cos v, a + b cos u sin v, b sin u The second fundamental form characterises the plane Theorem 54 Suppose the second fundamental form of a surface is identically zero Then the surface is a part of a plane Proof If L M N 0 then n u r u n u r v n v r u n v r v 0 so n u and n v are parallel to n Since n is of constant length we therefore have that n is constant

5 M435: INTRODUCTION TO DIFFERENTIAL GEOMETRY 5 54 Principal curvature We want to consider the curvature of the set of curves on Σ through a fixed point p Σ Let γt rut, vt such that ru0, v0 p Suppose furthermore that γ t ie γ is parametrised by arc length Let τ : γ t We have an orthonormal basis for T R 3 at each point of the image of γ {n, τ, n τ} Recall this frame from the spherical curves section in Chapter 2 The curvature is kt γ t Since γ γ we have so that γ τ 0 Therefore γ γ 0, γ κ n n + κ g n τ where κ n γ n is the normal curvature and κ g γ n τ is the geodesic curvature of γ at p Note that κ 2 n + κ 2 g k 2 Now, τ n 0 γ n so γ n + γ n 0 Thus By the chain rule we have So taking the dot product κ n γ n γ u r u + v r v n u n u + v n v κ n γ n Lu 2 + 2Mu v + Nv 2 We want to compute the principal curvatures and directions Definition 55 The principal curvatures of Σ at p Σ are the maximum and minimum values of the normal curvatures of curves through p Consider a plane Π through p which contains n, τ Then Π Σ is a curve γt through p, which we will call γt, parametrised as above by arc length Note that γ t n τ 0 since n τ is normal to Π Thus κ g 0 and κ n k So the principal curvatures are the maximum and minimum curvatures of curves Π Σ at p where Π ranges over all planes through p containing n p So, we can compute the principal curvatures as the maximum and minimum values of Lξ 2 + 2Mξη + Nη 2 for ξ, η R 2 such that Eξ 2 + 2F ξη + Gη 2 This latter condition is that γ t Since the first fundamental form is positive definite, this is the maximum/minimum of a function on R 2 restricted to an ellipse To find the maximum and minimum values, we use simultaneous diagonalisation Recall that the spectral theorem from linear algebra says that any real symmetric matrix A can be diagonalised over the real numbers This means that there is a

6 M435: INTRODUCTION TO DIFFERENTIAL GEOMETRY 6 orthogonal matrix P such that P AP T is a diagonal matrix Orthogonal means P P T Id Apply this to the first fundamental form s matrix There exists an orthogonal matrix R such that E F R R λ 0 0, λ 2 where λ, λ 2 are the eigenvalues of the first fundamental form The first fundamental form is positive definite so λ, λ 2 > 0 Let Then Q : λ 0 R 0 λ2 E F Q Q T 0 0 L M Now the matrix Q Q T is a symmetric matrix, so there exists another orthogonal matrix P such that L M P Q Q T P T κ 0 0 κ 2 is diagonal and E F P Q Q T P T P P T 0 0 We therefore change coordinates to X Y ξ P Q η In these coordinates Eξ 2 + 2F ξη + Gη 2 X 2 + Y 2 and Lξ 2 + 2Mξη + Nη 2 κ X 2 + κ 2 Y 2 The maximum and minimum values of the latter subject to X 2 + Y 2 are κ and κ 2 These are our principal curvatures The vectors, 0 and 0, in the X, Y coordinates are the principal directions

7 M435: INTRODUCTION TO DIFFERENTIAL GEOMETRY 7 E F L M It turns out that we need to find the eigenvalues and eigenvectors of To check this: E F L M P Q Q P E F L M P Q Q T P T E F P Q Q P L M P Q Q T P T E F P Q Q P L M P Q Q T P T E F P Q Q T P T L M P Q Q T P T 0 L M P Q Q 0 T P T 0 κ κ 2 κ 0 0 κ 2 E F L M The eigenvalues and eigenvectors are the same as above Note that coincides with the matrix which earlier we computed to represent the shape operator Therefore we can also define the principal curvatures and directions as follows Definition 56 The principal curvatures at p are the eigenvalues of the shape operator at p and the principal directions are the eigenvectors We also define two important quantities The Gaussian curvature in particular is of central importance in the geometry of surfaces Definition 57 The Gaussian curvature K is the product of the principal curvatures κ κ 2 which is equal to E F L M det LN M 2 EG F 2 Definition 58 The mean curvature H is the average κ +κ 2 2 of the principal curvatures, which is equal to E F L M 2 tr F G M N LG 2MF + NE 2EG F 2

8 M435: INTRODUCTION TO DIFFERENTIAL GEOMETRY 8 Example 59 a The Gaussian and mean curvatures of the plane are zero b The principal curvatures of the sphere of radius a are both /a, since Ldu 2 + 2Mdudv + Ndv 2 a Edu 2 + 2F dudv + Gdv 2 and therefore E F L M a 0 0 a All directions are principal directions The mean curvature is /a and the Gaussian curvature is a constant /a 2 c The torus with parametrisation ru, v a + b cos u cos v, a + b cos u sin v, b sin u u, v 0, 2π 0, 2π We compute: r u b sin u cos v, b sin u sin v, b cos u so that r v a + b cos u sin v, a + b cos u cos v, 0, r u r v ba + b cos u cos u cos v, ba + b cos u cos u sin v, ba + b cos u sin u We have so Also r u r v ba + b cos u n cos u cos v, cos u sin v, sin u r uu b cos u cos v, b cos u sin v, b sin u r uv b sin u sin v, b sin u cos v, 0 r vv a + b cos u cos v, a + b cos u sin v, 0 Therefore we take the relevant dot products to find: and E b 2 ; F 0; G a + b cos u 2 L b; M 0; N a + b cos u cos u Thus the Gaussian curvature is given by The mean curvature is given by H K LN M 2 EG F 2 cos u ba + b cos u LG 2MF + NE 2EG F 2 a + 2b cos u 2ba + b cos u Note that the curvatures are constant along lines of constant u Topologically the torus is S S, but geometrically the two copies of S are not symmetric In particular note that when u 0, K > 0, when u ±π/2, we have K 0, and when u π we have K < 0

9 M435: INTRODUCTION TO DIFFERENTIAL GEOMETRY 9 d For the catenoid, with parametrisation with u 0, 2π and v R, we have So Also Therefore ru, v cosh v cos u, cosh v sin u, v r u cosh v sin u, cosh v cos u, 0 r v sinh v, cos u, sinh v sin u, n cos u, sin u, sinh v cosh v E cosh 2 v; F 0; G cosh 2 v r uu cosh vcos u, sin u, 0 r uv sinh v sin u, cos u, 0 r vv cosh vcos u, sin u, 0 L ; M 0; N So the Gaussian curvature is given by K LN M 2 EG F 2 cosh 4 v The mean curvature is given by H LG 2MF + NE 2EG F 2 cosh2 v 0 + cosh v cosh 4 v 0 The catenoid has mean curvature zero; such surfaces are called minimal Definition 50 An asymptotic direction is a unit vector with respect to the first fundamental form in the tangent space along which the normal curvature vanishes To compute, find a, b such that a b L M a b 0 and scale so that E F a a b b An asymptotic curve through p Σ is a curve in Σ along which the normal curvature is zero So γ : t rut, vt, γ0 p, such that Lu 2 +2Mu v +Nv 2 0 Exercise 5 i Find the Gaussian and mean curvatures of a graph z fx, y ii Compute principal curvatures and principal directions of the saddle surface z xy at the origin, and of an ellipsoid x 2 /a 2 + y 2 /b 2 + z 2 /c 2 at each of the points where two of x, y and z are zero Also compute the Gaussian and mean curvatures everywhere of these surfaces

10 M435: INTRODUCTION TO DIFFERENTIAL GEOMETRY 0 iii Give an expression for the Gaussian curvature of a surface of revolution ru, v fu cos v, fu sin v, u iv Do Carmo 33 Exercises 7, 20 Question 6 on the pseudo sphere is a particularly good example: a surface with Gaussian curvature constant and equal to For question 20, an umbilical point is a point where the principal curvatures are equal Here is a cool interpretation of the Gaussian curvature in terms of how area is altered by the Gauss map Recall that the Gaussian curvature is a determinant, so we shouldn t be too surprised that it has an interpretation in terms of area Theorem 52 Let x Σ and let {U i } with U i x for all i be a collection of open sets in Σ contracting to x Then signed areanu areau tends to the Gaussian curvature Kx as U contracts to x Here the area of nu has a sign according to whether n preserves orientation in S 2, with respect to the standard orientation on S 2 C { } Proof Let r: V U Σ be a parametrisation The area of U is given by r u r v dudv EG F 2 dudv The area of nu is given by V V V n u n v dudv, however to find the signed area we instead compute n n u n v dudv V The vector n u n v points in the direction of n or in the opposite direction, so taking the indicated dot product with the unit vector n gives the correct sign Now n n u n v r u r v n u n v r u r v r u n u r v n v r u n v r v n u r u r v LN M 2 EG F 2 Thus the stated quotient is V V LN M 2 dudv EG F 2 EG F 2 dudv

11 M435: INTRODUCTION TO DIFFERENTIAL GEOMETRY In the limit, by continuity, as U shrinks to x, V shrinks to the preimage of x, and the integral tends to the value at x, which is as claimed LN M 2 EG F 2 K Corollary 53 Suppose Σ is convex and compact Then KdA 4π Σ Proof The Gauss map is a bijection in this case KdA areanda, Σ KdA areanda which is the area of the unit sphere Σ This is a remarkable corollary in itself Think of all the possible convex compact surfaces one could make by deforming a 2-sphere The curvature can be highly variable Nevertheless the integral of the curvature always conspires to be 4π Pretty-pretty-pretty good Said in Curb Your Enthusiasm style 55 The Theorema Egregium of Gauss The theorem of this section is so remarkable that even Gauss thought so, and he called it the Theorema Egregium, which is latin for remarkable theorem Recall that the principal curvatures are defined in terms of both the first and second fundamental form The second fundamental form was measuring how a surface curves away from its tangent plane near a point, and it seems like such a notion is vital for the study of curvature Amazingly, the product of the principal curvatures is intrinsic Theorem 54 Gauss The Gaussian curvature is an intrinsic quantity That is, it can be computed purely in terms of the coefficients of the first fundamental form and their derivatives Another way to say this is that the Gaussian curvature is invariant under local isometries The Gaussian curvature has less information than the two principal curvatures separately, but it has the great advantage that it only depends on the first fundamental form, so makes sense as a notion on an abstract smooth surface with a first fundamental form, ie if we defined a surface which is not embedded in R 3, but exists as an abstract notion Proof We begin by defining the Christoffel symbols Γ k ij Γk ji, where each of i, j, k is from {, 2} Recall that {r u, r v, n} is a basis for T R 3 at each point of a surface Σ Therefore we can write the second derivatives at each point in terms of this basis r uu Γ r u + Γ 2 r v + Ln r uv Γ 2r u + Γ 2 2r v + Mn r vv Γ 22r u + Γ 2 22r v + Nn Note the appearance of L, M and N as the coefficients of n, by their definitions Note that the Γ k ij are also functions of u, v Our proof is in two steps:

12 M435: INTRODUCTION TO DIFFERENTIAL GEOMETRY 2 Express the Christoffel symbols in terms of the coefficients E, F and G of the first fundamental form 2 Express the Gaussian curvature in terms of the Christoffel symbols Combining these two completes the proof of the theorem accomplished in the following lemma The first step is Lemma 55 The Christoffel symbols can be expressed in terms of the coefficients of the first fundamental form, as follows Γ E F Γ 2 2 E u F u 2 E v Γ 2 E F Γ 2 2 E v 2 2 G u Γ 22 E F Fv 2 G u Γ G v Proof of Lemma 55 Start with E r u r u Differentiating yields 2 E u r uu r u Γ E + Γ 2 F In a similar vein we compute the derivatives of the first fundamental form coefficients 2 E v r uv r u Γ 2E + Γ 2 2F F u r uu r v + r u r uv Γ F + Γ 2 G + Γ 2E + Γ 2 2F F v r uv r v + r u r vv Γ 2F + Γ 2 2G + Γ 22E + Γ 2 22F 2 G u r uv r v Γ 2F + Γ 2 2G 2 G v r vv r v Γ 22F + Γ 2 22G Rearranging yields the claimed equations: 2 E u E F Γ F u 2 E v Γ 2 2 E v E F Γ 2 G 2 u Γ 2 2 Fv 2 G u E F Γ 22 2 G v This completes the proof of the lemma We continue with the proof of the Theorema Egregium; we now need to show step 2, that the Gaussian curvature can be expressed in terms of the Christoffel symbols Γ 2 22

13 M435: INTRODUCTION TO DIFFERENTIAL GEOMETRY 3 Next we recall the matrix of the negative of the shape operator S : v v n, which acts on the tangent space of a surface, sending r u n u a r u + a 2 r v r v n v a 2 r u + a 22 r v So n u Sr u, n v Sr v The matrix of the negative of the shape operator with respect to the basis {r u, r v } is a a 2 a 2 a 22 We saw before that this matrix is E F L M Z Therefore and n u EG F 2 F M GL F N GM F L EM F M EN F M GLru EG F 2 + F L EMr v n v F N GMru EG F 2 + F M ENr v Now for the coup de grace We equate r uu v r uv u, in particular the coefficients of r u and r v in this equation Take r uu Γ r u + Γ 2 r v + Ln and differentiate: r uu v Γ v r u + Γ r uv + Γ 2 v r v + Γ 2 r vv + L v n + Ln v Γ v r u + Γ Γ 2r u + Γ 2 2r v + Mn + Γ 2 v r v + Γ 2 Γ 22r u + Γ 2 22r v + Nn + L v n + L Sr v This is equal to r uv u Γ 2 r u + Γ 2 2 r v + Mn u, which is: r uv u Γ 2 u r u + Γ 2r uu + Γ 2 2 u r v + Γ 2 2r uv + M u n + Mn u Γ 2 u r u + Γ 2Γ r u + Γ 2 r v + Ln + Γ 2 2 u r v + Γ 2 2Γ 2r u + Γ 2 2r v + Mn + M u n + M Sr u Recall that Sr u a r u + a 2 r v and Sr v a 2 r u + a 22 r v Equate coefficients of r v to get Γ Γ Γ 2 v + Γ 2 Γ La 22 Γ 2Γ 2 + Γ 2 2 u + Γ 2 2Γ Ma 2 Then note that La 22 Ma 2 LF M LEN MF L + M 2 E EG F 2 ELN M 2 EG F 2 EK So EK Γ 2 v Γ 2 2 u + Γ Γ Γ 2 Γ 2 22 Γ 2 2Γ 2 2 Γ 2Γ 2 This expresses the Gaussian curvature K in terms of the Christoffel symbols and E, so completely in terms of the first fundamental form This completes the proof

14 M435: INTRODUCTION TO DIFFERENTIAL GEOMETRY 4 Since, for any local isometry, we have coordinates with respect to which the first fundamental forms are the same, this proves that the Gaussian curvature is an invariant of local and therefore global isometries Remark 56 The Gauss equations can be derived in a similar fashion to the equation above, by equating coefficients of r u and r v in r uu v r uv v and r vv u r uv v They are, in addition to the equation above: F K Γ 2 u Γ v + Γ 2 2Γ 2 2 Γ 2 Γ 22 F K Γ 2 2 v Γ 2 22 u + Γ 2Γ 2 2 Γ 22Γ 2 GK Γ 22 u Γ 2 v + Γ Γ 22 + Γ 2Γ 22 Γ 2 2Γ 22 Γ 2Γ 2 Exercise 57 The formula for K becomes a lot simpler when F 0 Show that in this case K 2 Gu + Ev, EG u EG v EG and then use this formula to compute the Gaussian curvature of the sphere rθ, ϕ a cos θ cos ϕ, a sin θ cos ϕ, a sin ϕ Note on exam question 4: Saying that two surfaces have different first fundamental forms with some coordinates does not imply that the surfaces are not isometric One has to show that the surfaces could not have the same first fundamental forms after any change of coordinates This is where the Gaussian curvature comes in As an invariant, it is the same in all parametrisations Therefore is one surface has constant K 0 and another has constant K, those surfaces cannot be locally isometric Exercise 58 Do Carmo Section 4-3, Questions, 3, 6, 8, 9 References [DoCa] Manfredo P DoCarmo Differential Geometry of Curves and Surfaces [Hit] Nigel Hitchens Lecture notes for B3: Geometry of Surfaces Department of Mathematics, Indiana University, Bloomington, IN 47405, USA address: macp@indianaedu