5.1 Gauss Remarkable Theorem

Transcription

1 5.1 Gauss Remarkable Theorem Recall that, for a surface M, its Gauss curvature is defined by K = κ 1 κ 2 where κ 1, κ 2 are the principal curvatures of M. The principal curvatures are the eigenvalues of the shape-operator S P. The shape-operator S P v = D v np is defined by using the Gauss map np, which depends, not only the surface itself, but also on its position in R 3. Note that, the computational formula 4.19 which we derived for calculating the Gauss curvature depends both on the first and second fundamental form of Mthe second fundamental form of M depends on the Gauss map np, hence it is an extrinsic quantity. Gauss discovered that the Gauss curvature, in fact, depends only on the first fundamental form of M. Hence, it is an intrinsic propertyonly depends on the surface M itself. In other words, the Gauss curvature is unchanged when the surface is bent without stretchingfor the the first fundamental form, i.e. the measurement about the length, does not change, if the surface is bent without stretching. Gauss called this result egregium, and the Latin word for remarkable has remained attached to this theorem ever since. This discovery is one of the most important discoveries about surface and is the foundation for the theory of Riemannian Geometry. 1 Review of the definition of the Gauss curvature In this section, we review the definition of the Gauss curvature see section 3.4 and section 4.1. Recall that the Gauss curvature is defined by K = κ 1 κ 2 where κ 1, κ 2 are the principal curvatures of M. The principal curvatures are the eigenvalues of the shape-operator S P, The shape-operator S P is defined by S P v = D v np. Let σ : U R 3 be a parametrization for the surface M, then {σ u, σ v } is a basis of the vector space T P M. Then, the matrix of S P with respect to the basis {σ u, σ v }, denoted by A, is see A = FI 1 F II, 1

2 where E F e f F I =, F F G II =, f g Here {E, F, G} and {e, f, g} are the First and Second Fundamental forms of M. If we write a c A =, b d thensee ff eg a = EG F, 2 gf fg c = EG F, 2 ef fe b = EG F, 2 ff ge d = EG F. 2 The Gauss curvature is defined by K = κ 1 κ 2 where κ 1, κ 2 are the eigenvalues of S P. Hence, we have the following formula K = det A = ad bc = eg f 2 EG F 2. 2 The Christoffel symbols Let σ : U R 3 be a parametrization of a surface M, and let {E, F, G}, {e, f, g} be its first and second fundamental form. Let P M, note that {σ u, σ v } forms a basis of the vector space T P M, hence {σ u, σ v, n} forms a basis of R 3, where n is the unit normal to the surface M. Therefore, we can express σ uu, the partial derivatives of the vector σ u, as the linear combination of basis {σ u, σ v, n}, i.e.: σ uu = Γ 1 11σ u + Γ 2 11σ v + c 3 n, where Γ 1 11, Γ 2 11, and c 3 are some real numbers which depend on the point P hence they are, in fact, functions on M. Since σ u n = 0, σ v n = 0, we have c 3 = σ uu n. By the definition of the Second Fundamental Form see 3.4.4, σ uu n = e, where {e, f, g} is the second fundamental form of M. Hence c 3 = e. Therefore, we have σ uu = Γ 1 11σ u + Γ 2 11σ v + en. Similarly, we have σ uv = Γ 1 12σ u + Γ 2 12σ v + fn, σ vu = Γ 1 21σ u + Γ 2 21σ v + fn, 2

3 5.1.5 σ vv = Γ 1 22σ u + Γ 2 22σ v + gn. Using σ uv = σ vu, it is easy to check that Γ l ik = Γ l ki, 1 i, l, k 2. The six functions Γ l ik = Γ l ki, 1 i, l, k 2, are called the Christoffel symbols. The are the functions on the surface M. Note that the Christoffel symbols Γ l ik are determined by writing σ uu, σ uv and σ vv as a linear combination of {σ u, σ v, n}. Also, write n u = a 11 σ u + a 21 σ v, n v = a 12 σ u + a 22 σ v, Since n u = Dσ u n = S P σ u, n v = Dσ v n = S P σ v, we have that a11 a 21 a 12 a 22 = A, where A is the matrix of S P with respect to the basis {σ u, σ v }. Since A = FI 1 F II, we have a11 a 21 = FI 1 F a 12 a II. 22 So a 11 = ff eg EG F 2, a 21 = ef fe EG F 2, a 12 = gf fg EG F 2, a 22 = ff ge EG F 2. 3 The method of computing the Christoffel symbols To compute the Christoffel symbols, we take the inner product in 5.1.2, 5.1.3, 5.1.4, with σ u and σ v respectively, and notice that σ u n = 0, σ v n = 0. For example, taking the inner product on both sides in with σ u, we get σ uu σ u = Γ 1 11σ u σ u + Γ 2 11σ v σ u. Since, by the definition, E = σ u σ u, F = σ v σ u, we have σ uu σ u = Γ 1 11E + Γ 2 11F. 3

4 From σ u σ u = E, taking partial differentiation with respect to u, we get 2σ uu σ u = E u. Substituting σ uu σ u = 1 2 E u into yields Γ 1 11E + Γ 2 11F = 1 2 E u. Similarly, taking the inner product on both sides in with σ v, we get Γ 1 11F + Γ 2 11G =< σ uu, x v >= F u 1 2 E v. So, we can compute the Christoffel symbol by solving the above system of linear equations and jointly with unknowns Γ 1 11 and Γ 2 11 to get Γ 1 11 = GE u 2F F u + F E v 2EG F 2, Γ 2 11 = 2EF u EE v F E u. 2EG F 2 Other Christoffel symbols can be computed in a similar way, we have Γ 1 12 = Γ 1 21 = GE v F G u 2EG F 2, Γ 1 22 = 2GF v GG u F G v 2EG F 2 Γ2 12 = Γ 2 21 = EG u F E v 2EG F 2, Γ 2 22 = EG v 2F F v + F G u EG F 2 In the case that F = 0, then the above formula can be the following simple form Γ 1 11 = 1 ln E 2 u, Γ2 11 = 1 E 2G v Γ 1 12 = Γ 1 21 = 1 ln E 2 v, Γ2 12 = Γ 2 21 = 1 2 Γ 1 22 = 1 G 2E u, Γ2 22 = 1 ln G 2 v. ln G u Note that the Christoffel symbols depend only on the first fundamental form. Example. Let 2u σu, v = 1 + u 2 + v, 2v u 2 + v, u2 + v u 2 + v 2 Calculate its Christoffel symbols. Solution. Since σ u u, v = 21 u 2 + v u 2 + v 2, 4uv u 2 + v 2, 4u, u 2 + v 2 2 4

5 4uv σ v u, v = 1 + u 2 + v 2, 21 + u2 v u 2 + v 2, 4v, u 2 + v 2 2 we can calculate the first fundamental form for σ is E = G = u 2 + v 2 2, F = 0. Hence, using or the simplified version wince F = 0, we get, by a direct calculation, Γ 1 11 = Γ 2 12 = Γ 1 2u 21 = 1 + u 2 + v, 2 Γ 2 22 = Γ 1 12 = Γ 2 2v 21 = 1 + u 2 + v, 2 Γ 2 11 = 2v 1 + u 2 + v, 2u 2 Γ1 22 = 1 + u 2 + v. 2 4 Gauss equations, Codazzi-Mainardi equations We recall that, in Calculus III, we have the theorem that f xy = f yx, for any smooth functions f, i.e. it doesn t matter whether we take the partial with respect to x first and then take the partial w.r.t. y or, we take the partial w.r.t. y first and then take the partial w.r.t. x. For σ uu, σ uv, σ vv, we then have, by taking partials again, σ uuv = σ uvu, σ vvu = σ vuv. These relations give equations which will be refereed as Gauss equations and the Codazzi-Mainardi equations. We first consider , i.e. σ uuv = σ uvu. Using and 5.1.3, the equation σ uuv = σ uvu gives Hence, It yields Γ 1 11σ u + Γ 2 11σ v + en v = Γ 1 12σ u + Γ 2 12σ v + fn u. Γ 1 11 v σ u + Γ 1 11σ uv + Γ 2 11 v σ v + Γ 2 11σ vv + e v n + en v = Γ 1 12 u σ u + Γ 1 12σ uu + Γ 2 12 u σ v + Γ 2 12σ uv + f u n + fn u Γ 1 11 v Γ 1 12 u σu + Γ 2 11 v Γ 2 12 u σv + e v f u n = Γ 1 12σ uu + Γ 2 12 Γ 1 11 σuv Γ 2 11σ vv + fn u en v

6 Using to substitute σ uv, to substitute σ vv, 5.1.6, and to substitute n u and n v, we getwe only look at the coefficient before σ u for simplicity reason!, from Γ 1 11 v Γ 1 12 u σu + Γ 2 11 v Γ 2 12 u σv + e v f u n = Γ 1 12Γ Γ 2 12 Γ 1 11Γ 1 12 Γ 2 11Γ fa 11 ea 12 σu + = Γ 2 12Γ 1 12 Γ 2 11Γ f 2 egf σ EG F 2 u By comparing the coefficients before σ u appearing in , we get Since, by 5.1.1, Γ 1 11 v Γ 1 12 u = Γ 2 12Γ 1 12 Γ 2 11Γ f 2 egf EG F 2. K = eg f 2 EG F 2. Hence, the above equation becomes F K = Γ 2 12Γ 1 12 Γ 2 11Γ 1 22 Γ 1 11 v + Γ 1 12 u. Similarly, by comparing the coefficients before σ v appearing in , we get EK = Γ 2 12 u Γ 2 11 v + Γ 1 12Γ Γ 2 12Γ 2 12 Γ 2 11Γ 2 22 Γ 1 11Γ The equations and are called Gauss equations. Similarly, by comparing the coefficients before n appearing in , we obtain e v f u = eγ fγ 2 12 Γ 1 11 gγ 2 11, and, using and comparing the coefficients before n, f v g u = eγ fγ 2 22 Γ 1 12 gγ The equations and are called Codazzi-Mainardi equations. 5 Gauss s remarkable theorem We are now ready to state and prove the following Gauss s remarkable theorem: 6

7 TheoremGauss remarkable theorem Let M be a regular surfaceso EG F 2 0, then EG F 2 K = Γ 2 11 v Γ 2 12 u Γ 1 12Γ 2 11 Γ 2 12Γ Γ 2 11Γ Γ 1 11Γ 2 12 G Γ 2 12Γ 1 12 Γ 2 11Γ 1 22 Γ 1 11 v + Γ 1 12 u F In particular, the Gauss curvature depends only on the first fundamental form. Proof. Since M is a regular surface, σ u σ v 0, in particular, EG F 2 0. Multiplying F to and G to and then add them up, we can obtain the This proves the theorem. Q.E.D allows us to calculate the Gauss curvature in terms of the first fundamental form only. Assume F = 0, then, from the Gauss equations above, we can derive that K = 1 2 Ev Gu +. EG EG EG Example: For the surface of revolution σu, v = fu cos v, fu sin v, gu where f > 0 and f 2 + g 2 = 1. We found that E = 1, F = 0, G = fu 2. Hence, from the above formula, we get K = 1 2 Gu = 1 2 G == f G G G u 2 f. 6 Isometry u Gauss remarkable theorem can be interpreted in a slightly different way in terms of isometries. We discuss this point here briefly and informally. Let M, N be two surfaces. A one-to-one map f : M N that preserves lengths and angles is called an isometry. This may be understood at the level of tangent vectors: f takes curves to curves, and hence velocity vectors to velocity vectors. f is an isometry if and only if f preserves angle between velocity vectors and preserve length of velocity vectors. Theorem Gauss curvature is a bending invariant, i.e. is invariant under isometries, by which we mean: if f : M N is an isometry then K N fp = K M P, 7 v u

8 i.e. the Gauss curvature is the same at corresponding points. Proof: f preserves lengths and angles. Hence, with appropriate parameterizations, the first fundamental forms for M and N are the sameat corresponding points. Since the Gauss curvature pnly depends on the irst fundamental form, Gauss curvature will be the same at corresponding points. Application 1. The cylinder has the Gaussian curvature K = 0 because a plane has zero Gaussian curvature. Applicaition 2. No piece of a plane can be bent to a piece of a sphere without distorting lenghts becuase the Gauss curvature of plane is zero, the Gauss curvature of a sphere is 1/r 2 where r =radius. Theorem Riemann. Let M be a surface with vanishing Gaussian curvature, i.e. K 0, then each P M has a neighborhood which is isometric to an oepn set in the Euclidean place. The proof of Riemann s theorem is omitted here. 7 Recap The Christoffel symbols: Γ 1 11 = GE u 2F F u + F E v, Γ 2 2EG F 2 11 = 2EF u EE v F E u 2EG F 2 Γ 1 12 = Γ 1 21 = GE v F G u 2EG F 2, Γ 1 22 = 2GF v GG u F G v 2EG F 2 The Christoffel symbolsin the case F = 0: Γ2 12 = Γ 2 21 = EG u F E v 2EG F 2, Γ 2 22 = EG v 2F F v + F G u. 2EG F 2 Γ 1 11 = 1 ln E 2 u, Γ2 11 = 1 E 2G v Γ 1 12 = Γ 1 21 = 1 ln E 2 v, Γ2 12 = Γ 2 21 = 1 2 Γ 1 22 = 1 G 2E u, Γ2 22 = 1 ln G 2 v. ln G u 8

9 Gauss equations and Codazzi-Mainardi equations: Gauss equations: F K = Γ 2 12Γ 1 12 Γ 2 11Γ 1 22 Γ 1 11 v + Γ 1 12 u, EK = Γ 2 12 u Γ 2 11 v + Γ 1 12Γ Γ 2 12Γ 2 12 Γ 2 11Γ 2 22 Γ 1 11Γ Codazzi-Mainardi equations: Formula for Gauss curvature: e v f u = eγ fγ 2 12 Γ 1 11 gγ 2 11, f v g u = eγ fγ 2 22 Γ 1 12 gγ EG F 2 K = Γ 2 11 v Γ 2 12 u Γ 1 12Γ 2 11 Γ 2 12Γ Γ 2 11Γ Γ 1 11Γ 2 12 G In particular, when F = 0, then Γ 2 12Γ 1 12 Γ 2 11Γ 1 22 Γ 1 11 v + Γ 1 12 u F. K = 1 2 EG Ev EG v + Gu EG u. 9