Principal Curvatures

Transcription

1 Principal Curvatures Com S 477/577 Notes Yan-Bin Jia Oct 26, Definition To furtheranalyze thenormal curvatureκ n, we make useof the firstandsecond fundamental forms: Edu 2 +2Fdudv +Gdv 2 and Ldu 2 +2Mdudv +Ndv 2. For convenience, we introduce two symmetric matrices E F F 1 = and F F G 2 = L M M N The tangent vector of the unit-speed curve γt = σut,vt is γ = uσ u + vσ v. Introducing T = u, v t, where t denotes the transpose operator, the normal curvature equation can be rewritten as κ n = L u 2 +2M u v +N v 2 κ n = T t F 2 T. 1 Since σ u and σ v span the tangent plane, consider two tangent vectors: We obtain their inner product: ˆt 1 = ξ 1 σ u +η 1 σ v and ˆt 2 = ξ 2 σ u +η 2 σ v. ˆt 1 ˆt 2 = ξ 1 σ u +η 1 σ v ξ 2 σ u +η 2 σ v = Eξ 1 ξ 2 +Fξ 1 η 2 +ξ 2 η 1 +Gη 1 η 2 = T1 t E F T F G 2, 2 where T 1 = ξ1 η 1 and T 2 = ξ2 η The material is adapted from the book Elementary Differential Geometry by Andrew Pressley, Springer-Verlag, 1

2 The principal curvatures of the surface patch σ are the roots of the equation detf 2 κf 1 = L κe M κf M κf N κg = 0. 3 From the linear independence of σ u and σ v, it is easy to show that matrix F 1 is always invertible. Equation 3 essentially states that the principal curvatures are the eigenvalues of F1 1 F 2. Let κ be a principal curvature of F1 1 F 2, and T = ξ,η T the corresponding eigenvector. That F1 1 F 2T = κt implies F 2 κf 1 T = 0. 4 The unit tangent vector ˆt in the direction of ξσ u +ησ v is called the principal vector corresponding to the principal curvature κ. Theorem 1 Let κ 1 and κ 2 be the principal curvatures at a point p of a surface patch σ. Then i κ 1,κ 2 R; ii if κ 1 = κ 2 = κ, then F 2 = κf 1 and every tangent vector at p is a principal vector. iii if κ 1 κ 2, then the two corresponding principal vectors are perpendicular to each other For a proof of the theorem, we refer the reader to [2, ]. Intuitively, the principal vectors give the directions of maximum and minimum bending of the surface at the point, and the principal curvatures measure the bending rates. In case ii, the point is umbilic. In this case, the surface bends the same amount in all directions at p thus all directions are principal. Example 2. A sphere bends the same amount in every direction. Take the unit sphere in Example 9 in the notes Surfaces, for instance, with the parametrization We found previously that σθ,φ = cosθcosφ,cosθsinφ,sinθ, E = 1, F = 0, G = cos 2 θ. Since a sphere is a surface of revolution, we can plug in the result from Example 1, with fθ = cosθ and gu = sinθ: L = f g fġ = sinθ sinθ cosθcosθ = 1, M = 0, N = fġ = cos 2 θ. Hence the principal curvatures are the roots of detf 2 kf 1 = 1 κ 0 0 cos 2 θ κcos 2 θ = 0. Hence κ = 1. And every tangent direction is a principal vector. 2

3 Example 3. Consider a cylinder with the z-axis as its axis and circular cross sections of unit radius. The parametrization is given as ˆn ˆt 1 ˆt 2 p σu,v = cosv,sinv,u. The coefficients of the first and second fundamental forms can be computed as E = 1, F = 0, G = 1, L = 0, M = 0, N = 1. The principal curvatures are roots of 0 κ κ So we obtain κ 1 = 0 and κ 2 = 1. The eigenvectors T i = ξ i,η i, i = 1,2 of F 1 1 F 2 are found from solving the equation F 2 κ i F 1 T i = 0. The results are T 1 = λ 1 1,0 t and T 2 = λ 2 0,1 t for any non-zero λ 1,λ 2 R. Hence the principal vector ˆt 1 is along the direction of 1σ u +0σ v, i.e., ˆt 1 = 0,0,1. The principal vector ˆt 2 is along the direction of 0σ u +1σ v = sinv,cosv,0, i.e., ˆt 2 = sinv,cosv,0. A curve γ on the surface σ is a principal curve if its velocity γ always points in a principal direction, that is, the direction of a principal vector. At every point on a principal curve, the normal curvature is a maximum or minimum. The next figure shows some principal curves on the ellipsoid 2 Euler s Formula x y2 5 +z2 = 1. Supposethetwoprincipalcurvaturesκ 1 κ 2 atponthesurfaceσ. Then by Theorem 1iii, the two corresponding principal vectors ˆt 1 = ξ 1 σ u + η 1 σ v and ˆt 2 = ξ 2 σ u + η 2 σ v must be orthogonal to each other. Denote by T 1 = ξ 1,η 1 t and T 2 = ξ 2,η 2 t. Replace the T in 4 with T j, j = 1,2, multiply both sides of the equation by Ti t to the left, and move the second resulting term to the right hand side of the equation. This yields T t i F 2T j = κ j T t i F 1T j, i,j = 1,2. Meanwhile, the orthogonality of the two principal vectors implies that To summarize, we have ˆt 1 ˆt 2 = T t 1 F 1T 2 = 0, from 2. { TiF t 2 T j = κ i TiF t κi, if i = j, 1 T j = 0, otherwise With the principal curvatures and vectors at p, we can evaluate the normal curvature in any direction. 3 5

4 Theorem 2 Let κ 1,κ 2 be the principal curvatures, and ˆt 1,ˆt 2 the two corresponding principal vectors of a patch σ at p. The normal curvature of σ in the direction û = cosθˆt 1 +sinθˆt 2 is κ n = κ 1 cos 2 θ+κ 2 sin 2 θ. Proof Let û = ξσ u + ησ v and T = ξ,η t. We first look at the special case κ 1 = κ 2 = κ. By Theorem 1ii, û = ξσ u +ησ v is a principal vector. The normal curvature in the direction û is κ n = T t F 2 T by 1 = κt t F 1 T by 4 = κû û = κ. 6 Meanwhile, we have κ 1 cos 2 θ+κ 2 sin 2 θ = κcos 2 θ+sin 2 θ = κ. So the theorem holds when the point is umbilic. Assume κ 1 κ 2. Therefore by Theorem 1iii, ˆt 1 and ˆt 2 are perpendicular to each other. Let ˆt i = ξ i σ u +η i σ v, and T i = ξ i,η i t. Thus, So we have û = ξσ u +ησ v, where û = cosθξ 1 σ u +η 1 σ v +sinθξ 2 σ u +η 2 σ v = ξ 1 cosθ +ξ 2 sinθσ u +η 1 cosθ+η 2 sinθσ v. ξ = ξ 1 cosθ+ξ 2 sinθ, η = η 1 cosθ+η 2 sinθ, The above is written succinctly as T = cosθt 1 +sinθt 2. By equation 1 the normal curvature in the û direction is κ n = cosθt t 1 +sinθtt 2 F 2cosθT 1 +sinθt 2 = cos 2 θt t 1 F 2T 1 +cosθsinθt t 1 F 2T 2 +T t 2 F 2T 1 +sin 2 θt t 2 F 2T 2 = κ 1 cos 2 θ+κ 2 sin 2 θ. The last step above followed from the equation 5. Theorem 2 implies that κ 1 and κ 2 are the maximum and minimum of any normal curvatures at the point. Equivalently, among all tangent directions at the point, the geometry varies the most in one principal direction while the least in the other. 4

5 3 Geometric Interpretation of Principal Curvatures In this section, we look at how the local shape at a surface point can be approximated using its principal curvatures and direction. The values of the principal curvatures and vectors at a point p on a surface patch σ tell us about the shape near p. To see this, we apply a rigid motion followed by a reparametrization. 1 More specifically, we move the origin to p and let the tangent plane to σ at p be the xy-plane with the x-axis and y-axis along the directions of the two principal vectors, which correspond to principal curvatures κ 1 and κ 2, respectively. Furthermore, we let the values of both parameters at the origin be zero, that is, σ0,0 = 0. 7 Without any ambiguity, we still denote the new parametrization by σ. Let us determine the function z = zx,y that describes the local shape. The unit principal vectors can be expressed in terms of the partial derivatives: So can any point x,y,0 in the tangent plane: where 1,0,0 = ξ 1 σ u +η 1 σ v, 0,1,0 = ξ 2 σ u +η 2 σ v. x,y,0 = x1,0,0 +y0,1,0 = xξ 1 σ u +η 1 σ v +yξ 2 σ u +η 2 σ v = sσ u +tσ v, 8 s = xξ 1 +yξ 2 and t = xη 1 +yη 2. 9 Let us evaluate σs,t at the parameter values s and t, applying Taylor s theorem with higher order terms in s and t neglected: σs,t = σ0,0+sσ u +tσ v s2 σ uu +2stσ uv +t 2 σ vv = x,y, s2 σ uu +2stσ uv +t 2 σ vv, by 7 and 8 All derivatives are evaluated at the origin p. Neglecting the second order terms added to x and y, the coordinates of σs,t is x,y,z, where z = σs,t ˆn = 1 2 Ls2 +2Mst+Nt 2 = 1 2 s t L M M N s. t 1 The shape does not change under any rigid motion or reparametrization. 5

6 z Writing T 1 = ξ1 η 1 and T 2 = ξ2 η 2, x σ p u 1 u 2 y we have from 9: s = xt 1 +yt 2. t Thus, z = 1 2 xt 1 +yt 2 t F 2 xt 1 +yt 2 = 1 x 2 T 1 F 2 T 1 +xyt t 2 1F 2 T 2 +T2F t 2 T 1 +y 2 T2F t 2 T 2 = 1 2 κ 1x 2 +κ 2 y 2, since T t i F 2T j = κ i if i = j or 0 otherwise. Hence the shape of a surface near the point p has a quadratic approximation determined by its principal curvature κ 1 and κ 2. It is described by the equation z = 1 2 κ 1x 2 +κ 2 y 2. 4 Covariant Derivative Next, welook at how to characterizes therate of change of a vector definedon asurfacewith respect to a tangent vector. Let us slightly abuse the notation ˆn to represent a function that assigns to every point p on the surface S the normal ˆnp at the point. Since ˆn is continuous, it is a vector field on S, and referred to as the normal vector field. Similarly, ˆt 1 and ˆt 2 are also vector fields on S that continuously assign to every point two orthogonal principal vectors. At the point p, a vector field Z typically changes differently in different tangential directions. The rate of change along a tangent w is characterized by its covariant derivative along w. More specifically, we let αt be a curve on S that has initial velocity α 0 = w. Consider restriction of Z to α. Then, the covariant derivative of Z with respect to w is defined to be w Z = dzαt dt. t=0 In particular, consider the u-curve αu = σu,v 0 passing through p = σu 0,v 0 at velocity w = σ u u 0,v 0. We have w Z = dzαu du u=u0 = dzσu,v 0 du u=u0 = Z u u 0,v 0. Reparametrize αu as a unit-speed curve βs, where s is arc length. Clearly, ds du 0 = α u 0 = σ u u 0,v 0. 6

7 At p, let ˆx = β 0 = σ u u 0,v 0 / σ u u 0,v 0 be the unit velocity of the u-curve. The covariant derivative with respect to ˆx is ˆx Z = dzβs ds s=0 = dzαus/du ds/du = Z uu 0,v 0 σ u u 0,v 0. u=u0 In the Darboux frame T-V-U at p of a unit-speed surface curve, where T is the curve tangent, U the unit surface normal ˆn, and V = U T, it holds that U = κ n T τ g V, where κ n and τ g are the surface s normal curvature and curve s geodesic torsion at p. Meanwhile, U is the covariant derivative along T, i.e., U = T U. The normal curvature at the point in the direction T is equivalently defined to be k n T [1, p. 196], for we have κ n T = U T = T U T. The principal curvatures are the normal curvatures in the two principal directions, that is, the covariant derivatives of the normal with respect to the principal vectors: κ 1 = κ n ˆt 1 = ˆt 1 U ˆt 1 = ˆt 1ˆn ˆt 1, κ 2 = κ n ˆt 2 = ˆt 2 U ˆt 2 = ˆt 2ˆn ˆt 2. It can be shown that ˆt iˆn ˆt j = 0 if i j. References [1] B. O Neill. Elementary Differential Geometry. Academic Press, Inc., [2] A. Pressley. Elementary Differential Geometry. Springer-Verlag London,