Totally quasi-umbilic timelike surfaces in R 1,2
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1 Totally quasi-umbilic timelike surfaces in R 1,2 Jeanne N. Clelland, University of Colorado AMS Central Section Meeting Macalester College April 11, 2010
2 Definition: Three-dimensional Minkowski space R 1,2 is the vector space R 3 with coordinates x = (x 0, x 1, x 2 ) endowed with the indefinite inner product x, x = (x 0 ) 2 (x 1 ) 2 (x 2 ) 2.
3 Definition: Three-dimensional Minkowski space R 1,2 is the vector space R 3 with coordinates x = (x 0, x 1, x 2 ) endowed with the indefinite inner product x, x = (x 0 ) 2 (x 1 ) 2 (x 2 ) 2. A nonzero vector x R 1,2 is called: spacelike if x, x < 0; timelike if x, x > 0; lightlike (or null) if x, x = 0.
4 Figure: Spacelike sphere, timelike sphere, and light cone
5 A regular surface Σ R 1,2 is called: spacelike if the restriction of, to each tangent plane T x Σ is negative definite; timelike if the restriction of, to each tangent plane T x Σ is indefinite; lightlike if the restriction of, to each tangent plane T x Σ is degenerate.
6 Consider a timelike surface Σ R 1,2. We can use moving frames to compute invariants of timelike surfaces under the action of the Minkowski isometry group, which consists of all transformations of the form ϕ(x) = Ax + b, where A O(1, 2) and b R 1,2.
7 Consider a timelike surface Σ R 1,2. We can use moving frames to compute invariants of timelike surfaces under the action of the Minkowski isometry group, which consists of all transformations of the form ϕ(x) = Ax + b, where A O(1, 2) and b R 1,2. An adapted orthonormal frame at a point x Σ consists of three mutually orthogonal vectors (e 0, e 1, e 2 ) such that: e 0, e 0 = 1, e 1, e 1 = e 2, e 2 = 1; e 0, e 1 span the tangent plane T x Σ.
8 As in the Euclidean case, the normal vector field e 2 of an adapted orthonormal frame along Σ defines the Gauss map from Σ to the sphere S 1, and the differential de 2 of the Gauss map is a self-adjoint linear operator d(e 2 ) x : T x Σ T x Σ.
9 As in the Euclidean case, the normal vector field e 2 of an adapted orthonormal frame along Σ defines the Gauss map from Σ to the sphere S 1, and the differential de 2 of the Gauss map is a self-adjoint linear operator d(e 2 ) x : T x Σ T x Σ. The Gauss curvature K and mean curvature H at any point x Σ are defined to be: K(x) = det (d(e 2 ) x ), H(x) = 1 2 tr (d(e 2) x ).
10 But here things start to look different from the Euclidean case: the matrix representation of the self-adjoint linear operator d(e 2 ) x with respect to an orthonormal basis for the indefinite metric on T x Σ is skew-symmetric, rather than symmetric. This means that:
11 But here things start to look different from the Euclidean case: the matrix representation of the self-adjoint linear operator d(e 2 ) x with respect to an orthonormal basis for the indefinite metric on T x Σ is skew-symmetric, rather than symmetric. This means that: The quantity H 2 K can take on any real value. (In the Euclidean case, H 2 K 0.)
12 But here things start to look different from the Euclidean case: the matrix representation of the self-adjoint linear operator d(e 2 ) x with respect to an orthonormal basis for the indefinite metric on T x Σ is skew-symmetric, rather than symmetric. This means that: The quantity H 2 K can take on any real value. (In the Euclidean case, H 2 K 0.) If H 2 K > 0, then d(e 2 ) x has distinct real eigenvalues and an orthonormal basis of eigenvectors.
13 But here things start to look different from the Euclidean case: the matrix representation of the self-adjoint linear operator d(e 2 ) x with respect to an orthonormal basis for the indefinite metric on T x Σ is skew-symmetric, rather than symmetric. This means that: The quantity H 2 K can take on any real value. (In the Euclidean case, H 2 K 0.) If H 2 K > 0, then d(e 2 ) x has distinct real eigenvalues and an orthonormal basis of eigenvectors. If H 2 K < 0, then d(e 2 ) x has complex eigenvalues and no real eigenvectors.
14 But here things start to look different from the Euclidean case: the matrix representation of the self-adjoint linear operator d(e 2 ) x with respect to an orthonormal basis for the indefinite metric on T x Σ is skew-symmetric, rather than symmetric. This means that: The quantity H 2 K can take on any real value. (In the Euclidean case, H 2 K 0.) If H 2 K > 0, then d(e 2 ) x has distinct real eigenvalues and an orthonormal basis of eigenvectors. If H 2 K < 0, then d(e 2 ) x has complex eigenvalues and no real eigenvectors. If H 2 K = 0, then d(e 2 ) x has a repeated real eigenvalue, and it may have either a 1- or a 2-dimensional real eigenspace.
15 A point x Σ where H 2 K = 0 and d(e 2 ) x has a 2-dimensional eigenspace is called an umbilic point. Umbilic points are precisely those points where the second fundamental form II x is a multiple of the first fundamental form I x.
16 A point x Σ where H 2 K = 0 and d(e 2 ) x has a 2-dimensional eigenspace is called an umbilic point. Umbilic points are precisely those points where the second fundamental form II x is a multiple of the first fundamental form I x. Theorem: Suppose that every point of a timelike surface Σ R 1,2 is umbilic. (Such a surface is called totally umbilic.) Then Σ is contained in either a plane or a hyperboloid of one sheet.
17 Definition: A point x Σ where H 2 K = 0 and d(e 2 ) x has a 1-dimensional eigenspace will be called a quasi-umbilic point of Σ.
18 Definition: A point x Σ where H 2 K = 0 and d(e 2 ) x has a 1-dimensional eigenspace will be called a quasi-umbilic point of Σ. Problem: Classify the totally quasi-umbilic timelike surfaces in R 1,2.
19 Definition: A point x Σ where H 2 K = 0 and d(e 2 ) x has a 1-dimensional eigenspace will be called a quasi-umbilic point of Σ. Problem: Classify the totally quasi-umbilic timelike surfaces in R 1,2. We will approach this problem via the method of moving frames. It turns out to be more convenient to use null frames rather than orthonormal frames.
20 An adapted null frame (f 1 (x), f 2 (x), f 3 (x)) along a timelike surface Σ R 1,2 is a basis for the tangent space T x R 1,2 with the properties that: f 1 (x), f 2 (x) are null vectors (i.e., f 1 (x), f 1 (x) = f 2 (x), f 2 (x) = 0) which span the tangent space T x Σ at each point x Σ; f 1 (x), f 2 (x) = 1; f 3 (x) is orthogonal to T x Σ at each point x Σ, with f 3 (x), f 3 (x) = 1.
21 For instance, if (e 0 (x), e 1 (x), e 2 (x)) is an adapted orthonormal frame at x Σ, then the vectors f 1 (x) = 1 2 (e 0 (x) + e 1 (x)), f 2 (x) = 1 2 (e 0 (x) e 1 (x)), f 3 (x) = e 2 (x) form an adapted null frame at x.
22 Let η i, ηj i, 1 i, j 3 be the Maurer-Cartan forms associated to an adapted null frame along Σ. These forms are defined by the equations: dx = f i η i df i = f j η j i, and they satisfy the structure equations dη i = η i j η j dη i j = η i k ηk j.
23 Differentiating the inner product relations f 1, f 1 = 0, f 2, f 2 = 0, f 3, f 3 = 1, f 1, f 2 = 1, f 1, f 3 = 0, f 2, f 3 = 0 yields the following relations among the Maurer-Cartan forms: η2 1 = η1 2 = η3 3 = 0, η2 2 = η1, 1 η3 1 = η2, 3 η3 2 = η1. 3
24 From the equation dx = f 1 η 1 + f 2 η 2 + f 3 η 3 and the fact that dx takes values in T x Σ, it follows that η 3 = 0. Moreover, η 1 and η 2 are linearly independent 1-forms which form a basis for T xσ at each point x Σ.
25 Differentiating the equation η 3 = 0 yields: 0 = dη 3 = η 3 1 η 1 η 3 2 η 2. By Cartan s Lemma, there exist functions k ij = k ji on Σ such that [ ] [ ] [ ] η 3 1 k11 k 12 η 1 =. k 12 k 22 η 2 η 3 2
26 The differential of the Gauss map is df 3 = f 1 η f 2 η 2 3 = f 1 η f 2 η 3 1 = f 1 (k 12 η 1 + k 22 η 2 ) f 2 (k 11 η 1 + k 12 η 2 ).
27 The differential of the Gauss map is df 3 = f 1 η f 2 η 2 3 = f 1 η f 2 η 3 1 = f 1 (k 12 η 1 + k 22 η 2 ) f 2 (k 11 η 1 + k 12 η 2 ). Therefore, the matrix of df 3 with respect to the basis (f 1, f 2 ) for T x Σ is [ ] k12 k 22, k 11 k 12 and we have H = k 12, K = k 2 12 k 11 k 12, H 2 K = k 11 k 22.
28 Quasi-umbilic points are those points where k 11 k 22 = 0, but k 11, k 22 are not both zero. WLOG, we may assume that k 22 = 0, k 11 0.
29 Given one adapted null frame (f 1, f 2, f 3 ) on Σ, any other adapted null frame on Σ has the form e θ 0 0 ] [ ] [ f1 f2 f3 = f1 f 2 f 3 0 e θ ±1 for some function θ on Σ. Under such a transformation we have ] [ [ k11 k12 e 2θ ] k 11 k 12 = ±. k22 k 12 e 2θ k 22 k 12 Therefore, we can find an adapted null frame with k 11 = 1.
30 For such a frame, the first and second fundamental forms are I = dx, dx = 2η 1 η 2, II = df 3, dx = (η 3 1η 1 + η 3 2η 2 ) = (η 1 ) 2 + 2Hη 1 η 2.
31 For such a frame, the first and second fundamental forms are I = dx, dx = 2η 1 η 2, II = df 3, dx = (η 3 1η 1 + η 3 2η 2 ) = (η 1 ) 2 + 2Hη 1 η 2. Now let u, v be local null coordinates on Σ. There exist functions f, g on Σ such that η 1 = e f du, η 2 = e g dv, and therefore I = 2e f+g du dv, II = e 2f du 2 + 2He f+g du dv.
32 The associated Maurer-Cartan forms are: η 3 = η 1 2 = η 2 1 = η 3 3 = 0, η 1 = e f du, η 2 = e g dv, η 3 1 = η 2 3 = (e f du + He g dv), η 3 2 = η 3 1 = He f du, η 1 1 = η 2 2 = g u du f v dv. (The last equation comes from the structure equations for dη 1 and dη 2.)
33 The remaining structure equations imply that the functions H, f, g satisfy the PDE system: H v = 0, (1) H u = 2f v e f g, (2) (f + g) uv = H 2 e f+g. (3)
34 The remaining structure equations imply that the functions H, f, g satisfy the PDE system: H v = 0, (1) H u = 2f v e f g, (2) (f + g) uv = H 2 e f+g. (3) Equation (1) implies that H is a function of u alone. We will divide into cases based on whether H(u) is zero or nonzero.
35 Case 1: H(u) 0. In this case equations (2) and (3) reduce to: f v = 0, (f + g) uv = 0.
36 Case 1: H(u) 0. In this case equations (2) and (3) reduce to: f v = 0, (f + g) uv = 0. Thus f = f(u) is a function of u alone, and f + g = φ(u) + ψ(v) for some functions φ(u), ψ(v).
37 The first and second fundamental forms now take the form I = 2e φ(u) e ψ(v) du dv, II = e 2f(u) du 2.
38 The first and second fundamental forms now take the form I = 2e φ(u) e ψ(v) du dv, II = e 2f(u) du 2. By making a change of coordinates (u, v) (ũ, ṽ) with dũ = e φ(u) du, dṽ = e ψ(v) dv we can assume that φ(u) = ψ(v) = 0, and so I = 2 du dv, II = e 2f(u) du 2.
39 Case 2: H(u) 0. Let h(u) = ln H(u). Then we can write equation (3) as (f + g) uv = e f+g+2h(u), or equivalently, (f + g + 2h(u)) uv = e f+g+2h(u).
40 Case 2: H(u) 0. Let h(u) = ln H(u). Then we can write equation (3) as (f + g) uv = e f+g+2h(u), or equivalently, (f + g + 2h(u)) uv = e f+g+2h(u). Thus the function z(u, v) = f(u, v) + g(u, v) + 2h(u) is a solution to Liouville s equation z uv = e z.
41 The general solution to Liouville s equation is ( 2φ (u)ψ ) (v) z(u, v) = ln (φ(u) + ψ(v)) 2, where φ(u), ψ(v) are arbitrary functions with φ (u), ψ (v) 0.
42 The general solution to Liouville s equation is ( 2φ (u)ψ ) (v) z(u, v) = ln (φ(u) + ψ(v)) 2, where φ(u), ψ(v) are arbitrary functions with φ (u), ψ (v) 0. By making the change of variables ũ = φ(u), ṽ = ψ(v), we can assume that ( ) 2 z(u, v) = ln (u + v) 2.
43 The general solution to Liouville s equation is ( 2φ (u)ψ ) (v) z(u, v) = ln (φ(u) + ψ(v)) 2, where φ(u), ψ(v) are arbitrary functions with φ (u), ψ (v) 0. By making the change of variables ũ = φ(u), ṽ = ψ(v), we can assume that ( ) 2 z(u, v) = ln (u + v) 2. Thus we have ( 2 f + g = ln (u + v) ( 2 = ln ) 2h(u) 2 (H(u)) 2 (u + v) 2 ). (4)
44 Now we can write equation (2) as H (u) = 2f v e 2f e (f+g) = (e2f ) v (H(u)) 2 (f + g) uv (from equation (3)); therefore, H (u) (H(u)) 2 (f + g) uv = (e 2f ) v.
45 Now we can write equation (2) as therefore, H (u) = 2f v e 2f e (f+g) Integrate with respect to v: for some function k(u). = (e2f ) v (H(u)) 2 (f + g) uv (from equation (3)); H (u) (H(u)) 2 (f + g) uv = (e 2f ) v. H (u) (H(u)) 2 (f + g) u = (e 2f ) + k(u)
46 But from equation (4), we have ( H (u) (f + g) u = 2 H(u) + 1 ). u + v
47 But from equation (4), we have ( H (u) (f + g) u = 2 H(u) + 1 ). u + v Therefore, e 2f = H (u) (H(u)) 2 (f + g) u k(u) ( = 2H (u) H (u) (H(u)) 2 H(u) + 1 ) k(u) u + v 2H (u) = (H(u)) 2 (u + v) k(u), where k(u) = k(u) + 2 (H (u)) 2 (H(u)) 3.
48 The first and second fundamental forms now take the form I = 4 (H(u)) 2 du dv (u + v) 2 ( 2H ) (u) II = (H(u)) 2 (u + v) + k(u) du du dv. H(u)(u + v) 2
49 In order to find the corresponding parametrization x(u, v) for Σ, we have to solve the compatible, overdetermined PDE system given by the Maurer-Cartan equations: [ ] [ ] η 1 η1 1 η2 1 η 1 3 dx df1 df 2 df 3 = x f1 f 2 f 3. (5) η 2 η1 2 η2 2 η3 2 η 3 η 3 1 η 3 2 η 3 3
50 Case 1: H(u) = 0. In this case the Maurer-Cartan equations are equivalent to the PDE system: [ x f1 f 2 f 3 ]u = [ ] e f(u) f (u) 0 0 x f 1 f 2 f f (u) e f(u), 0 e f(u) [ x f1 f 2 f 3 ]v = [ ] x f 1 f 2 f 3. e f(u)
51 These equations can be integrated explicitly, and the general solution is x(u, v) =x 0 + ue f(0) f [ u 0 [ ( u1 0 v u ( u1 0 e 2f(u 2) du 2 ) du 1 ] f 0 3, 0 e 2f(u 2) du 2 ) 2 du 1 ] e f(0) f 0 2 where [ x(0, 0) f1 (0, 0) f 2 (0, 0) f 3 (0, 0) ] = [ x 0 f1 0 f2 0 f3 0 ].
52 Thus Σ is a cylindrical surface of the form x(u, v) = α(u) + vf 0 2, where the v-coordinate curves are null lines parallel to the null vector f2 0, and the generating curve α(u) is the null curve [ α(u) = x 0 + ue f(0) f [ u 0 ( u u ( u1 0 0 e 2f(u 2) du 2 ) du 1 ] f 0 3. e 2f(u 2) du 2 ) 2 du 1 ] e f(0) f 0 2
53 Thus Σ is a cylindrical surface of the form x(u, v) = α(u) + vf 0 2, where the v-coordinate curves are null lines parallel to the null vector f2 0, and the generating curve α(u) is the null curve [ α(u) = x 0 + ue f(0) f [ u 0 ( u u ( u1 0 0 e 2f(u 2) du 2 ) du 1 ] f 0 3. e 2f(u 2) du 2 ) 2 du 1 ] e f(0) f 0 2 We could reparametrize Σ (not by null coordinates) to arrange that [ u ( u1 ) ] α(u) = x 0 + ue f(0) f1 0 e 2f(u2) du 2 du 1 f3 0 ; then α is essentially the graph of a function F (u) with F (u) <
54 Case 2: H(u) 0. In this case the Maurer-Cartan equations must be integrated numerically in most cases.
55 Case 2: H(u) 0. In this case the Maurer-Cartan equations must be integrated numerically in most cases. However, it can be shown that there exists a parametrization x for Σ (not by null coordinates) of the form where: x(u, v) = α(u) + vf 2 (u), the generating curve α(u) is a null curve; the vectors f 1 (u) = α (u), f 2 (u) are linearly independent null vectors for all u; and f 1 (u), f 1 (u) are linearly independent for all u.
56 Case 2: H(u) 0. In this case the Maurer-Cartan equations must be integrated numerically in most cases. However, it can be shown that there exists a parametrization x for Σ (not by null coordinates) of the form where: x(u, v) = α(u) + vf 2 (u), the generating curve α(u) is a null curve; the vectors f 1 (u) = α (u), f 2 (u) are linearly independent null vectors for all u; and f 1 (u), f 1 (u) are linearly independent for all u. Therefore Σ is a ruled surface whose rulings are null lines. Moreover, every such surface is quasi-umbilic, possibly containing a curve of umbilic points.
57 Combining both cases, we have: Theorem: Every totally quasi-umbilic, timelike regular surface Σ R 1,2 is a ruled surface of the form where: α(u) is a null curve; x(u, v) = α(u) + vf 2 (u), the vectors f 1 (u) = α (u), f 2 (u) are linearly independent null vectors for all u; f 1 (u), f 1 (u) are linearly independent for all u. Moreover, every regular timelike surface Σ R 1,2 of this form is quasi-umbilic, possibly containing a curve of umbilic points. The mean curvature H(u) of Σ is zero if and only if the vectors f 2 (u), f 2 (u) are linearly dependent.
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