The First Fundamental Form

Transcription

1 The First Fundamental Form Outline 1. Bilinear Forms Let V be a vector space. A bilinear form on V is a function V V R that is linear in each variable separately. That is,, is bilinear if αu + β v, w αu, w + β v, w and u, αv + β w αu, v + β u, w for all u, v, w V and α, β R. Any bilinear form on R n can be written as n v, w P ij v i w j P 11 v 1 w 1 + P 12 v 1 w P nn v n w n. i,j1 This equation can be written as v, w v T P w. where P is an n n matrix of coefficients. The entries of P can be obtained by applying the bilinear form to the basis vectors: e 1, e 1 e 1, e 2 e 1, e n. e P 2, e e n, e 1 e n, e n In a similar way, we can define the matrix for any bilinear form on a vector space with respect to a given basis. 2. Inner Products Let V be a vector space. An inner product on V is a bilinear form, with the following properties: 1. Positive Definite: v, v > for all nonzero v V. 2. Symmetric: v, w w, v for all v, w V. For example, the dot product v, w v 1 w v n w n is an inner product on R n. More generally, a bilinear form v, w v T P w is an inner product if and only if the matrix P is symmetric and positive definite (see below). A vector space V together with an inner product, on V is called an inner product space. Every inner product space has an associated norm, defined by the formula v v, v. Thus we can talk about the lengths of vectors in an inner product space. We can also talk about the angle θ between nonzero vectors, defined by the formula v, w v w cos θ.

2 3. Positive Definite Matrices A symmetric n n matrix P is positive definite if it satisfies either of the following equivalent conditions: All the eigenvalues of P are positive. v T P v > for every nonzero vector v R n. Positive definite matrices can be used to define inner products on R n (see above). A 2 2 symmetric matrix is positive definite if and only if its trace and determinant are both positive. For a larger matrix, the following test can be used: Sylvester s Criterion. A symmetric matrix P is positive definite if and only if all of the leading principal minors of P are positive. Here the phrase leading principal minor refers to the determinant of any square submatrix lying in the upper-left corner of P. An n n matrix has n leading principal minors, the first of which is the upper-left entry, and the last of which is the determinant of the entire matrix. 4. The First Fundamental Form Let S be a smooth surface, and let p S. The first fundamental form of S at p is the inner product, p on T p S obtained by restricting the dot product on R 3 to pairs of tangent vectors. Given a surface patch σ: U S, the matrix for the first fundamental form with respect to the basis {σ u, σ v } is [ ] E F where E σ u σ u, F σ u σ v, and G σ v σ v. Given any pair of tangent vectors v and w at the same point satisfying F v v 1 σ u + v 2 σ v and w w 1 σ u + w 2 σ v The dot product of v and w is given by the formula v w [ ] [ ] [ ] E F w1 v 1 v 2. F G Using the first fundamental form, the infinitesimal distance ds between two nearby points σ(u, v) and σ(u + du, v + dv) on the surface satisfies the equation G w 2 ds 2 E du 2 + 2F dudv + Gdv 2. The expression on the right is classically referred to as the first fundamental form. Its integral gives the arc length of a path σ ( u(t), v(t) ) on the surface: E s u2 + 2F u v + G v 2 dt.

3 Practice Problems 1. Let V be a vector space, and let v, w V. Let, be a bilinear form on V, and suppose that v, v 2, v, w 4, w, v 3, and w, w 5. Compute v + w, 2v w. 2. Determine which of the following matrices are positive definite. [ ] [ ] Let, be the inner product on R 2 corresponding to the matrix [ ] (a) Find the length of the vector (1, 5) with respect to this inner product. (b) Find the angle between the vectors (1, ) and (1, 5) with respect to this inner product. (c) Find a nonzero vector orthogonal to (1, ) with respect to this inner product. 4. Let, be an inner product on R 2, and suppose that (4, ), (1, ) 12, (, 1), (, 1) 6, and (, 1), (1, 2) 14. Find the matrix for,. 5. Let S be a smooth surface, let σ: ( 2, 2) R S be a surface patch, and suppose that the first fundamental form of S is Let γ: [, 1] S be the curve γ(t) σ(t, t). ds 2 du 2 + ududv + dv 2. (a) Find the speed of γ at t and at t 1. (b) Find the length of γ from t to t 1. (c) Let γ: [, 1] S be the parameter curve γ(t) σ(t, 1). Find the angle between γ and γ at the point σ(1, 1). 6. Let σ: R 2 S be the surface patch σ(u, v) ( u, v, 1 2 (u2 v 2 ) ). (a) Compute the first fundamental form of S with respect to the basis {σ u, σ v }. (b) Let γ: [, 1] S be the curve γ(t) σ(sinh t, sinh t). Use the first fundamental form to compute the length of γ.

4 Solutions 1. Using bilinearity, v + w, 2v w v, 2v w + w, 2v w 2v, v v, w + 2w, v w, w 2(2) (4) + 2(3) (5) 1 2. (a) The first matrix is positive definite, since the diagonal entries and the determinant are both positive. (b) The second matrix is not positive definite, since the determinant is 1. (c) The third matrix is not positive definite, since <. 1 2 (d) The fourth matrix is positive definite, since 1 > and > and >. 3. Let v (1, 5), let w (1, ), and let M be the given matrix. (a) We have v v, v v T Mv (b) Since w 5 and v, w 2, we have Solving gives cos θ 1/ 2, so θ 45. (2) (4 1)( 5) cos θ. (c) Let u (u 1, u 2 ) be the desired vector. Then u T Mw, which reduces to the equation 5u 1 + 3u 2. Hence u (3, 5) is one such vector. 4. e 1 (1, ) and e 2 (, 1). The first equation tells us that e 1, e 1 3, and the second says that e 2, e 2 6. The third can be written e 2, e 1 + 2e 2, e 2 14, and it follows that [ ] 3 2 e 1, e 2 2. Thus, the matrix is This is the curve (u, v) (t, t), so u v 1, and therefore γ u 2 + u u v + v 2 (1) 2 + (t)(1)(1) + (1) 2 t + 2. (a) Using the formula above, γ() 2 and γ(1) 3. (b) The length is γ dt t + 2 dt 3.

5 (c) The matrix for the first fundamental form at the point σ(1, 1) is [ ] [ ] 1 u/2 1 1/2 I u/2 1 1/2 1 The two tangent vectors are (1, 1) and (1, ). We compute the norms and inner product: (1, 1) 3, (1, ) 1, and (1, 1), (1, ) 3/2. This gives us (3/2) ( 3)(1) cos θ. Then cos θ 3/2, so θ (a) We have σ u (1,, u) and σ v (, 1, v), so [ ] σu σ u σ u σ v I σ v σ u σ v σ v [ 1 + u 2 uv uv 1 + v 2 ] That is, ds 2 (1 + u 2 )du 2 2uvdudv + (1 + v 2 )dv 2. (b) The length is (1 + u2 ) u 2 2uv u v + (1 + v 2 ) v 2 dt (1 + sinh 2 t) cosh 2 t 2 sinh 2 t cosh 2 t + (1 + sinh 2 t) cosh 2 t dt 2 cosh 2 t dt 2 cosh t dt 2 sinh(1)