Pullbacks, Isometries & Conformal Maps

Transcription

1 Pullbacks, Isometries & Conformal Maps Outline 1. Pullbacks Let V and W be vector spaces, and let T : V W be an injective linear transformation. Given an inner product, on W, the pullback of, is the inner product T, on V defined by T v, v = T (v), T (v ). In terms of matrices, if T is defined by a matrix A, and, is defined by a matrix P, then the pullback T, is defined by the matrix A T P A. In particular, the pullback of the dot product corresponds to the matrix A T A. If σ: U R 3 is a surface patch, the first fundamental form at a point p U can be thought of as the pullback of the dot product on R 3 under the derivative D p σ. In particular, the matrix I for the first fundamental form is given by the formula I = (Dσ) T (Dσ). 2. Isometric Linear Transformations Now suppose that V and W are inner product spaces, with inner products, V, W. A linear transformation T : V W is said to be isometric if T (v), T (v ) = v, W v V and for all v, v V. Here are a few equivalent definitions of isometric: 1. T is isometric if and only if T (v) = v for all v V. 2. T is isometric if and only if the pullback T, W of the inner product on W is equal to the inner product, V on V. 3. Let P V and P W be the matrices for, V and, W with respect to a certain pair of bases, and let A be the matrix for T. Then T is isometric if and only if A T P W A = P V. (This is simply the matrix form of the previous definition.) 4. Let b 1,..., b n be a basis for V. Then T is isometric if and only if T (bi ), T (b j ) W = b i, b j V for all i and j. 5. Let e 1,..., e n be an orthonormal basis for V. Then T is isometric if and only if the vectors T (e 1 ),..., T (e n ) are orthonormal in W.

2 3. Isometries A smooth map f : S 1 S 2 between two surfaces is called a local isometry if it preserves the lengths of curves. That is, f is a local isometry if, for each curve γ on S 1, the image curve f γ on S 2 has the same length as γ. A local isometry f : S 1 S 2 is called an isometry if it is bijective. In this case, the surfaces S 1 and S 2 are said to be isometric. The following theorem characterizes local isometries: Theorem. A smooth map f : S 1 S 2 is a local isometry if and only if the derivative D p f : T p S 1 T f(p) S 2 at each point p S 1 is an isometric linear transformation. In light of this theorem, we can use any one of the following criteria to determine whether a map is a local isometry: 1. A map f is a local isometry if and only if D p f(t) = t for each point p S 1 and each tangent vector t T p S A map f is a local isometry if and only if (D p f), f(p) =, p for each point p S 1. That is, f is a local isometry if and only if the pullback under Df of the first fundamental form on S 2 is the first fundamental form on S A map f is a local isometry if and only if (Df) T I 2 (Df) = I 1, where I 1 and I 2 are the matrices for the first fundamental forms on S 1 and S Let t 1, t 2 be a basis of tangent vectors at each point of S 1. Then f is a local isometry if and only if Df(t 1 ) = t 1 and Df(t 2 ) = t 2 and Df(t 1 ), Df(t 2 ) = t 1, t In particular, suppose that t 1, t 2 is a basis of orthonormal tangent vectors at each point of S 1. Then f is a local isometry if and only if Df(t 1 ) and Df(t 2 ) are orthonormal. There is a special trick that can be used to simplify condition (3) further. Given a surface patch σ 1 : U S 1, the composition σ 2 = f σ can be thought of as a surface patch U S 2. With respect to these surface patches, the 2 2 derivative matrix D f is simply the identity matrix, so f is a local isometry if and only if matrices I 1 and I 2 for the first fundamental forms are equal. Conversely, suppose that σ 1 : U S 1 and σ 2 : U S 2 are surface patches with the same domain. If the matrices I 1 and I 2 for the first fundamental forms are equal, then the images σ 1 (U) and σ 2 (U) must be isometric, with the isometry being the composition σ 2 σ Conformal Linear Transformations Let V and W be inner product spaces, with inner products, V and, W. A linear transformation T : V W is said to be conformal if it preserves the angles between vectors. The following theorem characterizes conformal transformations:

3 Theorem. A linear transformation T : V W is conformal if and only if there exists a scalar λ > 0 so that T (v), T (v ) = λv, v for all v, v V. That is, a conformal linear transformation must be the composition of an isometric transformation and a dilation. Here are a few equivalent definitions of conformal: 1. T is conformal if and only T (v) and T (v ) are orthogonal for each pair of orthogonal vectors v, v V. 2. T is conformal if and only if T, W = λ, V for some scalar λ > Let P V and P W be the matrices for, V and, W with respect to a certain pair of bases, and let A be the matrix for T. Then T is isometric if and only if for some scalar λ > 0. A T P W A = λp V. 4. Let b 1,..., b n be a basis for V. Then T is isometric if and only if there exists a scalar λ > 0 so that T (bi ), T (b j ) W = λb i, b j V for all i and j. 5. Let e 1,..., e n be an orthonormal basis for V. Then T is isometric if and only if the vectors T (e 1 ),..., T (e n ) are orthogonal and all have the same length. 5. Conformal Maps A smooth map f : S 1 S 2 between two surfaces is said to be conformal if the derivative D p f at each point p S 1 is a conformal linear transformation. Such a map will preserve the angles between curves on the surface. As with local isometries, there are several ways to show that a map is conformal: 1. A map f is conformal if and only if Df p (t 1 ) and Df p (t 2 ) are orthogonal for each point p S 1 and each pair of orthogonal tangent vectors t 1, t 2 T p S A map f is a conformal if and only if there exists a scalar function λ: S 1 (0, ) such that (D p f), f(p) = λ(p), p for each point p S A map f is conformal if and only if (Df) T I 2 (Df) = λi 1, for some scalar function λ, where I 1 and I 2 are the matrices for the first fundamental forms on S 1 and S Let t 1, t 2 be a basis of tangent vectors at each point of S 1. Then f is conformal if and only if there exists a scalar function λ such that Df(t 1 ) = λ t 1, Df(t 2 ) = λ t 2, and Df(t 1 ), Df(t 2 ) = λt 1, t 2 5. In particular, suppose that t 1, t 2 is a basis of orthonormal tangent vectors at each point of S 1. Then f is conformal if and only if Df(t 1 ) and Df(t 2 ) are orthogonal and have the same length.

4 By the way, the scalar function λ appearing in these criteria is actually the Jacobian of the conformal transformation. (See the notes on Area & Jacobians). As with isometries, there is a special trick that can be used to simplify condition (3) further. Specifically, given surface patches σ 1 : U S 1 and σ 2 : U S 2 satisfying σ 2 = f σ 1, the map f is conformal if and only if I 2 = λi 1 for some scalar function λ, where I 1 and I 2 are the matrices for the first fundamental forms obtained from σ 1 and σ Special Surface Patches It is also possible to define the notions of local isometry and conformal map for surface patches. Given a surface patch σ: U S, We say that σ is isometric if the matrix I for the first fundamental form obtained from σ is the identity matrix. We say that σ is conformal if the matrix I for the first fundamental form obtained from σ is the product of the identity matrix with a scalar function. That is, σ is isometric if the first fundamental form on U is the same as the dot product, and σ is conformal if the first fundamental form on U is a scalar multiple of the dot product. If we think of U as a subset of the xy-plane in R 3, then these definitions for a surface patch reduce to the definitions given above for a map between surfaces. This makes it possible to use any of the criteria given before for maps between surfaces to show that a surface patch is isometric or conformal.

5 Practice Problems 1. Let P be the xy-plane in R 3, let C be the cylinder x 2 + y 2 = 1, and let f : P C be the map Show that f is a local isometry. f(x, y, 0) = (cos x, sin x, y). 2. Let C be the cylinder x 2 + y 2 = 1, and let f : C R 2 be the map f(x, y, z) = (xe z, ye z ). (a) What is the image of f? (b) Show that f maps the cylinder C conformally onto its image. 3. Let γ: R R 2 be a smooth unit-speed curve without self-intersections, and define a surface patch σ: R 2 R 3 by σ(u, v) = ( γ 1 (u), γ 2 (u), v ). Show that σ is isometric. 4. Let H 2 be the hyperbolic plane, and let σ: R (0, ) H 2 be a surjective surface patch with first fundamental form ds 2 = (du 2 + dv 2 )/v 2. (a) Is the surface patch σ conformal? Is it isometric? (b) Find a pair of vectors {t 1, t 2 } at each point (u, v) that are orthonormal with respect to the first fundamental form. (c) Show that the map f : H 2 H 2 defined by f(σ(u, v)) = σ(2u, 2v) is an isometry. 5. Let S be the cone x 2 + y 2 = z 2 with the origin removed, and let f : S S be the map f(x, y, z) = ( x/z 2, y/z 2, 1/z ). (a) Find an orthonormal basis of tangent vectors at each point (x, y, z) S. (b) Show that f is conformal. What is the Jacobian?