MATHEMATICS E-23a, Fall 2016 Linear Algebra and Real Analysis I Module #3, Week 4 Implicit functions, manifolds, tangent spaces, critical points Author: Paul Bamberg R scripts by Paul Bamberg Last modified: December 5, 2016 by Paul Bamberg (added the graph for HW problem 5) Reading • Hubbard, Section 3.1 (Implicit functions and manifolds) • Hubbard, Section 3.2 (Tangent spaces) • Hubbard, Section 3.6 (Critical points) • Hubbard, Section 3.7 through page 354 (constrained critical points) Recorded Lectures • November 24, 2015 (watch on November 22) • December 1, 2015 (watch on November 29) • December 3, 2015 (watch on December 1) Proofs to present in section or to a classmate who has done them. • 12.1 (Hubbard Theorem 3.2.4) Suppose that U ⊂ Rn is an open subset, F : U ! Rn−k is a C1 mapping, and manifold M can be described as the set of points that satisfy F(z) = 0. Use the implicit function theorem to show that if [DF(c)] is onto for c 2 M, then the tangent space TcM is the kernel of [DF(c)]. You may assume that the variables have been numbered so that when you row-reduce [DF(c)], the first n − k columns are pivotal. • 12.2(Hubbard, theorems 3.6.3 and 3.7.1) Let U 2 Rn be an open subset and let f : U ! R be a C1 (continuously differentiable) function. First prove, using a familiar theorem from single-variable calculus, that if x0 2 U is an extremum, then [Df(x0)] = [0]: Then prove that if M ⊂ Rn is a k-dimensional manifold, and c 2 M \ U is a local extremum of f restricted to M, then TcM ⊂ ker[Df(c)]: • 12.3(Special case of Hubbard, theorem 3.7.5) Let M be a manifold known by a real-valued C1 function F (x) = 0, where F goes from an open subset U of Rn to R and [DF (x)] is nowhere zero. Let f : U ! R be a C1 function. Prove that c 2 M is a critical point of f restricted to M if and only if there exists a Lagrange multiplier λ such that [Df(c)] = λ[DF (c)]. 1 R Scripts • Script 3.4A-ImplicitFunction.R Topic 1 - Three variables, one constraint Topic 2 - Three variables, two constraints • Script 3.4B-Manifolds2D.R Topic 1 - A one-dimensional submanifold of R2 { the unit circle Topic 2 - Interesting examples from the textbook Topic 3 - Parametrized curves in R2 Topic 4 - A two-dimensional manifold in R2 Topic 5 - A zero-dimensional manifold inR2 • Script 3.4C-Manifolds3D.R Topic 1 - A manifold as a function graph Topic 2 - Graphing a parametrized manifold Topic 3 - Graphing a manifold that is specified as a locus • Script 3.4D-CriticalPoints Topic 1 - Behavior near a maximum or minimum Topic 2 - Behavior near a saddle point • Script 3.5A-LagrangeMultiplier.R Topic 1 - Constrained critical points in R2 2 1 Executive Summary 1.1 Implicit functions { review of the linear case. We have n unknowns, n − k equations, e.g for n = 3; k = 1 2x + 3y − z = 0, 4x − 2y + 3z = 0 2 3 −1 Create an (n − k) × n matrix: T = 4 −2 3 If the matrix T is not onto, its rows (the equations) are linearly dependent. Otherwise, when we row reduce, we will find n − k = 2 pivotal columns and k = 1 nonpivotal columns. We assign values arbitrarily to the \active" variables that correspond to the nonpivotal columns, and then the values of the \passive" variables that corresponds to the pivotal column are determined. Suppose that we reorder the unknowns so that the \active" variables come last. Then, after we row reduce the matrix, the first n − k columns will be pivotal. So the first n − k columns will be linearly independent, and they form an invertible square matrix. The matrix is now of the form T = [AjB], where A is invertible. ~x The solution vector is of the form ~v = , where the passive variables ~x come ~y first, the active variables ~y come second. A solution to T~v = ~0 is obtained by choosing ~y arbitrarily and setting ~x = −A−1B~y: Our system of equations determines ~x \implicitly" in terms of ~y: 1.2 Implicit function theorem { the nonlinear case. We have a point c 2 Rn, a neighborhood W of c, and a function F : W ! Rn−k for which F(c) = 0 and [DF(c)] is onto. F imposes constraints. The variables are ordered so that the n − k pivotal columns in the Jacobian matrix, which correspond to the passive variables, come first. Let a denote the passive variables at c; let b denote the active variables at c. The implicit function g expresses the passive variables in terms of the active variables, and g(b) = a. For y near b, x = g(y) determines passive variables a such that F = 0. Tweak y, and g specifies how to tweak x so that the b constraints are still satisfied. Although we usually cannot find a formula for g, we can find its derivative at b by the same recipe that worked in simple cases. Evaluate the Jacobian matrix [DF(c)]. Extract the first n − k columns to get an invertible square matrix A. Let the inverse of this matrix act on the remaining k columns (matrix B) and change the sign to get the (n − k) × k Jacobian matrix for g. That is, [Dg(b)] = −A−1B. 3 1.3 Curves, Surfaces, Graphs, and Manifolds Manifolds are a generalization of smooth curves and surfaces. The simplest sort of manifold is a flat one, described by linear equations. An example is the line of slope 2 that passes through the point x = 0; y = −2: a one-dimensional submanifold of R2 There are three equivalent ways to describe such a manifold. • (The definition) As the graph of a function that expresses the passive vari- ables in terms of the active variables: either y = f(x) = −2 + 2x or 1 x = g(y) = 2 (y + 2). x • As a \locus" defined by a constraint equation F = 2x − y − 2 = 0. y 1 1 • By a parametrization function g(t) = + t . 0 2 Definition: A subset M ⊂ Rn is a smooth manifold if locally it is the graph of a C1 function (the partial derivatives are continuous). \Locally" means that for any point x 2 M we can find a neighborhood U of x such that within M \ U, there is a C1 function that expresses n − k passive variables in terms of the other k active variables. The number k is the dimension of the manifold. In R3 there are four possibilities: • k = 3. Any open subset M ⊂ R3 is a smooth 3-dimensional manifold. In this case k = 3, and the manifold is the graph of a function f : R3 ! f~0g, whose codomain is the trivial vector space f~0g that contains just a single point. Such a function is necessarily constant, and its derivative is zero. x • k = 2. The graph of z = f = x2 + y2 is a paraboloid. y x cos 2πz • k = 1. The graph of the function = ~f(z) = is a helix. y sin 2πz • k = 0. In this case the manifold consists of one of more isolated points. ~ ~ 3 Near any of these points x0, it is the graph of a function f : f0g ! R whose domain is a zero-dimensional vector space and whose image is the 3 point x0 2 R : There is no requirement that a manifold be the graph of a single function, or that the \active" variables be the same at every point on the manifold. The unit circle, the locus of x2 + y2 − 1 = 0, is the union of four function graphs, two of which have x as the active variable, two of which have y. By using a parameter t that is not one of the variables, we can represent it by the parametrization x cos t = g(t) = y sin t 4 1.4 Using the implicit function theorem Start with an open subset U ⊂ Rn and a C1 function F : U ! Rn−k. Consider the \locus," M \ U, the set of solutions of the equation F(z) = 0. If [DF(z)] is onto (surjective) for every z 2 M \ U, then M \ U is a smooth k-dimensional manifold embedded in Rn: Proof: the implicit function theorem says precisely this. The statement that [DF(z)] is onto guarantees the differentiability of the implicitly defined function. If [DF(z)] does not exist or fails to be onto, perhaps even just at a single point, the locus is not a manifold. We use the notation M \ U because F may define just part of a larger manifold M that cannot be described as the locus as a single function. To say that M itself is a manifold, we have to find an appropriate U and F for every point z in the manifold. 1.5 Parametrizing a manifold For a k-dimensional submanifold of Rn, the parametrization function is γ : U ! M, where U ⊂ Rk is an open set. The variables in Rk are called \parameters." The parametrization function γ must be C1 , one-to-one, and onto M. In other words, we want γ to give us the entire manifold. Finding a local parametrization that gives part of the manifold is of no particular interest, because there is, by definition, a function graph that does that. An additional requirement: The derivative of the parametrization function is one-to-one for all parameter values.