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LectureNotes1G - 1 Math 528 Geometry and Topology II Fall...

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Math 528 Jan 11, 2005 1 Geometry and Topology II Fall 2005, PSU Lecture Notes 1 1 Topological Manifolds The basic objects of study in this class are manifolds. Roughly speaking, these are objects which locally resemble a Euclidean space. In this section we develop the formal definition of manifolds and construct many examples. 1.1 The Euclidean space By R we shall always mean the set of real numbers, which is a well ordered field with the following property. Completeness Axiom . Every nonempty subset of R which is bounded above has a least upper bound . The set of all n -tuples of real numbers R n := { ( p 1 , . . . , p n ) | p i R } is called the Euclidean n-space. So we have p R n ⇐⇒ p = ( p 1 , . . . , p n ) , p i R . Let p and q be a pair of points (or vectors) in R n . We define p + q := ( p 1 + q 1 , . . . , p n + q n ). Further, for any scalar r R , we define rp := ( rp 1 , . . . , rp n ). It is easy to show that the operations of addition and scalar multiplication that we have defined turn R n into a vector space over the field of real numbers. Next we define the standard inner product on R n by p, q = p 1 q 1 + . . . + p n q n . Note that the mapping · , · : R n × R n R is linear in each variable and is sym- metric. The standard inner product induces a norm on R n defined by p := p, p 1 / 2 . If p R , we usually write | p | instead of p . Theorem 1.1.1. (The Cauchy-Schwartz inequality) For all p and q in R n | p, q | p q . The equality holds if and only if p = λ q for some λ R . 1 Last revised: October 2, 2006 1
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Proof I. If p = λ q it is clear that equality holds. Otherwise, let f ( λ ) := p - λ q, p - λ q . Then f ( λ ) > 0. Further, note that f ( λ ) may be written as a quadratic equation in λ : f ( λ ) = p 2 - 2 λ p, q + λ 2 q 2 . Hence its discriminant must be negative: 4 p, q 2 - 4 p 2 q 2 < 0 which completes the proof. Proof II. Again suppose that p = λ q . Then p, q = p q p p , q q . Thus it su ffi ces to prove that for all unit vectors p and q we have | p, q | 1 , and equality holds if and only if p = ± q . This may be proved by using the method of lagrangne multipliers to find the maximum of the function x, y subject to the constraints x = 1 and y = 1. More explicitly we need to find the critical points of f ( x, y, λ 1 , λ 2 ) := x, y + λ 1 ( x 2 - 1) + λ 2 ( y 2 - 1) = n i =1 ( x i y i + λ 1 x 2 i + λ 2 y 2 i ) - λ 1 - λ 2 . At a critical point we must have 0 = f/ x i = y i + 2 λ 1 x i , which yields that y = ± x . The standard Euclidean distance in R n is given by dist( p, q ) := p - q . The proof of the following fact is left as an exercise. Corollary 1.1.2. (The triangle inequality) For all p , q , r in R n . dist( p, q ) + dist( q, r ) dist( p, r ) By a metric on a set X we mean a mapping d : X × X R such that 1. d ( p, q ) 0, with equality if and only if p = q . 2
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2. d ( p, q ) = d ( q, p ). 3. d ( p, q ) + d ( q, r ) d ( p, r ). These properties are called, respectively, positive-definiteness, symmetry, and the triangle inequality. The pair ( X, d ) is called a metric space . Using the above exercise, one immediately checks that ( R n , dist) is a metric space. Geometry, in its broadest definition, is the study of metric spaces.
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