LectureNotes9G

LectureNotes9G - Math 598 Mar 2 2005 1 Geometry and...

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Unformatted text preview: Math 598 Mar 2, 2005 1 Geometry and Topology II Spring 2005, PSU Lecture Notes 9 3 Some Topics from Differential Topology 3.1 Regular points and values; Fundamental Theorem of Algebra Let f : M → N be a smooth map. We say that p ∈ M is a regular point of f provided that rank ( df p ) = dim ( N ); otherwise we say that p is a critical point or a singular point . We say q ∈ N is a regular value of f provided that every p ∈ f − 1 ( q ) is a regular point. If q is not a regular value of f , then we say that it is a critical value or singular value. Exercise 1. Show that if f : M m → N n has rank k at some point p ∈ M , then it has rank k on a neighborhood of p . Proposition 2. If f : M m → N n is a smooth map, q is a regular value of f , and f − 1 ( q ) 6 = ∅ , then f − 1 ( q ) is a smooth ( m − n )-dimensional submanifold of M . Proof. By definition, every p ∈ f − 1 ( q ) is a regular point. Thus, f has con- stant rank n on f − 1 ( q ). This implies that f has constant rank n on an open neighborhood of U of f − 1 ( q ). Since f : U → N has constant rank n , it now follows, as we had proved earlier, that f − 1 ( q ) is an ( m − n )-dimensional submanifold of U , and therefore of M . Exercise 3. Show that if f : M → N is a smooth map, dim ( M ) = dim ( N ), M is compact, and q is a regular value of f , then f − 1 ( q ) consists of a finite number of points. Further, show that if we denote the number of these points by # f − 1 ( q ), then # f − 1 ( q ) is locally constant, i.e., there is an open neighborhood U of q in N , such that # f − 1 ( q ) = # f − 1 ( q ), for all q ∈ U . 1 Last revised: March 8, 2005 1 The following proof demonstrates the elegant and powerful utility of basic techniques of Differential topology. Theorem 4 (Fundamental Theorem of Algebra). Every nonconstant com- plex polynomial has a zero. Proof(After Milnor). Let P : R 2 → R 2 be a complex polynomial, and π + : S 2 − { (0 , , 1) } → R 2 be the stereographic projection from the north pole. Define f : S 2 → S 2 by f ((0 , , 1)) := (0 , , 1), and f ( p ) := ( π + ) − 1 ◦ P ◦ π + ( p ) , if p 6 = (0 , , 1). We claim that f is smooth. This is obvious on S 2 − { (0 , , 1) } where f is the composition of three smooth functions. To see that f is smooth in a neighborhood of (0 , , 1) as well, let π − : S 2 −{ (0 , , − 1) } → R 2 be the stere- ographic projection from the south pole, U ⊂ R 2 be a small neighborhood of the origin o ∈ R 2 , and define Q : U → R 2 by Q ( z ) := π − ◦ f ◦ ( π − ) − 1 ( z ) . Note that if U is sufficiently small f ◦ ( π − ) − 1 ( z ) is close to f ◦ ( π − ) − 1 ( o ) = (0 , , 1) for all z ∈ U . In particular, if U is sufficiently small, f ◦ ( π − ) − 1 ( z ) 6 = (0 , , − 1) for all z ∈ U . Thus Q is well defined. Secondly, note that π + ◦ ( π − ) − 1 is inversion with respect to the unit circle in R 2 , i.e., π + ◦ ( π − ) −...
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LectureNotes9G - Math 598 Mar 2 2005 1 Geometry and...

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