LectureNotes7G

LectureNotes7G - Math 598 Geometry and Topology II Spring...

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Math 598 Feb 14, 2005 1 Geometry and Topology II Spring 2005, PSU Lecture Notes 7 2.7 Smooth submanifolds Let N be a smooth manifold. We say that M N m is an n -dimensional smooth submanifold of N , provided that for every p M there exists a local chart ( U, φ )o f N centered at p such that φ ( U M )= R n ×{ o } , where o denotes the origin of R m - k . Proposition 1. A smooth submanifold M N is a smooth manifold. Proof. Since M N , M is Hausdorf and has a countable basis. For every p M , let ( U, φ ) be a local chart of M with φ ( U M R n o } . Set U := U M , and φ := φ | U . Then φ : U R n o R n is a homeomorphism, and thus M is a toplogical manifold. It remains to show that M is smooth. To see this note that if ( V, ψ ) is the restriction of another local chart of N to M . Then ψ ( φ ) - 1 = ψ φ - 1 | φ ( U ) , which is smooth. The above proof shows how M induces a di±erential structure on N . Whenver we talk of a submanifold M as a smooth manifold in its own right, we mean that M is equipped with the di±erential structue which it inherits from N . Theorem 2. Let f : M n N m be a smooth map of constant rank k (i.e., rank ( df p k , for all p M ). Then, for any q N , f - 1 ( q ) is an ( n - k ) - dimensional smooth submanifold of M . Proof. Let p f - 1 ( q ). By the rank theorem there exists local neighborhoods ( U, φ ) and ( V,ψ f M and N centered at p and q respectively such that ˜ f ( x ):= ψ f φ - 1 ( x 1 ,...,x n )=( x 1 k , 0 ,..., 0) . 1 Last revised: February 19, 2005 1
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Next note that φ ( U f - 1 ( q )) = φ ( U ) φ f - 1 ψ - 1 ( o )= R n ˜ f - 1 ( o { o R n - k . Thus f - 1 ( q ) is a smooth submanifold of N (To be quite strict, we need to show that φ ( U f - 1 ( q )) = R n - k ×{ o } , but this is easily achieved if we replace ψ with θ ψ , where θ : R m R m is the diFeomorphism which switches the first k and last m - k cordinates). Exercise 3. Use the previous result to show that S n is smooth n -dimensional submanifold of R n +1 . Another application of the last theorem is as follows: Example 4. SL n is a smooth submanifold of GL n . To see this define f : GL n R by f ( A ) := det( A ). Then n = f - 1 (1), and thus it re- mains to show that f has constant rank on GL n . Since this rank has to be either 1 or 0 at each point (why?), it su±ces to show that the rank is not zero anywhere, i.e., it is enough to show that for every A GL n there exists X T A GL n such that df A ( X ) ± = 0. To see this, let X =[ α ] where α :( - (,( ) GL n is the curve given by α ( t ):=(1 - t ) A . Note that, since det is contiuous, det( α ( t )) ± = 0, for all t ( - (,( ), once we make sure that ( is small enough. Thus α is indeed well-defined. Now recall that A ( X ):=[ f α ] T f ( A ) R . ²urther recall that there is a canonical isomorphism θ : T f ( A ) R R given θ ([ γ ]) = γ ± (0). Thus θ A ( X )=( f α ) ± (0) = det( A ) ± =0 .
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This note was uploaded on 08/25/2011 for the course MATH 6456 taught by Professor Staff during the Spring '08 term at Georgia Tech.

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LectureNotes7G - Math 598 Geometry and Topology II Spring...

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