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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 ) of 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 ff 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 ff 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 ) of M and N centered at p and q respectively such that ˜ f ( x ) := ψ f φ - 1 ( x 1 , . . . , x n ) = ( x 1 , . . . , x 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 di ff eomorphism 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 SL 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 ffi 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 df A ( X ) := [ f α ] T f ( A ) R . Further recall that there is a canonical isomorphism θ : T f ( A ) R R given by θ ([ γ ]) = γ (0). Thus θ df A ( X ) = ( f α ) (0) = det( A ) = 0 .
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LectureNotes7G - Math 598 Geometry and Topology II Spring...

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