wk6_132sp17.pdf - Math 132 Topology II Smooth Manifolds...

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Math 132 - Topology II: Smooth Manifolds. Spring 2017. Week 6 Summary Reading § 1.5, 2.1, 2.2 GP References GP - Differential Topology , Guillemin & Pollack 1 Lecture 15, February 27 1.1 Transversality Question: Let f X Y be smooth, Z Y a submanifold. When is f - 1 ( Z ) a submani- fold? Example 1.1. 1. If Z = { z } Y , with z Y a regular value of f X Y . Then, f - 1 ( Z ) is a submanifold by the Preimage Theorem. 2. If Z Y is open then f - 1 ( Z ) X is open in X , hence a submanifold. Remark 1.2. The examples given above are the extreme cases dim Z 0 ? ? dim Y Z Y { z } reg. value ? ? Z open Definition 1.3. Let f X Y be smooth, Z Y manifolds. We say that f is transversal to Z , written f Z , if, for every x f - 1 ( Z ) , im ( df ) x + T f ( x ) Z = T f ( x ) Y. If j X Y is the inclusion of the submanifold X Y , then we say that X is transver- sal to Z , written if X Z , if j is transversal to Z . Example 1.4. 1. Let f R R 2 ,t ( t,t 2 ) , Z = {( x, 0 ) x R } . Then, Z is not transversal to f : we have f - 1 ( Z ) = { 0 } . Then, T ( 0 , 0 ) Z = span ( e 1 ) , im ( df ) 0 = colJac 0 f = span ( e 1 ) . Hence, T ( 0 , 0 ) Z + im ( df ) 0 = span ( e 1 ) R 2 . 2. Let f be as above, Z = {( x,x ) x R } . Then, f - 1 ( Z ) = { 0 , 1 } . ( t = 0 ) T ( 0 , 0 ) Z = span ( e 1 + e 2 ) , im ( df ) 0 = colJac 0 f = span ( e 1 ) , T ( 0 , 0 ) Z + im ( df ) 0 = span ( e 1 + e 2 ,e 1 ) = R 2 . ( t = 1 ) T ( 1 , 1 ) Z = span ( e 1 + e 2 ) , im ( df ) 1 = colJac 1 f = span ( e 1 + 2 e 2 ) , T ( 1 , 1 ) Z + im ( df ) 1 = R 2 . Hence, f Z . 1
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3. Let X = S 2 R 3 , Z = yz -plane. Then, X Z = {( 0 ,y,z ) y 2 + z 2 = 1 } . Let p = ( 0 ,y,z ) X Z . Then, T p Z = span ( e 2 ,e 3 ) , and T p X = { v = ( a,b,c ) v p = 0 } = {( a,b,c ) by + cz = 0 } span ( e 1 ) . Hence, dim T p X = dim T p Z = 2 and T p X + T p Z = T p R 3 T p X T p Z. Since e 1 T p X , for all p X Z , but e 1 T p Z , this condition always holds. Hence, X Z .
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