MATH
diagnostics132sp17.pdf

# diagnostics132sp17.pdf - Math 132 Topology II Smooth...

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Math 132 - Topology II: Smooth Manifolds. Spring 2017. Diagnostic Quiz. February 15, 1.30-2pm Name: SOLUTION 1. Let X = {( x,y,z ) z = x 2 + y 2 } R 3 . Consider the smooth map f X R 2 , ( x,y,z ) ( y,z ) . (a) Compute T ( a,b,c ) X , where ( a,b,c ) X . (b) Define what it means for f to be an immersion. (c) Determine the points ( a,b,c ) X where f is an immersion. Solution: (a) We choose a parameterisation φ R 2 X , ( u,v ) ( u,v,u 2 + v 2 ) . φ is smooth and has inverse ψ X R 2 , ( x,y,z ) ( x,y ) . As ψ extends to the smooth function Ψ R 3 R 2 , ( x,y,z ) ( x,y ) we get that ψ is smooth. Then, T ( a,b,c ) X = im ( ) ( a,b ) = col Jac ( a,b ) φ = col 1 0 0 1 2 a 2 b (b) f is an immersion if, for every x X , ( df ) x is injective. (c) Let ( a,b,c ) X . Let v = λ 1 1 0 2 a + λ 2 0 1 2 b T ( a,b,c ) X . The map f extends to smooth F R 3 R 2 , ( x,y,z ) ( y,z ) . Thus, ( df ) ( a,b,c ) ( v ) = ( Jac ( a,b,c ) F ) v = 0 1 0 0 0 1 λ 1 1 0 2 a + λ 2 0 1 2 b = λ 2 2 ( λ 1 a + λ 2 b ) f is an immersion at those ( a,b,c ) where ker ( df ) ( a,b,c ) = { 0 } : - If a = 0 then we see that span 1 0 2 a ker ( df ) ( a,b,c ) ; - Assume a 0, then ker ( df ) ( a,b,c ) = { 0 } .

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