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Unformatted text preview: 5.4. DEGREE OF A MAP 139 1. Let v be a vector field in R n + 1 defined by v ( x ) = ϕ ( x ) x . 2. Then v never points to the origin. So, Index( v ) = 1, and, therefore, ϕ has a fixed point in B . 3. Alternatively, suppose that ϕ does not have a fixed point. 4. Then v is nonvanishing. 5. Let ψ : B → ∂ B = S n be a map defined as follows. The point ψ ( x ) is the point on the sphere S n where the line from the ϕ ( x ) to the point x intersects S n . 6. Then ψ ( x ) = x for any x ∈ S n . 7. Let α be an nform on S n normalized by R S n α = 1. 8. Since ψ is an identity map on S n , the form ψ * α is an nform on B whose restriction to S n is equal to α . 9. Since also S n = ∂ B and d α = 0, we have 1 = Z S n α = Z ∂ B ψ * α = Z B d ( ψ * α ) = , which is a contradiction. • Problem. Let M be a closed ndimensional submanifold of R n + 1 Let v be a unit vector field on M . Let vol be the volume ( n + 1)form in R n + 1 . Let vol ( S n ) be the volume of the unit sphere...
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This note was uploaded on 11/26/2011 for the course MAT 4821 taught by Professor Wong during the Spring '10 term at FSU.
 Spring '10
 Wong
 Geometry

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