m425b-hw9soln-s08

m425b-hw9soln-s08 - MATH 425b ASSIGNMENT 9 SOLUTIONS SPRING...

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Unformatted text preview: MATH 425b ASSIGNMENT 9 SOLUTIONS SPRING 2007 Prof. Alexander Chapter 10: (16) Let σ ij = [ p ,..,p i- 1 ,p i ,..,p j- 1 ,p j +1 ,..,p k ] (the simplex with p i ,p j missing.) Then ∂σ = k X i =0 (- 1) i [ p ,..,p i- 1 ,p i +1 ,..,p k ] , ∂ 2 σ = k X i =0 (- 1) i X j<i (- 1) j σ ji + X j>i (- 1) j- 1 σ ij ! A given σ ij (= σ ji ), say σ 24 , appears once with i = 2 ,j = 4 and once with j = 2 ,i = 4, giving a contribution of (- 1) 2 σ 24 + (- 1) 3 σ 24 = 0 . Similarly we get, after switching the names of the indices in the second sum, ∂ 2 σ = k X i =0 (- 1) i X j<i (- 1) j σ ji + k X j =0 (- 1) j X i>j (- 1) i- 1 σ ji = X ( i,j ): j<i (- 1) i + j σ ji + X ( i,j ): j<i (- 1) i + j +1 σ ji = 0 For a chain Ψ = Φ 1 + .. + Φ r , we have ∂ 2 Ψ = ∑ r i =1 ∂ 2 Φ i so it’s enough to show ∂ 2 Φ = 0 for all surfaces Φ = T ◦ σ (where σ is affine and T is C 00 , as in 10.30.) Then ∂ Φ = T ( ∂σ ) and ∂ 2 Φ = T ( ∂ 2 σ ) = 0 since ∂ 2 σ = 0....
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m425b-hw9soln-s08 - MATH 425b ASSIGNMENT 9 SOLUTIONS SPRING...

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