Α 1 x p β 1 y p γ 1 z p i v p α 2 x p β 2 y p γ

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= α 1 ∂x p + β 1 ∂y p + γ 1 ∂z p , i * ∂v p ! = α 2 ∂x p + β 2 ∂y p + γ 2 ∂z p , for some constants α i , β i , γ i . Find ( α i , β i , γ i ) for i = 1 , 2. Proof: Note that x i = u , y i = v and zzzzz . Then applying the equations given above to x , y , and z respectively yields the following: α 1 = i * ∂u p ! ( x ) = ∂u p ( x i ) = ∂u p ( u ) = 1 β 1 = i * ∂u p ! ( y ) = ∂u p ( y i ) = ∂u p ( v ) = 0 γ 1 = i * ∂u p ! ( z ) = ∂u p ( z i ) = ∂u p ( 1 - u 2 - v 2 ) = - 2 u 2 1 - u 2 - v 2 p = - a c α 2 = i * ∂v p ! ( x ) = ∂v p ( x i ) = ∂v p ( u ) = 0 β 2 = i * ∂v p ! ( y ) = ∂v p ( y i ) = ∂v p ( v ) = 1 γ 2 = i * ∂v p ! ( z ) = ∂v p ( z i ) = ∂v p ( 1 - u 2 - v 2 ) = - 2 v 2 1 - u 2 - v 2 p = - b c As a result, we have determined the coefficients in the expressions i * ∂u p ! = ∂x + - a c ∂z and i * ∂v p ! = ∂y + - b c ∂z , as desired. Problem 11.5: Prove that if N is a compact manifold, then a one-to-one immersion f : N M is an embedding. Proof: Note that f is C because f is an immersion. As N is compact and M is Hausdorff, it follows from the continuity of f that f will be a closed map, which implies that f - 1 is continuous. Recall that a function is always surjective onto its image. As a result, we have that f is continuous in both directions and bijective onto its image, so f is a homeomorphism from N to f ( N ). Because f is also an immersion, we have then shown that f is an embedding. 48
Chapter 12 Problem 12.2: Let ( U, φ ) = ( U, x 1 , . . . , x n ) and ( V, ψ ) = ( V, y 1 , . . . , y n ) be overlapping coordinate charts on a manifold M . They induce coordinate charts ( TU, ˜ φ ) and ( TV, ˜ ψ ) on the total space TM of the tangent bundle (see equation (12.1)), with transition function ˜ ψ ˜ φ - 1 : ( x 1 , . . . , x n , a 1 , . . . , a n ) 7→ ( y 1 , . . . , y n , b 1 , . . . , b n ) . (a) Compute the Jacobian matrix of the transition function ˜ ψ ˜ φ - 1 at φ ( p ). Proof: From the given information we know that the Jacobian of ˜ ψ ˜ φ - 1 will be of the form J ( ˜ ψ ˜ φ - 1 ) = ∂y 1 ∂x 1 · · · ∂y 1 ∂x n ∂y 1 ∂a 1 · · · ∂y 1 ∂a n . . . . . . . . . . . . ∂y n ∂x 1 · · · ∂y n ∂x n ∂y n ∂a 1 · · · ∂y n ∂a n ∂b 1 ∂x 1 · · · ∂b 1 ∂x n ∂b 1 ∂a 1 · · · ∂b 1 ∂a n . . . . . . . . . . . . ∂b n ∂x 1 · · · ∂b n ∂x n ∂b n ∂a 1 · · · ∂b n ∂a n . We also know that ˜ φ - 1 ( x 1 , . . . , x n , a 1 , . . . , a n ) = ( p, a p ), where p = φ - 1 ( x 1 , . . . , x n ) and a p = n X i =1 a i ∂x i p . Furthermore, we have that ∂x i p = n X j =1 ∂y j ∂x i ∂y i p . As a result, we now have that ˜ ψ ( p, a p ) = y 1 ( x 1 , . . . , x n ) , . . . , y n ( x 1 , . . . , x n ) , n X i =1 ∂y 1 ∂x i a i , . . . , n X i =1 ∂y n ∂x i a i ! . 49
It then follows from our above statements that J ( ˜ ψ ˜ φ - 1 ) = ∂y 1 ∂x 1 · · · ∂y 1 ∂x n ∂y 1 ∂a 1 · · · ∂y 1 ∂a n .

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