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By Prof. P. M. Gadea, Prof. J. Muñoz Masqué (auth.)

ISBN-10: 904813563X

ISBN-13: 9789048135639

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Applying Gauss’s method of decomposition in squares, one associated to H(0,0,0) has: 2x y + 2xz + 2y z = 2((x + z)(y + z) − z2 ) =2 1 1 (x + y + 2z)2 − (x − y)2 − z2 4 4 1 1 = (x + y + 2z)2 − (x − y)2 − 2z2 . 2 2 As two negative signs appear, the index of f at (0, 0, 0) is 2. 5. Consider the C∞ manifold Rn and a submanifold L given by a vector subspace of Rn with dim L n − 1. Prove that L has zero measure. Solution. Let dim L = k n − 1. Consider the map f : Rk → Rn , f (x1 , x2 , . . , xk ) = xi ei , where {ei } is a basis of L.

5 Immersions, Submanifolds, Embeddings and Diffeomorphisms Thus, 35 ϕ (R3 ) = (x, y, z) ∈ R3 : x > 0, x + y > 0, x e2z > y . The map ϕ is injective, since the above formulae give the unique point (x, y, z) having (x , y , z ) as its image by ϕ . In order to see that ϕ is a diffeomorphism from R3 to ϕ (R3 ), it suffices to prove that the determinant of its Jacobian matrix Jϕ never vanishes. We have ⎛ 0 ⎜ 2x det Jϕ = det ⎜ ⎝2e 1 2e2y 0 −1 2e2z ⎞ ⎟ −2e2z ⎟ ⎠ 0 = −4(e2y+2z + e2x+2z ) = 0. 8. Consider the C∞ function f : R3 → R3 defined by f (x, y, z) = (x cos z − y sin z, x sin z + y cos z, z).

As f (p) ∈ U, we have f∗ = ∂ ∂θ = p ∂ (θ ◦ f ) ∂ (p) ∂θ ∂θ ◦ Φ −1 ) ∂ (θ ◦ f ∂u (Φ (p)) ∂ ∂θ + f (p) + f (p) ∂ (ϕ ◦ f ) ∂ (p) ∂θ ∂ϕ ◦ Φ −1 ) ∂ (ϕ ◦ f ∂u f (p) (Φ (p)) ∂ ∂ϕ , f (p) ( ) f∗ = ∂ ∂ϕ = p ∂ (θ ◦ f ) ∂ (p) ∂ϕ ∂θ ◦ Φ −1 ) ∂ (θ ◦ f ∂v (Φ (p)) ∂ ∂θ + f (p) + f (p) ∂ (ϕ ◦ f ) ∂ (p) ∂ϕ ∂ϕ ◦ Φ −1 ) ∂ (ϕ ◦ f ∂v f (p) (Φ (p)) ∂ ∂ϕ . 3 Differentiable Functions and Mappings 27 √ = arccos 2 (− sin u cos v + cos u) , 2 (ϕ ◦ f ◦ Φ −1 )(u, v) = arctan √ 2 sin u sin v . ) Then, we obtain by calculating and substituting the four partial derivatives in ( ) above: f∗ f∗ ∂ ∂θ ∂ ∂ϕ = p = p sin θ0 + cos θ0 cos ϕ 0 ∂ ∂θ 1 + sin2 θ0 sin2 ϕ 0 + sin 2θ0 cos ϕ 0 √ 2 sin ϕ 0 ∂ + 2 2 1 + sin θ0 sin ϕ 0 + sin 2θ0 cos ϕ 0 ∂ ϕ − sin θ0 sin ϕ 0 ∂ ∂θ 1 + sin2 θ0 sin2 ϕ 0 + sin 2θ0 cos ϕ 0 √ 2(sin2 θ0 + 12 sin 2θ0 cos ϕ 0 ) ∂ + 2 2 1 + sin θ0 sin ϕ 0 + sin 2θ0 cos ϕ 0 ∂ ϕ f (p) , f (p) f (p) .

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