By P M Gadea; J Muñoz Masqué; I V Mikiti︠u︡k

Differentiable Manifolds -- Tensor Fields and Differential types -- Integration on Manifolds -- Lie teams -- Fibre Bundles -- Riemannian Geometry -- a few formulation and Tables

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Ii) Calculate the index of f at (0, 0, 0). Solution (i) f∗(0,0,0) ≡ (sin y + z cos x, x cos y + sin z, y cos z + sin x)(0,0,0) = (0, 0, 0). Thus rank f∗(0,0,0) = 0, so (0, 0, 0) is a critical point. The Hessian matrix of f at (0, 0, 0) is ⎛ ⎞ ⎛ ⎞ −z sin x cos y cos x 0 1 1 f −x sin y cos z ⎠ = ⎝ 1 0 1⎠ . H(0,0,0) = ⎝ cos y cos x cos z −y sin z (0,0,0) 1 1 0 f Since det H(0,0,0) = 2 = 0, the point (0, 0, 0) is non-degenerate. f (ii) The index of f at (0, 0, 0) is the index of H(0,0,0) , that is, the number of negative signs in a diagonal matrix representing the quadratic form 2(x y + x z + y z) f associated to H(0,0,0) .

Prove that L has zero measure. 36 1 Differentiable Manifolds Fig. 18 The graph of the map t → (t 2 , t 3 ) Solution Let dim L = k n − 1. Consider the map f x 1 , x 2 , . . , x k = x i ei , f : R k → Rn , where {ei } is a basis of L. By virtue of Sard’s Theorem, f (Rk ) = L has zero measure. 56 Let M1 and M2 be two C ∞ manifolds. Give an example of differentiable mapping f : M1 → M2 such that all the points of M1 are critical points and the set of critical values has zero measure. Solution Let f : M1 → M2 defined by f (p) = q, for every p ∈ M1 and q a fixed point of M2 .

The plane of R3 containing L and passing through the point (0, 0, 1) must be parallel to the director vector of L, which is (v, 0), and also to the vector (a, 0) − (0, 0, 1) = (a, −1), (where we use the notation (x, b) ∈ R3 , for x = (x1 , x2 ) ∈ R2 , to denote the point (x1 , x2 , b)). Since these two vectors are linearly independent, we have f −1 (L) = f −1 r(a, v) = π (v, 0) ∧ (a, −1) = π(−v2 , v1 , −v2 a1 + v1 a2 ) = π J v, J v, a . Conversely, let π(w) = π(w1 , w2 , w3 ) ∈ P0 and, to be short, write w¯ = (w1 , w2 ) ∈ R2 , so that w = (w, ¯ w3 ).