By Prof. P. M. Gadea, Prof. J. Muñoz Masqué (auth.)
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Extra info for Analysis and Algebra on Differentiable Manifolds: A Workbook for Students and Teachers
Let pr1 : M × N → M, pr2 : M × N → N denote the projection maps, and iq : M → M × N, x → (x, q), i p : N → M × N, y → (p, y), the inclusion maps. The map ϕ : T(p,q) (M × N) → Tp M ⊕ Tq N, v → (pr1∗ v, pr2∗ v), is an isomorphism. In fact, it is immediate that it is linear. Moreover, letting ψ : Tp M ⊕ Tq N → T(p,q) (M × N), (v1 , v2 ) → iq∗ v1 + i p∗ v2 , we have (ϕ ◦ ψ )(v1 , v2 ) = pr1∗ (iq∗ v1 + i p∗ v2 ), pr2∗ (iq∗ v1 + i p∗ v2 ) .
We have σ (t) ≡ (− sin t, cos t), thus √ 2 2 , σ (π /4) ≡ − 2 2 √ √ 2 ∂ 2 ∂ ≡− + 2 ∂ x (√2/2,√2/2) 2 ∂y √ . √ √ ( 2/2, 2/2) √ Hence σ (π /4) f = − 2/4. 5. Consider the curve in R2 given by σ (t) = (x(t), y(t)) = (t 2 − 1,t 3 − t). Find σ (t) and σ (t) for t = 1 and t = −1. Compare σ (1) with σ (−1) and σ (1) with σ (−1). Fig. 15 The curve σ (t) = (x(t), y(t)) = (t 2 − 1,t 3 − t). 3 Differentiable Functions and Mappings 25 Solution. We have σ (1) = (0, 0), σ (−1) = (0, 0), and σ (1) ≡ (2t, 3t 2 − 1)t=1 = (2, 2), σ (−1) ≡ (−2, 2).
Vk ). As the vectors v1 , . . , vk are linearly independent, we can complete them up to a basis (v1 , . . , vk , vk+1 , . . , vn ) in Rn . Let k v j,h = ∑ aij,h vi + i=1 n ∑ bij,h vi , 1 j k, ( ) i=k+1 be the expression of v j,h in this basis. As lim h→∞ v j,h = v j , we obtain lim h→∞ aij,h = δi j , for i, j = 1, . . , k, and lim h→∞ bij,h = 0, for k + 1 i n, 1 j k. Set X = (vk+1 , . . ,k 1 j k . ,k , Then, ( ) can be rewritten as Xh = XAh + XBh ; hence Xh Zh = XAh Zh + XBh Zh , and passing to the limit we obtain XZ = X lim (Ah Zh ) + X lim (Bh Zh ).
Analysis and Algebra on Differentiable Manifolds: A Workbook for Students and Teachers by Prof. P. M. Gadea, Prof. J. Muñoz Masqué (auth.)