By Pedro M. Gadea, Jaime Muñoz Masqué, Ihor V. Mykytyuk

This is the second one version of this most sensible promoting challenge ebook for college students, now containing over four hundred thoroughly solved routines on differentiable manifolds, Lie idea, fibre bundles and Riemannian manifolds.

The workouts cross from basic computations to fairly subtle instruments. some of the definitions and theorems used all through are defined within the first portion of every one bankruptcy the place they appear.

A 56-page choice of formulae is integrated which might be worthwhile as an aide-mémoire, even for lecturers and researchers on these topics.

In this 2d edition:
• 76 new difficulties
• a part dedicated to a generalization of Gauss’ Lemma
• a brief novel part facing a few houses of the power of Hopf vector fields
• an improved number of formulae and tables
• an prolonged bibliography

Audience

This e-book can be worthy to complex undergraduate and graduate scholars of arithmetic, theoretical physics and a few branches of engineering with a rudimentary wisdom of linear and multilinear algebra.

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Additional info for Analysis and Algebra on Differentiable Manifolds: A Workbook for Students and Teachers

Example text

13 A set with a C ∞ atlas, whose induced topology is not Hausdorff The classes of the quotient set S = E/∼ are represented by the elements (x, 0) for x < 0, and the elements (x, 0) and (x, 1) for x 0 (see Fig. 13). Prove that S admits a C ∞ atlas, but S is not Hausdorff with the induced topology. The relevant theory is developed, for instance, in Brickell and Clark [1]. Solution Denote by [(x, y)] the class of (x, y). We can endow S with a manifold structure by means of the charts (U1 , ϕ1 ) and (U2 , ϕ2 ), where U1 = (x, 0) : x ∈ R , ϕ1 (x, 0) = x, U2 = ϕ2 (x, 0) (x, 0) : x < 0 ∪ = ϕ2 (x, 1) (x, 1) : x 0 , = x.

In fact, let p and q be different points of S. If they belong to the domain of some chart (U, ϕ) of S, we can choose disjoint open subsets V1 , V2 of Rn (assuming dim S = n), contained in ϕ(U ), and such that ϕ(p) ∈ V1 , ϕ(q) ∈ V2 . Since ϕ is continuous, ϕ −1 (V1 ) and ϕ −1 (V2 ) are disjoint open subsets of S containing p and q, respectively. If p and q do not belong to the domain of a given chart of S, there must be a chart whose domain U1 contains p but not q, and one chart whose domain U2 contains q but not p.

Hence f ψ1−1 (x) = r x2 (1, x1 ) , (x1 , −1) . 1 + x12 46 1 Differentiable Manifolds Now, f (ψ1−1 (x)) belongs to the domain Up of ϕp if and only if the straight line r x2 (1, x1 ) , (x1 , −1) 1 + x12 does not contain p. e. if p− x2 (1, x1 ) 1 + x12 is parallel to (x1 , −1), or equivalently, if and only if it is orthogonal to J (x1 , −1), or even if and only if p− x2 (1, x1 ) , (1, x1 ) = 0. 1 + x12 This equation is polynomial in the components of x, so that the set of points satisfying the condition is a closed subset of R2 , as we wanted.

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