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# Algebraic Topology, Homotopy and Homology - download pdf or read online By R. Switzer

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Regular Surface 23 chart X(s1 , s2 ) = (s1 , s2 , 0)T , where (s1 , s2 )T ∈ (0, 1) × (0, 1), has a boundary. Indeed, ∂ S = [{0} × (0, 1)] ∪ [{1} × (0, 1)] ∪ [(0, 1) × {0}] ∪ [(0, 1) × {1}] . A closed surface is a surface that is compact and without boundary. 20) for some constant c. If G is at least C 1 in a neighborhood ⊂ 3 of some point in Γ , and ∂i G = 0 in (for some fixed i = 1, 2, 3), then one can use the implicit function theorem to solve for one of the variables x1 , x2 , x3 . For example, if ∂3 G = 0, then there exists an open set U and an F : U → such that x3 = F (x1 , x2 ), where (x1 , x2 , x3 )T ∈ ∩Γ.

Thus, we say, the plane 2 is orientable. In fact, any plane is orientable, with orientation determined by selecting one of the two normal 40 Chapter 3. 2. Illustration of orientability (arrows depict unit normal vector ν). (a) The unit sphere can be covered with a differentiable unit normal vector field (here a positive orientation is chosen). (b) The Möbius strip is a nonorientable surface. It is not possible to choose a normal vector field that is differentiable (or even continuous) over the entire surface.

51) with respect to q (while keeping s fixed). Because kn (s, q) = kn (s, C q) for any nonzero constant C , we can just maximize (minimize) over all q such that qT g q = 1. Since kn (s, q) is continuous (recall that the denominator is positive definite), from basic analysis we know that there exists a q1 and q2 that achieves the maximum and minimum, respectively, of kn (s, q). , κ1 = max −qT hq, q∈ 2 , qT g q=1 κ2 = min −qT hq. 53) Because g and h are symmetric and g is positive definite, it can be shown that κ1 and κ2 are the maximum and minimum eigenvalues of the generalized eigenvalue problem: −hq = λg q [12, 68, 70, 81].