Topology

Get Algebraic and Geometric Topology, Part 1 PDF

By Milgram R. (ed.)

ISBN-10: 082181432X

ISBN-13: 9780821814321

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Therefore the integer j∗ (α) can be any multiple of 2. August 26, 2009 16:21 9in x 6in 32 b789-ch02 M. Kervaire and J. Milnor Let us study the effect of replacing ε by εα = ε+j(α)ε on the homology of the modified manifold. 6, where i carries ε into an element λ of order l > 1. Evidently lε must be a multiple of ε , say: lε + l ε = 0. Since ε is not a torsion element, these two elements can satisfy no other relation. Since εα = ε + j∗ (α)ε it follows that lεα + (l − lj(α))ε = 0. Now using the sequence ε i α α Hk M 0 → Hk Mα → 0, Z→ we see that the inclusion homomorphism iα carries ε into an element λα ∈ Hk Mα of order |l − lj(α)|.

Adams [1], [2]. Proof. Let Σ be a homotopy n-sphere. Then the only obstruction to the triviality of τ ⊕ ε1 is a well-defined cohomology class on (Σ) ∈ H n (Σ; πn−1 (SOn+1 )) = πn−1 (SOn+1 ). The coefficient group may be identified with the stable group πn−1 (SO). But these stable groups have been computed by Bott [4], as follows, for n ≥ 2: The mod 8 residue class: 0 πn−1 (SO) 1 2 3 4 5 6 7 Z Z2 Z2 0 Z 0 0 0. ) Case 1. n ≡ 3, 5, 6 or 7 (mod 8). Then πn−1 (SO) = 0, so that on (Σ) is trivially zero. Case 2.

Definition. The linking number L(λ, µ) is the rational number modulo 1 defined by L(λ, µ) = ν · µ. This linking number is well defined, and satisfies the symmetry relation L(µ, λ) + (−1)pq L(λ, µ) = 0 (compare Seifert and Threlfall [23]). 4. The ration l /l modulo 1 is, up to sign, equal to the selflinking number L(λ, λ). Proof. Since lε + l ε = 0 in Hk M0 , we see that the cycle lε + l ε on bM0 bounds a chain c on M0 . Let c1 = ϕ(x0 ×Dk+1 ) denote the cycle in ϕ(S k ×Dk+1 ) ⊂ M with boundary ε . Then the chain c − l c1 , has boundary lε; hence (c − l c1 )/l has boundary ε, representing the homology class λ in Hk M .

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Algebraic and Geometric Topology, Part 1 by Milgram R. (ed.)


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