Download e-book for iPad: An extension of Casson's invariant by Kevin Walker

By Kevin Walker

ISBN-10: 0691025320

ISBN-13: 9780691025322

ISBN-10: 0691087660

ISBN-13: 9780691087665

This booklet describes an invariant, l, of orientated rational homology 3-spheres that's a generalization of labor of Andrew Casson within the integer homology sphere case. allow R(X) denote the distance of conjugacy periods of representations of p(X) into SU(2). allow (W, W, F) be a Heegaard splitting of a rational homology sphere M. Then l(M) is asserted to be an accurately outlined intersection variety of R(W) and R(W) within R(F). The definition of this intersection quantity is a fragile activity, because the areas concerned have singularities. A formulation describing how l transforms below Dehn surgical procedure is proved. The formulation comprises Alexander polynomials and Dedekind sums, and will be used to provide a slightly user-friendly facts of the lifestyles of l. it's also proven that once M is a Z-homology sphere, l(M) determines the Rochlin invariant of M

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Fig. 2 2. Let S = {1, 2, 3, 4} and R 2 be defined on S by a R2 b ~ a + b :::; 5. The set of arcs is given by A = {(1, 1), (1, 2), (1, 3), (1, 4), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (4, 1)}. (See Fig. ) 4 Fig. 3 51 GUIDE TO ABSTRACT ALGEBRA 3. Let S = {1, 2, 3, 4, 5, 6} and R 3 be defined on S by a R3 b 4 ¢> a = 2b. 2 ~ 6 • 5 3 Fig. 4 The set of arcs is given by A = {(2, 1), (4, 2), (6, 3)}. (See Fig. 5a I Draw graphs to illustrate the following relations on the set s ={1, 2, 3, 4, 5, 6}.

Hence from (i) and (ii), ii = b. • It is easy to show that the converse of this result also holds. 2 a = b =>a ~ b, 'I a, b E S. Proof Suppose Zi =b. We know a E a, by the reflexive property of Hence a E b, since a= b. But a E b => a ~ b, by definition of b. • ~. 2 gives that for any equivalence relation ~ on any set S, This result justifies our earlier assumption that any element of an equivalence class can equally well serve as class representative. Note that we 42 RELATIONS have had to use all three properties of an equivalence relation in the proof.

B = 0 =>A B = A s B'. Is B sA' also true? n B'. 5 Prove that A' n B' = (A U B)'. Hence show that A' n (B U C)' = (A U B)' n C'. 6 Prove that (A' n B)'. (b) A (a) (A')' =A; Hence prove that (A' n B)' U C = A U C if and only if B' sA U C. ) s 7 Prove that (A n B) U (B n C) U (C n A) = (A U B) n (B U C) n (C U A). What identity is obtained by putting C = 0 in this result? 4b In the following problems, A, B, C, D are subsets of a set 'fo. 1 Put the correct sign =>, <= or ditions on a real number x: ¢> between the following pairs of con- 1 (a) x E Z+;- E 0.

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