By J. F. Davis
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Additional info for A Survey of Spherical Space Form Problem
DAVIS AND R. J. MILGRAM 246 type has a non-zero finiteness obstruction and hence is not the homotopy type of a manifold. D. The Swan Obstruction and Reidemeister Torsion A basic philosophy is to avoid doing computations in Ko at all costs-do them in Kt instead! This is accomplished by means of pullback diagrams and Mayer-Vietoris type sequences in K-theory. Another philosophy is the Hasse principle-to solve a problem over Z, solve it first locally at all primes, and then globally. 29 Here Zp = lim Z/ pn is the p-adic completion of Z and Qp is its quotient field.
1 (Wall ) Let p' = jimage 4J1 for G = A(m, pt, 4J). Then G has period 2 p s and a2P'( G) = O. Proof Suppose first ps < pt, then there is a free linear representation in O(2 p S) defined by inducing up a faithful 2-dimensional representation of Z/ m . pl-s. This gives the desired complex on triangulating the orbit space of S(2 p '-I) under the action. Now, consider the case ps = pt. Then there is an eVIdent surjection S: A(m, pS+', 4J1T) ~ A(m, ps, 4J) (where 1T: Z/ ps+1 ~ Z/ ps is the surjection).
Let E (K) denote the units in the ring of integers of a number field K. Let E 2 (K) denote the squares in E(K). Let E+(K) denote the elements of E(K) which are positive at all real places. 11 Let K/Q be totally real and Galois of degree 2'. (a) E+(K) = E 2 (K) if and only if there is a u E E(K) such that, Nu = -1. ) (b) Let H be a subgroup of E (K) offinite index. If there is a u EH with Nu = -1, then IE(K)/ HI is odd. THEOREM Let An = 'n +,~I. 12 (~) = -1, If K is Kp' or K pq with the quadratic or the maximal 2 -extension of Q in Q[A 2 ", Apl with p ~ 1 (mod 8) then E 2 (K) = E+(K) and the cyclotomic units have odd index in U(K).