By J. F. Davis
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Additional resources for A Survey of Spherical Space Form Problem
MILGRAM TP'-I). Then set R equal to Now consider the pull-back diagram where the gluing automorphisms are the identity on Qq ® C and Qp ® V® C, while at Qp ® R the ReidemeisterDeRham torsions T of the gluing automorphisms are I at Qp(~pJ), j2:: 1 and pp' at the trivial representation Qp. But this torsion T satisfies a( T) = X( C q ). On the other hand, T is pS[p x- q1m Iq] at the trivial representation, and 1 at all the other representations. 5: aps[p x - Iq] = psa[p x - Iq] = O. 1 follows. qlm 0 In particular 0"4(D2r ) = 0 if r is odd.
But (T4( Q(24, 13, I)) = O. When n > 3 there are several major differences. First, Q(2 n , p, I) is not a subgroup of the unit quaternions S3, while Q(8, p, I) = Q(8p) is. Second, the Swan obstruction (T4( Q(2 n, p, I)) depends on the units in Z[A z"·', Ap ]. Here it is possible for non-cyclotomic units to influence the Swan obstruction (at least if p == I (mod 8)). 15  Let p be a prime not congruent to modulo 8. O ifp~ I (mod2 n-'). THEOREM In particular (T4( Q( 16,3, I)) ;£. O. This is a group of order 48 which gives the smallest possible group with a nonzero Swan obstruction.
Suppose A ~ A is a mapping of rings. Now suppose E is a finite free complex over A, not necessarily based or acyclic. Then if A ® A E is acyclic we define where A®AE is given a basis induced by an A-basis of E. Let A be the kernel of the augmentation map Then and are well defined, with the class of [4J ] equal to 'T( D) - 'T( C). One can further show that 'T(D) is trivial and 'T( C) is in fact an invariant of the group G. 30 in greater detail. Since ZpG is a semilocal ring, Ko(ZpG) is torsion free, as is Ko(QpG).
A Survey of Spherical Space Form Problem by J. F. Davis