A Survey of Spherical Space Form Problem by J. F. Davis

By J. F. Davis

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In particular if g*(k*) == 0 then the surgery obstruction is zero. At this point we note that since k* is primitive (2g )*( k*) == 0 in mod 2 cohomology, so the surgery problem corresponding to (2g) must have trivial surgery obstruction. Moreover after doing surgery the total space of the universal cover will be some homotopy sphere, and for these addition in the group of homotopy spheres r "-1 and addition of maps give the same result. Hence, since r" _I is finite, there is some finite multiple 2JL .

18 and 1*( a) ¥ 0 in L;(Z(Z/2)). REVIEW OF SURGERY THEORY 267 Here are a few further results on the L;(Z7T) for 7T finite. Perhaps the most basic result is the theorem of A. 19 For 7T a finite group where K runs over the conjugacy classes in 7T of 2-hyperelementary subgroups, and the maps are the restrictions. 19 implies the surgery obstruction is zero for a particular problem if and only if its restriction is zero for each 2-hyperelementary subgroup. It also reduces the study of L;n+I(Z7T) to the same question for 2-hyperelementary groups only.

Thomas, and C. T. C. Wall [27). 19 s( 4» = 0 if and only if s( r4» = 0 (where r is the restriction from H to 'TT) for every 2-hyperelementary He 'TT. 12 and our assumption s( r4> ) = 0 if and only if s( ror4» = 0 where ro : H z -i> H is restriction to the 2-Sylow subgroup. But from this we have s( 4» = 0 if and only if s( rz4» = 0 where rz: 'TTz -i> 'TT is restriction to the 2-Sylow subgroup. For n == 2 (mod 4) we have that 'TTz = Z/2 e, but from [18] L~(Z(Z/2e)) = 0 so this case is correct. Hence we may assume n == 0 (mod 4).

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