Which abelian groups can be fundamental groups of regions in euclidean spaces?

Abstract

Let Cn denote the collection of all abelian groups that can be fundamental groups of regions in It is clear that Ck c Ck+ v It is also easy to see that Cx and C2 each consist of just two groups—the trivial groups 1 and the infinite cyclic group Z. We shall see in this paper that actually Ck = Ck+1 for k ^ 4, so we shall be concerned mainly with the difference between regions in and regions in S. If a region A in S is not S* itself, we may assume that A c R, and that there is a point e of A that is at a distance ^ 1 from R A. Using barycentric subdivision Tk of JR W of mesh converging to zero, where 7J is a refinement of Tk if / < fc, let Uk be the interior of the union of those simplexes that lie in A and are at a distance rgfc from e. Take Ak to be the component of Uk that contains e(. It is easy to see that A{ ^ Ak if / < /c, and that U*°= i ^* = A ; thus n(A) is equal to the direct limit of the sequence {n(Ak)}. Since each n(Ak) is finitely generated, n(A) must be countable. Now suppose that G = n(A) is abelian. Since Gt = n(A^ is finitely generated, the image Kt of Gt in some Gs = n(As) of the inclusion Gt -• Gs must be abelian. Replacing the sequence {Gt} by a subsequence if necessary, we may assume that the image Kt of Gt in Gi+1 is abelian. The calculation of C3 is closely related to the following problem : Which elements of a link group commute? In fact, if we use brick subdivision instead of barycentric subdivision of R in the construction of Ak, we may assume that each Ak is the union of a finite number of handle-bodieswith-knotted-holes, semilinearly imbedded in S. Since each Gk is finitely generated, so is its abelianized group Gk = Hx(Ak). We can find nonsingular loops {xl9...9xp} that generate H^A^. By the Alexander duality theorem and the fact that Ak is a manifold, we* can also find nonsingular loops {yu..., yp} in 3 Ak which are dual to {x l 5 . . . , xp} in the sense that the linking number (xh yt) between xt and yt is equal to Sij9 where otj is the Kronecker delta. The image of any two elements of Gk _ j in Gk must commute in the complement of the link y 1 u y 2 U ' U y p . The following theorem (cf. [6] and [7]) makes it possible to deal with arbitrary links.