From: rusin@vesuvius.math.niu.edu (Dave Rusin)
Subject: Re: Cell complex
Date: 2 Jul 1999 03:54:35 GMT
Newsgroups: sci.math
Keywords: Every set of abelian groups can be the homology of a complex

In article <7l83g6$ad2$1@metro.ucc.usyd.edu.au>,
Kundan Misra  <"kundan"@maths.usyd.edu.au, kundanmisra@hotmail.com> wrote:
>Let H_i be a finitely generated abelian group, for 1 <= i <= n.
>
>Show that there is a connected finite cell complex X (with all cells of
>dimension at most n+1) such that H_q(X) isomorphic to H_q, for 1 <= q <=n.

Odd -- no response has yet shown up here.

First note that it suffices to assume  H_i = {0}  for all 0<i<n.  For if
we can construct a connected finite cell complex  X_n  in this simple case, 
then in the general case we need only let  X  be the wedge-product 
(one-point union) of all the corresponding  X_i .

Likewise note that if  H_n = G1 x G2  (a direct product of groups) then a
similar argument shows that it suffices to construct complexes with
H_n(X_i) = G_i  and let  X   be their wedge-product. By induction on the
number of generators, this reduces the problem to the construction of
a cell complex with a single cyclic homology group.

To obtain a cell complex with H_n(X) = Z  (infinite cyclic group) take
X  to be [a cellular decomposition of] the  n-sphere; as you probably know,
you may construct  X  with a single 0-cell and a single  n-cell  E  attached
with the only possible attaching map  (f | bdy(E) = constant).

To obtain a cell complex with  H_n(X) = Z/kZ  (cyclic of order  k), construct
a space  X  with one 0-cell, one  n-cell, and one (n+1)-cell as follows:
attach the  n-cell as above so that the  n-skeleton is the  n-sphere  S^n;
then attach the  (n+1)-cell  E by using as the attaching map any function
f : boundary(E) -> S^n  of degree k. Here you'll need to recognize the
boundary of  E  as being homeomorphic to  S^n  as well; with this
identification made you can explicitly give such a map  f  with formulas
if that appeals to you -- take the points in  S^n  as being those in
R^(n+1) = C x R^(n-1)  whose distance from the origin is  1, and let
f be the restriction of  F : C x R^(n-1) -> C x R^(n-1)  to the unit 
sphere, where  F(z,v) = (wz,v)  and  w  is a primitive  k-th  root of unity.
[As I said: "if that appeals to you" ...].

If the term "suspension" means anything to you, you may simply perform
this construction with  n=1, and then suspend  n-1  times.

The same approach works with homotopy groups instead of homology groups,
except we need to use direct products instead of wedge products. This
reduces to the need to construct spaces  K(C,n)  with only one
nonzero homotopy group  pi_n(K) = C. These spaces exist, but are almost
never finite-dimensional. They're called Eilenberg-MacLane spaces.
K(Z,1) is the circle, for example, but K(Z,2) is not the two-sphere; as
noted recently in this group  pi_3(S^2) isn't {0}. Be thankful you're
working in the simpler homology setting!

dave