TITLE: Which Abelian Group sequences arise as the Homology of Embedded CW Complexes?
QUESTION [7 upvotes]: Background
Let $\mathcal{A} = \lbrace A_0, \ldots, A_M \rbrace$ be an arbitrary sequence of finitely generated Abelian groups. It is well-known that a finite CW complex $X_\mathcal{A}$ may be constructed so that its $m$-th homology group $H_m(X_\mathcal{A})$ equals $A_m$ for all $0 \leq m \leq M$. In particular, one can build wedge sums of Moore spaces.
On the other hand, any CW complex is an Euclidean neighborhood retract (see Hatcher corollary A.10) and hence embeds into some Euclidean space. By a geometric complex of dimension $n$ I mean a CW complex whose minimal Euclidean embedding dimension is $n$.
Question
What $\mathcal{A}$'s arise as homology sequences for $n$-dimensional geometric complexes as a function of $n$?
Notes and Considerations
Some obstructions to having a CW complex embed in Euclidean space are outlined in the answers to this older question and may be helpful although I have not been able to use them efficiently. What one can say immediately about the sequences $\mathcal{A}$ arising from an $n$-dimensional geometric complex is that all but the first $n-1$ groups must be trivial by dimension considerations, and that the $(n-1)$-st group must be torsion-free. Alexander duality imposes constraints as well, but it is unlikely that this list of conditions characterizes the $\mathcal{A}$'s that are possible homology groups of geometric complexes. Is there a complete answer available somewhere?

REPLY [10 votes]: This is simpler than that other question, because it's only the homology and not the homotopy type that is being prescribed.
A Moore space with $H_1$ finite cyclic can be embedded in $4$-space, so by suspending this a Moore space with $H_m$ finite cyclic can be embedded in $n$-space if $1\le m\le n-3$. And of course an $m$-sphere can be embedded as long as $0\le m\le n-1$. Wedging together such Moore spaces and spheres you can achieve any sequence of finitely generated homology groups such that 
$A_m=0$ for $m\ge n$
$A_{n-1}$ and $A_{n-2}$ are torsionfree
$A_0$ is torsionfree and nontrivial.
Of course you can also achieve $A_m=0$ for all $m$ by taking $X$ to be empty. The other trivial case not satisfying the conditions above is a point embedded in $0$-space.
Conversely, the homology of any finite complex $X$ in $n$-space (other than those two trivial cases) must satisfy the conditions above.
The fact that $A_{n-2}$ is torsionfree follows from duality. Without loss of generality $X$ is a compact $n$-manifold-with-boundary in $n$-space. The map $H_1(X)\to H_1(X,\partial X)=H^{n-1}(X)$ is zero because it factors through the map $H_1(B)\to H^{n-1}(B,\partial B)$ for a closed ball $B$ containing $X$. This makes $H_1(X,\partial X)$ inject into $H_0(X)$, which is impossible if there is torsion in $H_{n-2}(X)$ and therefore in $H^{n-1}(X)=H_1(X,\partial X)$.