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## 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?

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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 $H^{n-1}(X)=H_1(X,\partial X)$.

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 Tom, thank you very much for this answer. But how does it follow that $X$ is a compact $n$-manifold with boundary in $n$-space? For example, the minimal CW decomposition of the $2$-sphere embeds in $\mathbb{R}^3$ but surely not as a compact $3$-manifold, so I must have misunderstood something in the "without loss of generality" part. – Vidit Nanda Oct 26 at 4:43 @Vel Nias: I think Tom is using the fact that a finite complex embedded in a manifold has a regular neighbourhood which deformation retracts to the complex. This is, by definition, a manifold with boundary, and it is compact if $X$ is finite. – Mark Grant Oct 26 at 7:42 I don't need the complex to be a deformation retract of a compact manifold neighborhood, just a retract. (A retract of a torsionfree group is torsionfree.) To get such a neighborhood, use that a finite complex is an ENR, and that every neighborhood of a bounded set in $n$-space contains a compact manifold neighborhood – Tom Goodwillie Oct 26 at 10:13 Of course. Thanks, Mark and Tom! – Vidit Nanda Oct 26 at 17:30