# How can I construct a closed manifold from a finite CW complex?

If I start with a, say, 3-CW complex $$X$$ which can be embedded in $$\mathbb{R}^5$$, I can get a neighbourhood $$U$$ of $$X$$ which has the same homotopy type of $$X$$. Then $$U$$ is a $$5-$$ dimensional open manifold. Can I get a close manifold (compact without a boundary) $$M$$, of dimension $$6$$ (or some higher dimension) such that $$M$$ and $$X$$ have the same homotopy type?

• As the answer indicates, this is impossible in general. For something a little weaker, you might be interested in Mike Davis' "reflection group trick". For any aspherical complex $X$, say, I think this enables you to produce a closed aspherical manifold $M$ such that $\pi_1(M)$ (virtually?) retracts to $\pi_1(X)$. I'm not certain if this group-theoretic retraction can be improved to a topological retraction. This should be explained in Davis' book "The geometry and topology of Coxeter groups".
– HJRW
Aug 8, 2021 at 16:11
• You are almost always going to have to attach some cells to get the homotopy-type of a manifold. Perhaps a productive way to rephrase your question would be through the lens of starting with a compact (boundaryless) manifold. If you puncture that manifold, you get the homotopy-type of a lower-dimensional CW complex. Which CW complexes do you get? Aug 9, 2021 at 0:43

Take $$X=S^3$$. Then no closed manifold of dimension at least 6 has the same homotopy type.
More generally, suppose $$n \le m$$ are non-negative integers, $$X$$ is a CW complex of dimension $$\le n$$, $$M$$ is a non-empty, closed $$m$$-manifold, and $$X$$ and $$M$$ have the same homotopy type.
It is well known that a non-empty closed $$m$$-manifold has non-trivial mod 2 homology in degree $$m$$, whereas a CW complex of dimension $$\le n$$ has no homology above dimension $$n$$.
Since $$X$$ and $$M$$ have the same homotopy type, they have isomorphic homology, so $$H_m(X;\Bbb Z/2) \ne 0$$. So the only possibility is that $$m=n$$.
The mod-2 homology groups of any closed $$n$$-manifold are `$$n$$-palindromic' by Poincare duality, i.e., $$H_m(M;\mathbb{Z}/2)\cong H_{n-m}(M;\mathbb{Z}/2)$$ for each $$m$$. If the mod-2 homology of $$X$$ is not $$n$$-palindromic for any $$n$$, there cannot be any closed manifold $$M$$ having the same mod-2 homology as $$X$$.
The smallest such $$X$$ is the one-point union of two circles.