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?
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3$\begingroup$ 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". $\endgroup$– HJRWCommented Aug 8, 2021 at 16:11
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$\begingroup$ 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? $\endgroup$– Ryan BudneyCommented Aug 9, 2021 at 0:43
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3 Answers
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Take $X=S^3$. Then no closed manifold of dimension at least 6 has the same homotopy type.
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6$\begingroup$ More generally, let X be an n dimensional complex and let m > n and let Y be a closed m manifold. Then by poincare duality Y has non vanishing cohomology in dimension m for aproppriate coefficients. But X has vanishing cohomology in dimension m for any coefficients. So X and Y are not homotopy equivalent. $\endgroup$ Commented Aug 8, 2021 at 14:46
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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 $m=n$.
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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.
