homology $n$-sphere in $\mathbb R^{n+1}$ is a sphere?

Question

Recall that a homology sphere of dimension $n$ is an $n$-dimensional manifold $X$ for which $$ H_0(X,\mathbb Z) = H_n(X,\mathbb Z) = \mathbb Z $$ and $$H_k(X,\mathbb Z) = \{0\}, \quad k \neq 0, n.$$

Does the requirement that $X$ is embedded in $\mathbb R^{n+1}$ eliminate any (compact) homology spheres? In other words, can we get rid of some (or all) non-sphere homology $n$-spheres if the ambient space is $\mathbb R^{n+1}$?

What kind of embedding? Do you assume smoothness? See also my answer here — Moishe Kohan, Apr 29 '24 at 08:33
@MoisheKohan embedding in the sense of topological embedding, and whatever extra structure is required. — shuhalo, Apr 29 '24 at 10:51

Moishe Kohan · Answer 1 · 2024-04-30T05:56:12.443

Here is what I know:

Every integer homology sphere $X$ of dimension $n\ne 4$ admits a locally flat embedding in $\mathbb R^{n+1}$. The proof is quite involved: One first proves that $X$ is smoothable (mostly Kirby and Siebenmann, and Kervaire, see references here). Then one constructs a smooth embedding in $\mathbb R^{n+1}$ using (mostly) results of Kervaire ($n\ge 5$, see my answer here), but also the h-cobordism theorem, and Freedman ($n=3$):

M Freedman, The topology of four–dimensional manifolds, J. Diff. Geom. 17 (1982), p. 357–453.

Namely, Freedman proves that every homology 3-sphere $X$ bounds a contractible topological 4-manifold $W$. Doubling $W$ along its boundary one obtains a locally flat topological embedding of $X$ in a homotopy 4-sphere $S$. Freedman also proved that every homotopy 4-sphere $S$ is homeomorphic to $S^4$. Hence, we obtain a locally flat embedding of $X$ in $S^4$.

The situation in the case $n=4$ is unclear: Smooth homology 4-spheres all embed in $S^5$. However, it is unclear which topological homology 4-spheres are smoothable (possibly all) and if every topological 4-manifold which is a homology 4-sphere embeddable in $S^5$ is smoothable.

homology $n$-sphere in $\mathbb R^{n+1}$ is a sphere?

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