Is the following problem equivalent to the embedding problem?

Question

What is the smallest $n\in\Bbb N$ such that a connected closed oriented manifold $M^m$ separates the $\Bbb R^n$ into two connected component? i.e. $\Bbb R^n \setminus {M}=M_1\sqcup M_2$? in other words: $H_0(\Bbb R^n\setminus M)=\Bbb R\oplus\Bbb R.$

How to solve this problem? at least in simple cases? Does "smallest" make sense here? i.e. Is $n$ unique?

I know that for hypersurfaces $n=m+1$. By strong Whitney embedding theorem $n\leq 2m$. (Right?)

This is similar to Nash's problem. – C.F.G Apr 04 '20 at 12:38 — C.F.G, Apr 04 '20 at 12:38
To be clear, is $M$ supposed to be fixed at the start? – Eric Wofsey Apr 04 '20 at 14:29 — Eric Wofsey, Apr 04 '20 at 14:29

Eric Wofsey · Accepted Answer · 2020-04-04T15:24:43.493

It follows from Alexander duality that if $M\subset\mathbb{R}^n$ is a closed manifold, then $\tilde{H}_0(\mathbb{R}^n\setminus M)\cong H^{n-1}(M)$. It follows immediately that for $M$ connected, $M$ separates $\mathbb{R}^n$ iff $M$ has dimension $n-1$ and is orientable.

So, starting with a connected closed orientable manifold $M$ of dimension $m$, then $M$ can never separate $\mathbb{R}^n$ for any $n$ unless $M$ embeds in $\mathbb{R}^{m+1}$.

Is the following problem equivalent to the embedding problem?

1 Answers1