# Is the following problem equivalent to the embedding problem?

Is the following problem equivalent to the embedding problem?

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. Apr 4 '20 at 12:38
• To be clear, is $M$ supposed to be fixed at the start? Apr 4 '20 at 14:29

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 $$\mathbb{R}^{m+1}$$.