# Embedding of exotic manifold

While solving some problems on Differential Topology I asked myself this question: Suppose a smooth manifold having atleast two exotic smooth structures can be embedded in some Euclidean space under one smooth structure; will the same thing happen for the other smooth structure?

In other words,will the smallest dimensional Euclidean spaces in which the manifold embeds under different smooth structures be the same?

Ps. I am asking this question just out of my curiosity.

Every exotic sphere (of dimension $$\ge 7$$) is an example. Indeed, suppose $$\Sigma$$ is an exotic $$n$$-dimensional sphere and that it embeds (smoothly) in $$R^{n+1}$$. Then it bounds a contractible compact submanifold $$W$$ of $$R^{n+1}$$. Now, remove a small ball $$B$$ from $$W$$. The result is an h-cobordism between $$\Sigma$$ and the boundary of $$B$$, which is the usual sphere $$S^n$$. Hence, by the smooth h-coboridism theorem, $$\Sigma$$ is diffeomorphic to $$S^n$$.

• Nice! Thank you. – Paladin May 12 '19 at 14:49
• How do you show that $\Sigma$ bounds a compact submanifold? Does this follow from some sort of Schoenflies type theorem? – Michael Albanese Feb 16 at 0:54
• @MichaelAlbanese: This is simpler than Schoenflies. It is a consequence of the Alexander Duality that $\Sigma$ separates $R^{n+1}$ in two components. One complementary component will be unbounded, the other bounded. The closure of the latter will be the compact submanifold $W$ bounded by $\Sigma$. – Moishe Kohan Feb 16 at 1:05

Not necessarily. Here's one such (non-compact) example.

There are uncountably many smooth structures on $$\mathbb{R}^4$$. An exotic $$\mathbb{R}^4$$ is said to be small if it can be smoothly embedded in the standard $$\mathbb{R}^4$$, and large otherwise; both small and large $$\mathbb{R}^4$$'s exist. A small $$\mathbb{R}^4$$ (or standard $$\mathbb{R}^4$$) embeds in $$\mathbb{R}^4$$, large $$\mathbb{R}^4$$'s do not. So the smallest $$n$$ for which a large $$\mathbb{R}^4$$ embeds into $$\mathbb{R}^n$$ satisfies $$n > 4$$ (and $$n \leq 8$$ by the Whitney Embedding Theorem). In fact, every large $$\mathbb{R}^4$$ embeds in $$\mathbb{R}^5$$, see this question.

• Presumably you should mention whether it's been shown that there are both small and large $\Bbb R^4$s. :) – John Hughes May 12 '19 at 1:49
• @John Hughes: Yes, of course. – Michael Albanese May 12 '19 at 1:50
• Thank you. It helped. – Paladin May 12 '19 at 14:50