# Are all manifolds in $\mathbb R^n$ homeomorphic to a smooth manifold?

My question is whether all manifolds that can be embedded in $$\mathbb R^n$$ are homeomorphic to a smooth manifold?

I know that every smooth manifold can be triangulated which I think is a result of Whitehead and I think every manifold in $$\mathbb R^n$$ can be triangulated so this lends plausibility I think. (If the dimension 4 E8 manifold can be embedded in $$\mathbb R^n$$ then we have a counterexample but I'm not sure if it can be. All manifolds of dimension up to 3 can be triangulated).

I appreciate if this is a basic question but it doesn't seem to be spelt out explicitly anywhere. Most textbooks define a manifold and then a smooth manifold but don't say in which cases these concepts may be equivalent.

• Any manifold can be embedded in $\mathbb R^n$ for some $n$, see e.g. here – Wojowu Aug 30 '19 at 9:17
• @Wojowu Thank you. Any chance you can add your response as an answer that I can accept? – Ivan Meir Aug 30 '19 at 9:31

The answer is negative. As you observe yourself, if we allow abstract manifolds, then the E8 manifold provides a counterexample. However, it turns out that any abstract manifold can be embedded into $$\mathbb R^n$$ for a suitable $$n$$ (for instance, twice the dimension plus one, see e.g. here), so in particular the E8 manifold is a submanifold of $$\mathbb R^n$$ (for $$n=9$$) which is not homeomorphic to a smooth manifold.
• @debabratachakraborty Nash embedding theorem concerns Riemannian manifolds, which ought to be $C^1$ for the Riemannian structure to even make sense. – Wojowu Aug 30 '19 at 10:35