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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.

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    $\begingroup$ Any manifold can be embedded in $\mathbb R^n$ for some $n$, see e.g. here $\endgroup$
    – Wojowu
    Commented Aug 30, 2019 at 9:17
  • $\begingroup$ @Wojowu Thank you. Any chance you can add your response as an answer that I can accept? $\endgroup$
    – Ivan Meir
    Commented Aug 30, 2019 at 9:31

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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.

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    $\begingroup$ Thank you. It's perhaps worth noting that the result is trivially true for manifolds of dimension 1,2 and 3 simply because all manifolds in these lower dimensions have a unique differential and hence smooth structure. $\endgroup$
    – Ivan Meir
    Commented Aug 30, 2019 at 9:52
  • $\begingroup$ This is Nash embedding theorem $\endgroup$
    – debabrata chakraborty
    Commented Aug 30, 2019 at 10:21
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    $\begingroup$ @debabratachakraborty Nash embedding theorem concerns Riemannian manifolds, which ought to be $C^1$ for the Riemannian structure to even make sense. $\endgroup$
    – Wojowu
    Commented Aug 30, 2019 at 10:35
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    $\begingroup$ @IvanMeir I don't think one can call the uniqueness of smooth structures on three manifolds "trivial". I think it is the main result of an influential Annals paper. $\endgroup$
    – ssx
    Commented Aug 30, 2019 at 10:52
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    $\begingroup$ @Spiritofperpetualnegation I meant that my result in dims 1,2 and 3 was a trivial deduction from the proven fact that all manifolds in these dimensions have smooth structure. $\endgroup$
    – Ivan Meir
    Commented Aug 30, 2019 at 11:13

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