Does there exist a vector space $V$ which is isomorphic to $\mathbb{R}^m$, but is not a topological/smooth manifold of dimension $m$? Does there exist a vector space $V$ which is isomorphic to $\mathbb{R}^m$, but is not a topological/smooth manifold of dimension $m$?



Suppose $V$ is isomorphic to $\mathbb{R}^m$.

In the category of vector spaces, $V$ is equivalent to $\mathbb{R}^m$, but $V$ may not have a topology on it, so it may not be a topological manifold, and hence it can definitely not be a smooth manifold (since all smooth manifolds are just topological manifolds endowed with a smooth structure).

If $V$ is homeomorphic to $\mathbb{R}^m$, then we can conclude in that $V$ is equivalent to $\mathbb{R}^m$ in the category of topological spaces, and $V$ must also be a topological manifold, however since every topological manifold is not a smooth manifold (see the Exotic Sphere as an example), we can't necessarily conclude that $V$ is diffeomorphic to $\mathbb{R}^m$.



So my question is the following, does an isomorphism between $V$ and $\mathbb{R}^m$ induce a homeomorphism between $V$ and $\mathbb{R}^m$, and furthermore does a homeomorphism between $V$ and $\mathbb{R}^m$ (along with the isomorphism between the two vector spaces) induce a diffeomorphism between $V$ and $\mathbb{R}^m$?

If not could someone provide me with an example of a vector space $V$ which is  isomorphic to $\mathbb{R}^m$, but not homeomorphic, and also of an example of a vector space $V$ which is homeomorphic and isomorphic to $\mathbb{R}^m$ but not diffeomorphic to $\mathbb{R}^m$?



Note: Isomorphism here means isomorphism of vector spaces.
You can give whatever topology/smooth structure to $\mathbb R^n$ as you like without affecting it's vector space structure.

Consider $X=\mathbb R^n$ with the trivial topology, i.e. only two open subsets: the empty set and $X$. Then, this is not homeomorphic to the $Y=\mathbb R^n$ with the usual Euclidean topology. Then these are not homeomorphic spaces for many reasons (eg: $X$ is compact but $Y$ is not).

Again, let $X'=\mathbb R^n$ and find a bijection with $\mathbb R$ and use this bijection to make $X'$ a $1$ dimensional manifold. Of course this cannot be homeomorphic to $Y=\mathbb R^n$ which is an $n$-dimensional manifold for $n&gt;1$.
No. Change of base in V gives you a smooth function in $\mathbb{R}^n$. So you have a smooth structure which is diffeomorphic to $\mathbb{R}^n$