# Smooth Vector Space

1. Sep 24, 2014

### Geometry_dude

Let $V$ be a real vector space and assume that $V$ (together with a topology and smooth structure) is also a smooth manifold of dimension $n$ with $0 < n < \infty$, not necessarily diffeomorphic or even homeomorphic to $\mathbb R^n$.

Here's my question: Does this imply that addition and scalar multiplication is smooth?

I tried to find a counterexample and thought about exotic $\mathbb R^4$, but my knowledge about that is quite limited.

2. Sep 24, 2014

### dextercioby

Any real vector space of dimension n<infinity is necessarily isomorphic to R^n, as vector spaces. I think the manifold you're trying to imagine is necessarily diffeomorphic to R^n.

3. Sep 24, 2014

### Geometry_dude

That there is a vector space isomorphism to $\mathbb R^n$ is not disputed, yet this does not necessarily mean that it is a homeomorphism or diffeomorphism when we consider $\mathbb R^n$ with the standard smooth structure and topology.

EDIT: Maybe group theory holds the answer?

4. Sep 24, 2014

### dextercioby

Alright then, let's assume I am wrong: offer me an example of a vector space endowed with a topology (what type ?) and a smooth structure that is not diffeomorphic to R^n with the usual topology and differential structure.

5. Sep 24, 2014

### Geometry_dude

You know very well that this is not the way mathematics works.

I do know one though, the one in the opening post:
http://en.wikipedia.org/wiki/Exotic_R4