Real Vector Space: Is Addition & Scalar Multiplication Smooth?

Geometry_dude
Messages
112
Reaction score
20
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.
 
Physics news on Phys.org
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.
 
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?
 
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.
 
Hello! There is a simple line in the textbook. If ##S## is a manifold, an injectively immersed submanifold ##M## of ##S## is embedded if and only if ##M## is locally closed in ##S##. Recall the definition. M is locally closed if for each point ##x\in M## there open ##U\subset S## such that ##M\cap U## is closed in ##U##. Embedding to injective immesion is simple. The opposite direction is hard. Suppose I have ##N## as source manifold and ##f:N\rightarrow S## is the injective...

Similar threads

Replies
4
Views
886
Replies
21
Views
3K
Replies
4
Views
2K
Replies
6
Views
2K
Replies
36
Views
4K
Replies
13
Views
3K
Back
Top