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It is not hard to see that $(\mathbb R^2,+)$ with this product

$ {r\cdot(x,y)=(rx,ry) } $

is vector space over field $\mathbb R$.

I'm looking for another product that $(\mathbb R^2,+)$ is vector space over $\mathbb R$. I know

$n*(x,y)=\underbrace{(x,y)\oplus (x,y) \oplus \cdots \oplus (x,y)}_n=(nx,ny)$ but I have no idea for arbitrary element of $\mathbb R$. Any Suggestion. Thanks

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You are trying to find other definitions for scalar multiplications? –  Asaf Karagila Jun 23 '12 at 18:23
    
Aren't the two products equivalent, at least for natural $n$? –  talmid Jun 23 '12 at 18:30
    
@AsafKaragila yes I'm trying find another definitions for scalar multiplications –  Babak Miraftab Jun 23 '12 at 18:36
    
You point out that $n\cdot (x,y) = (n\cdot x, n\cdot y)$ for $n\in \Bbb N$. It's worth pointing out that you can extend this argument to show that $q\cdot (x,y) = (q\cdot x, q\cdot y)$ for $q\in \Bbb Q$. And, if you have just a bit more structure (like say, considering inner product space structures of $(\Bbb R^2,+)$ over $\Bbb R$), you can extend the argument further to any $r\in \Bbb R$. –  Managu Jun 23 '12 at 18:53

1 Answer 1

If you are allowed to use the axiom of choice, then the abelian groups $(\mathbb R, +), (\mathbb R^2, +), (\mathbb R^3,+), \ldots$ are isomorphic. In fact they are all isomorphic to a direct sum of continuum-many copies of $(\mathbb Q,+)$. So giving a vector space structure to any of them is the same thing. In particular, we know how to give $(\mathbb R^3,+)$ a structure of $\mathbb R$-vector space which is of dimension $3$, thus by transporting it to $\mathbb R^2$, we can give a vector space structure to $\mathbb R^2$ making it into a 3-dimensional $\mathbb R$-vector space (and similarly for any dimension you wish, as long as it's not more than the continuum cardinality).

In fact, even without the axiom of choice, giving a finite dimensional $\mathbb R$-vector space structure to any abelian group $G$ is the same as finding a group isomorphism from $G$ to $\mathbb R^n$ and transporting the natural vector space structure of $\mathbb R^n$ back to $G$. So the real question is how to find those group isomorphisms.

Sadly, I don't think it is possible to find any nontrivial one without the axiom of choice, so all the nontrivial vector space structure you can put on $\mathbb R^2$ need you to use it, and are not constructive.

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"not too big"=cardinality at most that of the continuum? –  Managu Jun 23 '12 at 19:03
    
@ Managu : yes, I wasn't sure so I needed to check that a direct sum of $\kappa$-many direct sums of $\kappa'$-many copies of $\mathbb Q$ is a direct sum of max $(\kappa,\kappa')$-many copies, and of cardinal max $(\kappa,\kappa',\aleph_0)$ as a result. –  mercio Jun 23 '12 at 19:27

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