# Can an abelian group be a real vector space in more than one way?

We know that any $$\mathbb{Z}$$ module structure on an abelian group is unique, and furthermore the same is true for $$\mathbb{Q}$$.

Any complex vector space structure is not unique, we can just compose with an automorphism of $$\mathbb{C}$$ (for example the conjugation map).

However, $$\mathbb{R}$$ has trivial automorphism group, see:

Is an automorphism of the field of real numbers the identity map?

So my question is, given an abelian group made into an $$\mathbb{R}$$ vector space, is this the only way we can do this?

We would get another structure if we can embed $$\mathbb{R}$$ into itself, but I'm not sure if this is possible.

If not, then of all the different $$\mathbb{R}$$ structures, must they all have the same dimension?

I know that both of these are not true for general fields.

Sure. Note that instead of using an automorphism of $$\mathbb{R}$$, you can use an automorphism of the abelian group. Given $$\mathbb{R}$$-vector spaces $$V$$ and $$W$$ any abelian group isomorphism $$f:V\to W$$, you can define a new $$\mathbb{R}$$-vector space structure on $$V$$ with scalar multiplication $$\mathbb{R}\times V\to V$$ given by $$(r,v)\mapsto f^{-1}(rf(v))$$. This will not be the same as the original scalar multiplication on $$V$$ unless $$f$$ is $$\mathbb{R}$$-linear.

In particular, as an abelian group, an $$\mathbb{R}$$-vector space $$V$$ of dimension $$n$$ is just a $$\mathbb{Q}$$-vector space of dimension $$n\cdot 2^{\aleph_0}$$. If $$n>0$$, this gives loads of automorphisms of $$V$$ that are not $$\mathbb{R}$$-linear (for instance, for any nonzero $$v\in V$$ and any irrational $$\alpha\in\mathbb{R}$$, $$v$$ and $$\alpha v$$ are linearly independent over $$\mathbb{Q}$$ so there is an automorphism that exchanges $$v$$ and $$\alpha v$$, and this cannot be $$\mathbb{R}$$-linear), and thus loads of isomorphic but distinct $$\mathbb{R}$$-vector space structures on $$V$$. Or, if $$0, then $$n\cdot 2^{\aleph_0}=m\cdot 2^{\aleph_0}=2^{\aleph_0}$$, so $$V$$ is actually isomorphic as an abelian group to any $$\mathbb{R}$$-vector space $$W$$ of dimension $$m$$, and so $$V$$ can also be given an $$\mathbb{R}$$-vector space structure of dimension $$m$$.

• Excellent, thanks! :) – James Jul 12 at 22:42

$$\Bbb{R}$$ and $$\Bbb{R}^2$$ are isomorphic as vector spaces over $$\Bbb{Q}$$ and hence (because they have dimensions $$c$$ and $$c^2$$ over $$\Bbb{Q}$$ respectively, where $$c$$ is the cardinality of the continuum. and $$c^2 = c$$). Hence $$\Bbb{R}$$ and $$\Bbb{R}^2$$ are certainly isomorphic as abelian groups but they not isomorphic as real vector spaces because they have different dimensions over $$\Bbb{R}$$.

• Thanks, and are of different $\mathbb{R}$ dimension. – James Jul 12 at 22:27