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.

Thanks in advance.

(Related: number of differents vector space structures over the same field $\mathbb{F}$ on an abelian group)


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 $0<n,m \leq 2^{\aleph_0}$, 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 $m$.

  • $\begingroup$ Excellent, thanks! :) $\endgroup$ – 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 $\Bbb{R}$.

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

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