Consider the abelian group ($\mathbb{R}$,+) of real numbers with the usual addition. Is there a scalar multiplication \begin{equation} \cdot : \mathbb{R} \times \mathbb{R} \rightarrow \mathbb{R}, \end{equation} other than the usual multiplication, which makes ($\mathbb{R}$,+,$ \cdot $) a real vector space?


Let us denote by $m \colon \mathbb{R} \times \mathbb{R} \rightarrow \mathbb{R}$ the multiplication we are looking for and assume that $(\mathbb{R}, +, m)$ is a real vector space. Assuming the axiom of choice, every real vector space has a basis and so there must be an isomorphism of real vector spaces $\varphi \colon (\bigoplus_{i \in I} \mathbb{R}, +, \cdot) \rightarrow (\mathbb{R}, +, m)$. Then

$$ \varphi(\alpha \cdot v) = m(\alpha, \varphi(v)) $$

and so the scalar multiplication is given by

$$m(\alpha, x) = \varphi(\alpha \cdot \varphi^{-1}(x)). $$

Now, we can try and reverse this process. Let $\varphi \colon (\mathbb{R}^2, +) \rightarrow (\mathbb{R}, +)$ be an isomorphism of abelian groups (see this answer) and define a scalar multiplication on $\mathbb{R}$ by the formula

$$m(\alpha, x) = \varphi(\alpha \cdot \varphi^{-1}(x)) $$

where $\cdot$ is the standard scalar multiplication $\cdot \colon \mathbb{R} \times \mathbb{R}^2 \rightarrow \mathbb{R}^2$. Intuitively, we are transferring the standard scalar multiplication on $(\mathbb{R}^2, +, \cdot)$ to $(\mathbb{R}, +)$ using $\varphi$. You can easily check that $m$ satisfies the relevant four axioms to be a scalar product. Finally, if $m(\alpha, x) = \alpha x$ then

$$ \alpha x = \varphi(\alpha \cdot \varphi^{-1}(x)) \implies \varphi^{-1}(\alpha x)= \alpha \cdot \varphi^{-1}(x)$$

so $m$ is the standard multiplication if and only if $\varphi^{-1}$ is not only an additive isomorphism but also an isomorphism of vector spaces. This is not possible for dimension reasons so this indeed defines a different vector space structure on $(\mathbb{R}, +)$ and all such structures come from this construction. You can also take an additive non-linear isomorphism $\varphi \colon (\mathbb{R}, +) \rightarrow (\mathbb{R}, +)$ and apply the construction using this $\varphi$. In any case, after performing the construction the additive isomorphism $\varphi$ becomes by definition an isomorphism of real vector spaces.

  • 1
    $\begingroup$ This is really a nice solution! Your last suggestion seems to give the more direct construction. I only add a brief proof here of the existence of a non-linear additive bijection $\phi : \mathbb{R} \rightarrow \mathbb{R}$. Take a Hamel basis $\{ v_i \}_{i \in I}$ of $\mathbb{R}$ (with the usual addition and multiplication) considered as a vector space over $\mathbb{Q}$. Let $i, j \in I$, with $i \neq j$ and define $\phi(v_i)= v_j$, $\phi(v_j)=v_i$ and $\phi(v_k)=v_k$ for all other $k \in I$. Extend $\phi$ on $\mathbb{R}$ by linearity over $\mathbb{Q}$ and you're done. Thank you very much! $\endgroup$ – Maurizio Barbato Oct 14 '15 at 13:51
  • $\begingroup$ Thank! You are right that it is easier to use $\varphi \colon \mathbb{R} \rightarrow \mathbb{R}$ but for some reason I tried in the beginning to put a vector space structure on $(\mathbb{R},+)$ that is not one-dimensonal so I ended up using $\mathbb{R}^2$ instead of $\mathbb{R}$. $\endgroup$ – levap Oct 14 '15 at 14:00

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