It is common knowledge that the abelian group $\mathbb{R}$ forms a vector space over the field $\mathbb{R}$ with scalar multiplication coming from the product in $\mathbb{R}$. My question is: is there any other scalar multiplication $\cdot: \mathbb{R} \times \mathbb{R} \to \mathbb{R}$ making $\mathbb{R}$ into a different vector space over itself with the same additive structure? I am aware that the action of $\mathbb{Q}\subset\mathbb{R}$ by scaling is entirely determined by the vector space axioms, and that $\mathbb{R}$ has no non-identity automorphisms as a ring. It still seems possible, perhaps using the axiom of choice somehow, that there could be such an "exotic" action. Another more general question I would like to resolve is this: given an $\mathbb{F}$-vector space $(V, +, \cdot)$, must all possible scalar multiplications be related by an automorphism of $\mathbb{F}$? That is, if I have two different ring maps $\cdot: \mathbb{F} \to \operatorname{End}(V)$ and $\cdot': \mathbb{F} \to \operatorname{End}V$ serving as scalar multiplications that make $V$ a vector space with the same additive structure, must there be some ring automorphism $\sigma: \mathbb{F} \to \mathbb{F}$ for which $\cdot' = \cdot \circ \sigma$?
2 Answers
Exotic multiplications over $\Bbb R$ as a vector space do exist. One approach to producing such a multiplication is to apply transport of structure with a bijective, discontinuous solution to Cauchy's functional equation.
To spell that out: first, we observe that for any bijective $f:S \to V$ (for a set $S$ and vector space $V$), we can produce an alternative vector-space structure over $S$ using transport of structure: for $x,y \in S$ and $k \in \Bbb F$, define $$ x \oplus y = f^{-1}(f(x) + f(y))\\ k \odot x = f^{-1}(kf(x)). $$ It is straightforward to show that these operations satisfy the vector space axioms.
Incidentally, vector spaces that can be constructed in this fashion commonly appear in linear algebra textbooks in problems where one is expected to show that an exotic vector space satisfies the required axioms. Two popular instances of these problems can be constructed using the maps $f:\Bbb R_+ \to \Bbb R$ given by $f(x) = \log(x)$ and $f:\Bbb R^n \to \Bbb R^n$ given by $f(\mathbf x) = \mathbf x - \mathbf v$ for some "base vector" $\mathbf v$.
Now, let $f:\Bbb R \to \Bbb R$ be a bijective, discontinuous function satisfying $f(x + y) = f(x) + f(y)$ for all $x,y$ in $\Bbb R$ (to put that another way, $f$ is bijective and $\Bbb Q$-linear but not $\Bbb R$-linear). Use the operations $\oplus$ and $\odot$ defined above. Note however that the $\oplus$ that you obtain in this way is the usual addition over $\Bbb R$: $$ x \oplus y = f^{-1}(f(x) + f(y)) = f^{-1}(f(x + y)) = x + y. $$
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4Note that the existence of a discontinuous bijective additive function on $\mathbb R$ in the last paragraph requires some kind of axiom of choice. See https://math.stackexchange.com/q/492751 – user1551 Feb 08 '25 at 18:32
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@Paprika Good catch – Ben Grossmann Feb 08 '25 at 20:41
Another more general question I would like to resolve is this: given an $\mathbb{F}$-vector space $(V, +, \cdot)$, must all possible scalar multiplications be related by an automorphism of $\mathbb{F}$? That is, if I have two different ring maps $\cdot: \mathbb{F} \to \operatorname{End}(V)$ and $\cdot': \mathbb{F} \to \operatorname{End}V$ serving as scalar multiplications that make $V$ a vector space with the same additive structure, must there be some ring automorphism $\sigma: \mathbb{F} \to \mathbb{F}$ for which $\cdot' = \cdot \circ \sigma$?
If $F$ is a prime field $\mathbb{Q}$ or $\mathbb{F}_p$ it's not hard to see that if such a ring map exists it is necessarily unique (exercise). So in this case the answer is yes.
In all other cases the answer is no, and we can already see this with finite extensions of $\mathbb{Q}$ and $\mathbb{F}_p$. For example, suppose $V = \mathbb{Q}^2$ and $F = \mathbb{Q}(i)$. Then $F$-vector space structures on $V$ (not up to isomorphism) correspond to maps
$$\varphi : \mathbb{Q}(i) \to M_2(\mathbb{Q})$$
which in turn correspond to matrices $\varphi(i) \in M_2(\mathbb{Q})$ satisfying $\varphi(i)^2 = -1$ (linear complex structures). There are infinitely many of these, but $\mathbb{Q}(i)$ only has $2$ automorphisms. (However all such matrices are conjugate.)
More generally let $F$ be any field which is not prime. Then $V = F^2$ has a natural $F$-vector space structure
$$\varphi(f)(a, b) = (fa, fb)$$
given by the usual scalar multiplication. But since $F$ is not prime, it has dimension at least $2$ over its prime subfield, so it has at least one nontrivial automorphism $g : F \to F$ as an abelian group (eg pick a basis and permute it) which is not multiplication by an element of $F$. This lets us construct a second $F$-vector space structure
$$\varphi'(f)(a, b) = (fa, g(f(g^{-1} b)))$$
which cannot be related to the first one by an automorphism of $F$ (or even an endomorphism) since any such structure would act the same on the two coordinates.
In the above examples, and in Ben Grossmann's example, the two vector space structures are still related by conjugation by an element of $\text{Aut}(V)$; equivalently, they are still isomorphic as $F$-vector spaces. But it's possible to construct worse examples where this doesn't happen, using the fact that there exist fields which embed properly into themselves. The simplest example is that $F = \mathbb{Q}(x)$ embeds properly into itself as the subfield $\mathbb{Q}(x^2)$, and pulling back along this embedding gives, for example, two $F$-vector space structures on $V = \mathbb{Q}(x)$ which have different dimensions $1$ and $2$. We can construct even worse examples where one dimension is finite and another is infinite, etc.
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