Automorphism group of the group $(\mathbb{R}^n,+)$

Question

As a vector space, it is clear that the automorphism group of $\mathbb{R}^n$ is $\mathsf{GL}_n(\mathbb{R})$.

My question is: What is the $\text{Aut}((\mathbb{R}^n,+))$ ($n \ge 2$) as a group? Clearly, $\text{Aut}((\mathbb{R}^n,+))$ contains $\mathsf{GL}_n(\mathbb{R})$.

Given that the automorphism group of $\mathbb{R}$ is enormous, to simplify matters, I would greatly appreciate an answer with respect to continuous automorphisms (if this is known).

The question above is related to this question concerning additive maps that do not preserve scalar multiplication. In particular, are there continuous isomorphisms of $\mathbb{R}^n$ ($n \ge 2$) that do not preserve scalar multiplication? (Clearly, this would indicate that the containment above is strict.)

Answer 1

Continuous homomorphisms are adressed in @JyrkiLahtonen's comment, so I'll adress non-continuous ones.

Note that $\Bbb R$ is a $\mathfrak{c}$-dimensional $\Bbb Q$ vector space. ($\mathfrak{c}$ is the cardinality of $\Bbb R$)

So as $\Bbb Q$-vector spaces and in particular as groups we have:

$\Bbb R^n \cong \left(\Bbb Q^{(\mathfrak{c})}\right)^n \cong \Bbb Q^{(n\mathfrak{c})} \cong \Bbb{Q}^{(\mathfrak{c})} \cong \Bbb R$ (This isomorphism is not continuous!)

Thus $\operatorname{Aut}(\Bbb R^n) \cong \operatorname{Aut}(\Bbb R)$, which is adressed in the linked wikipedia page.

Note that this also illustrates how large $\operatorname{Aut}(\Bbb R)$ is: it contains a copy of $\operatorname{GL}_n(\Bbb R)$ for every $n\in \Bbb N$.

Answer 2

The set of continuous automorphisms of $\mathbb R^n$ is $\mathrm{GL}_n(\mathbb R)$.

To see this assume $f\colon\mathbb R^n \to \mathbb R^n$ is an additive homomorphism, i.e. $f(u + v) = f(u) + f(v)$ and $f(0) = 0$, but we know nothing about whether it respects scalar multiplication. If $n \geq 0$ is an integer then $f(nu) = nf(u)$ because multiplication by $n$ is repeated addition and $f$ respects addition. If $n \leq 0$ is an integer then $f(nu) = -f(-nu) = -(-n)f(u) = nf(u)$. So $f$ respects scalars from $\mathbb Z$. Now $nf\left(\frac{1}{n}u\right) = f\left(n\frac{1}{n}u\right) = f(u)$ so multiply both sides by $\frac{1}{n}$ to get $f\left(\frac{1}{n}u\right) = \frac{1}{n}f(u)$. Since $f$ already respects $\mathbb Z$ we get now that $f$ is $\mathbb Q$-linear.

None of this used continuous so far. The automorphism group of $\mathbb R^n$ is $\mathrm{End}_{\mathbb Q}(\mathbb R^n)$. If we ask that $f$ be continuous and $r$ is a real number then take rationals $a_i$ such that $a_i \to r$ as $i \to \infty$. We have $a_if(u) = f(a_iu)$. Taking the limit as $i \to \infty$ on both sides (and using that $f$ is continuous to pass this limit to the inside of $f$) we get $rf(u) = f(ru)$. So if $f$ is continuous then $f$ is $\mathbb R$-linear.