Let ($\mathbb{R}^{\times}, \cdot$) be the multiplicative group of non-zero real numbers. I want to find the automorphism group of ($\mathbb{R}^{\times}, \cdot$). My guess is that it is the group of all rational numbers $p/q$ such that $p, q$ are odd integers which are co-prime.

We know that any automorphism of the additive group of real numbers ($\mathbb{R}, +$) is of the form $x \mapsto \lambda x$. Also, ($\mathbb{R}_{>0},\cdot $) is isomorphic to ($\mathbb{R}, +$), where ($\mathbb{R}_{>0},\cdot $) denotes the multiplicative group of positive reals. $\require{AMScd}$ \begin{CD} (\mathbb{R}_{>0}, \cdot) @>{\phi}>> (\mathbb{R}_{>0}, \cdot)\\ @V{\log(x)}VV @VV{\log(x)}V\\ (\mathbb{R},+) @>{\lambda x}>> (\mathbb{R}, +) \end{CD}

By the diagram above, any automorphism $\phi$ of $(\mathbb{R}_{>0}, \cdot)$ is of the form $$ \phi(x) = \exp(\lambda \log(x)) = x^\lambda$$where $\lambda \in \mathbb{R}^{\times}$. As any automorphism $\psi$ of ($\mathbb{R}^{\times}, \cdot$) should take positive numbers to positive numbers, $\psi$ must restrict to an automorphism on ($\mathbb{R}_{>0},\cdot $). The only automorphisms of ($\mathbb{R}_{>0},\cdot $) which extend to the automorphisms of ($\mathbb{R}^{\times}, \cdot$) are $x \mapsto x^\lambda$ such that $\lambda = \frac{p}{q}$ such that $p,q$ are odd integers which are co-prime. This is because:

  1. For irrational $\lambda$'s, the map is not well defined for negative reals.
  2. For $\lambda = \frac{p}{q}$, and $q$ even, the map is not well defined for negative reals.
  3. For $\lambda = \frac{p}{q}$, and $p$ even, the inverse of this map does not exist.

From here on, I don't know how to proceed in proving my guess, if it is right in the first place.

  • 4
    $\begingroup$ It's not true that the automorphism group of the additive group of real numbers is only multiplication by scalars. There are also horrible $\mathbb{Q}$-vector space automorphisms that are also group automorphisms. $\endgroup$ – Matt Samuel Feb 23 at 11:47
  • $\begingroup$ Are there automorphisms which are discontinuous? $\endgroup$ – Ajay Kumar Nair Feb 23 at 11:50
  • 1
    $\begingroup$ The only continuous automorphisms are multiplications by scalars in the usual topology. But yes, there are more discontinuous automorphisms than there are continuous ones. $\endgroup$ – Matt Samuel Feb 23 at 11:52
  • $\begingroup$ Thanks! Can we conclude from here that the ones mentioned above are all the continuous automorphisms of $ ( \mathbb{R}^{\times}, \cdot )$? $\endgroup$ – Ajay Kumar Nair Feb 23 at 11:55
  • 1
    $\begingroup$ Use this question and this one. $\endgroup$ – Dietrich Burde Feb 23 at 11:58

We have that $(\mathbb{R}^*,\cdot)\simeq \{-1,1\}\times (\mathbb{R}^+,\cdot)$. For any automorphism $f:(\mathbb{R}_{>0},\cdot)\to(\mathbb{R}_{>0},\cdot)$ we can define $$f'(\epsilon,x) = (\epsilon,f(x))$$ and this is an automorphism of $(\mathbb{R}^*,\cdot)$. Does this define all automorphisms? Suppose conversely that $$g(\epsilon,x) = (g_1(\epsilon,x),g_2(\epsilon,x))$$ is an automorphism. If we compose with the absolute value, we obtain $|g|:(\mathbb{R}^*,\cdot)\to (\mathbb{R}^+,\cdot)$ given by $$|g|(\epsilon,x) = g_2(\epsilon,x)$$ The kernel is exactly $\{g^{-1}(-1,1), 1\}$, and since $(-1,1)$ is the only element of order $2$, $g(-1,1) = (-1,1)$, hence $g_2(\epsilon,x)$ is independent of $\epsilon$ and hence arises from an automorphism $g_2:(\mathbb{R}_{>0},\cdot)\to (\mathbb{R}_{>0},\cdot)$. Thus $$g(\epsilon,x) = (g_1(\epsilon,x),g_2(x))$$ If $g_1(\epsilon,x) = (-1,g_2(x))$, then $g(\epsilon,x) = (-1,1)g(-\epsilon,x)$. Since $g(\epsilon,x)^2 = g(-\epsilon,x^2) = g(-\epsilon,x)^2$, it follows that $-\epsilon = 1$. Hence $$g(\epsilon,x) = (\epsilon,g_2(x))$$ This characterizes the automorphisms as being automorphisms of $(\mathbb{R}_{>0},\cdot)$ linearly extended to be able to pull out minus signs. These will usually no longer be given by a single formula involving exponents, even if they are continuous. The continuous automorphisms end up being $$f(r) = \begin{cases} r^a&\mbox{ if }r>0\\ -r^a&\mbox{ if }r<0 \end{cases}$$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.