2

I'm trying to find the automorphism group of $(\mathbb R_{>0}, \cdot)$.

What I have so far:

$x \mapsto x^r$ for any $r \in \mathbb R - \{0\}$ is an automorphism.

But I can't think of any others. I have found $|\mathbb R|$ many but because I think $\text{Aut} (\mathbb R_{>0})$ should be much larger than $\mathbb R$ I believe that I didn't find them all.

Which automorphisms am I missing?

a student
  • 4,365
  • 3
    $\log$ is an isomorphism from $(\mathbb{R}_{>0},\cdot)$ to $(\mathbb{R},+)$, so this is essentially a duplicate of http://math.stackexchange.com/questions/115486/what-is-operatornameaut-mathbbr – Eric Wofsey Jan 19 '16 at 02:00

1 Answers1

4

$(\mathbb R_{>0}, \cdot) \cong (\mathbb R, +)$ via $\log$.

So the question reduces to finding the automorphisms of $(\mathbb R, +)$, that is the additive functions.

If you restrict yourself to functions that are either continuous, monotonic on any interval, or bounded on any interval, then the only solutions are the scalings $x\mapsto cx$, which gives you the multiplicative $x\mapsto x^r$.

If you do not impose restrictions on the functions, there are too many and they can be weird. For instance, bijections between two bases of $\mathbb R$ as vector space over $\mathbb Q$ induce additive automorphisms of $\mathbb R$. Note that such bases are uncountable...

lhf
  • 216,483
  • I suggest that the OP did not realize that by taking logs he would arrive at the Q of additive real functions, which gets asked here about once a week. – DanielWainfleet Jan 19 '16 at 02:56
  • Thank you. Could you include a proof that these permutations of bases are the only automorphisms? I looked at the other threads and a proof is mentioned nowhere. – a student Jan 21 '16 at 00:58
  • See https://en.wikipedia.org/wiki/Cauchy%27s_functional_equation#Proof_of_the_existence_of_other_solutions for instance. – lhf Jan 21 '16 at 01:03