How to decide the cardinality of $\{\text{all group isomorphisms from }(\mathbb{R},+)\text{ onto }(\mathbb{R}^+,\cdot)\}$?

Question

The additive group of reals $(\mathbb{R},+)$ and the multipilicative group of positive reals $(\mathbb{R}^+,\cdot)$ are isomorphic, and $x \mapsto \exp(x)$ is one isomorphism from $(\mathbb{R},+)$ onto $(\mathbb{R}^+,\cdot)$. This is probably an elementary example of group isomorphism that every beginning learner knows. But I wonder something more.

Problem:
How to decide the cardinality of $\{\text{all group isomorphisms from }(\mathbb{R},+)\text{ onto }(\mathbb{R}^+,\cdot)\}$?

• Every function of the form $x \mapsto c^x$ in which $c \in (0,\infty) \backslash \{1\}$ is one continuous group isomorphism from $(\mathbb{R},+)$ onto $(\mathbb{R}^+,\cdot)$. But is every continuous group isomorphism from $(\mathbb{R},+)$ onto $(\mathbb{R}^+,\cdot)$ is exactly of this form?

• How to consider those group isomorphisms that contains discontinuities?

I have got hinted with something like that the/an answer to this problem depends on whether we have the axiom of choice or NOT. I got people hinted me on the freenode IRC in channel ##math, but that place (IRC) is not a good place to elaborate on things about this problem to a beginning learner. So I am asking more detailed elaboration on things about this problem here.

Thank you in advance for any possible help you may give!

Answer 1

Note that I am just going to consider homomorphisms $(\mathbb{R},+) \to (\mathbb{R},+)$ as you can translate everything to $(\mathbb{R},+) \to (\mathbb{R}_{>0}, \cdot)$ by composing with your favorite isomorphism(which is continuous so that it exists in all models of ZF).

This is somewhat related to the idea of "automatic continuity" which asks(under various restrictions): when are group homomorphisms forced to be continuous? There is a really nice note Automatic continuity of group homomorphisms by Christian Rosendal which talks about various aspects of this.

As was pointed out in ##math this has some relation to the axiom of choice in the case of $(\mathbb{R},+)$: it is consistent with ZF(ZFC without axiom of choice) that all group homomorphisms from $(\mathbb{R},+)$ are continuous and it is consistent that there are discontinuous homomorphisms from $(\mathbb{R},+)$ to itself.

This answer gives a simple proof that all continuous homomorphism $(\mathbb{R},+) \to (\mathbb{R},+)$ look like $t \mapsto rt$ for some $r$. In particular there are $| \mathbb{R}|=\mathfrak{c}$ continuous homomorphisms.

If you allow the axiom of choice there are $2^\mathfrak{c}$($=\mathfrak{c}^\mathfrak{c})$ homomorphisms. The axiom of choice gives us that any vector space has a basis, so we can think of $\mathbb{R}$ as a vector space of the rationals. The cardinality of such a basis will be $\mathfrak{c}$ since the rationals are countable. Any permutation of the basis gives a homomoprhism of $(\mathbb{R},+)$, and there are $\mathfrak{c}^\mathfrak{c}$ permuations. Note that this is the same number of functions $\mathbb{R} \to \mathbb{R}$ so we get the full theoretical cardinality upper bound.

Without the full axiom of choice there is the Solovay model(and I think some work of Shelah, I am not a set theory expert) that shows it is consistent that all sets of reals are Baire measurable. It is a (not hard)theorem of Pettis, which is Theorem 2.2 in Rosendal's notes, that all Baire measurable homomorphisms between Polish groups(the reals are an example) are continuous. So, in this Solovay model all homomorphisms from the reals to itself are continuous. Which means we only get the $t \mapsto rt$ maps discussed above.

Question(which I don't know the answer to): Adding choice is a lot, so can you have a ZF model of set theory where there are exactly $\kappa$ homomorphisms $(\mathbb{R},+)$ to itself where $\mathfrak{c} < \kappa < \mathfrak{c}^\mathfrak{c}$? If there is cardinal gaps, can you arrange for any such $\kappa$?