# Value groups of ultrapowers of $\mathbb{R}$.

$$\DeclareMathOperator{\Noo}{No}$$ $$\DeclareMathOperator{\ee}{e}$$

Let $$U$$ be a non principal ultrafilter on $$\mathbb{N}$$, let $$^*\mathbb{R}$$ denote the corresponding hyperreal field.

Let $$^*\Gamma$$ denote the value group of $$^*\mathbb{R}$$ seen as an ordered field with natural convex valuation $$\nu$$. If $$R_f$$ is the ring of finite elements of $$^*\mathbb{R}$$ (including infinitesimals), then $$\Gamma:= (^*\mathbb{R})^{\times} / (R_f)^{\times}$$ ordered by $$a{R_f}^{\times} \preceq b{R_f}^{\times} \Leftrightarrow aR_f \subseteq bR_f$$, and $$\nu$$ is the quotient map.

Is $$^*\Gamma$$ isomorphic as an ordered group to $$^*\mathbb{R}$$?

This is consistent because in ZFC + CH, $$^*\mathbb{R}$$ is isomorphic to the field $$\Noo(\omega_1)$$ of surreal numbers born before $$\omega_1$$, and this field has its underlying group (and therefore $$^*\mathbb{R}$$) as value group.

In the general case, it is tempting to try to define an "$$\omega$$-map" like section of $$\nu$$ by defining $$\omega^{[x]^U}:= [(0,1,2^{x(2)},3^{x(3)},...)]^U$$ but this does not work because for instance $$[(1,2,\frac{1}{\ln(3)},\frac{1}{\ln(4)},...)]^U$$ is sent to $$\ee$$ when it should be infinite.

Now, looking for general properties of $$^*\Gamma$$:

-We know that it is divisible since $$^*\mathbb{R}$$ has $$n^{th}$$ roots for any positive element

-We know that it is $$\aleph_1$$-saturated as a linear order (if $$\nu X = A < B = \nu Y$$ are countable subsets of $$^*\Gamma$$ where $$0 < X,Y$$ then the valuation of any element of $$^*\mathbb{R}$$ strictly between $$\mathbb{N}.X$$ and $$\frac{1}{\mathbb{N}+1}.Y$$ is strictly between $$A$$ and $$B$$.), and thus as an ordered group. (divisible ordered groups have quantifier elimination)

Does this go further?

edit: a partial anwser: Consider a positive infinite element $$\Omega$$ of$$\ ^*\mathbb{R}$$, and write$$\ ^*\mathbb{R}^{\preceq \Omega^{-1}}$$ for the additive group of hyperreal numbers $$a$$ with $$v a\geq - v \Omega$$. I claim that $$\Gamma$$ is naturally isomorphic to$$\ ^*\mathbb{R}/^*\mathbb{R}^{\preceq \Omega^{-1}}$$.

Indeed, fix a positive representative $$\omega$$ of $$\Omega$$ and for $$u \in \mathbb{R}^{\mathbb{N}}$$, set $$\varphi([u]):= v [({\operatorname{e}}^{\omega(n)u(n)})_{n \in \mathbb{N}}]$$. So $$\varphi$$ is a non-decreasing morphism $$\mathbb{R} \longrightarrow \Gamma$$. Its kernel is the convex subgroup$$\ ^*\mathbb{R}^{\preceq \Omega^{-1}}$$, so we obtain a strictly increasing bijective quotient map$$\ ^*\mathbb{R}/^*\mathbb{R}^{\preceq \Omega^{-1}} \rightarrow \Gamma$$.

• As a group, they're both $\mathbb{Q}$-vector spaces, so it's just a matter of cardinality (assuming choice). I don't know about the cardinality of ${}^*\Gamma$. More is needed, I suppose, if you mean to ask the same question about totally ordered $\mathbb{Q}$-vector spaces.
– user14972
Commented May 23, 2017 at 13:40
• Yes, that's what I meant, sorry. As for cardinals, the $\nu \ \omega^x$ for $x \in \ ^*\mathbb{R}$ real are pairwise distinct, so $|^*\Gamma| = |^*\mathbb{R}|$. Commented May 23, 2017 at 13:53