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

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$\ ^*\mathbb{R}/^*\mathbb{R}^{\preceq \Omega^{-1}} \rightarrow \Gamma$.

  • $\begingroup$ 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. $\endgroup$
    – user14972
    May 23, 2017 at 13:40
  • $\begingroup$ Yes, that's what I meant, sorry. As for cardinals, the $\nu \ \omega^x$ for $x \in \ ^*\mathbb{R}$ real are pairwise distinct, so $|^*\Gamma| = |^*\mathbb{R}|$. $\endgroup$
    – nombre
    May 23, 2017 at 13:53


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