# Constructing the surreal numbers as iterated Hahn series

A theorem due to N. Alling (Foundations of Analysis over Surreal Number Fields, §6.55) states that the surreal numbers are isomorphic, as an ordered and valued field, to the field of Hahn series with real coefficients and value group the surreal numbers themselves. There is also a restricted version, which I'll refer to in order to avoid the (IMHO uninteresting) foundational difficulties in dealing with classes: if $\kappa$ is a regular uncountable cardinal, the set $\mathrm{No}_\kappa$ of surreal numbers with birth date $<\kappa$ is isomorphic to the field of Hahn series of length $<\kappa$ with real coefficients and exponents in $\mathrm{No}_\kappa$ itself (in the indeterminate $\frac{1}{\omega}$).

Upon reading this, I thought to myself, “well, this is nice, this means the surreal numbers can be given a construction as iterated Hahn series, something along the lines of: start with the reals, take the Hahn series over that, then take the Hahn series over that (as value group), repeat transfinitely, and voilà, surreal numbers”. Unfortunately, it seems I was being a bit naïve there.

Let us define $F_0 = \mathbb{R}$ and inductively $F_{\alpha+1}$ to be the field of Hahn series of length $<\kappa$ with real coefficients and exponents in $F_\alpha$ (the indeterminate being written $\frac{1}{\omega}$); and for $\delta$ a limit let $F_\delta = \bigcup_{\alpha<\delta} F_\alpha$ with the obvious embeddings. Then if I am not mistaken, $F_\kappa$ is indeed isomorphic to Hahn series of length $<\kappa$ over itself, it is indeed an $\eta_\xi$ field for $\kappa=\omega_\xi$, of cardinality $2^{<\kappa}$, just like $\mathrm{No}_\kappa$, and it is quite conceivable (I didn't check) that the two are isomorphic (as ordered—and valued—fields over $F_0 = \mathbb{R}$). But this can't possibly respect the map $x \mapsto \omega^x$ because in $\mathrm{No}_\kappa$ the latter has plenty of fixed points whereas in $F_\kappa$ it has none. So this construction is “wrong” in that it doesn't explain surreal numbers properly.

Thus, my question is: is there some variant of this construction that will succeed in constructing $\mathrm{No}_\kappa$, including its map $x \mapsto \omega^x$? Perhaps the answer depends on what is done at limit ordinal steps, but I'm rather confused so I wish someone could clear up the confusion.

• For an iteration of Hahn series, you may consider transseries. [ For example G. Edgar, "Transseries for beginners", Real Analysis Exchange 35 (2010) 253--310 ] But again there is no known "canonical" correspondence with surreal numbers. Oct 30, 2011 at 19:28

On page 183 of “Fields of Surreal Numbers and Exponentiation”, Fundamenta Mathematica 167 (2001), pp. 173-188, Lou van den Dries and Philip Ehrlich prove the following result that I believe provides an answer to the author’s queries. The result is an improvement of a result the author attributes to Norman Alling, which was established independently by Ehrlich in “An Alternative Construction of Conway’s Ordered Field No,” Algebra Universalis 25 (1988), pp. 7-16. Ehrlich’s proof actually slightly predates Alling’s’s, but it appeared later.

Proposition. Let $\lambda$ be an epsilon number $\le On$ and let $\tau =1/\omega$. Then

(i) $${\bf{No}}\left( \lambda \right) = \bigcup\nolimits_\mu \mathbb{R}{\left( {\left( {\tau ^{{\bf{No}}\left( \mu \right)} } \right)} \right)_\lambda }$$ where $\mu$ ranges over the additively indecomposable ordinals $< \lambda$;

(ii) ${\bf{No}}\left( \lambda \right)$ is real-closed;

(iii) $${\bf{No}}\left( \lambda \right) = \mathbb{R}\left( {\left( {\tau ^{{\bf{No}}\left( \lambda \right)} } \right)} \right)_\lambda$$ if and only if $\lambda$ is a regular cardinal;

(vi) For all $y \in {\bf{No}}$, $y \in {\bf{No}}\left( \lambda \right)$ if and only if $\omega ^y \in {\bf{No}}\left( \lambda \right)$.

• Thanks for pointing out this nice result, which answers a few questions I had in mind. However, I don't think it answers my original question, because the serious problem is for surreals whose length is precisely an epsilon number (e.g., $x\in\mathbf{No}(\varepsilon_0^\omega)$ such that $x=\omega^-x$), and which cannot be "explained" in terms of Hahn series using simpler surreals (Conway calls such surreals "irreducible"). But maybe there is no satisfactory answer to my question. Nov 6, 2011 at 12:28

I'm a novice in this area so I don't think my reply should be taken too seriously, but I recently read Alling's Foundations of Analysis over Surreal Number Fields and have been corresponding with Philip Ehrlich about this topic. I've seen Ehrlich's proof of the isomorphism between Conway Names and the Hahn Series, which I find particularly interesting. Ehrlich also, this year, published the document Conway names, the simplicity hierarchy and the surreal number tree Journal of Logic & Analysis 3:1 (2011) 1–26.

What I want to draw your attention to is Conway's original 1976 On numbers and games. In the section where he introduces us to these Conway names he includes a caveat. Every surreal number can be expressed in the form of a Conway name, but the use of the Conway name to generate surreal numbers is problematic because it involves feedback loops between surreal numbers of the same generation so is not amenable to use for the generation of the surreals by induction. This differs from the definition of the surreal numbers using a Dedekind-like cut which does generate all the surreal numbers by induction.

To see the problem. Consider the two expressions $\exp(\omega)$ and $\sin(\omega)$. Neither of these can be expressed by Conway names without feedback. The expression $\exp(\omega)$ as a Conway name appears as $\omega^{\omega/\ln(\omega)}$ only because $\omega/\ln(\omega)$ is a surreal number. On the other hand, on the surreals the expression $\sin(x)=\{x-x^3/3!,\ldots|x,x-x^3/3!+x^5/5!,\ldots\}$ evaluates to $\sin(\omega)=0$. As a Conway name $\sin(\omega)$ appears as $\omega^{\ln(\sin(\omega))/\ln(\omega)}$. You see the feedback problem?

The isomorphism to Hahn series involved an interesting change of the sign of the power. So $\exp(\omega)$ on the surreals translates to $1/\exp(\omega)$ on Hahn series.