2
$\begingroup$

I recently found the paper of Berarducci + Mantova [1, 2] saying that surreal numbers are equivalent to trans-series. These are very different objects:

Has anyone checked this equivalence? Is it correct?

$\endgroup$
  • $\begingroup$ I can't propose an edit that doesn't affect the body of the question, but you may want to add the surreal-numbers tag (and maybe the surreal-analysis tag?) and perhaps drop the combinatorial-game-theory tag $\endgroup$ – Mark S. Aug 12 '17 at 16:31
  • $\begingroup$ @MarkS. : I added the surreal numbers tag. $\endgroup$ – Timothy Chow Aug 12 '17 at 19:09
  • 3
    $\begingroup$ @johnmangual : Usually a question that simply asks whether a paper is correct, without pointing out a specific step that is causing difficulty or otherwise indicating any reason why one might doubt its correctness, is considered a poor MO question. See for example this meta discussion: meta.mathoverflow.net/a/2332/3106 I recommend that you give more details if you can. $\endgroup$ – Timothy Chow Aug 12 '17 at 19:23
  • $\begingroup$ Please allow me some time to add more details from the two logic papers of Berarducci and Mantova. If the logic is straightforward enough I could start turning transseries arguments into combinatorial games and vice-versa. $\endgroup$ – john mangual Aug 12 '17 at 19:45
  • 4
    $\begingroup$ You might be interested in the book "Asymptotic Differential Algebra and Model Theory of Transseries" and the article "The Surreal Numbers as a Universal H-field" of Aschenbrenner, van den Dries, and van der Hoeven which clarify in what sense they are equivalent. Note that the articles you cite do not claim that they are isomorphic (transseries usually denote a set-sized structure) or that surreal numbers are known to exhibit every major property of the field of transseries. $\endgroup$ – nombre Aug 12 '17 at 22:36
4
$\begingroup$

It seems there might be some confusion about the term "equivalent". The two structures in question are "equivalent" in a technical sense, i.e. elementarily equivalent (in the language of ordered differential rings).

Let $\mathcal{L_d} = \{+,\times,0,1,<,\partial\}$ be the language of ordered differential fields. The field of Trans-series is naturally an "ordered differential field". A priori, the Surreal numbers are an ordered field and not naturally a differential field. On the other hand, Berarducci and Mantova constructed a formal differential operator over the Surreal numbers (which allows one to view the field as a differential field).

In the paper The Surreal Numbers as a Universal H-field (as pointed out above by nombre), the authors show that the two structures are elementarily equivalent as ordered differential fields (Actually, they show something much stronger. They show that the field of Trans-series is an elementary substructure of the field of Surreal numbers). In particular, for any sentence $\varphi$ in the language $\mathcal{L_d}$, we have that $\mathbf{No} \models \varphi \iff \mathbb{T} \models \varphi$, i.e. for any sentence in this language, it is true in the (ordered -differential) field of Surreal numbers iff it is true in the (o.d.) field of Trans-series.

To reiterate: These results only applies to first order sentences in the language of ordered differential rings. This does not apply to second order properties, or non-trivial extensions of these languages.

$\endgroup$
  • $\begingroup$ This is fascinating -- the surreals naturally form a real-closed field, and it sounds like you're saying that they also form a 'transseries-closed field'? $\endgroup$ – Alec Rhea Aug 14 '17 at 7:13
  • 1
    $\begingroup$ @AlecRhea: It depends on definition of "real-closed field" you use :). If you use the first one on this page, then one could phrase it as follows: With the correct choice of derivation, the Surreal numbers form an ordered differential field which is a 'transeries-closed' field. $\endgroup$ – Kyle Aug 14 '17 at 16:23

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.