# Automorphism group of the class of surreal numbers

Do we know the group (Group) of automorphisms of the ordered Field of surreal numbers?

I feel the different ways to see the surreal numbers should provide us with several ways to define interesting automorphisms. However, I can't think of any except the identity.

• @QiaochuYuan It preserves the ordering anyway, I believe--since if $x = y^2$, $\phi(x) = \phi(y)^2 \ge 0$. Commented Sep 24, 2015 at 17:35
• Yes field automorphisms are ordered field automorphisms because No is euclidean. Commented Sep 24, 2015 at 17:42
• I would expect the surreals are saturated as an ordered field, so the automorphism group is huge (using Grothendieck universes to avoid worrying about size issues, for instance). Commented Sep 24, 2015 at 19:23

To avoid having to worry about things like proper classes, let us suppose $\kappa$ is an inaccessible cardinal and that by "the surreal numbers" $No$ we mean the surreal numbers of rank $<\kappa$. Then the automorphism group of $No$ is truly enormous: it has cardinality $2^\kappa$, and is what Conway calls an "improper class". This essentially follows from Theorems 28 and 29 of ONAG. Specifically, if $f\subseteq g\subset No$ and $\bar{f}\subset No$ are small subfields (i.e. of cardinality $<\kappa$) and $\varphi:f\to \bar{f}$ is an isomorphism of ordered fields, then Theorem 28 says there is a subfield $\bar{g}$ such that $\bar{f}\subseteq \bar{g}\subset On$ and an isomorphism $\varphi':g\to \bar{g}$ extending $\varphi$. Furthermore, note that for any such $f$ and $\varphi$, there is an $g$ such that the extension $\varphi'$ is not unique (for instance, if $g=f(x)$ where $x$ is greater than every element of $f$, then $\varphi'$ can send $x$ to any $x\in No$ which is greater than every element of $\bar{f}$). As in the proof of Theorem 29, we can construct automorphisms of $No$ by iterating this extension property back and forth by an induction of length $\kappa$, and we can do so such that at $\kappa$ many steps of the iteration, we have two different ways to choose the extension. Thus for every function $\kappa\to\{0,1\}$ (telling us how to make the choices), we get a different automorphism of $No$.
Regarding automorphisms of $$\mathsf{No}$$ considered as an ordered field, there is, in fact, a proper class of such. Or, put in a (perhaps) more palatable way, given any set of ordered field automorphisms of $$\mathsf{No},$$ there is an ordered field automorphism of $$\mathsf{No}$$ not contained in the given set.
However, the identity map is the only ordered field automorphism that respects the tree structure inherent in the construction of $$\mathsf{No}.$$ That is, if we define the relation $$<_S$$ by $$x<_S y$$ (read "$$x$$ is simpler than $$y$$") iff $$x\in L_y\cup R_y$$ (where $$L_y,R_y$$ are the respective sets of left and right options of $$y$$), then the identity map is the only automorphism on $$\mathsf{No}$$ that is $$<_S$$-preserving (this is the first theorem of $$\S5$$).