I was studying combinatorical group theory recently, and I came across the infinite regular rooted binary tree and its automorphism group $Aut(T^{(2)})$with the Grigorchuk subgroup.
Let me now elaborate more on a different topic, the surreal numbers. They are constructed similarly to the binary tree at first, but in the $\omega$-th generation, infinitesimal and transfinite numbers start to appear.
I wonder - if we remove, from the surreal number "tree", all numbers and order relations, thus focusing solely on the "transfinite binary tree" (in a graph-theoretic sense), what would be the automorphism group (I allow proper classes to be groups) of that tree?
For a better vsiualization of what I mean by such a tree: https://upload.wikimedia.org/wikipedia/commons/thumb/4/49/Surreal_number_tree.svg/532px-Surreal_number_tree.svg.png


Each connected component of the tree is either rooted at $0$ or at a number which birthday is a limit ordinal, these are exactly the vertices of degree $2$. Let $A$ be the class of all limit ordinals, with addition of $0$. For an element $\alpha \in A$ let $s_\alpha = \{+, -\}^\alpha$ be the set of its sign expansions, and $\mathcal{A} = \cup_{\alpha \in A} s_{\alpha}$ the class of "roots". In a way, $\mathcal{A}$ is a quotient of the class of surreal numbers $\mathbf{On}$ and $\mathfrak{c} = 2^{\aleph_0}$ (the size of a component).

The automorphism group is now the semidirect product of $G^{\mathcal{A}}$ and $\mathrm{Sym}(\mathcal{A})$, where $G$ is the group of automorpisms of the infinite tree of depth $\omega$. Of course, you need to adopt a powerful enough set theory for any of this to make sense.

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    $\begingroup$ the "Grigorchuk group" refers to some countable subgroup of the automorphism group of a binary rooted tree, not the whole automorphism group. $\endgroup$ – YCor Oct 19 '17 at 23:01
  • $\begingroup$ Indeed, I was misleaded by the OP statement. Thanks for pointing that out. $\endgroup$ – Mikhail Tikhomirov Oct 20 '17 at 5:33
  • $\begingroup$ Possibly you understood what the OP means by transfinite binary tree (I didn't), could you define it (or give a reference)? $\endgroup$ – YCor Oct 20 '17 at 7:15
  • $\begingroup$ I think the structure is evident from the notion of surreal numbers: en.wikipedia.org/wiki/Surreal_number. To give a formal definition, each vertex of the tree corresponds to a pair of an ordinal $\alpha$ and a map $f: \alpha \to \{+, -\}$, and an edge connects $(\alpha, f)$ with $(\alpha + 1, f')$, where $f'$ is an extension of $f$. $\endgroup$ – Mikhail Tikhomirov Oct 20 '17 at 7:31
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    $\begingroup$ @EmilJeřábek True, the question does not specify whether we should treat the tree as a graph or as a partial order; the two trees have different automorphisms, and the answer only applies in the first case. Judging from that the answer is accepted, maybe I guessed the OP's intention right? A clarification of the question would be nice though. $\endgroup$ – Mikhail Tikhomirov Oct 20 '17 at 9:29

Although this question already has an accepted answer, which is correct for the question as stated, I posit that the surreal number tree is best viewed as a tree in the order-theoretic rather than the graph-theoretic sense.

In that case, the automorphism group is isomorphic to the direct product of a class of two-element groups indexed by the surreal numbers:

$$ G := \prod_{\alpha \in \textbf{No}} C_2 $$

To formalise this, consider an automorphism $\phi : \textbf{No} \rightarrow \textbf{No}$. We describe the surreal numbers inductively, where element is either:

  • The zero (root) number $0$;
  • The left successor (child) $l(\alpha)$ of an existing number $\alpha \in \textbf{No}$;
  • The right successor (child) $r(\alpha)$ of an existing number $\alpha \in \textbf{No}$;
  • A limit of an ascending chain (with respect to the partial order induced by the tree, not numerical order) of numbers $\sup_{i \in I} \alpha_i$.

We shall construct such a $\phi$ naturally from an arbitrary $\psi \in G$ viewed for convenience as a function $\psi : \textbf{No} \rightarrow C_2$. Let $C_2$ act on the two-element set $\{ l, r \}$ in the obvious way, and define:

  • $ \phi(0) := 0 $
  • $ \phi(l(\alpha)) := (\psi(\alpha)(l))(\phi(\alpha)) $
  • $ \phi(r(\alpha)) := (\psi(\alpha)(r))(\phi(\alpha)) $
  • $ \phi(\sup_{i \in I} \alpha_i) := \sup_{i \in I} \phi(\alpha_i) $

We have described a map $\Theta : G \rightarrow \textrm{Aut}(\textbf{No})$ which sends $\psi$ to $\phi$, and it is straightforward to check that this is an isomorphism.

EDIT: I should probably clarify that in order to manipulate the surreals set-theoretically without pain, we can think of them as being a subset of a Grothendieck universe within our ambient set theory. It is henceforth safe to treat the surreals of that Grothendieck universe as being a set within the larger ambient set theory, and perform usual set-theoretic constructions such as powersets.


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