Welcome to MathOverflow

Question

3

1

I have a general-type question: Suppose $M$ is a countable structure that is ultrahomogeneous, i.e. every (partial) isomorphism between finitely generated substructures of $M$ extends to an automorphism of $M$. Also, let $Aut(M)$ be the group of automorphisms of $M$ equipped with the pointwise convergence topology.

I want to know necessary and/or sufficient conditions under which $Aut(M)$ preserves a linear ordering on $\mathbb{N}$, in the sense that $Aut(M)$ preserves $(\mathbb{N},\prec)$ iff for every $n,m\in\mathbb{N}$ and for every $g\in Aut(M)$, $$n\prec m\Leftrightarrow g(n)\prec g(m).$$

flag	asked Oct 31 2011 at 16:54 Ioannis Souldatos 815●2●10

1

Conditions on $M$ or conditions on $\mathrm{Aut}(M)$? A necessary condition is that $\mathrm{Aut}(M)$ is torsion-free. – Emil Jeřábek Oct 31 2011 at 17:06

Are you assuming that there already is a definable ordering in your setup and asking for conditions to have an infinite ascending chain, or are you asking about situations where you can prove that there is such a definable ordering? – Scott McKuen Oct 31 2011 at 17:19

If you're interested in the usual ordering on $\mathbb N$, not a general one, then that happens if and only if, for each element, there is a formula that uniquely picks it out. – Will Sawin Oct 31 2011 at 18:13

1

@Will: I don’t think so. Let $M$ be $\mathbb Z$ (as a set) with all unary predicates of the form $x\equiv a\pmod b$ for natural numbers $0\le a< b$. Then every two elements of $M$ are distinguishable by an atomic formula, hence $M$ is ultrahomogeneous with a trivial automorphism group (in particular, it preserves whatever you please), but no element of $M$ is definable (because every reduct of $M$ to a finite language has a nontrivial automorphism, namely a suitable shift). – Emil Jeřábek Oct 31 2011 at 18:39

1

Even more generally, if $g\in\mathrm{Aut}(M)$ and $a\in M$ are such that $g^n(a)=a$ for some $n>0$, then $g(a)=a$. If not, then WLOG $a\prec g(a)$. Since $g$ preserves $\prec$, we have $g^k(a)\prec g^{k+1}(a)$ for every $k< n$, which gives $a\prec g^n(a)=a$ by transitivity, a contradiction. – Emil Jeřábek Nov 2 2011 at 10:47

show 4 more comments

When Aut(M) preserves a linear order?

Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)

You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you.