6
$\begingroup$

My apologies: There were a couple of typos in the original question. Hope I got them all.

Let $\kappa$ be an uncountable cardinal of cofinality $\omega$ and $M$ a model of size $\kappa$. We equip $\kappa^\kappa$ with the product topology and let $\operatorname{Aut}(M)\subset \kappa^\kappa$ denote the topological group of automorphisms of $M$. The closure of $\operatorname{Aut}(M)$ under the product topology is denoted by $\overline{\operatorname{Aut}(M)}$ and $\partial \operatorname{Aut}(M)=\overline{\operatorname{Aut}(M)}\setminus\operatorname{Aut}(M)$ denotes the boundary of $\operatorname{Aut}(M)$.

My question is: If we know that (EDIT)$|\partial\operatorname{Aut}(M)|\ge\kappa^+$(EDIT), can we say anything about the cardinality of $\operatorname{Aut}(M)$, other than (EDIT)$|\operatorname{Aut}(M)|\le|\partial\operatorname{Aut}(M)|$(EDIT)?

Side note: If $f$ is in $\partial\operatorname{Aut}(M)$, then $f$ is 1-1, not onto, and for every formula $\phi$ and every finite $\vec{a}$, $M\models\phi[\vec{a}]$ iff $M\models\phi[f(\vec{a})]$. I.e. $f$ is an elementary embedding.

$\endgroup$
7
  • $\begingroup$ For the product topology on $\kappa^\kappa$, you place the discrete topology on each factor? $\endgroup$ Aug 2, 2012 at 19:21
  • 1
    $\begingroup$ Well, we can say that $|Aut(M)| \le \kappa^+$ because, choosing $f \in \partial Aut(M)$, the map $g \mapsto f \circ g$ is an injection $Aut(M) \to \partial Aut(M)$. So if $|Aut(M)| \ge \kappa^+$ (this isn't immediately apparent to me) then $|Aut(M)| = \kappa^+$. $\endgroup$ Aug 2, 2012 at 19:23
  • $\begingroup$ I was assuming that you meant the product of the discrete topologies on $\kappa$, by the way. $\endgroup$ Aug 2, 2012 at 19:24
  • $\begingroup$ At both Joel and Trevor: Yes, the topology on $\kappa$ is the discrete topology. $\endgroup$ Aug 2, 2012 at 21:24
  • 1
    $\begingroup$ So to be clear, $\partial$ here is not the usual $\partial$ from point-set topology? Or does it coincide for some reason? It doesn't seem like it should. $\endgroup$ Aug 2, 2012 at 23:25

2 Answers 2

1
$\begingroup$

By Trevor's comment, $\DeclareMathOperator{\Aut}{Aut}|\Aut(M)|\leq|\partial \Aut(M)|$. From purely topological considerations we get this:

Let $\lambda=\Aut(M)$.
The closure of $\Aut(M)$ cannot have more than $2^{2^\lambda}$ elements. If $\kappa^+\leq|\partial \Aut(M)|$, then we must have $2^{2^\lambda}\geq\kappa^+$. So in particular, if GCH or something similar holds below $\kappa$, we have $|\Aut(M)|\geq\kappa$.

$\endgroup$
1
  • $\begingroup$ Stefan, this is a nice observation making use of the fact that $\kappa$ is a limit cardinal. My original motivation stemmed from the fact that for $\kappa$ countable, the mere existence of one element in $\partial Aut(M)$ implies the existence of $2^\omega$ many automorphisms. This follows from a theorem of Su Gao in "On automorphism groups of countable structures", J. Symb. Log. 63, No.3, 891-896 (1998). $\endgroup$ Aug 8, 2012 at 15:36
0
$\begingroup$

I was able to prove the following for $\kappa$ of cofinality $\omega$:

Assume $M$ is a model of size $\kappa$, $cf(\kappa)=\omega$ and for all $\alpha<\beta<\kappa^+$, there exist functions $j_{\beta,\alpha}$ in $\DeclareMathOperator{\Aut}{Aut}\overline{\Aut(M)}^{T}\setminus \Aut(M)$, such that for $\alpha<\beta<\gamma<\kappa^+$,

$$(*) j_{\gamma,\beta}\circ j_{\beta,\alpha}=j_{\gamma,\alpha},$$ where $\overline{\Aut(M)}^{T}$ is the closure of $\Aut(M)$ under the product topology in $\kappa^\kappa$.

Then there are at least $\kappa^\omega$ automorphisms of $M$. The proof is using Infinitary Logic and the assumption that $cf(\kappa)=\omega$ is fundamental.

I do not know of any way to get the above result directly.

(link to arXiv: arXiv:1211.7145).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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