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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 $Aut(M)\subset \kappa^\kappa$ denote the topological group of automorphisms of $M$. The closure of $Aut(M)$ under the product topology is denoted by $\overline{Aut(M)}$ and $\partial Aut(M)=\overline{Aut(M)}\setminus Aut(M)$ denotes the boundary of $Aut(M)$.

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

Side note: If $f$ is in $\partial Aut(M)$, 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.

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For the product topology on $\kappa^\kappa$, you place the discrete topology on each factor? –  Joel David Hamkins Aug 2 '12 at 19:21
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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 $|Aut(M)| = \kappa^+$. –  Trevor Wilson Aug 2 '12 at 19:23
    
I was assuming that you meant the product of the discrete topologies on $\kappa$, by the way. –  Trevor Wilson Aug 2 '12 at 19:24
    
At both Joel and Trevor: Yes, the topology on $\kappa$ is the discrete topology. –  Ioannis Souldatos Aug 2 '12 at 21:24
    
@ Trevor: Your argument establishes $|Aut(M)|\le|\partial Aut(M)|$. My apologies there are a couple of typos in the question. It seems it has been a long day! –  Ioannis Souldatos Aug 2 '12 at 22:17

2 Answers 2

By Trevor's comment, $|Aut(M)|\leq|\partial Aut(M)|$. 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 $|Aut(M)|\geq\kappa$.

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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). –  Ioannis Souldatos Aug 8 '12 at 15:36

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 $\overline{Aut(M)}^{T}\setminus Aut(M)$, 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: http://arxiv.org/abs/1211.7145).

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