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Hello, I have the following question: Is it possible for a complete $L_{\omega_1,\omega}$ sentence $\phi$ to satisfy (a) the (unique) countable model of $\phi$ has $2^{\aleph_0}$ many automorphisms and (b) there is a model $N$ of $\phi$, $N$ of size $\aleph_\omega$ and $N$ has $\le\aleph_\omega$ many automorhphisms?

So, $\phi$ will have "many" automorphisms in the countable case and "not very many" automorphisms for a model in power $\aleph_\omega$. Is this possible and if so, can you reference any examples?

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up vote 1 down vote accepted

It appears that we can already do this in first-order logic, without making any use of the infinitary language.

Namely, the unique countable endless dense linear order $\langle\mathbb{Q},\lt\rangle$ has continuum many order automorphisms, but Shelah has reportedly proved that for every uncountable cardinality $\lambda$, there is a rigid endless dense linear order of cardinality $\lambda$, that is, having a trivial automorphism group. (See the remarks on page 347 of this article; perhaps someone can give a better reference.) In particular, there will be an endless dense linear order of size $\aleph_\omega$ with no nontrivial automorphisms at all.

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One last step in this answer: the formula DLO in this example is in fact a complete infinitary (well, finitary, but finitary$\subseteq$infinitary) formula, since it is $\aleph_0$-categorical; see Corollary 2.12 at the end of homepages.math.uic.edu/~marker/inf.pdf. –  Noah S May 29 '13 at 23:27
    
Yes, that's right. (But I would say EDLO, since with just DLO, you haven't settled the endpoint question.) –  Joel David Hamkins May 29 '13 at 23:30
    
Yeah, I think EDLO is better than DLO for this sentence. But don't Marker/others tend to use DLO for "dense linear order without endpoints?" –  Noah S May 29 '13 at 23:33
    
That is a common convention; the competing convention is that DLO means just "dense linear order", which has four completions, depending on how you settle the endpoint question. –  Joel David Hamkins May 29 '13 at 23:40
    
@Joel: Thank you for the answer. So, it turns out that we can impose more restrictive conditions (first-order vs infinitary $\phi$; 1 vs $\le\aleph_\omega$ automorphisms) and still get a positive result. I was a little curious about Shelah's result. I sent him an email asking about it. If he gives me a reference, I will post it here. –  Ioannis Souldatos Jun 3 '13 at 18:20
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