## Automorphisms of locally finite countable posets

I rephrase my last question. Given a locally finite countable connected (as a graph) poset which satisfies the following further condition: the intersection of any antichain with the set of elements preceding (or succeding) any element is a finite set.

1. Is the automorphism group of this poset countable or uncountable?

2. If the group is uncountable, is there a suitable topology AND a measure on the group? Thank you

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 Re 2: If $M$ is an arbitrary countable structure (in the model-theoretic sense, i.e., a set equipped with a bunch of finitary operations and relations), then $\mathrm{Aut}(M)$ carries a canonical topology making it a Polish group, namely the one inherited from the inclusion of $\mathrm{Aut}(M)$ in the product space $M^M$, where $M$ is given the discrete topology. – Emil Jeřábek Sep 19 2011 at 16:47 Emil:thanks for the information, I assume this is not locally compact? – Math-player Sep 19 2011 at 16:57 (I guess you are asking because of the Haar measure.) For general structures, no. It is compact if and only if every $a\in M$ has a finite orbit, which I think implies that it is locally compact if and only if there exists a finite subset $X\subseteq M$ such that every $a\in M$ has a finite orbit under the subgroup of those automorphisms of $M$ that fix $X$ pointwise. I don’t know whether this holds in your situation; in Joel’s example below, the group is compact, but I suspect this may not hold for other posets. – Emil Jeřábek Sep 19 2011 at 17:16

For question 1, the automorphism group needn't be countable. To see this, consider the poset consisting of infinitely many diamonds stacked on top of one another.

     .
.
.

3
/ \
*   *
\ /
2
/ \
*   *
\ /
1
/ \
*   *
\ /
0


This is locally finite, countable and connected in the sense you have requested. Your additional property is satisfied because every antichain (in the sense of pairwise incomparable elements) has size at most two. The automorphism group is uncountable, however, because we can swap the two intermediate points of any diamond independently, and so any infinite binary sequence determines a distinct manner of performing these swaps.

There is a natural topology on the automorphism group here, as in your other question, whose basic open sets are determined by a finite piece of the automorphism. Indeed, the group is essentially the same as Cantor space $2^{\mathbb{N}}$, since there is a one-to-one correspondence between the binary sequences and the automorphisms, just by considering the diamonds in which the swap is made or not. This space has a natural probability measure, treating the swaps as coin flips. Thus, the collection of automorphisms that make $n$ many prescribed swaps or non-swaps has measure $2^{-n}$. This measure interacts well with the topology, because they are both the standard concepts on Cantor space.

Finally, to address Gerhard's idea of building a poset satisfying the properties but having a countably infinite automorphism group, while having diamonds, let us simply join the diamonds horizontally:

         *   *   *   *
/ \ / \ / \ / \
...  -2  -1   0   1   2  ...
\ / \ / \ / \ /
*   *   *   *


This poset is locally finite, countable and connected, and it exhibits the extra property because every point has only finitely many successors and predecessors. Meanwhile, the automorphism group is countably infinite because it is precisely the infinite dihedral group: we have exactly the horizontal translations and reflections. (And to satisfy Gerhard, we have many diamonds!)

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