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Given is a locally finite countable connected poset which satisfies further the following properties:

  1. Let $C$ be any maximal chain ( i.e. inextendible chain) and $A$ be any antichain. Then $A$ is covered by both the sets ${\rm Past}(x)$ and ${\rm Future}(x)$ for $x$ running over $C$ i.e. $A \subset \bigcup_{x \in C} ({\rm Past}(x) \cup {\rm Future}(x))$ and where ${\rm Past}(x):=\lbrace y\mid y \leq x\rbrace$ and ${\rm Future}(x):=\lbrace y \mid x \leq y\rbrace$.

  2. For any $x$, the intersection of ${\rm Past}(x)$ with any antichain is finite. Similarily for ${\rm Future}(x)$.

    Questions: 

       1. Is the orbit of any point $x$ by an automorphism $f$ of the poset finite?

   2. Is the group of automorphisms of this poset countable?

   3. As a polish group, is the group of automorphisms of the poset locally compact?

    Thank you

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  • $\begingroup$ I dont understand the title of the question: what are we quantizing? $\endgroup$
    – André Henriques
    Sep 28, 2011 at 18:53
  • $\begingroup$ It is related to a quantization procedure on a family of isomorphic causal sets, which are locally finite countable connected posets. $\endgroup$
    – user16974
    Sep 28, 2011 at 19:21
  • $\begingroup$ The quantifiers in your hypothesis are not clear. Do you mean to assume that your partial order, in addition to being locally finite, countable and connected, has the covering property for every pair (A,C) of a maximal chain and maximal antichain? And when you say A is covered by Past(x) and Future(x), do you mean that the family of pasts alone covers A and also the family of futures alone covers A? Or do you mean that one might possibly need both futures and pasts from C to cover A? $\endgroup$
    – Joel David Hamkins
    Sep 29, 2011 at 0:46
  • $\begingroup$ Joel: I edited the question. $\endgroup$
    – user16974
    Sep 29, 2011 at 4:50
  • $\begingroup$ I still do not quite understand it. Every singleton is an antichain, so is condition 1 simply saying that every maximal chain is upwards and downwards cofinal in the poset? Since the poset is countable, isn’t this equivalent to the poset being upwards and downwards directed? $\endgroup$
    – Emil Jeřábek
    Sep 29, 2011 at 9:25

2 Answers 2

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If I've understood the hypotheses correctly, the covering relation gives a connected locally finite directed graph: starting from a point $x$, the maximal elements underneath it form a finite antichain, as do the elements covering it. The automorphism group is therefore a second-countable totally disconnected locally compact group under the topology given by declaring the stabiliser of each point to be open. The automorphism group as a whole will be countable if and only if every point stabiliser is finite and compact if and only if every orbit of the group is finite.

There's no reason for orbits of even individual automorphisms to be finite: for instance, your poset could be $\mathbb{Z}$ and the automorphism could be $x \mapsto x+1$.

Edit: Here's a simple example of how the automorphism group can be large. Start with $\mathbb{Z}$, and replace every odd number $n$ with $|n|$ incomparable copies of itself. The automorphism group is $H \rtimes \mathbb{Z}/2\mathbb{Z}$, where $H$ is a Cartesian product of symmetric groups of unbounded degree. In particular the automorphism group is uncountable, and contains a copy of every countably-based profinite group.

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  • $\begingroup$ If it is possible, please let me know of any example where the automorphism group is not countable. For the moment, I am satisfied with the result that the group is locally compact. $\endgroup$
    – user16974
    Sep 29, 2011 at 13:01
  • $\begingroup$ If we ask whether the orbit of an automorphism which sends $x$ to an element incomparable with $x$ that is $x \mapsto y$ and not $y \leq x$ or $x \leq y$ would the orbit of $x$ be finite? $\endgroup$
    – user16974
    Sep 29, 2011 at 13:24
  • $\begingroup$ Colin: where can I find proofs of what you said? Thank you $\endgroup$
    – user16974
    Sep 29, 2011 at 19:49
  • $\begingroup$ Standard fact (can be found in a textbook on profinite groups): an inverse limit of finite groups is compact and totally disconnected. Everything else should become clear as soon as you try to prove it. $\endgroup$
    – Colin Reid
    Sep 30, 2011 at 21:24
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Claim: The poset is a chain.

Proof: Consider an unrelated pair a, b in the poset. Let A = {a}. Then a \in Past(b) by condition 1, which leads to a contradiction.

I imagine that your definitions are unclear. Please carefully define 'covered by', 'maximal chain', poset?, etc.

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  • $\begingroup$ Judging by previous questions, I think the OP is feeling around for axioms that will ensure a reasonably tame automorphism group. My suggestion would be to change 1. to 'every finite set lies in an interval' or something like that. $\endgroup$
    – Colin Reid
    Sep 29, 2011 at 15:30
  • $\begingroup$ David, while the OP’s terminology is not quite clear, I think you are a bit overdoing it. Poset and maximal chain are standard terms. It seems that the intended reading of 1 is $A\subseteq\bigcup_{x\in C}\mathrm{Past}(x)$ and $A\subseteq\bigcup_{x\in C}\mathrm{Future}(x)$. $\endgroup$
    – Emil Jeřábek
    Sep 29, 2011 at 15:48
  • $\begingroup$ I appologize for not being clear. My intention is $A \subset \bigcup_{x \in C} {\rm Past}(x) \cup \bigcup_{x \in C} {\rm Future}(x)$. $\endgroup$
    – user16974
    Sep 29, 2011 at 17:21

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