Hi, Is the automorphism group of a countable locally finite connected poset finite or countable? If not, is there a way to equipp it (the uncountable group) with a topology and a measure? Need this information for a work in progress on causal set quantization. Thank you
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$\begingroup$ It is not necessarily countable, especially in some examples having countably many antichains. Of course one can add a toplogy and measure to the group, but I do not know enough to tell you how to do so meaningfully. Gerhard "Ask Me About System Design" Paseman, 2011.09.18 $\endgroup$– Gerhard PasemanCommented Sep 18, 2011 at 23:39
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$\begingroup$ If you add more detail about your situation, I might be able to post an answer that will shine more light on that situation. Gerhard "Ask Me About System Design" Paseman, 2011.09.18 $\endgroup$– Gerhard PasemanCommented Sep 19, 2011 at 1:02
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The poset consisting of a single countable antichain, with all elements incomparable, is locally finite, but the automoprhism group consists of any permutation of the elements, which is an uncountable set.
But you also said "connected". If by this you mean that the poset is linearly ordered, which is also commonly described as the connectedness axiom, then the poset must be either a finite linear order, the integers, the positive integers or the negative integers, and each of these posets has a trivial automorphism group.
If instead you mean that the graph relation underlying the poset is a connected graph, then consider an infinite antichain with a single top point above them all. This is still locally finite, since all intervals have at most two elements, and it is connected as a graph, but the automorphism group is again any permutation of the elements of the antichain, which is uncountably many automorphisms.
Meanwhile, the automorphism group of any structure admits a natural topology, where the basic open sets are determined by a finite piece of the automorphism. That is, for any finite partial automorphism $p$ of the structure, one may consider the set of all automorphisms that extend $p$, and call this a basic open set. This topology is useful in diverse contexts, but perhaps you haven't really given us enough information about your context to determine if it might be useful for you.
answered Sep 19, 2011 at 1:11
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$\begingroup$ Although the original poster has not said so, I am interpreting connected locally finite to mean essentially that (there exists something which is) the smallest connected component of the poset containing the finitely many chosen elements (and) is itself finite as well as convex within the given poset. Even under this restriction, one does not need large antichains but only infinitely many of them to construct examples with uncountable automorphism group. Gerhard "Ask Me About System Design" Paseman, 2011.09.18 $\endgroup$ Commented Sep 19, 2011 at 1:27
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$\begingroup$ I am not clear on what is meant. In most contexts I am familiar with, a "connected" poset means a linear order. But it seems unlikely that this is what the OP means, since as I mention the question trivializes in this case. Meanwhile, there are many other interpretations, and so I think the issues needs clarification. $\endgroup$ Commented Sep 19, 2011 at 1:34
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$\begingroup$ Joel: Thank you for the enlightning answer. Gerhard: I will ask a more precise question on the same subject. $\endgroup$– user16974Commented Sep 19, 2011 at 7:44
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$\begingroup$ Is the application that assigns to every basic open set $U_p$ where $p$ is a finite piece of the automorphism the value $2^{-|p|}$ a measure? $|p|$ is the cardinality of $p$ assumed a minimal generating peice for the basic open set $U_p$. $\endgroup$– user16974Commented Sep 19, 2011 at 9:36
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$\begingroup$ Hollowdead, that particular function does not seem to be a measure, since imagine that a particular point not mentioned by $p$ had three possible images in automorphisms extending $p$, so $U_p$ wouuld be the disjoint union of $U_{p_0}$, $U_{p_1}$ and $U_{p_2}$, but by your proposal these would each have half the measure of $U_p$. So it isn't additive. But in principle, it would seem that there could often be a similar such kind of measure... $\endgroup$ Commented Sep 19, 2011 at 10:58