Thursday, 20 December 2007

co.combinatorics - Is there a poset with 0 with countable automorphism group?

It seems unlikely (once you assume d.c.c.). Define the height of an element x in P to be the length of the shortest unrefinable chain from x to 0.



Let Pn denote the elements of P whose height is at most n. Since each element has a finite number of covers, the number of elements in Pn is finite.



By d.c.c., every element of P is in some Pn.



Let G denote the automorphisms of P and let Gn denote the automorphisms of Pn. G is the inverse limit of the system Gn. Let Hn denote the image of G inside Gn. (Note that this might not be all of Gn, since there could be automorphisms of Pn that don't extend to P.) G is also the inverse limit of the system Hn.



If the system Hn stabilizes, then G is finite. On the other hand, if Hn doesn't stabilize, then the cardinality of G is an infinite product, i.e. uncountable.

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