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 xx in PP to be the length of the shortest unrefinable chain from xx to 00.



Let PnPn denote the elements of PP whose height is at most nn. Since each element has a finite number of covers, the number of elements in PnPn is finite.



By d.c.c., every element of PP is in some PnPn.



Let GG denote the automorphisms of PP and let GnGn denote the automorphisms of PnPn. GG is the inverse limit of the system GnGn. Let HnHn denote the image of GG inside GnGn. (Note that this might not be all of GnGn, since there could be automorphisms of PnPn that don't extend to PP.) GG is also the inverse limit of the system HnHn.



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

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