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Aug 9, 2021 at 12:32 vote accept Andrew Uzzell
S Aug 7, 2021 at 11:03 history suggested Jukka Kohonen
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Aug 7, 2021 at 10:44 comment added YCor Remark: the statement that among isomorphism classes, the number of "rigid" ones is larger, is stronger. Indeed, the counting is then different since in counting orders, rigid ones occur $n!$ times while non-rigid ones occur less times.
Aug 7, 2021 at 10:28 review Suggested edits
S Aug 7, 2021 at 11:03
Aug 7, 2021 at 10:23 answer added Jukka Kohonen timeline score: 3
Mar 7, 2014 at 18:24 comment added Kevin P. Costello @AndrewUzzell, and that indeed is what I was missing. Sorry about the mistake.
Mar 7, 2014 at 12:27 comment added Andrew Uzzell @KevinP.Costello Could you explain why one almost always gets an automorphism? (It's clear that not just any pair of adjacent elements of $L$ can be swapped: for example, adjacent elements may have different degrees in the comparability graph that corresponds to the underlying partial order.)
Mar 6, 2014 at 19:23 comment added Kevin P. Costello If you let $L$ be a linear extension of your partial order, then it seems you can almost always construct at least some automorphisms by swapping adjacent elements of $L$. The only time this wouldn't work is if your partial order is already a total order. Am I missing something here?
Mar 6, 2014 at 17:11 comment added The Masked Avenger Not for partial orders, but for many classes of structures, check out Ralph Freese and his 1990 article "On the two kinds of probability in algebra". It talks about labeled versus unlabeled structures, and notes how probabilities for them are similar since most automorphism groups will be trivial. You might find something useful in his bibliography.
Mar 6, 2014 at 15:58 history asked Andrew Uzzell CC BY-SA 3.0