1
$\begingroup$

I am playing with some questions concerning connections between certain poset partitions and their linear extensions. This is not my usual playground, I just happened to stumble upon something.

When I was playing with these things, I came up with a very simple construction that I cannot find anywhere.I suspect that this construction is a minor particular case of something well known and much more general. It reminds of an equivariant map, but the group is not fixed here.

Let $n>2$, let $X$ be a finite set of cardinality $n$. Let $\oplus_n$ be an action of $Z_n$ on $X$ such that the action of $1$ is a cycle of length $n$. Pick $x\in X$ and let $\pi$ be a partition of the set $X$ such that $\{x,x\oplus_n 1\}$ is the only nonsingleton block of $\pi$.

In a very simple way, this gives rise to an action $\oplus_{n-1}$ of $Z_{n-1}$ on $\pi$ by ,,squeezing the action'' at $x$:

Put

  • $\{x,x\oplus_n 1\}\oplus_{n-1} 1=\{x\oplus_n 2\}$
  • $\{x\oplus_n(n-1)\}\oplus_{n-1} 1=\{x,x\oplus_n 1\}$
  • $\{y\}\oplus_{n-1} 1=\{y\oplus_n 1\}$ for any $y\neq x\oplus_n(n-1)$.

It can be visualized in a simple way by a digraph construction: if we identify the action of $1$ on $X$ with an oriented cycle, this construction corresponds to a contraction of an edge.

Has anyone seen this construction before? Is there any name for it?

Improve this question
$\endgroup$
2
  • $\begingroup$ Seems to me this could fail if n is not prime power. A faithful action corresponds to a set of cycles with LCM equal to n, not necessarily a single cycle. $\endgroup$
    – Omer
    Commented Apr 9, 2011 at 14:55
  • $\begingroup$ You are right, of course. I have mixed up "regular" and "faithful". I have edited the question. $\endgroup$
    – Gejza Jenča
    Commented Apr 9, 2011 at 21:05

1 Answer 1

Reset to default
2
$\begingroup$

I might be missing something, but it seems to me that there is not much going on in your construction. In fact, your original action of $Z_n$ on $X$ does nothing more than putting a cyclic ordering on your set $X$, and your complicated-looking construction creates a new cyclically ordered set $Y$ obtained from $X$ by removing one element (and retaining the ordering).

In any case, the construction is rather artificial from a group-theoretical point of view, so I don't expect a more general construction (e.g. for arbitrary group actions) in that sense. It is faintly reminiscent of the construction of a primitive action from an imprimitive permutation group by considering the induced action on the blocks of imprimitivity.

Improve this answer
$\endgroup$
1
  • $\begingroup$ Yes, it seems that the most natural way how to work with this construction would be to deal solely with the cyclical orderings and leave the whole action of Zn out. But it is simply not true. The group action is actively used in my proof and, as far as I can see, there is no elegant way how to get rid of it. Moreover, the proof using this notion is relatively simple (less than 4 pages), and I prove a new result about posets and their linear extensions, which are popular and well understood things. Hmm, maybe there is a mistake in my proof. $\endgroup$
    – Gejza Jenča
    Commented Apr 23, 2011 at 20:45

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .