6
$\begingroup$

We know that integer partitions form a poset (actually we can define more than one partial orders on it), and so we can have some kind of Möbius function on it and consequently Möbius Inversion Formula (correct me if I am wrong).

I am looking for some examples of applications of such kind of formula if it exists, and whether we can view Hook length formula as a manifestation of it or not. Either you can give an example or point out some literature to look into...

$\endgroup$
2
  • $\begingroup$ If you want to compute dimensions using the poset of partitions with containment, you need only to count the number of maximal chains ending at a partition. As it happens, the Mobius function for this poset is not too hard. $\endgroup$ Commented Dec 9, 2011 at 3:16
  • $\begingroup$ @John It'll be helpful if you can point out an example in literature $\endgroup$ Commented Dec 10, 2011 at 9:25

2 Answers 2

13
$\begingroup$

The Möbius function of the poset $P_n$ of partitions of the integer $n$ ordered by refinement is not well-behaved, as is discussed e.g. in:

Günter Ziegler, On the poset of partitions of an integer, J. Combin. Theory Ser. A 42 (1986), no. 2, 215--222.

where the poset is shown not to be Cohen-Macaulay for $n\ge 19$ and not to have Möbius function on intervals that alternates in sign for $n\ge 111$. Thus, its order complex (nerve) on intervals is not always homotopy equivalent to a wedge of top dimensional spheres. Wanting nonetheless to understand the topology of the order complex for this poset and more general posets of multiset partitions was the original motivation for my work with Eric Babson on discrete Morse theory for posets, but we only obtained partial results in this direction, focusing mainly on $\mu_{P_n} (\hat{0},\hat{1})$ rather than arbitrary intervals.

If you are (more likely) interested in partially ordering the number partitions by shape containment rather than by refinement, then this is Young's lattice, and it is indeed well-behaved, as described e.g. at the wikipedia article on "Young's lattice". Young's lattice is a distributive lattice, hence each interval is shellable, and each interval has Möbius function equaling $0, 1, $ or $-1$. Specifically, it is 0 except for $\mu (\lambda_1,\lambda_2 )$ where the skew shape $\lambda_2 \setminus \lambda_1$ consists of single boxes touching at most at corners.

As far as applications, I would be looking in the theory of symmetric functions and in representation theory. There seems to be some discussion of this and references in the wikipedia article on "Young's lattice".

$\endgroup$
3
  • 1
    $\begingroup$ One reason I gave a partial answer to this question is that I'd also be interested to know more about whether the Möbius function helps in proofs regarding symmetric functions, has representation theoretic "meaning" etc., so hoped others might take a look at this question. I haven't been able to figure out how to fix my broken wikipedia link and would be grateful for assistance -- I wonder if my difficulty is related to "Young's" having an apostrophe in it. $\endgroup$ Commented Oct 17, 2012 at 20:22
  • $\begingroup$ Yes, the apostrophe has to be URL-encoded (it's "%27"). Fixed this (Firefox does this automatically when you copy and paste a URL). $\endgroup$ Commented Oct 18, 2012 at 1:39
  • $\begingroup$ I don't want to edit the post again and bump it again to the top, but focusing mainly on $\mu_{P_n}(\hat{0},\hat{1})$'' should really say focusing mainly on $\mu_{P_{\lambda }}(\hat{0},\hat{1})$. $\endgroup$ Commented Oct 19, 2012 at 0:24
5
$\begingroup$

If you mean the poset you get with the relation "refinement" then its Möbius function is not known (see Exercise 122 in Chapter 3 of Stanley's Enumerative Combinatorics, 2nd edition).

$\endgroup$
3
  • $\begingroup$ I don't have the second edition. It'll be helpful if you can attach a snippet of it... $\endgroup$ Commented Dec 10, 2011 at 9:26
  • 1
    $\begingroup$ Second edition of EC1: www-math.mit.edu/~rstan/ec/ec1 $\endgroup$ Commented Oct 18, 2012 at 1:37
  • $\begingroup$ In the edition currently posted on Stanley's website (the July 15 2011 version), this is actually problem 135, rather than 122. It is tagged with [5], which means unsolved. $\endgroup$ Commented Oct 18, 2012 at 20:26

You must log in to answer this question.

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