MathOverflow will be down for maintenance for approximately 3 hours, starting Monday evening (06/24/2013) at approximately 9:00 PM Eastern time (UTC-4).
2

2

Is there an intuitive way to define what the p-quotient of an integer partition is?

In response to the comments: my vague understanding is that the p-quotient is somehow division with remainder generalized to partitions. They turn up in the classic by Littlewood that I was reading: http://rspa.royalsocietypublishing.org/content/209/1098/333.abstract See also Macdonalds book on symmetric functions and Hall polynomials.

flag
2 
 I guess if we don't already know what the p-quotient of an integer partition is, we're not invited to take part. – Gerry Myerson Oct 18 2010 at 0:38
It would help to give a link to a paper or web page defining or using the term so that if we know it under another name, we can recognize it. – Douglas Zare Oct 18 2010 at 1:03
An intuitive description of the core-quotient bijection, illustrations, and precise definitions of the notions involved, can be found in section 2 of my paper "Edge sequences, ribbon tableaux, and an action of affine permutations" Europ. J. Combinatorics 20 (1999). It is available from my home page www-math.univ-poitiers.fr/~maavl. – Marc van Leeuwen Apr 27 at 20:05

1 Answer

5

I learned the following perspective in part from Bill Doran.

Redefine a partition through its boundary, extended along the axes in both directions:

A partition is a bi-infinite sequence of vertical $(0,1)$ and horizontal $(1,0)$ steps $(... S_{-2},S_{-1},S_0,S_1,S_2,...)$ whose prefix is all vertical steps, and whose suffix is all horizontal steps. A square of the partition is a pair of steps so that the first step is horizontal and the second step is vertical. The hook length of that square is the distance between the steps.

Then the $p$-quotient of a partition is the $p$-tuple of partitions you get by the steps in each congruence class mod $p$: $P_i = (...S_{-2p+i},S_{-p+i},S_i,S_{p+i},S_{2p+i},...)$ for $i \in \{0,..., p-1\}$.

Each square of a partition of the $p$-quotient corresponds to a square of the original partition whose hook length is divisible by $p$.

$p$-cores are partitions whose $p$-quotient is a $p$-tuple of empty partitions. They can be parametrized by an equivalence class of $p$-tuples of values $k_i$ so that $S_{p k + i}$ is vertical iff $k \le k_i$, i.e., the offsets of the empty partitions.

If you strip $p$-rimhook after $p$-rimhook off of a partition, this always results in the same $p$-core, and the choices don't matter. The choices correspond to taking one corner square (hook length 1) at a time off of a partition in the $p$-quotient.

Although in some applications, $p$ is prime, that is not necessary.

link|flag
1 
 There is a bijection between the squares of the $p$-quotient of hook length $h$ and squares of the original of hook length $ph$. Rimhooks of size $p$ or squares of hook length $p$ correspond to corners of the partitions of the quotient. – Douglas Zare Oct 18 2010 at 5:21
1 
 I think it may be clearer if you do an example for $p=2$: Take a large partition, write the hook lengths into each square, and highlight the even hook lengths. Compare the even hook lengths with the $2$-quotient with hook lengths written in. – Douglas Zare Oct 18 2010 at 7:59
Thanks! Now I understand. In what sense do the p-core and p-quotient determine the partition? For example (3) and (2,1) seem to have the same 3-core and the same 3-quotient – Roland van der Veen Oct 18 2010 at 9:35
1 
 $(3)$ and $(2,1)$ lead to different ordered triples of partitions as $3$-quotients. The map to the $p$-core and $p$-quotient is a bijection, which leads to some nice generating function identities such as $P(x) = P(x^2)^2 (1 + x + x^3 + x^6 + x^{10} + ...)$ since the last term is the generating function for $2$-cores. – Douglas Zare Oct 18 2010 at 13:25
1 
 You can't set $S_0$ that way. You should get different ordered triples which would be the same without order. There is some ambiguity I glossed over, that ...vhhhh.... is the same partition as ...vvvhh.... . There is an equivalence relation on bi-infinite sequences whose Vs have a maximum index and whose Hs have a minimum index so that a sequence is equivalent to one shifted over. Given any bi-infinite sequence, there is a unique shifted version so that there are just as many Hs with negative indices as Vs with nonnegative indices. – Douglas Zare Oct 19 2010 at 9:41
show 1 more comment

Your Answer

Get an OpenID
or

Not the answer you're looking for? Browse other questions tagged or ask your own question.