Hook-length in a Ferrer’s Diagram
Point of Post: In this post we discuss the notion of hook-length in a Ferrer’s diagram and give a few characterizations of the product of the hook-length over every square in a Ferrer’s diagram in preparation for the hook length formula.
In our last post we let slip the deal with looking at the combinatorial objects we have been looking at. In particular, we noted that we will associate to each -frame an irrep of . What we mentioned though about this association was that . Accordingly, it would be great if there was some formula that could compute . In fact, believe it or not there is such a formula. That said, it involves a somewhat strange idea–the hook-length of a square in a Ferrer’s diagram. Intuitively, the hook-length is just just the number of squares to the right of a square, below the square, and for the square itself. The reason the hook-length gets its name is that because if one imagines the hook-length it makes a ‘hook’ (see below) at the square in the sense that it looks like a line starting from the bottom of the column the square in question sits in, extends up to that square, and then makes a right turn and continues to the end of the row. So, after we define the hook-length we find certain characterizations of the product of the hook-length over all squares in a given Ferrer’s diagram since this is what shows up in the formula for .
Convenient Labeling for Ferrer’s Diagrams
Up until this point we’ve gotten by without defining an explicit labeling for the elements of a given Ferrer’s diagram, but to define the hook-length it is more convenient to do so. That said, the labeling is very intuitive. Namely, given a -frame we define the labeling in the matrix style fashion. Namely, the square in (we will often just say ) is the square in row and the . For example the square marked in
is in position .
We now can describe the hook-length of the square in a Ferrer’s diagram. Namely, given a -frame and some square we define the hook-length of this square to be the number . It’s clear that this counts the number of squares to the right of (i.e. ) plus the number of squares below (i.e. ) and an extra for the square itself. Pictorially this counts the size of the ‘hook’ formed by the squares mentioned. So in our above Ferrer’s diagram with the square the ‘hook’ formed is marked below
Product of Hook-lengths
It shall soon become clear that to us the product of over all for some -frame . Thus, we would like to give a quick equivalent of this product which shall be more convenient in the future. Namely:
Theorem: Let be a -frame, then
Proof: The proof is somewhat surprising. Namely, we first note that for each we have that
is a permutation of . Indeed, it’s first clear by definition that and evidently using the column decreasing conditions we see that and so for and and thus if we can conclude that for and we’ll be done because there are of them. To do this we examine two cases. Namely, first assume that then we have that
from where the conclusion follows by previous remark. From this though we get the fact that
finishes the argument.
1. Aigner, Martin. “Symmetric Functions.” A Course in Enumeration. Berlin: Springer, 2007. Print.