# Abstract Nonsense

## 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.

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Motivation

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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 $n$-frame $\mathcal{F}$ an irrep $\rho^{(\mathcal{F})}$ of $S_n$. What we mentioned though about this association was that $\deg\rho^{(\mathcal{F})}=f_{\text{st}}\left(\mathcal{F}\right)$. Accordingly, it would be great if there was some formula that could compute $f_{\text{st}}\left(\mathcal{F}\right)$. 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 $1$ 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 $\text{f}_{\text{st}}\left(\mathcal{F}\right)$.

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Convenient Labeling for Ferrer’s Diagrams

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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 $n$-frame $\mathcal{F}$ we define the labeling in the matrix style fashion. Namely, the square $(i,j)$ in $\mathcal{F}$ (we will often just say $(i,j)\in\mathcal{F}$) is the square in $i^{\text{th}}$ row and the $j^{\text{th}}$. For example the square marked $\times$ in

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$\begin{array}{|c|c|c|c|}\cline{1-4}\hphantom{\times} &\hphantom{\times} &\hphantom{\times}&\hphantom{\times}\\ \cline{1-4}&&\\ \cline{1-3}&&&\multicolumn{1}{c}{}\\ \cline{1-3}&&\times&\multicolumn{1}{c}{}\\ \cline{1-3}&&\multicolumn{2}{c}{}\\ \cline{1-2}&\multicolumn{3}{c}{}\\ \cline{1-1}\end{array}$

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is in position $(4,3)$.

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Hook Length

We now can describe the hook-length of the square in a Ferrer’s diagram. Namely, given a $n$-frame $\mathcal{F}=(m_1\cdots,m_r)$ and some square $(i,j)\in\mathcal{F}$ we define the hook-length of this square to be the number $h_{i,j}=m_i+r-i-j+1$. It’s clear that this counts the number of squares to the right of $(i,j)$ (i.e. $m_i-i$) plus the number of squares below $(i,j)$ (i.e. $r-j$) and an extra $1$ for the square $(i,j)$ itself. Pictorially this counts the size of the ‘hook’ formed by the squares mentioned. So in our above Ferrer’s diagram with the square $(3,2)$ the ‘hook’ formed  is marked below

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$\begin{array}{|c|c|c|c|}\cline{1-4}\hphantom{\times} &\hphantom{\times}&\hphantom{\times}&\hphantom{\times}\\ \cline{1-4}&&\\ \cline{1-3}&\times&\times&\multicolumn{1}{c}{}\\ \cline{1-3}&\times&&\multicolumn{1}{c}{}\\ \cline{1-3}&\times&\multicolumn{2}{c}{}\\ \cline{1-2}&\multicolumn{3}{c}{}\\ \cline{1-1}\end{array}$

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Product of Hook-lengths

It shall soon become clear that to us the product of $h_{i,j}$ over all $(i,j)\in\mathcal{F}$ for some $n$-frame $\mathcal{F}$. Thus, we would like to give a quick equivalent of this product which shall be more convenient in the future. Namely:

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Theorem: Let $\mathcal{F}=(m_1,\cdots,m_r)$ be a $n$-frame, then

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$\displaystyle \prod_{(i,j)\in\mathcal{F}}h_{i,j}=\frac{\displaystyle \prod_{i=1}^{r}(h_{i,1})!}{\displaystyle \prod_{1\leqslant i

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Proof: The proof is somewhat surprising. Namely, we first note that for each $i\in[r]$ we have that

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$h_{i,1},h_{i,2}\cdots,h_{i,m_i},h_{i,1}-h_{i+1,1},h_{i,1}-h_{i+2,1},\cdots,h_{i,1}-h_{r,1}$

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is a permutation of $\{1,\cdots,h_{i,1}\}$. Indeed, it’s first clear by definition that $h_{i,1}>h_{i,2}>\cdots>h_{i,m_i}\geqslant 1$ and evidently using the column decreasing conditions we see that $1\leqslant h_{i,1}-h_{i+1,1}<\cdots and so $h_{i,k},h_{i,1}-h_{i+\ell,1}\in[h_{i,1}]$ for $k=1,\cdots,m_i$ and $\ell=i+1,\cdots,r$ and thus if we can conclude that $h_{i,k}\ne h_{i,1}-h_{i+\ell,1}$ for $k\in[r]$ and $\ell=i+1,\cdots,r$ we’ll be done because there are $m_i+r-i=h_{i,1}=\#[h_{i,1}]$ of them. To do this we examine two cases. Namely, first assume that $k\leqslant m_\ell$ then we have that

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$h_{i,k}\geqslant (m_i-m_\ell)+(\ell-i)+1>(m_i-m_\ell)+(\ell-i)=h_{i,1}-h_{\ell,1}$

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If $k>m_\ell$ then

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$h_{i,k}\leqslant (m_i-m_\ell-1)+(\ell-1-i)+1<(m_i-m_\ell)+(\ell-i)=h_{i,1}-h_{i,\ell}$

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from where the conclusion follows by previous remark. From this though we get the fact that

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$\displaystyle (h_{i,1})!\prod_{i

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noting that

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$\displaystyle \prod_{(i,j)\in\mathcal{F}}h_{i,j}=\prod_{i=1}^{r}\prod_{j=1}^{m_r}h_{i,j}$

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finishes the argument. $\blacksquare$

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References:

1. Aigner, Martin. “Symmetric Functions.” A Course in Enumeration. Berlin: Springer, 2007. Print.