# Abstract Nonsense

## Subordinate and Superordinate Frames

Point of Post: In this post we define the notion of a subordinate frame and superordinate frame and discuss equivalent ways of defining them.

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Motivation

It’s clear that in our definition of $n$-frames that sitting inside each $n$-frame $\mathcal{F}$ is a lot of $n-1$-frames which can be gotten simply by removing a single box from $\mathcal{F}$. These $n-1$-frames ‘sitting’ inside $\mathcal{F}$ shall be what we call the $n-1$-frames ‘subordinate’ to $\mathcal{F}$. Of course, there is a dual notion where given an $n$-frame $\mathcal{F}$ we see that $\mathcal{F}$ sits subordinately inside a lot of $n+1$-frames $\mathcal{G}$, we shall say in this case that $\mathcal{G}$ is ‘superordinate’ to $\mathcal{F}$. Said slightly differently the $n+1$-frames superordinate to $\mathcal{F}$ are the $n+1$-frames which can be obtained from $\mathcal{F}$ by adding a single box to $\mathcal{F}$. The interesting thing is that given $f_{\text{st}}\left(\mathcal{G}\right)$ (the number of standard Young tableaux) for each $\mathcal{G}$ subordinate to $\mathcal{F}$ we can calculate $f_{\text{st}}\left(\mathcal{F}\right)$ and dually given $f_{\text{st}}\left(\mathcal{G}\right)$ for all $n+1$-frames $\mathcal{G}$ superordinate to $\mathcal{F}$ we can calculate $f_{\text{st}}\left(\mathcal{F}\right)$. That will be the topic of our next post

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Subordinate Frames

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Let $\mathcal{F}$ be an $n$-frame, then we say that the $n-1$-frame $\mathcal{G}$ is subordinate to $\mathcal{F}$, written $\mathcal{G}\triangleleft\mathcal{F}$, if $\mathcal{G}$ can be obtained from $\mathcal{F}$ by removing one box. For example,

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

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Given a frame $\mathcal{F}$ we denote, when convenient, the set of all $n-1$ frames subordinate to $\mathcal{F}$ by $\text{Sub}\left(\mathcal{F}\right)$. If given a specified block $b\in\mathcal{F}$ we denote the collection of squares obtained by removing $b$ from $\mathcal{F}$ by $\mathcal{F}-b$. It’s clear then that

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$\text{Sub}\left(\mathcal{F}\right)=\left\{\mathcal{F}-b:b\text{ is a block and }\mathcal{F}-b\text{ is a Ferrer's diagram}\right\}$

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A square in a $n$-frame $\mathcal{F}$ is called a bottom right corner of $\mathcal{F}$ if the square has no square to the right or below it. If $\mathcal{F}$ is given in the row structure notation as $\mathcal{F}=(m_1,\cdots,m_r)$ then a square is a bottom right corner if it occurs as the $m_j^{\text{th}}$ square in the $j^{\text{th}}$ row (for some $j\in[r]$) and either $j=r$ or $m_{j+1}. We denote the set of all bottom right corners of $\mathcal{F}$ by $\text{BRC}\left(\mathcal{F}\right)$. What we note is that the only removable squares occur as bottom right corners in the sense that if a square is removable then there must be no squares to the right or below that square otherwise the resulting collection of squares (after removing the square in question) shall not be a Ferrer’s diagram. Moreover, it’s clear that if a square occurs as a bottom right corner in $\mathcal{F}$ then removing that square results in a Ferrer’s diagram and so we arrive at the following theorem:

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Theorem: Let $\mathcal{F}$ be an $n$-frame. Then,

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$\text{Sub}\left(\mathcal{F}\right)=\left\{\mathcal{F}-b:b\in\text{BRC}\left(\mathcal{F}\right)\right\}$

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In particular, $\#\text{Sub}\left(\mathcal{F}\right)=\#\text{BRC}\left(\mathcal{F}\right)$.

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Superordinate Frames

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We now define the dual notion of a subordinate frame. Namely, if $\mathcal{F}$ is a $n$-frame and $\mathcal{G}$ is a $n+1$-frame which can be obtained by adding one block to $\mathcal{F}$ we say that $\mathcal{G}$ is superordinate to $\mathcal{F}$, and denote this $\mathcal{G}\triangleright \mathcal{F}$. We denote, when convenient, the set of all $n+1$-frames superordinate to a $n$-frame $\mathcal{F}$ by $\text{Super}\left(\mathcal{F}\right)$. It’s evident that if $\mathcal{F}$ is a $n$-frame that

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$\displaystyle \text{Super}\left(\mathcal{F}\right)=\left\{\mathcal{G}:\mathcal{G}\text{ is a }n+1\text{-frame}\text{ and }\mathcal{F}\triangleleft\mathcal{G}\right\}$

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Let $\mathcal{F}$ be a $n$-frame with row structure $(m_1,\cdots,m_r)$. It’s evident that we may insert a square at the end of the $j^{\text{th}}$ row if and only if $j=1$ or $m_{j-1}>m_j$. It thus makes sense that

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$\text{Super}\left(\mathcal{F}\right)=\left\{(m_1,\cdots,m_j+1,\cdots,m_r):j=1\text{ or }m_{j-1}>m_j\right\}\cup\left\{(m_1,\cdots,m_r,1)\right\}$

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where the last term represents the $n+1$-frame obtained from $\mathcal{F}$ by forming a new row. Note though that each $\mathcal{G}\in\text{Super}\left(\mathcal{F}=(m_1,\cdots,m_r\right)$ which is not the $n+1$-frame obtained by starting a new row corresponds uniquely to a block at the end of a row $j$ of $\mathcal{F}$ with $j=1$ or $m_{j-1}>m_j$. But, each of these blocks is either an element of $\text{BCR}\left(\mathcal{F}\right)$ or lies directly above an element of $\text{BCR}\left(\mathcal{F}\right)$. For example, the block marked $\times$ in the following $16$-frame

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

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is a block for which one may add a block beside it to obtain a $17$-frame superordinate to $\mathcal{F}$ and yet this block is not an element of $\text{BCR}\left(\mathcal{F}\right)$ we see that it corresponds (uniquely) to the bottom right corner marked $\times$ below

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

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it thus follows that

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$\#\text{Super}\left(\mathcal{F}\right)=\#\text{BCR}\left(\mathcal{F}\right)+1=\text{Sub}\left(\mathcal{F}\right)+1$

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References:

1. Simon, Barry. Representations of Finite and Compact Groups. Providence, RI: American Mathematical Society, 1996. Print.