# Abstract Nonsense

## Relation Between the Number of Standard Young Tableaux on a Frame and the Number of Young Tableaux on the Frame’s Subordinate/Superordinate Frames

Point of Post: In this post we find a relation between the number of standard Young tableaux on a frame $\mathcal{F}$ and the number of Young tableaux all the subordinate and superordinate frames to $\mathcal{F}$.

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Motivation

As was stated in our last post we can find a very interesting way to calculate the number, $f_{\text{st}}\left(\mathcal{F}\right)$, of standard Young tableaux $\mathcal{T}$ with $\text{Frame}\left(\mathcal{T}\right)=\mathcal{F}$. In this post we actually prove this claim. The intuitive idea is clear, by construction of Young tableaux we see that if $\mathcal{T}$ is a Young Tableaux such that $\text{Frame}\left(\mathcal{T}\right)$ is a $n$-frame then the number $n$ must lie in a bottom right corner of $\text{Frame}\left(\mathcal{T}\right)$ and then fixing $n$ in that position $b$  we see that the possible Young tableaux are just the Young tableaux of $\mathcal{F}-b$ and thus it makes sense then that $f_{\text{st}}\left(\mathcal{F}\right)$ is some sort of sum of $\text{f}_{\text{st}}\left(\mathcal{G}\right)$ where $\mathcal{G}$ is taken over the subordinate frames to $\mathcal{F}$. The other theorem which has to do with finding $f_{\text{st}}\left(\mathcal{F}\right)$ given the values $f_{\text{st}}\left(\mathcal{H}\right)$ where $\mathcal{H}$ is taken over the frames superordinate to $\mathcal{F}$.

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The Relation Between $f_{\text{st}}\left(\mathcal{F}\right)$ and $f_{\text{st}}\left(\mathcal{G}\right)$ for $\mathcal{G}\triangleleft\mathcal{F}$

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We jump right into the claim:

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Theorem: Let $\mathcal{F}$ be a $n$-frame, then

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$\displaystyle f_{\text{st}}\left(\mathcal{F}\right)=\sum_{\mathcal{G}\triangleleft \mathcal{F}}f_{\text{st}}\left(\mathcal{G}\right)$

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Proof: Let $\text{Tab}\left(\mathcal{F}\right)$ be the set of all Young tableaux with frame $\mathcal{F}$. Let $b\in\mathcal{F}$ be the block with $n\in b$ (by this I merely mean the block which contains the value $n$). Note that since the row $b$ is increasing we must have that it is the last in its row since any block to its right would have to have a lower value than $n$. Similarly, $b$ is the bottom of its columns for the same reason, and thus $b\in\text{BCR}\left(\mathcal{F}\right)$. It clearly then follows that if we let for every $b\in\text{BCR}\left(\mathcal{F}\right)$, $\text{Tab}_b\left(\mathcal{F}\right)$ be the elements of $\text{Tab}\left(\mathcal{F}\right)$ with $n\in b$ then

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$\displaystyle \text{Tab}\left(\mathcal{F}\right)=\bigsqcup_{b\in\text{BCR}\left(\mathcal{F}\right)}\text{Tab}_b\left(\mathcal{F}\right)\quad\mathbf{(1)}$

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where the disjointness is obvious since elements from different $\text{Tab}_b\left(\mathcal{F}\right)$‘s have the value of $n$ in different boxes. That said, it’s clear that every standard Young tableau in $\text{Tab}_b\left(\mathcal{F}\right)$ is really just a standard Young tableau of $\mathcal{F}-b$ (which we know is really a legitimate $n-1$-frame) with then just the extra condition that $n\in b$ and moreover it’s clear that every standard Young tableau of $\mathcal{F}-b$ can be extended to an element of $\text{Tab}_b\left(\mathcal{F}\right)$ by taking the standard Young tableau, adding in the box $b$ at it’s correct point and then filling it in with $b$. It clearly follows then that $\#\text{Tab}_b\left(\mathcal{F}\right)=f_{\text{st}}\left(\mathcal{F}-b\right)$. Thus, by $\mathbf{(1)}$ and our previous characterization of $\text{Sub}\left(\mathcal{F}\right)$ we see that

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$\displaystyle f_{\text{st}}\left(\mathcal{F}\right)=\#\text{Tab}\left(\mathcal{F}\right)=\sum_{b\in\text{BCR}\left(\mathcal{F}\right)}\#\text{Tab}_b\left(\mathcal{F}\right)=\sum_{b\in\text{BCR}\left(\mathcal{F}\right)}f_{\text{st}}\left(\mathcal{F}-b\right)=\sum_{\mathcal{G}\triangleleft\mathcal{F}}f_{\text{st}}\left(\mathcal{G}\right)$

$\blacksquare$

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The Relation Between $f_{\text{st}}\left(\mathcal{F}\right)$ and $f_{\text{st}}\left(\mathcal{H}\right)$ for $\mathcal{H}\triangleright\mathcal{F}$

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Of course, as usual in algebra, there is a dual notion to the above relation which we dive right into

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Theorem: Let $\mathcal{F}$ be a $n$-frame, then

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$\displaystyle (n+1)f_{\text{st}}\left(\mathcal{F}\right)=\sum_{\mathcal{H}\triangleright\mathcal{F}}f_{\text{st}}\left(\mathcal{H}\right)$

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Proof: We proceed by induction on the size of the frames. For $1$-frames this obvious since the unique $1$-frame is $\begin{array}{|c|}\cline{1-1}\hphantom{\times}\\ \cline{1-1}\end{array}$. So, assume that the result is true for every $n$-frame and let $\mathcal{F}$ be a $n+1$-frame. Then, using the previous theorem we note that

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$\displaystyle \sum_{\mathcal{H}\triangleright\mathcal{F}}f_{\text{st}}\left(\mathcal{H}\right)=\sum_{H\triangleright\mathcal{F}}\sum_{\mathcal{K}\triangleleft\mathcal{H}}f_{\text{st}}\left(\mathcal{K}\right)$

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In this second sum we are summing over subordinates of superordinates of $\mathcal{F}$ and it’s clear that except the case when the superordinate is the addition of the new row at the bottom of $\mathcal{F}$ every such subordinaet of a superordinate is really the superordinate of a subordinate of $\mathcal{F}$. Said differently

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$\displaystyle \sum_{\mathcal{H}\triangleright\mathcal{F}}f_{\text{st}}\left(\mathcal{F}\right)=\sum_{\mathcal{H}\triangleright\mathcal{F}}\sum_{\mathcal{K}\triangleleft\mathcal{H}}f_{\text{st}}\left(\mathcal{K}\right)=\sum_{\mathcal{G}\triangleleft\mathcal{F}}\sum_{\mathcal{J}\triangleright\mathcal{G}}f_{\text{st}}\left(\mathcal{J}\right)+f_{\text{st}}\left(\mathcal{F}\right)$

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But, since each $\mathcal{J}$ is a $n$-frame (being subordinate to $\mathcal{F}$) we may apply the induction hypothesis and the first theorem to get

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\displaystyle \begin{aligned}\sum_{\mathcal{H}\triangleright\mathcal{F}}f_{\text{st}}\left(\mathcal{H}\right)& =\sum_{\mathcal{G}\triangleleft\mathcal{F}}\sum_{\mathcal{J}\triangleright\mathcal{G}}f_{\text{st}}\left(\mathcal{J}\right)+f_{\text{st}}\left(\mathcal{F}\right)\\ &=\sum_{\mathcal{G}\triangleleft\mathcal{F}}(n+1)f_{\text{st}}\left(\mathcal{G}\right)+f_{\text{st}}\left(\mathcal{F}\right)\\ &=(n+1)f_{\text{st}}\left(\mathcal{F}\right)+f_{\text{st}}\left(\mathcal{F}\right)\\ &=(n+2)f_{\text{st}}\left(\mathcal{F}\right)\end{aligned}

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and thus the induction is complete. $\blacksquare$

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References:

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

May 11, 2011 -