Abstract Nonsense

A Weird Condition on Tableaux

Point of Post: In this post we discuss an interesting property between two tableaux which will ultimately help us construct the irreps of $S_n$ associated to each $n$-frame.

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Motivation

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So, enough being cryptic. I promised that we will create a bijection $\text{Frame}_n\to\widehat{S_n}$ in such a way that $\deg\rho^{(\mathcal{F})}=f_{\text{st}}\left(\mathcal{F}\right)$–it’s about time I explained roughly how. So, in our last post we created this interesting function $E:\text{Tab}\left(\mathcal{F}\right)\to\mathbb{C}\left[S_n\right]$. Our main goal to the construction is to show that up to normalization $E\left(\mathcal{T}\right)$ is a minimal projection from where we shall get our corresponded irrep. In the journey to prove this we will need a strange, un-motivated concept which has to do with the relationship between the rows of one tableau $\mathcal{T}$ and another tableau $\mathcal{T}'$.Luckily, the motivation and usefulness will become apparent shortly. That said, we can at least give a glance of why anyone would even care about this condition. In particular, we shall use this condition to prove that the irreps associated to two different $n$-frames are different.

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The Condition

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We now define the condition we stated above. In particular, let $\mathcal{T},\mathcal{T}'$ be two $n$– tableaux. We say that $\mathcal{T}'\multimap\mathcal{T}$ if there exists $i,j\in[n]$ such that $i,j$ are in the same row of $\mathcal{T}$ and the same column of $\mathcal{T}'$. For example:

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$\begin{array}{|c|c|c|}\cline{1-3} 4 & 6 & 3\\ \cline{1-3} 5 & 2 & \multicolumn{1}{c}{}\\ \cline{1-2} 1 & \multicolumn{2}{c}{}\\ \cline{1-1}\end{array}\multimap\;\; \begin{array}{|c|c|c|}\cline{1-3} 1 & 2 & 3\\ \cline{1-3} 4 & 5 &\multicolumn{1}{c}{} \\ \cline{1-2} 6 &\multicolumn{2}{c}{}\\ \cline{1-1}\end{array}$

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since $4,5$ are both in the same row of the right tableau and the same column of the left tableau. Note that this is really saying that the transposition $(i,j)$ is in $\mathcal{R}\left(\mathcal{T}\right)\cap\mathcal{C}\left(\mathcal{T}'\right)$. If $\mathcal{T}'\multimap\mathcal{T}$ we obviously write $\mathcal{T}'\not\multimap\mathcal{T}$. To get the first reason why this condition would ever come up in relation to our last post we first note that:

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Theorem: Let $\mathcal{T},\mathcal{T}'$ be two $n$-tableaux such that $\mathcal{T}'\multimap\mathcal{T}$. Then, $Q\left(\mathcal{T}'\right)P\left(\mathcal{T}\right)=0$ (where the product of these two elements of $\mathbb{C}\left[S_n\right]$ is meant to be interpreted as convolution as per the identification of $\mathbb{C}\left[S_n\right]$ with $\mathcal{A}\left(S_n\right)$ as previously discussed).

Proof: We begin by noticing that

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$\displaystyle p_0P\left(\mathcal{T}\right)=p_0\sum_{p\in\mathcal{R}\left(\mathcal{T}\right)}p=\sum_{p\in\mathcal{R}\left(\mathcal{T}\right)}p_0 p=\sum_{p\in\mathcal{T}\left(\mathcal{T}\right)}p=P\left(\mathcal{T}\right)$

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where we’ve used the fact that $\mathcal{R}\left(\mathcal{T}\right)\leqslant S_n$, $p_0p=\delta_{p_0p}$, and $p\mapsto p_0p$ is a bijection $\mathcal{R}\left(\mathcal{T}\right)\to\mathcal{R}\left(\mathcal{T}\right)$. Similarly one can show that

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$P\left(\mathcal{T}\right)p_0=P\left(\mathcal{T}\right)\quad\text{and}\quad q_0 Q\left(\mathcal{T}'\right)=Q\left(\mathcal{T}'\right)q_0=\text{sgn}(q_0)Q\left(\mathcal{T}'\right)$

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where $q_0\in\mathcal{C}\left(\mathcal{T}\right)$ (the only [really quite obvious] extra trick for the $Q$ case is the fact that $\text{sgn}$ is a homomorphism). With this observation the proof becomes easy. In particular, since $(i,j)\in\mathcal{R}\left(\mathcal{T}\right)\cap\mathcal{C}\left(\mathcal{T}'\right)$ (where $i,j$ are the guaranteed elements of $[n]$ in the definition of $\multimap$) we see that

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\begin{aligned}Q\left(\mathcal{T}'\right)P\left(\mathcal{T}\right) &=Q\left(\mathcal{T}'\right)\left((i,j)P\left(\mathcal{T}\right)\right)\\ &=\left(Q\left(\mathcal{T}'\right)(i,j)\right)P\left(\mathcal{T}\right)\\ &=\text{sgn}\left((i,j)\right)Q\left(\mathcal{T}'\right)P\left(\mathcal{T}\right)\\ &=-Q\left(\mathcal{T}'\right)P\left(\mathcal{T}\right)\end{aligned}

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from where the conclusion follows. $\blacksquare$

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Corollary: Let $\mathcal{T},\mathcal{T}'$ be two $n$-tableaux with $\mathcal{T}'\multimap\mathcal{T}$, then $E\left(\mathcal{T}'\right)E\left(\mathcal{T}\right)=0$.

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A Characterization of when $\mathcal{T}'\not\multimap\mathcal{T}$

We actually now find a characterization of when two $n$-tableaux $\mathcal{T}',\mathcal{T}$ satisfy $\mathcal{T}'\not\multimap\mathcal{T}$ and they have the same frame. Indeed:

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Theorem: Let $\mathcal{T},\mathcal{T}'$ be two $n$-tableaux with $\text{Frame}\left(\mathcal{T}\right)=\text{Frame}\left(\mathcal{T}'\right)$, then $\mathcal{T}'\multimap\mathcal{T}$ if and only if  there exists $p\in\mathcal{R}\left(\mathcal{T}\right)$ and $q\in\mathcal{C}\left(\mathcal{T}'\right)$ such that

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$p\mathcal{T}=q\mathcal{T}'$

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recalling the $S_n$-action on tableaux.

Proof: Suppose first that $p\mathcal{T}=q\mathcal{T}'$ as described above. Next, suppose that $i,j$ were in the same row in $\mathcal{T}$, then by definition $i,j$ are in the same row of $p\mathcal{T}$. That said, we also have by assumption then that $i,j$ are in the same row of $q\mathcal{T}'$ and in particular not in the same column. But, this implies by definition that (since $\mathcal{C}\left(\mathcal{T}'\right)\leqslant S_n$) that $i,j$ are in different rows of $q^{-1}q\mathcal{T}'=\mathcal{T}'$. Since $i,j$ were arbitrary the conclusion follows.

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Conversely, suppose that $\mathcal{T}'\not\multimap\mathcal{T}$. Let $i,j$ be two elements in the same column of $\mathcal{T}'$, since $\mathcal{T}'\not\multimap\mathcal{T}$ we have that there exists a permutation $\sigma\in \mathcal{R}\left(\mathcal{T}\right)$ such that $i,j$ are in the same column of $\sigma\mathcal{T}$. And, so there exists a permutation $\tau\in \mathcal{C}\left(\mathcal{T}'\right)$ such that $i,j$ lie in the same position in $\tau\mathcal{T}'$ as they do in $\sigma\mathcal{T}$. Noting that we can repeat this process and not screw any of the previous permutations up (since the permutations that shift rows and columns can move the things that are of interest now say some $k,\ell$ different than $i,j$ and leave $i,j$ fixed) we may conclude that there exists $p\in\mathcal{R}\left(\mathcal{T}\right)$ and $s\in\mathcal{C}\left(\mathcal{T}'\right)$ such that

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$p\mathcal{T}=s\mathcal{T}'$

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The conclusion follows.

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Corollary: Let $\mathcal{T},\mathcal{T}'$ are $n$-frames with $\text{Frame}\left(\mathcal{T}\right)=\text{Frame}\left(\mathcal{T}'\right)$ and $\mathcal{T}'\not\multimap\mathcal{T}$, then there exists $p\in\mathcal{R}\left(\mathcal{T}\right)$ and $q\in\mathcal{C}\left(\mathcal{T}\right)$ such that $\mathcal{T}'=pq\mathcal{T}$

Proof: This doesn’t follow immediately from the previous theorem since the order of the $S_n$ elements is different. But, we note that if $p,s$ are the elements of $\mathcal{R}\left(\mathcal{T}\right)$ and $\mathcal{C}\left(\mathcal{T}'\right)$ guaranteed by the previous theorem such that $p\mathcal{T}=s\mathcal{T}'$ then we see that $\mathcal{T}'=pq\mathcal{T}$ where $q=p^{-1}s^{-1}p=\left(p^{-1}s\right)s^{-1}\left(p^{-1}s\right)^{-1}$ but we have by previous theorem then that

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$s\in \left(p^{-1}s\right)\mathcal{C}\left(\mathcal{T}'\right)\left(p^{-1} s\right)^{-1}=\mathcal{C}\left(p^{-1}s\mathcal{T}'\right)=\mathcal{C}\left(p^{-1}\mathcal{T}\right)=\mathcal{C}\left(\mathcal{T}\right)$

$\blacksquare$

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References:

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