# Abstract Nonsense

## Two Technical Lemmas for the Construction of the Irreps of S_n

Point of Post: In this post we prove two technical lemmas in relation to the row and column stabilizer functions which will ultimately help us construct the irreps of $S_n$.

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Motivation

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We are at the penultimate post before carrying through with our long-ago promised goal of constructing the irreps of $S_n$ in a way for which they are naturally labeled by $n$-frames. In this post we just need to prove two technical lemmas before this.

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The Lemmas

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We dive right into it:

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Lemma 1: Let $\mathcal{T}$ be a fixed $n$-tableaux and suppose that $\sigma\in S_n$ cannot be written as $\sigma=pq$ where $p\in\mathcal{R}\left(\mathcal{T}\right)$ and $q\in\mathcal{C}\left(\mathcal{T}\right)$ (the row and column stabilizers and the product is the product in $\mathbb{C}[S_n]$ not in $S_n$ itself). Then, there exists some $p_0\in\mathcal{R}\left(\mathcal{T}\right)$ and $q_0\in\mathcal{C}\left(\mathcal{T}\right)$ such that $q_0$ is odd, and $\sigma=p_0\sigma q_0$.

Proof: Note that since $\sigma\mathcal{T}$ cannot be written as $pq\mathcal{T}$ by assumption we have by a previous theorem that $\sigma\mathcal{T}\multimap\mathcal{T}$. Thus, there exists $i,j\in[n]$ which are in the same row of $\mathcal{T}$ and the same column of $\sigma\mathcal{T}$. We have by definition that $p_0=(i,j)\in\mathcal{R}\left(\mathcal{T}\right)$ and so consider $q_0=\sigma^{-1}(i,j)\sigma$, we have by a previous theorem that $q_0\in\mathcal{C}\left(\sigma^{-1}\sigma\mathcal{T}\right)=\mathcal{C}\left(\mathcal{T}\right)$. Moreover, it’s clear that $\text{sgn}(q_0)=\text{sgn}(\sigma^{-1}(i,j)\sigma)=\text{sgn}((i,j))=-1$ and so $q_0$ is odd. To finish we note that

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$p_0\sigma q_0=(i,j)\sigma\sigma^{-1}(i,j)\sigma=(i,j)^2\sigma=\sigma$

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from where there the conclusion follows. $\blacksquare$

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Lemma 2: Let $\mathcal{T}$ be a fixed $n$-tableau and $\sigma\in\mathbb{C}[S_n]$ be such that for all $p\in\mathcal{R}\left(\mathcal{T}\right)$ and $q\in\mathcal{C}\left(\mathcal{T}\right)$ it’s true that  $p\sigma q=\text{sgn}(q)\sigma$. Then, there exists some $z\in\mathbb{C}$ such that $\sigma=z E(\mathcal{T})$.

Proof: We know by definition that there exists $z_\pi\in\mathbb{C}$ for each $\pi\in S_n$ such that

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$\displaystyle \sigma=\sum_{\pi\in S_n}z_\pi \pi$

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But, this then implies by assumption that $z_{p\pi q}=\text{sgn}(q)z_\pi$ for each $p\in\mathcal{R}\left(\mathcal{T}\right)$ and $q\in\mathcal{C}\left(\mathcal{T}\right)$. But, in particular we see that this implies that $z_{pq}=\text{sgn}(q)z_e$. But by the previous lemma if we choose $s\ne pq$ with $q$ odd then $z_s=z_{p_0sq_0}=-z_s$ and so $z_s=0$. It thus follows that most of the terms in the above sum fall out and we are left with

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$\displaystyle \sigma=z_e\sum_{p\in\mathcal{R}\left(\mathcal{T}\right)}\sum_{q\in\mathcal{C}\left(\mathcal{T}\right)}\text{sgn}(q)pq=z_e E\left(\mathcal{T}\right)$

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the conclusion follows.

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References:

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