Abstract Nonsense

Crushing one theorem at a time

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.

\text{ }

Motivation

\text{ }

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.

\text{ }

The Lemmas

\text{ }

We dive right into it:

\text{ }

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

\text{ }

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

\text{ }

from where there the conclusion follows. \blacksquare

\text{ }

\text{ }

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

\text{ }

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

\text{ }

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

\text{ }

\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)

\text{ }

the conclusion follows.

\text{ }

\text{ }

References:

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

Advertisements

May 23, 2011 - Posted by | Algebra, Algebraic Combinatorics, Representation Theory | , , , , ,

No comments yet.

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s

%d bloggers like this: