Lectures in Representation Theory

Arun Ram
Department of Mathematics and Statistics
University of Melbourne
Parkville, VIC 3010 Australia
aram@unimelb.edu.au

Last update: 27 August 2013

Lecture 15

Theorem 2.45 $$p_r\left(x\right)s_{\lambda }\left(x\right)=\sum _{\mu }{(-1)}^{k-1}{s}_{\mu }\left(x\right),$$ where the sum is over partitions $\mu \supseteq \lambda$ such that $\mu /\lambda$ is a border strip of length $r,$ and $k$ is the number of rows in $\mu /\lambda \text{.}$

		Proof.
	Since the function $p_r\left(x\right)$ is symmetric, we have $$\begin{array}{ccc}{p}_r\left(x\right)a_{\lambda +\delta }\left(x\right)& =& \sum _{w\in W}\epsilon \left(w\right)w{p}_r\left(x\right)x^{\lambda +\delta }\\ & =& \sum _{w\in W}\epsilon \left(w\right)w\sum _{i=1}^n{x}_i^r{x}_1^{{\lambda }_1+n-1}{x}_2^{{\lambda }_2+n-2}\cdots x_n^{{\lambda }_n}\\ & =& \sum _{i=1}^n\sum _{w\in W}\epsilon \left(w\right)w{x}_1^{{\lambda }_1+n-1}{x}_2^{{\lambda }_2+n-2}\cdots x_i^{{\lambda }_i+n-i+r}\cdots x_n^{{\lambda }_n}\\ & =& \sum _{i=1}^n{a}_{\lambda +\delta +r{\epsilon }_i}\left(x\right),\end{array}$$ where ${\epsilon }_i=(0,0,\dots ,1,0,\dots ,0)$ has a 1 in the $i\text{th}$ slot and 0 everywhere else. The function $a_{\lambda +\delta +r{\epsilon }_i}\left(x\right)$ is alternating, so $$a_{\lambda +\delta +r{\epsilon }_i}\left(x\right)=0,$$ if there is a repeated entry in the exponents. That is, it is zero if $$(n-i)+{\lambda }_i+r=(n-j)+{\lambda }_j$$ for some $1\le j\le i\text{.}$ If this is not the case, then there exists some $j$ with $1\le j<i$ so that $w=s_j{s}_{j+1}\cdots s_{i-1}$ satisfies $$w{x}^{\lambda +{\delta }^r{\epsilon }_i}=x^{\mu +\delta }$$ for some partition $\mu \text{.}$ In other words, $w$ rearranges the exponents so that they are in the form $\mu +\delta \text{.}$ Then we have $$a_{\lambda +\delta +r{\epsilon }_i}\left(x\right)={(-1)}^{\ell \left(w\right)}{a}_{\mu +\delta }\left(x\right)$$ To see the relation to border strips, consider the case where $\lambda =(3,3,2,1,1,1,0),$ $r=6,$ and $i=4\text{.}$ Then $$\lambda +\delta +r{\epsilon }_i=\begin{array}{ccccccccccc}\square & \square & \square & \square & \square & \square & |& \square & \square & \square & \\ & \square & \square & \square & \square & \square & |& \square & \square & \square & \\ & & \square & \square & \square & \square & |& \square & \\ & & & \square & \square & \square & |& \square & •& •& •& •& •& •\\ & & & & \square & \square & |& \square & \\ & & & & & \square & |& \\ & & & & & & |& \end{array}\text{.}$$ Since $a_{\lambda +\delta +r{\epsilon }_i}\left(x\right)$ is alternating, we can apply $s_3,$ as long as we multiply by $-1\text{.}$ Multiplying by $s_3$ switches row three and row four. If we do this and then write the result with $\delta$ left of the wall, we get $$\begin{array}{ccccccccccc}\square & \square & \square & \square & \square & \square & |& \square & \square & \square & \\ & \square & \square & \square & \square & \square & |& \square & \square & \square & \\ & & \square & \square & \square & \square & |& \square & •& •& •& •& •\\ & & & \square & \square & \square & |& \square & •& \\ & & & & \square & \square & |& \square & \\ & & & & & \square & |& \\ & & & & & & |& \end{array}\text{.}$$ Continuing in this way, i.e., multiplying by $s_2$ and then $s_1,$ gives the diagram $$\begin{array}{ccccccccccc}\square & \square & \square & \square & \square & \square & |& \square & \square & \square & •\\ & \square & \square & \square & \square & \square & |& \square & \square & \square & •\\ & & \square & \square & \square & \square & |& \square & •& •& •\\ & & & \square & \square & \square & |& \square & •& \\ & & & & \square & \square & |& \square & \\ & & & & & \square & |& \\ & & & & & & |& \end{array},$$ which is a border strip of length 6 that has 4 rows. For each $i,$ we add a border strip of length $r$ at row $i,$ and then sum over $i\text{.}$ If adding such a strip does not produce a partition, then at some point in the process, we have violated the condition that all of the exponents (thus the length of each row) must be distinct, and we get 0. Thus, we have shown $$p_r\left(x\right)a_{\lambda +\delta }\left(x\right)=\sum _{\mu }{(-1)}^{k-1}{a}_{\mu +\delta }\left(x\right)\text{.}$$ Dividing both sides by $a_{\delta }\left(x\right)$ proves the theorem. $\square$

Corollary 2.46 $p_1\left(x\right)s_{\lambda }\left(x\right)={\displaystyle \sum _{\underset{\mu /\lambda =\square }{\mu \supseteq \lambda }}}{s}_{\mu }\left(x\right)\text{.}$

Corollary 2.47 $p_{\left(1^m\right)}\left(x\right)={\displaystyle \sum _{\lambda }}{f}_{\lambda }{s}_{\lambda }\left(x\right)$ for some positive integer $f_{\lambda }\text{.}$

		Proof.
	Use induction and the previous corollary. $\square$

Since $f_{\lambda }>0,$ we have finished the proof of the Frobenius formula 2.29.

Theorem 2.48 [Murnaghan-Nakayama Rule] Let $\mu =({\mu }_1,\dots ,{\mu }_k)$ and $\hat{\mu }=({\mu }_1,\dots ,{\mu }_{k-1})\text{.}$ Then $${\chi }^{\lambda }\left(\mu \right)=\sum _{\nu }{(-1)}^{s-1}\left(\hat{\mu }\right),$$ where the sum is over $\nu \supseteq \lambda$ such that $\lambda /\nu$ is a border strip of length ${\mu }_k,$ and $s$ is the number of rows in this border strip.

		Proof.
	From the Frobenius formula, we have $$\begin{array}{ccc}\sum _{\lambda }{\chi }^{\lambda }\left(\mu \right)s_{\lambda }\left(x\right)& =& p_{\mu }\left(x\right)\\ & =& p_{{\mu }_1}\left(x\right)p_{{\mu }_2}\left(x\right)\cdots p_{{\mu }_k}\left(x\right)\\ & =& p_{\hat{\mu }}\left(x\right)p_{{\mu }_k}\left(x\right)\\ & =& \left(\sum _{\nu }{\chi }^{\nu }\left(\hat{\mu }\right)s_{\nu }\left(x\right)\right)p_{{\mu }_k}\left(x\right)\\ & =& \sum _{\nu }{\chi }^{\nu }\left(\hat{\mu }\right)\sum _{\lambda }{(-1)}^{s-1}{s}_{\lambda }\left(x\right)\\ & =& \sum _{\lambda }{s}_{\lambda }\left(x\right)\sum _{\nu }{(-1)}^{s-1}{\chi }^{\nu }\left(\hat{\mu }\right)\end{array}$$ where $\lambda \supseteq \nu$ and $\lambda /\nu$ is a border strip of length ${\mu }_k$ with $s$ rows. Comparing coefficients of $s_{\lambda }$ gives the theorem. $\square$

Definition 2.49 The weight wt of a border strip $\lambda /\mu$ is given by $\text{wt}(\lambda /\mu )={(-1)}^{s-1},$ where $s$ is the number of rows in $\lambda /\mu \text{.}$

Definition 2.50 Let $\mu =({\mu }_1,\dots ,{\mu }_k)$ be a partition. A $\mu \text{-border}$ strip tableau of shape $\lambda$ is a sequence $T$ of partitions $$T=(\varnothing ={\lambda }^{\left(0\right)}\subseteq {\lambda }^{\left(1\right)}\subseteq {\lambda }^{\left(2\right)}\subseteq \cdots \subseteq {\lambda }^{\left(k\right)}=\lambda )$$ such that ${\lambda }^{\left(j\right)}/{\lambda }^{(j+1)}$ is a border strip of length ${\mu }_j\text{.}$ The weight of $T$ is given by $$\text{wt}\left(T\right)=\prod _{j=1}^k\text{wt}({\lambda }^{\left(j\right)}/{\lambda }^{(j+1)})\text{.}$$

With these definitions, the Murnaghan-Nakayama rule can be written in the following form

Theorem 2.51 [Murnaghan-Nakayama Rule II] $${\chi }^{\lambda }\left(\mu \right)=\sum _T\text{wt}\left(T\right),$$ where the sum is over all μ-border strip tableaux $T\text{.}$

Now let $\mu =\left(1^m\right)\text{.}$ We can compute all of the $\mu \text{-border}$ strip tableaux as follows. $$\begin{array}{ccccccc}\multicolumn{6}{c}{}& \varnothing \\ \multicolumn{6}{c}{}& \downarrow \\ \multicolumn{6}{c}{}& \\ \multicolumn{5}{c}{}& \swarrow & & \searrow \\ \multicolumn{4}{c}{}& & \multicolumn{3}{c}{}& \\ \multicolumn{3}{c}{}& \swarrow & & \searrow & & \swarrow & & \searrow \\ \multicolumn{2}{c}{}& & \multicolumn{3}{c}{}& & \multicolumn{3}{c}{}& \\ \multicolumn{1}{c}{}& \swarrow & & \searrow & & \swarrow & \downarrow & \searrow & & \swarrow & & \searrow \\ & \multicolumn{3}{c}{}& & & & & & \multicolumn{3}{c}{}& \\ ⋮& \multicolumn{3}{c}{}& ⋮& & ⋮& & ⋮& \multicolumn{3}{c}{}& ⋮\end{array}$$ This graph is called the Young lattice. It is characterized by the following properties:

(1)	On the $m\text{th}$ level, the nodes are labeled by partitions of $m\text{.}$
(2)	A node on the $m\text{th}$ level is connected to a node on the $(m+1)\text{st}$ level if the partitions differ by a single box.

Fix a partition $\lambda \text{.}$ The paths from $\varnothing$ to $\lambda$ that are always increasing in level are the $\left(1^m\right)\text{-border}$ strip tableaux of shape $\lambda \text{.}$ Moreover, from the Murnaghan-Nakayama rule, we see that the number of these paths is equal to ${\chi }^{\lambda }\left(1^m\right)$ which is the dimension of the irreducible $S_m\text{-module}$ labeled by $\lambda \text{.}$

Definition 2.52 A standard tableau of shape $\lambda ⊢m$ is a filling in of the boxes in the Ferrers diagram of $\lambda$ with the integers $\{1,2,\dots ,m\}$ such that numbers increase in both the rows (moving left to right) and the columns (moving top to bottom).

The increasing paths in the Young diagram correspond exactly to standard tableaux. For example, the path $$\begin{array}{ccccccccc}\varnothing & \to & 1 & \to & 1 2 & \to & 1 2 3 & \to & 1 2 3 4 \end{array},$$ is completely determined by the standard tableau $$\begin{array}{c} 1 2 3 4 \end{array},$$ and vice-versa.

Recall that from the semisimplicity of $ℂ\left[S_m\right]$ we have $$\left|S_m\right|=\sum _{\lambda ⊢m}\text{dim}{\left(V^{\lambda }\right)}^2=\sum _{\lambda ⊢m}{\left(f_{\lambda }\right)}^2,$$ where $f_{\lambda }$ denotes the number of standard tableaux of shape $\lambda \text{.}$ For example, in the 4th row of the Young diagram, we have $$4!=1+3^2+2^2+3^2+1\text{.}$$

Notes and References

This is a copy of lectures in Representation Theory given by Arun Ram, compiled by Tom Halverson, Rob Leduc and Mark McKinzie.

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