Lectures in Representation Theory

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

Last update: 20 August 2013

Lecture 11

Definition 2.16 For an integer $r>0$ we define the power symmetric function $p_r\left(x\right)$ to be $$p_r\left(x\right)=p_r(x_1,x_2,\dots ,x_n)=x_1^r+x_2^r+\cdots x_n^r,$$ and we extend the definition to sequences $\mu =({\mu }_1,{\mu }_2,\dots ,{\mu }_k)$ of positive integers by $$p_{\mu }\left(x\right)=p_{{\mu }_2}\left(x\right)p_{{\mu }_2}\left(x\right)\cdots p_{{\mu }_k}\left(x\right)\text{.}$$

As an immediate corollary of 2.14 and 2.15, we have

Corollary 2.17 If $\sigma \in S_m$ has cycle type $\mu ⊢m,$ then $\text{wtr}\left(\sigma \right)=p_{\mu }\left(x\right)\text{.}$

2.2$$Symmetric Functions

Let $W=S_n$ (we use $W$ for Weyl group), and define an action of $W$ on polynomials in $ℂ[x_1,x_2,\dots ,x_n]$ in the following way. Let $w\in W$ act on the monomial $x_{i_1}{x}_{i_2}\cdots x_{i_n}\in ℂ[x_1,x_2,\dots ,x_n]$ by $$w{x}_{i_1}{x}_{i_2}\cdots x_{i_n}=x_{i_{w\left(1\right)}}{x}_{i_{w\left(2\right)}}\cdots x_{i_{w\left(n\right)}},$$ and extend the action linearly to all of $ℂ[x_1,x_2,\dots ,x_n]\text{.}$ Notice that the action has the property that $$w{x}_{i_1}^{{\lambda }_1}{x}_{i_2}^{{\lambda }_2}\cdots x_{i_n}^{{\lambda }_n}=x_{i_{w\left(1\right)}}^{{\lambda }_1}{x}_{i_{w\left(2\right)}}^{{\lambda }_2}\cdots x_{i_{w\left(n\right)}}^{{\lambda }_n}=x_{i_1}^{{\lambda }_{w^{-1}\left(1\right)}}{x}_{i_2}^{{\lambda }_{w^{-1}\left(2\right)}}\cdots x_{i_n}^{{\lambda }_{w^{-1}\left(n\right)}}\text{.}$$ Moreover, we have for all $w\in W$ $$w{p}_r\left(x\right)=p_r\left(x\right)$$ and $$w{p}_{\mu }\left(x\right)=p_{\mu }\left(x\right)\text{.}$$

Definition 2.18 A polynomial $f\left(x\right)\in ℂ[x_1,x_2,\dots ,x_n]$ is a symmetric polynomial if it satisfies $$wf\left(x\right)=f\left(x\right)\text{for all}\hspace{0.17em}w\in W,$$ and a polynomial $g\left(x\right)\in ℂ[x_1,x_2,\dots ,x_n]$ is an alternating polynomial or a skew-symmetric polynomial if it satisfies $$wg\left(x\right)=\epsilon \left(w\right)g\left(x\right)\text{for all}w\in W\text{.}$$

Let $\lambda =({\lambda }_1,\cdots ,{\lambda }_n)$ be a sequence with each ${\lambda }_i\in \mathbb{N},$ and let $x^{\lambda }$ denote the monomial $x_1^{{\lambda }_1}{x}_2^{{\lambda }_2}\cdots x_n^{{\lambda }_n}\text{.}$ We construct a symmetric function by “symmetrizing” $x^{\lambda }\text{.}$ Let $\text{Re}\left(\lambda \right)$ denote the set of all sequences in ${\mathbb{N}}^n$ that are rearrangements of the sequence $\lambda \text{.}$ Note that $\text{Re}\left(\lambda \right)$ is the $W\text{-orbit}$ $W_{\lambda }$ of $\lambda$ in ${\mathbb{N}}^n\text{.}$ Then define the symmetric function $m_{\lambda }\left(x\right)$ by $$m_{\lambda }\left(x\right)=\sum _{\mu \in \text{Re}\left(\lambda \right)}{x}^{\mu }=\sum _{\mu \in W_{\lambda }}{x}^{\mu },$$ Notice that if $\nu \in \text{Re}\left(\lambda \right),$ then $$m_{\nu }\left(x\right)=m_{\lambda }\left(x\right)\text{.}$$ Moreover, there is a unique $\nu \in \text{Re}\left(\lambda \right)$ such that $${\nu }_1\ge {\nu }_2\ge \cdots \ge {\nu }_n\ge 0\text{.}$$ The polynomials $$\left\{m_{\nu }\left(x\right)\hspace{0.17em}\right|\hspace{0.17em}\nu =({\nu }_1,{\nu }_2,\dots ,{\nu }_n),\hspace{0.17em}{\nu }_1\ge {\nu }_2\ge \cdots \ge {\nu }_n\ge 0\}$$ are called the monomial symmetric polynomials.

The symmetric polynomials in $ℂ[x_1,\dots ,x_n]$ form a $ℂ\text{-vector}$ space which we denote by ${\Lambda }_n\text{.}$ If $f\left(x\right)$ is a symmetric polynomial, let $c_{\nu }$ denote the coefficient of $x_{\nu }$ in $f\left(x\right)\text{.}$ Then, since $f\left(x\right)$ is symmetric, if $\lambda \in \text{Re}\left(\nu \right),$ $c_{\nu }$ is also the coefficient of $x^{\lambda }$ in $f\left(x\right)\text{.}$ Therefore, $$f\left(x\right)=\sum _{{\nu }_1\ge {\nu }_2\ge \cdots \ge {\nu }_n}{c}_{\nu }{m}_{\nu }\left(x\right),$$ and the monomial symmetric polynomials form a basis of ${\Lambda }_n\text{.}$

Analogously, the alternating polynomials in $ℂ[x_1,\dots ,x_n]$ form a $ℂ\text{-vector}$ space which we denote by $A_n\text{.}$ To find a basis for $A_n,$ we anti-symmetrize the monomial $x^{\lambda }\text{.}$ That is, we define $$a_{\lambda }\left(x\right)=\sum _{w\in W}\epsilon \left(w\right)w{x}^{\lambda },$$ where, as before, $\epsilon \left(w\right)$ is the sign of $w\text{.}$ If $v\in W,$ then $$\begin{array}{ccc}v{a}_{\lambda }\left(x\right)& =& \sum _{w\in W}\epsilon \left(w\right)vw{x}^{\lambda }\\ & =& \epsilon \left(v\right)\sum _{vw\in W}\epsilon \left(vw\right)vw{x}^{\lambda }\\ & =& \epsilon \left(v\right)a_{\lambda }\left(x\right),\end{array}$$ and, therefore, $a_{\lambda }$ is alternating.

Lemma 2.19 Let $\lambda =({\lambda }_1,{\lambda }_2,\dots ,{\lambda }_n)$ with ${\lambda }_i\in \mathbb{N},$ and suppose that ${\lambda }_i={\lambda }_j$ for some $i\ne j\text{.}$ Then, $a_{\lambda }\left(x\right)=0\text{.}$

		Proof.
	Let $t_{ij}\in W$ be the transposition that switches $i$ and $j\text{.}$ Then we have $$\begin{array}{ccc}{a}_{\lambda }\left(x\right)& =& \sum _{w\in W}\epsilon \left(w\right)w{x}_1^{{\lambda }_1}\cdots x_i^{{\lambda }_i}\cdots x_j^{{\lambda }_j}\cdots x_n^{{\lambda }_n}\\ & =& \sum _{w\in W}\epsilon \left(w\right)w{t}_{ij}{x}^{\lambda }\\ & =& \epsilon \left(t_{ij}\right)\sum _{t_{ij}w\in W}\epsilon \left(t_{ij}w\right)w{t}_{ij}{x}^{\lambda }\\ & =& \epsilon \left(t_{ij}\right)a_{\lambda }\left(x\right)\\ & =& -a_{\lambda }\left(x\right),\end{array}$$ and, therefore, $a_{\lambda }\left(x\right)=0\text{.}$ $\square$

If $\nu \in \text{Re}\left(\lambda \right),$ then $\pm a_{\lambda }\left(x\right)=a_{\nu }\left(x\right),$ so, as in the case of the symmetric polynomials, we see that $$\left\{a_{\nu }\left(x\right)\hspace{0.17em}\right|\hspace{0.17em}\nu =({\nu }_1,{\nu }_2,\dots ,{\nu }_n),\hspace{0.17em}{\nu }_1>{\nu }_2>\cdots >{\nu }_n\ge 0\}$$ forms a basis of $A_n\text{.}$

Our next goal is to show that the alternating symmetric functions and the symmetric functions are exactly the same. In fact, we will describe a bijection between ${\Lambda }_n$ and $A_n\text{.}$ To do this we let $$\delta =(n-1,n-2,\dots ,2,1,0)\text{.}$$ Then if $\lambda \in {\mathbb{N}}^n$ with ${\lambda }_1>{\lambda }_2>\cdots >{\lambda }_n\ge 0,$ and $$\mu =\lambda -\delta ({\lambda }_1-(n-1),{\lambda }_2-(n-2),\dots {\lambda }_{n-2}-2,{\lambda }_{n-1}-1,{\lambda }_n),$$ we have ${\mu }_1\ge {\mu }_2\ge \cdots {\mu }_n\ge 0\text{.}$

For example, suppose that $n=8,$ $\delta =(7,6,5,4,3,2,1,0),$ and $\lambda =(13,10,9,8,3,2,1,0)\text{.}$ Then $\mu =(6,4,4,4,0,0,0,0)\text{.}$ We can picture this as follows. $$\begin{array}{cccccccccccccc}\square & \square & \square & \square & \square & \square & \square & |& \square & \square & \square & \square & \square & \square \\ & \square & \square & \square & \square & \square & \square & |& \square & \square & \square & \square \\ & & \square & \square & \square & \square & \square & |& \square & \square & \square & \square \\ & & & \square & \square & \square & \square & |& \square & \square & \square & \square \\ & & & & \square & \square & \square & |\\ & & & & & \square & \square & |\\ & & & & & & \square & |\\ & & & & & & & |\end{array}$$ The sequence $\delta$ is pictured to the left of the wall, the sequence $\mu$ is pictured to the right of the wall, and the sequence $\lambda$ is the entire picture. In this way we get a bijection between the index sets of the symmetric polynomials and the alternating polynomials.

Now we define a map between ${\Lambda }_n$ and $A_n\text{.}$ To this end, we note that $a_{\lambda }\left(x\right)={\displaystyle \sum _{w\in S_n}}\epsilon \left(w\right)w{x}_1^{{\lambda }_1}\cdots x_n^{{\lambda }_n}=\text{det}\left(x_i^{{\lambda }_j}\right),$ where by $\left(x_i^{{\lambda }_j}\right)$ we mean the $n$ by $n$ matrix whose $i,j\text{-entry}$ is given by $x_i^{{\lambda }_j}\text{.}$ This is called the Vandermonde determinant, and it satisfies the following

Theorem 2.20 [Weyl’s Denominator Formula] $$a_{\lambda }\left(x\right)=\text{det}\left(x_i^{{\lambda }_j}\right)=\prod _{1\le i<j\le n}(x_i-x_j)\text{.}$$

		Proof.
	We proceed in several steps. Step 1. ${\displaystyle \prod _{i<j}}(x_i-x_j)$ divides $a_{\lambda }\left(x\right)$ for all ${\lambda }_1>{\lambda }_2>\cdots >{\lambda }_n\text{.}$ Proof. We use the evaluation map to send both $x_i$ and $x_j$ to $\alpha \in ℂ\text{.}$ Then the $i\text{th}$ and $j\text{th}$ row of the matrix $\left(x_i^{{\lambda }_j}\right)$ are identical, and so $a_{\lambda }\left(x\right)=\text{det}\left(x_i^{{\lambda }_j}\right)=0\text{.}$ This holds for all $\alpha \in ℂ,$ so $a_{\lambda }\left(x\right)$ is divisible by $(x_i-x_j)\text{.}$ This argument holds for any pair $i<j,$ so the product ${\prod }_{i<j}(x_i-x_j)$ divides $a_{\lambda }\left(x\right)\text{.}$ $\square$ Step 2. The polynomial ${\displaystyle \prod _{i<j}(x_i-x_j)}$ is alternating. Proof. Write the product ${\displaystyle \prod _{i<j}(x_i-x_j)}$ as follows $$\begin{array}{c}(x_1-x_2)(x_1-x_3)\cdots (x_1-x_{i-1})(x_1-x_i)(x_1-x_i)\cdots \cdots (x_1-x_{n-1})(x_1-x_n)·\\ (x_2-x_3)(x_2-x_4)\cdots (x_2-x_i)(x_2-x_{i+1})(x_2-x_{i+2})\cdots \cdots (x_2-x_n)·\\ ⋮\\ (x_i-x_{i+1})(x_i-x_{i+2})(x_i-x_{i+3})\cdots (x_i-x_{n-1})(x_i-x_n)·\\ (x_{i+1}-x_{i+2})(x_{i+1}-x_{i+3})(x_{i+1}-x_{i+4})\cdots (x_{i+1}-x_n)·\\ ⋮\\ (x_{n-1}-x_n)\text{.}\end{array}$$ (This is a trick of Littlewood.) Then consider the action of the simple transposition $s_i$ on this product. We see that $s_i$ preserves all rows except the $i\text{th}$ and the $(i+1)\text{st.}$ Moreover, the second element of the $i\text{th}$ row is changed with the 1st element of the $(i+1)\text{st}$ row, the third element of the $i\text{th}$ row is changed with the 2nd element of the $(i+1)\text{st}$ row, and so on. The first element of the $i\text{th}$ row stays the same with a sign change. Therefore, $$s_i·\prod _{i<j}(x_i-x_j)=-\prod _{i<j}(x_i-x_j)=\epsilon \left(s_i\right)\prod _{i<j}(x_i-x_j)\text{.}$$ Since the result holds for each simple transposition $s_i,$ it holds for the entire symmetric group $S_n\text{.}$ $\square$ Continued next lecture. $\square$

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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