10 Representation of Symmetric Groups

10.1 Preliminary for Symmetric Groups

Parity and Presentation of Symmetric Groups

parity

Each permutation in $S_n$ can be presented as a diagram permutation said to be odd (rep. even) if the number of crossing in this diagram is odd (rep. even). Define $\mathrm{sgn}:S_n\to \{\pm 1\}$ as $\mathrm{sgn}(\sigma )=1$ if $\sigma$ is even and $\mathrm{sgn}(\sigma )=-1$ if $\sigma$ is odd.

Equivalently, the parity of permutation $\sigma$ is the number of inversions $(i,j)$ in $\sigma$: $1⩽i<j⩽n$ such that $\sigma (i)<\sigma (j)$.

The adjacent transposition $s_i:=(ii+1)$ for all $1⩽i⩽n-1$, then $S_n$ is generated by $\{s_i:1⩽i⩽n-1\}$.

Proposition

The symmetric group has the following presentation

$$S_n=⟨s_1,\cdots ,s_{n-1}:s_i^2=1,s_i{s}_j=s_j{s}_i\text{mbox}with|i-j|>2,s_i{s}_{i+1}{s}_i=s_{i+1}{s}_i{s}_{i+1}⟩.$$

\begin{proof} Define the LHS equals ${\Gamma }_n$. We have a surjective homomorphism ${\Gamma }_n\to S_n,s_i\mapsto (ii+1)$. It suffices to show $|{\Gamma }_n|⩽n!$. Note that the case of $n=1$ holds. Suppose that $|{\Gamma }_n|⩽n!$, and consider the subgroup $H<{\Gamma }_{n+1}$ generated by $s_1,\cdots ,s_{n-1}$where $|H|⩽n!$ by induction hypothesis. Note that

$${\Gamma }_{n+1}=H\cup s_{n}H\cup \cdots \cup s_1\cdots s_{n}H$$

and this composition is stable by the action of $n$. Therefore, $|{\Gamma }_{n+1}|⩽(n+1)!$ and so ${\Gamma }_n\simeq S_n$. \end{proof}

Definition

Any element of $S_n$ can be written as product of $s_i$‘s, and for $\sigma =s_{i_1}\cdots s_{i_{\ell }}$, then minimum value of $\ell$ is called the length of $\sigma$.

Cycle Types and Partitions

Definition

Any permutation $\sigma$ can be written as product of disjoint cycles. The cycle type is the set of length of cycles. For example, the cycle type of $(1234)(567)$ in $S_8$ is $\{1,3,4\}$.

As two permutations with the same cycle type belong to the same conjugacy class, the conjugacy classes of $S_n$ can be described as $\lambda =({\lambda }_1,\cdots ,{\lambda }_{\ell })$ with ${\lambda }_i>0,{\lambda }_i\in \mathbb{Z}$ and ${\lambda }_1+\cdots +{\lambda }_{\ell }=n$.

There are some properties of partition functions. Write partition in the multiplicity from $\lambda =\{1^{m_1},\cdots ,n^{m_n}\}$ with $n=m_1+2m_2+\cdots +n{m}_n$, then we have the following property.

Proposition

The cardinality of conjugacy class $C_{\lambda }$ of $S_n$ corresponding to partition $\lambda =\{1^{m_1},\cdots ,n^{m_n}\}$ is

$$|C_{\lambda }|=\frac{n!}{1^{m_1}{m}_1!\cdots n^{m_n}{m}_n!}.$$

\begin{proof} Note that

• $(123),(231),(312)$ are same cycles and there are $1^{m_1}\cdots n^{m_n}$ repetitions.
• $(123)(456)$ and $(456)(123)$ are same cycles and there are $m_1!\cdots m_n!$ repetitions.

Now we finish the proof. \end{proof}

For any given $n\in {\mathbb{N}}_{+}$, define $p(n)$ as the number of partitions of $n$. Then we can get its generating function.

Euler formula

The generating function for partition is

$$\sum _{n⩾0}p(n)t^n=\frac{1}{1-t}\frac{1}{1-t^2}\cdots \frac{1}{1-t^n}\cdots$$

where $p(n)$ is the number of partitions of $n$.

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10.2 Tableaux, Tabloids and Specht Module

Definition

Let $\lambda ⊢n$ be a partition of $n$ with $\lambda =({\lambda }_1,\cdots ,{\lambda }_{\ell })$, then define corresponding Young subgroup as

$$S_{\lambda }=S_{\{1,\cdots ,{\lambda }_1\}}\times \cdots \times S_{\{n-{\lambda }_{\ell },\cdots ,n\}}.$$

For a trivial representation of $S_{\lambda }$, we have that

$$\dim \mathrm{tr}{\uparrow }_{S_{\lambda }}^{S_n}=\frac{|S_n|}{|S_{\lambda }|}=\frac{n!}{{\lambda }_1!\cdots {\lambda }_{\ell }!}.$$

tableau

Let $\lambda ⊢n$. A Young tableau of shape $\lambda$ is an array obtained by putting $1,\cdots ,n$ into $\lambda$-tableau.

For example, see here. Note that $\tau \in S_n$ can naturally act on $\lambda$-tableaus.

Definition

Two tableaux are equivalent if they are of the same $\lambda$-shape and they have the same element in each row. Here is an example.

tabloid

Given a tableau $t$, define the tabloid or $\lambda$-tabloid as the set $\{t^{\prime }:t^{\prime }\sim t\}$.

the Corresponding Module $M^{\lambda }$

Definition

For every partition $\lambda ⊢n$ the corresponding module $M^{\lambda }$ is the linear space of all tabloids of shape $\lambda$, $M^{\lambda }\simeq \mathbb{C}[\{t_1\},\cdots .,\{t_n\}]$ where $\dim M^{\lambda }=\frac{n!}{\lambda !}$ and $\lambda !={\lambda }_1!\cdots {\lambda }_{\ell }!$.

Examples.

• Let $\lambda =(n)$, then $M^{(n)}=\mathbb{C}[\{t\}]$ and $\dim M^{(n)}=1$.
• Let $\lambda =(1,1,\cdots ,1)$, then $M^{\lambda }=\mathbb{C}[\{t_1\},\cdots ,\{t_{n!}\}]$ and this action is regular, as $G$ acting on $\mathbb{C}G$.
• Let $\lambda =(n-1,1)$, then $M^{\lambda }=\mathbb{C}[\{t_1\},\cdots ,\{t_n\}]$ and this action deduces the permutation representation of $S_n$.

Proposition

Let $\lambda ⊢n$ be a partition, and let $S_{\lambda }<S_n$ be the subgroup corresponding to $\lambda$. Then $\mathrm{tr}{\uparrow }_{S_{\lambda }}^{S_n}\simeq M^{\lambda }$.

\begin{proof} Define $\phi :{\pi }_i\otimes W\mapsto \{{\pi }_i{t}^{\lambda }\}$ where $W$ is the trivial $S_{\lambda }$-module and $t^{\lambda }$ is the tabloid whose $\ell$ rows are $\{1,\cdots ,{\lambda }_1\}$, $\{{\lambda }_1+1,\cdots ,{\lambda }_2\}$, …, $\{n-{\lambda }_{\ell },\cdots ,n\}$. Then $\phi$ is an isomorphism of $S_n$-modules. \end{proof}

the Row/Column Stabilizer and the Associated Polytabloid

Definition

Let $\lambda ⊢n$ be a partition, and let $t$ be a tableau in $\lambda$-shape. Suppose $t$ has rows $R_1,\cdots ,R_{\ell }$ and columns $C_1,\cdots ,C_m$. Define subgroups of $S_n$ as $R_t=S_{R_1}\times \cdots \times S_{R_{\ell }}$ and $C_t=S_{C_1}\times \cdots \times S_{C_m}$, and they are called the row-stabilizer and the column-stabilizer of $t$.

Remark. With this definition, we have that $\{t\}=R_t\cdot t$.

associated polytabloid

For any subset $H\subseteq S_n$, define the following elements of $\mathbb{C}{S}_n$

$$H^{+}=\sum _{\pi \in H}\pi ,H^{-}=\sum _{\pi \in H}\mathrm{sgn}\pi \cdot \pi =\sum _{\pi \in H}{\pi }^{-},$$

where ${\pi }^{-}=\mathrm{sgn}\pi \cdot \pi$.

For any tableau $t$ with $C_1,\cdots ,C_m$, define

$$K_{C_i}=C_i^{-}=\sum _{\pi \in S_{C_i}}\mathrm{sgn}\pi \cdot \pi ,K_t=K_{c_1}\times \cdots \times K_{C_m}\text{mbox}and{e}_t=K_t\{t\},$$

where $e_t$ is called the associated polytabloid of $\lambda$-tableau $t$.

Remark. For any $\lambda$-tableau $t$, the associated polytabloid $e_t=K_t\{t\}$ is an element of permutation module $M^{\lambda }$.

Example. Here is an example of $S_5$.

Lemma

Let $t$ be a tableau and $\sigma \in S_n$. Then:

• $R_{\sigma t}=\sigma R_t{\sigma }^{-1}$;
• $C_{\sigma t}=\sigma C_t{\sigma }^{-1}$;
• $K_{\sigma t}=\sigma K_t{\sigma }^{-1}$;
• $e_{\sigma t}=\sigma e_t$, where $e$ is the associated polytabloid of $t$.

\begin{proof} Note that $\pi \in R_{\sigma t}$ iff $\pi \{\sigma t\}=\{\sigma t\}$ iff $\pi \sigma \{t\}=\sigma \{t\}$ iff ${\sigma }^{-1}t\sigma \{t\}=\{t\}$ iff ${\sigma }^{-1}t\sigma \in R_t$, and the proof of ii) and iii) are similar. Furthermore, $e_{\sigma t}=K_{\sigma t}\{\sigma t\}=\sigma K_t{\sigma }^{-1}\{\sigma t\}=\sigma e_t$. \end{proof}

Specht Module

Definition

Let $\lambda ⊢n$. The Specht module $S^{\lambda }$ is the submodule of $M^{\lambda }$ spanned by polytabloids $e_t$ where $t$ is of shape $\lambda$.

Remark. Since $e_{\sigma t}=\sigma e_t$, any $S^{\lambda }$ is a cyclic module (a module that is generated by a single element), i.e., it is generated by any tabloid $t$.

Examples.

• Let $\lambda =(n)$. Then $M^{(n)}=\mathbb{C}[\{t\}]$, $C_t=\{e\}$ and $K_t=ee\cdots e$. Note that $S^{\lambda }$ is spanned by $e_t=\{t\}$ and $S^{\lambda }=M^{\lambda }$.
• Let $\lambda =(1,1,\cdots ,1)$. Then $M^{\lambda }=\mathbb{C}[\{t_1\},\cdots ,\{t_{n!}\}]$ and $K_t={\sum }_{\pi \in S_n}\mathrm{sgn}\pi \cdot \pi$. It follows that $e_t=K_t\{t\}$ and $\sigma e_t=\mathrm{sgn}\sigma \cdot e_t$. Thus, $S^{\lambda }$ is isomorphic to the sign representation.
• Let $\lambda =(n-1,1)$. Then $M^{\lambda }=\mathbb{C}[\{t_1\},\cdots ,\{t_n\}]$ and $K_t=e-(1n)$. It follows that $e_t=K_t\{t\}=\{t\}-(1n)\{t\}$ and $\sigma e_t=\overline{\sigma (\mathbf{n})}-\overline{\sigma (1)}$ where $\overline{\mathbf{k}}$ is the tabloid whose second line is $k$. Thus, $S^{\lambda }=\{c_1\overline{1}+\cdots +c_n\overline{\mathbf{n}}:c_1+\cdots +c_n=0\}$ is the standard $S_n$ module.

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10.3 Ordering of Partitions

Definition

Let $\lambda =({\lambda }_1,\cdots ,{\lambda }_{\ell })$ and $\mu =({\mu }_1,\cdots ,{\mu }_k)$ be $\lambda ⊢n$ and $\mu ⊢n$. Then $\lambda <\mu$ in lexicographic order if for some $i$, ${\lambda }_j={\mu }_j$ for all $j<i$ and ${\lambda }_i<{\mu }_i$.

Definition

For any $\lambda ⊢n$ and $\mu ⊢n$, if ${\lambda }_1+\cdots +{\lambda }_i⩽{\mu }_1+\cdots +{\mu }_i$ for any $i⩾1$, then we say $\mu$ dominates $\lambda$, written as $\mu ⊵\lambda$.

Example. Hasse diagram for dominance order for $6$, see here.

dominance lemma

Let $t$ and $s$ be tableaux of shape $\lambda$ and $\mu$ respectively. If for each index $i$, the elements of row $i$ of $s$ are in different columns in $t$, then $\lambda ⊵\mu$.

\begin{proof} We will arrange the tableau $t$ to new form $t^{\prime }$ by permuting entries within each column. Take elements of the first row of $s$ all in different columns of $t$ and move them to the first row in $t$. Continue to do it for the second row for each $i$. The number of elements in the first $i$ rows of $s$ is ${\mu }_1+\cdots +{\mu }_i$ and all elements of first $i$ rows of $\mu$ appears in first $i$ rows of $t^{\prime }$. Thus for any $i⩾1$, ${\lambda }_1+\cdots +{\lambda }_i⩾{\mu }_1+\cdots +{\mu }_i$ and so $\lambda ⊵\mu$. As an example, see here. \end{proof}

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10.4 Complete List of Irreducible Modules

Definition

For each $M^{\lambda }=\mathbb{C}[\{t_1\},\cdots ,\{t_k\}]$, define an inner product as $⟨\{t_i\},\{t_i\}⟩={\delta }_{ij}$. Note that this inner produce is $S_n$-invariant, i.e., $⟨\pi \{t_i\},\pi \{t_j\}⟩=⟨\{t_i\},\{t_j\}⟩$.

sign lemma

Let $H$ be a subgroup of $S_n$.

• If $\pi \in H$, then $\pi H^{-}=H^{-}\pi =(\mathrm{sgn}\pi )H^{-}$.
• For any $u,v\in M^{\lambda }$, $⟨H^{-}u,v⟩=⟨u,H^{-}v⟩$.
• If $(bc)\in H$, then $H^{-}=\kappa (1-(bc))$ for some $\kappa \in \mathbb{C}{S}_n$.
• If $b$ and $c$ are in the same row of $t$, then $H^{-}\{t\}=0$.

\begin{proof} Easy. See here. \end{proof}

Corollary

Let $t$ be tableau of shape $\lambda$ and $s$ be tableau of shape $\mu$, where $\lambda ⊢n$ and $\mu ⊢n$. If $K_t\{s\}\ne 0$, then $\lambda ⊳\mu$. Moreover, if $\lambda =\mu$ and $K_t\{s\}\ne 0$, then $K_t\{s\}=\pm e_t$.

\begin{proof} Suppose $b$ and $c$ are in the same row of $s$, then they belong to different columns of $t$. Otherwise $K_t\{s\}=\kappa (1-(bc))\{t\}=0$ by ^fae9c5. Then by ^3e06ba $\lambda ⊳\mu$. If $\lambda =\mu$ and $K_t\{s\}\ne 0$, then $\{s\}=\pi \{t\}$ for some $\pi \in C_t$ (by ^fae9c5, each pair of numbers in the same column does not appear in the same row of $\{s\}$). It follows that $K_t\{s\}=K_t\{\pi t\}=\mathrm{sgn}\pi K_t\{t\}=\pm e_t$ by ^fae9c5. \end{proof}

Corollary

If $u\in M^{\mu }$ and $t$ is tableau of shape $\mu$, then $K_{t}u$ is multiple of $e_t$.

\begin{proof} Since $u$ is a linear combination of tabloid of shape $\mu$, it follows from ^2fe718. \end{proof}

submodule theorem

Let $U$ be submodule of $M^{\lambda }$, then either $U\supseteq S^{\lambda }$ on or $U\subseteq (S^{\lambda }{)}^{\perp }$. In particular, $S^{\lambda }$ is irreducible.

\begin{proof} Take $u\in M^{\lambda }$, then $K_{t}u=c{e}_t$ for some $c\in \mathbb{C}$. Suppose that there exists $t$ such that $c\ne 0$. But then $c{e}_t\in U$. As $S^{\lambda }$ is cyclic, $\pi (c{e}_t)$ generates $S^{\lambda }$ and is inside of $U$, which yields that $S^{\lambda }\subseteq U$. Otherwise, for any tableau $t$ and $u\in M^{\lambda }$, we have $K_{t}u=0$ and so

$$⟨u,e_t⟩=⟨u,K_t\{t\}⟩=⟨K_{t}u,\{t\}⟩=⟨0,\{t\}⟩=0.$$

Therefore, $U\subseteq (S^{\lambda }{)}^{\perp }$. \end{proof}

Proposition

If there is a non-zero element $\theta \in \mathrm{Hom}(S^{\lambda },M^{\mu })$, then $\lambda ⊳\mu$. If $\lambda =\mu$, then $\theta$ is multiplication by a scalar.

\begin{proof} Since $\theta$ is non-zero, there exists polytabloid $e_t$ such that $\theta (e_t)\ne 0$. Moreover, as $M^{\lambda }$ can be decomposed as $M^{\lambda }=S^{\lambda }\oplus (S^{\lambda }{)}^{\perp }$, we extend $\theta$ to a homomorphism

$$\theta :M^{\lambda }\to M^{\mu },(S^{\lambda }{)}^{\perp }\mapsto 0.$$

Since $0\ne \theta (e_t)=\theta (K_t\{t\})=K_t(\theta \{t\})=K_t{\sum }_i{\alpha }_i\{s_i\}$ for some $\mu$-tabloids $\{s_i\}$, there is $\lambda ⊳\mu$ by ^2fe718.

If $\lambda =\mu$, then $\theta (e_t)=K_t{\sum }_i{\alpha }_i\{s_i\}=c{e}_t$. Then for any permutation $\pi$,

$$\theta (e_{\pi t})=\theta (\pi e_t)=\pi (\theta (e_t))=c\pi e_t=c{e}_{\pi t}$$

and so $\theta$ is a multiplication by $c$. \end{proof}

Theorem

The Specht modules $S^{\lambda }$ for $\lambda ⊢\mu$ form a complete list of irreducible modules of $S_n$.

\begin{proof} Since the number of $S^{\lambda }$ equals the number of $\lambda ⊢n$ and so the number of conjugacy classes of $S_n$, it suffices to show $S^{\lambda }\ne S^{\mu }$ for any $\lambda \ne \mu$.

Otherwise, suppose that $S^{\lambda }=S^{\mu }$ with $\lambda \ne \mu$. Since $S^{\mu }\hookrightarrow M^{\mu }$ is non-trivial, we have non-zero $\theta \in \mathrm{Hom}(S^{\lambda },M^{\mu })$ and so $\lambda ⊳\mu$ by ^376fda. Similarly we have $\mu ⊳\lambda$ and so $\mu =\lambda$, contradiction. Now we finish the proof. \end{proof}

Corollary

The irreducible decomposition of the permutation module $M^{\lambda }$ has the form

$$M^{\mu }={\oplus }_{\lambda ⊳\mu }{\kappa }_{\lambda \mu }{S}^{\lambda },$$

where ${\kappa }_{\lambda \mu }$ is called Kostka numbers. Moreover, ${\kappa }_{\mu \mu }=1$.

Here is an example.

\begin{proof} Note that ${\kappa }_{\lambda \mu }=\dim {\mathrm{Hom}}_{S_n}(S^{\lambda },M^{\mu })=⟨{\chi }^{\lambda },{\psi }^{\mu }⟩$ where ${\chi }^{\lambda }$ and ${\psi }^{\mu }$ are characters of $S^{\lambda }$ and $M^{\mu }$, respectively. If ${\kappa }_{\lambda \mu }\ne 0$, then $\lambda ⊳\mu$ by ^376fda. Furthermore, since $\theta \in \mathrm{Hom}(S^{\mu },M^{\mu })$ is multiplication by scalar by ^376fda, we have $\dim {\mathrm{Hom}}_{S_n}(S^{\lambda },M^{\lambda })={\kappa }_{\mu \mu }=1$. \end{proof}

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10.5 Basis of Specht Modules

Definition

A tableau $t$ is called standard if the rows and the columns of $t$ are increasing sequences.

In this section, we aim to prove the following theorem. The proof is written here.

Theorem

The set $\{e_t:t\text{mbox}isastandard\lambda \text{mbox}-tableau\}$ is a basis of Specht module $S^{\lambda }$.

We call a sequence $\lambda =({\lambda }_1,\cdots ,{\lambda }_k)$ with ${\lambda }_1+\cdots +{\lambda }_k=n$ is a composition of $n$, and ${\lambda }_i$ is called part of $\lambda$. (Compare with partition: composition with ${\lambda }_1⩾\cdots ⩾\cdots {\lambda }_k$.) We say that composition $\lambda$ dominates $\mu$, if for any $k⩾1$ there is ${\lambda }_1+\cdots +{\lambda }_k⩾{\mu }_1+\cdots +{\mu }_k$.

Suppose $\{t\}$ is a tabloid of shape $\lambda ⊢n$. For each $i=1,\cdots ,n$, define $\{t^i\}$ as the tabloid formed by all elements $⩽i$ in $\{t\}$, and define ${\lambda }^i$ as the composition which is the shape of $t^i$. Here is an example:

There is a way to order tabloids in the same shape.

Definition

Let $\{t\}$ and $\{s\}$ be two tabloids in $\lambda$-shape with the corresponding composition ${\lambda }^i$ and ${\mu }^i$. We say that $\{s\}$ dominates $\{t\}$, denoted by $\{s\}⊳\{t\}$ if ${\mu }^i⊳{\lambda }^i$ for all $i$.

Remark. Not all tabloids can be compared. For example, take

\begin{matrix} 1 & 3 & 4\\ 2 & 5 \end{matrix}\right\}\mbox{ and } \{s\}=\left\{ \begin{matrix} 1 & 2 & 5 \\ 3 & 4 \end{matrix}\right\}.$$ Note that $\lambda^2=(1,1)\lhd \mu^2=(2,0)$ and $\lambda^4=(3,1)\rhd\mu^4=(2,2)$. > [!lemma] > > If $k<\ell$ and $k$ occurs in a lower row than $\ell$ in $\{t\}$, then $\{t\}\lhd(k,\ell)\{t\}$. ^rswrnz `\begin{proof}` Let $\lambda^i$ and $\mu^i$ be composition sequences for $\{t\}$ and $(k,\ell)\{t\}$, respectively. If $k\leqslant i\leqslant \ell$, and suppose $\ell$, $k$ is in row $q$, $r$ of $\{t\}$ respectively, then

(\lambda^i)_q=(\mu^i)_q-1,(\lambda^i)_r=(\mu^i)_r+1

and so $\lambda^i\lhd \mu^i$ for all $i$. [[Pasted image 20241128202606.png|Here]] is a illustration. `\end{proof}` > [!corollary] > > If $v=\sum_i c_i\{t_i\}\in M^\lambda$, we say that $\{t_i\}$ appears in $v$ if $c_i\neq 0$. If $t$ is a [[10.5 Basis of Specht Modules#^8pq74m|standard tableau]] and $\{s\}$ appears in $e_t$, then $\{t\}\rhd \{s\}$. `\begin{proof}` Write $s=\pi t$ with $\pi\in C_t$. We do induction on number of "inversions" in $s$, that is, the number of pairs $(k,\ell)$ such that $k\leqslant \ell$ and $k,\ell$ are in the same column but $k$ is in lower row than $\ell$. Then $\{s\}\lhd (k,\ell) \{s\}$ for each inversion and thus $\{s\}\lhd \{t\}$. Here is an example. ![[Pasted image 20241125150516.png|500]] For the last term $\{s\}$, $\{1,2\}$ and $\{3,4\}$ are two inversions. By [[#^rswrnz|^rswrnz]], $\{s\}\lhd (12)\{s\}\lhd (12)(34)\{s\}=\{t\}$. `\end{proof}` Now we are ready to prove [[#^030a4e|^030a4e]]. ^goswcn `\begin{proof}` First we prove that they are linear independent. Suppose $c_1e_{t_1}+\cdots+c_ne_{t_n}=0$, and we label $\{t_1\}$ in such way that there is no $i>1$ with $\{t_1\}\rhd \{t_i\}$. Then by corollary $\{t_1\}$ only appears in $e_{t_1}$ and so $c_1=0$. By induction all $c_i=0$, contradiction. Next we prove that standard polytabloids of shape $\lambda$ span $S^\lambda$. For a polytabloid $e_t$ with tableau $t$, we may assume that column of $t$ are increasing by replacing $e_t$ with $e_{\sigma t}=\sigma e_t=\mathrm{sgn}(\sigma)e_t$ for some suitable $\sigma\in C_t$. Then if $t$ is not standard, we will find two adjacent elements in one row with $t_{i,j}>t_{i,j+1}$. Now we aim to find a linear combination of polytabloids where this row descent is eliminated. This algorithm uses [Garnir relations](https://en.wikipedia.org/wiki/Garnir_relations), as the following example shows. > [!note] Example > > Consider the following Young tableau $t$: > > ![[Pasted image 20241128213455.png|75]] > > There is a row descent in the second row, so we choose the subsets $A$ and $B$ as indicated. > > Consider all partitions of $A\sqcup B=A_i\sqcup B_i$ with $|A_i|=|A|$ and $|B_i|=|B|$, that is > > $$ > \{5,6\}\sqcup\{2,4\},\{4,6\}\sqcup\{2,5\},\{2,6\}\sqcup\{4,5\},\{4,5\}\sqcup\{2,6\},\{2,5\}\sqcup\{4,6\},\{2,4\}\sqcup\{5,6\}, > $$ > > and they corresponds the following polytabloid $t-t_2+t_3+t_4-t_5+t_6$ > > ![[Pasted image 20241128213634.png|500]] > > and the Garnir element $g_{A, B}=1-(45)+(245)+(465)-(2465)+(25)(46)$. One may [[#^791bdb|check]] $g_{A,B}e_t=0$ and so > > $$ > e_t=e_{t_2}-e_{t_3}-e_{t_4}+e_{t_5}-e_{t_6}. > $$ > > Therefore, the row descent in the second row is removed. One can repeatedly apply this procedure to straighten a polytabloid, eventually writing it as a linear combination of standard polytabloids. With this example, it suffices to show $g_{A,B}e_t=0$. > [!proposition] > > For a tableau $t$ with a row descent and the corresponding sets $A$ and $B$, we have $g_{A,B}e_t=0$. > ^791bdb `\begin{proof}` We claim that $S^-_{A\sqcup B}e_t=\{0\}$. Since $|A|+|B|$ is greater than the size of column containing $A$, for any $\sigma\in C_t$ there exist $a,b\in A\sqcup B$ which are in the same row of $\sigma t$. Then by [[10.4 Complete List of Irreducible Modules#^fae9c5|^fae9c5]], $(ab)\in S_{A\sqcup B}$ yields that $S^-_{A\sqcup B}\{\sigma t\}=\{0\}$. Since $e_t=\sum_{\sigma\in C_t}\mathrm{sgn}\sigma\{\sigma t\}$, there is $S^-_{A\sqcup B}e_t=\{0\}$. Then consider the coset decomposition of $S_{A\sqcup B}$. Note that

S_{A\sqcup B}=\pi_1(S_A\times S_B)\sqcup\cdots\sqcup \pi_\ell(S_A\times S_B)

where $g_{A,B}=\sum_{k=1}^\ell\mathrm{sgn}\pi_i\cdot \pi_i$. Hence, $S_{A\sqcup B}^-=g_{A,B}(S_A\times S_B)^-$. For any element $\tau\in S_A\times S_B$, $\tau e_t=\mathrm{sgn}\tau\cdot e_t$ and so $(S_A\times S_B)^- e_t=|S_A\times S_B|e_t$. Therefore,

{0}=S^-{A\sqcup B}e_t=g{A,B}(S_A\times S_B)^- e_t=g_{A,B}|S_A\times S_B|e_t

yields that $g_{A,B}e_t=0$. `\end{proof}` Now we finish the proof of [[#^030a4e|^030a4e]]. `\end{proof}` Now we get a basis of Specht module $S^\lambda$. The corresponding representation of it, is Young natural representation. Recall that $S_n$ is generated by transpositions $(k\;k+1)$, then we denote $s_k=(k\; k+1)$. For $e_t$ be a polytabloid corresponding to standard tableau, there is - if $k$ and $k+1$ are in the same column, then $s_ke_t=-e_t$; - if $k$ and $k+1$ are in the same row, then $(k\;k+1)e_t$ will have row descent and then apply Garnir element; - if $k$ and $k+1$ are in different columns and different rows, then $(k\; k+1)e_t$ is another polytabloid corresponding to standard tableau. Therefore, each transposition $(k\;k+1)$ induces a linear transformation on the subspace spanned by s polytabloids corresponding to standard tableaus. We refer to this representation as *Young natural representation*. **Remark.** Every irreducible representation of $S_n$ has integer values, in particular defined over $\mathbb{R}$. Hence, all representations are of [[7 Character Table, Frobenius-Schur Indicator & Complexification#^df16b3|symmetric type]].Link to original

10.6 Frobenius Character Formula

Frobenius character formula

Let $\lambda ⊢n$ with $\lambda =({\lambda }_1,\cdots ,{\lambda }_{\ell })$, and let ${\lambda }_{\ell +1}=\cdots ={\lambda }_n=0$. The value of ${\chi }^{\lambda }$ of $S^{\lambda }$ on conjugacy class $C_{\mu }$ given by $\mu =({\mu }_1,\cdots ,{\mu }_k)$ with ${\mu }_1⩾\cdots ⩾{\mu }_k⩾1$ equals to the coefficient of the monomial $x_1^{{\ell }_1}\cdots x_n^{{\ell }_n}$ in polynomial

$$\prod _{i=1}^k(x_1^{{\mu }_i}+\cdots +x_n^{{\mu }_i})\prod _{i<j}(x_i-x_j),$$

where ${\ell }_i={\lambda }_i+n-i$.

Corollary

The dimension of Specht module $S^{\lambda }$ is

$$\dim S^{\lambda }=f^{\lambda }=\frac{n!}{{\ell }_1!\cdots {\ell }_n!}\prod _{1⩽i<j⩽n}({\ell }_i-{\ell }_j).$$

\begin{proof} Note that $\dim S^{\lambda }={\chi }^{\lambda }(1)$, then by ^4b468a

$$\dim S^{\lambda }=\text{mbox}coefficientof{x}_1^{{\ell }_1}\cdots x_n^{{\ell }_n}\text{mbox}in(x_1+\cdots +x_n{)}^n\prod _{1⩽i<j⩽n}(x_i-x_j).$$

By Vandermonde determinant,

\vdots & \vdots & & \vdots \\ x_1 & x_2 & \cdots & x_n \\ 1 & 1 & \cdots & 1\end{matrix}\right|=\prod_{1\leqslant i< j\leqslant n}(x_i-x_j)$$ and so the coefficient of $x_1^{\ell_1}\cdots x_n^{\ell_n}$ in

(x_1+\cdots+x_n)^n\prod_{1\leqslant i<j\leqslant n}(x_i-x_j)=\left(\sum_{k_1+\cdots+k_n=n}\frac{n!}{k_1!\cdots k_n!}x_1^{k_1}\cdots x_n^{k_n}\right)\left(\sum_{\sigma\in S^n}\mathrm{sgn}\sigma \cdot x_1^{n-\sigma(1)}x_2^{n-\sigma(2)-1}\cdots x_n^{n-\sigma(n)}\right)

is $$\sum_{\sigma\in S_n}\mathrm{sgn}\sigma\frac{n!}{(\ell_1-n+\sigma(1))!\cdots(\ell_n-n+\sigma(n))!}=n!\det\left|\begin{matrix} \dfrac{1}{(\ell_1-n+1)!} & \cdots & \dfrac{1}{(\ell_n-n+1)!} \\ \vdots & & \vdots \\ \dfrac{1}{\ell_1!} & \cdots & \dfrac{1}{\ell_n!} \end{matrix}\right|=\frac{n!}{\ell_1!\cdots\ell_n!}\prod_{1\leqslant i<j\leqslant n}(\ell_i-\ell_j).$$ Now we finish the proof. `\end{proof}` # Hook Length Formula For a Young tableau $\lambda$ and the element on $i$ th row, $j$ th column, define $h(i,j)=\lambda_i+\lambda_j'-i-j+1$, where $\lambda_j'$ is the number of $j$ th column. The hook length formula tells us that

\dim S^\lambda=f^\lambda=\frac{n!}{\prod_{(i,j)\in\lambda}h(i,j)}.

[[Pasted image 20241202154117.png|Here]] is an example, and [[List 4 (Oral Exam)#^xmv91w|here]] is a proof. # Young-Fibonacci Lattice The number of standard tableaux for $\lambda$ is the number of path from $\emptyset$ to $\lambda$ diagram in Young diagram. Here is an example. ![[Pasted image 20241204154913.png|300]].Link to original

10.7 Branching Rule

Definition

A box of $\lambda$ is called removable, if its removal leaves a diagram, which is denoted by ${\lambda }^{-}$.

Definition

A box of $\lambda$ is called addable to $\lambda$ if the union of $\lambda$ and this box is a diagram.

Lemma

We have $\dim S^{\lambda }=f^{\lambda }={\sum }_{{\lambda }^{-}}{f}^{{\lambda }^{-}}$.

\begin{proof} Recall that basis of $S^{\lambda }$ is polytabloids of standard tableaux. Notice every standard tableau consist of $n$ in some removable box and a standard tableau for some ${\lambda }^{-}$ and then we finish the proof. \end{proof}

branching rule

If $\lambda ⊢n$ is a partition, then

S^\lambda\downarrow_{S_{n-1}}=\oplus_{\lambda^-}S^{\lambda^-}\mbox{ and }S^\lambda\uparrow^{S_{n+1}}=\oplus_{\lambda^+}S^{\lambda^+}. $$ ^5voxbj

\begin{proof} Suppose that the removable boxes appear in rows ${\tau }_1<\cdots <{\tau }_k$, for each $i$, denote by ${\lambda }^i$ diagram when you remove box in row $r_i$. For standard tableau $t$ with $n$ in $i$th two denote by $t^i$ tableau obtained by removing box with $n$. Since $S_n$ is a finite group, for any $S_n$-modules $V,W$ with $W\subseteq V$, we have $V\simeq V/W\oplus W$ by Maschke’s theorem.

To prove the theorem, it suffices to construct chain of $S_{n-1}$-modules

$$\{0\}=V^{(0)}\subseteq V^{(1)}\subseteq \cdots \subseteq V^{(k)}=S^{\lambda }\text{\hspace{1em}(*)}$$

such that $V^{(i)}/V^{(i-1)}\simeq S^{{\lambda }^i}$.

Define $V^i$ as submodule in $S^{\lambda }$ spanned by all polytabloids corresponding to standard tableaux with $n$ being in the rows $⩽i$, that is, $n$ being in rows $r_1,\cdots ,r_i$. Then we get $(\ast )$, the chain of modules what we desire. Define

$${\theta }_i:M^{\lambda }\to M^{{\lambda }^i},\{t\}\mapsto \{\begin{array}{ll}\{t^i\}& \text{mbox}ifn\text{mbox}isintherow{r}_i\text{mbox}of\{t\},\\ 0& \text{mbox}otherwise.\end{array}$$

Then ${\theta }_i$ is $S_{n-1}$-homomorphism. Since $n$ occupies a removable box, we have

$${\theta }_i:e_t\mapsto \{\begin{array}{ll}{e}_{t^i}& \text{mbox}ifn\text{mbox}isintherow{r}_i\text{mbox}of\{t\},\\ 0& \text{mbox}otherwise.\end{array}$$

and so ${\theta }_i{V}^{(i)}\simeq S^{{\lambda }^i}$ and $\ker {\theta }_i=V^{(i-1)}$. It deduces that

$$\dim V^i/V^i\cap \ker {\theta }_i=\dim {\theta }_i{V}^i=\dim S^{{\lambda }^i}=f^{{\lambda }^i}$$

and $(\ast )$ can be written as

$$\{0\}=V^{(0)}=V^{(1)}\cap \ker {\theta }_1\subseteq V^{(1)}=V^{(2)}\cap \ker {\theta }_2\subseteq \cdots \subseteq V^{(k)}=S^{\lambda }.$$

Therefore, we have

$$\sum _{i=1}^k\dim V^{(i)}/V^{(i)}\cap \ker {\theta }_i=\sum _{i=1}^k{f}^{{\lambda }^i}=\dim S^{\lambda }.$$

Furthermore, by Frobenius reciprocity, there is

$$m_{\mu }=⟨{\chi }^{\lambda }{\uparrow }^{S_{n+1}},{\chi }^{\mu }⟩=⟨{\chi }^{\lambda },{\chi }^{\mu }{\downarrow }_{S_n}⟩=⟨{\chi }^{\lambda },\sum _{{\mu }^{-}}{\chi }^{{\mu }^{-}}⟩$$

and so $m_{\mu }=0$ if $\lambda \ne {\mu }^{-}$ and $m_{\mu }=1$ if $\lambda ={\mu }^{-}$. Note that $\lambda ={\mu }^{-}$ iff ${\lambda }^{+}=\mu$, thus ${\chi }^{\lambda }{\uparrow }^{S^{n+1}}={\sum }_{{\lambda }^{+}}{\chi }^{{\lambda }^{+}}$. Now we finish the proof. \end{proof}

Corollary

The restriction $S^{\lambda }{\downarrow }_{S_{n-1}}$ is irreducible iff the diagram $\lambda =(m^{\ell })$, that is, $\lambda$ is a rectangle.

Link to original

10.8 Symmetric Polynomials

Some Symmetric Polynomials: p, e and h

Definition

A polynomial in $m$ variables is called symmetric if it is stable under all permutations $\sigma \in S_m$, that is, $\sigma (p(x_1,\cdots ,x_n))=p(x_{\sigma (1)},\cdots ,x_{\sigma (n)})$.

The algebra of symmetric polynomials is denoted by ${\Lambda }_m$.

Examples. They are symmetric polynomials.

• $p_n=x_1^n+\cdots +x_m^n$ for all $n\in {\mathbb{N}}_{+}$, where the corresponding generated function is $P(t)={\sum }_{n=1}^{\infty }{p}_n{t}^{n-1}={\sum }_{i=1}^m\frac{x_i}{1-x_{i}t}$.
• The elementary symmetric polynomials $e_n={\sum }_{i_1<\cdots <i_n}{x}_{i_1}\cdots x_{i_n}$ for $n⩾1$. Furthermore, define $e_0=1$. Note that $e_n=0$ when $n⩾m$, and the corresponding generated function is $E(t)={\sum }_{n=0}^t{e}_n{t}^n={\prod }_{i=1}^m(1+x_{i}t)$.
• The complete symmetric polynomials $h_n={\sum }_{i_1⩽\cdots ⩽i_n}{x}_{i_1}\cdots x_{i_n}$ for $n⩾1$. We also define $h_0=1$, and the corresponding generated function is $H(t)={\sum }_{n=0}^{\infty }{h}_n{t}^n={\prod }_{i=1}^{\infty }\frac{1}{1-x_{i}t}$.

Since $H(t)E(-t)=1$, we have ${\sum }_{n=0}^{\infty }(-1{)}^r{e}_r{h}_{n-r}=0$. Next, notice that $p(t)=d/dt\ln {\prod }_{i=1}^{\infty }\frac{1}{1-x_{i}t}$, then

P(t)=d/dt\ln(H(t))=\frac{H'(t)}{H(t)}. $$ ^skztji Similarly $P(-t)=E'(t)/E(t)$. It deduces the Newton identity. > [!lemma] the Newton identity > > With the definition above, we have $nh_n=\sum_{r=1}^\infty p_rh_{n-r}$ and $ne_n=\sum_{r=1}^n(-1)^{r-1}p_re_{n-r}$. ^sbpikb > [!proposition] > > If $A$ is an $m \times m$ matrix, then the coefficients of the characteristic polynomial $\operatorname{det}(A-t I)$ are given by the elementary symmetric functions of the eigenvalues $\left\{\lambda_1, \lambda_2, \ldots, \lambda_m\right\}$, with alternating signs $\pm 1$ depending on the degree of each term. The power sums of the eigenvalues coincide with $\operatorname{tr}A^n=\lambda_1^n+\cdots+\lambda_m^n$. **Remark.** There is some connection between symmetric polynomials and the rooted solution of polynomial equations. See [here](https://math.stackexchange.com/a/96329/1445401). When the degree of polynomial is less than or equal to $4$, the roots can be obtained by operating the coefficients by combination of adding, subtracting, multiplication, division, and taking roots. # Schur Polynomial > [!definition] > > Suppose $\lambda=(\lambda_1,\cdots,\lambda_m)$ is a partition of $n$ of length at most $m$. The Schur polynomial $\mathscr s_\lambda$ is defined by > > $$\mathscr s_\lambda=\frac{\left|\begin{matrix} x_1^{\lambda_1+m-1} & \cdots & x_m^{\lambda_1+m-1} \\ x_1^{\lambda_2+m-2} & \cdots & x_m^{\lambda_2+m-2} \\ \vdots & & \vdots \\ x_1^{\lambda_m} & \cdots & x_m^{\lambda_m} \end{matrix}\right|}{\left|\begin{matrix} x_1^{m-1} & \cdots & x_m^{m-1} \\ x_1^{m-2} & \cdots & x_m^{m-2} \\ \vdots & & \vdots \\ 1 & \cdots & 1 \end{matrix}\right|}.$$ > > ^i6y60w For example, when $\lambda=(n)$, $\mathscr s_\lambda=h_n$. When $\lambda=1^n$ and $m=n$, $\mathscr s_\lambda=x_1\cdots x_n$. We can prove them by computing directly, or by the following proposition. > [!proposition] Jacobi-Trudi formula > > We have > > $$ > \mathscr s_\lambda=\left|\begin{matrix}h_{\lambda_1} & h_{\lambda_1+1} & \cdots & h_{\lambda_1+m-1} \\ h_{\lambda_2-1} & h_{\lambda_2} & \cdots & h_{\lambda_2+m-2}\\ \vdots & \vdots & & \vdots \\ h_{\lambda_m-m+1} & h_{\lambda_m-m+2} & \cdots & h_{\lambda_m}\end{matrix}\right|. > $$ ^jrsjoc `\begin{proof}` Let $\alpha=(\alpha_1,\cdots,\alpha_m)$ be a composition. Put $$A_\alpha=\begin{bmatrix} x_1^{\alpha_1} & \cdots & x_m^{\alpha_1} \\ \vdots & & \vdots \\ x_1^{\alpha_m} & \cdots & x_m^{\alpha_m} \end{bmatrix}\mbox{ and }H_\alpha =\begin{bmatrix} h_{\alpha_1-m+1} & \cdots & h_{\alpha_1} \\ \vdots & & \vdots \\ h_{\alpha_m-m+1} & \cdots & h_{\alpha_m}\end{bmatrix}.$$ Consider the elementary symmetric polynomial $e_n^{(k)}$ in variables $x_1,\cdots,\hat x_k,\cdots,x_m$ with $1\leqslant k\leqslant m$ and write them to $m\times m$ matrix $$M=\begin{bmatrix} (-1)^{m-1}e_{m-1}^{(1)} & \cdots & (-1)^{m-1}e_{m-1}^{(m)} \\ \vdots & & \vdots \\ (-1)e_1^{(1)} & \cdots & (-1)e_1^{(m)} \\ 1 & \cdots & 1\end{bmatrix}.$$ We claim that $A_\alpha=H_\alpha M$. Consider generating function for $e_n^{(k)}$

E^{(k)}(t)=\sum_{n=0}^{m-1}e_n^{(k)}t^n=\prod_{i\neq k}(1+x_it),

then by $H(t)E(-t)=1$ there is

H(t)E^{(k)}(t)=\frac{1}{1-x_kt}=1+x_kt+\cdots+(x_kt)^\ell+\cdots.

Take the coefficient of $t^{\alpha_i}$ and we have

\sum_{j=1}^m h_{\alpha_i-m+j}(-1)^{m-j}e_{m-j}^{(k)}=x_k^{\alpha_i}\tag{*}.

With $(*)$, we can prove the claim. Thus, $\det A_\alpha=\det H_\alpha\det M$ and so $\det H_\alpha=\det A_\alpha/\det M$. For $\delta=(m-1,m-2,\cdots,1,0)$, the matrix $H_\delta=1$ and so $\det A_\delta=\det M$. It follows that

\det H_\alpha=\frac{\det A_\alpha}{\det A_\delta}=\mathscr s_\lambda,

where $\alpha=\lambda+\delta=(\lambda_1+m-1,\lambda_2+m-2,\cdots,\lambda_m)$. Now we finish the proof. `\end{proof}` We have a more direct formula for $\mathscr s_\lambda$ by considering a generalized tableau of shape $\lambda$. > [!definition] > > We place numbers $\{1,\cdots,n\}$ into tableau $\lambda\vdash n$ allowing repetition. Such tableau is called *semistandard* if rows are weakly increasing (not decreasing) while columns are strictly increasing sequences. > > If $T$ is semistandard with $\{1,\cdots,n\}$, set > > $$ > \mathscr x^T=\prod_{i=1}^n(x_i)^{\operatorname{number of occurrence of }i \in T}. > $$ > [!proposition] > > For $\lambda\vdash n$ with length $m$, the Schur polynomial in $m$ variables is $\mathscr s_\lambda=\sum_T \mathscr x^T$. # Basis of $\Lambda_m$ Now we consider algebra of symmetric polynomials in $m$ variables. First define $\Lambda_m^n=\mathrm{span}\{\mathscr s_\lambda:\lambda\vdash n\}$, the subspace of homogeneous symmetric polynomial of degree $n$, and then define $\Lambda_m=\oplus_{n\geqslant 0}\Lambda_m^n$. For example, when $m=3$, we have $\Lambda_m^0=\mathrm{span}\{1\}$, $\Lambda_m^1=\mathrm{span}\{x_1+x_2+x_3\}$,

\begin{aligned} \Lambda_m^2&=\mathrm{span}{\mathscr s_{(1,1)},\mathscr s_{(2,0)}}=\mathrm{span}{x_1x_2+x_1x_3+x_2x_3,x_1^2+x_2^2+x_3^2+x_1x_2+x_1x_3+x_2x_3}, \ \Lambda_m^3&=\mathrm{span}{s_{(3)},s_{(2,1)},s_{(1,1,1)}}=\mathrm{span}{x_1x_2x_3,x_1^2x_2+x_1^2x_3+x_1x_2^2+2x_1x_2x_3+x_1x_3^2+x_2^2x_3+x_2x_3^2,h_3}. \end{aligned}

> [!proposition] > > $\mathscr s_\lambda$ is a basis for $\Lambda_m$ for all $\lambda\vdash n$ with $n\geqslant 0$. (For a fixed $n$, $\lambda\vdash n$ running through all partitions $\{\mathscr s_\lambda:\lambda\vdash n\}$ is a basis for $\Lambda_m^k$.) Then we introduce a few families of symmetric polynomials for algebra $\Lambda_m$. - For any partition $\lambda=(\lambda_1,\cdots,\lambda_m)$, define $p_\lambda=p_{\lambda_1}\cdots p_{\lambda_m}$, $e_\lambda=e_{\lambda_1}\cdots e_{\lambda_m}$ and $h_\lambda=h_{\lambda_1}\cdots h_{\lambda_m}$. For example, when $m=3$ and $n=2$, $p_{(2)}=p_2=x_1^2+x_2^2+x_3^2$ and $p_{(1,1)}=p_1p_1=(x_1+x_2+x_3)^2$, $e_{(2)}=e_2=x_1x_2+x_2x_3+x_1x_3$ and $e_{(1,1)}=e_1e_1=(x_1+x_2+x_3)^2$, $h_{(2)}=h_2=x_1^2+x_2^2+x_3^2+x_1x_2+x_1x_3+x_2x_3$ and $h_{(1,1)}=h_1h_1=(x_1+x_2+x_3)^2$. - Let $\lambda=(\lambda_1,\cdots,\lambda_m)$. Another family of symmetric polynomials is defined as the sum of all different monomials obtained from $x_1^{\lambda_1}\cdots x_m^{\lambda_m}$ under permutation. For example, when $m=3$ and $n=2$, we have $m_{(2)}=x_1^2+x_2^2+x_3^2$, $m_{(1,1)}=2(x_1x_2+x_2x_3+x_1x_3)$. > [!theorem] > > Suppose that $\lambda$ runs over all partitions of $n$ of length at most $m$. Then each of the following families is basis of the space $\Lambda_m^n$: > > $$ > m_\lambda,\mathscr s_\lambda,e_{\lambda'},h_{\lambda'},p_{\lambda'},e_\lambda,h_\lambda,p_\lambda > $$ > > In particular, $\dim \Lambda_m^n=p(n)$ with $m\geqslant n$. `\begin{proof}` We first prove $\{m_\lambda:\lambda\vdash n\}$ is a basis. Note that they are linearly independent: If $\lambda\vdash n$ and $\mu\vdash n$ with $\lambda\neq \mu$, then $m_\lambda$ and $m_\mu$ have no common monomials. Hence, if $\sum_i \alpha_i m_{\lambda_i}=0$, then $\alpha_i=0$. Suppose $f\in \Lambda_m^n$, and we aim to show $f\in\mathrm{span}\{m_\lambda:\lambda\vdash n\}$. We do induction on the order of the element. Take the greatest element $x_1^{n_1}\cdots x_m^{n_m}$ in [[10.3 Ordering of Partitions#^0ujtr4|lexicographic]] order with $\mu=(n_1,\cdots,n_m)$. Then $f-m_\mu\in\Lambda_m^n$ and $f-m_\mu<f$. By induction, $f$ can be written as linear combination of $\{m_\lambda:\lambda\vdash n\}$. To show $\mathscr s_\lambda$ form a basis, define $A_m$ as the space of skew-symmetric polynomials in $x_1,\cdots,x_m$. We say $g$ is skew-symmetric if $(ij)g=-g$. It is easy to see that any skew-symmetric polynomials is divisible by $\prod_{1\leqslant i<j\leqslant m}(x_i-x_j)$. We have that $\varphi:\Lambda_m\to A_m,f\mapsto f\prod_{1\leqslant i<j\leqslant m}(x_i-x_j)$, where $\varphi$ defines an isomorphism between $\Lambda_m$ and $A_m$. One can [check](https://math.uchicago.edu/~may/REU2020/REUPapers/Graham.pdf?utm_source=chatgpt.com) that basis of $A_m$ can be given by $$b_\lambda=\left|\begin{matrix} x_1^{\lambda_1+m-1} & \cdots & x_m^{\lambda_1+m-1} \\ x_1^{\lambda_2+m-2} & \cdots & x_m^{\lambda_2+m-2} \\ \vdots & & \vdots \\ x_1^{\lambda_m} & \cdots & x_m^{\lambda_m} \end{matrix}\right|$$ where $\lambda$ runs over through partitions of $n$ of length $\leqslant m$. Notice that $\mathscr s_\lambda=b_\lambda/\prod_{1\leqslant i<j\leqslant m}(x_i-x_j)$ and $\varphi(\mathscr s_\lambda)=b_\lambda$. Furthermore, as $\dim\Lambda_m^n=\dim A_m^{n+{m\choose 2}}$, it deduces that $\mathscr s_\lambda$ form a basis. Moreover, recall that [[10.4 Complete List of Irreducible Modules#^i3a31d|^i3a31d]] and similarly we have $\mathscr s_\mu=\sum_{\lambda}\kappa_{\mu\lambda}m_\lambda=m_\mu+\sum_{\lambda\neq \mu}\kappa_{\mu\lambda}m_\lambda$, and it yields that $\left\langle \mathscr s_\mu:\mu\vdash n\right\rangle=\left\langle m_\mu:\mu\vdash n\right\rangle=\Lambda_m^n$. To show $e_{\lambda'}$ form a basis, suppose that $\lambda\vdash n$ with $\lambda=(\lambda_1,\cdots,\lambda_k)$ and $\lambda'=(\rho_1,\cdots,\rho_k)$, and then

e_{\lambda’}=e_{\rho_1}\cdots e_{\rho_k}=(x_1\cdots x_{\rho_1}+\cdots)(x_1\cdots x_{\rho_2}+\cdots)\cdots(x_1\cdots x_{\rho_k}+\cdots)=x_1^{\lambda_1}\cdots x_m^{\lambda_m}+\cdots+\mbox{smaller terms}=m_\lambda+\sum_{\mu\lhd\lambda}m_\mu.

For $h_\lambda$, we can do it by [[#^jrsjoc|^jrsjoc]], as it deduces that $\left\langle \mathscr s_\lambda:\lambda\vdash n\right\rangle\subseteq\left\langle h_\lambda:\lambda\vdash n\right\rangle$. `\end{proof}` > [!definition] > > For each $m$ we have > > $$ > \Lambda_{m+1}\to \Lambda_m,f(x_1,\cdots,x_m,x_{m+1})\mapsto f(x_1,\cdots,x_m,0) > $$ > > Define a [projective limit](https://en.wikipedia.org/wiki/Inverse_limit) algebra $\Lambda=\oplus_{n\geqslant 0}\Lambda^n$ where $f=(f_m:m\geqslant 0)$ is a sequence of $f_m\in\Lambda_m^n$ and $f_{m+1}\mapsto f_m$ for all $m\geqslant 0$. Then $f$ is a formal series in infinitely many variables, and there are several examples: - $p_n=\sum_{i=1}^\infty x_i^n$ - $e_n=\sum_{1\leqslant i_1<\cdots< i_n}x_{i_1}\cdots x_{i_n}$ - $h_n=\sum_{1\leqslant i_1\leqslant\cdots\leqslant i_n}x_{i_1}\cdots x_{i_n}$ For more details of projective limit, see page 489 of [[Algebra chapter 0 - 2009 - Aluffi.pdf]]. The sequence of Schur polynomials $\mathscr s_\lambda$ defines Schur functions, and the sequence of $m_\lambda$ defines the monomial symmetric functions. > [!lemma] > > We have > > $$ > h_n=\sum_{\lambda\vdash n}\frac{p_\lambda}{z_\lambda}\mbox{ and }e_n=\sum_{\lambda\vdash n}\frac{\epsilon_\lambda p_\lambda}{z_\lambda} > $$ > > where $\lambda=(1^{m_1},\cdots,n^{m_n})$, $z_\lambda=1^{m_1}m_1! \cdots n^{m_n}m_n!$ and $\epsilon_\lambda=(-1)^{n-\ell(\lambda)}$. ^dbe3dc `\begin{proof}` Recall that ![[10.8 Symmetric Polynomials#^skztji|^skztji]] It follows

H(t)=\exp(\sum_{k=1}^\infty \frac{p_kt^k}{k})=\prod_{k=1}^\infty\exp(\frac{p_kt^k}{k})=\left( 1+p_1t+\frac{(p_1t)^2}{2!}+\cdots\right)\left( 1+\frac{p_2t}{2}+\frac{(p_1t)^2}{2^2\cdot2!}+\cdots\right)\left( 1+\frac{p_3t}{3}+\frac{(p_3t)^2}{3^2\cdot 2!}+\cdots\right)

and it deduces what we desire. Similarly we can prove $e_n=\sum_{\lambda\vdash n}{\epsilon_\lambda p_\lambda}/{z_\lambda}$. `\end{proof}` > [!proposition] orthogonal properties > > For two families of variables $x=(x_1,\cdots,x_k,\cdots)$ and $y=(y_1,\cdots,y_k,\cdots)$, we have the following properties > > $$ > \begin{aligned} > > \prod_{i,j\geqslant 1}(1-x_iy_j)^{-1}&=\sum_{\lambda}z_\lambda^{-1}p_\lambda(x)p_\lambda(y),\\ > \prod_{i,j\geqslant 1}(1-x_iy_j)^{-1}&=\sum_\lambda h_\lambda(x)m_\lambda(y)=\sum_\lambda m_\lambda(x)h_\lambda(y)=\sum_\lambda \mathscr s_\lambda(x)\mathscr s_\lambda(y). > > \end{aligned} > $$ > `\begin{proof}` By [[#^dbe3dc|^dbe3dc]], we have

\prod_{i\geqslant 1}(1-x_it)^{-1}=\sum_\lambda z_\lambda^{-1}p_\lambda t^{|\lambda|}\tag{*}

where $|\lambda|$ is the length of $\lambda$. Note that $\sum_{i,j}(x_iy_j)^k=p_k(x)p_k(y)$. Use $(*)$ for $x_iy_j$ and set $t=1$, we have

\prod_{i,j\geqslant 1}(1-x_i y_j)^{-1}=\sum_\lambda z_\lambda^{-1} p_\lambda(x_iy_j)=\sum_\lambda z_\lambda^{-1}p_\lambda(x_i)p_\lambda(y_j).

Equip $\Lambda$ with the following form $\left\langle h_\lambda, m_\mu\right\rangle=\delta_{\lambda\mu}$, and it deduces that $\left\langle p_\lambda,p_\mu\right\rangle=z_\lambda \delta_{\lambda\mu}$ and $\left\langle \mathscr s_\lambda,\mathscr s_\mu\right\rangle=\delta_{\lambda\mu}$. With respect to this inner product, $\{\mathscr s_\lambda:\lambda\vdash n\}$ is orthonormal basis of $\Lambda^n$. `\end{proof}`Link to original

10.9 Characteristical Map

For a finite group $G$, its irreducible characters form basis for the space of class functions $F_c(G,\mathbb{C})$. Define $R_n=F_c(S_n,\mathbb{C})$ and $R={\oplus }_{n⩾0}{R}_n$ with $R_0=\mathbb{C}$.

We introduce on the $R$ product. For any $\varphi \in R_n$ and $\psi \in R_m$, where $\varphi$ (rep. $\psi$) is a character of $S_n$ (rep. $S_m$), define $\varphi \cdot \psi \in R_{n+m}$ as

$$\varphi \cdot \psi =(\varphi \otimes \psi ){\uparrow }_{Sn\times S_m}^{S_{n+m}}.$$

Theorem

The product $\cdot$ on $R$ is commutative and associative.

\begin{proof} Define $\rho =\left(\begin{array}{cccccc}1& \cdots & n& n+1& \cdots & m\\ m+1& \cdots & n+m& 1& \cdots & m\end{array}\right)$. Then there is a bijection

$$S_n\times S_m\to S_m\times S_n,\omega \mapsto {\rho }^{-1}\omega \rho .$$

For associativity, use $\left(W{\uparrow }_K^H\right){\uparrow }_H^G=W{\uparrow }_K^G$. \end{proof}

Furthermore, we can equip the algebra $R$ with an inner product

$$⟨f,g⟩=\sum _{n⩾0}{⟨f_n,g_n⟩}_{S_n}.$$

Definition

Define the character map

$$\mathrm{ch}:R\to \Lambda ,f\mapsto \frac{1}{n!}\sum _{\omega \in S_n}f(\omega )p_{\omega }.$$

Remark. For any $f\in R_n$ and any conjugacy class $C_p$ with representative $p$, we can write the character map as

$$\mathrm{ch}:f\to \frac{1}{n!}\sum _{p⊢n}f(C_p)p_p|C_p|=\sum _{p⊢n}{z}_p^{-1}f(C_p)p_p.$$

Theorem

The characteristic map $\mathrm{ch}:R\to \Lambda$ is an isometry and an algebra isomorphism. Moreover, $\mathrm{ch}({\chi }^{\lambda })=s_{\lambda }$ and $\mathrm{ch}({\varphi }^{\lambda })=h_{\lambda }$, where ${\varphi }^{\lambda }$ is the character of permutation module $M^{\lambda }$.

\begin{proof}

A Generalization

Let $G$ be a finite group, and let $A$ be an associative algebra. If $\Phi ,\Psi :G\to A$ are functions, we may define a bilinear form with values in $A$

$$⟨\Phi ,\Psi ⟩=\frac{1}{|G|}\sum _{g\in G}\Phi (g)\Psi (g^{-1}).$$

If $\Phi$ is a class function on $H$ and $\Psi$ is a class function on $G$, the generalized Frobenius reciprocity still holds, that is

$$⟨\Phi {\uparrow }^G,\Psi ⟩=⟨\Phi ,\Psi {\downarrow }_H⟩$$

where $\Phi {\downarrow }_H(h)=\Phi (h)$ and $\Psi {\uparrow }^G(g)=\frac{1}{|H|}{\sum }_{x\in G}\Psi (x^{-1}gx)$ with the convention that $\Psi (g^{\prime })=0$ if $g^{\prime }\notin H$.

For any $f,g\in R_n$, there is

$$\begin{array}{rl}⟨\mathrm{ch}(f),\mathrm{ch}(g)⟩& =⟨\sum _{p⊢n}{z}_p^{-1}f(C_p)p_p,\sum _{\lambda ⊢n}{z}_{\lambda }^{-1}g(C_{\lambda })p_{\lambda }⟩\\ & =\sum _{\lambda ⊢n}{z}_{\lambda }^{-1}{z}_{\lambda }^{-1}f(C_{\lambda })g(C_{\lambda })z_{\lambda }\\ & =\sum _{\lambda ⊢n}|C_{\lambda }|\frac{1}{n!}f(C_{\lambda })g(C_{\lambda })\\ & =⟨f,g⟩\end{array}$$

Also note that $g({\omega }^{-1})=g(\omega )$.

Since $p_{\omega }=p_{{\omega }^{-1}}$ for $\omega \in S_n$ and $\mathrm{ch}(f)=⟨f,p⟩$, we have

$$\begin{array}{rl}\mathrm{ch}(fg)& =⟨fg,p⟩=⟨(f\otimes g){\uparrow }_{S_n\times S_m}^{S_{m\times n}},p⟩\\ & =\frac{1}{n!m!}\sum _{\sigma \in S_n,\tau \in S_m}(f\otimes g)(\sigma \tau )p_{\sigma \tau }\\ & =\frac{1}{n!m!}\sum _{(\sigma ,\tau )\in S_n\times S_m}f(\sigma )g(\tau )p_{\sigma }{p}_{\tau }\\ & =\left(\frac{1}{n!}\sum _{\sigma \in S_n}f(\sigma )p_{\sigma }\right)\left(\frac{1}{m!}\sum _{\tau \in S_m}g(\tau )p_{\tau }\right)\\ & =\mathrm{ch}(f)\mathrm{ch}(g)\end{array}$$

and so $\mathrm{ch}$ is a homomorphism.

Let ${\eta }_n$ be the trivial representation of $S_n$, then

$$\mathrm{ch}({\eta }_n)=⟨{\eta }_n,p⟩=\sum _{\lambda ⊢n}{z}_{\lambda }^{-1}{p}_{\lambda }=h_{\lambda }.$$

Recall that the permutation module $M^{\lambda }=1{\uparrow }_{S_{\lambda }}^{S_n}$ with $S_{\lambda }=S_{{\lambda }_1}\times \cdots \times S_{{\lambda }_k}$. Then ${\varphi }^{\lambda }={\eta }_{{\lambda }_1}\cdots {\eta }_{{\lambda }_k}$ yields that

$$\mathrm{ch}({\varphi }^{\lambda })=\mathrm{ch}({\eta }_{{\lambda }_1})\cdots \mathrm{ch}({\eta }_{{\lambda }_k})=h_{{\lambda }_1}\cdots h_{{\lambda }_k}=h_{\lambda }.$$

To prove that $\mathrm{ch}({\chi }^{\lambda })=s_{\lambda }$, note that

\eta_{\lambda_1} & \cdots & \eta_{\lambda_1+m-1} \\ \vdots & & \vdots \\ \eta_{\lambda_m-m+1} & \cdots & \eta_{\lambda_m} \end{matrix}\right|$$ and $\mathrm{ch}(\widetilde {\chi^\lambda})=\mathscr s_\lambda$. Now we finish the proof, as $\widetilde {\chi^\lambda}$ is induced $\chi^\lambda$. `\end{proof}` > [!corollary] > > The character table $\chi_{\mathscr p}^\lambda$ is the transition matrix between bases $p_\lambda$ and $\mathscr s_\lambda$ of $\Lambda^n$. ^7ed3d1 `\begin{proof}` It suffices to show $p_{\mathscr p}=\sum_{\lambda}\chi_{\mathscr p}^\lambda \mathscr s_\lambda$. By [[#^492b55|^492b55]] we have

\mathscr s_\lambda=\mathrm{ch}(\chi^\lambda)=\sum_{\mathscr p}z_{\mathscr p}^{-1}\chi_{\mathscr p}^\lambda p_{\mathscr p}

and so $\chi_{\mathscr p}^\lambda=\left\langle \mathscr s_\lambda, p_\mathscr p\right\rangle$. It deduces that

\left\langle \mathscr s_\lambda,p_\mu\right\rangle =\sum_{\mathscr p} z_\mathscr p^{-1}\chi_{\mathscr p}^{\lambda}\left\langle p_\mathscr p,p_\mu\right\rangle =z_{\mu}^{-1}\chi_{\mu}^\lambda z_{\mu}=\chi_{\mu}^\lambda.

Now we finish the proof. `\end{proof}` > [!corollary] Frobenius character formula > > If $\rho=(\rho_1,\cdots,\rho_m)$ with $\rho_1\geqslant \rho_2\geqslant\cdots\geqslant \rho_m\geqslant 1$, then for any $k\geqslant\ell(\lambda)$ the value $\chi_\rho^\lambda$ is the coefficient of $x_1^{\ell_1}\cdots x_k^{\ell_k}$ in the monomial > > $$ > \prod_{i=1}^m(x_1^{\rho_i}+\cdots+x_k^{\rho_i})\prod_{1\leqslant i< j\leqslant k}(x_i-x_j) > $$ > > where $\ell_i=\lambda_i+k-i$ for $i=1,\cdots,k$. **Remark.** It has been proved in [[10.6 Frobenius Character Formula#^4b468a|^4b468a]]. `\begin{proof}` Recall that $p_{\rho}=\prod_{i=1}^m(x_1^{\rho_i}+\cdots+x_k^{\rho_i})$ and ![[10.8 Symmetric Polynomials#^i6y60w|^i6y60w]] By [[#^7ed3d1|^7ed3d1]], we have $p_{\mathscr p}=\sum_{\lambda}\chi_{\mathscr p}^\lambda \mathscr s_\lambda$ and it deduces that $$\mathrm{LHS}=\sum_{\mu}\chi_{\rho}^\mu\left|\begin{matrix} x_1^{\mu_1+k-1} & ... & x_k^{\mu_1+k-1} \\ \vdots & & \vdots \\ x_1^{\mu_l} & \cdots & x_k^{\mu_k} \end{matrix}\right|.$$ Therefore, the coefficient of $x_1^{\ell_1}\cdots x_k^{\ell_k}$ with $\ell_i=\lambda_i+k-i$ of right hand side is $\chi_{\rho}^\lambda$. `\end{proof}`Link to original