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]].