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

$$\{t\}=\left\{\begin{array}{ccc}1& 3& 4\\ 2& 5& \end{array}\right\}\text{mbox}and\{s\}=\left\{\begin{array}{ccc}1& 2& 5\\ 3& 4& \end{array}\right\}.$$

Note that ${\lambda }^2=(1,1)⊲{\mu }^2=(2,0)$ and ${\lambda }^4=(3,1)⊳{\mu }^4=(2,2)$.

Lemma

If $k<\ell$ and $k$ occurs in a lower row than $\ell$ in $\{t\}$, then $\{t\}⊲(k,\ell )\{t\}$.

\begin{proof} Let ${\lambda }^i$ and ${\mu }^i$ be composition sequences for $\{t\}$ and $(k,\ell )\{t\}$, respectively. If $k⩽i⩽\ell$, 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 ${\lambda }^i⊲{\mu }^i$ for all $i$. 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 $c_i\ne 0$. If $t$ is a standard tableau and $\{s\}$ appears in $e_t$, then $\{t\}⊳\{s\}$.

\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 $k⩽\ell$ and $k,\ell$ are in the same column but $k$ is in lower row than $\ell$. Then $\{s\}⊲(k,\ell )\{s\}$ for each inversion and thus $\{s\}⊲\{t\}$. Here is an example.

For the last term $\{s\}$, $\{1,2\}$ and $\{3,4\}$ are two inversions. By ^rswrnz, $\{s\}⊲(12)\{s\}⊲(12)(34)\{s\}=\{t\}$. \end{proof}

Now we are ready to prove ^030a4e.

\begin{proof} First we prove that they are linear independent. Suppose $c_1{e}_{t_1}+\cdots +c_n{e}_{t_n}=0$, and we label $\{t_1\}$ in such way that there is no $i>1$ with $\{t_1\}⊳\{t_i\}$. 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, as the following example shows.

Example

Consider the following Young tableau $t$:

There is a row descent in the second row, so we choose the subsets $A$ and $B$ as indicated.

Consider all partitions of $A\bigsqcup B=A_i\bigsqcup B_i$ with $|A_i|=|A|$ and $|B_i|=|B|$, that is

$$\{5,6\}\bigsqcup \{2,4\},\{4,6\}\bigsqcup \{2,5\},\{2,6\}\bigsqcup \{4,5\},\{4,5\}\bigsqcup \{2,6\},\{2,5\}\bigsqcup \{4,6\},\{2,4\}\bigsqcup \{5,6\},$$

and they corresponds the following polytabloid $t-t_2+t_3+t_4-t_5+t_6$

and the Garnir element $g_{A,B}=1-(45)+(245)+(465)-(2465)+(25)(46)$. One may 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 $g_{A,B}{e}_t=0$.

\begin{proof} We claim that $S_{A\bigsqcup B}^{-}{e}_t=\{0\}$. Since $|A|+|B|$ is greater than the size of column containing $A$, for any $\sigma \in C_t$ there exist $a,b\in A\bigsqcup B$ which are in the same row of $\sigma t$. Then by ^fae9c5, $(ab)\in S_{A\bigsqcup B}$ yields that $S_{A\bigsqcup B}^{-}\{\sigma t\}=\{0\}$. Since $e_t={\sum }_{\sigma \in C_t}\mathrm{sgn}\sigma \{\sigma t\}$, there is $S_{A\bigsqcup B}^{-}{e}_t=\{0\}$.

Then consider the coset decomposition of $S_{A\bigsqcup B}$. Note that

$$S_{A\bigsqcup B}={\pi }_1(S_A\times S_B)\bigsqcup \cdots \bigsqcup {\pi }_{\ell }(S_A\times S_B)$$

where $g_{A,B}={\sum }_{k=1}^{\ell }\mathrm{sgn}{\pi }_i\cdot {\pi }_i$. Hence, $S_{A\bigsqcup B}^{-}=g_{A,B}(S_A\times S_B{)}^{-}$. 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,

$$\{0\}=S_{A\bigsqcup B}^{-}{e}_t=g_{A,B}(S_A\times S_B{)}^{-}{e}_t=g_{A,B}|S_A\times S_B|e_t$$

yields that $g_{A,B}{e}_t=0$. \end{proof}

Now we finish the proof of ^030a4e. \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 $(kk+1)$, then we denote $s_k=(kk+1)$. For $e_t$ be a polytabloid corresponding to standard tableau, there is

• if $k$ and $k+1$ are in the same column, then $s_k{e}_t=-e_t$;
• if $k$ and $k+1$ are in the same row, then $(kk+1)e_t$ will have row descent and then apply Garnir element;
• if $k$ and $k+1$ are in different columns and different rows, then $(kk+1)e_t$ is another polytabloid corresponding to standard tableau.

Therefore, each transposition $(kk+1)$ 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 symmetric type.