Given a positive integer $n$ and a $\lambda-$tableau $T$ for $\lambda \vdash n$, I defined in my last post two subgroups of $S_n$ : $R(T)$ and $C(T)$. We now associate to these subgroups certain elements of the group algebra $\mathbb{C}S_n$ : define

$a_T := \displaystyle\sum_{\sigma \in R(T)} \sigma$   and   $b_T := \displaystyle\sum_{\tau \in C(T)} \text{sgn} (\tau) \tau$

Also define the Young symmetrizer

$c_T = b_Ta_T = \displaystyle\sum_{\sigma \in R(T) \atop \tau \in C(T)} \text{sgn} (\tau) \tau \sigma.$

What we will end up showing is that for any tableau $T$, the element $c_T$ is a scalar multiple of a primitive idempotent of $\mathbb{C}S_n$ so that $\mathbb{C}S_nc_T$ is an irreducible representation of the symmetric group. In fact, we will show that this process is exhaustive : that two irreducible representation obtained in this way are isomorphic if and only if the shape of the underlying tableaux are the same. Thus the set $\{ c_{\lambda} \mid \lambda \vdash n\}$, where $c_{\lambda}$ means $c_T$ for any $\lambda-$tableau $T$, provide us with a clear, constructive bijection between the conjugacy classes of $S_n$ and its irreps.

Remark 1: By Lemma 1 from last post, if $\pi$ is any permutation of $S_n$, we have $a_{\pi \cdot T} = \pi a_T \pi^{-1}$ and $b_{\pi \cdot T} = \pi b_T \pi^{-1}$ hence $c_{\pi \cdot T} = \pi c_T \pi^{-1}$. Note also that for $\sigma \in R(T)$ and $\tau \in C(T)$, we have $\sigma a_T = a_T = a_T \sigma$ and $\tau b_T = (\text{sgn}\tau)b_T = b_T \tau.$

First, we will see a way to connect these idempotents with Young tableaux and it will permit us to think about them in a really concrete and visual way. It is important to realize that choosing a fixed $\lambda-$tableau $T_0$ permits us to think of the group algebra $\mathbb{C}S_n$ as the $\mathbb{C}S_n-$module of all $\lambda-$tableaux

$N^{\lambda} := \mathbb{C} \{T \mid T \text{ is a } \lambda \text{-tableau} \}$

where the action of $S_n$ is the obvious one. Indeed, the correspondence $\varphi(\sigma) = \sigma \cdot T_0$ is easily seen to be an isomorphism of $\mathbb{C}S_n-$module. We can also consider the $\mathbb{C}S_n-$module of all tabloids

$M^{\lambda} := \mathbb{C} \{ [T] \mid [T] \text{ is a } \lambda \text{-tabloid} \}.$

There clearly is a copy of $M^{\lambda}$ sitting inside of $N^{\lambda}$. We can describe it as

$M^{\lambda} \simeq \mathbb{C} \left\{ \displaystyle\sum_{\sigma \in R(T)} \sigma \cdot T \; \middle| \; T \text{ is a } \lambda \text{-tableau} \right\} \subset N^{\lambda}$

Using the fact that $S_n$ acts transitively on the set of tableaux in $N^{\lambda},$ this is

$M^{\lambda} \simeq \mathbb{C} \left\{ \displaystyle\sum_{\sigma \in R(\pi \cdot T_0)} \sigma \pi T_0 \; \middle| \; \pi \in S_n \right\}.$

In the group algebra, this corresponds to

$M^{\lambda} \simeq \mathbb{C} \left\{ \displaystyle\sum_{\sigma \in R(\pi \cdot T_0)} \sigma \pi \; \middle| \; \pi \in S_n \right\} \subset \mathbb{C}S_n.$

But by remark 1 , we have

$\displaystyle\sum_{\sigma \in R(\pi \cdot T_0)} \sigma \pi = a_{\pi \cdot T_0} \pi = \pi a_{T_0} \pi^{-1} \pi = \pi a_{T_0}$

so we showed that

$M^{\lambda} \simeq \mathbb{C} \{\pi a_{T_0} \mid \pi \in S_n\} = \mathbb{C}S_n a_{T_0}.$

We now introduce a submodule of $M^{\lambda}$, the so-called Specht modules

$S^{\lambda} := \mathbb{C} \{ b_T \cdot [T] \mid [T] \text{ is a } \lambda \text{-tabloid}\}.$

Using the above identifications we find that

$S^{\lambda} \simeq \mathbb{C} \left\{ b_T \cdot \displaystyle\sum_{\sigma \in R(T)} \sigma \cdot T \; \middle| \; T \text{ is a } \lambda \text{-tableau} \right\} \subset N^{\lambda}$

$\simeq \mathbb{C} \left\{ b_{\pi T_0} \cdot \displaystyle\sum_{\sigma \in R(\pi \cdot T_0)} \sigma \pi \; \middle| \; \pi \in S_n\right\} \subset \mathbb{C}S_n$

$= \mathbb{C} \{b_{\pi T_0} a_{\pi T_0} \pi \mid \pi \in S_n \}$

$= \mathbb{C}S_n b_{T_0} a_{T_0}$

where the last equality is because $c_{\pi T} = \pi c_T \pi^{-1}$. So a concrete way to view our potential irreducible representations of $S_n$ are the Specht modules, which can be thought of as submodules of the $M^{\lambda}$‘s by this identification. It is easy to see that the only choice we made in the identification, ie. the tableau $T_0$, is immaterial. Indeed, we obtain isomorphic copies of everything if we choose another tableau of the same shape as $T_0$.

Remark 2: The $M^{\lambda}$‘s and $S^{\lambda}$‘s are cyclic modules. Indeed, for any tabloid $[T] \in M^{\lambda}$, the set of $\pi \cdot [T]$ as $\pi$ ranges over $S_n$ is the whole set of tabloids of that shape. To see that $S^{\lambda}$ is cyclic, we do a little calculation: by identifying a tabloid $[T] \in M^{\lambda}$ with $\sum_{\sigma \in R(T)} \sigma \cdot T = a_T \cdot T \in N^{\lambda}$, we see that

$b_{\pi \cdot T}[\pi T] = b_{\pi \cdot T} \cdot (a_{\pi \cdot T} \cdot \pi T) = \pi b_T \pi^{-1} \pi a_T \pi^{-1} \pi T =\pi b_T[T].$

So the set of $\pi \cdot b_T[T]$ when $\pi$ ranges over $S_n$ is the full set of $b_T[T]$‘s.

Let’s look at some examples that illustrates the usefulness of changing point of views.

Example 1: Suppose $\lambda = (n)$. Then the associated diagram is completely horizontal so for any $\lambda$-tableau $T$, we have $R(T) = S_n$. So $a_T = \sum_{\sigma \in S_n} \sigma$ and therefore $\sigma a_T = a_T$ for all $\sigma \in S_n.$ This tells us that $\mathbb{C}S_na_T = \mathbb{C}a_T$ is the trivial representation. It is way easier to see this working with tabloids: for such a $\lambda$, there is only one $\lambda$-tabloid and it is fixed by every permutation. So we see that $M^{\lambda}$ is the trivial representation directly from the definition. Since $S^{\lambda}$ is a submodule, it is clear that it is also the trivial representation. In particular it is irreducible.

Example 2: Suppose now $\lambda = (1^n)$. ie. the associated diagram is the completely vertical one. Now $R(T) = {e}$ is the trivial group so $a_T$ is the identity element in the group algebra so $\mathbb{C}S_na_T = \mathbb{C}S_n$. From the point of view of tabloids, we see that every arrangement of entries in a tableau $T$ corresponds to a different tabloid $[T]$. Thus, we are led to the same identification as above and identify these tabloids with different permutations of $S_n$, corresponding to the different arrangements of their entries. Then $M^{\lambda} = \mathbb{C}\{[T_{\sigma}] \mid \sigma \in S_n\} \simeq \mathbb{C}S_n$. We almost got forced to think about this situation in terms of the group algebra.

What about $S^{\lambda}$? Clearly, if $T$ is of shape $\lambda$, then $C(T) = S_n$. So in this case we have $b_T = \sum_{\sigma \in S_n} (\text{sgn}\sigma) \sigma.$ So if we hit $b_T$ with a permutation $\pi$, we find

$\pi \cdot b_T = \displaystyle\sum_{\sigma \in S_n} (\text{sgn} \sigma) \pi \sigma$

$= \displaystyle\sum_{\tau \in S_n} (\text{sgn} \pi^{-1}\tau) \tau$

$= (\text{sgn}\pi^{-1}) \displaystyle\sum_{\tau \in S_n} (\text{sgn}\tau) \tau$

$= (\text{sgn} \pi) \cdot b_T$

so $\mathbb{C}S_nb_T = \mathbb{C}b_T$. Since $a_T = e$, we have $\mathbb{C}S_nc_T = \mathbb{C}b_T$ with the action given by $\pi \cdot b_T = (\text{sgn} \pi) b_T$. ie $S^{\lambda}$ is the sign representation, which is also irreducible. The same result follows from the point of view of tabloids when considering that for any $\lambda$-tableau $T_0$, the set of $\pi b_{T_0}[T_0]$ when $\pi$ ranges over all permutations generate $S^{\lambda}$ (this is remark 2). Indeed, since $\pi b_{T_0} = (\text{sgn}\pi) b_{T_0}$, we also find that $S^{\lambda} = \mathbb{C}b_{T_0}[T_0]$ with the sign action.

Example 3:

Now let $\lambda = (n-1,1)$. Then the tableaux have a long horizontal row of $n-1$ boxes and 1 lonely box in the second row. Since now the groups $R(T)$ and $C(T)$ depend on the choice of tableau, let’s choose the tableau with entries in ascending order from left to right and from the top down:

So $R(T)$ is the group of permutations fixing $n$ in $S_n$, $R(T) \simeq S_{n-1} \subset S_n$. We thus have

$a_T = \displaystyle\sum_{\sigma \in S_n \atop \sigma(n) = n} \sigma.$

For $\pi \in S_n$, we have

$\pi \cdot a_T = \displaystyle\sum_{\sigma \in S_n \atop \sigma(n) = n} \pi \sigma = \displaystyle\sum_{\tau \in S_n \atop \tau(n) = \pi(n)} \tau$

so the only real thing that distinguishes the action of one permutation compared to another when multiplied by $a_T$ is their action on the number $n$. We would therefore like to group together permutations by their action on $n$ and say that $\mathbb{C}S_na_T \simeq \mathbb{C}\{e_1, e_2, \ldots, e_n\}$ is the defining representation.

From the point of view of tabloids, this is obvious. Indeed, since a tabloid is uniquely determined by the element in its second row, we can identify the tabloid with this number, ie :

then it is clear that $\pi$ sends the tabloid $k$ to the tabloid $k'$ iff $\pi(k) = k'$. ie. $M^{\lambda} \simeq \mathbb{C}\{e_1,e_2,\ldots, e_n\}.$

In this case it is difficult to make sense of $S^{\lambda}$ directly in terms of $\mathbb{C}S_nb_Ta_T$ but let’s take a look at it from the point of view of tabloids. Choose the $\lambda$-tableau $T_0$ with entries in ascending order. Then $C(T) = \{e, (1,n)\}$ where $(1,n)$ is the transposition exchanging $1$ and $n$. Thus, still identifying a tabloid with its entry in the second row, we have $b_{T_0}[T_0] = (e-(1,n)) \cdot n = n-1 \in M^{\lambda}$. Hence

$S^{\lambda} = \mathbb{C}\{\pi \cdot b_{T_0}[T_0] \mid \pi \in S_n\}$

$= \mathbb{C}\{\pi \cdot n - \pi \cdot 1 \mid \pi \in S_n\}$

$= \mathbb{C}\{i-j \mid 1 \leq i,j \leq n \; , \; i \neq j\}.$

Fixing the basis $\{j-(j-1) \mid 2 \leq j \leq n\}$ for this space, we see that $S^{\lambda}$ is the $(n-1)$-dimensional subspace $\{\sum_{i=1}^n \alpha_i e_i \mid \sum_{i=1}^n \alpha_i = 0 \; , \; \alpha_i \in \mathbb{C}\}$ which one might recognize as the standard representation sitting inside the defining representation $\mathbb{C}\{e_1, \ldots, e_n\} \simeq \mathbb{C}^n$. Again, this is indeed an irreducible representation.

In the case of $S_3$, it is easy to see directly that these are all the irreducible representations. The bijection between partitions of 3 and irreps of $S_3$ reads

$(3)$                  $(2,1)$            $(1,1,1)$

$\updownarrow$

$\updownarrow$

trivial                standard              sign