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Young symmetrizer, Specht modules and examples.

January 22, 2013

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.flat_D 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. upward_DNow 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:n-1_1_D

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 :(n-1,1)_id

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)




trivial                standard              sign

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