# Young symmetrizer, Specht modules and examples.

Given a positive integer and a tableau for , I defined in my last post two subgroups of : and . We now associate to these subgroups certain elements of the group algebra : define

and

Also define the *Young symmetrizer*

What we will end up showing is that for any tableau , the element is a scalar multiple of a primitive idempotent of so that 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 , where means for any tableau , provide us with a clear, constructive bijection between the conjugacy classes of and its irreps.

**Remark 1: **By Lemma 1 from last post, if is any permutation of , we have and hence . Note also that for and , we have and

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 tableau permits us to think of the group algebra as the module of all tableaux

where the action of is the obvious one. Indeed, the correspondence is easily seen to be an isomorphism of module. We can also consider the module of all tabloids

There clearly is a copy of sitting inside of . We can describe it as

Using the fact that acts transitively on the set of tableaux in this is

In the group algebra, this corresponds to

But by remark 1 , we have

so we showed that

We now introduce a submodule of , the so-called *Specht modules*

Using the above identifications we find that

where the last equality is because . So a concrete way to view our potential irreducible representations of are the Specht modules, which can be thought of as submodules of the ‘s by this identification. It is easy to see that the only choice we made in the identification, ie. the tableau , is immaterial. Indeed, we obtain isomorphic copies of everything if we choose another tableau of the same shape as .

**Remark 2:** The ‘s and ‘s are cyclic modules. Indeed, for any tabloid , the set of as ranges over is the whole set of tabloids of that shape. To see that is cyclic, we do a little calculation: by identifying a tabloid with , we see that

So the set of when ranges over is the full set of ‘s.

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

**Example 1:** Suppose . Then the associated diagram is completely horizontal so for any -tableau , we have . So and therefore for all This tells us that is the trivial representation. It is way easier to see this working with tabloids: for such a , there is only one -tabloid and it is fixed by every permutation. So we see that is the trivial representation directly from the definition. Since is a submodule, it is clear that it is also the trivial representation. In particular it is irreducible.

**Example 2:** Suppose now . ie. the associated diagram is the completely vertical one. Now is the trivial group so is the identity element in the group algebra so . From the point of view of tabloids, we see that every arrangement of entries in a tableau corresponds to a different tabloid . Thus, we are led to the same identification as above and identify these tabloids with different permutations of , corresponding to the different arrangements of their entries. Then . We almost got forced to think about this situation in terms of the group algebra.

What about ? Clearly, if is of shape , then . So in this case we have So if we hit with a permutation , we find

so . Since , we have with the action given by . ie is the sign representation, which is also irreducible. The same result follows from the point of view of tabloids when considering that for any -tableau , the set of when ranges over all permutations generate (this is remark 2). Indeed, since , we also find that with the sign action.

**Example 3:**

Now let . Then the tableaux have a long horizontal row of boxes and 1 lonely box in the second row. Since now the groups and 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 is the group of permutations fixing in , . We thus have

For , we have

so the only real thing that distinguishes the action of one permutation compared to another when multiplied by is their action on the number . We would therefore like to group together permutations by their action on and say that 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 sends the tabloid to the tabloid iff . ie.

In this case it is difficult to make sense of directly in terms of but let’s take a look at it from the point of view of tabloids. Choose the -tableau with entries in ascending order. Then where is the transposition exchanging and . Thus, still identifying a tabloid with its entry in the second row, we have . Hence

Fixing the basis for this space, we see that is the -dimensional subspace which one might recognize as the standard representation sitting inside the defining representation . Again, this is indeed an irreducible representation.

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

trivial standard sign