We have a number of immediate consequences of the submodule theorem. First, and most important, the Specht modules form a complete list of irreducible modules for the symmetric group . We know that they’re irreducible, and that there’s one of them for each partition , which is the number of modules we’re looking for. But we need to show that the Specht modules corresponding to distinct partitions are themselves distinct. For this, we’ll use a lemma.
If is a nonzero intertwinor, then . Further, if , then must be multiplication by a scalar. Indeed, since there must be some polytabloid with . We decompose , and extent to all of by sending every vector in to . That is:
where the are -tableaux. Now, the can’t all be zero, so we must have at least one -tableau so that . But then our corollary of the sign lemma tells us that , as we asserted!
Further, if , then our other corollary shows us that for some scalar . We can thus calculate
and so multiplies every vector by .
As a consequence, the must be distinct for distinct permutations, since if then there is a nonzero homomorphism , and thus . But the same argument shows that , and thus .
More particularly, we have a decomposition
where the diagonal multiplicities are . The rest of these multiplicities will eventually have a nice interpretation.
Sorry for the delay.
Let be a submodule of one of the Young tabloid modules. Then I say that either contains the Specht module , or it is contained in the orthogonal complement . In particular, each Specht module is irreducible, since any nontrivial submodule cannot be contained in the orthogonal complement, and so it must also contain — and thus be equal to — .
To see this, let be any vector, and let be a -tableau. Our last result showed us that for some . First, we’ll assume that there is some pair of and so that . In this case, . Since generates all of — the Specht modules are cyclic — we must have .
On the other hand, what if there is no such pair of and ? That is, for all vectors and -tableaux . We can calculate
So each vector is orthogonal to each polytabloid . Since the polytabloids span , we must have .
The results we showed last time have a few immediate consequences we will have use of.
To see this, let and be two entries in the same row of . They cannot be in the same column of , since if they were then the swap would be in the column-stabilizer . Then we could conclude that , which we assumed not to be the case. But if no two entries from the same row of are in the same column of , the dominance lemma tells us that .
Now if it turns out that it’s not surprising that . Luckily in that situation we can say something interesting:
Indeed, we must have for some , basically by the same reasoning that led to the dominance lemma in the first place. Indeed, the thing that would obstruct finding such a is having two entries in some column of needing to go on the same row of , which we know doesn’t happen. And so we calculate
Now if is any vector in the Specht module, and if is a tableau of shape , then is some multiple of . Indeed, we can write
were the are -tableaux. For each one of these, we have . Thus we find
which is a multiple of , as asserted.
As we move towards proving the useful properties of Specht modules, we will find the following collection of results helpful. Through them all, let be a subgroup, and also consider the -invariant inner product on for which the distinct Young tabloids form an orthonormal basis.
First, if , then
Next, for any vectors we have
Indeed, we can calculate
where we have used the facts that , and that as runs over a group, so does .
Next, if the swap , then we have the factorization
for some . To see this, consider the subgroup , and pick a transversal. That is, write as a disjoint union:
but then we can write the alternating sum
as we stated.
Finally, if is some tableau with and in the same row, and if the swap , then
Our hypothesis tells us that . We can thus use the above factorization to write
Let’s look at a few examples of Specht modules.
First, let . The only polytabloid is
on which acts trivially. And so is a one-dimensional space with the trivial group action. This is the only possibility anyway, since , and we’ve seen that is itself a one-dimensional vector space with the trivial action of .
Next, consider — with parts each of size . This time we again have one polytabloid. We fix the Young tableau
Since every entry is in the same column, the column-stabilizer is all of . And so we calculate the polytabloid
We use our relations to calculate
We conclude that is a one-dimensional space with the signum representation of . Unlike our previous example, there is a huge difference between and ; we’ve seen that is actually the left regular representation, which has dimension .
Finally, if , then we can take a tableau and write a tabloid
where the notation we’re using on the right is well-defined since each tabloid is uniquely identified by the single entry in the second row. Now, the polytabloid in this case is , since the only column rearrangement is to swap and . It’s straightforward to see that these polytabloids span the subspace of where the coefficients add up to zero:
As a basis, we can pick . We recognize this pattern from when we calculated the invariant subspaces of the defining representation of . And indeed, is the defining representation of , which contains as the analogous submodule to what we called before.
Now we have everything in place to define the representations we’re interested in. For any partition , the Specht module is the submodule of the Young tabloid module spanned by the polytabloids where runs over the Young tableaux of shape .
To see that the subspace spanned by the polytabloids is a submodule, we must see that it’s invariant under the action of . We can use our relations to check this. Indeed, if is a polytabloid, then is another polytabloid, so the subspace spanned by the polytabloids is invariant under the action of .
The most important fact about the Specht modules is that they’re cyclic. That is, we can generate one just by starting with a single vector and hitting it with all the elements in the group algebra . Not all of the resulting vectors will be different, but among them we’ll get the whole Specht module. The term “cyclic” comes from group theory, where the cyclic groups are those from modular arithmetic, like . Many integers give the same residue class modulo , but every residue class comes from some integer.
Anyway, in the case of Specht modules, we will show that the action of can take one vector and give a whole basis for . Then any vector in the Specht module can be written as a sum of basis vectors, and thus as the action of some algebra element from on our starting vector. But which starting vector will we choose? Well, any polytabloid will do. Indeed, if and are polytabloids, then there is some (not unique!) permutation so that . But then , and so is in the -orbit of . Thus starting with we can get to every vector in by the action of .
We’ve defined a bunch of objects related to polytabloids. Let’s see how they relate to permutations.
First of all, I say that
Indeed, what does it mean to say that ? It means that preserves the rows of the tableau . And therefore it acts trivially on the tabloid . That is: . But of course we know that , and thus we rewrite , or equivalently . This means that , and thus , as asserted.
Similarly, we can show that . This is slightly more complicated, since the action of the column-stabilizer on a Young tabloid isn’t as straightforward as the action of the row-stabilizer. But for the moment we can imagine a column-oriented analogue of Young tabloids that lets the same proof go through. From here it should be clear that .
Finally, I say that the polytabloid is the same as the polytabloid . Indeed, we compute
Given any collection of permutations, we define two group algebra elements.
Notice that doesn’t have to be a subgroup, though it often will be. One particular case that we’ll be interested in is
so we have a nice factorization of this element.
Now if is a tableau, we define the associated “polytabloid”
Now, as written this doesn’t really make sense. But it does if we move from just considering Young tabloids to considering the vector space of formal linear combinations of Young tabloids. This means we use Young tabloids like basis vectors and just “add” and “scalar multiply” them as if those operations made sense.
As an example, consider the tableau
Our factorization lets us write
And so we calculate
Now, the nice thing about is that if we hit it with any permutation , we get .
Every Young tableau with shape gives us two subgroups of , the “row-stabilizer” and the “column-stabilizer” . These are simple enough to define, but to write them succinctly takes a little added flexibility to our notation.
Given a set , we’ll write for the group of permutations of that set. For instance, the permutations that only mix up the elements of the set make up
Now, let’s say we have a tableau with rows . Any permutation that just mixes up elements of leaves all but the first row alone when acting on . Since it leaves every element on the row where it started, we say that it stabilizes the rows of . These permutations form the subgroup . Of course, there’s nothing special about here; the subgroups also stabilize the rows of . And since entries from two different subgroups commute, we’re dealing with the direct product:
We say that is the row-stabilizer subgroup, since it consists of all the permutations that leave every entry in on the row where it started. Clearly, this is the stabilizer subgroup of the Young tabloid .
The column-stabilizer is defined similarly. If has columns , then we define the column-stabilizer subgroup
Now column-stabilizers do act nontrivially on the tabloid . The interaction between rearranging rows and columns of tableaux will give us the representations of we’re looking for.
The biggest problem with the dominance order is that it’s only a partial order. That is, there are some pairs of partitions and so that neither nor . We sometimes need a total order, and that’s where the lexicographic order comes in.
The lexicographic order gets its name because it’s just like the way we put words in order in the dictionary. We compare the first parts of each partition. If they’re different we use their order, while if they’re the same we break ties with the tails of the partitions. More explicitly, if for some we have for all , and .
As an example, here’s the lexicographic order on the partitions of :
Now, comparing this order to the dominance order, we notice that they’re almost the same. Specifically, every time , we find as well.
If , then this is trivially true. But if there must be at least one with . Let be the first such index. Now we find that
This inequality must go in this direction since . We conclude that , and so in the lexicographic order.