V(λ) as Schur Functor
Again, we will denote throughout this article. In the previous article, we saw that the Schur-Weyl duality can be described as a functor:
- given a -module M, the corresponding -module is set as
Definition. The construction
is functorial in and is called the Schur functor when M is fixed.
Here, functoriality means that any linear map induces a linear .
For example, when , the functor is the identity functor. By Schur-Weyl duality, when M is irreducible as an -module, the resulting is either 0 or irreducible. We will see the Schur functor cropping up in two other instances.
Following the reasoning as in -modules, we have for partitions and ,
Since the only common irrep between the two representations is , any non-zero G-equivariant must induce an isomorphism between those two components. We proceed to construct such a map.
For illustration, take and pick the following filling:
To construct the map, we will take and as subspaces of Thus:
Let us map according to the above filling, i.e. goes into components 1, 4, 2 of while goes into component 3. Similarly, we map by mapping components 1, 3 to , components 4 and 2 to the other two copies of V. In diagram, we have:
This construction is clearly functorial in V. Hence, if is a linear map of vector spaces, then this induces a linear map
Young Symmetrizer Revisited
Another means of defining the Schur functor is by the Young symmetrizer. Here we shall let act on on the left and act on it on the right via:
Now given any (left) -module M, consider:
a left -module. We shall prove that corresponds to the Schur-Weyl duality, i.e. Once again, by additivity, we only need to consider the case . This gives where T is any filling of shape λ and thus:
From here, it is clear that and so is yet another expression of the Schur functor.
Recall that the irreducible -module can be written as where is the Young symmetrizer for a fixed filling of shape λ. Hence, the irrep can be written as:
For d=3, and , let us take the Young symmetrizer:
If is the standard basis for , then is spanned by elements of the form:
These satisfy the following:
By the first relation, we only include those with . By the second relation, we may further restrict to the case since if we have and if we replace We claim that the resulting spanning set forms a basis. Indeed the number of such triplets (i, j, k) is:
On the other hand, we know that has one copy of , one copy of and two copies of so
Thus is the cardinality of the set and we are done. ♦
Observe that the set corresponds to the set of all SSYT with shape (2, 1) and entries in [n] (by writing i, j in the first row and k below i). This is an example of our earlier claim that a basis of can be indexed by SSYT’s with shape and entries in [n]. For that, we will explore as a quotient module of in the next article. This corresponds to an earlier article, which expressed -irrep as a quotient of .