## Schur-Weyl Duality

Throughout the article, we denote for convenience.

So far we have seen:

- the Frobenius map gives a correspondence between symmetric polynomials in of degree
*d*and representations of ; - there is a correspondence between symmetric polynomials in and polynomial representations of .

Here we will describe a more direct relationship between representations of and polynomial representations of Recall from earlier, that and act on as follows:

and the two actions commute, so as endomorphisms of

Lemma. The subspace of all elements fixed by every is spanned by

**Proof**

Use induction on *d*; the case *d*=1 is trivial so suppose *d*>1. For integers , consider the binomial expansion in :

We claim: for large *k*, the matrix with (*i*, *j*)-entry (where ) has rank *d*+1.

- Indeed, otherwise there are , not all zero, such that for all large
*k*, which is absurd since this is a polynomial in*k*.

Hence, we can find a linear combination summing up to:

Thus lies in the subspace spanned by all . By induction hypothesis, the set of all spans the whole space. Hence, the set of all spans . ♦

This gives:

Proposition. If is an -equivariant map, then it is a linear combination of the image of .

**Proof**

Note that since we have Hence from the given condition

By the above lemma, *f* is a linear combination of for all Since is dense, *f* is also a linear combination of for . ♦

## Main Statement

Now let *U* be any complex vector space and consider the complex algebra Recall: if is any subset,

is called the **centralizer** of *A*. Clearly is a subalgebra and we have

Theorem (Schur-Weyl Duality). Let be a subalgebra which is semisimple. Then:

- is semisimple;
- ; (
double centralizer theorem)- U decomposes as , where are respectively complete lists of irreducible A-modules and B-modules.

**Proof**

Since *A* is semisimple, we can write it as a finite product . Each simple *A*-module is of the form for some As an *A*-module, we can decompose: Here since as *A*-modules we have:

By Schur’s lemma if and 0 otherwise. This gives:

which is also semisimple. Now each simple *B*-module has dimension . From the action of *B* on *U*, we can write where *A* acts on the and *B* acts on the . Expressed as a sum of simple *B*-modules, we have ; thus repeating the above with *A* replaced by *B* gives:

From we thus have This proves all three properties. ♦

**Note**

From the proof, we see that

- as complex vector spaces,
- acts on the , and
- acts on the .

Thus the correspondence between and works as follows:

The nice thing about this point-of-view is that the construction is now *functorial*, i.e. for any *A*-module *M*, we can define the corresponding: This functor is *additive*, i.e. , since the Hom functor is bi-additive.

## The Case of *S*_{d} and GL_{n}**C**

_{d}

Now for our main application.

Consider and acting on ; their actions span subalgebras . Now *A* is semisimple since it is a quotient of . From the lemma, we have *B* = *C*(*A*) so Schur-Weyl duality says *A* = *C*(*B*), *B* is semisimple and

where are complete lists of simple *A*– and *B*-modules respectively. Since *A* is a quotient of , the are also irreps of so they can be parametrized by .

Proposition. If is the irrep for isomorphic to , then is the irrep for corresponding to

**Proof**

It suffices to show: if corresponds to via the functor in the above note, then

By definition Recall that is a transitive -set; picking a point , any map which is -equivariant is uniquely defined by the element , as long as this element is invariant under the stabilizer group:

Thus, the coefficients of in remain invariant when acted upon by . So we have an element of ♦

Theorem. The set of irreps of occurring in is:

**Proof**

The following is the complete set of -irreps of degree *d*:

We claim that this is also the set of all irreps in Clearly, each irrep in is of degree *d*; conversely, has

Clearly so contains all of degree *d*. Now apply the above proposition. ♦

**Example**

The simplest non-trivial example follows from the decomposition

The action of is trivial on the first component and alternating on the second.