Properties of the Young Symmetrizer
Recall that for a filling , we have the subgroup of elements which take an element of the i-th row (resp. column) of T to the i-th row (resp. column) of T. Then:
where is the Young symmetrizer. Recall the following results from earlier:
The following is obvious.
Lemma 1. If , then and Thus
This follows from and E.g. the latter gives:
The following generalizes the Young symmetrizer.
Proposition 1. Consider the element for fillings of shape Note that since is contained in both and it is either 0 or isomorphic to
The following are equivalent.
- no two distinct lie in the same row of and same column of ;
- there exist and such that
The third condition says: we can change T to T’ by permuting elements in each row, then elements in each column.
Suppose but two distinct lie in the same row of T and same column of T’. Let swap those two elements; then Letting C run through a set of coset representatives of we have But now we have so and we have c=0.
Suppose the second condition holds. Elements of the first row of T are in different columns of T’. Bringing those in T’ to the first row, there exist and such that and have identical first rows.
Likewise, since the second row of are in different columns of there exist and such that and have the same first and second rows. Repeating, we eventually get the desired and
Finally, suppose the third condition holds; let . By lemma 1,
Lemma 2. If are distinct SYT of the same shape, they do not satisfy the conditions of proposition 1.
Order all the fillings of shape as follows: given T, T’, let k be the largest number occurring in different squares of T and T’; we write T’ > T if k occurs earlier in the word w(T’) than in w(T). For example, we have:
Note that for any SYT T we have:
since the largest entry of T which is moved by g must move to its left in w(T), and the largest entry moved by g’ moves to its right in w(T). Thus if T, T’ are SYT and g(T) = g’(T’) for some , we have so T = T’. ♦
Lemma 3. For any , is a multiple of if and only if:
- for any and , we have
⇒ is left as an easy exercise, by proving
For ⇐, write where Then for any , and Taking x=e, it suffices to show: if is not of the form for , then
For that, consider the filling . We claim that satisfy the conditions of proposition 1.
- Indeed, if for some and then
- So and we have which is a contradiction.
Hence there exist distinct i, j in the same row of T and same column of T’. Let t be the transposition (i, j); then so by the given condition. Since so satisfies Now consider the coefficient of tx in v. We have:
Thus as desired. ♦
Here is one immediate application of lemma 2.
Proposition 2. For any , is a scalar multiple of
In the case we have where
Indeed if and we have:
so by lemma 2, is a scalar multiple of
For , consider right-multiplication on by and let A be its trace. This map takes where the coefficient of g is 1; taking the basis comprising of elements of G, we have On the other hand, the image of this map is and takes each Thus the map is a scalar map on and we have ♦
This gives the following.
Theorem. We have, as a direct sum of irreps:
Let us show that is a direct sum, where T runs over all SYT of shape Indeed, if not we have:
for some distinct SYT and some Note that if then by lemma 2, T and T’ do not satisfy the conditions in proposition 1, so we can find distinct i, j in the same row of T and same column of T’; using the same technique as the first paragraph of proof of proposition 1, we have and thus
Right-multiplying (*) by thus gives which gives , a contradiction. Hence, the inner sum is a direct sum. The entire sum is a direct sum since there is no common irrep between distinct
Hence, the RHS is a subspace of the LHS. Now we count the dimension: on LHS we get d!; on RHS we get so the two sides are equal. ♦