Hall Inner Product
Let us resume our discussion of symmetric polynomials. First we define an inner product on d-th component of the formal ring. Recall that the sets
are both -bases of .
Definition. The Hall inner product
is defined by setting and to be dual, i.e. where is 1 if and 0 otherwise.
The introduction of the Hall inner product may seem random and uninspired, but it has implications in representation theory, which we will (hopefully) see much later. The following properties of the inner product are easy to prove.
- The inner product is symmetric, i.e. for any
- The involution is unitary with respect to the inner product, i.e.
For , we have so By definition for any partitions and . Thus the Hall inner product is symmetric.
Next, has bases given by and over all We get:
which is equal to since the inner product is symmetric and By linearity, for any ♦
In the next section, we will see that the inner product is positive-definite. Even better, we will explicitly describe an orthonormal basis which lies in
Schur Polynomials – An Orthonormal Basis
Recall that we have, in ,
where is the Kostka number, i.e. number of SSYT with shape and type
Definition. For each partition , the Schur polynomial is defined as:
Written vectorially, this gives and thus Note that where
For each n>0, the image of in is also called the Schur polynomial; we will take care to avoid any confusion.
Consider the case of d=3. We have:
We then have:
Proposition. The polynomials form an orthonormal basis of i.e. , the Kronecker delta function (which takes 1 when and 0 otherwise).
From and we have:
Treating as a matrix , we get ; since is invertible, so the are orthonormal. ♦
Corollary. The Hall inner product on is positive-definite.
Further Results on Schur Polynomials
Since form an orthonormal basis of so do Now apply the following.
Lemma. Suppose is a finite free abelian group with orthonormal basis , i.e. If is another orthonormal basis of then there is a permutation of such that:
Thus, unlike vector spaces over a field, an orthonormal basis of a finite free abelian group is uniquely defined up to permutation and sign.
Fix and we have for some We get:
Thus for some depending on , and for all The map gives a function Since the form a basis, is a bijection. ♦
Thus for some permutation of We write this as where has exactly one non-zero entry in each row and each column, and such an entry is Thus
Now we saw earlier that is upper-triangular with all 1’s on the main diagonal where is a permutation matrix swapping with (so ). Hence:
Since all entries of are non-negative, it follows that is a permutation matrix. And since is also upper-triangular with 1’s on the main diagonal, we have .
Thus we have shown:
The second and third relations can also be written as:
Also since and J is a permutation matrix, we have and
Consider the case d=3. Check that :