In 1988 Macdonald introduced a new class of symmetric functions depending on two parameters and in Seminaire Lotharingian.
His definition requires
- Orthogonality with respect to new scalar product
- Lower unitriangularity with respect to monomial symmetric functions
The new scalar product in our language is
Since plethystic substitution is self-adjoint, this definition is symmetric in and
Remark: By the cauchy indentiy we have that and are - dual if and only if and are - dual. So we have to check either of the equations
Remark: The non-plethystic expression for is
The strategy is to write them as eigenfunctions of an operator.
Define where is the operator that picks the coefficient of (i.e. the constant term) in the expansion.
Also define so for example for we have . With the convention that for all , we have the alternative description
The Macdonald polynomial is the eigenfunction of the operator
normalized so that
A proposition we will use below
Proposition: A symmetric function is an eigenfuncion of with eigenvalue if and only off is an eigenfunction of with eigenvalue . Here
Proof: It is straightforward to check that which are equal to respectively, and they are both equal since changing by its negative doesn't change the constant term
Our main job is to show the Macdonald symmetric functions exist and are well-defined. First we settle the existence of such eigenfunctions with the lower triangularity property.
Lower triangularity of Macdonald symmetric functions
Proposition: is lower triangular with respect to the basis of monomial symmetric functions. More precisely
Proof: The first step of the proof is to go back to the schur function section and read about the Bernstein operator to get a feeling of what's going on. This follows the same ideas:
Since the schur functions are lower unitriangular with respect to monomials (remember the Kostka coefficients) it suffices to show
where
We work with finitely many variables . Then by partial fraction expansion we have
Note that in the numerator we have when and for the rest we can write
where is the vandermonde determinant, and means we replace the variable with (note that we could have used the plethystic notation ). This is true since all the terms not involving will cancel out, and the terms involving will have the same sign bottom and top so we can consider all of them with first. So we have
Now recall that . So
Recall that the coefficient of in a symmetric function is equal so the coefficient of in . Consider the coefficient in the first term, call it . It is not the Kostka number, since they express schur in terms monomials and we need monomial in terms of schur, but it is an entry of the inverse to the Kostka matrix. Since the Kostka matrix is lower unitriangular its inverse is too, so . Thus to find the coefficient of in it suffices to compute the second summand of
But on each summand the dominant exponent is , given by in and in v(X). An further analysis reveals that all possible appearing will be smaller in dominance order. Therefore the largest partition than can come up is and we have the triangularity. Now lets analyze the diagonal elements, i.e. the coefficient of .
On each product the coefficient of goes with a coefficient and the first summand contributes a so the diagonal elements are
recall that for all , so that the sum is independent of and finally
By using the relation discussed above.
Summarizing: has distinct eigenvalues and its corresponding eigenfunctions have lowertriangularity with respect to the monomial basis and nonzero coefficient of . So we have the first requirement for the Macdonald polynomials, we still need to say stuff about their inner products.
Macdonald - Kostka polynomials
Recall where are rational functions
Conjecture/Theorem
Macdonald / Garsia, Remmel, Noumi, Kiriloov, Sahi, Knop
is a polynomial in , called the integral form. Here stand for the arm and the leg of the cell in the diagram of .
Conjecture/Theorem
Macdonald / Haiman
Define
Then
The plethystic substution seems to appear out of no where. There is a more natural expression involving transformed Macdonald symmetric functions.
Remark: Haiman's proof is based on the conjecture of Garsia and Haiman and the geometry of Hilber schemes. For an excellent, inspiring, mind-blowing, survey see
Mark Haiman Combinatorics, symmetric functions, and Hilbert schemes Current Developments in Mathematics 2002, no. 1 (2002), 39–111.
Remark: the Kostka-Foulkes polynomial. A combinatorial interpretation of along the lines of Lascoux and Schutzenberger formula is still missing.
Transformed Macdonald Symmetric Functions
For many reasons that we shall not discuss here, the following transformed Macdonald polynomials play a fundamental role in the theory
where
So now we have that is an eigenfunction of . More precisely
and
where
By the symmetry of the operator and checking the eigenvalues we get that
and
Remark: as proved by Macdonald. So the coefficient of in is 1.
We have the following important proposition
Proposition: The transformed Macdonald symmetric functions are uniquely characterized by
Proof: and hence are homogeneous of degree . So
is a scalar multiple of and hence of .
Since and the involution transposes the lambdas and transposing reverses the dominance ordering we have
Hence we have number 2 and by the symmetry mentioned before the proposition we have also number 1. Number 3 is the last remark.
For uniqueness: Assume is another solution to the conditions 1 and 2. The first condition implies
Which implies
Similarly the second condition implies
Implying that is a scalar multiple of , but the third condition fixes the scalar equal to 1.
As a corollary we have
which implies also
To prove this, first observe that satisfies conditions 1 and 2 from the proposition, hence it is a scalar multiple of . The scalar is 1 since
a non trivial identity proved by Macdonald.
Orthogonality of the Macdonald symmetric functions
Replacing we are to show
Or equivalently
because so we pull out the and give the minus to .
Since is a scalar multiple of what we need to show in the transformed Macdonald polynomials is
But up to sign we have so now using the first two conditions on the characterization of the transformed Macdonald symmetric functions we're done.
