Subset Representations and Eigenvalues of the Universal Intertwining Matrix

We solve a combinatorial question concerning eigenvalues of the universal intertwining endomorphism of a subset representation. This is then applied to justify the evaluation of the Eisenbud-Levine-Khimshiashvili (ELK) signature formula for the gradient index at a degenerate star in arXiv:2001.10882


Introduction
The symmetric group S n acts on the set C n k of subsets of k elements of a set of n elements. This defines a representation, known as a subset representation. Subset representations are related to Young Tableaux with 2 rows. We consider the universal intertwining matrix B (n−k,k) for a subset representation and use Schur's lemma and Young's rule to show that the eigenvalues are Z-linear in the natural parameters of the intertwining matrix (Proposition 1). Next we compute the eigenvalues (Theorem 1). In the terminology that is customary in algebraic combinatorics, what we are doing is recomputing the "eigenmatrix P of a Johnson scheme". This eigenmatrix was determined much earlier by Delsarte [1]. (Around this time Philips was developing the compact disc.) Theorem 1 is then applied to justify the evaluation of the Eisenbud-Levine-Khimshiashvili (ELK) signature formula for the gradient index at a degenerate star in [6]. For this we also need the package MultiSum [7], in order to perform a summation of complicated hypergeometric terms.

The question
Let n be a positive integer and 0 ≤ k ≤ n/2 . Consider combinations C n k of k elements (unordered) out of a set of n elements. Take an arbitrary tuple of complex numbers b 0 , · · · , b k . We constitute a matrix B = B (n−k,k) , where the rows and columns are indexed (in lexicographic order) by elements of C n k . The matrix elements are defined as follows: We want to compute the eigenvectors and eigenvalues. The result is needed in [6, §3] for the computation of a 'gradient index'.

Intertwining
The symmetric group S n acts on C n k and thus defines a permutation representation known as ). Use the standard basis of M (n−k,k) to view N as a matrix. Then N is intertwining if and only if the matrix elements N σ1,τ1 and N σ2,τ2 are equal as soon as σ 1 ∩ τ 1 and σ 1 ∩ τ 1 have the same cardinality. In particular, B is the 'universal intertwining matrix'.
Proof. This is easy and essentially Theorem 2.51 in Prasad [4].

Specht modules
With n, k as above, let ν be the two part partition (n − k, k) of n. We define T k to be the maximal standard tableau [3, pp. 84-85] of shape ν. One could call its numbering lexicographic. See figure 2 for examples.  Recall that the row subgroup R(T k ) of T k is the subgroup of S n which consists of those permutations that permute the entries of each row among themselves. Similarly the 2 column subgroup C(T k ) of T k is the subgroup of S n which consists of those permutations that permute the entries of each column among themselves. One now puts where the product is taken in the group ring C[S n ]. The c k are known as Young symmetrizers.
The Specht module S ν may now be defined as the image of the endomorphism of C[S n ] that is right multiplication by c k [3, p. 119]. The Specht module S ν is an irreducible S n module of dimension

Diagonalizing
Let 0 ≤ m ≤ n/2 . Choose a basis d = d 1 , · · · , d ( n m ) of M (n−m,m) which is the union of bases of the m + 1 irreducible submodules. Then the basis d diagonalizes all intertwining maps M (n−m,m) → M (n−m,m) simultaneously, by Schur's Lemma. As recalled after Lemma 5 below, the d i may be chosen in the Q-span of the standard basis. In particular, B (n−m,m) transforms to a diagonal matrix D with Q-linear combinations of the b i on the diagonal. If one specializes b i = 1 and puts the other b j equal to zero, then the eigenvalues become algebraic integers because they are roots of the characteristic polynomial of a matrix with integer entries. We have proved: The matrix B (n−m,m) has the properties: • The eigenspaces are independent of a (generic) choice of b 0 , · · · b m , • The eigenvalues are Z-linear combinations of b 0 , · · · b m .

Mapping Specht modules to a subset representation
Let Ω be the last element of C n m . So Ω = n − m + 1, n , the set of integers in [n − m + 1, n]. We define an S n -linear map π : C[S n ] → M (n−m,m) by π(p) = p(Ω).
Our strategy is now as follows. We know already the eigenspaces of B = B (n−m,m) and want to compute eigenvalues. Below we take the eigenvector π(c k ) and compare it with its image Bπ(c k ). It is sufficient to consider only one of the coordinates, in fact the 'last coordinate' will do. Note that c k is a double sum of signed products ±qp. We first look at the effect of π on each term.
Let 0 ≤ k ≤ m. We will focus on the sets Ω ∩ π(qp), where p ∈ R(T k ), q ∈ C(T k ). Notice that p, q permute the elements of 1, n , not the boxes in a tableau. Nevertheless a diagram-figure makes it easier to follow the actions of p and q . See figure 3. We write Ω = Ω 1 ∪ Ω 2 , where Ω 1 = n − m + 1, n − k and Ω 2 = n − k + 1, n .  n-k Notice that sgn(q) = (−1) #V , where #X denotes the cardinality of a set X.
Lemma 3. Given V ⊆ 1, k , there is a unique q ∈ C(T k ) such that equation 1 holds.