Appendix to: Generalized Stability of Kronecker Coefficients. John R. Stembridge. 14 August 2014

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Appendix to: Generalized Stability of Kronecker Coefficients John R. Stembridge 14 August 2014 Contents A. Line reduction B. Complementation C. On rectangles D. Kronecker coefficients and Gaussian coefficients In this appendix to [S], we provide some complementary remarks and observations that didn t make it into the paper. We follow the same terminology and notation. External references to numbered equations etc. are pointers to [S]. A. Line reduction Since I m is the trivial representation of S m, one knows that g(α,β,m) = δ α,β. By taking V 3 to be one-dimensional and γ = (m), we may deduce from (2.1) that S m (V 1 V 2 ) = V 1 (α) V 2 (α) α (as gl(v 1 ) gl(v 2 )-modules), where α ranges over partitions with l(α) min(dimv 1,dim V 2 ). In particular, S m (V 1 V 2 V 3 ) = (V 1 V 2 )(γ) V 3 (γ) (A.1) as gl(v 1 ) gl(v 2 ) gl(v 3 )-modules. This yields a well-known reformulation of (2.1). Proposition A.1. If l(α) dim V 1 and l(β) dim V 2, then the Kronecker coefficient g(αβγ) is the multiplicity of V 1 (α) V 2 (β) in the gl(v 1 ) gl(v 2 )-module (V 1 V 2 )(γ). If V is m-dimensional, then V (1 m ) is the one-dimensional gl(v )-module carried by the trace map gl(v ) C. It follows easily that V (n m ) = V (1 m ) n and more generally γ V (µ) V (1 m ) = V (µ + 1 m ) (A.2) for all partitions µ with at most m parts.

Proposition A.2. Assume l(α) a and l(β) b. (a) (Well known.) If g(αβγ) > 0, then l(γ) ab. (b) (Vallejo [V1].) We have g(αβγ) = g(α + b a,β + a b,γ + 1 ab ). (c) If l(γ) = ab and g(αβγ) > 0, then α has at least b columns of length a and β has at least a columns of length b. Proof. Choose vector spaces V 1 and V 2 of dimensions a and b. (a) All of the nonzero summands in (A.1) have l(γ) ab. (b) We may assume l(γ) ab; otherwise (a) implies that both Kronecker coefficients are zero. One can check that (V 1 V 2 )(1 ab ) = V 1 (b a ) V 2 (a b ), (A.3) so the claimed formula follows by repeated application of (A.2). (c) If l(γ) = ab then γ = ˆγ + 1 ab for some partition ˆγ with at most ab parts. It follows that all summands V 1 (α) V 2 (β) in the irreducible decomposition of (V 1 V 2 )(γ) are obtained by tensoring the summands of (V 1 )V 2 )(ˆγ) by (V 1 V 2 )(1 ab ). The claim now follows via (A.2) and (A.3). Thus if the length of one partition in a triple αβγ matches the product of the lengths of the other two, then the first column of that partition and appropriate numbers of columns of the other two may be deleted without changing the Kronecker coefficient. By conjugation invariance, similar reductions are also possible if (say) l(γ) = α 1 β 1 or γ 1 = α 1 l(β). We call any such operation a line reduction. B. Complementation If l(µ) m = dimv, then the dual gl(v )-module V (µ) is isomorphic to V (µ ), where µ := ( µ m,..., µ 2, µ 1 ). Of course µ will generally not be a partition, but (A.2) implies V (µ ) V (n m ) = V (n m + µ ), and ν = n m + µ will be a partition as long as n µ 1. Indeed, ν is the partition one obtains by removing the diagram of µ from an m n rectangle and rotating the result by 180 degrees. We call ν the n m -complement of µ. In the following result, the dual weights α, β, and γ should be understood as associated with vector spaces of dimensions a, b, and ab (respectively). Proposition B.1. If l(α) a, l(β) b, and γ 1 n, then g(αβγ) = g((bn) a + α,(an) b + β,n ab + γ ). Furthermore, α (bn) a, β (an) b, and γ n ab, or else g(αβγ) = 0. 2

1 2 3 4 5 6 7 8 9 10 11 12 1 1 1 1 1 1 1 1 1 1 1 1 1 2 0 1 0 1 0 1 0 1 0 1 0 1 3 0 1 1 2 1 3 2 4 3 5 4 7 4 0 1 1 5 4 16 21 67 118 307 630 1495 5 0 0 1 6 21 216 1890 18371 167596 1437657 6 0 0 1 13 158 9309 445442 20969042 7 0 0 0 14 1456 438744 125250433 8 0 0 1 18 9854 17957625 9 0 0 1 14 44852 Table C.1: The Kronecker coefficients G(m,n) = g(n m,n m,n m ). Proof. To see the necessity of the rectangle-fitting, note that if γ failed to fit in the rectangle n ab, then g(αβγ) = 0 by Proposition A.2. If (say) α failed to fit in the rectangle (bn) a, then α would have length greater than bn, and hence g(αβγ) = g(βγ α ) = 0 by conjugation invariance and Proposition A.2. Now choose vector spaces V 1 and V 2 of dimensions a and b. If we dualize Proposition A.1, we see that g(αβγ) is the multiplicity of V 1 (α ) V 2 (β ) in (V 1 V 2 )(γ ). If we take the tensor product of the latter with (V 1 V 2 )(n ab ), one sees that the claimed formula is a consequence of (A.3). C. On rectangles Consider the Kronecker coefficients G(m,n) := g(n m,n m,n m ) (see Table C.1). Remark C.1. (a) The first row of the table is trivial. (b) For a proof that the pattern in the second row persists, see Corollary D.2. (c) We have computed G(3,n) (i.e., the quantities in the third row) for n 16, and have found that the data is consistent with G(3,n) q n = n 0 1 (1 q 2 )(1 q 3 )(1 q 4 ). If true, then G(3,n) is asymptotic to n 2 /48, has a quasi-period of 12, and a partial fraction analysis of the right hand side would imply { [n(n + 6)/48] if n is odd, G(3,n) = [(n + 4)(n + 8)/48] if n is even, where [x] denotes the integer nearest to x. (d) The conjugation invariance g(n m,n m,n m ) = g(m n,m n,n m ) and Proposition A.2 imply that G(m,n) can be nonzero only for m n 2. 3

(e) Noting that the (n 2 ) n -complement of an m n rectangle is an (n 2 m) n rectangle, the complementation formula in Proposition B.1 implies G(m,n) = g(n n2 m,n n2 m,(n 2 m) n ) = G(n 2 m,n). (C.1) That is, the n-th column of the matrix [G(m,n)] is symmetric for 0 m n 2. Table C.1 and the symmetry (C.1) suggest that the distribution of G(m,n) for fixed (large) n is bell-shaped. In particular, it seems likely that the vanishing of G(2,n) and G(n 2 n,n) for n odd are the only zeroes of G(m,n) in the feasible range 0 m n 2. D. Kronecker coefficients and Gaussian coefficients Let p k (m,n) denote the number of partitions of k whose Young diagrams fit inside an m n rectangle. It is well known that these quantities are the coefficients of the Gaussian polynomials. More precisely, we have [ ] m + n m = (1 qn+m ) (1 q n+1 ) (1 q m ) (1 q) = k p k (m,n)q k. Also well-known is that p k (m,n) is unimodal with respect to k, and there are many proofs, some quite elementary. The following result, noticed earlier by Vallejo [V2] and Pak and Panova [PP], shows that nonnegativity of Kronecker coefficients also implies unimodality. Proposition D.1. We have g(n m,n m,(mn k,k)) = p k (m,n) p k 1 (m,n). Proof. In the language of symmetric functions, we seek to evaluate s ρ s (mn k,k),s ρ, where ρ denotes the rectangle partition n m. By the Jacobi-Trudi identity one knows that s (mn k,k) = h mn k h k h mn k+1 h k 1, so it suffices to show that s ρ h mn k h k,s ρ = p k (m,n). By the combinatorial rule for s α h β (see Exercise 7.84.a in [EC2]), one knows that s ρ h mn k h k is the sum of s α s ρ/α, where α ranges over all partitions of size k contained in ρ. Moreover, since ρ = n m is a rectangle, it follows that ρ/α is the 180 degree rotation of the n m -complement of α. One knows that skew Schur functions are invariant under rotations, so we have s ρ h mn k h k = s α s α, where α := n m + α = (n α m,...,n α 2,n α 1 ). Recalling the basic adjoint relationship s µ s ν,s λ = s ν,s λ/µ, we obtain s ρ h mn k h k,s ρ = s α,s ρ/α = s α,s α = p k (m,n), since the Schur functions are orthonormal. 4

As far as we are aware, there is no explicit positive combinatorial description known for the quantities p k (m,n) p k 1 (m,n). However, one case where it is relatively easy to inject the partitions of k 1 into the partitions of k occurs when k n. In this case, the leftmost empty column of the diagram is inside the rectangle, and so adding a column of length 1 is an injective map. The leftover partitions of k are those with no columns of length 1, so in the case k = n we obtain Corollary D.2. We have g(n (1 m,1 m,(m 1,1)))q n = m 0 and in the special case m = 2, { 1 if n even, g(n 2,n 2,n 2 ) = 0 if n odd. 1 (1 q 2 )(1 q 3 ) (1 q m ), Considering the special case m = 4 of the above result, we see that the conjecture in Remark C.1(c) would imply the peculiar identity g(n 4,n 4,(3n,n)) = g(n 3,n 3,n 3 ). Remark D.3. Recall that the irreducible sl 2 (C)-module of highest weight m is the symmetric power S m (C 2 ). It is easy to show that p k (m,n) is the dimension of the subspace of weight mn 2k in the plethystic composition S m (S n (C 2 )), so a further consequence of Proposition D.1 is that the Kronecker coefficient g(n m,n m,(mn k,k)) is the multiplicity of S mn 2k (C 2 ) in S m (S n (C 2 )). References [PP] I. Pak and G. Panova, Strict unimodality of q-binomial coefficients, C. R. Math. Acad. Sci. Paris 351 (2013), 415 418. [EC2] R. P. Stanley, Enumerative Combinatorics, Vol. 2, Cambridge Univ. Press, Cambridge, 1999. [S] J. R. Stembridge, Generalized stability of Kronecker coefficients, preprint. [V1] E. Vallejo, A stability property for coefficients in Kronecker products of complex S n characters, Electron. J. Combin. 16 (2009), Note 22, 8 pp. [V2] E. Vallejo, A diagrammatic approach to Kronecker squares, preprint. Department of Mathematics, University of Michigan, Ann Arbor MI 48109 1043 USA E-mail address: jrs@umich.edu 5