Table of Contents
Fetching ...

All Kronecker coefficients are reduced Kronecker coefficients

Christian Ikenmeyer, Greta Panova

Abstract

We settle the question of where exactly the reduced Kronecker coefficients lie on the spectrum between the Littlewood-Richardson and Kronecker coefficients by showing that every Kronecker coefficient of the symmetric group is equal to a reduced Kronecker coefficient by an explicit construction. This implies the equivalence of a question by Stanley from 2000 and a question by Kirillov from 2004 about combinatorial interpretations of these two families of coefficients. Moreover, as a corollary, we deduce that deciding the positivity of reduced Kronecker coefficients is $NP$-hard, and computing them is $\#P$-hard under parsimonious many-one reductions.

All Kronecker coefficients are reduced Kronecker coefficients

Abstract

We settle the question of where exactly the reduced Kronecker coefficients lie on the spectrum between the Littlewood-Richardson and Kronecker coefficients by showing that every Kronecker coefficient of the symmetric group is equal to a reduced Kronecker coefficient by an explicit construction. This implies the equivalence of a question by Stanley from 2000 and a question by Kirillov from 2004 about combinatorial interpretations of these two families of coefficients. Moreover, as a corollary, we deduce that deciding the positivity of reduced Kronecker coefficients is -hard, and computing them is -hard under parsimonious many-one reductions.
Paper Structure (10 sections, 7 theorems, 67 equations, 3 figures)

This paper contains 10 sections, 7 theorems, 67 equations, 3 figures.

Key Result

Theorem 1

For all partitions $\lambda$, $\mu$, $\nu$ of equal sizes, we have

Figures (3)

  • Figure 1: An example of the situation in Lemma \ref{['lem:shift']} with $\lambda=(4,2,1)$, $\mu=(3,2,1,1)$, $\nu=(3,3,1)$, $l=3$, and $m=4$.
  • Figure 2: An example of the proof of Lemma \ref{['lem:walls']} with $\lambda=(5,2)$, $\mu=(3,3,1)$ and $\nu=(4,3)$, with $l=2$, $m=3$ and $c=4$. The red boxes are the addition from the first application of Lemma \ref{['lem:shift']} and the blue boxes are the second application.
  • Figure 3: Lemma \ref{['lem:extremecase']} for $a=5$, $b=4$, $c=2$, $h=4$. A gray cube represents a forced 1 in the contingency array. Absence of color represents a forced 0 in the contingency array. The blue box shows the area where both zeros and ones are possible.

Theorems & Definitions (17)

  • Theorem 1
  • Corollary 1: settles conj. in PPr
  • proof
  • Lemma 1
  • Lemma 2
  • proof
  • Theorem 2
  • Lemma 3
  • proof
  • Lemma 4
  • ...and 7 more