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All Kronecker coefficients are reduced Kronecker coefficients

Published online by Cambridge University Press:  18 November 2024

Christian Ikenmeyer
Affiliation:
Department of Computer Science, Mathematics Institute, Zeeman Building, University of Warwick, Coventry, CV4 7AL, United Kingdom; E-mail: christian.ikenmeyer@warwick.ac.uk
Greta Panova*
Affiliation:
Mathematics Department, University of Southern California, 3620 S. Vermont Ave., Los Angeles, CA 90089, USA
*
E-mail: gpanova@usc.edu (corresponding author)

Abstract

We settle the question of where exactly do 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 an open problem by Stanley from 2000 and an open problem 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 ${\textsf {NP}}$-hard, and computing them is ${{{\textsf {#P}}}}$-hard under parsimonious many-one reductions. Our proof also provides an explicit isomorphism of the corresponding highest weight vector spaces.

Information

Type
Discrete Mathematics
Creative Commons
Creative Common License - CCCreative Common License - BY
This is an Open Access article, distributed under the terms of the Creative Commons Attribution licence (https://creativecommons.org/licenses/by/4.0/), which permits unrestricted re-use, distribution, and reproduction in any medium, provided the original work is properly cited.
Copyright
© The Author(s), 2024. Published by Cambridge University Press
Figure 0

Figure 1 An example of the situation in Lemma 1 with $\lambda =(4,2,1)$, $\mu =(3,2,1,1)$, $\nu =(3,3,1)$, $l=3$ and $m=4$.

Figure 1

Figure 2 An example of the proof of Lemma 2 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 1, and the blue boxes are the second application.

Figure 2

Figure 3 Lemma 4 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.