English

The hyperdeterminant vanishes for all but two Schur Functors

Algebraic Geometry 2014-10-24 v1

Abstract

We recall the notion of hyperdeterminant of a multidimensional matrix (tensor). We prove that if we restrict the hyperdeterminant to a skew-symmetric tensor pVVp\wedge^p V\subseteq V^{\otimes p} with p3p \geq 3 then it vanishes. The hyperdeterminant also vanishes when we restrict it to the space ΓλVSλVVp\Gamma^\lambda V\otimes S_\lambda V\subseteq V^{\otimes p} where λ\lambda is a Young diagram with p boxes and λ22\lambda_2\geq 2 or λ31\lambda_3\geq 1.

Keywords

Cite

@article{arxiv.1410.6190,
  title  = {The hyperdeterminant vanishes for all but two Schur Functors},
  author = {Alicia Tocino Sánchez},
  journal= {arXiv preprint arXiv:1410.6190},
  year   = {2014}
}

Comments

6 pages

R2 v1 2026-06-22T06:33:22.574Z