The hyperdeterminant vanishes for all but two Schur Functors
  We recall the notion of hyperdeterminant of a multidimensional matrix (tensor). We prove that if we restrict the hyperdeterminant to a skew-symmetric tensor $\wedge^p V\subseteq V^{\otimes p}$ with $p \geq 3$ then it vanishes. The hyperdeterminant also vanishes when we restrict it to the space $\Gamma^\lambda V\otimes S_\lambda V\subseteq V^{\otimes p}$ where $\lambda$ is a Young diagram with p boxes and $\lambda_2\geq 2$ or $\lambda_3\geq 1$. 
Alicia Tocino S\'anchez
math.AG