Noncommutativity makes determinants hard

Markus Bläser

Abstract: We consider the complexity of computing the determinant over arbitrary finite-dimensional algebras. We first consider the case that A is fixed. In this case, we obtain the following dichotomy: If A/radA<math><mi is="true">A</mi><mo stretchy="false" is="true">/</mo><mi mathvariant="normal" is="true">rad</mi><mspace width="0.2em" is="true"></mspace><mi is="true">A</mi></math> is noncommutative, then computing the determinant over A is hard. “Hard” here means #P<math><mi mathvariant="normal" is="true">#</mi><mi mathvariant="sans-serif" is="true">P</mi></math>-hard over fields of characteristic 0 and ModpP<math><msub is="true"><mrow is="true"><mi mathvariant="sans-serif" is="true">Mod</mi></mrow><mrow is="true"><mi is="true">p</mi></mrow></msub><mi mathvariant="sans-serif" is="true">P</mi></math>-hard over fields of characteristic p>0<math><mi is="true">p</mi><mo is="true">&gt;</mo><mn is="true">0</mn></math>. If A/radA<math><mi is="true">A</mi><mo stretchy="false" is="true">/</mo><mi mathvariant="normal" is="true">rad</mi><mspace width="0.2em" is="true"></mspace><mi is="true">A</mi></math> is commutative and the underlying field is perfect, then we can compute the determinant over A in polynomial time.We also consider the case when A is part of the input. Here the hardness is closely related to the nilpotency index of the commutator ideal of A.Our work generalizes and builds upon previous papers by Arvind and Srinivasan (STOC 2010) [1] as well as Chien et al. (STOC 2011) [5].