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Paper:

TR12-142 | 3rd November 2012 00:22

Noncommutativity makes determinants hard

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TR12-142
Authors: Markus Bläser
Publication: 3rd November 2012 21:38
Downloads: 1813
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Abstract:

We consider the complexity of computing the determinant over arbitrary finite-dimensional algebras. We first consider the case that $A$ is fixed. We obtain the following dichotomy: If $A/rad(A)$ is noncommutative, then computing the determinant over $A$ is hard. ``Hard'' here means $\#P$-hard over fields of characteristic $0$ and $ModP_p$-hard over fields of characteristic $p > 0$. If $A/rad(A)$ 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 $A$. The commutator ideal $com(A)$ of $A$ is the ideal generated by all elements of the form $xy - yx$ with $x,y \in A$. We prove that if the nilpotency index of $com(A)$ is linear in $n$, where $n \times n$ is the format of the given matrix, then computing the determinant is hard. On the other hand, we show the following upper bound: Assume that there is an algebra $B \subseteq A$ with $B = A/ rad(A)$. (If the underlying field is perfect, then this is always true.) The center $Z(A)$ of $A$ is the set of all elements that commute with all other elements. It is a commutative subalgebra. We call an ideal $J$ a complete ideal of noncommuting elements if $B + Z(A) + J = A$. If there is such a $J$ with nilpotency index $o(n/log(n))$, then we can compute the determinant in subexponential time. Therefore, the determinant cannot be hard in this case, assuming the counting version of the exponential time hypothesis.

Our results answer several open questions posed by Chien et al.



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