As everyone knows, SAT is complete for $\mathsf{NP}$ w.r.t. polynomial-time many-one reductions. It is still complete w.r.t. $\mathsf{AC^0}$ many-one reductions.
My questions is what is the minimum required depth for the reductions? More formally,
What is the least $d$ such that SAT is $\mathsf{NP}$-hard w.r.t $\mathsf{AC^0_d}$ many-one reductions?
It seems to me that $\mathsf{AC^0_2}$ should be sufficient? Does anyone know a reference?
"Theorem 10. (Gap Theorem) Let C be any proper complexity class. The sets hard for C under non-uniform AC0 reductions are hard for C under non-uniform NC0 reductions." and
"Corollary 4. For every proper complexity class C, every set complete for C under NC0 reductions is complete under reductions computable by depth two AC0 circuits and invertible by depth three AC0 circuits." where proper means "closed under DLogTime-uniform NC1 reductions".
Would you like to post it as an answer so I can accept it?
– Kaveh Jun 11 '13 at 16:34