1 Evidence that Graph Isomorphism is not NP - Complete

Theorem 1 If GI is NP-complete, then the polynomial hierarchy collapses (specifically, PH = Σ2). Proof We first observe that AM ⊆ Π2 (why?). Now, assume GI is NP-complete. Then GNI is coNP-complete and hence (since GNI ∈ AM) we have coNP ⊆ AM. We show that this implies Σ2 ⊆ AM ⊆ Π2 and hence PH = Σ2. Say L ∈ Σ2. Then by definition of Σ2, there is a language L′ ∈ Π1 = coNP such that: (1) if x ∈ L then there exists a y such that (x, y) ∈ L′, but (2) if x 6∈ L then for all y we have (x, y) 6∈ L′. This immediately suggests the following proof system for L: 1. Merlin sends y to Arthur. 2. Arthur and Merlin then run an AM protocol that (x, y) ∈ L′ (this is possible precisely because L′ ∈ coNP ⊆ AM). The above is an MAM proof system for L. But, as we have seen, this means there is an AM proof system for L. Since L ∈ Σ2 was arbitrary this means Σ2 ⊆ AM, completing the proof.