# Is it known that $NEXP = \Sigma_2 \implies NEXP = MA$?

Is it known whether the implication $\mathsf{NEXP} = \Sigma_2 \implies \mathsf{NEXP} = \mathsf{MA}$ holds?

(The question is inspired by well-known $\mathsf{NEXP} \subseteq \mathsf{P/poly} \Leftrightarrow \mathsf{NEXP} = \mathsf{MA}$.)

• Crosspost from cs.se where it did not get an answer. – sdcvvc May 16 '13 at 8:01
• Interesting question! All I can say is that I don't know of any result of this kind, nor do I know of any existing techniques that seem up to it. Granted, the hypothesis is extremely strong, but in some sense a collapse to Sigma_2 seems "still too high above MA's head" for MA to reach up and grab it. – Scott Aaronson May 16 '13 at 14:18
• Answered on CS.SE: cs.stackexchange.com/q/11973/755 – D.W. Nov 9 '17 at 18:56