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

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

CC BY-SA 3.0
2013-05-16T08:01:38.833Z
Crosspost from [cs.se](http://cs.stackexchange.com/questions/11973) where it did not get an answer.
sdcvvc
CC BY-SA 3.0
2013-05-16T14:18:53.797Z
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
CC BY-SA 3.0
2017-11-09T18:56:51.647Z
Answered on CS.SE: https://cs.stackexchange.com/q/11973/755
D.W.
<cc.complexity-theory><complexity-classes><conditional-results>
Is it known that $NEXP = \Sigma_2 \implies NEXP = MA$?
sdcvvc
# Is it known that $NEXP = \Sigma_2 \implies NEXP = MA$?

(Asked by: username_0 on 2013-05-16)

Is it known whether the implication $\mathsf{NEXP} = \Sigma\_2 \implies \mathsf{NEXP} = \mathsf{MA}$ holds?

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

**Tags:** cc.complexity-theory, complexity-classes, conditional-results
> **Comments:**
> **username_0**: Crosspost from [cs.se](http://cs.stackexchange.com/questions/11973) where it did not get an answer.  
> **username_1**: 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.  
> **username_2**: Answered on CS.SE: https://cs.stackexchange.com/q/11973/755  

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