From $PIT\in P$ to $P=BPP$

If PIT over a finite field $F$ is in P, then there is a family of multilinear polynomials whose graph is decidable in $\mathsf{NE}$ but which does not have poly-size $F$-algebraic circuits (Carmosino-Impagliazzo-Kabanets-Kolokolova APPROX-RANDOM '15). (And a similar result holds over $\mathbb{Z}$.) This is an algebraic circuit lower bound on $\mathsf{NE} \subseteq \mathsf{NEXP}$.

To get $\mathsf{P} = \mathsf{BPP}$ out of, e.g., Nisan-Wigderson (& Impagliazzo-Wigderson), one needs a Boolean (sub)exponential circuit lower bound on functions in (deterministic) $\mathsf{EXP}$.

So, on the one hand, an algebraic circuit lower bound on $\mathsf{NEXP}$ is a lot closer to $\mathsf{P} = \mathsf{BPP}$ than we are today, but still seems pretty far in some absolute sense: algebraic -> Boolean*, super-poly lower bound -> exponential, and $\mathsf{NEXP}$ -> $\mathsf{EXP}$.

(Of course, as discussed by Emil Jerabek in the comments, if one uses Impagliazzo-Kabanets to get PIT in $\mathsf{quasiP}$ by giving an exponential algebraic circuit lower bound, then the "super-poly -> exponential" gap goes away. But I took the question to mean "Suppose $PIT \in \mathsf{P}$ by any arbitrary method...")

*Over a fixed finite field this is probably the easiest one to overcome.

Solved by user129 (35,655 )
Mar 19, 2018 at 05:33