Does $VP \neq VNP$ imply $P \neq NP$? As far as I understand, the geometric complexity theory program attempts to separate $VP \neq VNP$ by proving that the permament of a complex-valued matrix is much harder to compute than the determinant.

The question I had after skimming through the GCT Papers: Would this immediately imply $P \neq NP$, or is it merely a major step towards this goal?
The short answer is 'no'. No such implication is known. There are two main obstacles: Going from arithmetic circuit complexity to boolean complexity (VP≠VNP implies P/poly≠NP/poly) and then going from boolean circuit complexity (P/poly≠NP/poly) to uniform complexity (P≠NP). Neither of these steps is known. I believe that P/poly≠NP/poly implies VP≠VNP, however.