Is integrability necessary for the Amplituhedron? It is well known that there exist mappings between operators in N = 4 Super Yang–Mills and spin chain states making the theory Bethe Ansatz integrable. Is integrability a necessity for the Amplituhedron? Most QFTs are not integrable. 
Does Nima Arkani-Hamed, et al. claim to be able to extend the Amplituhedron to generic QFTs? Is then maybe integrability somehow hidden in the structure of all QFTs and the ones which are not integrable emerge as certain limits of the underlying Amplituhedron? 
 A: This question is answered by Nima Arkani-Hamed in his Simons Center talk, at about 112 minutes in.
His answer is that the structure of the amplituhedron itself does not directly use integrability of the theory in any way. It is only when you come to do the integrals themselves that integrability makes it possible. The amplituhedron itself is more linked to locality and unitarity conditions, while the positivity of the grassmannian is linked to the planar limit.
Since they only have the amplituhedron fully working for the N=4 theory at present it is impossible to say that integrability is not necessary to make the structure work, but he would be very surprised if it is not possible to extend the amplituhedron itself to N < 4 and perhaps even to ordinary N=0 Yang-Mills.
The part which should change with N is the form defined on the amplituhedron which is integrated to give its "volume". This form is simple for N=4 with just logarithmic singularities on the boundary. For N < 4 you need to multiply by another factor that has singularities elsewhere corresponding to unltraviolet divergences, Although this extension of the theory is not in such a complete state as the N=4 case, they are optimistic that it still works without the integrability so there is no reason to think that integrability is hidden in all QFTs
By the way they also hope to go beyond the planar limit by replacing positivity with some more general structure and it sounds like that work is progressing well.
