is the crucial component in the scheme described above; in particular, it should be strong enough such that (\[decaykg\]) holds, and also weak enough such that the “second iteration”, namely the output function $f_{\sigma}$ of (\[duhamel\]) assuming the input functions $f_{\mu}$ and $f_{\nu}$ are Schwartz and independent of time, has bounded $Z$ norm.

Following [@IP12], let $$\|f\|_{Z}=\sup_{j,k}\|f_{jk}\|_{Z_{jk}},$$ where $f_{jk}$ is obtained by localizing $f$ at frequency scale $\sim 2^k$ and space scale $\sim 2^{j}$. In [@IP12] it turns out that, in order for (\[decay98\]) to hold, the $Z$ norm has to incorporate strong space localization, and has to be (almost) as strong as $$\|f\|_{Z_{jk}}=2^{j}\|f\|_{L^{2}}.$$ Moreover, we can see from a simple volume counting argument that, to guarantee that the second iteration belongs to this space, it is necessary that the phase $\Phi$ is nondegenerate, namely $\det(\nabla_{\eta}^{2}\Phi_{\sigma\mu\nu})\neq 0$ on the spacetime resonance set $\mathcal{R}$. This is why (\[assm1\]) has to be assumed in [@IP12]; it is a convenient algebraic condition that implies this non-degeneracy. In the absence of (\[assm1\]), the best control we can prove for the output function of (\[duhamel\]) (i.e., the strongest $Z$ norm possible) is uniform bound in the norm $\|f\|_{Z_{jk}'}=2^{5j/6}\|f\|_{L^{2}}$, which is not strong enough to imply 