$$S_{\alpha\beta}^{jk}(u,\partial u)=\sum_{\gamma=1}^{d}\sum_{l=1}^{3}\big(A_{\alpha\beta\gamma}^{jk}u_{\gamma}+B_{\alpha\beta\gamma}^{jkl}\partial_{l}u_{\gamma}\big).$$Note that in the whole time interval $[0,T]$ the $H^{N}$ based norms are small, we thus have$$\mathcal{E}(t)\sim\|\Gamma^{\mu}u(t)\|_{L^{2}}^{2}+\|\Gamma^{\mu}(\partial_{x},\partial_{t})u(t)\|_{L^{2}}^{2}.$$ Now using (\[nlkg\]) and the symmetry assumption of $S$, and integrating by parts, we may compute that $$\begin{aligned}
\frac{1}{2}\partial_{t}\mathcal{E}(t)&=&\frac{1}{2}\sum_{|\mu|\leq N}\sum_{\alpha,\beta=1}^{d}\sum_{j,k=1}^{3}\int_{\mathbb{R}^{3}}\partial_{t}S_{\alpha\beta}^{jk}(u,\partial u)\cdot\partial_{j}\Gamma^{\mu}u_{\alpha}\cdot\partial_{k}\Gamma^{\mu}u_{\beta}\nonumber\\
&-&\sum_{|\mu|\leq N}\sum_{\alpha,\beta=1}^{d}\sum_{j,k=1}^{3}\int_{\mathbb{R}^{3}}\partial_{j}S_{\alpha\beta}^{jk}(u,\partial u)\cdot \partial_{t}\Gamma^{\mu}u_{\alpha}\cdot\partial_{k}\Gamma^{\mu}u_{\beta}\nonumber\\
&+&\sum_{|\mu|\leq N}\sum_{\alpha,\beta=1}^{d}\sum_{j,k=1}^{3}\partial_{t}\Gamma^{\mu}u_{\alpha}\cdot\big[\Gamma^{\mu}(S_{\alpha\beta}^{jk}(u,\partial u)\partial_{j}\partial_{k}u_{\beta})-S_{\alpha\beta}^{jk}(u,\partial u)\Gamma^{\mu}\partial_{j}\partial_{k}u_{\beta}\big]\nonumber\\
&+&\sum_{|\mu|\leq N}\sum_{\alpha=1}^{d}\partial_{t}\Omega^{\mu}u_{\alpha}\cdot\Gamma^{\mu}\mathcal{Q}_{\alpha}'(u,\partial u)\nonumber.\end{aligned}$$ Now, using the equation (\[nlkg\]) again to eliminate $\partial_{t}^{2}$ terms and using Leibniz rule, we can bound the time derivative by $$|\partial_{t}\mathcal{E}(t)|\lesssim\mathcal{E}(t)\cdot\bigg(\sup_{|\mu|\leq N/2}\|\Gamma^{\mu}u(t)\|_{L^{\infty}}+\sup_{|\mu|\leq N/2}\|\Gamma^{\mu}(\partial_{t},\partial_{x})u(t)\|_{L^{\infty}}\bigg).$$ Next, since we have$$\label{recover}\partial_{t}u_{\alpha}=\frac{v_{\alpha}+v_{-\alpha}}{2};\qquad u_{\alpha}=\frac{i}{2}\Lambda_{\alpha}^{-1}(v_{\alpha}-v_{-\alpha}),$$ and also $v_{\sigma}(t)=e^{-{i}t\Lambda_{\sigma}}f_{\sigma}(t)$ for $\nu\in\mathcal{P}$ with the bound $\|f_{\sigma}(t)\|_{X}\lesssim \varepsilon_{1}$, we can use Corollary \[cor\] to deduce that$$\label{decay00}\sup_{|\mu|\leq N/2}\|\Gamma^{\mu}u(t)\|_{L^{\infty}}+\sup_{|\mu|\leq N/2}\|\Gamma^{\mu}(\partial_{t},\partial_{x})u(t)\|_{L^{\infty}}\lesssim\frac{\varepsilon_{1}}{(1+|t|)\log^{20}(2+|t|)},$$ and hence (note that $\mathcal{E}(t)\lesssim\varepsilon_{1}^{2}$)$$|\partial_{t}\mathcal{E}(t)|\lesssim\frac{\varepsilon_{1}^{3}}{(1+|t|)\log^{20}(2+|t|)}.$$ Since $\mathcal{E}(0)\lesssim\varepsilon^{2}$ and the weight $(1+|t|)^{-1}(\log(2+|t|))^{-20}$ is integrable in $t$, this proves (\[resenergy\]).

Duhamel formula, and control of $Z$ norm
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By definition of $v$ and (\[nlkg\]), we know that $(\partial_{t}-{i}\Lambda_{\sigma})v_{\sigma}$ equals a (constant coefficient) quadratic form involving at most the second derivative of $u$. Using also (\[recover\]) and Duhamel formula, we obtain the equation $$\label{duhamelforf}\widehat{f_{\sigma}}(t,\xi)-\widehat{f_{\sigma}}(0,\xi)=\sum_{\mu,\nu\in\mathcal{P}}\int_{0}^{t}\int_{\mathbb{R}^{3}}e^{{i}s\Phi_{\sigma\mu\nu}(\xi,\eta)}{m}_{\sigma\mu\nu}(\xi,\eta)\widehat{f_{\mu}}(s,\xi-\eta)\widehat{f_{\nu}}(s,\eta)\,\mathrm{d}\eta\mathrm{d}s$$ for each $\sigma\in\mathcal{P}$. The weight $$\label{weightm}{m}_{\sigma\mu\nu}=\sum_{i=1}^{20}\sum_{k,k_{1},k_{2}}(1+2^{\max(k,k_{1},k_{2})})\psi_{kk_{1}k_{2}}^{\sigma\mu\nu,i,0}\bigg(\frac{\xi}{2^{k}}\bigg)\psi_{kk_{1}k_{2}}^{\sigma\mu\nu,i,1}\bigg(\frac{\xi-\eta}{2^{k_{1}}}\bigg)\psi_{kk_{1}k_{2}}^{\sigma\mu\nu,i,2}\bigg(\frac{\eta}{2^{k_{2}}}\bigg),$$ where the $\psi$’s have the same compact support and are bounded uniformly in $\mathcal{G}_{6}$.

The proof of Proposition \[bootstrap\] is now reduced to the following $Z$ norm estimate.

\[main01\] Fix a choice of $(\sigma,\mu,\nu)$. 