Proposition: The Cauchy pairing is the leading onset coefficient

Let ${k_1}^{\mu }=(1,\mathbf{n})$, ${k_2}^{\mu }=(1,-\mathbf{n})$, $|\mathbf{n}|=1$, $\lambda >0$, ${{\gamma }_a}^{\lambda }={ϵ}_a{e}^{i\lambda k_a\cdot y}$ with constant symmetric ${ϵ}_a$, and $\varphi$ smooth. Put ${\bar{\beta }}_{\mu \nu }(t):={\int }_{{\mathbb{R}}^3}\beta [f,g,\varphi {]}_{\mu \nu }(t,\mathbf{x}){\mathrm{d}}^{3}x$, smooth on $[0,T]$ and vanishing near $T$. Then

with $B_{\Sigma }({{\gamma }_1}^{\lambda },{{\gamma }_2}^{\lambda };\varphi )={\lambda }^2{\mathsf{q}}^{\mu \nu }{\bar{\beta }}_{\mu \nu }(0)$. The left side is a class invariant of $[u{]}_{\mathrm{sol}}$.

Proof. $J_3({{\gamma }_1}^{\lambda },{{\gamma }_2}^{\lambda })={\lambda }^2\mathsf{q}{e}^{i\lambda (k_1+k_2)\cdot y}={\lambda }^2\mathsf{q}{e}^{2i\lambda t}$, independent of $\mathbf{x}$; insert in (eq:upair) and integrate over $\mathbf{x}$ first. Two integrations by parts in $t$, using $\bar{\beta }(T)={\bar{\beta }}^{\prime }(T)=0$, give ${{\int }_0}^T{\overline{\beta e}}^{2i\lambda t}\mathrm{d}t=-\bar{\beta }(0)/(2i\lambda )+O({\lambda }^{-2})$, and $-1/(2i)=i/2$. The value of $B_{\Sigma }$ on the pair is (eq:Bsig) with $J_3{|}_{\Sigma }={\lambda }^2\mathsf{q}$. Invariance is Theorem thm:criterion(a).