2.1 Proof of Lemma 2.1

We first present the proof of the sharp estimate (2.4), then that of formula (2.5).

2.1.1 Proof of formula (2.4)

Let I denote the integral in formula (2.4). Decompose the four factors into Littlewood–Paley pieces so that

$$ \begin{align*} I=\sum_{M_{1},M_{2},M_{3},M}I_{M_{1},M_{2},M_{3},M}, \end{align*} $$

where

$$ \begin{align*} I_{M_{1},M_{2},M_{3},M}=\iint_{x,t}u_{1,M_{1}}u_{2,M_{2}}u_{3,M_{3}}g_{M}dxdt, \end{align*} $$

with $u_{j,M_{j}}=P_{M_{j}}u_{j}$ and $g_{M}=P_{M}g$. As $M_{2}$, $M_{3}$ and M are symmetric, it suffices to take care of the $M_{1}\sim M_{2}\geq M_{3}\geq M$ case. Decompose the $M_{1}$ and $M_{2}$ dyadic spaces into $ M_{3}$-size cubes; then

$$ \begin{align*} I_{1A} &\lesssim \sum_{\substack{ M_{1},M_{2},M_{3},M \\ M_{1}\sim M_{2}\geq M_{3}\geq M}}\sum_{Q}\left\lVert P_{Q}u_{1,M_{1}}P_{Q_{c}}u_{2,M_{2}}u_{3,M_{3}}g_{M}\right\rVert _{L_{t,x}^{1}} \\ &\lesssim \sum_{\substack{ M_{1},M_{2},M_{3},M \\ M_{1}\sim M_{2}\geq M_{3}\geq M}}\sum_{Q}\left\lVert P_{Q}u_{1,M_{1}}\right\rVert _{L_{t,x}^{\frac{ 10}{3}}}\left\lVert P_{Q_{c}}u_{2,M_{2}}\right\rVert _{L_{t,x}^{\frac{10}{3} }}\left\lVert u_{3,M_{3}}\right\rVert _{L_{t,x}^{\frac{10}{3}}}\left\lVert g_{M}\right\rVert _{L_{t,x}^{10}}. \end{align*} $$

Using formulas (2.12) and (2.13),

$$ \begin{align*} &\lesssim \sum_{\substack{ M_{1},M_{2},M_{3},M \\ M_{1}\sim M_{2}\geq M_{3}\geq M}}\sum_{Q}M_{3}^{\frac{2}{5}}\left\lVert P_{Q}u_{1,M_{1}}\right\rVert _{Y^{0}}\left\lVert u_{3,M_{3}}\right\rVert _{Y^{0}}M_{3}^{\frac{1}{5}}\left\lVert P_{Q_{c}}u_{2,M_{2}}\right\rVert _{Y^{0}}M^{\frac{7}{5}}\left\lVert g_{M}\right\rVert _{Y^{0}} \\ &\lesssim \sum_{\substack{ M_{1},M_{2},M_{3},M \\ M_{1}\sim M_{2}\geq M_{3}\geq M}}M_{3}^{\frac{3}{5}}M^{\frac{7}{5}}\left\lVert g_{M}\right\rVert _{Y^{0}}\left\lVert u_{3,M_{3}}\right\rVert _{Y^{0}}\sum_{Q}\left\lVert P_{Q}u_{1,M_{1}}\right\rVert _{Y^{0}}\left\lVert P_{Q_{c}}u_{2,M_{2}}\right\rVert _{Y^{0}}. \end{align*} $$

Applying Cauchy–Schwarz to sum in Q, we have

$$ \begin{align*} &\lesssim \sum_{\substack{ M_{1},M_{2},M_{3},M \\ M_{1}\sim M_{2}\geq M_{3}\geq M}}M_{3}^{\frac{3}{5}}M^{\frac{7}{5}}\left\lVert u_{1,M_{1}}\right\rVert _{Y^{0}}\left\lVert u_{2,M_{2}}\right\rVert _{Y^{0}}\left\lVert u_{3,M_{3}}\right\rVert _{Y^{0}}\left\lVert g_{M}\right\rVert _{Y^{0}} \\ &\lesssim \sum_{\substack{ M_{1},M_{2} \\ M_{1}\sim M_{2}}} M_{2}M_{1}^{-1}\left\lVert u_{1,M_{1}}\right\rVert _{Y^{-1}}\left\lVert u_{2,M_{2}}\right\rVert _{Y^{1}}\sum_{\substack{ M_{3},M \\ M_{1}\sim M_{2}\geq M_{3}\geq M}}M_{3}^{-\frac{2}{5}}M^{\frac{2}{5}}\left\lVert u_{3,M_{3}}\right\rVert _{Y^{1}}\left\lVert g_{M}\right\rVert _{Y^{1}}. \end{align*} $$

We are done, by Schur’s test.

2.1.2 Proof of formula (2.5)

We reuse the setup from the proof of formula (2.4). However, due to the symmetry assumption $M_{1}\geq M_{2}\geq M_{3}$ on the frequencies in the proof of formula (2.4), we cannot simply assume that $P_{\leq M_{0}}$ lands on $u_{2}$. The worst-gain (least-gain) case here would be that $u_{1}$ is still put in $Y^{-1}$ and $ P_{\leq M_{0}}$ is applied to $u_{3}$. Thus we will prove estimate (2.5) subject to the extra localisation that $ P_{\leq M_{0}}$ be applied on $u_{3}$. By symmetry in $M_{2}$ and M, it suffices to take care of two cases: A, $M_{1}\sim M_{2}\geq M_{3}\geq M$, and B, $M_{1}\sim M_{2}\geq M\geq M_{3}$. We will get a $ T^{\frac {1}{4}}M_{0}^{\frac {3}{5}}$ in case A and a $T^{\frac {1}{7}}M_{0}^{ \frac {3}{7}}$ in case B. Since formula (2.5) is nowhere near optimal, and we just need it to hold with some powers of T and $M_{0}$, there is no need to match these powers or pursue the best power in these cases.

Case A: $M_{1}\sim M_{2}\geq M_{3}\geq M$.

Decompose the $M_{1}$ and $M_{2}$ dyadic spaces into $M_{3}$-size cubes:

$$ \begin{align*} I_{M_{1},M_{2},M_{3},M} &\leq \sum_{Q}\left\lVert P_{Q}u_{1,M_{1}}P_{Q_{c}}u_{2,M_{2}}\left( P_{\leq M_{0}}u_{3,M_{3}}\right) g_{M}\right\rVert _{L_{t,x}^{1}} \\ &\leq \sum_{Q}\left\lVert P_{Q}u_{1,M_{1}}g_{M}\right\rVert _{L_{t,x}^{2}}\left\lVert P_{Q_{C}}u_{2,M_{2}}\right\rVert _{L_{t,x}^{4}}\left\lVert P_{\leq M_{0}}u_{3,M_{3}}\right\rVert _{L_{t,x}^{4}}, \end{align*} $$

where

$$ \begin{align*} \left\lVert P_{\leq M_{0}}u_{3,M_{3}}\right\rVert _{L_{t,x}^{4}}\leq T^{\frac{1}{4}}M_{0}^{\frac{3}{5}}M_{3}^{\frac{2}{5}}\left\lVert P_{\leq M_{0}}u_{3,M_{3}}\right\rVert _{L_{t}^{\infty }L_{x}^{2}}\lesssim T^{\frac{1}{ 4}}M_{0}^{\frac{3}{5}}M_{3}^{\frac{2}{5}}\left\lVert P_{\leq M_{0}}u_{3,M_{3}}\right\rVert _{Y^{0}}. \end{align*} $$

Using formulas (2.12) and (2.13),

$$ \begin{align*} I_{M_{1},M_{2},M_{3},M}\lesssim T^{\frac{1}{4}}M_{0}^{\frac{3}{5} }\sum_{Q}\left( M\left\lVert P_{Q}u_{1,M_{1}}\right\rVert _{Y^{0}}\left\lVert g_{M}\right\rVert _{Y^{0}}\right) M_{3}^{\frac{1}{2}}\left\lVert P_{Q_{C}}u_{2,M_{2}}\right\rVert _{Y^{0}}M_{3}^{\frac{2}{5}}\left\lVert P_{\leq M_{0}}u_{3,M_{3}}\right\rVert _{Y^{0}}. \end{align*} $$

Note that we actually used a bilinear estimate for the first factor, but did not record or use the bilinear gain factor. Using Cauchy–Schwarz to sum in Q, we have

$$ \begin{align*} I_{M_{1},M_{2},M_{3},M}\lesssim T^{\frac{1}{4}}M_{0}^{\frac{3}{5}}\left\lVert u_{1,M_{1}}\right\rVert _{Y^{0}}\left\lVert g_{M}\right\rVert _{Y^{1}}\left\lVert u_{2,M_{2}}\right\rVert _{Y^{0}}M_{3}^{\frac{9}{10} }\left\lVert P_{\leq M_{0}}u_{3,M_{3}}\right\rVert _{Y^{0}}. \end{align*} $$

Thus, summing in M nonoptimally gives

$$ \begin{align*} I_{1A} &\lesssim T^{\frac{1}{4}}M_{0}^{\frac{3}{5}}\left\lVert g\right\rVert _{Y^{1}}\sum_{\substack{ M_{1},M_{2},M_{3} \\ M_{1}\sim M_{2}\geq M_{3} }}\left\lVert u_{1,M_{1}}\right\rVert _{Y^{0}}\left\lVert u_{2,M_{2}}\right\rVert _{Y^{0}}M_{3}^{\frac{9}{10}}\log M_{3}\left\lVert P_{\leq M_{0}}u_{3,M_{3}}\right\rVert _{Y^{0}} \\ &\lesssim T^{\frac{1}{4}}M_{0}^{\frac{3}{5}}\left\lVert g\right\rVert _{Y^{1}}\sum_{\substack{ M_{1},M_{2},M_{3} \\ M_{1}\sim M_{2}\geq M_{3} }}\left\lVert u_{1,M_{1}}\right\rVert _{Y^{0}}\left\lVert u_{2,M_{2}}\right\rVert _{Y^{0}}\frac{M_{3}^{\frac{9}{10}}\log M_{3}}{M_{3}} \left\lVert P_{\leq M_{0}}u_{3,M_{3}}\right\rVert _{Y^{1}}. \end{align*} $$

Again, summing in $M_{3}$ nonoptimally and swapping a derivative between $ u_{1}$ and $u_{2}$ give

$$ \begin{align*} I_{1A} &\lesssim T^{\frac{1}{4}}M_{0}^{\frac{3}{5}}\left\lVert g\right\rVert _{Y^{1}}\left\lVert P_{\leq M_{0}}u_{3}\right\rVert _{Y^{1}}\sum _{\substack{ M_{1},M_{2} \\ M_{1}\sim M_{2}}}\left\lVert u_{1,M_{1}}\right\rVert _{Y^{-1}}\left\lVert u_{2,M_{2}}\right\rVert _{Y^{1}} \\ &\lesssim T^{\frac{1}{4}}M_{0}^{\frac{3}{5}}\left\lVert u_{1}\right\rVert _{Y^{-1}}\left\lVert u_{2}\right\rVert _{Y^{1}}\left\lVert P_{\leq M_{0}}u_{3}\right\rVert _{Y^{1}}\left\lVert g\right\rVert _{Y^{1}}. \end{align*} $$

Case B: $M_{1}\sim M_{2}\geq M\geq M_{3}$.

Sum up $M_{3}$ first. We then consider

$$ \begin{align*} I_{M_{1},M_{2},M}=\iint_{x,t}u_{1,M_{1}}u_{2,M_{2}}\left( P_{\leq M}P_{\leq M_{0}}u_{3}\right) g_{M}dxdt. \end{align*} $$

Decompose the $M_{1}$ and $M_{2}$ dyadic spaces into M-size cubes:

$$ \begin{align*} I_{M_{1},M_{2},M} &\leq \sum_{Q}\left\lVert P_{Q}u_{1,M_{1}}P_{Q_{c}}u_{2,M_{2}}\left( P_{\leq M}P_{\leq M_{0}}u_{3}\right) g_{M}\right\rVert _{L_{t,x}^{1}} \\ &\leq \sum_{Q}\left\lVert P_{Q}u_{1,M_{1}}\right\rVert _{L_{t,x}^{\frac{7 }{2}}}\left\lVert P_{Q_{C}}u_{2,M_{2}}\right\rVert _{L_{t,x}^{\frac{7}{2} }}\left\lVert P_{\leq M}P_{\leq M_{0}}u_{3}\right\rVert _{L_{t,x}^{7}}\left\lVert g_{M}\right\rVert _{L_{t,x}^{\frac{7}{2}}}, \end{align*} $$

where

$$ \begin{align*} \left\lVert P_{\leq M}P_{\leq M_{0}}u_{3}\right\rVert _{L_{t,x}^{7}} &\leq T^{\frac{1}{7}}\left\lVert P_{\leq M}P_{\leq M_{0}}u_{3}\right\rVert _{L_{t}^{\infty }L_{x}^{7}} \\ &\lesssim T^{\frac{1}{7}}M_{0}^{\frac{3}{7}}\left\lVert P_{\leq M}P_{\leq M_{0}}u_{3}\right\rVert _{L_{t}^{\infty }H_{x}^{1}} \\ &\lesssim T^{\frac{1}{7}}M_{0}^{\frac{3}{7}}\left\lVert P_{\leq M}P_{\leq M_{0}}u_{3}\right\rVert _{Y^{1}}. \end{align*} $$

Applying formulas (2.12) and (2.13),

$$ \begin{align*} I_{M_{1},M_{2},M}\lesssim T^{\frac{1}{7}}M_{0}^{\frac{3}{7}}\left\lVert P_{\leq M}P_{\leq M_{0}}u_{3}\right\rVert _{Y^{1}}\sum_{Q}M^{\frac{2 }{7}}\left\lVert P_{Q}u_{1,M_{1}}\right\rVert _{Y^{0}}M^{\frac{2}{7} }\left\lVert P_{Q_{C}}u_{2,M_{2}}\right\rVert _{Y^{0}}M^{\frac{2}{7} }\left\lVert g_{M}\right\rVert _{Y^{0}}. \end{align*} $$

Applying Cauchy–Schwarz to sum in Q, we have

$$ \begin{align*} I_{M_{1},M_{2},M}\lesssim T^{\frac{1}{7}}M_{0}^{\frac{3}{7}}\left\lVert P_{\leq M}P_{\leq M_{0}}u_{3}\right\rVert _{Y^{1}}M^{-\frac{1}{7} }\left\lVert u_{1,M_{1}}\right\rVert _{Y^{0}}\left\lVert u_{2,M_{2}}\right\rVert _{Y^{0}}\left\lVert g_{M}\right\rVert _{Y^{1}}. \end{align*} $$

Thus, swapping a derivative between $u_{1}$ and $u_{2}$ gives

$$ \begin{align*} I_{1B}\lesssim T^{\frac{1}{7}}M_{0}^{\frac{3}{7}}\left\lVert P_{\leq M_{0}}u_{3}\right\rVert _{Y^{1}}\sum_{\substack{ M_{1},M_{2},M \\ M_{1}\sim M_{2}\geq M}}M^{-\frac{1}{7}}\left\lVert u_{1,M_{1}}\right\rVert _{Y^{-1}}\left\lVert u_{2,M_{2}}\right\rVert _{Y^{1}}\left\lVert g_{M}\right\rVert _{Y^{1}}. \end{align*} $$

Burning that $\frac {1}{7}$-derivative to sum in M and then applying Cauchy–Schwarz in $M_{1}$, we have

$$ \begin{align*} I_{1B}\lesssim T^{\frac{1}{7}}M_{0}^{\frac{3}{7}}\left\lVert P_{\leq M_{0}}u_{3}\right\rVert _{Y^{1}}\left\lVert u_{1}\right\rVert _{Y^{-1}}\left\lVert u_{2}\right\rVert _{Y^{1}}\left\lVert g\right\rVert _{Y^{1}}, \end{align*} $$

as needed.

3.1 Proof of Lemma 3.3 and uniform-in-time frequency localisation for the NLS

We provide a direct proof using the equation, even though one could look for a more abstract proof. By substituting the equation, we compute

$$ \begin{align*} \left\lvert \partial _{t}\left\lVert \nabla P_{\leq M}u\right\rVert _{L_{x}^{2}}^{2}\right\rvert & =2\left\lvert \operatorname{Im} \int P_{\leq M}\nabla u\cdot P_{\leq M}\nabla \left(\lvert u\rvert^{2}u\right)dx\right\rvert \\ & \leq 2\left\lVert P_{\leq M}\nabla u\right\rVert _{L^{4}}\left\lVert P_{\leq M}\nabla \left(\lvert u\rvert^{2}u\right)\right\rVert _{L^{4/3}} \\ & =2M^{2}\left\lVert \tilde{P}_{\leq M}u\right\rVert _{L^{4}}\left\lVert \tilde{P}_{\leq M}\left(\lvert u\rvert^{2}u\right)\right\rVert _{L^{4/3}}, \end{align*} $$

where if the symbol associated to $P_{\leq M}$ is $\chi (\xi /M)$, then the symbol associated to $\tilde {P}_{\leq M}$ is $\tilde {\chi }(\xi /M)$, with $ \tilde {\chi }(\xi )=\xi \chi (\xi )$. By the $L^{p}\rightarrow L^{p}$ boundedness of the Littlewood–Paley projections (see, for example, [Reference Herr and Sohinger35, Appendix]),

$$ \begin{align*} \left\lvert \partial _{t}\left\lVert \nabla P_{\leq M}u\right\rVert _{L_{x}^{2}}^{2}\right\rvert \lesssim M^{2}\lVert u\rVert _{L^{4}}^{4}. \end{align*} $$

By Sobolev embedding,

$$ \begin{align*} \left\lvert \partial _{t}\left\lVert \nabla P_{\leq M}u\right\rVert _{L_{x}^{2}}^{2}\right\rvert \lesssim M^{2}\lVert u\rVert _{H^{1}}^{4}. \end{align*} $$

Hence there exists $\delta ^{\prime }>0$ – depending on M, $\lVert u\rVert _{L_{\left [0,T\right ]}^{\infty }H^{1}}$ and $\varepsilon $ – such that for any $ t_{0}\in \lbrack 0,T]$, it holds that for any $t\in (t_{0}-\delta ^{\prime },t_{0}+\delta ^{\prime })\cap \lbrack 0,T]$,

(3.11)$$ \begin{align} \left\lvert \left\lVert \nabla P_{\leq M}u(t)\right\rVert _{L_{x}^{2}}^{2}-\left\lVert \nabla P_{\leq M}u(t_{0})\right\rVert _{L_{x}^{2}}^{2}\right\rvert \leq \tfrac{1}{16} \varepsilon ^{2}. \end{align} $$

On the other hand, since $u\in C_{\left [0,T\right ]}^{0}H_{x}^{1}$, for each $t_{0}$ there exists $\delta ''>0$ such that for any $t\in (t_{0}-\delta '',t_{0}+\delta '')\cap \lbrack 0,T]$,

(3.12)$$ \begin{align} \left\lvert \left\lVert \nabla u(t)\right\rVert _{L_{x}^{2}}^{2}-\left\lVert \nabla u(t_{0})\right\rVert _{L_{x}^{2}}^{2}\right\rvert \leq \tfrac{1}{16}\varepsilon ^{2}. \end{align} $$

Note that $\delta ''$ depends on u itself (or the ‘modulus of continuity’ of u), unlike $ \delta '$, which depends only on M, $\lVert u\rVert _{L_{\left [0,T\right ]}^{\infty }H^{1}}$ and $\varepsilon $. Now let $\delta =\min (\delta ',\delta '')$. Then by formulas (3.11) and (3.12), we have that for any $t\in (t_{0}-\delta ,t_{0}+\delta )\cap \lbrack 0,T]$,

$$ \begin{align*} \left\lvert \left\lVert \nabla P_{>M}u(t)\right\rVert _{L_{x}^{2}}^{2}-\left\lVert \nabla P_{>M}u(t_{0})\right\rVert _{L_{x}^{2}}^{2}\right\rvert \leq \tfrac{1}{4}\varepsilon ^{2}. \end{align*} $$

For each $t\in \lbrack 0,T]$, there exists $M_{t}$ such that

$$ \begin{align*} \left\lVert \nabla P_{>M_{t}}u(t)\right\rVert _{L_{x}^{2}}\leq \tfrac{1}{2}\varepsilon. \end{align*} $$

By the foregoing, there exists $\delta _{t}>0$ such that on $(t-\delta _{t},t+\delta _{t})$, we have

$$ \begin{align*} \left\lVert \nabla P_{>M_{t}}u\right\rVert _{L_{\left(t-\delta _{t},t+\delta _{t}\right)}^{\infty }L_{x}^{2}}\leq \varepsilon. \end{align*} $$

Here, $\delta _{t}>0$ depends on u and $M_{t}$. The collection of intervals $(t-\delta _{t},t+\delta _{t})$ as t ranges over $[0,T]$ is an open cover of $[0,T]$. Let

$$ \begin{align*} \left(t_{1}-\delta _{t_{1}},t_{1}+\delta _{t_{1}}\right),\dotsc, \left(t_{J}-\delta _{t_{J}},t_{J}+\delta _{t_{J}}\right) \end{align*} $$

be an open cover of $[0,T]$. Letting

$$ \begin{align*} M=\max \left(M_{t_{1}},\dotsc ,M_{t_{J}}\right), \end{align*} $$

we have established formula (3.3).

Now conversely suppose that $u\in C_{\left [0,T\right ]}^0H_{x,\text {weak}}^1\cap C_{\left [0,T\right ]}^1H_{x,\text {weak}}^{-1}$ and satisfies formula (3.3). Then we claim that $u\in C_{\left [0,T\right ]}^0H_x^1 \cap C_{\left [0,T\right ]}^1 H_x^{-1}$. Let $t_0\in [0,T]$ be arbitrary. If u is not strongly continuous at $t_0$, then there exist $\epsilon>0$ and a sequence $t_k \to t_0$ such that $\lVert u(t_k) - u(t_0) \rVert _{H_x^1}> 2\epsilon $. Then for each k, there exists $\phi _k \in H_x^{-1} $ with $\lVert \phi _k \rVert _{H_x^{-1}} \leq 1$ and

(3.13)$$ \begin{align} \lvert\langle u(t_k) - u(t_0) , \phi_k \rangle\rvert> 2\epsilon. \end{align} $$

Get M as in formula (3.3). Then

(3.14)$$ \begin{align} \left\lvert\left\langle u(t_k)-u(t_0) , P_{>M} \phi_k \right\rangle \right\rvert \leq \epsilon. \end{align} $$

On the other hand, by the Rellich–Kondrachov compactness theorem, there exists a subsequence such that $P_{\leq M} \phi _k \to \phi $ in $H_x^{-1}$. This combined with the assumption that u is weakly continuous implies that

(3.15)$$ \begin{align} \left\langle u(t_k) - u(t_0) , P_{\leq M} \phi_k \right\rangle \to 0. \end{align} $$

But formulas (3.14) and (3.15) contradict formula (3.13). The proof that $\partial _t u$ is strongly continuous is similarly straightforward.

4.1 A more elaborated proof of Lemma 3.6

Let us first give a brief review of the original KM board game, which since its invention has been used in every paper involving the analysis of Gross–Pitaevskii hierarchies. Recall the notation of $ \mu $ in Lemma 3.6: $\left \{ \mu \right \} $ is a set of maps from $\{2,\dotsc ,k+1\}$ to $\{1,\dotsc ,k\}$ satisfying $\mu (2)=1$ and $\mu (l)<l$ for all $l,$ and

$$ \begin{align*} J_{\mu}^{(k+1)}\left(f^{(k+1)}\right)\left(t_{1},\underline{t}_{k+1}\right) &=U^{(1)}(t_{1}-t_{2})B_{1,2}U^{(2)}(t_{2}-t_{3})B_{\mu(3),3}\dotsm \\ &\dotsm U^{(k)}(t_{k}-t_{k+1})B_{\mu(k+1),k+1}\left(f^{(k+1)}(t_{k+1})\right). \end{align*} $$

Example 1. An example of $\mu $ when $k=5$ is

$$ \begin{align*} \begin{array}{c|c@{\kern10pt}c@{\kern10pt}c@{\kern10pt}c@{\kern10pt}c} j & 2 & 3 & 4 & 5 & 6 \\ \hline \mu & 1 & 1 & 3 & 2 & 1 \\ \end{array}. \end{align*} $$

If $\mu $ satisfies $\mu (j)\leq \mu (j+1)$ for $2\leq j\leq k$ in addition to $\mu (j)<j$ for all $2\leq j\leq k+1$, then it is in upper-echelon form Footnote ¹⁶ in the terminology of [Reference Klainerman and Machedon47].

Let $\mu $ be a collapsing map as already defined and $\sigma $ a permutation of $\{2,\dotsc ,k+1\}$. A Klainerman–Machedon acceptable move, which we denote $\text {KM}(j,j+1)$, is allowed when $\mu (j)\neq \mu (j+1)$ and $ \mu (j+1)<j$, and is the following action: $(\mu ',\sigma ')=\text {KM}(j,j+1)(\mu ,\sigma )$:

$$ \begin{align*} \mu' & =(j,j+1)\circ \mu \circ (j,j+1), \\ \sigma' & =(j,j+1)\circ \sigma. \end{align*} $$

A key observation of Klainerman and Machedon [Reference Klainerman and Machedon47] is that if $(\mu ',\sigma ')=\text {KM}(j,j+1)(\mu ,\sigma )$ and $f^{(k+1)}$ is a symmetric density, then

(4.1)$$ \begin{align} J_{\mu'}^{(k+1)}\left(f^{(k+1)}\right)\left(t_{1},\sigma^{\prime\,-1}\left(\underline{t}_{k+1}\right)\right)=J_{\mu }^{(k+1)}\left(f^{(k+1)}\right)\left(t_{1},\sigma ^{-1}\left(\underline{t}_{k+1}\right)\right), \end{align} $$

where for $\underline {t}_{k+1}=(t_{2},\dotsc ,t_{k+1})$ we define

$$ \begin{align*} \sigma ^{-1}\left(\underline{t}_{k+1}\right)=\left(t_{\sigma ^{-1}(2)},\dotsc ,t_{\sigma ^{-1}(k+1)}\right). \end{align*} $$

Associated to each $\mu $ and $\sigma $, we define the Duhamel integrals

(4.2)$$ \begin{align} I\left(\mu ,\sigma ,f^{(k+1)}\right)(t_{1})=\int_{t_{1}\geq t_{\sigma (2)}\geq \dotsb \geq t_{\sigma (k+1)}}J_{\mu }^{(k+1)}\left(f^{(k+1)}\right)\left(t_{1},\underline{t}_{k+1}\right)d\underline{t}_{k+1}. \end{align} $$

It follows from equation (4.1) that

$$ \begin{align*} I\left(\mu',\sigma',f^{(k+1)}\right)=I\left(\mu ,\sigma ,f^{(k+1)}\right). \end{align*} $$

It is clear that we can combine Klainerman–Machedon acceptable moves as follows: If $\rho $ is a permutation of $\{2,\dotsc ,k+1\}$ such that it is possible to write it as a composition of transpositions

$$ \begin{align*} \rho =\tau _{1}\circ \dotsb \circ \tau _{r}, \end{align*} $$

for which each operator $\text {KM}\left (\tau _{j}\right )$ on the right side of

$$ \begin{align*} \text{KM}(\rho )\overset{\text{def}}{=}\text{KM}(\tau _{1})\circ \dotsb \circ \text{KM}(\tau _{r}) \end{align*} $$

is an acceptable action, then $\text {KM}(\rho )$, defined by this composition, is acceptable as well. In this case, $(\mu ',\sigma ')=\text {KM}(\rho )(\mu ,\sigma )$ and

$$ \begin{align*} \mu' & = \rho \circ \mu \circ \rho^{-1}, \\ \sigma' & =\rho \circ \sigma, \end{align*} $$

and equations (4.1) and (4.2) hold. If $\mu $ and $\mu ^{\prime }$ are such that there exists $\rho $ for which $(\mu ',\sigma ')=\text {KM}(\rho )(\mu ,\sigma )$, then we say that $\mu '$ and $\mu $ are KM-relatable. This is an equivalence relation that partitions the set of collapsing maps into equivalence classes.

In short, one can describe the KM board game in [Reference Klainerman and Machedon47] which combines the $k!$ many terms in $J^{(k+1)}\left (f^{(k+1)}\right )$ as follows:

The key take away in Algorithm 1 is that, although it is very much not obvious, quite a few of the summands in $J^{(k+1)}\left (f^{(k+1)}\right )$ actually have the same integrand if one switches the variable labellings in a clever way. Algorithm 1 leaves only one ambiguity – the time integration domain $D_{m}$ – which is obviously very complicated for large $k,$ as it is a union of a very large number of simplexes in high dimension under the action of a proper subset of the permutation group $ S_{k} $ depending on the integrand. So far, for the analysis of GP hierarchies on $\mathbb {R}^{d}/$$\mathbb {T}^{d}$, $d\leq 3$, knowing $ D_{m}\subset \left [ 0,1\right ] ^{k}$ has been enough, as the related $ L_{t}^{1}H^{s}$ estimates are true. $\mathbb {T}^{4}$ appears to be the first domain on which one has to know what $D_{m}$ is so that one can at least have a chance to use space-time norms like $X_{s,b}$ and U-V, as the related $L_{t}^{1}H^{s}$ estimates are difficult to prove and may not even be true.

It turns out that $D_{m}$ is in fact simple, as we will see. We now present a more elaborated proof of Lemma 3.6, in which $D_{m}$ is computed in a clear way. Given a $\mu $, and hence a summand inside $J^{(k+1)}\left (f^{(k+1)}\right )$, we construct a binary tree with the following algorithm:

Example 2.

Let us work with the following exampleFootnote ¹⁷:

$$ \begin{align*} \begin{array}{c|c@{\kern10pt}c@{\kern10pt}c@{\kern10pt}c@{\kern10pt}c} j & 2 & 3 & 4 & 5 & 6 \\ \hline \mu_{\text{out}} & 1 & 1 & 1 & 2 & 3\\ \end{array} \end{align*} $$

We start with $j=2$, and note that $\mu _{\text {out}}(2)=1$, so we need to find minimal $a>2, b>2$, such that $\mu (a)=1$ and $\mu (b)=2$. In this case, it is $a=3$ and $b=5$, so we put those as the left and right children of $2$, respectively, in the tree shown at left.

Now we move to $j=3$. Since $\mu _{\text {out}}(3)=1$, we find minimal a and b so that $a>3$, $b>3$, $\mu (a)=1$ and $\mu (b)=3$. We find that $ a=4$ and $b=6$, so we put these as the left and right children of $3$, respectively, in the tree shown at left. Since all indices appear in the tree, it is complete.

For example, the skeleton of the tree in Example 2 is shown at left.

By the hierarchy structure, Algorithm 2, which produces a tree from a $\mu $, produces only admissible trees. As we have made a distinction between left and right children in the algorithm, the procedure is reversible – given an admissible binary tree, we can uniquely reconstruct the $\mu $ that generated it.

Example 3. Suppose we are given the following tree:

Using the fact that for every right child, $\mu $ maps the child value to the parent value, we fill in the following values in the $\mu $ table:

$$ \begin{align*} \begin{array}{c|c@{\kern10pt}c@{\kern10pt}c@{\kern10pt}c@{\kern10pt}c@{\kern10pt}c@{\kern10pt}c@{\kern10pt}c} j & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \\ \hline \mu & 1 & & & 2 & 3 & 4 & 6 & \\ \end{array} \end{align*} $$

Now we use the left-child rule and note that since $3$ is a left child of $2$ and $\mu (2)=1$, we must have $\mu (3)=1$, and so on, to recover the following table:

$$ \begin{align*} \begin{array}{c|c@{\kern10pt}c@{\kern10pt}c@{\kern10pt}c@{\kern10pt}c@{\kern10pt}c@{\kern10pt}c@{\kern10pt}c} j & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \\ \hline \mu & 1 & 1 & 1 & 2 & 3 & 4 & 6 & 6 \\ \end{array} \end{align*} $$

One can show that in the tree representation of $\mu $, an acceptable move defined in [Reference Klainerman and Machedon47] is the operation which switches the labels of two nodes with consecutive labels on an admissible tree, provided that the outcome is still an admissible tree, by writing out the related trees on [Reference Klainerman and Machedon47, pp. 180–182]. For example, interchanging the labelling of 5 and 6 in the tree in Example 2 is an acceptable move. That is, acceptable moves in [Reference Klainerman and Machedon47] preserve the tree structures but permute the labelling under the admissibility requirement. Two collapsing maps $\mu $ and $\mu '$ are KM-relatable if and only if the trees corresponding to $\mu $ and $\mu '$ have the same skeleton.

Given k, we would like to have the number of different binary tree structures of k nodes. This number is exactly defined as the Catalan number and is controlled by $4^{k}$. Hence, we have just provided a proof of the original KM board game, neglecting the trees showing the effects of acceptable moves on a tree.

But now let us get to the main ‘elaborate’ part, namely, how to compute $D_{m}$ for a given upper-echelon class. To this end, we need to define what is an upper-echelon form. Though the requirement $\mu (j)\leq \mu (j+1)$ for $2\leq j\leq k$ is good enough, we give an algorithm which produces the upper-echelon tree given the tree structure, as the tree representation of an upper-echelon form is in fact labelled in sequential order (see, for example, the tree in Example 2).

We define a map $T_{D}$ which maps an upper-echelon tree to a time integration domain (a set of inequality relations) by

(4.3)$$ \begin{align} T_{D}(\alpha )=\left\{ t_{j}\geq t_{k}:j,k\text{ are labels on nodes of }\alpha \text{ such that the }k\text{ node is a child of the }j\text{ node} \right\}, \end{align} $$

where $\alpha $ is an upper-echelon tree. We then have the integration domain as follows:

Proposition 4.3. Given a $\mu _{m}$ in upper-echelon form, we have

$$ \begin{align*} \sum_{\mu \sim \mu _{m}}\int_{t_{1}\geq t_{2}\geq t_{3}\geq \dotsb\geq t_{k+1}}J_{\mu }^{(k+1)}\left(f^{(k+1)}\right)\left(t_{1},\underline{t}_{k+1}\right)d \underline{t}_{k+1}=\int_{T_{D}\left(\mu _{m}\right)}J_{\mu _{m}}^{(k+1)}\left(f^{(k+1)}\right)\left(t_{1},\underline{t}_{k+1}\right)d\underline{t}_{k+1}. \end{align*} $$

Here, $\mu \sim \mu _{m}$ means that $\mu $ is equivalent to $\mu _{m}$ under acceptable moves (the trees representing $\mu $ and $\mu _{m}$ have the same structure) and $T_{D}(\mu _{m})$ is the domain defined in equation (4.3).

Proof. We prove by an example, as the notation is already heavy. For the general case, one merely needs to rewrite $\Sigma _{1}$ and $\Sigma _{2}$, to be defined in this proof. The key is the admissible condition or the simple requirement that the child must carry a larger label than the parent.

Recall the upper-echelon tree in Example 2 and denote it with $\alpha $. Here are all the admissible trees equivalent to $\alpha $:

We first read by definition that

$$ \begin{align*} T_{D}(\alpha )=\{t_{1}\geq t_{2},t_{2}\geq t_{3},t_{3}\geq t_{4},t_{3}\geq t_{6},t_{2}\geq t_{5}\}. \end{align*} $$

Let $\sigma $ denote some composition of acceptable moves. We then notice the equivalence of the two sets

$$ \begin{align*} \Sigma _{1} &=\left\{ \sigma :\sigma ^{-1}(1)<\sigma ^{-1}(2)<\sigma ^{-1}(3)<\sigma ^{-1}(4),\sigma ^{-1}(2)<\sigma ^{-1}(5),\sigma ^{-1}(3)<\sigma ^{-1}(6)\right\} , \\ \Sigma _{2} &=\left\{ \sigma :\sigma \text{ takes input tree to }\alpha \text{ where the input tree is admissible}\right\}, \end{align*} $$

both generated by the requirement that the child must carry a larger label than the parent. That is, both $\Sigma _{1}$ and $\Sigma _{2}$ classify the whole upper-echelon class represented by $\alpha $.

Hence,

$$ \begin{align*} \bigcup _{\sigma \in \Sigma _{1}}\left\{ t_{1}\geq t_{\sigma(2)}\geq t_{\sigma (3)}\dotsb\geq t_{\sigma \left( 6\right) }\right\} =\{t_{1}\geq t_{2}\geq t_{3}\geq t_{4},t_{2}\geq t_{5},t_{3}\geq t_{6}\}=T_{D}(\alpha ), \end{align*} $$

and we are done.

4.2 Signed KM acceptable moves

While Proposition 4.3 shows that summing over an entire KM upper-echelon class yields a time integration domain with clean structure, it is not sufficient for our purposes. We prove an extended KM board game in Sections 4.2–4.5. Recall the key observation of the KM board game: Many summands in $J^{(k+1)}\left (f^{(k+1)}\right )$ actually have the same integrand if one switches the variable labellings, and thus one can take the acceptable moves to combine them. In fact, one can combine them even more after the acceptable moves to get a larger integration domain $D_{m}$.Footnote ²⁰ Instead of aiming to reduce the number of summands in $J^{(k+1)}\left (f^{(k+1)}\right )$ even more, our goal this time is to enlarge the integration domain when estimating $J_{\mu ,\operatorname {sgn}}^{(k+1)}\left (f^{(k+1)}\right )\left (t_{1},\underline {t}_{k+1}\right )$, so that U-V techniques can actually apply. Depending on the sign combination in $J_{\mu _{m},\operatorname {sgn}}^{(k+1)}\left (f^{(k+1)}\right )\left (t_{1},\underline {t}_{k+1}\right )$, one could run into the problem of needing to estimate the x part and the $x'$ part using the same time integral. This problem is another obstacle stopping U-V space techniques from being used in the analysis of $GP$ hierarchies, separate from the other obstacle that $D_{m}$ was previously unknown.

From here on out, we denote the already unioned or combined integrals in one echelon class as a upper-echelon class integral, and we use Proposition 4.3 for its integration limits. We also put a $+$ or $ -$ sign at the corresponding node of a tree, as we are dealing with $J_{\mu ,\operatorname {sgn}}^{(k+1)}\left (f^{(k+1)}\right )\left (t_{1},\underline {t}_{k+1}\right )$, in which there are $ B^{+}$ and $B^{-}$ at each coupling. We start with the following example:

Example 4. Let us consider the two upper-echelon trees:

They have the upper-echelon class integrals

$$ \begin{align*} I_{1} &= \int_{D_{1}}U^{(1)}(t_{1}-t_{2})B_{1,2}^{-}U^{(2)}(t_{2}-t_{3})B_{1,3}^{+}U^{(3)}(t_{3}-t_{4})B_{2,4}^{+}\left(f^{(4)}\right)\left(t_{1},\underline{t}_{4}\right)d\underline{t}_{4}, \\ I_{2} &= \int_{D_{2}}U^{(1)}(t_{1}-t_{2})B_{1,2}^{+}U^{(2)}(t_{2}-t_{3})B_{1,3}^{-}U^{(3)}(t_{3}-t_{4})B_{3,4}^{+} \left(f^{(4)}\right)\left(t_{1},\underline{t}_{4}\right)d\underline{t}_{4}, \end{align*} $$

where $D_{1}=\{t_{2}\leq t_{1},t_{3}\leq t_{2},t_{4}\leq t_{2}\}$ and $D_{2}=\{t_{2}\leq t_{1},t_{3}\leq t_{2},t_{4}\leq t_{3}\}$ following from our discussions in Section 4.1.

$I_{1}$ and $I_{2}$ actually have the same integrand if one does a $t_{2}\leftrightarrow t_{3}$ swap in $I_{1}$, despite the fact that the trees corresponding to $I_{1}$ and $I_{2}$ have different skeletons. In fact, shortening $e^{i\left ( t_{i}-t_{j}\right ) \triangle } $ as $U_{i,j}$, we have

$$ \begin{align*} I_{1} &=\int_{D_{1}}U_{1,3}\left(\left\lvert U_{3,4}\phi \right\rvert ^{2}U_{3,4}\phi \right)(x_{1})U_{1,2}\left(\overline{U_{2,4}\phi }\overline{U_{2,4}\phi }U_{2,4}\left( \left\lvert \phi \right\rvert ^{2}\phi \right) \right)\left(x^{\prime}_{1}\right)d\underline{t}_{4} \\ &=\int_{D^{\prime}_{1}}U_{1,2}\left(\left\lvert U_{2,4}\phi \right\rvert ^{2}U_{2,4}\phi \right)(x_{1})U_{1,3}\left(\overline{U_{3,4}\phi }\overline{U_{3,4}\phi }U_{3,4}\left( \left\lvert \phi \right\rvert ^{2}\phi \right) \right)\left(x^{\prime}_{1}\right)d\underline{t}_{4} \\ &=\int_{D^{\prime}_{1}}U^{(1)}(t_{1}-t_{2})B_{1,2}^{+}U^{(2)}(t_{2}-t_{3})B_{1,3}^{-}U^{(3)}(t_{3}-t_{4})B_{3,4}^{+}\left(f^{(4)}\right)\left(t_{1},\underline{t}_{4}\right)d\underline{t}_{4}, \end{align*} $$

where $D^{\prime }_{1}=\{t_{3}\leq t_{1},t_{2}\leq t_{3},t_{4}\leq t_{3}\}$ and we have put in $f^{(4)}=\left ( \left \lvert \phi \right \rangle \left \langle \phi \right \rvert \right ) ^{\otimes 4}$ for simplicity.Footnote ²¹ Hence,

(4.4)$$ \begin{align} I_{1}+I_{2} &= \int_{D}U^{(1)}(t_{1}-t_{2})B_{1,2}^{+}U^{(2)}(t_{2}-t_{3})B_{1,3}^{-}U^{(3)}(t_{2}-t_{3})B_{3,4}^{+}\left(f^{(4)}\right)\left(t_{1},\underline{t}_{4}\right)d\underline{t}_{4} \\ &= \int_{t_{4}=0}^{t_{1}}\int_{t_{2}=0}^{t_{1}} \int_{t_{3}=t_{4}}^{t_{1}}U^{(1)}(t_{1}-t_{2})B_{1,2}^{+}U^{(2)}(t_{2}-t_{3})B_{1,3}^{-}U^{(3)}(t_{2}-t_{3})B_{3,4}^{+}\left(f^{(4)}\right)\left(t_{1},\underline{t}_{4}\right)d\underline{t}_{4}, \notag \end{align} $$

where $D=D^{\prime }_{1}\cup D_{2}=\{t_{3}\leq t_{1},t_{2}\leq t_{1},t_{4}\leq t_{3}\}$.

Example 4 is an easy example of what we will call the wild moves, and shows that one could indeed further combine the summands in $J^{(k+1)}\left (f^{(k+1)}\right )$ after the original KM board game has been performed. We will explain why our U-V techniques apply to $I_{1}+I_{2}$ but not $I_{1}$ and $I_{2}$ individually in Section 5.1. Despite the fact that Example 4 uses the already-combined upper-echelon integrals, our extended KM board game actually starts from scratch – that is, it starts from $\gamma ^{(1)}(t_{1})$ instead of already-combined upper-echelon integrals – as not all upper-echelon integrals act so nicely under the wild moves. However, it is still a multistep process. We will first switch the terms in $\gamma ^{(1)}(t_{1})$ into their tamed form via signed KM acceptable moves in Sections 4.2 and 4.3, and then categorise the tamed forms into tamed classes via the wild moves in Sections 4.4 and 4.5.

We now explain the program as follows: As before, start by expanding $\gamma ^{(1)}(t_{1})$ to coupling level k, which generates a sum expansion of $k!$ terms. But now for each of these $k!$ terms, expand the collapsing operators $B_{\mu \left (j\right ),j}^{\left (j\right )}$ into $+$ and $-$ components, which introduces $2^{k}$ terms. Thus, in all, we have $2^{k}k!$ terms, each of which has sign-dependent collapsing operators

(4.5)$$ \begin{align} \gamma ^{(1)}=\sum_{\mu ,\operatorname{sgn}}I\left(\mu ,\text{id},\operatorname{sgn},\gamma ^{(k+1)}\right), \end{align} $$

where $\text {id}$ is the identity permutation on $\{2,\dotsc ,k+1\},$

$$ \begin{align*} I(\mu ,\sigma ,\operatorname{sgn})=\int_{t_{1}\geq t_{\sigma (2)}\geq \dotsb \geq t_{\sigma (k+1)}}J_{\mu ,\operatorname{sgn}}^{(k+1)}\left(\gamma ^{(k+1)}\right)\left(t_{1},\underline{t}_{k+1}\right)d\underline{t}_{k+1} \end{align*} $$

and $J_{\mu ,\operatorname {sgn}}^{(k+1)}$ is defined as in equation (3.10). Equation (4.5) is a sum over all admissible $\mu $ – that is, collapsing maps that satisfy $\mu (j)<j$ – of which there are $k!$. It is also a sum over all $\operatorname {sgn}$ maps, of which there are $2^{k}$.

We define a signed version of the KM acceptable moves, still denoted $\text {KM}(j,j+1)$, which is defined provided $\mu (j)\neq \mu (j+1)$ and $\mu (j+1)<j$. It is defined as the following action on a triple $(\mu ,\sigma ,\operatorname {sgn})$:

$$ \begin{align*} (\mu',\sigma',\operatorname{sgn}')=\text{KM}(j,j+1)(\mu ,\sigma ,\operatorname{sgn}), \end{align*} $$

where

$$ \begin{align*} \mu' & = (j,j+1) \circ \mu \circ (j,j+1), \\ \sigma' & = (j,j+1) \circ \sigma, \\ \operatorname{sgn}' & = \operatorname{sgn} \circ (j,j+1). \end{align*} $$

Graphically, this means that nodes j and $j+1$ belong to different left branches and correspond to switching nodes j and $j+1$, leaving the signs in place on the tree – in other words, the node previously labelled $ j $ is relabelled $j+1$, and the node previously labelled $j+1$ is relabelled $ j $, but the signs are left in place.

A slight modification of the arguments in [Reference Klainerman and Machedon47] shows that, analogous to equation (4.1), if $(\mu ',\sigma ',\operatorname {sgn}')=\text {KM}(j,j+1)(\mu ,\sigma ,\operatorname {sgn})$ and $f^{(k+1)}$ is a symmetric density, then

(4.6)$$ \begin{align} J_{\mu',\operatorname{sgn}'}^{(k+1)}\left(f^{(k+1)}\right)\left(t_{1},{\sigma'}^{-1}\left(\underline{t}_{k+1}\right)\right)=J_{\mu ,\operatorname{sgn}}^{(k+1)}\left(f^{(k+1)}\right)\left(t_{1},{\sigma }^{-1}\left(\underline{t}_{k+1}\right)\right). \end{align} $$

It follows from equation (4.6) that

(4.7)$$ \begin{align} I\left(\mu',\sigma',\operatorname{sgn}',f^{(k+1)}\right)=I\left(\mu ,\sigma ,\operatorname{sgn},f^{(k+1)}\right). \end{align} $$

As in the sign-independent case (or more accurately, the combined-sign case), we can combine KM acceptable moves as follows: If $\rho $ is a permutation of $\{2,\dotsc ,k+1\}$ such that it is possible to write $\rho $ as a composition of transpositions

$$ \begin{align*} \rho =\tau _{1}\circ \dotsb \circ \tau _{r} \end{align*} $$

for which each operator $\text {KM}\left (\tau _{j}\right )$ on the right side of

$$ \begin{align*} \text{KM}(\rho )\overset{\mathrm{def}}{=}\text{KM}(\tau _{1})\circ \dotsb \circ \text{KM}(\tau _{r}) \end{align*} $$

is an acceptable action, then $\text {KM}(\rho )$, defined by this composition, is acceptable as well. In this case, $(\mu ',\sigma ',\operatorname {sgn}')=\text { KM}(\rho )(\mu ,\sigma ,\operatorname {sgn})$, and

$$ \begin{align*} \mu' & =\rho \circ \mu \circ \rho ^{-1}, \\ \sigma' & =\rho \circ \sigma, \\ \operatorname{sgn}' & =\operatorname{sgn}\circ \rho ^{-1}. \end{align*} $$

Of course, equations (4.6) and (4.7) hold as well. If $(\mu ,\operatorname {sgn})$ and $(\mu ',\operatorname {sgn}')$ are such that there exists $\rho $ for which $(\mu ',\sigma ',\operatorname {sgn}')=\text {KM}(\rho )(\mu ,\sigma ,\operatorname {sgn})$, then we say that $(\mu ',\operatorname {sgn}')$ and $(\mu ,\operatorname {sgn})$ are KM-relatable. This is an equivalence relation that partitions the set of collapsing map/sign map pairs into equivalence classes. In the graphical representation, two such pairs are KM-relatable if and only if they have the same signed skeleton tree.

Whereas we could use the signed KM acceptable moves to convert an arbitrary admissible $\mu $ to an upper-echelon $\mu '$, this will no longer suit our purpose. Instead, our program will be to convert each pair $(\mu ,\operatorname {sgn})$ to a tamed form, which we define in the next section. The reason for our preference of tamed form over upper-echelon form is that it is invariant under wild moves, to be introduced in Section 4.4.

4.3 Tamed form

In this section, we define what it means for a pair $(\mu ,\operatorname {sgn})$ and its corresponding tree representation to be tamed, in Definition 4.4. Then through an example, we present an algorithm for producing the tamed enumeration of a signed skeleton. The general algorithm is then stated in Algorithm 5. Notice that it produces a different enumeration from Algorithm 4. Compared with Algorithm 4, the tamed-form enumeration deals not just with left branches first, it also deals with $+$ first.Footnote ²² In Section 4.3.1, we exhibit how to reduce a signed tree with the same skeleton but different enumeration into the tamed form using signed KM acceptable moves.

We will now give a nongraphical set of conditions on $\mu $ and $\operatorname {sgn}$ that determine whether or not $(\mu ,\operatorname {sgn})$ is tamed. First, we define the concept of a tier. We say that $j\geq 2$ is of tier q if

$$ \begin{align*} \mu^q(j) = 1 \quad \text{but} \quad \mu^{q-1}(j)>1, \end{align*} $$

where $\mu ^q = \mu \circ \dotsb \circ \mu $, the composition taken q times. We write $t(j)$ for the tier value of j.

Note that the statement $\mu ^2(\ell ) = \mu ^2(r)$ means graphically that the parents of $\ell $ and r belong to the same left branch. Conditions (2), (3) and (4) specify the ordering for $\ell $ and r belonging to the same tier, and the rule depends upon whether or not the parents of $\ell $ and r belong to the same left branch. If they do, rule (3) says that a positive parent dominates over a negative parent, but rule (2) says that if the parents are of the same sign, then the ordering follows the parental ordering. Finally, if the parents do not belong to the same left branch, rule (4) says that the ordering follows the parental ordering regardless of the signs of the parents.

Example 5. The $(\mu ,\operatorname {sgn})$ pair with tier properties indicated in the following chart is tamed:

$$ \begin{align*} \begin{array}{c|c@{\kern10pt}c@{\kern10pt}c@{\kern10pt}c@{\kern10pt}c@{\kern10pt}c@{\kern10pt}c@{\kern10pt}c@{\kern10pt}c@{\kern10pt}c@{\kern10pt}c@{\kern10pt}c@{\kern10pt}c} j & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 & 13 & 14 \\ \hline \mu (j) & 1 & 1 & 1 & 1 & 5 & 5 & 2 & 2 & 7 & 7 & 9 &9 & 8 \\[3pt] \operatorname{sgn}(j) & - & - & + & + & - & + & - & + & - & + & + & - & + \\[3pt] t(j) & 1 & 1 & 1 & 1 & 2 & 2 & 2 & 2 & 3 & 3 & 3 & 3 & 3 \\ \end{array} \end{align*} $$

All four conditions in Definition 4.4 can be checked from the chart. This is in fact the $(\mu ,\operatorname {sgn})$ pair that appears in the example that follows.

In the following example, we illustrate an algorithm for determining the unique tamed enumeration of a signed skeleton tree. After the example is completed, we give the general form of the algorithm.

For the example, we start with the following skeleton, with only the signs indicated. (Recall that KM acceptable moves will leave the signs in place in the tree and change just the numbering of the nodes.) Start by considering all nodes mapping to $1$ (the universal ancestor) – this is the left branch attached to $1$ that is four nodes long in the order $--++$, and we enumerate it in order as $2,3,4,5$.

We then put this full left branch in the (empty) queue, listing the $+$ nodes first and then the $-$ nodes:

$$ \begin{align*} \text{Queue: } 4+, 5+, 2-, 3-. \end{align*} $$

Then we start working along the queue from left to right. Since $4+$ has no right child, we skip it and move to $5+$. Since $5+$ does have a right child, we label it with the next available number ($6$) and completely enumerate the entire left branch that starts with this $6$ node (that means, in this case, labelling $6-$ and $7+$ as shown on the next graph).

Then we add this entire left branch to the queue, putting the $+$ nodes before the $-$ nodes. We also pop $4+$ and $5+$ from the (left of the) queue, since we have already dealt with them. The queue now reads

$$ \begin{align*} \text{Queue: } 2-, 3-, 7+, 6-. \end{align*} $$

Now we come to the next node in the queue (reading from the left), which is $2-$. The node $2$ does have a right child. We label it as $8$ (the next available number) and completely enumerate the left branch that starts with $8$, which means labelling $8-$ and $9+$as shown.

From the queue, we pop $2$ and add the $8-$, $9+$ left branch – all $+$ nodes first and then all $-$ nodes:

$$ \begin{align*} \text{Queue: } 3-, 7+, 6-, 9+, 8-. \end{align*} $$

Since $3-$ does not have a right child, we pop it and proceed to $7+$, which does have a right child, which is labelled with $10$, and the left branch starting at $10$ is enumerated as $10-, 11+$, as shown.

The queue is updated:

$$ \begin{align*} \text{Queue: } 6-, 9+, 8-, 11+, 10-. \end{align*} $$

By now the procedure is probably clear, so we will jump to the fully enumerated tree.

Here is the general algorithm. Recall that a queue is a data structure where elements are added on the right and removed (dequeued) on the left.

4.3.1 Reducing to tamed forms via the signed KM board game

We will now explain how to execute a sequence of signed KM acceptable moves that will bring the example tree from the previous section, with some other enumeration, into the tamed form. This tree corresponds to the following $\mu $ and $\operatorname {sgn}$ functions:

$$ \begin{align*} \begin{array}{c|c@{\kern10pt}c@{\kern10pt}c@{\kern10pt}c@{\kern10pt}c@{\kern10pt}c@{\kern10pt}c@{\kern10pt}c@{\kern10pt}c@{\kern10pt}c@{\kern10pt}c@{\kern10pt}c@{\kern10pt}c} j & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 & 13 & 14\\ \hline \operatorname{sgn}(j) & - & - & + & - & + & + & + & - & - & + & + & - & +\\[3pt] \mu(j) & 1 & 1 & 1 & 2 & 2 & 1 & 6 & 7 & 6 & 7 & 5 & 11 & 11 \\ \end{array} \end{align*} $$

We are going to start with the enumeration at left, which is not tamed, and explain how to execute KM acceptable moves in order to convert this tree into tamed form. Of course, this is quite similar to what Klainerman and Machedon described, with just a modification to prioritise plusses over minuses.

We will keep a queue that right now includes only the node $1$:

$$ \begin{align*} \text{Queue: }1. \end{align*} $$

Following the queue, we move all nodes (all j) for which $\mu (j)=1$ all the way to left using KM moves. Since $\mu (7)=1$, although $\mu (5)=2$ and $\mu (6)=2$, we apply the KM moves $\text {KM}(6,7)$ and then $\text {KM}(5,6)$.

The $\text {KM}(6,7)$ move is

$$ \begin{align*} \mu & \mapsto (6,7) \circ \mu \circ (6,7), \\ \operatorname{sgn} & \mapsto \operatorname{sgn} \circ (6,7). \end{align*} $$

The $\text {KM}(5,6)$ move is

$$ \begin{align*} \mu & \mapsto (5,6) \circ \mu \circ (5,6), \\ \operatorname{sgn} & \mapsto \operatorname{sgn} \circ (5,6), \end{align*} $$

and together these result in the following:

$$ \begin{align*} \begin{array}{c|c@{\kern10pt}c@{\kern10pt}c@{\kern10pt}c@{\kern10pt}c@{\kern10pt}c@{\kern10pt}c@{\kern10pt}c@{\kern10pt}c@{\kern10pt}c@{\kern10pt}c@{\kern10pt}c@{\kern10pt}c} j & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 & 13 & 14\\ \hline \operatorname{sgn}(j) & - & - & + & + & - & + & + & - & - & + & + & - & +\\[3pt] \mu(j) & 1 & 1 & 1 & 1 & 2 & 2 & 7 & 5 & 7 & 5 & 6 & 11 & 11 \\ \end{array} \end{align*} $$

These two moves have been implemented in the revised graph at left.

Inspecting the $\mu $ chart, we see that all output $1$s have been moved to the left, and the complete list of j for which $\mu (j)=1$ is $2-, 3-, 4+, 5+$. We add these numbers to our queue, but first add all plusses and then all minuses:

Since we have completed $1$ on the queue, we next move to $4$, but there are no j for which $\mu (j)=4$, so we proceed to $5$. As we can see from the $\mu $ table or from the tree, $\mu (9)=5$ and $\mu (11)=5$, so we execute KM moves to bring these all the way to the left (but to the right of the $1$s):

The next step is therefore to implement moves $\text {KM}(8,9)$, $\text {KM} (7,8)$ and $\text {KM}(6,7)$, which brings the $\mu $ table to the following:

$$ \begin{align*} \begin{array}{c|c@{\kern10pt}c@{\kern10pt}c@{\kern10pt}c@{\kern10pt}c@{\kern10pt}c@{\kern10pt}c@{\kern10pt}c@{\kern10pt}c@{\kern10pt}c@{\kern10pt}c@{\kern10pt}c@{\kern10pt}c} j & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 & 13 & 14 \\ \hline \operatorname{sgn}(j) & - & - & + & + & - & - & + & + & - & + & + & - & + \\[3pt] \mu(j) & 1 & 1 & 1 & 1 & 5 & 2 & 2 & 8 & 8 & 5 & 7 & 11 & 11 \\ \end{array} \end{align*} $$

This is followed by the moves $\text {KM}(11,10)$, $\text {KM}(10,9)$, $\text {KM}(9,8)$ and $\text {KM}(8,7)$, which bring the $\mu $ table to the following:

$$ \begin{align*} \begin{array}{c|c@{\kern10pt}c@{\kern10pt}c@{\kern10pt}c@{\kern10pt}c@{\kern10pt}c@{\kern10pt}c@{\kern10pt}c@{\kern10pt}c@{\kern10pt}c@{\kern10pt}c@{\kern10pt}c@{\kern10pt}c} j & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 & 13 & 14 \\ \hline \operatorname{sgn}(j) & - & - & + & + & - & + & - & + & + & - & + & - & + \\[3pt] \mu(j) & 1 & 1 & 1 & 1 & 5 & 5 & 2 & 2 & 9 & 9 & 8 & 7 & 7 \\ \end{array} \end{align*} $$

At this point, the tree takes the form as pictured to the left. All $5$s have been moved to their proper position in the $\mu $ table. The complete list of j for which $\mu (j)=5$ is $6-, 7-$, so we add these numbers to the queue, adding the plusses first and then the minuses:

Since we have addressed $5$ in the queue, we move to the next item, which is $2$. This means we have to move all j for which $\mu (j)=2$ all the way to the left (just to the right of $5$). Examining the $\mu $ table, we see that these j are already in place, at positions $8-, 9+$. So no KM moves are needed, and we add to the queue:

Next on the queue is $3$, but there are no j for which $\mu (j)=3$, so we proceed to $7$ on the queue. From the $\mu $ table or the tree, we see there are two j for which $\mu (j)=7$, namely $13$ and $14$. We therefore execute KM moves to bring these to the left in the $\mu $ table, just to the right of $2$.

Specifically, we do $\text {KM}(12,13)$, $\text {KM}(11,12)$ and $\text {KM}(10,11)$, which brings us to the following $\mu $ table:

$$ \begin{align*} \begin{array}{c|c@{\kern10pt}c@{\kern10pt}c@{\kern10pt}c@{\kern10pt}c@{\kern10pt}c@{\kern10pt}c@{\kern10pt}c@{\kern10pt}c@{\kern10pt}c@{\kern10pt}c@{\kern10pt}c@{\kern10pt}c} j & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 & 13 & 14 \\ \hline \operatorname{sgn}(j) & - & - & + & + & - & + & - & + & - & + & - & + & + \\[3pt] \mu(j) & 1 & 1 & 1 & 1 & 5 & 5 & 2 & 2 & 7 & 9 & 9 & 8 & 7 \\ \end{array} \end{align*} $$

After that, we do $\text {KM}(13,14)$, $\text {KM}(12,13)$ and $\text {KM} (11,12)$, which brings us to the following $\mu $ table:

$$ \begin{align*} \begin{array}{c|c@{\kern10pt}c@{\kern10pt}c@{\kern10pt}c@{\kern10pt}c@{\kern10pt}c@{\kern10pt}c@{\kern10pt}c@{\kern10pt}c@{\kern10pt}c@{\kern10pt}c@{\kern10pt}c@{\kern10pt}c} j & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 & 13 & 14 \\ \hline \operatorname{sgn}(j) & - & - & + & + & - & + & - & + & - & + & + & - & + \\[3pt] \mu(j) & 1 & 1 & 1 & 1 & 5 & 5 & 2 & 2 & 7 & 7 & 9 & 9 & 8 \\ \end{array} \end{align*} $$

Now that the seven outputs are in place, we take the set of j for which $ \mu (j) = 7$, which is $10-, 11+$, and put them in the queue with plusses first, followed by minuses:

There are no j for which $\mu (j)=6$, so we proceed in the queue to $9$. However, the two $9$s are already in place, and the next item in the queue is $8$; the one $8$ is already in place. So this completes the example.

We now describe the preceding algorithm in general.

Algorithm 6. Given $(\mu ,\operatorname {sgn})$, start with a queue Q that initially contains $1$ and a marker j, which is initially set to $j=2$. Repeat the following steps:

1. 1. Dequeue the leftmost entry $\ell $ of the queue. If the queue is empty, then stop. Clear the temporary ordered list L.

2. 2. If $\mu (j)=\ell $, add j to the right of L, then increment the marker j by $1$ (so now j is the old $j+1$). If (the new marker) j is out of range, jump to step 4. If $\mu (j)\neq \ell $, then proceed to step 3; otherwise, repeat step 2.

3. 3. Find the smallest $r\geq j+1$ such that $\mu (r) = \ell $ (if there is no such r, jump to step 4). Execute signed KM moves $\text {KM}(r-1,r)$, followed by $\text {KM}(r-2,r-1)$, $\dotsc $, until $\text {KM}(j+1,j)$. Now $ \mu (j)=\ell $. Return to step 2.

4. 4. Take all elements of the temporary ordered list L, read all $+$ entries in order (from left to right) and add them to the (right end of the) queue Q; then read all $-$ entries in order (from left to right) and add them to the (right end of the) queue Q. Return to step 1.

We have the following adaptation of Proposition 4.3, revised to include sign maps and to reference tamed forms in place of upper-echelon forms.

Proposition 4.5. Within a signed KM-relatable equivalence class of collapsing map/sign map pairs $(\mu ,\operatorname {sgn})$, there is a unique tamed $(\mu _{\ast },\operatorname {sgn}_{\ast })$. Moreover,

(4.8)$$ \begin{align} \sum_{\left(\mu ,\operatorname{sgn}\right)\sim \left(\mu _{\ast },\operatorname{sgn}_{\ast }\right)}I\left(\mu ,{\mathop{\textrm{{id}}}\nolimits},\operatorname{sgn},\gamma ^{(k+1)}\right)=\int_{T_{D}\left(\mu _{\ast }\right)}J_{\mu _{\ast }, \operatorname{sgn}_{\ast }}\left(\gamma ^{(k+1)}\right)\left(t_{1},\underline{t}_{k+1}\right)d\underline{ t}_{k+1}, \end{align} $$

where $T_{D}(\mu _{\ast })$ is defined in equation (4.3).

To proceed with our program, we divide the expansion (4.5) into sums over signed KM-relatable equivalence classes, and apply equation (4.8) for the sum over each equivalence class. Thus we obtain

(4.9)$$ \begin{align} \gamma ^{(1)}(t_{1})=\sum_{\left(\mu _{\ast },\operatorname{sgn}_{\ast }\right)\text{ tamed} }\int_{T_{D}\left(\mu _{\ast }\right)}J_{\mu _{\ast },\operatorname{sgn}_{\ast }}\left(\gamma ^{(k+1)}\right)\left(t_{1},\underline{t}_{k+1}\right)d\underline{t}_{k+1}. \end{align} $$

The next step will be to round up the tamed pairs $(\mu _{\ast },\operatorname {sgn}_{\ast })$ via wild moves, as defined and discussed in the next section. This will produce a further reduction of equation (4.9).

4.4 Wild moves

Definition 4.6. A wild move $\text {W}(\rho )$ is defined as follows: Suppose $(\mu ,\operatorname {sgn})$ is a collapsing operator/sign map pair in tamed form, and $\{\ell , \dotsc , r\}$ is a full left branch – that is,

$$ \begin{align*} z \overset{\mathrm{def}}{=} \mu(\ell)=\mu(\ell+1)=\dotsb = \mu(r), \end{align*} $$

but $\mu (\ell -1)\neq z$ (or is undefined) and $\mu (r+1) \neq z$ (or is undefined).

Let $\rho $ be a permutation of $\{\ell ,\ell +1, \dotsc , r \}$ that satisfies the following condition: If $\ell \leq q<s\leq r$ and $\operatorname {sgn}(q)=\operatorname {sgn}(s)$, then q appears before s in the list $\left (\rho ^{-1}(\ell ), \dotsc , \rho ^{-1}(r)\right )$ – or equivalently, $\rho (q)<\rho (s)$.

Then the wild move $\text {W}(\rho )$ is defined as an action on a triple $ (\mu ,\sigma , \operatorname {sgn})$, where

$$ \begin{align*} (\mu',\sigma',\operatorname{sgn}') = W(\rho)(\mu,\sigma,\operatorname{sgn}), \end{align*} $$

provided

$$ \begin{align*} \mu' & = \rho \circ \mu = \rho \circ \mu \circ \rho^{-1}, \\ \sigma' & = \rho \circ \sigma, \\ \operatorname{sgn}' & = \operatorname{sgn} \circ \rho^{-1}. \end{align*} $$

We note that W is an action

$$ \begin{align*} W(\rho_1)W(\rho_2) = W(\rho_1 \circ\rho_2). \end{align*} $$

It is fairly straightforward to show the following, using the definition of a tamed form. It is important to note that the analogous statement for upper-echelon forms does not hold, which is the reason for introducing the tamed class.

Proposition 4.7. Suppose $(\mu ,\operatorname {sgn})$ is a collapsing operator/sign map pair in tamed form, and $W(\rho )$ is a wild move. Letting $(\mu ', \operatorname {sgn}')$ be the output,

$$ \begin{align*} (\mu',\sigma', \operatorname{sgn}') = W(\rho) (\mu,\sigma, \operatorname{sgn}), \end{align*} $$

then $(\mu ',\operatorname {sgn}')$ is also tamed.

Thus wild moves preserve the tamed class, and we can say that two tamed forms $(\mu ,\operatorname {sgn})$ and $(\mu ',\operatorname {sgn}')$ are wildly relatable if there exists $\rho $ as in Definition 4.6 such that

$$ \begin{align*} (\mu',\sigma', \operatorname{sgn}') = W(\rho)(\mu,\sigma, \operatorname{sgn}). \end{align*} $$

This is an equivalence relation, and in the sum (4.9) we can partition the class of tamed pairs $(\mu ,\operatorname {sgn})$ into equivalence classes of wildly relatable forms (we pursue this in the next section).

The main result of this section is the following:

Proposition 4.8. Suppose that $\rho $ is as in Definition 4.6 and

$$ \begin{align*} (\mu', \sigma', \operatorname{sgn}') = W(\rho)(\mu, \sigma, \operatorname{sgn}). \end{align*} $$

Then for any symmetric density $f^{(k+1)}$,

$$ \begin{align*} J_{\mu',\operatorname{sgn}'}\left( f^{(k+1)}\right)\left(t_1, {\sigma'}^{-1}\left(\underline{t}_{k+1}\right)\right) = J_{\mu,\operatorname{sgn}}\left( f^{(k+1)}\right)\left(t_1, \sigma^{-1}\left(\underline{t}_{k+1}\right)\right). \end{align*} $$

Consequently, the Duhamel integrals are preserved, after adjusting for the time permutations

$$ \begin{align*} \int_{\sigma'\left[T_D\left(\mu'\right)\right]} J_{\mu',\operatorname{sgn}'}\left( \gamma^{(k+1)}\right)\left(t_1, \underline{t}_{k+1}\right) d\underline{t} _{k+1} = \int_{\sigma\left[T_D\left(\mu\right)\right]} J_{\mu,\operatorname{sgn}}\left( \gamma^{(k+1)}\right)\left(t_1, \underline{t}_{k+1}\right) d\underline{t}_{k+1}, \end{align*} $$

where $\sigma [T_D(\mu )]$ is defined by modifying equation (4.3) so that node labels are pushed forward by $\sigma $:

$$ \begin{align*} \sigma[T_D(\mu)] = \left\{ t_{\sigma\left(j\right)}\geq t_{\sigma(k)} : \right. & j,k\text{ are labels on nodes of }\alpha \text{ such} \\ & \left. \vphantom{t_{\sigma\left(j\right)}}\text{that the }k\text{ node is a child of the }j\text{ node} \right\}. \end{align*} $$

Proof. A permutation $\rho $ of the type described in Definition 4.6 can be written as a composition of permutations

$$ \begin{align*} \rho = \tau_1 \circ \dotsb \circ \tau_s, \end{align*} $$

with the property that each $\tau = (i,i+1)$ for some $i\in \{\ell , \dotsc , \ell +r\}$ and $\operatorname {sgn}(i) \neq \operatorname {sgn}(i+1)$. Thus it suffices to prove

$$ \begin{align*} U^{(i-1)}(-t_i) B_{\mu(i),i}^- U^{(i)} (t_i -t_{i+1}) B_{\mu(i+1),i+1}^+ &U^{(i+1)}(t_{i+1}) \\ &= U^{(i-1)}(-t_{i+1}) B_{\mu(i),i}^+ U^{(i)} (t_{i+1} -t_i) B_{\mu(i+1),i+1}^- U^{(i+1)}(t_i) \end{align*} $$

when the two sides act on a symmetric density. Recall that $ z=\mu (i)=\mu (i+1) $. Without loss, we might as well take $z=1$ and $i=2$ so that this becomes

(4.10)$$ \begin{align} U^{(1)}(-t_2) B_{1,2}^- U^{(2)} (t_2 -t_3) B_{1,3}^+ U^{(3)}(t_3)= U^{(1)}(-t_3) B_{1,2}^+ U^{(2)} (t_3 -t_2) B_{1,3}^- U^{(3)}(t_2). \end{align} $$

To prove equation (4.10), on the left side we proceed as follows: First we plug in

$$ \begin{align*} U^{(1)}(-t_2) & = U^1_{-2} U^{1'}_{2} \\ U^{(2)}(t_2-t_3) & = U_2^1 U_{-3}^1 U_{-2}^{1'} U_{3}^{1'} U_2^2 U_{-3}^2 U_{-2}^{2'} U_3^{2'} \\ U^{(3)}(t_3) & = U_3^1 U_{-3}^{1'} U_3^2 U_{-3}^{2'} U_3^3 U_{-3}^{3'}, \end{align*} $$

where the subscript indicates the time variable and the superscript indicates the spatial variable. Then we note that for the two collapsing operators on the left side of equation (4.10), the following hold:

• • $B_{1,2}^-$ acts only on the $2$, $2'$ and $1'$ coordinates, so we can move all $U^1$ operators in the middle to the left.

• • $B_{1,3}^+$ acts only on the $3$, $3'$ and $1$ coordinates, so we can move all $U^2$, $U^{2'}$ and $U^{1'}$ operators in the middle to the right.

This results in

(4.11)$$ \begin{align} \text{left side of equation~} (4.10) = U_{-3}^1 U_2^{1'} B_{1,2}^- B_{1,3}^+ U_3^1 U_{-2}^{1'} U_2^2 U_{-2}^{2'} U_3^3 U_{-3}^3. \end{align} $$

Similarly, on the right side of equation (4.10), plug in

$$ \begin{align*} U^{(1)}(-t_3) & = U_{-3}^1 U_3^{1'} \\ U^{(2)}(t_3-t_2) & = U_3^1 U_{-2}^1 U_{-3}^{1'} U_{2}^{1'} U_3^2 U_{-2}^2 U_{-3}^{2'} U_2^{2'} \\ U^{(3)}(t_2) & = U_2^1 U_{-2}^{1'} U_2^2 U_{-2}^{2'} U_2^3 U_{-2}^{3'}. \end{align*} $$

Then we note that for the two collapsing operators on the right side of equation (4.10), the following are true:

• • $B_{1,2}^+$ acts only on the $2$, $2'$ and $1$ coordinates, so we can move all $U^{1'}$ operators in the middle to the left.

• • $B_{1,3}^-$ acts only on the $3$, $3'$ and $1'$ coordinates, so we can move all $U^2$, $U^{2'}$ and $U^{1}$ operators in the middle to the right.

This results in

(4.12)$$ \begin{align} \text{right side of equation~} (4.10) = U_{-3}^1 U_2^{1'} B_{1,2}^+ B_{1,3}^- U_3^1 U_{-2}^{1'} U_3^2 U_{-3}^{2'} U_2^3 U_{-2}^{3'}. \end{align} $$

Since equations (4.11) and (4.12) are equal when applied to a symmetric density, this proves equation (4.10). In particular, one just needs that to permute

$$ \begin{align*} \left(x_2,x_2',x_3,x^{\prime}_3\right) \leftrightarrow \left(x_3,x^{\prime}_3,x_2,x^{\prime}_2\right).\end{align*} $$

Example 6. The pair $(\mu _1, \operatorname {sgn}_1)$ is defined as follows:

$$ \begin{align*} \begin{array}{c|c@{\kern10pt}c@{\kern10pt}c@{\kern10pt}c@{\kern10pt}c@{\kern10pt}c} & 2 & 3 & 4 & 5 & 6 & 7 \\ \hline \mu _{1} & 1 & 1 & 1 & 2 & 4 & 4 \\[3pt] \operatorname{sgn}_{1} & + & + & - & - & + & - \\ \end{array} \end{align*} $$

There are five nontrivial wild moves for $j = 2, \dotsc , 6$,

$$ \begin{align*} \left(\mu_j,\sigma_j, \operatorname{sgn}_j\right) = W\left(\rho_j\right)(\mu_1, \text{id}, \operatorname{sgn}_1), \end{align*} $$

as indicated in the following table:

$$ \begin{align*} \begin{array}{@{\ \ \ \ }c@{\ \ \ \ }|@{\ \ \ \ }c@{\ \ \ \ }c@{\ \ \ \ }c@{\ \ \ \ }|@{\ \ \ \ }c@{\ \ \ \ }c@{\ \ }|@{\ \ }|@{\ \ \ \ }c@{\ \ \ \ }|@{\ \ \ \ }c@{\ \ \ \ }c@{\ \ \ \ }c@{\ \ \ \ }|@{\ \ \ \ }c@{\ \ \ \ }c} & 2 & 3 & 4 & 6 & 7 & & 2 & 3 & 4 & 6 & 7 \\ \hline \rho _{1} & 2 & 3 & 4 & 6 & 7 & \rho _{1}^{-1} & 2 & 3 & 4 & 6 & 7 \\[2pt] \rho _{2} & 2 & 4 & 3 & 6 & 7 & \rho _{2}^{-1} & 2 & 4 & 3 & 6 & 7 \\[2pt] \rho _{3} & 3 & 4 & 2 & 6 & 7 & \rho _{3}^{-1} & 4 & 2 & 3 & 6 &7 \\[2pt] \rho _{4} & 2 & 3 & 4 & 7 & 6 & \rho _{4}^{-1} & 2 & 3 & 4 & 7 & 6 \\[2pt] \rho _{5} & 2 & 4 & 3 & 7 & 6 & \rho _{5}^{-1} & 2 & 4 & 3 & 7 & 6 \\[2pt] \rho _{6} & 3 & 4 & 2 & 7 & 6 & \rho _{6}^{-1} & 4 & 2 & 3 & 7 & 6 \\ \end{array} \end{align*} $$

Notice that each $\rho _j^{-1}$ preserves the order of $2, 3$, as in Definition 4.6 – meaning that $2$ appears before $3$ in the list $\left (\rho _j^{-1}(2), \rho _j^{-1}(3), \rho _j^{-1}(4)\right )$; equivalently, $\rho (2)<\rho (3)$. Thus the action of $\rho ^{-1}$ on $\{2,3,4\}$ is completely determined by where $4$ appears in the list $\left (\rho _j^{-1}(2), \rho _j^{-1}(3), \rho _j^{-1}(4)\right )$.

The corresponding trees and explicit mappings $\left (\mu _{j},\operatorname {sgn}_{j}\right )$ are indicated in the following. We notice that all $\left (\mu _{j},\operatorname {sgn}_{j}\right )$ are tamed (in accordance with Proposition 4.7) and that wild moves, unlike KM moves, do change the tree skeleton, but this change is restricted to shuffling nodes along a left branch, subject to the restrictions (indicated in Definition 4.6) that the ordering of the plus nodes and minus nodes remain intact.

4.5 Reference forms and tamed integration domains

For example, the tree $(\mu _1,\operatorname {sgn}_1)$ in Example 6 is a reference pair. If $(\ell , \dotsc , r)$ is a full left branch of $\hat \mu $, then the definition of a reference pair means that there is some intermediate position m such that the $\operatorname {sgn}$ map looks like the following:

$$ \begin{align*} \begin{array}{c|c@{\kern10pt}c@{\kern10pt}c@{\kern10pt}c@{\kern10pt}c@{\kern10pt}c@{\kern10pt}c} j & \ell & \dotsb & m-1 & m & m+1 & \dotsb & r \\ \hline \operatorname{sgn} & + & + & + & - & - & - & - \\ \end{array} \end{align*} $$

However, we note that it is possible that they are all plusses $(m=r+1)$ or all minuses $(m=\ell )$. With this notation, we can say that $\rho $ is allowable if $\rho (\ell )< \dotsb < \rho (m-1)$ and $\rho (m) < \dotsb < \rho (r) $ – or equivalently, if in the list

$$ \begin{align*} \left(\rho^{-1}(\ell), \dotsc, \rho^{-1}(r)\right), \end{align*} $$

the values $(\ell , \dotsc , m-1)$ appear in that order and the values $(m, \dotsc , r)$ appear in that order.

Proposition 4.11. An equivalence class of wildly relatable tamed pairs

$$ \begin{align*} Q = \{ (\mu, \operatorname{sgn}) \} \end{align*} $$

contains a unique reference pair $\left (\hat \mu , \hat {\operatorname {sgn}}\right )$. By the definition of being wildly relatable, for every $(\mu ,\operatorname {sgn})\in Q$ there is a unique permutation $\rho $ of $\{2, \dotsc , k+1\}$ such that

$$ \begin{align*} (\mu,\operatorname{sgn}) = W(\rho)\left(\hat \mu, \hat{\operatorname{sgn}}\right), \end{align*} $$

and this $\rho $ is allowable. The collection P of all $\rho $ arising in this way from Q is exactly the set of all allowable $\rho $ with respect to the reference pair $\left (\hat \mu , \hat {\operatorname {sgn}}\right )$.

Now, recall equation (4.9):

$$ \begin{align*} \gamma ^{(1)}(t_{1})=\sum_{\left(\mu ,\operatorname{sgn}\right)\text{ tamed}}\int_{T_{D}(\mu )}J_{\mu ,\operatorname{sgn}}\left(\gamma ^{(k+1)}\right)\left(t_{1},\underline{t}_{k+1}\right)d \underline{t}_{k+1}. \end{align*} $$

In this sum, group together equivalence classes Q of wildly relatable $(\mu ,\operatorname {sgn})$:

(4.13)$$ \begin{align} \gamma ^{(1)}(t_{1})=\sum_{\text{classes }Q}\sum_{(\mu ,\sigma )\in Q}\int_{T_{D}(\mu )}J_{\mu ,\operatorname{sgn}}\left(\gamma ^{(k+1)}\right)\left(t_{1},\underline{t}_{k+1}\right)d\underline{t}_{k+1}. \end{align} $$

Each class Q can be represented by a unique reference $\left (\hat {\mu },\hat {\operatorname {sgn}}\right )$, and as in Proposition 4.11, for each $(\mu ,\operatorname {sgn})\in Q$, there is an allowable $\rho \in P$ (with respect to $\left (\hat {\mu },\hat {\operatorname {sgn}}\right )$) such that

$$ \begin{align*} (\mu ,\operatorname{sgn})=W(\rho )\left(\hat{\mu},\hat{\operatorname{sgn}}\right). \end{align*} $$

Since W is an action, we can write

$$ \begin{align*} \left(\hat{\mu},\hat{\operatorname{sgn}}\right)=W\left(\rho ^{-1}\right)(\mu ,\operatorname{sgn}). \end{align*} $$

Into the action $W\left (\rho ^{-1}\right )$, let us input the identity time permutation and define $\sigma $ as the output time permutation:

$$ \begin{align*} \left(\hat{\mu},\sigma ,\hat{\operatorname{sgn}}\right)=W\left(\rho ^{-1}\right)(\mu ,\text{id},\operatorname{sgn}), \end{align*} $$

where, in accordance with Definition 4.6, $\sigma =\rho ^{-1}$. Since $\rho $ is allowable, this implies that for each left brach $(\ell ,\dotsc ,r)$ with m as already defined, $\sigma ^{-1}(\ell )<\dotsb <\sigma ^{-1}(m-1)$ and $\sigma ^{-1}(m)<\dotsb <\sigma ^{-1}(r)$. In other words, $ (\ell ,\dotsc ,m-1)$ and $(m,\dotsc ,r)$ appear in order inside the list of values $(\sigma (\ell ),\dotsc ,\sigma (r))$. By Proposition 4.8,

$$ \begin{align*} \int_{T_{D}(\mu )}J_{\mu ,\operatorname{sgn}}\left(\gamma ^{(k+1)}\right)\left(t_{1},\underline{t}_{k+1}\right)d\underline{t}_{k+1}=\int_{\sigma \left\lbrack T_{D}\left(\hat{\mu}\right)\right]}J_{\hat{\mu},\hat{\operatorname{sgn}}}\left(\gamma ^{(k+1)}\right)\left(t_{1},\underline{t}_{k+1}\right)d\underline{t}_{k+1}. \end{align*} $$

Now as we sum this over all $(\mu ,\operatorname {sgn})\in Q$, we are summing over all $\rho \in P$ and hence over all $\sigma =\rho ^{-1}$ meeting the condition already mentioned. Hence the integration domains on the right side union to a set that we will denote

$$ \begin{align*} T_{R}\left(\hat{\mu},\hat{\operatorname{sgn}}\right)\overset{\text{def}}{=}\bigcup_{\rho \in P}\sigma \left(T_{D}\left(\hat{\mu}\right)\right), \end{align*} $$

which can be described as follows: For each left branch $(\ell ,\dotsc ,r)$, with

$$ \begin{align*} z=\mu (\ell )=\dotsb =\mu (r) \end{align*} $$

and m the division index between plus and minus nodes, $T_{R}\left (\hat {\mu },\hat {\operatorname {sgn}}\right )$ is described by the inequalities

(4.14)$$ \begin{align} t_{m-1}\leq \dotsb \leq t_{\ell }\leq t_{z}\quad \text{and}\quad t_{r}\leq \dotsb \leq t_{m}\leq t_{z}. \end{align} $$

Plugging into equation (4.13), we obtain the following:

Proposition 4.12. The Duhamel expansion to coupling order k can be grouped into at most $8^{k}$ terms:

(4.15)$$ \begin{align} \gamma ^{(1)}(t_{1})=\sum_{\text{reference }\left(\hat{\mu},\hat{\operatorname{sgn}}\right)}\int_{T_{R}\left(\hat{\mu},\hat{\operatorname{sgn}}\right)}J_{\mu ,\operatorname{sgn}}\left(\gamma ^{(k+1)}\right)\left(t_{1},\underline{t}_{k+1}\right)d\underline{t}_{k+1}, \end{align} $$

where each integration domain $T_{R}\left (\hat {\mu },\hat {\operatorname {sgn}}\right )$ is as defined in formula (4.14).

A quick example of formulas (4.14) and (4.15) is Example 4, in which the reference tree is the one corresponding to $I_2$. Reading from that tree, formula (4.14) becomes the set D and the combined integral is equation (4.4).

Returning to Example 6, $(\mu _1,\operatorname {sgn}_1)$ is the reference pair. To combine the Duhamel integrals, we convert all five other tamed forms $\left (\mu _j,\operatorname {sgn}_j\right )$ to $(\mu _1,\operatorname {sgn}_1)$ via wild moves. The resulting combined time integration set will be read off from the $(\mu _1,\operatorname {sgn}_1)$ tree as

$$ \begin{align*} t_3\leq t_2 \leq t_1, \qquad t_4 \leq t_1, \qquad t_5 \leq t_2, \qquad t_6 \leq t_4, \qquad t_7 \leq t_4. \end{align*} $$

Proposition 4.12 and the integration domain (4.14) are compatible with the U-V space techniques we proved in Section 2. This fact may not be so clear at the moment, as they are written with much shorthand. We will prove this fact in Section 5.2.

5.1 An example of how to estimate

We estimate the integral in Example 4:

$$ \begin{align*} I=\int_{t_{4}=0}^{t_{1}}\int_{t_{2}=0}^{t_{1}} \int_{t_{3}=t_{4}}^{t_{1}}U^{(1)}(t_{1}-t_{2})B_{1,2}^{+}U^{(2)}(t_{2}-t_{3})B_{1,3}^{-}U^{(3)}(t_{3}-t_{4})B_{3,4}^{+}\gamma ^{(4)}d \underline{t}_{4}, \end{align*} $$

where the integration limits have already been computed in Section 4.2. Its reference tree is exactly the tree corresponding to $I_{2}$ in Example 4.

Plugging in equation (3.6), we find that the integrand is in fact

$$ \begin{align*} I&=\int_{t_{4}=0}^{t_{1}}dt_{4}\int d\mu _{t_{4}}\left( \phi \right) \int_{t_{2}=0}^{t_{1}}dt_{2}\int_{t_{3}=t_{4}}^{t_{1}}U_{1,2}\left(\left\lvert U_{2,4}\phi \right\rvert ^{2}U_{2,4}\phi \right)(x_{1})\\ &\qquad {}\times U_{1,3}\left(\overline{U_{3,4}\phi }\overline{U_{3,4}\phi }U_{3,4}\left( \left\lvert \phi \right\rvert ^{2}\phi \right) \right)\left(x^{\prime}_{1}\right)dt_{3}. \end{align*} $$

We denote the cubic term $\left \lvert \phi \right \rvert ^{2}\phi $ generated in the innermost coupling with $\mathcal {C}_{R}^{(4)}$, where the subscript R stands for ‘rough’, as it has no propagator inside to smooth things out. That is,

$$ \begin{align*} I=\int_{t_{4}=0}^{t_{1}}dt_{4}\int d\mu _{t_{4}}\left( \phi \right) \int_{t_{2}=0}^{t_{1}}dt_{2}\int_{t_{3}=t_{4}}^{t_{1}}U_{1,2}\left(\left\lvert U_{2,4}\phi \right\rvert ^{2}U_{2,4}\phi \right)(x_{1})U_{1,3}\left(\overline{U_{3,4}\phi }\overline{U_{3,4}\phi }U_{3,4}\mathcal{C}_{R}^{(4)}\right)\left(x^{\prime}_{1}\right)dt_{3}. \end{align*} $$

For expression (3.9) with a general k, we will use $\mathcal {C}_{R}^{(k+1)}$ to denote this innermost cubic term. Notice that $\mathcal {C}_{R}^{(k+1)}$ is always independent of time and is hence qualified to be an $f_{j}$ in estimates (5.1)–(5.4).

In the second coupling, if we denote

$$ \begin{align*} D_{\phi ,R}^{(3)}=U_{-t_{3}}\left(\overline{U_{3,4}\phi }\overline{U_{3,4}\phi } U_{3,4}\mathcal{C}_{R}^{(4)}\right)\left(x^{\prime}_{1}\right), \end{align*} $$

we have

$$ \begin{align*} \int_{t_{3}=t_{4}}^{t_{1}}U_{1,3}\left(\overline{U_{3,4}\phi }\overline{ U_{3,4}\phi }U_{3,4}\mathcal{C}_{R}^{(4)}\right)\left(x^{\prime}_{1}\right)dt_{3}=\int_{t_{3}=t_{4}}^{t_{1}}U_{1}D_{\phi ,R}^{(3)}dt_{3}. \end{align*} $$

In general, let us use $D^{(l+1)}$, which is $D_{\phi ,R}^{(3)}$ here, to denote the cubic term together with the $U(-t_{l+1})$ during the lth coupling where $l<k$. We add a $\phi $ subscript if the cubic term generated at the lth coupling has contracted a $U\phi $. We add an R subscript if the cubic term generated at the lth coupling has contracted the rough cubic term $\mathcal {C}_{R}^{(k+1)}$ or a $ D_{R}^{\left (j+1\right )} $ for some j. The coupling process makes sure that every time integral corresponds to one and only one cubic term, and thus the notation of D is well defined. We suppress all $t_{k+1}$-dependence, which is the $t_{4}$-dependence here, in all the D markings, as we will not explore any smoothing given by the $dt_{k+1}$ integral. Finally, notice that $ D^{(l+1)}$ always carries the $t_{l+1}$ variable and will make a Duhamel term whenever it is hit by a $U\left (t_{j}\right )$, where $j\neq l+1$.

Then, using the same marking strategy at the first coupling, we reach

$$ \begin{align*} I=\int_{t_{4}=0}^{t_{1}}dt_{4}\int d\mu _{t_{4}}\left( \phi \right) \left( \int_{t_{2}=0}^{t_{1}}U_{1}D_{\phi }^{(2)}\left( x_{1}\right) dt_{2}\right) \left( \int_{t_{3}=t_{4}}^{t_{1}}U_{1}D_{\phi ,R}^{(3)}\left(x^{\prime}_{1}\right)dt_{3}\right). \end{align*} $$

We can now start estimating. Taking the norm inside,

$$ \begin{align*} &\left\lVert \left\langle \nabla _{x_{1}}\right\rangle ^{-1}\left\langle \nabla _{x^{\prime}_{1}}\right\rangle ^{-1}I\right\rVert _{L_{t_{1}}^{\infty }L_{x,x'}^{2}} \\ \leq &\int_{0}^{T}\int dt_{4}d\left\lvert \mu _{t_{4}}\right\rvert \left( \phi \right) \left\lVert \left(\left\langle \nabla _{x_{1}}\right\rangle ^{-1}\int_{t_{2}=0}^{t_{1}}U_{1}D_{\phi }^{(2)}\left( x_{1}\right) dt_{2}\right)\left(\left\langle \nabla _{x^{\prime}_{1}}\right\rangle ^{-1}\int_{t_{3}=t_{4}}^{t_{1}}U_{1}D_{\phi ,R}^{(3)}\left(x^{\prime}_{1}\right)dt_{3}\right)\right\Vert _{L_{t_{1}}^{\infty }L_{x,x'}^{2}}, \end{align*} $$

the $L_{t_{1}}^{\infty }L_{x,x'}^{2}$ norm ‘factors’ in the sense that

$$ \begin{align*} &\left\lVert \left\langle \nabla _{x_{1}}\right\rangle ^{-1}\left\langle \nabla _{x^{\prime}_{1}}\right\rangle ^{-1}I\right\rVert _{L_{t_{1}}^{\infty }L_{x,x'}^{2}} \\ &\qquad\qquad\leq \int_{0}^{T}\int \left\lVert \int_{t_{2}=0}^{t_{1}}U_{1}D_{\phi }^{(2)}\left( x_{1}\right) dt_{2}\right\rVert _{L_{t_{1}}^{\infty }H_{x}^{-1}}\left\lVert \int_{t_{3}=t_{4}}^{t_{1}}U_{1}D_{\phi ,R}^{(3)}\left(x^{\prime}_{1}\right)dt_{3}\right\rVert _{L_{t_{1}}^{\infty }H_{x'}^{-1}}dt_{4}d\left\lvert \mu _{t_{4}}\right\rvert \left( \phi \right). \end{align*} $$

The term $D_{\phi }^{(2)}$ carries no R subscript, so we can bump it to $ H^{1} $ and then use the embedding (2.1), which gives

$$ \begin{align*} &\left\lVert \left\langle \nabla _{x_{1}}\right\rangle ^{-1}\left\langle \nabla _{x^{\prime}_{1}}\right\rangle ^{-1}I\right\rVert _{L_{t_{1}}^{\infty }L_{x,x'}^{2}} \\ &\qquad\qquad\qquad \quad \leq \int_{0}^{T}\int \left\lVert \int_{t_{3}=t_{4}}^{t_{1}}U_{1}D_{\phi ,R}^{(3)}\left(x^{\prime}_{1}\right)dt_{3}\right\rVert _{X^{-1}}\left\lVert \int_{t_{2}=0}^{t_{1}}U_{1}D_{\phi }^{(2)}\left( x_{1}\right) dt_{2}\right\rVert _{X^{1}}dt_{4}d\left\lvert \mu _{t_{4}}\right\rvert \left( \phi \right). \end{align*} $$

Applying formula (5.2) to the first coupling and replacing all $\left \lVert U\phi \right \rVert _{X^{s}}$ by $\left \lVert \phi \right \rVert _{H^{s}}$, we have

$$ \begin{align*} &\left\lVert \left\langle \nabla _{x_{1}}\right\rangle ^{-1}\left\langle \nabla _{x^{\prime}_{1}}\right\rangle ^{-1}I\right\rVert _{L_{t_{1}}^{\infty }L_{x,x'}^{2}} \\ \leq &C\int_{0}^{T}\int \left\lVert \int_{t_{3}=t_{4}}^{t_{1}}U_{1}D_{\phi ,R}^{(3)}\left(x^{\prime}_{1}\right)dt_{3}\right\rVert _{X^{-1}}\left\lVert \phi \right\rVert _{H^{1}}^{2}\left( T^{\frac{1}{7}}M_{0}^{\frac{3}{5}}\left\lVert P_{\leq M_{0}}\phi \right\rVert _{H^{1}}+\left\lVert P_{>M_{0}}\phi \right\rVert _{H^{1}}\right) ^{1}dt_{4}d\left\lvert \mu _{t_{4}}\right\rvert \left( \phi \right). \end{align*} $$

Using formula (5.1) with the second coupling and replacing all $\left \lVert U\phi \right \rVert _{X^{s}}$ by $\left \lVert \phi \right \rVert _{H^{s}}$, we have

$$ \begin{align*} \left\lVert \left\langle \nabla _{x_{1}}\right\rangle ^{-1} \vphantom{\left\langle \nabla _{x^{\prime}_{1}}\right\rangle ^{-1}} \right. & \left. \left\langle \nabla _{x^{\prime}_{1}}\right\rangle ^{-1}I\right\rVert _{L_{t_{1}}^{\infty }L_{x,x'}^{2}} \\ &\leq C^{2}\int_{0}^{T}\int \left\lVert \phi \right\rVert _{H^{1}}^{3}\left( T^{\frac{1}{7}}M_{0}^{\frac{3}{5}}\left\lVert P_{\leq M_{0}}\phi \right\rVert _{H^{1}}+\left\lVert P_{>M_{0}}\phi \right\rVert _{H^{1}}\right) ^{2}\left\lVert \mathcal{C}_{R}^{(4)}\right\rVert _{H^{-1}}dt_{4}d\left\lvert \mu _{t_{4}}\right\rvert \left( \phi \right) \\ &= C^{2}\int_{0}^{T}\int \left\lVert \phi \right\rVert _{H^{1}}^{3}\left( T^{ \frac{1}{7}}M_{0}^{\frac{3}{5}}\left\lVert P_{\leq M_{0}}\phi \right\rVert _{H^{1}}+\left\lVert P_{>M_{0}}\phi \right\rVert _{H^{1}}\right) ^{2}\left\lVert \left\lvert \phi \right\rvert ^{2}\phi \right\rVert _{H^{-1}}dt_{4}d\left\lvert \mu _{t_{4}}\right\rvert \left( \phi \right). \end{align*} $$

Using the 4D Sobolev

(5.5)$$ \begin{align} \left\lVert \left\lvert \phi \right\rvert ^{2}\phi \right\rVert _{H^{-1}}\leq C\left\lVert \phi \right\rVert _{H^{1}}^{3} \end{align} $$

on the rough coupling, we get to

$$ \begin{align*} \left\lVert \left\langle \nabla _{x_{1}}\right\rangle ^{-1}\left\langle \nabla _{x^{\prime}_{1}}\right\rangle ^{-1}I\right\rVert _{L_{t_{1}}^{\infty }L_{x,x'}^{2}}\leq C^{3}\int_{0}^{T}dt_{4}\int d\left\lvert \mu _{t_{4}}\right\rvert \left( \phi \right) \left\lVert \phi \right\rVert _{H^{1}}^{6}\left( T^{\frac{1}{7}}M_{0}^{\frac{3}{5}}\left\lVert P_{\leq M_{0}}\phi \right\rVert _{H^{1}}+\left\lVert P_{>M_{0}}\phi \right\rVert _{H^{1}}\right) ^{2}. \end{align*} $$

Plugging in the support property of the measure (see equation (3.5)) yields

(5.6)$$ \begin{align} \left\lVert \left\langle \nabla _{x_{1}}\right\rangle ^{-1}\left\langle \nabla _{x^{\prime}_{1}}\right\rangle ^{-1}I\right\rVert _{L_{t_{1}}^{\infty }L_{x,x'}^{2}} &\leq C^{3}C_{0}^{6}\left( T^{\frac{1}{7} }M_{0}^{\frac{3}{5}}C_{0}+\varepsilon \right) ^{2}\int_{0}^{T}dt_{4}\int d\left\lvert \mu _{t_{4}}\right\rvert \left( \phi \right) \\ &\leq C^{3}C_{0}^{6}\left( T^{\frac{1}{7}}M_{0}^{\frac{3}{5} }C_{0}+\varepsilon \right) ^{2}2T, \notag \end{align} $$

and we are done.

5.2 The extended KM board game is compatible

In Section 5.1, the U-V estimates worked perfectly with the integration limits obtained via the extended KM board game from Section 4. One certainly wonders whether the extended KM board game is necessary, and whether it is compatible with the estimates in the general case.

In the beginning of Section 4.2, we briefly mentioned the problem one would face without the extended KM board game. We can now explain by a concrete example. For comparison, rewrite $I_{1}$ in Example 4 with the notation

$$ \begin{align*} I_{1} &=\int_{t_{4}=0}^{t_{1}}\int_{t_{2}=t_{4}}^{t_{1}} \int_{t_{3}=0}^{t_{2}}U^{(1)}(t_{1}-t_{2})B_{1,2}^{-}U^{(2)}(t_{2}-t_{3})B_{1,3}^{+}U^{(3)}(t_{3}-t_{4})B_{3,4}^{-}\gamma ^{(4)}d \underline{t}_{4} \\ &=\int_{t_{4}=0}^{t_{1}}dt_{4}\int d\mu _{t_{4}}\left( \phi \right) \int_{t_{2}=t_{4}}^{t_{1}}\left( U_{1}D_{\phi }^{(2)}\left( x_{1}\right) \left[ \int_{t_{3}=0}^{t_{2}}U_{1}D_{\phi ,R}^{(3)}\left(x^{\prime}_{1}\right)dt_{3} \right] \right) dt_{2}. \end{align*} $$

One sees that the $dt_{3}$ integral is encapsulated inside the $dt_{2}$ integral, or the x and $x'$ parts do not factor, even with the carefully worked-out time integration limits in the original KM board game. Hence, one cannot apply U-V estimates. To be very precise for readers who are curious about this, since there are only two integrals that got entangled, $I_{1}$ could in fact be estimated using [Reference Koch, Tataru and Vişan50, (4.25), p. 60], based on the idea of integration by parts. However, if one allows the coupling level to be large, it is not difficult to find, at any stage of a long coupling, multiple encapsulations which have more than three factors entangled together and cannot be estimated by the ideas of integration by parts. We are not presenting such a construction, as the formula would be unnecessarily long and does not give new ideas. Finally, we remark that such an entanglement problem, generated by the time-integral reliance of the U-V space techniques, does not show up in the couplings with only $B^{+}$ or only $B^{-}$, and does not have to emerge in the $\mathbb {R}^{3}/\mathbb {R}^{4}/\mathbb {T}^{3}$ cases in which U-V spaces are not necessary.

We now prove how the extended KM board game is compatible with the U-V techniques. Given a reference tree, we will create a Duhamel tree (we write ‘D-tree’ for short) to supplement the reference tree. The D-tree supplements the given reference tree in the sense that the D-tree completely shows the arrangement of the cubic terms $D^{\left (j\right )}$, defined in Section 5.1, and one could also read off the integration limits from it as in the given reference tree. The whole point of the D-tree is to get these two pieces of information in the same picture, as the proof of compatibility then follows trivially. Of course, from now on, we assume that equation 3.6) has already been plugged in and we are doing the $dt_{k+1}$ integral, which is from $0$ to $t_{1}$, last.

Algorithm 7. In the D-tree, we will write each node prefaced by a D. Each node $D^{(j)}$ will have a left child, middle child and right child:

The labelling of ls, r+ and r$-$ for the left, middle and right children, respectively, is a shorthand mnemonic for the procedure for determining the children of $D^{\left (j\right )}$ by inspecting the reference tree. Apply the following steps for $j=1$ (with no left child), then repeat the steps for all $D^{\left (j\right )}$ that appear as children; continue to repeat the steps until all vertices without children are F:

1. 1. To determine the left child of $D^{\left (j\right )}$, locate node j in the reference tree and apply the ‘left same’ rule. If node j in the reference tree is $+$, and $j+$ has a left child $\ell +$ (of the same sign $+$), then place $D^{(l)}$ as the left child in the D-tree. If the j node in the reference tree is $-$, and $j-$ has a left child $\ell -$ (of the same sign $-$), then place $D^{(l)}$ as the left child in the D-tree. If node j in the reference tree does not have a left child of the same sign, then place F as the left child of $D^{\left (j\right )}$ in the D-tree.

2. 2. To determine the middle and right children of $D^{\left (j\right )}$, locate node j in the reference tree. Examine the right child of j (if it exists), and consider its full left branch

   $$ \begin{align*} p_{1}+,\dotsc ,p_{\alpha }+,n_{1}-,\dotsc ,n_{\beta }-. \end{align*} $$
   It is possible here that $\alpha =0$ (no $+$ nodes on this left branch), and it is also possible that $\beta =0$ (no $-$ nodes on this left branch). In the D-tree, as the middle child of $D^{\left (j\right )}$ place $D^{\left (p_{1}\right )}$, and as the right child of $D^{\left (j\right )}$ place $D^{(n_{1})}$. If either or both is missing ($\alpha =0$ or $\beta =0$, respectively), place F instead.

A quick and simple example is the D-tree for the integral in Section 5.1:

Here is a longer example:

Example 7. Consider the following reference tree:

Its supplemental D-tree is as follows:

Every bottom node of the form $D^{\left (j\right )}$ (as opposed to F) has implicitly three F children, except for the $D^{(k+1)}$ node, which is special (in our case here, it is $D_{9}$). In this case, the D-tree was generated as follows. Take $D^{(2)}$, for example, in the reference tree.

• • To determine the left child of $D^{(2)}$ in the D-tree, we look at the reference tree and follow the ‘left same’ rule. The left child of $2+$ is $3+$, so we place $ D^{(3)}$ as the left child of $D^{(2)}$ in the D-tree. (If it were $3-$ instead, we would place F in the D-tree, since the signs are different.)

• • To determine the middle child of $D^{(2)}$ in the D-tree, we look at the reference tree and follow the ‘right $+$’ rule. That is, we take the right child and consider its left branch: $5+$, $6-$, $7-$. We note the first $+$ node, which is $ 5+ $, and assign $D^{(5)}$ as the middle child of $D^{(2)}$. If there were no $+$ node in the left branch, we would have assigned F.

• • To determine the right child of $D^{(2)}$ in the D-tree, we look at the reference tree and follow the ‘right $-$’ rule. That is, we take the right child and consider its left branch: $5+$, $6-$, $7-$. We note the first $-$ node, which is $6-$, and assign $D^{(6)}$ as the right child of $D_{2}$. If there were no $-$ node in the left branch, we would have assigned F.

Proof of compatibility.

With the D-tree, we can now read formula (4.14) better. This is because the rule for assigning upper limits of time integration is actually the same rule for constructing children in the D-tree. By the construction of the D-tree, we can write the form of each $D^{\left (j\right )}$, $j\neq k+1$, and the integration limit for $t_{j}$. If $D^{\left (j\right )}$ has children $L, M, R$ (for left, middle and right) and has parent $D^{(l)}$ in the $ D $-tree, then (ignoring the role of complex conjugates)

$$ \begin{align*} D^{\left(j\right)}\left(t_{j}\right)=U\left(-t_{j}\right)\left[\left( U_{j}L\right) \left( U_{j}M\right) \left( U_{j}R\right) \right], \end{align*} $$

and the integration of $t_{j}$ is exactly from $0$ to $t_{l}$. One can directly see from the picture in Example 7 that all Duhamel terms inside a $D^{\left (j\right )}$ must have the same integration limit, and they factor. Therefore, there is no entanglement in each stage of the coupling process. An induction then shows that there is no entanglement for any coupling of finite length or stages. Or in other words, the extended KM board game is compatible with the U-V techniques.

For completeness, we finish Example 7 with the integration limits:

Example 8. Continuing Example 7, we have

(5.7)$$ \begin{align} D^{(2)}=U(-t_{2})\left[U_{2}D^{(3)}\cdot U_{2}D^{(5)}\cdot U_{2}D^{(6)}\right]. \end{align} $$

The three terms inside this expression are

$$ \begin{align*} D^{(3)} & =U(-t_{3})[U_{3}F(t_{9})\cdot U_{3}F(t_{9})\cdot U_{3}F(t_{9})], \\ D^{(5)} & =U(-t_{5})[U_{5}F(t_{9})\cdot U_{5}F(t_{9})\cdot U_{5}F(t_{9})], \\ D^{(6)} & =U(-t_{6})\left[U_{6}D^{(7)}\cdot U_{6}D^{(9)}\cdot U_{6}F(t_{9})\right], \end{align*} $$

where $F(t_{i})=U(-t_{i})\phi $. On the other hand, we have

(5.8)$$ \begin{align} D^{(4)}=U(-t_{4})\left[U_{4}F(t_{9})\cdot U_{4}D^{(8)}\cdot U_{4}F(t_{9})\right]. \end{align} $$

Now, read the time integration limits from the reference tree or the D -tree; $t_{2}$ and $t_{4}$ have upper limit $t_{1}$, while $t_{3}$, $ t_{5}$ and $t_{6}$ all have upper limit $t_{2}$, and so on. Start by writing $ \int _{t_{9}=0}^{t_{1}}$ on the outside. Notice that this makes the inner $t_6$ integral start at $t_9$ in order to retain the condition $t_9\leq t_6$ from formula (4.14) and the tree reading. Take all $t_{j}$ integrals for $j=2$ or for which $D^{\left (j\right )}$ is a descendant of $D^{(2)}$. This is

(5.9)$$ \begin{align} \int_{t_{2}=0}^{t_{1}}\int_{t_{3}=0}^{t_{2}}\int_{t_{5}=0}^{t_{2}} \int_{t_{6}=t_9}^{t_{2}}\int_{t_{7}=0}^{t_{6}}. \end{align} $$

Then collect all $t_{j}$ integrals for $j=4$ or for which $D^{\left (j\right )}$ is a descendant of $D^{(4)}$. This is

(5.10)$$ \begin{align} \int_{t_{4}=0}^{t_{1}}\int_{t_{8}=0}^{t_{4}}. \end{align} $$

Notice that expressions (5.9) and (5.10) split, by Fubini, since none of the limits of integration in expression (5.9) appear in expression (5.10), and vice versa. So we can write this piece of $\gamma ^{(1)}$ as

(5.11)$$ \begin{align} \gamma ^{(1)}(t_{1})=\int_{t_{9}=0}^{t_{1}}\left[ \int_{t_{2}=0}^{t_{1}} \int_{t_{3}=0}^{t_{2}}\int_{t_{5}=0}^{t_{2}}\int_{t_{6}=t_9}^{t_{2}} \int_{t_{7}=0}^{t_{6}}U_{1}D^{(2)}(t_{2},x_{1})\right] \left[ \int_{t_{4}=0}^{t_{1}}\int_{t_{8}=0}^{t_{4}}U_{1}D^{(4)}\left(t_{4},x^{\prime}_{1}\right)\right]. \end{align} $$

Write out $D^{(2)}$ as in equation (5.7) and $D^{(4)}$ as in equation (5.8). Notice that we can distribute the integrals $\int _{t_{3}=0}^{t_{2}} \int _{t_{5}=0}^{t_{2}}\int _{t_{6}=0}^{t_{2}}$ onto the $D^{(3)}$, $D^{(5)}$ and $D^{(6)}$ terms, respectively:

$$ \begin{align*} \int_{t_{2}=0}^{t_{1}}\int_{t_{3}=0}^{t_{2}}\int_{t_{5}=0}^{t_{2}} &\int_{t_{6}=t_9}^{t_{2}}\int_{t_{7}=0}^{t_{6}}U_{1}D^{(2)}(t_{2},x_{1}) \\ &=\int_{t_{2}=0}^{t_{1}}U_{1,2}\left[ \left( \int_{t_{3}=0}^{t_{2}}U_{2}D^{(3)}(t_{3})\right) \cdot \left( \int_{t_{5}=0}^{t_{2}}U_{2}D^{(5)}(t_{5})\right) \cdot \left( \int_{t_{6}=t_9}^{t_{2}}\int_{t_{7}=0}^{t_{6}}U_{2}D^{(6)}(t_{6})\right) \right]. \end{align*} $$

We have kept the $t_{7}$ integral together with $t_{6}$ because $D^{(7)}$ is a child of $D^{(6)}$ in the D-tree. We can see that all the Duhamel structures are fully compatible with the U-V techniques. The rest is similar, and we omit further details.

5.3 Estimates for general k

As the compatiblity between the extended KM board game and the U-V techniques has been proved in Section 5.2, we can now apply the U-V techniques from Section 5.1 to the general case. We see from Section 5.1 that estimates (5.1) and (5.2) provide gains whenever the lth coupling contracts a $U\phi $. For large k, at least $\frac {2}{3}k$ of the couplings carry such a property and thus allow gains.

Proof. Assume there are j congested couplings; then there are $(k-1-j)$ unclogged couplings. Before the $(k-1)$th coupling, there are $2k-1$ copies of $U\phi $ available. After the first coupling, all of these $2k-1$ copies of $U\phi $ except one must be inside some Duhamel term. Since the j congested couplings do not consume any $U\phi $, to consume all $2k-2$ copies of $ U\phi $ we have to have

(5.12)$$ \begin{align} 2k-2\leq 3(k-1-j), \end{align} $$

because a unclogged coupling can consume at most three copies of $U\phi $. This inequality certainly holds only if $j< \frac {k}{3}$. Hence, there are at least $\frac {2k}{3}$ unclogged couplings.

We can now present the algorithm which proves the general case:

1. Step 0 Plug equation (3.6) into expression (3.9). Mark $\mathcal {C}_{R}^{(k+1)}$ and all $ D^{(l+1)}$ for $l=1,\dotsc ,k-1$ per the general rule given in the example and Section 5.1. We obtain

   $$ \begin{align*} &\left\lVert \left\langle \nabla _{x_{1}}\right\rangle ^{-1}\left\langle \nabla _{x^{\prime}_{1}}\right\rangle ^{-1}\int_{I_{2}}\dotsi\int_{I_{k}}J_{\mu _{m},\operatorname{sgn}}^{(k+1)}\left(\gamma ^{(k+1)}\right)\left(t_{1},\underline{t}_{k+1}\right)d\underline{t}_{k+1}\right\rVert _{L_{t_{1}}^{\infty }L_{x,x'}^{2}} \\ &\qquad\leq \int_{0}^{T}dt_{k+1}\int d\left\lvert \mu _{t_{k+1}}\right\rvert \left( \phi \right) \left\lVert \left( \left\langle \nabla _{x_{1}}\right\rangle ^{-1}f^{(1)}(t_{1},x_{1})\right) \left( \left\langle \nabla _{x^{\prime}_{1}}\right\rangle ^{-1}g^{(1)}\left(t_{1},x^{\prime}_{1}\right)\right) \right\rVert _{L_{t_{1}}^{\infty }L_{x,x'}^{2}}, \end{align*} $$
   which ‘factors’ into
   $$ \begin{align*} &\leq \int_{0}^{T}dt_{k+1}\int d\left\lvert \mu _{t_{k+1}}\right\rvert \left( \phi \right) \left\lVert \left\langle \nabla _{x_{1}}\right\rangle ^{-1}f^{(1)}(t_{1},x_{1})\right\rVert _{L_{t_{1}}^{\infty }L_{x}^{2}}\left\lVert \left\langle \nabla _{x^{\prime}_{1}}\right\rangle ^{-1}g^{(1)}\left(t_{1},x^{\prime}_{1}\right)\right\rVert _{L_{t_{1}}^{\infty }L_{x'}^{2}} \\ &\leq \int_{0}^{T}dt_{k+1}\int d\left\lvert \mu _{t_{k+1}}\right\rvert \left( \phi \right) \left\lVert f^{(1)}\right\rVert _{X^{-1}}\left\lVert g^{(1)}\right\rVert _{X^{-1}} \end{align*} $$
   for some $f^{(1)}$ and $g^{(1)}$. Of course, only one of $f^{(1)}$ and $ g^{(1)}$ can carry the cubic rough term $\mathcal {C}_{R}^{(k+1)}$, as there is only one, so bump the other one into $X^{1}$. Go to step 1.

2. Step 1 Set a counter $l=1$ and go to step 2.

3. Step 2 If $D^{(l+1)}$ is a $D_{\phi ,R}^{(l+1)}$, apply estimate (5.1), put the factor carrying $\mathcal {C} _{R}^{(k+1)}$ – which will be a $U\mathcal {C}_{R}^{(k+1)}$ or a $ D_{R}^{\left (j+1\right )}$ for some j – in $X^{-1}$ and replace the $X^{1}$ norm of $U\phi $ by the $H^{1}$ norm of $\phi $; if the ending estimate includes $ \left \lVert U\mathcal {C}_{R}^{(k+1)}\right \rVert _{X^{-1}}$, replace it by $ \left \lVert \mathcal {C}_{R}^{(k+1)}\right \rVert _{H^{-1}}$. Then go to step 6. If $D^{(l+1)}$ is not a $D_{\phi ,R}^{(l+1)}$, go to step 3.

4. Step 3 If $D^{(l+1)}$ is a $D_{\phi }^{(l+1)}$, apply estimate (5.2) and replace the $X^{1}$ norm of $U\phi $ by the $H^{1}$ norm of $\phi $. Then go to step 6. If $D^{(l+1)}$ is not a $ D_{\phi }^{(l+1)}$, go to step 4.

5. Step 4 If $D^{(l+1)}$ is a $D_{R}^{(l+1)}$, apply estimate (5.3), put the factor carrying $\mathcal {C} _{R}^{(k+1)}$ – which will be a $U\mathcal {C}_{R}^{(k+1)}$ or a $ D_{R}^{\left (j+1\right )}$ for some j – in $X^{-1}$ and replace the $X^{1}$ norm of $U\phi $ by the $H^{1}$ norm of $\phi $; if the ending estimate includes $ \left \lVert U\mathcal {C}_{R}^{(k+1)}\right \rVert _{X^{-1}}$, replace it by $ \left \lVert \mathcal {C}_{R}^{(k+1)}\right \rVert _{H^{-1}}$. Then go to step 6. If $D^{(l+1)}$ is not a $D_{R}^{(l+1)}$, go to step 5.

6. Step 5 If $D^{(l+1)}$ is a $D^{(l+1)}$, apply estimate (5.4) and replace the $X^{1}$ norm of $U\phi $ by the $H^{1}$ norm of $\phi $. Then go to step 6.

7. Step 6 Set the counter $l=l+1$. If $l<k$, go to step 2; otherwise go to step 7.

8. Step 7 Replace all the leftover $\left \lVert U\phi \right \rVert _{X^{1}} $ by $\left \lVert \phi \right \rVert _{H^{1}}$. There is actually at most one leftover $\left \lVert U\phi \right \rVert _{X^{1}}$, which is exactly $ f^{(1)}$ or $g^{(1)}$ from the beginning and only happens when the sign $ J_{\mu _{m},\operatorname {sgn}}^{(k+1)}$ under consideration is all $+$ or all $-$. As it is not inside any Duhamel, it is not taken care of by steps 1–6. Go to step 8.

9. Step 8 We are now at the kth coupling, and have applied formulas (5.1) and (5.2) at least $\frac {2}{3}k$ times; thus we are looking at

   $$ \begin{align*} &\left\lVert \left\langle \nabla _{x_{1}}\right\rangle ^{-1}\left\langle\nabla _{x^{\prime}_{1}}\right\rangle^{-1}\int_{I_{2}}\dotsi\int_{I_{k}}J_{\mu _{m},\operatorname{sgn}}^{(k+1)}\left(\gamma ^{(k+1)}\right)\left(t_{1},\underline{t}_{k+1}\right)d\underline{t}_{k+1}\right\rVert_{L_{t_{1}}^{\infty }L_{x,x'}^{2}} \\ \leq &C^{k-1}\int_{0}^{T}dt_{k+1}\int d\left\lvert \mu_{t_{k+1}}\right\rvert \left( \phi \right) \left\lVert \phi \right\rVert _{H^{1}}^{\frac{4}{3}k-1}\left( T^{\frac{1}{7}}M_{0}^{\frac{3}{5}}\left\lVert P_{\leq M_{0}}\phi \right\rVert _{H^{1}}+\left\lVert P_{>M_{0}}\phi \right\rVert_{H^{1}}\right) ^{\frac{2}{3}k}\left\lVert \left\lvert \phi \right\rvert^{2}\phi \right\rVert _{H^{-1}}. \end{align*} $$
   Applying the 4D Sobolev (5.5) to the rough factor yields
   $$ \begin{align*} \leq C^{k}\int_{0}^{T}dt_{k+1}\int d\left\lvert \mu _{t_{k+1}}\right\rvert \left( \phi \right) \left\lVert \phi \right\rVert _{H^{1}}^{\frac{4}{3}k+2}\left( T^{\frac{1}{7}}M_{0}^{\frac{3}{5}}\left\lVert P_{\leq M_{0}}\phi \right\rVert _{H^{1}}+\left\lVert P_{>M_{0}}\phi \right\rVert _{H^{1}}\right) ^{\frac{2}{3}k}. \end{align*} $$
   Putting in the support property (3.5) gives
   $$ \begin{align*} &\leq \int_{0}^{T}dt_{k+1}\int d\left\lvert \mu _{t_{k+1}}\right\rvert \left( \phi \right) C^{k}C_{0}^{\frac{4}{3}k+2}\left( T^{\frac{1}{7}}M_{0}^{ \frac{3}{5}}C_{0}+\varepsilon \right) ^{\frac{2}{3}k} \\ &\leq 2TC^{k}C_{0}^{\frac{4}{3}k+2}\left( T^{\frac{1}{7}}M_{0}^{\frac{3 }{5}}C_{0}+\varepsilon \right) ^{\frac{2}{3}k} \\ &\leq 2TC_{0}^{2}\left( CC_{0}^{3}T^{\frac{1}{7}}M_{0}^{\frac{3}{5} }+CC_{0}^{2}\varepsilon \right) ^{\frac{2}{3}k}, \end{align*} $$
   as claimed.