One observes that for a given equation we have constructed a rather large renormalisation group $\mathfrak{G}_-$. In our setting, compared to the usual setting of translation invariant SPDEs in flat space, one a priori obtains a family of solutions indexed by the (infinite-dimensional) space of sections of certain vector bundles over $M$.
Fortunately, one can often still single out a canonical, finite dimensional family of renormalised solutions. Let us informally describe the corresponding procedure: Typically one is working with a mollification scheme which depends on a scale $\epsilon$. If this mollification scheme is “sufficiently covariant”  it is possible to decompose the corresponding family of renormalisation differential operators
$$g_\epsilon (T)= \sum_{i} f_i(\epsilon) \cdot g^i(T)  \ ,$$

* each $f_i: (0,1)\to \mathbb{R}$ typically
It can sometimes be useful to also include $f_i$, which just converge to a finite limit in the above decomposition.
converges to $\pm \infty$ as $\epsilon \to 0$.

* the $g^i(T)$ are differential operators which depend explicitly on the curvature and the differential operator of the linear part of the equation in a local manner  and thus are arguably “canonical”.

Thus, for such regularisation schemes it is then natural to work with only the finite dimensional subgroup ${\mathfrak{G}}'_-\subset \mathfrak{G}_-$ containing the differential operators $g_\epsilon (T)$. One thus obtains a canonical (finite dimensional) solution theory.

We note that, when working with an arbitrary mollification scheme, one in general needs elements of $g_\epsilon \in \mathfrak{G}_-$ which are not contained in ${\mathfrak{G}}'_-$ in order to converge to one of the  “canonical" solutions. Thus, in this case the canonical solutions are only reached by considering the cosets  ${\mathfrak{G}}'_- g_\epsilon \subset \mathfrak{G}_-$.

Note that in general it non-trivial to extract the operators $g^i(T)$ described above explicitly. When working for example with generalised Laplacians and white noise this is a rather straightforward consequence of the following:

* Explicit expansions of the heat kernel, c.f. Theorem <ref>, which can formally be written as

\begin{equation}\label{eq:heat_kernel_asymptotics}
G_t(p,q)=  \frac{1}{(4\pi t)^{d/2}}e^{-d(p,q)/4t} \sum_{i=0}^N t^i \Phi_i(p,q) +O(t^{N-n/2}) \ ,
\end{equation}

for $0<t\ll 1$.

* Explicit expansion of the density in exponential coordinates, c.f. <cit.>. Fur us the following will be sufficient

\begin{equation}\label{eq:volume_asymptotics}
\dVol_p= \left(1-\frac{1}{6}\sum_{i,j} R_{i,j} x_ix_ j  \right) dx +\mathcal{O}(|x|^3) dx \ .
\end{equation}

This explains, why for many equations of interest when working with a heat kernel regularisation, renormalisation constants are sufficient.
More precisely, when working with a generalised Laplacian and white noise, as long as the most negative tree has homogeneity higher than $-2$, one does not expect any geometric counterterms to appear.
In Section <ref> we treat an equation where this condition is violated and indeed a logarithmic divergence proportional to the scalar curvature $s(x)$ appears.

§ CONCRETE APPLICATIONS

Finally, we turn to the three examples described in the introduction. Recall that for a vector bundle $E\to M$, $E$-valued (spatial) white noise can be characterised as the unique random distributions $\xi\in \mathcal{D}'(E)$,
such that for any $\psi, \psi' \in \mathcal{D}(E)$,

* $\xi  ( \psi)$ is a centred Gaussian random variable,

* $E[\xi  ( \psi), \xi  ( \psi')]= \int_M \langle \psi, \psi' \rangle_E \dVol_g$.

We speak of $E$-valued space-time white noise, when the bundle above is replaced by the pull-back bundle $\pi^*E$, where $\pi: \mathbb{R}\times M \to M$ is the canonical projection.

We shall often writhe $\langle \xi  , \psi \rangle = \xi  ( \psi)$, where the absence of a subscript means the canonical pairing of $\mathcal{D}'(E)\times \mathcal{D}(E)$ or $\mathcal{D}'(\pi^*E)\times \mathcal{D}(\pi^*E)$ respectively.

We shall be working with the following two regularisation procedures. Let $G_t\in \mathcal{C}^\infty (E \hat{\otimes} E^*)$ be the heat kernel of the Laplacian in the equation.

* If $\xi\in \mathcal{D}'(E)$ is spatial white noise, set
$$\xi_\epsilon (p)= \langle \xi, G_{{\epsilon}^2} (p,\cdot )\rangle  $$

* If $\xi\in \mathcal{D}'(\pi^*E)$ is space time white noise, furthermore fix
an additional even, smooth compactly supported function
$\phi: \mathbb{R}\to \mathbb{R}$ integrating to $1$.
$\phi^\epsilon(t) :=  \frac{1}{\epsilon^2}\phi \left( \frac{t}{\epsilon^2} \right) $,
we shall set

\begin{equation}\label{eq:mollified space-time white noise}
\xi_\epsilon (t,p)= \langle \xi, \phi^\epsilon(t-\cdot) G_{{\epsilon}^2} (p,\cdot )\rangle \ ,
\end{equation}

which can informally be written as $\xi_\epsilon (t,p)= \int_{\mathbb{R}\times M} \phi^\epsilon(t-s) G_{{\epsilon}^2} (p,q )\xi(ds, dq)$.

Furthermore, fix a cut-off $\kappa: \mathbb{R}\to [0,1]$ such that the following properties are satisfied

* $\kappa(t)=0$ for $t<0$ or $t>2$,

* $\kappa(t)=1$ for $t\in [0,1)$.

* $\kappa|_{\mathbb{R}_+}$ is smooth.

For fixed $N \in \mathbb{N}$ large enough[This number is conveniently fixed depending on the equation, for the examples discussed here $N=10$ is sufficent.] we denote by $\bar{r}= \frac{\min_{p \in M} r_p}{N}$, where $r_p$ is the injectivity radius at $p\in M$.
We introduce the following notation for the leading terms of the scalar heat kernel $G$, c.f. (<ref>)

\begin{align}\label{eq:diverging parts of kernel}
Z_t(p,q):&=  \frac{1}{(4\pi t)^{d/2}}e^{-d(p,q)/4t} \kappa(d(p,q)/\bar{r}) \kappa(t)\ ,\\
Z^1_t(p,q):& = t \, Z_t(p,q)\Phi_1(p,q) \,
\end{align}

as well
$Z^E_t(p,q)=  Z_t (p,q) \tau_{p,q} $
for the $E$-valued kernel, where we recall that $\tau_{p,q}: E|_q\to E|_p$ is the parallel transport map.

Lastly, we use the following notation for the corresponding truncation of the heat kernel in Euclidean space

$$\bar{Z}_t (x)= \frac{1}{(4\pi t)^{d/2}}e^{-|x|^2 /4t} \kappa(t) .$$

§.§ The g-PAM equation

For $\epsilon>0$, we define the following counterterms:

\begin{align}
{C_\epsilon}&=\int_{\mathbb{R}} \int_{B_{r}(0)} \bar{Z}_{t-s+\epsilon^2}(x) \bar{Z}_{\epsilon^2}(x) \ dx\
ds ,\label{pam counterterm1} \\
{C_\epsilon'}&=\int_{\mathbb{R}^2}	\int_{B_{r}(0)} \partial_i \bar{Z}_{t-s+\epsilon^2}(x ) \partial_i \bar{Z}_{t-s'+\epsilon^2}(x ) dx\
% \kappa(t-s)\kappa(t-s')
ds ds' \  \label{pam counterterm2}
\end{align}

where the latter is the same for $i\in \{1,2\}$.
In this section we proof the following theorem.

Let $M$ be a compact two dimensional Riemannian manifold, $\xi$ be white noise on $M$, and $\xi_\epsilon$ its regularisation as described above.
Furthermore, let $f\in \cC^{\infty}(\RR)$, $A(\cdot, \cdot)$ be a smooth bilinear form on the tangent bundle $TM$ and let $u_0\in \cC^{\alpha}(M) $ for $\alpha>0$.
For $C_\epsilon, C'_\epsilon$ as in (<ref>)& (<ref>) and for $ u_\epsilon$ solving

\begin{equation}\label{gpam}
\partial_t u_\epsilon + \Delta u_\epsilon =
A ( \nabla u_\epsilon, \nabla u_\epsilon)- C_\epsilon' f^2(u_\epsilon)(\tr A) +f(u_\epsilon)\left(\xi_\epsilon- C_\epsilon f'(u_\epsilon)\right), \qquad u_\epsilon(0)=u_0 \ ,
\end{equation}

there exists a (random) $T>0$ and $u: [0,T]\times M \to \mathbb{R}$, such that $u_\epsilon \to u$ uniformly as $\epsilon\to 0$ in probability.

The proof of the theorem follows along the usual lines, when applying the theory of regularity structures.

First, introduce the following types:

* kernel types $\mathcal{E}_+= \{\mathcal{I}, \nabla\mathcal{I} \}$,

* noise types $\mathcal{E}_-= \{ \Xi \}$,

* jet types
$\mathcal{E}^0_0 = \mathcal{E}^A_0 \cup \mathcal{E}^f_0 \ ,$
It is notationally convenient to treat the right-hand side as two different non-linearities and introduce the edge types
$\delta^{f, n}$,  $\delta^{A, n}$.
Comparing to (<ref>), the $\delta^{f, n}$ edges correspond to a derivative in the $\xi $ direction.
] $\mathcal{E}^f_0=\{ \delta^{(f,n)} \ : \ n\in \mathbb{N} \}$ and $\mathcal{E}^A_0= \{ \delta^{(A,n )}\  :\  n\in \mathbb{N} \}$ to encode the non-linearities,

* further jet types $\mathcal{E}^+_0= \{\delta^{\mathcal{I}}, \delta^{\nabla \mathcal{I}}\}$. We fix the injection $\iota:\mathcal{E}_+\to  \mathcal{E}^+_0$ from (<ref>) to be

$$\mathcal{I}\mapsto \delta^{\mathcal{I}},\qquad \nabla \mathcal{I} \mapsto \delta^{\nabla \mathcal{I}} \ .$$

Since the equation is rather simple, we do not introduce the vector bundle assignments and the indexing maps explicitly.

$$|\mathcal{I}|\in (2-{1}/{10}, 2), \qquad  |\nabla \mathcal{I}| \in (1-{1}/{10}, 1), \qquad |\Xi|\in (-1-{1}/{10}, -1)\ $$

and define the naive rule

$$\mathring{R}(\mathcal{I})= \mathring{R}(\nabla \mathcal{I})= \{ [\delta^{f,n}] \sqcup[\mathcal{I}]^n\sqcup [\Xi] ,\  [\delta^{A,n}] \sqcup [\nabla \mathcal{I}]^2 \ : n\geq 0 \} $$

as well as its normalisation $R$.
Let us introduce the the analogue of the usual graphical notation. We write $\Xi= \<X>$ (using the common convention of not writing it as an edge) and denote an edge of type $\mathcal{I}$ by $\<I>$ as well as an edge of type $\nabla \mathcal{I}$ by $\<bI>$.
Thus, we for example write
$\mathcal{I}\Xi= \<1>$. Denote edges of type $\delta^{\mathcal{I}}$ by $\<1'>$.
Since it suffices to solve the abstract in $\mathcal{D}^\gamma$ for some $\gamma>1$,
we can work with

$$\mathfrak{T}^{\mathcal{I}}:= \{ \<1>,\ \<1'> \}$$
to describe the abstract solution

\begin{align*}
\mathfrak{T}^r
%&:= \{ \delta^{f,0}\Xi,\  \ \delta^{f,1} \mathcal{I}(\Xi) \Xi,\  \delta^{f,1}  \delta^{\mathcal{I}}\Xi,\  \delta^{A,0} (\nabla\mathcal{I}(\Xi))^2 \}\\
&_= \{ \delta^{f,0}\Xi,\  \ \delta^{f,1} \<Xi2> ,\  \delta^{f,1}  \<1'Xi> , \  \delta^{A,0} \<b2> \}
\end{align*}

to describe the right-hand side.
We observe that the only negative subtrees appearing are given by

$$\mathfrak{T}_-=\{  \<Xi2> ,\ \<b2> \} \ .$$

Thus, we next fix $\delta_0> 1$ and work with the regularity structure
constructed form the above trees. We quotient it, so that the product of jets agrees with the canonical product, c.f. Remark <ref> and
denote by $Z^\epsilon=(\Pi^\epsilon, \Gamma^\epsilon)$ the canonical model for the smooth noise $\xi_\epsilon$, such that the abstract Integration map $\mathcal{I}$ realises $K_t(p,q):=G_t(p,q) \kappa(t)\kappa(d(p,q)/\bar{r})$ and $\nabla\mathcal{I}$ realises $K^\nabla_t (p,q):= (\nabla_1 G_t(p,q)) \kappa(t)\kappa(d(p,q)/\bar{r})$.
Next, in order to identify the correct two dimensional solution family,

we restrict attention to the subgroup $\mathfrak{G}'_-$ of elements $\mathfrak{g}_{c,c' }\in \mathfrak{G}_-$such that

$$\mathfrak{g}_{c,c' }( \<X> ) = 0,\qquad
\mathfrak{g}_{c,c' }( \<Xi2>) = c,\qquad   \mathfrak{g}_{c,c' }( \<b2>)= c' g^{\sharp},$$

for some $c,c'\in \mathbb{R}$ and where
where $g^{\sharp}\in \mathcal{C}(TM^{\otimes 2})$ denotes the adjoint tensor field of the metric $g$.
To lighten notation, we denote the renormalised model $(Z^\epsilon)^{\mathfrak{g}_{C_\epsilon, C'_\epsilon}}$ by
$\hat{Z}^\epsilon= (\hat{\Pi}^\epsilon, \hat{\Gamma}^\epsilon)$.

In the setting described above, let $\eta, \gamma\in \mathbb{R}$ such that
$0<\eta<\alpha$ and $-|\Xi|<\gamma<\delta_0$.
It follows from
Theorem <ref> there exists a unique abstract solution $\hat{U}^\epsilon\in \mathcal{D}^{\gamma, \eta}(\mathcal{T}^{\mathcal{I}})$
of the abstract lift of

\partial_t u + \Delta u = A( \nabla u, \nabla u) +f(u)\xi,

for the renormalised model $\hat{Z}^\epsilon$.
Then, it is simple special case of Proposition <ref> and the observation $A(g^{\sharp})= \tr A$, that
$\hat{u}= \mathcal{R} U^{\epsilon}$ solves (<ref>).
Thus, the proof is complete once Proposition <ref> below is shown.

The renormalised models ${\hat{Z}}_\epsilon$ converge to a limiting model $\hat{Z}$
as $\epsilon\to 0$.

It follows from Proposition <ref> below combined with a Kolmogorov type criterion, obtained by going to charts,
that for each $p,q\in \mathbb{R}\times M$ , $\tau_p \in \mathcal{T}^r$ there exist random distributions

$$ \hat{\Pi}_p \tau_p \qquad \hat{\Pi}_q \hat{\Gamma}_{q,p} \tau_p,  $$

such that[Here we use this notation to denote the restriction of the corresponding expressions for $\|\ \cdot \  \|_{[-1,T+1]\times M} $ and $\llbracket\  \cdot \   \rrbracket_{[-1,T+1] \times M} $  in Definition <ref> to only $\tau\in\mathcal{T}^r $.]

$$\|\  \hat{\Pi}^\epsilon-\hat{\Pi}\  | \ {\mathcal{T}^r }   \|_{[-1,T+1]\times M} + \llbracket \hat{Z}^\epsilon  \ ;\hat{Z}\ | \ {\mathcal{T}^r}  \rrbracket_{[-1,T+1] \times M} \to 0 \ .$$

Then, it follows from Theorem <ref> that
$\|\hat{Z}^\epsilon;\hat {Z}\|_{[-1,T+1]\times M} \to 0$.

§.§.§ Stochastic estimates

This section is dedicated to proving the following.

There exists a $\kappa>0$, such that for any $p,q\in \mathbb{R}\times M$, $\tau_p \in {\mathcal{T}}^r_{\alpha, \delta}$ and $N\in \mathbb{N}$, it holds that uniformly over $\lambda\in (0,1]$ and $\phi_q^\lambda\in  \mathcal{B}^{r, \lambda}_q$ that

\begin{equation}\label{eq: stochastic bound on model}
\E\left[\left(\hat \Pi_q^\epsilon \tau_q) (\phi_q^\lambda)\right)^N \right]^{1/N}\ \lesssim \lambda^{\alpha+2\kappa} \ , \qquad
\E\left[\left((\hat{\Pi}^\epsilon_q\hat{\Gamma}^\epsilon_{q,p} \tau_p -\hat{\Pi}_p^\epsilon \tau_p) (\phi_q^\lambda)\right)^N \right]^{1/N}\ \lesssim \lambda^{\delta+2\kappa}
\end{equation}

as well as the same estimate for $ \hat \Pi_p^\epsilon$ and $\hat{ \Gamma}^\epsilon_{p,q}$ replaced by $ \hat \Pi_p$ and $\hat{ \Gamma}_{p,q}$.
Furthermore, it holds that

\begin{align*}
&\E\left[\left(\hat \Pi_p^\epsilon \tau_p- \hat \Pi_p \tau_p) (\phi_q^\lambda)\right)^N \right]^{1/N}\ \lesssim \epsilon^{\kappa} \lambda^{\alpha+\kappa} \\
&\E\left[\left((\hat{\Pi}^\epsilon_q\hat{\Gamma}^\epsilon_{q,p} \tau_p -\hat{\Pi}_p^\epsilon \tau_p) (\phi_q^\lambda)-(\hat{\Pi}_q\hat{\Gamma}_{q,p} \tau_p +\hat{\Pi}_p \tau_p) (\phi_q^\lambda)\right)^N \right]^{1/N}\ \lesssim \epsilon^{\kappa} \lambda^{\delta+\kappa} \ .
\end{align*}

We first introduce similar notation to [38], <cit.>.
Recall, that a random variable belonging to the $k$th homogeneous Wiener chaos can
be described by a kernel in $L^2( M)^{\otimes k}$, i.e. the
space of square-integrable functions in $k$ space variables via the correspondence
$f \mapsto I_k(f)$ given in <cit.>.

We shall graphically represent $\bigl(\Pi_\star^\epsilon \tau\bigr)(\phi^\lambda_\star)$,
where $\phi^\lambda_\star\in \mathcal{B}^{r,\lambda}_\star$ denotes a test function that is localised around the
point $\star \in \RR\times M$. Nodes in our graph will represent variables in $M$ or $\mathbb{R}\times M$, with one
special node [baseline=-3] [root] ; representing the point $\star\in \mathbb{R}\times M$.
The nodes [baseline=-3] [var] ; represent the arguments of our kernel,
so that a random variable in the $k$th (homogeneous) Wiener chaos is represented by
a graph with exactly $k$ such nodes. The remaining nodes, which we draw as [baseline=-3] [dot] ;, represent dummy space time variables that are to be integrated out.

Each line then represents a space or space-time kernel, with
[baseline=-0.1cm] [kernel] (0,0) to (1,0) ;
representing the kernel $K$,
[baseline=-0.1cm] [rho] (0,0) to (1,0);
representing the kernel $\rho_\epsilon=G_{{\epsilon}^2}$ and
[baseline=-0.1cm] [testfcn] (1,0) to (0.0,0) ;
representing a generic test function $\phi_\star^\lambda$. Note that since
$\rho_\epsilon (p,q)= \rho_\epsilon (q,p)$, the notation [baseline=-0.1cm] [rho] (0.0,0) to (1,0);
is unambiguous.

Whenever we draw a barred arrow
[baseline=-0.1cm] [kernel1] (0.0,0) to (1,0);
this represents a factor $K((t,p), (s,q)) - K(\star,  (s,q))$, where
$(s,q)\in \mathbb{R}\times M$ and $(t,p)\in \mathbb{R}\times M$ are the coordinates of the starting and end
point respectively.

§.§.§ A first stochastic term

Using the notation introduced above, one can write

\begin{equation}
\bigl(\Pi_\star^{\epsilon} \<Xi2>\bigr)(\phi_\star^\lambda) = \;
\begin{tikzpicture}[scale=0.35,baseline=0.3cm]
\node at (0,-1)  [root] (root) {};
\node at (-2,1)  [dot] (left) {};
\node at (-2,3)  [dot] (left1) {};
\node at (0,1) [var] (variable1) {};
\node at (0,3) [var] (variable2) {};

\draw[testfcn] (left) to  (root);

\draw[kernel1] (left1) to (left);
\draw[rho] (variable2) to (left1);
\draw[rho] (variable1) to (left);
\end{tikzpicture}\;
+ \;
\begin{tikzpicture}[scale=0.35,baseline=0.3cm]
\node at (0,-1)  [root] (root) {};
\node at (-2,1)  [dot] (left) {};
\node at (-2,3)  [dot] (left1) {};
\node at (0,2) [dot] (variable1) {};
\node at (0,2) [dot] (variable2) {};

\draw[testfcn] (left) to (root);

\draw[kernel1] (left1) to (left);
\draw[rho] (variable2) to (left1);
\draw[rho] (variable1) to (left);
\end{tikzpicture}\;
\begin{tikzpicture}[scale=0.35,baseline=0.3cm]
\node at (0,-1)  [root] (root) {};
\node at (-2,1)  [dot] (left) {};
\node at (-2,3)  [dot] (left1) {};
\node at (0,1) [var] (variable1) {};
\node at (0,3) [var] (variable2) {};

\draw[testfcn] (left) to  (root);

\draw[kernel1] (left1) to (left);
\draw[rho] (variable2) to (left1);
\draw[rho] (variable1) to (left);
\end{tikzpicture}\;
\begin{tikzpicture}[scale=0.35,baseline=0.3cm]
\node at (0.0,-1)  [root] (root) {};
\node at (-2,1)  [dot] (left) {};
\node at (-2,3)  [dot] (left1) {};
\node at (0,2) [dot] (variable1) {};
\node at (0,2) [dot] (variable2) {};

\draw[testfcn] (left) to (root);

\draw[kernel] (left1) to (left);
\draw[rho] (variable2) to (left1);
\draw[rho] (variable1) to (left);
\end{tikzpicture}\;
- \;
\begin{tikzpicture}[scale=0.35,baseline=0.3cm]
\node at (0.0,-1)  [root] (root) {};
\node at (-1,1)  [dot] (left) {};
\node at (0,3)  [dot] (top) {};
\node at (1,1) [dot] (right) {};

\draw[testfcn] (left) to  (root);

\draw[kernel] (right) to (root);
\draw[rho] (top) to (right);
\draw[rho] (top) to (left);
\end{tikzpicture}\;
\end{equation}

We further introduce the graphical notation

\begin{tikzpicture}[scale=0.35,baseline=0.6cm]
\node at (-2,1)  [root] (left) {};
\node at (-2,3)  [reddot] (left1) {};
\node at (0,2) [reddot] (variable) {};

\draw[red, kernel] (left1) to (left);
\draw[red, rho] (variable) to (left1);
\draw[red, rho] (variable) to (left);
\end{tikzpicture}\; :=

and decompose the function

\begin{equation}\label{eq:decom_counterterm1}
\begin{tikzpicture}[scale=0.35,baseline=0.6cm]
\node at (-2,1)  [root] (left) {};
\node at (-2,3)  [dot] (left1) {};
\node at (0,2) [dot] (variable) {};

\draw[kernel] (left1) to (left);
\draw[rho] (variable) to (left1);
\draw[rho] (variable) to (left);
\end{tikzpicture}\;
= \;
\begin{tikzpicture}[scale=0.35,baseline=0.6cm]
\node at (-2,1)  [root] (left) {};
\node at (-2,3)  [reddot] (left1) {};
\node at (0,2) [reddot] (variable) {};

\draw[red, kernel] (left1) to (left);
\draw[red, rho] (variable) to (left1);
\draw[red, rho] (variable) to (left);
\end{tikzpicture}\; +\;
\begin{tikzpicture}[scale=0.35,baseline=0.6cm]
\node at (-2,1)  [root] (left) {};
\node at (-2,3)  [bluedot] (left1) {};
\node at (0,2) [bluedot] (variable) {};

\draw[blue, kernel] (left1) to (left);
\draw[blue, rho] (variable) to (left1);
\draw[blue, rho] (variable) to (left);
\end{tikzpicture}\; ,
\end{equation}

where the blue term is determined by (<ref>).

The function  $
\begin{tikzpicture}[scale=0.35,baseline=0.6cm]
\node at (-2,1)  [root] (left) {};
\node at (-2,3)  [bluedot] (left1) {};
\node at (0,2) [bluedot] (variable) {};

\draw[blue, kernel] (left1) to (left);
\draw[blue, rho] (variable) to (left1);
\draw[blue, rho] (variable) to (left);
\end{tikzpicture}\;
is uniformly bounded for $\epsilon\in (0,1]$ and converges uniformly as $\epsilon \to 0$.

Recall that

$$\pmb{\Pi}^{\epsilon} \<1>(t,p)=\int_{\mathbb{R}}\int_{M} K_{t-s}(p,q) \xi_{\epsilon}(q) \dVol_q ds
= \int_{\mathbb{R}} \langle \xi, G_{t-s+\epsilon^2} (p,\cdot )\rangle \kappa(t-s) ds + r^\epsilon(t,p)$$

and thus

\begin{align*}
\begin{tikzpicture}[scale=0.35,baseline=0.6cm]
\node at (-2,1)  [root] (left) {};
\node at (-2,3)  [dot] (left1) {};
\node at (0.0,2) [dot] (variable) {};
\draw[kernel] (left1) to (left);
\draw[rho] (variable) to (left1);
\draw[rho] (variable) to (left);
\end{tikzpicture}\;
&=\E\left[\pmb{\Pi}^{\epsilon} \<Xi2> (t,p)\right]\\
&= \E\left[ \int_{\mathbb{R}} \langle \xi, G_{t-s+\epsilon^2}(p,\cdot )\rangle \kappa(t-s)  \langle \xi, G_{\epsilon^2}(p,\cdot )\rangle  ds\right] +\E\left[ r^\epsilon (t,p)\xi_{\epsilon}(p)\right] \ .
\end{align*}

Since the conclusion of the lemma holds for the map $p\mapsto \E\left[ R^\epsilon_1(t,p)\xi_{\epsilon}(p)\right]$ we proceed by noting that

\begin{align*}
& \E\left[ \int_{\mathbb{R}} \langle \xi, G_{t-s+\epsilon^2}(p,\cdot )\rangle \kappa(t-s)  \langle \xi, G_{\epsilon^2}(p,\cdot )\rangle  ds\right] \\
&= \int_{\mathbb{R}} \int_M G_{t-s+\epsilon^2}(p,z) G_{\epsilon^2}(p,z) \dVol_z \kappa(t-s) ds \\
&= \int_{\mathbb{R}} \int_{B_{r}(p)} Z_{t-s+\epsilon^2}(p,z) Z_{\epsilon^2}(p,z) \dVol_z \kappa(t-s) ds\\
&+ \int_{\mathbb{R}} \int_{M} \left(G_{t-s+\epsilon^2}(p,z) -Z_{t-s+\epsilon^2}(p,z) \right) G_{\epsilon^2}(p,z) \dVol_z \kappa(t-s) ds\\
&+\int_{\mathbb{R}} \int_{M} Z_{t-s+\epsilon^2}(p,z) \left( G_{\epsilon^2}(p,z)-Z_{\epsilon^2}(p,z)\right)  \dVol_z \kappa(t-s) ds \ , \\
%- \int_{\mathbb{R}} \int_{M} Z_{t-s+\epsilon^2}(p,z) \left( G_{\epsilon^2}(p,z)-Z_{\epsilon^2}(p,z)\right)  \dVol_z \kappa(t-s) ds\\
% &= \int_{\mathbb{R}} \int_{B_{r}(p)} G_{t-s+\epsilon^2}(p,z) G_{\epsilon^2}(p,z) \dVol_z \kappa(t-s) ds
%&= \redc{C} |\log(\epsilon)| + \mathcal{O}(1)  \
\end{align*}

where if follows from Theorem <ref> that the latter two summands converge uniformly as $\epsilon \to 0$.
Finally, if follows from(<ref>) and (<ref>) that

$$ \int_{\mathbb{R}} \int_{B_{r}(p)} Z_{t-s+\epsilon^2}(p,z) Z_{\epsilon^2}(p,z) \dVol_z \kappa(t-s) ds- C_\epsilon$$

converges uniformly to a limiting function.

Thus, after renormalisation we find

\begin{equation}\label{eq:renormalised_chaos_dec_tree1}
\bigl(\hat \Pi_\star^{\epsilon} \<Xi2>\bigr)(\phi_\star^\lambda) = \;
\begin{tikzpicture}[scale=0.35,baseline=0.3cm]
\node at (0,-1)  [root] (root) {};
\node at (-2,1)  [dot] (left) {};
\node at (-2,3)  [dot] (left1) {};
\node at (0,1) [var] (variable1) {};
\node at (0,3) [var] (variable2) {};

\draw[testfcn] (left) to  (root);

\draw[kernel1] (left1) to (left);
\draw[rho] (variable2) to (left1);
\draw[rho] (variable1) to (left);
\end{tikzpicture}\;
- \;
\begin{tikzpicture}[scale=0.35,baseline=0.3cm]
\node at (0.0,-1)  [root] (root) {};
\node at (-1,1)  [dot] (left) {};
\node at (0,3)  [dot] (top) {};
\node at (1,1) [dot] (right) {};

\draw[testfcn] (left) to  (root);

\draw[kernel] (right) to (root);
\draw[rho] (top) to (right);
\draw[rho] (top) to (left);
\end{tikzpicture}\;
\begin{tikzpicture}[scale=0.35,baseline=0.3cm]
\node at (0.0,-1)  [root] (root) {};
\node at (-2,1)  [dot] (left) {};
\node at (-2,3)  [bluedot] (left1) {};
\node at (0,2) [bluedot] (variable1) {};

\draw[testfcn] (left) to (root);

\draw[blue,kernel] (left1) to (left);
\draw[blue, rho] (variable1) to (left1);
\draw[blue, rho] (variable1) to (left);
\end{tikzpicture}\;.
\end{equation}

§.§.§ A second stochastic term

Here we use a slight variation of the above notation and use
[baseline=-0.1cm] [keps] (0,0) to (1,0) ;
to represent the kernel

$$ \left( K^{\nabla}_t * \rho_\epsilon\right) (p, q)= \int_M   K^{\nabla}_t (p,z) \rho_\epsilon (z,q) \dVol_z \ ,  $$
[baseline=-0.1cm] [testfcn] (1,0) to (0,0) ;
represents a localised test function $\phi_\star^\lambda\in \mathcal{C}^\infty (TM^{\otimes 2})$. So

\begin{equation*}
\bigl(\Pi_\star^{\epsilon}\<b2>\bigr)(\phi_\star^\lambda)
\begin{tikzpicture}[scale=0.35,baseline=0.3cm]
\node at (0,-1)  [root] (root) {};
\node at (0,1)  [dot] (int) {};
\node at (-1,3.5)  [var] (left) {};
\node at (1,3.5)  [var] (right) {};

\draw[testfcn] (int) to  (root);

\draw[keps] (left) to (int);
\draw[keps] (right) to (int);
\end{tikzpicture}\;
+ \;
\begin{tikzpicture}[scale=0.35,baseline=0.3cm]
\node at (0,-1)  [root] (root) {};
\node at (0,1)  [dot] (int) {};
\node at (0,3.5) [dot] (top) {};

\draw[testfcn] (int) to  (root);

\draw[keps] (top) to[bend left=60] (int);
\draw[keps] (top) to[bend right=60] (int);
\end{tikzpicture}\;.
\end{equation*}

Similarly to above, we write

\begin{tikzpicture}[scale=0.35,baseline=0.3cm]
\node at (0,0)  [root] (root) {};
\node at (0,2.5) [reddot] (top) {};
\draw[red, keps] (top) to[bend left=60] (root);
\draw[red, keps] (top) to[bend right=60] (root);
\end{tikzpicture}\; =  C'_\epsilon g^{\sharp}
\qquad \text{and} \qquad
\begin{tikzpicture}[scale=0.35,baseline=0.3cm]
\node at (0,0)  [root] (root) {};
\node at (0,2.5) [bluedot] (top) {};
\draw[blue,keps] (top) to[bend left=60] (root);
\draw[blue,keps] (top) to[bend right=60] (root);
\end{tikzpicture}\; = \;
\begin{tikzpicture}[scale=0.35,baseline=0.3cm]
\node at (0,0)  [root] (root) {};
\node at (0,2.5) [dot] (top) {};
\draw[keps] (top) to[bend left=60] (root);
\draw[keps] (top) to[bend right=60] (root);
\end{tikzpicture}\; - \;
\begin{tikzpicture}[scale=0.35,baseline=0.3cm]
\node at (0,0)  [root] (root) {};
\node at (0,2.5) [reddot] (top) {};
\draw[red, keps] (top) to[bend left=60] (root);
\draw[red, keps] (top) to[bend right=60] (root);
\end{tikzpicture}\;  \ .

The section $\begin{tikzpicture}[scale=0.35,baseline=0.3cm]
\node at (0,0)  [root] (int) {};
\node at (0,2.5) [bluedot] (top) {};
\draw[blue, keps] (top) to[bend left=60] (int);
\draw[blue, keps] (top) to[bend right=60] (int);
\end{tikzpicture}\;$ of $\mathcal{C}^\infty(T^*M^{\otimes_s 2})$
is uniformly bounded for $\epsilon\in (0,1]$ and converges uniformly.

Let $\{e_{p,i}\}_{i=1,2}$ be a basis of $T_p M$.
We proceed exactly as in Lemma <ref> and find that

\begin{align*}
\begin{tikzpicture}[scale=0.35,baseline=0.3cm]
\node at (0.0,0)  [root] (int) {};
\node at (0,2.5) [dot] (top) {};
\draw[keps] (top) to[bend left=60] (int);
\draw[keps] (top) to[bend right=60] (int);
\end{tikzpicture}\;
\E\left[ \int_{\mathbb{R}} \langle \xi, \nabla G_{t-s+\epsilon^2}(p,\cdot )\rangle \kappa(t-s) ds \otimes \int_{\mathbb{R}} \langle \xi, \nabla G_{t-s'+\epsilon^2}(p,\cdot )\rangle \kappa(t-s') ds'\right] + R^\epsilon_1 (p)
\\
%&=  \int_{\mathbb{R}^2}	\int_M	\nabla G_{t-s+\epsilon^2}(p,z )\otimes \nabla G_{t-s'+\epsilon^2}(p,z )	\dVol_z	 \kappa(t-s)\kappa(t-s') ds ds' + R^\epsilon_1 (p)\\
%&=  \int_{\mathbb{R}^2}	\int_{B_{r}(p)}	\nabla Z_{t-s+\epsilon}(p,z )\otimes \nabla Z_{t-s'+\epsilon}(p,z )	\dVol_z	ds ds' +R^\epsilon_2 (p)\\
&= \sum_{i,j} e_{p,i}\otimes e_{p,j} \int_{\mathbb{R}^2}	\int_{B_{r}(p)} \partial_i Z_{t-s+\epsilon^2}(p,z )\otimes \partial_j Z_{t-s'+\epsilon^2}(p,z )\dVol_z ds ds' +R^\epsilon_2\\
&=   \sum_{i,j}  {C}^{(i,j)}_\epsilon e_i\otimes e_j   + R^\epsilon_3\ ,
\end{align*}

{C}^{(i,j)}_\epsilon:=\int_{\mathbb{R}^2}	\int_{B_{r}(0)} \partial_i \bar{Z}_{t-s+\epsilon^2}(z )\otimes \partial_j \bar{Z}_{t-s'+\epsilon^2}(z ) dz ds ds' \

and the sections $R^\epsilon_1, \, R^\epsilon_2, \, R^\epsilon_3$ of $TM^{\otimes 2}$ converge uniformly as $\epsilon \to 0$.
Noting that $C'_\epsilon={C}^{(1,1)}_\epsilon={C}^{(2,2)}_\epsilon$ and ${C}^{(1,2)}_\epsilon={C}^{(2,1)}_\epsilon=0$, this concludes the proof.

Thus, after renormalisation,

\begin{equation}\label{eq:renormalised_chaos_dec_tree2}
\bigl(\hat \Pi_\star^{\epsilon}\<b2>\bigr)(\phi_\star^\lambda)
\begin{tikzpicture}[scale=0.35,baseline=0.3cm]
\node at (0,-1)  [root] (root) {};
\node at (0,1)  [dot] (int) {};
\node at (-1,3.5)  [var] (left) {};
\node at (1,3.5)  [var] (right) {};

\draw[testfcn] (int) to  (root);

\draw[keps] (left) to (int);
\draw[keps] (right) to (int);
\end{tikzpicture}\;
%+\; 2\;
%	\node at (0,-1)  [root] (root) {};
%	\node at (0,1)  [dot] (int) {};
%	\node at (0,3.5) [dot] (top) {};
%	\draw[testfcn] (int) to  (root);
%	\draw[red, keps] (top) to[bend left=60] (int);
%	\draw[blue, keps] (top) to[bend right=60] (int);
\begin{tikzpicture}[scale=0.35,baseline=0.3cm]
\node at (0,-1)  [root] (root) {};
\node at (0,1)  [dot] (int) {};
\node at (0,3.5) [bluedot] (top) {};

\draw[testfcn] (int) to  (root);

\draw[blue, keps] (top) to[bend left=60] (int);
\draw[blue, keps] (top) to[bend right=60] (int);
\end{tikzpicture}\; .
\end{equation}

Note that it follows from (<ref>) and  (<ref>) that

\begin{equation}
\E \bigl|\bigl(\hat \Pi_\star^{\epsilon} \<Xi2>\bigr)(\phi_\star^\lambda)\bigr|^2 \le 2\;
\begin{tikzpicture}[scale=0.35,baseline=0.3cm]
\node at (0,-1)  [root] (root) {};
\node at (-1.5,1)  [dot] (left) {};
\node at (-1.5,3)  [dot] (left1) {};
\node at (1.5,1) [dot] (variable1) {};
\node at (1.5,3) [dot] (variable2) {};

\draw[testfcn] (left) to  (root);
\draw[testfcn] (variable1) to  (root);

\draw[kernel1] (left1) to (left);
\draw[kernel1] (variable2) to (variable1);
\draw[rho] (variable2) to node[dot] {} (left1);
\draw[rho] (variable1) to node[dot] {} (left);
\end{tikzpicture}\; + \; 2 \;
\left(
\begin{tikzpicture}[scale=0.35,baseline=0.3cm]
\node at (0,-1)  [root] (root) {};
\node at (-1,1)  [dot] (left) {};
\node at (0,3)  [dot] (top) {};
\node at (1,1) [dot] (right) {};

\draw[testfcn] (left) to  (root);

\draw[kernel] (right) to (root);
\draw[rho] (top) to (right);
\draw[rho] (top) to (left);
\end{tikzpicture}
\right)^2\; +\; 2 \;
\left(
\begin{tikzpicture}[scale=0.35,baseline=0.3cm]
\node at (0,-1)  [root] (root) {};
\node at (-2,1)  [dot] (left) {};
\node at (-2,3)  [bluedot] (left1) {};
\node at (0,2) [bluedot] (variable1) {};
\node at (0,2) [bluedot] (variable2) {};

\draw[testfcn] (left) to (root);

\draw[blue,kernel] (left1) to (left);
\draw[rho] (variable2) to (left1);
\draw[rho] (variable1) to (left);
\end{tikzpicture}
\right)^2\;
\end{equation}

\begin{equation}
\E |\bigl(\hat \Pi_\star^{\epsilon}\<b2>\bigr)(\phi_\star^\lambda)|^2
\le 2\;
\begin{tikzpicture}[scale=0.35,baseline=0.25cm]
\node at (0.0,-1)  [root] (root) {};
\node at (-2,1)  [dot] (intl) {};
\node at (2,1)  [dot] (intr) {};
\node at (0,1)  [dot] (top1) {};
\node at (0,3)  [dot] (top2) {};

\draw[testfcn] (intl) to  (root);
\draw[testfcn] (intr) to  (root);

\draw[keps] (top1) to (intl);
\draw[keps] (top1) to (intr);
\draw[keps] (top2) to (intl);
\draw[keps] (top2) to (intr);
\end{tikzpicture}\;.
%+\; 8\; \left(
%	\node at (0,-1)  [root] (root) {};
%	\node at (0,1)  [dot] (int) {};
%	\node at (0,3.5) [dot] (top) {};
%	\draw[testfcn] (int) to  (root);
%	\draw[red, keps] (top) to[bend left=60] (int);
%	\draw[blue, keps] (top) to[bend right=60] (int);
%\end{tikzpicture}\;\right)^2 \;
+ \; \left(
\begin{tikzpicture}[scale=0.35,baseline=0.3cm]
\node at (0.0,-1)  [root] (root) {};
\node at (0,1)  [dot] (int) {};
\node at (0,3.5) [bluedot] (top) {};

\draw[testfcn] (int) to  (root);

\draw[blue, keps] (top) to[bend left=60] (int);
\draw[blue, keps] (top) to[bend right=60] (int);
\end{tikzpicture}
\; \right)^2\  .
\end{equation}

The last term of each has already been bounded in Lemma <ref>, resp. Lemma <ref>, while the rest can be estimated using the bounds of Section <ref> exactly as in the proof of <cit.>.
Combining this
with the fact that for
$\tau_p =\<Xi2>$ and $\<2>$ one has

$$ \hat{\Pi}^\epsilon_q \hat{\Gamma}^\epsilon_{q,p} \tau_p -\hat{\Pi}^\epsilon_p \tau_p= 0 \ ,$$

implies that for these elements (<ref>) is satisfied.
Since every element $\tau_p\in {\mathcal{T}}^r$ is the product of a jet with either $\<Xi2>$ or $ \<2>$ (<ref>) follows generally.
In order to conclude the remainder of Proposition <ref> one proceeds as usual, edge by edge replacing
$\rho_\epsilon$ by $\rho_\epsilon (p,q)- \delta_p(q)$.

§.§ The $\Phi^4_3(E)$ equation

\begin{align}
C_\epsilon &= \int_\mathbb{R^2} \int_{B_r(0)}  \phi^\epsilon (s- r )\phi^\epsilon (s'- r )
\bar{Z}_{t-s+\epsilon^2}(z)
\bar{Z}_{t-s'+\epsilon^2}( z)
ds  ds' dr dz \label{phi4 counterterm1} \\
C'_\epsilon &= \int_{\mathbb{R}\times B_r(0)} \left(\bar{Z}^{(\epsilon)}(z)\right)^2 \bar{Z} (z) dz \label{phi4 counterterm2}
\end{align}

where $\bar{Z}^{(\epsilon)}(t,x)=\int_\mathbb{R^2} \int_{B_r}   \phi^\epsilon (s- r )\phi^\epsilon (s'- r )
\bar{Z}_{t-s-s'+2\epsilon^2}(x)
ds  ds' dr dx

We shall prove the following theorem.

Let $E\to M$ be a vector bundle over a compact 3-manifold equipped with with a metric $h$ and compatible connection $\nabla^{E}$. Let $|\cdot|_h$ be the norm on $E$ induced by $h$ and $\Delta^E$
a generalised Laplacian on $E$ differing form the connection Laplacian only by a $0$-th order term.
Let $\xi$ be an $E$-valued space-time white noise, and denote by $\xi_\epsilon$ its regularisation as in (<ref>).
Then, let $u_0\in \cC^{\alpha}(E)$ for $\alpha>-\frac{2}{3}$, let $C_\epsilon$, $C'_\epsilon$ as in (<ref>) and (<ref>) and let $u_\epsilon$ solve

\begin{equation}\label{phi4}
\partial_t u_\epsilon + \Delta^E u_\epsilon= -u_\epsilon |u_\epsilon |^2_h+ 3 C_\epsilon u_\epsilon - 9 C_\epsilon' u_\epsilon +\xi_\epsilon\ , \qquad
u_\epsilon(0) =u_0 \ .
\end{equation}

Then, there exists a (random) $T>0$ and $u\in  \mathcal{D}'((0,T)\times E)$, where we canonically identify $(0,T)\times E$ with a subset of $\pi^* E$, such that $u_\epsilon \to u$ in $\mathcal{D}'((0,T)\times E)$ as $\epsilon\to 0$ in probability.

The condition that $\Delta^E$
differs form the connection Laplacian by only a $0$-th order term is not a real loss of generality, c.f. Remark <ref>.

It is well known that one can take $T= \infty$ a.s. in Theorem <ref>, see <cit.>, <cit.>, <cit.>, <cit.><cit.> &  <cit.>.
Furthermore, it is well understood that $u$ can actually be realised as a continuous function $u : [0,\infty) \to \mathcal{D}'(E)$.

First, we define a new symmetric nonlinearity

$$H: E^{\otimes 3}\to E, \qquad (u_1,u_2,u_3)\mapsto \frac{1}{3!}\sum_{\sigma\in S_3} u_{\sigma(1)} h(u_{\sigma(2)}, u_{\sigma(3)}) \ , $$

where $S_3$ denotes the permutation group of the set $\{1, 2, 3\}\subset \NN$
and study

\begin{equation}\label{eq:phi}
\partial_t u + \Delta^E u =  -H(u,u,u) +\xi\ .
\end{equation}

We define the edge types:

* kernel types $\mathcal{E}_+= \{\mathcal{I} \}$,

* noise types $\mathcal{E}_-= \{ \Xi \}$,

* jet types $\mathcal{E}^0_0 =\{ \delta^{(3)} \}$
and $\mathcal{E}^+_0= \{\delta^{\mathcal{I}}\}$ with the injection $\iota:\mathcal{E}_+\to  \mathcal{E}^+_0$.

We fix homogeneities $|\mathcal{I}|\in (2-\frac{1}{10}, 2)$, $|\Xi|\in (-\frac{5}{2}-\frac{1}{10}, -\frac{5}{2})$
the naive rule characterised by $\mathring{R}(\mathcal{I})= \{ [\mathcal{I}]^3,\ [\Xi] \}$. Denote by $R$ its normalisation.
Using graphical notation we write $\Xi= \<X>$,  denote an edge of type $\mathcal{I}$ by $\<I>$ and edges of type $\delta^{\mathcal{I}}$ by $\<1'>$.
Since the nonlinearity is given by a parallel section of $H\in L(E^{\otimes 3}, E)$, we shall suppress the $\delta^{(3)} $ edges in the notation and use the convention that it is implicitly present, whenever a node has three incident edges from above. For example, we write

$$\delta^{(3)}(\mathcal{I}\Xi)^3= \<3>\ , \qquad \delta^{(3)} \delta^{\mathcal{I}} (\mathcal{I}\Xi)^2= \<3'>\ , \qquad \delta^{(3)} (\delta^{\mathcal{I}})^2 \mathcal{I}\Xi = \<3''>, \qquad  \delta^{(3)} (\delta^{\mathcal{I}})^3=  \<3'''>\ .$$

By exactly the same power counting as for the usual $\phi^4$ equation, it suffices to solve the equation in $\mathcal{D}^\gamma$ for some $\gamma>1$.
Thus, the relevant trees to describe the solution are

$$\mathfrak{T}^\mathcal{J}:=\left\{\<1'>, \; \<1>, \; \<30>, \; \<3'0> \right\} $$

and to describe the right-hand side we use

$${\mathfrak{T}}^r:=\left\{ \<X>, \; \<3>, \;\<3'>, \; \<3''>, \;  \<3'''>  ,\; \<32>,\; \<3'2>,\; \<32'>   \right\} \ .$$

For $\delta_0\in (1,2)$ we work with the regularity structure constructed in Section <ref>. Note that this time, as we are working with a multi-linear nonlinearity, we do not need to identify products of jets, c.f. Section <ref>. Denote by $Z^\epsilon=(\Pi^\epsilon, \Gamma^\epsilon)$ the canonical model for the smooth noise $\xi_\epsilon$, such that the abstract Integration map $\mathcal{I}$ realises $K_t(p,q):=G_t(p,q) \kappa(t)\kappa(d(p,q)/\bar{r})$, where
$G$ denotes the heat kernel of $\Delta^E$.

Finally, we shall work with the subgroup $\mathfrak{G}'_- \subset \mathfrak{G}_-$ consisting of elements  $\mathfrak{g}_{c,c'} \in \mathfrak{G}_-$
such that

$$\mathfrak{g}(\<2>)= c h^{\sharp}, \qquad \mathfrak{g}( \<22>)= 2c' \id_E \   $$

for some $c, c'\in \mathbb{R}$
and such that for other $\tau \in \mathfrak{T}_-^{(-)}$ one has $\mathfrak{g}(\tau)= e(\tau)$, where $e\in \mathfrak{G}_-$ denotes the unit element.
We again write  $\hat{Z}^\epsilon:= (Z^\epsilon)^{\mathfrak{g}_{C_\epsilon, C'_\epsilon}}$ and
$\hat{Z}^\epsilon= (\hat{\Pi}^\epsilon, \hat{\Gamma}^\epsilon)$.

Note that we have incorporated another simplification due to the fact that $H$ is a parallel non-linearity. A priory $\mathfrak{g}( \<22>)$ is a $0$-th order differential operator

$$\mathfrak{g}( \<22>): \mathcal{C}^\infty\left( (E\otimes L(E^{\otimes_s 3}, E))\times L(E^{\otimes_s 3}, E) \right) \to \mathcal{C}^\infty(E)\ .$$
Only after contracting both sets of indices corresponding to $L(E^{\otimes_s 3}, E)$ with $H$ one obtains a $0$-th order differential operator $\mathcal{C}^\infty(E)\to \mathcal{C}^\infty(E)$.

In the setting described above, let $\eta, \gamma\in \mathbb{R}$ such that
$\big| 2|\Xi| + 2|\mathcal{I}|\big|<\gamma <\delta_0$ and
$-{2}/3<\eta<\left( |\Xi| + |\mathcal{I}|\right)\wedge \alpha$ as well as $\eta\leq \alpha$.
Note that the homogeneity of $|\Xi|$ is a priori too low in order to apply the  Theorem <ref> directly, but this can be resolved exactly as in <cit.> for the classical $\Phi^4_3$ equation.
One then finds that there exists a unique abstract (local) solution $\hat{U}^\epsilon\in \mathcal{D}^{\gamma, \eta}(\mathcal{T}^{\mathcal{I}})$
of the abstract lift of
for the renormalised models $\hat{Z}^\epsilon$.
Then, it follows from Proposition <ref> that
$\hat{u}= \mathcal{R} \hat{U}^{\epsilon}$ solves

$$\partial_t \hat{u} + \Delta^E \hat{u}= -H(\hat{u}) + 3C_\epsilon H (\hat{u}\odot h^{\sharp})  - 9 C'_\epsilon H(\hat{u} \odot h^{\sharp}) +\xi_\epsilon$$

which is exactly (<ref>).
The proof is thus complete once Proposition <ref> below is shown.

The restriction of the renormalised model ${\hat{Z}}_\epsilon$ to $\mathcal{T}^\mathcal{I}_{\leq 1, :}$ and $\mathcal{T}^r_{\leq 0, :}$ converges
as $\epsilon\to 0$.

This follows exactly as in the proof of Proposition <ref> from Proposition <ref> below.

A difference between Propositions <ref> and <ref> is that the latter only claims convergence of the model restricted to objects relevant for the equation at hand. This is done purely for convenience, since this reduces the number of stochastic estimates we need to show below. An appropriate modification of the general results in <cit.>, <cit.> would be desirable and is expected to hold in our setting as well.

§.§.§ Stochastic estimates

This section is dedicated to proving the following Proposition, which is the combination of Lemma <ref>, Lemma <ref> and Lemma <ref> below.

The conclusion of Proposition <ref> holds for $\tau_p \in {\mathcal{T}}^r_{\leq 1, :}$.

We use similar notation as in Section <ref>, where
this time a random variable belonging to the $k$th homogeneous Wiener chaos can
be described by a kernel in $L^2(\pi^* E )^{\hat{\otimes} k}$, where $\pi: \mathbb{R}\times M \to M$ is the canonical projection. That is, the space of of square-integrable sections of $k$ space-time variables.

As previously, we introduce convenient graphical notation:
Nodes in our graph will represent variables in $\RR\times M$, with one
special node [baseline=-3] [root] ; representing the point $\star\in \mathbb{R}\times M$.
As previously, the nodes [baseline=-3] [var] ; represent the arguments of our kernel and the the nodes which we draw as [baseline=-3] [dot] ;, represent dummy variables that are to be integrated out.
Each line then represents a kernel, with
[baseline=-0.1cm] [kernel] (0,0) to (1,0) ;
representing the kernel  $K((t,p),(s,q)):=G_{t-s}(p,q) \kappa(t-s)\kappa(d(p,q)/\bar{r})$,
[baseline=-0.1cm] [keps] (0,0) to (1,0) ;
representing the kernel

\begin{equation}\label{eq:space_time mollified kernel}
\left(K * \rho_\epsilon\right) (z_1, z_2)= \int_{\mathbb{R}\times M}   K (z_1,z') \rho_\epsilon (z',z_2) \dVol_z
\end{equation}

for $\rho_\epsilon(z_1, z_2) = \phi^\epsilon(t_1-{t}_2) G_{{\epsilon}^2} (p_1 , {p}_2 )$
[baseline=-0.1cm] [testfcn] (1,0) to (0,0) ;
representing a representing a localised test function $\phi_\star^\lambda\in \mathcal{C}^\infty (\pi^*E^{\otimes i})$, for some $i\in \{1,2,3\}$ depending on the context.
Whenever we draw a barred arrow
[baseline=-0.1cm] [kernel1] (0,0) to (1,0) ;
this represents a factor $K(z_1, z_2) - R \left(Q_{\leq 0} j_\star K(\cdot, z_2)\right) (z_1)$, where
$z_1$ and $z_2$ are the coordinates of the starting and end
point respectively.
Similarly[baseline=-0.1cm] [kernel2] (0,0) to (1,0) ;
represents a factor $K(z_1, z_2) -R \left(Q_{\leq 1} j_\star K(\cdot, z_2)\right) (z_1)$.
In the calculations that follow it will be convenient to recall that the heat kernel $G_t(p,q)\in  E|_p \otimes E^*|_q$ satisfies
$G^{\star}_t(q,p)= G_t(p,q)$, c.f. Remark <ref>.

Similarly to above we can write

\begin{equation}
\bigl(\Pi_\star^{(\epsilon)}\<2>\bigr)(\phi_\star^\lambda)
\begin{tikzpicture}[scale=0.35,baseline=0.3cm]
\node at (0,-1)  [root] (root) {};
\node at (0,1)  [dot] (int) {};
\node at (-1,3.5)  [var] (left) {};
\node at (1,3.5)  [var] (right) {};

\draw[testfcn] (int) to  (root);

\draw[keps] (left) to (int);
\draw[keps] (right) to (int);
\end{tikzpicture}\;
+ \;
\begin{tikzpicture}[scale=0.35,baseline=0.3cm]
\node at (0,-1)  [root] (root) {};
\node at (0,1)  [dot] (int) {};
\node at (0,3.5) [dot] (top) {};

\draw[testfcn] (int) to  (root);

\draw[keps] (top) to[bend left=60] (int);
\draw[keps] (top) to[bend right=60] (int);
\end{tikzpicture}\;.
\end{equation}

as well as

\begin{tikzpicture}[scale=0.35,baseline=0.3cm]
\node at (0,0)  [root] (root) {};
\node at (0,2.5) [reddot] (top) {};
\draw[red, keps] (top) to[bend left=60] (root);
\draw[red, keps] (top) to[bend right=60] (root);
\end{tikzpicture}\; =   C_\epsilon h^{\sharp}
\qquad \text{and } \qquad
\begin{tikzpicture}[scale=0.35,baseline=0.3cm]
\node at (0,0)  [root] (root) {};
\node at (0,2.5) [bluedot] (top) {};
\draw[blue,keps] (top) to[bend left=60] (root);
\draw[blue,keps] (top) to[bend right=60] (root);
\end{tikzpicture}\; = \;
\begin{tikzpicture}[scale=0.35,baseline=0.3cm]
\node at (0,0)  [root] (root) {};
\node at (0,2.5) [dot] (top) {};
\draw[keps] (top) to[bend left=60] (root);
\draw[keps] (top) to[bend right=60] (root);
\end{tikzpicture}\; - \;
\begin{tikzpicture}[scale=0.35,baseline=0.3cm]
\node at (0,0)  [root] (root) {};
\node at (0,2.5) [reddot] (top) {};
\draw[red, keps] (top) to[bend left=60] (root);
\draw[red, keps] (top) to[bend right=60] (root);
\end{tikzpicture}\;  .

§.§.§ A first counterterm.

The section $\begin{tikzpicture}[scale=0.35,baseline=0.3cm]
\node at (0,0)  [root] (int) {};
\node at (0,2.5) [bluedot] (top) {};
\draw[blue, keps] (top) to[bend left=60] (int);
\draw[blue, keps] (top) to[bend right=60] (int);
\end{tikzpicture}\;$ of $\mathcal{C}^\infty(\pi^* E^{\otimes 2})$
is uniformly bounded for $\epsilon\in (0,1]$ and converges uniformly.

We proceed as in the proof of Lemma <ref>, the only slight difference being the fact the $\xi$ is an $E$-valued space time white noise, instead of a scalar valued spatial white noise.
First, fix a local frame $\{e_{p,i}\}_{i=1}^{\dim E} \in E|_p$ and let
$\xi_{i,p}= \langle \tau_{z,p}e_{p,i}, \xi\rangle_E\in \mathcal{D}'(M)$.

Thus, we find writing $\left(\phi^\epsilon\right)^{\star 2}(s-s')= \int \phi^\epsilon (s- r )\phi^\epsilon (s'- r )  dr$ (recall that $\phi$ is even)

\begin{align*}
\begin{tikzpicture}[scale=0.35,baseline=0.3cm]
\node at (0,0)  [root] (root) {};
\node at (0,2.5) [dot] (top) {};
\draw[keps] (top) to[bend left=60] (root);
\draw[keps] (top) to[bend right=60] (root);
\end{tikzpicture}
&= \sum_i \int_\mathbb{R^2}  \left(\phi^\epsilon\right)^{\star 2}(s-s') \\
&\qquad \times \left(\int_{B_r}
\left( Z^E_{t-s+\epsilon^2}(p, z)(\tau_{z,p}e_{p,i}) \right) \otimes
\left(Z^E_{t-s'+\epsilon^2}(p, z)(\tau_{z,p}e_{p,i})  \right)
\dVol_z\right)  ds  ds' + R^\epsilon\\
&= \sum_i e_{p,i}\otimes e_{p,i} \int_\mathbb{R^2} \int_{B_r(p)}  \left(\phi^\epsilon\right)^{\star 2}(s-s')
{Z}_{t-s+\epsilon^2}(p, z)
{Z}_{t-s'+\epsilon^2}(p, z)
ds  ds'  \dVol_z+ R^\epsilon \ ,
\end{align*}

where we used that the $\xi_{i,p}(r,z)$ for $i=1,\ldots,d$ are independent. Finally, note the conclusion of the Lemma holds for $R^\epsilon$ and

\begin{align*}
\int_\mathbb{R^2} \int_{B_r(p)}  \phi^\epsilon (s- r )\phi^\epsilon (s'- r )&
{Z}_{t-s+\epsilon^2}(p, z)
{Z}_{t-s'+\epsilon^2}(p, z)
ds  ds' dr \dVol_z  -C_\epsilon
\end{align*}

converges uniformly in $p$ as $\epsilon \to 0$.

Thus, we find as previously,

\begin{equation}
\bigl(\hat \Pi_\star^{(\epsilon)}\<2>\bigr)(\phi_\star^\lambda)
\begin{tikzpicture}[scale=0.35,baseline=0.3cm]
\node at (0,-1)  [root] (root) {};
\node at (0,1)  [dot] (int) {};
\node at (-1,3.5)  [var] (left) {};
\node at (1,3.5)  [var] (right) {};

\draw[testfcn] (int) to  (root);

\draw[keps] (left) to (int);
\draw[keps] (right) to (int);
\end{tikzpicture}\;
%+\; 2\;
%	\node at (0,-1)  [root] (root) {};
%	\node at (0,1)  [dot] (int) {};
%	\node at (0,3.5) [dot] (top) {};
%	\draw[testfcn] (int) to  (root);
%	\draw[red, keps] (top) to[bend left=60] (int);
%	\draw[blue, keps] (top) to[bend right=60] (int);
\begin{tikzpicture}[scale=0.35,baseline=0.3cm]
\node at (0,-1)  [root] (root) {};
\node at (0,1)  [dot] (int) {};
\node at (0,3.5) [bluedot] (top) {};

\draw[testfcn] (int) to  (root);

\draw[blue, keps] (top) to[bend left=60] (int);
\draw[blue, keps] (top) to[bend right=60] (int);
\end{tikzpicture}\; .
\end{equation}

\begin{align}
\bigl(\Pi_\star^{(\epsilon)}\<3>\bigr)(\phi_\star^\lambda)&=
\begin{tikzpicture}[scale=0.35,baseline=0.25cm]
\node at (0,1)  [int] (int) {};
\node at (-1.5,3.5)  [var] (left) {};
\node at (0,3.8)  [var] (middle) {};
\node at (1.5,3.5)  [var] (right) {};
\node at (0,-1)  [root] (root) {};
\draw[keps] (left) to  (int);
\draw[keps] (right) to (int);
\draw[keps] (middle) to (int);
\draw[testfcn] (int) to  (root);
\end{tikzpicture}\;
+ 3\;
\begin{tikzpicture}[scale=0.35,baseline=0.25cm]
\node at (0,-1)  [root] (root) {};
\node at (0,1)  [int] (int) {};
\node at (-1,3.6)  [int] (top) {};
\node at (1,3.6)  [var] (topr) {};
\draw[keps,bend right = 60] (top) to  (int);
\draw[keps,bend left = 60] (top) to (int);
\draw[keps,bend left = 60] (topr) to (int);
\draw[testfcn] (int) to  (root);
\end{tikzpicture}\;,
\end{align}

we also find that

\begin{align}
\bigl(\hat \Pi_\star^{(\epsilon)}\<3>\bigr)(\phi_\star^\lambda)&=
\begin{tikzpicture}[scale=0.35,baseline=0.25cm]
\node at  (0,1)  [int] (int) {};
\node at (-1.5,3.5)  [var] (left) {};
\node at (0,3.8)  [var] (middle) {};
\node at (1.5,3.5)  [var] (right) {};
\node at (0,-1)  [root] (root) {};
\draw[keps] (left) to  (int);
\draw[keps] (right) to (int);
\draw[keps] (middle) to (int);
\draw[testfcn] (int) to  (root);
\end{tikzpicture}\;
+ 3\;
\begin{tikzpicture}[scale=0.35,baseline=0.25cm]
\node at (0,-1)  [root] (root) {};
\node at (0,1)  [int] (int) {};
\node at (-1,3.6)  [bluedot] (top) {};
\node at (1,3.6)  [var] (topr) {};
\draw[blue, keps,bend right = 60] (top) to  (int);
\draw[blue, keps,bend left = 60] (top) to (int);
\draw[keps,bend left = 60] (topr) to (int);
\draw[testfcn] (int) to  (root);
\end{tikzpicture}\;.
\end{align}

The next lemma summarises what we can conclude so far, arguing exactly as in the proof of Proposition <ref>.

The conclusion of Proposition <ref> holds for $\tau_p\in \Func_W(T)$ for

$$T\in\left\{ \<3>, \ \<3'>, \  \<3''>, \  \<3'''>   \right\}\ . $$

§.§.§ A second counterterm

We introduce the kernel of two variables

\begin{equation}
Q_\epsilon =
\begin{tikzpicture}[scale=0.35,baseline=-0.1cm]
\node at (4,0)  [root] (root) {};
\node at (0,0)  [root] (middle) {};
\node at (2,-1.5)  [int] (left) {};
\node at (2,1.5)  [int] (right) {};

\draw[keps,bend right=30] (left) to  (root);
\draw[keps,bend left=30] (left) to  (middle);
\draw[keps,bend left=30] (right) to  (root);
\draw[keps,bend right=30] (right) to  (middle);
\draw[kernel] (middle) to (root);
\end{tikzpicture}\;.
\end{equation}

with the convention that the left green dot represents the variable $\star_1\in \mathbb{R}\times M$ and the right one $\star_2\in \mathbb{R}\times M$.
Further set

\begin{equation}\label{eq:renormalised sunset}
\begin{tikzpicture}[scale=0.35,baseline=-0.1cm]
\node at (4,0)  [root] (root) {};
\node at (0,0)  [root] (middle) {};
\node at (2,-1.5)  [reddot] (left) {};
\node at (2,1.5)  [reddot] (right) {};

\draw[red,keps,bend right=30] (left) to  (root);
\draw[red,keps,bend left=30] (left) to  (middle);
\draw[red,keps,bend left=30] (right) to  (root);
\draw[red,keps,bend right=30] (right) to  (middle);
\draw[red,kernel] (middle) to (root);
\end{tikzpicture}\; := C_\epsilon \delta_{\star_2} (\star_1)
\qquad  \textit{ and } \qquad
\begin{tikzpicture}[scale=0.35,baseline=-0.1cm]
\node at (4,0)  [root] (root) {};
\node at (0,0)  [root] (middle) {};
\node at (2,-1.5)  [bluedot] (left) {};
\node at (2,1.5)  [bluedot] (right) {};

\draw[blue,keps,bend right=30] (left) to  (root);
\draw[blue,keps,bend left=30] (left) to  (middle);
\draw[blue,keps,bend left=30] (right) to  (root);
\draw[blue,keps,bend right=30] (right) to  (middle);
\draw[blue,kernel] (middle) to (root);
\end{tikzpicture}\; = Q_\epsilon (\star_1, \star_2) -C'_\epsilon \delta_{\star_2} (\star_1) \ .
\end{equation}

where $\delta_{\star_2} ( \ \cdot \ )$ denotes the Dirac distribution at $\star_2$ acting on $\phi\in \mathcal{D}(\pi^* E)$ by
$\langle \delta_{\star_2} , \phi\rangle= \phi(\star_2)$.

One has
\node at (4,0)  [root] (root) {};
\node at (0,0)  [root] (middle) {};

\node at (2,-1.5)  [bluedot] (left) {};
\node at (2,1.5)  [bluedot] (right) {};

\draw[blue,keps,bend right=30] (left) to  (root);
\draw[blue,keps,bend left=30] (left) to  (middle);
\draw[blue,keps,bend left=30] (right) to  (root);
\draw[blue,keps,bend right=30] (right) to  (middle);
\draw[blue,kernel] (middle) to (root);
\end{tikzpicture}\; = \mathcal{R}Q_\epsilon (\star_1, \star_2) -k_\epsilon(\star_2) \delta_{\star_2} (\star_1) \ $$

where the $k_\epsilon\in \mathcal{C}^\infty\left( L(\pi^*E, \pi^*E)\right)$ is uniformly bounded for $\epsilon \in (0,1]$ and converges uniformly as $\epsilon\to 0$.

Let us first calculate, denoting the first green node as $\star_1=(t,p)$ and the second as $\star_2=(t',q)$

\begin{align*}
\node at (4,0)  [root] (root) {};
\node at (0,0)  [root] (middle) {};
\node at (2,1.5)  [int] (right) {};
\draw[keps,bend left=30] (right) to  (root);
\draw[keps,bend right=30] (right) to  (middle);
\end{tikzpicture}\;\\
&= \sum_i e_{p,i}\otimes \tau_{q,p}e_{p,i} \int_\mathbb{R^2} \int_{B_r(p)}   \phi^\epsilon (s- r )\phi^\epsilon (s'- r )
Z_{t+t'-s-s'+2\epsilon^2}(p, q)
ds  ds' dr \dVol_z\\
&\qquad + R^\epsilon_1
\end{align*}

$$\tilde{Z}^{(\epsilon)}((t,p),(t',q)):=\int_\mathbb{R^2} \int_{B_r}   \phi^\epsilon (s- r )\phi^\epsilon (s'- r )
Z_{t+t'-s-s'+2\epsilon^2}(p, q)
ds  ds' dr \dVol_z$$

we define
$S_\epsilon(\star_1, \star_2) = \tilde{Z}^{(\epsilon)} (\star_1, \star_2)^2 {Z}(\star_1, \star_2)$ and $S^E_\epsilon (\star_1, \star_2)=S_\epsilon (\star_1, \star_2) \tau_{q,p}$ and find

\begin{align*}
\begin{tikzpicture}[scale=0.35,baseline=-0.1cm]
\node at (4,0)  [root] (root) {};
\node at (0,0)  [root] (middle) {};
\node at (2,-1.5)  [int] (left) {};
\node at (2,1.5)  [int] (right) {};
\draw[keps,bend right=30] (left) to  (root);
\draw[keps,bend left=30] (left) to  (middle);
\draw[keps,bend left=30] (right) to  (root);
\draw[keps,bend right=30] (right) to  (middle);
\draw[kernel] (middle) to (root);
\end{tikzpicture}\;&
=  H\left( \tau_{q,p} H(\cdot, e_{i,p}, e_{i,p}), \tau_{q,p} e_{i,p}, \tau_{q,p} e_{i,p}\right) S_\epsilon (\star_1, \star_2) +R_2^\epsilon(\star_1, \star_2) \\
&=  S^E_\epsilon (\star_1, \star_2) +R_2^\epsilon(\star_1, \star_2)  \ .
\end{align*}

where $R_2^\epsilon(p,q)$ satisfies $\vertiii{R_2^\epsilon}_{-|\fraks|+1/2, m}<\infty$ uniformly over $\epsilon$.
Since $\mathcal{R} S^E_\epsilon= S^E_\epsilon (\star_1, \star_2) - \int S_\epsilon(\star_1, z) dz \cdot \delta_{\star_1} (\star_2)$,
we conclude by observing that
$\int S_\epsilon(\star_1, z) dz - C'_\epsilon$ converges uniformly as $\epsilon \to 0$.

§.§.§ Larger trees

Note that

\begin{align*}
\bigl( \Pi_\star^{(\epsilon)}\<32>\bigr)(\phi_\star^\lambda) &=
\begin{tikzpicture}[scale=0.35,baseline=0.25cm]
\node at (0,-2)  [root] (root) {};
\node at (0,0)  [int] (int) {};
\node at (0,2.5)  [int] (middle) {};
\node at (-1.6,2)  [var] (left) {};
\node at (1.6,2)  [var] (right) {};
\node at (-1.5,4.5)  [var] (tl) {};
\node at (1.5,4.5)  [var] (tr) {};
\node at (0,5)  [var] (tm) {};
\draw[keps] (left) to  (int);
\draw[keps] (right) to (int);
\draw[keps] (tl) to  (middle);
\draw[keps] (tr) to (middle);
\draw[keps] (tm) to  (middle);
\draw[kernel1] (middle) to (int);
\draw[testfcn] (int) to  (root);
\end{tikzpicture}\; +
\begin{tikzpicture}[scale=0.35,baseline=0.25cm]
\node at (0,-2)  [root] (root) {};
\node at (0,-0)  [int] (int) {};
\node at (0,2.5)  [int] (middle) {};
\node at (-1.6,1.25)  [int] (left) {};
\node at (1.6,2)  [var] (right) {};
\node at (1.5,4.5)  [var] (tr) {};
\node at (0,6)  [var] (tm) {};
\draw[keps,bend right=40] (left) to  (int);
\draw[keps] (right) to (int);
\draw[keps,bend left=40] (left) to  (middle);
\draw[keps] (tr) to (middle);
\draw[keps] (tm) to  (middle);
\draw[kernel1] (middle) to (int);
\draw[testfcn] (int) to  (root);
\end{tikzpicture}\; +
\begin{tikzpicture}[scale=0.35,baseline=0.25cm]
\node at (0,-2)  [root] (root) {};
\node at (0,0)  [int] (int) {};
\node at (0,2.5)  [int] (middle) {};
\node at (-1.6,2)  [var] (left) {};
\node at (1.6,2)  [var] (right) {};
\node at (-1,4.6)  [int] (tl) {};
\node at (1,4.6)  [var] (tr) {};
\draw[keps] (left) to  (int);
\draw[keps] (right) to (int);
\draw[keps,bend right = 60] (tl) to  (middle);
\draw[keps,bend left = 60] (tl) to (middle);
\draw[keps,bend left = 60] (tr) to (middle);
\draw[kernel1] (middle) to (int);
\draw[testfcn] (int) to  (root);
\end{tikzpicture}\; + \;
\begin{tikzpicture}[scale=0.35,baseline=0.25cm]
\node at (0,-2)  [root] (root) {};
\node at (0,0)  [int] (int) {};
\node at (1,2.6)  [int] (middle) {};
\node at (-1,2.6)  [int] (left) {};
\node at (-0.5,4.5)  [var] (tl) {};
\node at (2.5,4.5)  [var] (tr) {};
\node at (1,5)  [var] (tm) {};
\draw[keps,bend right = 60] (left) to  (int);
\draw[keps,bend left = 60] (left) to (int);
\draw[kernel1,bend left = 60] (middle) to (int);
\draw[keps] (tl) to  (middle);
\draw[keps] (tr) to (middle);
\draw[keps] (tm) to  (middle);
\draw[testfcn] (int) to  (root);
\end{tikzpicture}\\
&\qquad +
\begin{tikzpicture}[scale=0.35,baseline=0.25cm]
\node at (0,-2)  [root] (root) {};
\node at (0,0)  [int] (int) {};
\node at (1,2.6)  [int] (middle) {};
\node at (-1,2.6)  [int] (left) {};
\node at (0,4.6)  [int] (tl) {};
\node at (2,4.6)  [var] (tr) {};
\draw[keps,bend right = 60] (left) to  (int);
\draw[keps,bend left = 60] (left) to (int);
\draw[kernel1,bend left = 60] (middle) to (int);
\draw[keps,bend right = 60] (tl) to  (middle);
\draw[keps,bend left = 60] (tl) to (middle);
\draw[keps,bend left = 60] (tr) to (middle);
\draw[testfcn] (int) to  (root);
\end{tikzpicture}\; + 6\;
\begin{tikzpicture}[scale=0.35,baseline=0.25cm]
\node at (0,-2)  [root] (root) {};
\node at (0,0)  [int] (int) {};
\node at (0,2.5)  [int] (middle) {};
\node at (-1.6,1.25)  [int] (left) {};
\node at (1.6,2)  [var] (right) {};
\node at (0,4.5)  [int] (top) {};
\draw[keps,bend right=40] (left) to  (int);
\draw[keps] (right) to (int);
\draw[keps,bend left=40] (left) to  (middle);
\draw[keps,bend right = 60] (top) to (middle);
\draw[keps,bend left = 60] (top) to  (middle);
\draw[kernel1] (middle) to (int);
\draw[testfcn] (int) to  (root);
\end{tikzpicture}\; + 6\;
\begin{tikzpicture}[scale=0.35,baseline=0.25cm]
\node at (0,-2)  [root] (root) {};
\node at (0,0)  [int] (int) {};
\node at (0,2.5)  [int] (middle) {};
\node at (-1.6,1.25)  [int] (left) {};
\node at (1.6,1.25)  [int] (right) {};
\node at (0,5)  [var] (tm) {};
\draw[keps,bend right=40] (left) to  (int);
\draw[keps,bend left=40] (left) to  (middle);
\draw[keps,bend left=40] (right) to  (int);
\draw[keps,bend right=40] (right) to  (middle);
\draw[keps] (tm) to  (middle);
\draw[kernel1] (middle) to (int);
\draw[testfcn] (int) to  (root);
\end{tikzpicture}\;,
\end{align*}

from which it follows that

\begin{align}\label{eq:renormalised 32}
\bigl( \Pi_\star^{(\epsilon)}\<32>\bigr)(\phi_\star^\lambda) &=
\begin{tikzpicture}[scale=0.35,baseline=0.25cm]
\node at (0,-2)  [root] (root) {};
\node at (0,0)  [int] (int) {};
\node at (0,2.5)  [int] (middle) {};
\node at (-1.6,2)  [var] (left) {};
\node at (1.6,2)  [var] (right) {};
\node at (-1.5,4.5)  [var] (tl) {};
\node at (1.5,4.5)  [var] (tr) {};
\node at (0,5)  [var] (tm) {};
\draw[keps] (left) to  (int);
\draw[keps] (right) to (int);
\draw[keps] (tl) to  (middle);
\draw[keps] (tr) to (middle);
\draw[keps] (tm) to  (middle);
\draw[kernel1] (middle) to (int);
\draw[testfcn] (int) to  (root);
\end{tikzpicture}\; +
\begin{tikzpicture}[scale=0.35,baseline=0.25cm]
\node at (0,-2)  [root] (root) {};
\node at (0,-0)  [int] (int) {};
\node at (0,2.5)  [int] (middle) {};
\node at (-1.6,1.25)  [int] (left) {};
\node at (1.6,2)  [var] (right) {};
\node at (1.5,4.5)  [var] (tr) {};
\node at (0,6)  [var] (tm) {};
\draw[keps,bend right=40] (left) to  (int);
\draw[keps] (right) to (int);
\draw[keps,bend left=40] (left) to  (middle);
\draw[keps] (tr) to (middle);
\draw[keps] (tm) to  (middle);
\draw[kernel1] (middle) to (int);
\draw[testfcn] (int) to  (root);
\end{tikzpicture}\; +
\begin{tikzpicture}[scale=0.35,baseline=0.25cm]
\node at (0,-2)  [root] (root) {};
\node at (0,0)  [int] (int) {};
\node at (0,2.5)  [int] (middle) {};
\node at (-1.6,2)  [var] (left) {};
\node at (1.6,2)  [var] (right) {};
\node at (-1,4.6)  [bluedot] (tl) {};
\node at (1,4.6)  [var] (tr) {};
\draw[keps] (left) to  (int);
\draw[keps] (right) to (int);
\draw[blue, keps,bend right = 60] (tl) to  (middle);
\draw[blue, keps,bend left = 60] (tl) to (middle);
\draw[keps,bend left = 60] (tr) to (middle);
\draw[kernel1] (middle) to (int);
\draw[testfcn] (int) to  (root);
\end{tikzpicture}\; + \;
\begin{tikzpicture}[scale=0.35,baseline=0.25cm]
\node at (0,-2)  [root] (root) {};
\node at (0,0)  [int] (int) {};
\node at (1,2.6)  [int] (middle) {};
\node at (-1,2.6)  [bluedot] (left) {};
\node at (-0.5,4.5)  [var] (tl) {};
\node at (2.5,4.5)  [var] (tr) {};
\node at (1,5)  [var] (tm) {};
\draw[blue, keps,bend right = 60] (left) to  (int);
\draw[blue, keps,bend left = 60] (left) to (int);
\draw[kernel1,bend left = 60] (middle) to (int);
\draw[keps] (tl) to  (middle);
\draw[keps] (tr) to (middle);
\draw[keps] (tm) to  (middle);
\draw[testfcn] (int) to  (root);
\end{tikzpicture}\\
&\qquad +
\begin{tikzpicture}[scale=0.35,baseline=0.25cm]
\node at (0,-2)  [root] (root) {};
\node at (0,0)  [int] (int) {};
\node at (1,2.6)  [int] (middle) {};
\node at (-1,2.6)  [bluedot] (left) {};
\node at (0,4.6)  [bluedot] (tl) {};
\node at (2,4.6)  [var] (tr) {};
\draw[blue, keps,bend right = 60] (left) to  (int);
\draw[blue, keps,bend left = 60] (left) to (int);
\draw[kernel1,bend left = 60] (middle) to (int);
\draw[blue, keps,bend right = 60] (tl) to  (middle);
\draw[blue, keps,bend left = 60] (tl) to (middle);
\draw[keps,bend left = 60] (tr) to (middle);
\draw[testfcn] (int) to  (root);
\end{tikzpicture}\; + 6\;
\begin{tikzpicture}[scale=0.35,baseline=0.25cm]
\node at (0,-2)  [root] (root) {};
\node at (0,0)  [int] (int) {};
\node at (0,2.5)  [int] (middle) {};
\node at (-1.6,1.25)  [int] (left) {};
\node at (1.6,2)  [var] (right) {};
\node at (0,4.5)  [bluedot] (top) {};
\draw[keps,bend right=40] (left) to  (int);
\draw[keps] (right) to (int);
\draw[keps,bend left=40] (left) to  (middle);
\draw[blue, keps,bend right = 60] (top) to (middle);
\draw[blue, keps,bend left = 60] (top) to  (middle);
\draw[kernel1] (middle) to (int);
\draw[testfcn] (int) to  (root);
\end{tikzpicture}\; + 6\;
\begin{tikzpicture}[scale=0.35,baseline=0.25cm]
\node at (0,-2)  [root] (root) {};
\node at (0,0)  [dot] (int) {};
\node at (0,2.5)  [bluedot] (middle) {};
\node at (-1.6,1.25)  [bluedot] (left) {};
\node at (1.6,1.25)  [bluedot] (right) {};
\node at (0,5)  [var] (tm) {};
\draw[blue, keps,bend right=40] (left) to  (int);
\draw[blue, keps,bend left=40] (left) to  (middle);
\draw[blue, keps,bend left=40] (right) to  (int);
\draw[blue, keps,bend right=40] (right) to  (middle);
\draw[keps] (tm) to  (middle);
\draw[blue, kernel] (middle) to (int);
\draw[testfcn] (int) to  (root);
\end{tikzpicture}\;
+ 6\;
\begin{tikzpicture}[scale=0.35,baseline=0.25cm]
\node at (0,-2)  [root] (root) {};
\node at (0,0)  [int] (int) {};
\node at (0,2.5)  [int] (middle) {};
\node at (-1.6,1.25)  [int] (left) {};
\node at (0,1.25)  [int] (right) {};
\node at (0,5)  [var] (tm) {};
\node[above] at (2,0) {${\scriptstyle 0}$};
\draw[keps,bend right=40] (left) to  (int);
\draw[keps,bend left=40] (left) to  (middle);
\draw[keps] (right) to  (int);
\draw[keps] (right) to  (middle);
\draw[keps] (tm) to  (middle);
\draw[kernel, bend left= 90] (middle) to (root);
\draw[gray, kernel, bend right= 70] (root) to (int);
\draw[testfcn] (int) to  (root);
\end{tikzpicture}\;. \nonumber
\end{align}

where the gray arrow and the $0$ in the last digram are to be read as follows:  For $n\in \mathbb{N}$

\begin{tikzpicture}[baseline=-0.1cm]
\draw[kernel] (0,0) to (0.95,0); \node at (1,0)  [root] (root) {}; \draw[gray, kernel] (1.1,0) to (2,0);
\node[above] at (0.5,0) {${\scriptstyle n}$};
\end{tikzpicture}

represents $\left(R Q_{\leq n} j_\star K(\cdot,z_1)\right) (z_2)$ where $z_1$ is the variable where the left arrow starts, and $z_2$ is the variable where the right gray arrow ends.

The conclusion of Proposition <ref> holds for $\tau_p\in \Func_W(T)$ for
$T\in\left\{ \<32>, \<32'>   \right\} $.

Equation <ref> together with $ \hat{\Pi}_q^{(\epsilon)} \hat{\Gamma}_{q,p} \<32>= \hat{\Pi}_p^{(\epsilon)} \<32>$ imply the conclusion of the lemma for $T=\<32>$ by repeated application of the Lemmas in Section <ref> exactly as in the proof of <cit.>.

In order to treat the case $T=\<32'>$, first note that $\<31>$ satisfies by a similar argument as above the conclusion of Proposition <ref> and then note that elements of $\Func_W(\<32'>)$ are obtained from it by multiplication with a jet.

§.§.§ A last tree

We introduce the following notation. For $v\in \bigoplus_{p\in M} JE$, we write

\begin{equation}
\begin{tikzpicture}[scale=0.35,baseline=-0.1cm]
\node at (4,0)  [root] (root) {};
\node [left] at (0.2,0) {$v$};
\draw[kernelBig] (0,0) to (root);
\end{tikzpicture}\;.
\end{equation}

for the function $R v (\star)$.
Then, writing  $\<3'2>_{(e_p)} \in \Func_W \<3'2>$ for the element canonically isomorphic to $e_p\in E|_p \subset J_p E$, one finds that

\begin{align*}
\bigl( \Pi_\star^{(\epsilon)}\<3'2>_{(e_\star)}\bigr)(\phi_\star^\lambda) &=
\begin{tikzpicture}[scale=0.35,baseline=0.25cm]
\node at (0,-2)  [root] (root) {};
\node at (0,0)  [int] (int) {};
\node at (0,2.5)  [int] (middle) {};
\node at (-1.6,2)  [var] (left) {};
\node at (1.6,2)  [var] (right) {};
\node at (-1.5,4.5)  [var] (tl) {};
\node at (1.5,4.5)  [var] (tr) {};
\node [above] at (0,4.8) {$e_\star$};
%	\node at (0,5)  [var] (tm) {};
\draw[keps] (left) to  (int);
\draw[keps] (right) to (int);
\draw[keps] (tl) to  (middle);
\draw[keps] (tr) to (middle);
\draw[kernelBig] (0,5) to  (middle);
\draw[kernel1] (middle) to (int);
\draw[testfcn] (int) to  (root);
\end{tikzpicture}\; +
\begin{tikzpicture}[scale=0.35,baseline=0.25cm]
\node at (0,-2)  [root] (root) {};
\node at (0,-0)  [int] (int) {};
\node at (0,2.5)  [int] (middle) {};
\node at (-1.6,1.25)  [int] (left) {};
\node at (1.6,2)  [var] (right) {};
\node at (1.5,4.5)  [var] (tr) {};
%	\node at (0,6)  [var] (tm) {};
\node [above] at (0,4.8) {$e_\star$};
\draw[keps,bend right=40] (left) to  (int);
\draw[keps] (right) to (int);
\draw[keps,bend left=40] (left) to  (middle);
\draw[keps] (tr) to (middle);
\draw[kernelBig] (0,5) to  (middle);
\draw[kernel1] (middle) to (int);
\draw[testfcn] (int) to  (root);
\end{tikzpicture}\; +
\;
\begin{tikzpicture}[scale=0.35,baseline=0.25cm]
\node at (0,-2)  [root] (root) {};
\node at (0,0)  [int] (int) {};
\node at (0,2.5)  [int] (middle) {};
\node at (-1.6,2)  [var] (left) {};
\node at (1.6,2)  [var] (right) {};
\node at (-1,4.6)  [int] (tl) {};
%	\node at (1,4.6)  [var] (tr) {};
\node [above] at (1.1,4.4) {$e_\star$};
\draw[keps] (left) to  (int);
\draw[keps] (right) to (int);
\draw[keps,bend right = 60] (tl) to  (middle);
\draw[keps,bend left = 60] (tl) to (middle);
\draw[kernelBig,bend left = 60] (1,4.6)  to (middle);
\draw[kernel1] (middle) to (int);
\draw[testfcn] (int) to  (root);
\end{tikzpicture}\; + \;
\begin{tikzpicture}[scale=0.35,baseline=0.25cm]
\node at (0,-2)  [root] (root) {};
\node at (0,0)  [int] (int) {};
\node at (1,2.6)  [int] (middle) {};
\node at (-1,2.6)  [int] (left) {};
\node at (-0.5,4.5)  [var] (tl) {};
%	\node at (2.5,4.5)  [var] (tr) {};
\node at (1,5)  [var] (tm) {};
\node [above] at (2.8,4.2) {$e_\star$};
\draw[keps,bend right = 60] (left) to  (int);
\draw[keps,bend left = 60] (left) to (int);
\draw[kernel1,bend left = 60] (middle) to (int);
\draw[keps] (tl) to  (middle);
\draw[kernelBig] (2.5,4.5)  to (middle);
\draw[keps] (tm) to  (middle);
\draw[testfcn] (int) to  (root);
\end{tikzpicture}\\
&\qquad +
\;
\begin{tikzpicture}[scale=0.35,baseline=0.25cm]
\node at (0,-2)  [root] (root) {};
\node at (0,0)  [int] (int) {};
\node at (1,2.6)  [int] (middle) {};
\node at (-1,2.6)  [int] (left) {};
\node at (0,4.6)  [int] (tl) {};
%	\node at (2,4.6)  [var] (tr) {};
\node [above] at (2.1,4.4) {$e_\star$};
\draw[keps,bend right = 60] (left) to  (int);
\draw[keps,bend left = 60] (left) to (int);
\draw[kernel1,bend left = 60] (middle) to (int);
\draw[keps,bend right = 60] (tl) to  (middle);
\draw[keps,bend left = 60] (tl) to (middle);
\draw[kernelBig,bend left = 60] (2,4.6) to (middle);
\draw[testfcn] (int) to  (root);
\end{tikzpicture}\; +  \;
2\; \begin{tikzpicture}[scale=0.35,baseline=0.25cm]
\node at (0,-2)  [root] (root) {};
\node at (0,0)  [int] (int) {};
\node at (0,2.5)  [int] (middle) {};
\node at (-1.6,1.25)  [int] (left) {};
\node at (1.6,1.25)  [int] (right) {};
%	\node at (0,5)  [var] (tm) {};
\node [above] at (0,4.8) {$e_\star$};
\draw[keps,bend right=40] (left) to  (int);
\draw[keps,bend left=40] (left) to  (middle);
\draw[keps,bend left=40] (right) to  (int);
\draw[keps,bend right=40] (right) to  (middle);
\draw[kernelBig] (0,5) to  (middle);
\draw[kernel1] (middle) to (int);
\draw[testfcn] (int) to  (root);
\end{tikzpicture}\;.
\end{align*}

and thus

\begin{align}\label{eq:renormalised 3'2}
\bigl( \hat \Pi_\star^{(\epsilon)}\<3'2>_{(e_\star)}\bigr)(\phi_\star^\lambda) &=
\begin{tikzpicture}[scale=0.35,baseline=0.25cm]
\node at (0,-2)  [root] (root) {};
\node at (0,0)  [int] (int) {};
\node at (0,2.5)  [int] (middle) {};
\node at (-1.6,2)  [var] (left) {};
\node at (1.6,2)  [var] (right) {};
\node at (-1.5,4.5)  [var] (tl) {};
\node at (1.5,4.5)  [var] (tr) {};
%	\node at (0,5)  [var] (tm) {};
\node [above] at (0,4.8) {$e_\star$};
\draw[keps] (left) to  (int);
\draw[keps] (right) to (int);
\draw[keps] (tl) to  (middle);
\draw[keps] (tr) to (middle);
\draw[kernelBig] (0,5) to  (middle);
\draw[kernel1] (middle) to (int);
\draw[testfcn] (int) to  (root);
\end{tikzpicture}\; +
\begin{tikzpicture}[scale=0.35,baseline=0.25cm]
\node at (0,-2)  [root] (root) {};
\node at (0,-0)  [int] (int) {};
\node at (0,2.5)  [int] (middle) {};
\node at (-1.6,1.25)  [int] (left) {};
\node at (1.6,2)  [var] (right) {};
\node at (1.5,4.5)  [var] (tr) {};
%	\node at (0,6)  [var] (tm) {};
\node [above] at (0,4.8) {$e_\star$};
\draw[keps,bend right=40] (left) to  (int);
\draw[keps] (right) to (int);
\draw[keps,bend left=40] (left) to  (middle);
\draw[keps] (tr) to (middle);
\draw[kernelBig] (0,5) to  (middle);
\draw[kernel1] (middle) to (int);
\draw[testfcn] (int) to  (root);
\end{tikzpicture}\; +
\;
\begin{tikzpicture}[scale=0.35,baseline=0.25cm]
\node at (0,-2)  [root] (root) {};
\node at (0,0)  [int] (int) {};
\node at (0,2.5)  [int] (middle) {};
\node at (-1.6,2)  [var] (left) {};
\node at (1.6,2)  [var] (right) {};
\node at (-1,4.6)  [bluedot] (tl) {};
%	\node at (1,4.6)  [var] (tr) {};
\node [above] at (1.1,4.4) {$e_\star$};
\draw[keps] (left) to  (int);
\draw[keps] (right) to (int);
\draw[blue, keps,bend right = 60] (tl) to  (middle);
\draw[blue, keps,bend left = 60] (tl) to (middle);
\draw[kernelBig,bend left = 60] (1,4.6)  to (middle);
\draw[kernel1] (middle) to (int);
\draw[testfcn] (int) to  (root);
\end{tikzpicture}\; + \;
\begin{tikzpicture}[scale=0.35,baseline=0.25cm]
\node at (0,-2)  [root] (root) {};
\node at (0,0)  [int] (int) {};
\node at (1,2.6)  [int] (middle) {};
\node at (-1,2.6)  [bluedot] (left) {};
\node at (-0.5,4.5)  [var] (tl) {};
%	\node at (2.5,4.5)  [var] (tr) {};
\node at (1,5)  [var] (tm) {};
\node [above] at (2.8,4.2) {$e_\star$};
\draw[blue, keps,bend right = 60] (left) to  (int);
\draw[blue, keps,bend left = 60] (left) to (int);
\draw[kernel1,bend left = 60] (middle) to (int);
\draw[keps] (tl) to  (middle);
\draw[kernelBig] (2.5,4.5)  to (middle);
\draw[keps] (tm) to  (middle);
\draw[testfcn] (int) to  (root);
\end{tikzpicture}\\
&\qquad +
\;
\begin{tikzpicture}[scale=0.35,baseline=0.25cm]
\node at (0,-2)  [root] (root) {};
\node at (0,0)  [int] (int) {};
\node at (1,2.6)  [int] (middle) {};
\node at (-1,2.6)  [bluedot] (left) {};
\node at (0,4.6)  [bluedot] (tl) {};
%	\node at (2,4.6)  [var] (tr) {};
\node [above] at (2.1,4.4) {$e_\star$};
\draw[blue, keps,bend right = 60] (left) to  (int);
\draw[blue, keps,bend left = 60] (left) to (int);
\draw[kernel1,bend left = 60] (middle) to (int);
\draw[blue, keps,bend right = 60] (tl) to  (middle);
\draw[blue, keps,bend left = 60] (tl) to (middle);
\draw[kernelBig,bend left = 60] (2,4.6) to (middle);
\draw[testfcn] (int) to  (root);
\end{tikzpicture}\; +  \;
\begin{tikzpicture}[scale=0.35,baseline=0.25cm]
\node at (0,-2)  [root] (root) {};
\node at (0,0)  [int] (int) {};
\node at (0,2.5)  [bluedot] (middle) {};
\node at (-1.6,1.25)  [bluedot] (left) {};
\node at (1.6,1.25)  [bluedot] (right) {};
%	\node at (0,5)  [var] (tm) {};
\node [above] at (0,4.8) {$e_\star$};
\draw[blue, keps,bend right=40] (left) to  (int);
\draw[blue, keps,bend left=40] (left) to  (middle);
\draw[blue, keps,bend left=40] (right) to  (int);
\draw[blue, keps,bend right=40] (right) to  (middle);
\draw[kernelBig] (0,5) to  (middle);
\draw[blue, kernel] (middle) to (int);
\draw[testfcn] (int) to  (root);
\end{tikzpicture}\; + \;
2\; \begin{tikzpicture}[scale=0.35,baseline=0.25cm]
\node at (0,-2)  [root] (root) {};
\node at (0,0)  [int] (int) {};
\node at (0,2.5)  [int] (middle) {};
\node at (-1.6,1.25)  [int] (left) {};
\node at (0,1.25)  [int] (right) {};
%	\node at (0,5)  [var] (tm) {};
\node [above] at (0,4.8) {$e_\star$};
\node[above] at (2,0) {${\scriptstyle 0}$};
\draw[keps,bend right=40] (left) to  (int);
\draw[keps,bend left=40] (left) to  (middle);
\draw[keps] (right) to  (int);
\draw[keps] (right) to  (middle);
\draw[kernelBig] (0,5) to  (middle);
\draw[kernel, bend left= 90] (middle) to (root);
\draw[testfcn] (int) to  (root);
\draw[gray, kernel, bend right= 70] (root) to (int);
\end{tikzpicture}\;. \nonumber
\end{align}

Next we find that for the canonical model

\begin{align*}
\Pi^\epsilon_q  \Gamma^\epsilon_{q,p} \<3'0>_{(e_p)}- \Pi^\epsilon_p \<3'0>_{(e_p)} =
& \pmb{\Pi}^\epsilon\left( \<3'0>_{(\Gamma_{q,p} e_p)} -\<3'0>_{(e_p)}\right) - R \left( Q_{\leq 1} j_q K(\pmb{\Pi}^\epsilon \<3'>_{(Q_0\Gamma_{q,p} e_p)}-\pmb{\Pi}^\epsilon\<3'>_{(e_p)})\right) \\
&- \left( \Pi^\epsilon_q  \Gamma^\epsilon_{q,p} - \Pi_q^\epsilon\right) Q_{\leq 0} j_{p} K( \pmb{\Pi}^\epsilon \<3'>_{(e_p)} )(p)\\
=& \pmb{\Pi}^\epsilon\left( \<3'0>_{(\Gamma_{q,p} e_p)} -\<3'0>_{(e_p)}\right) - R \left( Q_{\leq 1} j_q K(\pmb{\Pi}\epsilon \<3'>_{(Q_0\Gamma_{q,p} e_p)}-\pmb{\Pi}^\epsilon \<3'>_{(e_p)})\right) \\
&- R \left(  \Gamma_{q,p} - \id\right)  \left( Q_{\leq 0} j_{p}K( \pmb{\Pi}^\epsilon \<3'>_{(e_p)}) \right)(p)
\end{align*}

And therefore

\begin{align*}
\Pi^\epsilon_q  \Gamma_{q,p}^\epsilon \<3'2>_{(e_p)}- \Pi^\epsilon_p \<3'2>_{(e_p)} =&\underbrace{
\pmb{\Pi}^\epsilon\left( \<3'2>_{(\Gamma_{q,p} e_p)} -\<3'2>_{(e_p)}\right)
- R \left( Q_{\leq 1} j_q K(\pmb{\Pi}^\epsilon \<3'>_{(Q_0\Gamma_{q,p} e_p -e_p)})\right) \cdot \left(\pmb{\Pi}\<2>\right))}_{=:\bigl( \Pi^\epsilon_q  \Gamma_{q,p}^\epsilon \<3'2>_{(e_p)}- \Pi^\epsilon_p \<3'2>_{(e_p)}\bigr)_A} \\
&-  \underbrace{\left(\pmb{\Pi}^\epsilon \<2>\right) \cdot R\left(  \Gamma_{q,p} - \id\right) Q_{\leq 0} j_{p}K( \pmb{\Pi}^\epsilon \<3'>_{(e_p)} )}_{=:\bigl( \Pi^\epsilon_q  \Gamma_{q,p}^\epsilon \<3'2>_{(e_p)}- \Pi^\epsilon_p \<3'2>_{(e_p)}\bigr)_B}
\end{align*}

We suggestively define

\begin{equation}\label{eq:renormalised error bound 3'2 B}
\bigl( \hat{\Pi}^\epsilon_q  \hat{\Gamma}_{q,p}^\epsilon \<3'2>_{(e_p)}- \hat{\Pi}^\epsilon_p  \<3'2>_{(e_p)} \bigr)_{\hat{B}}:=
\left(\hat{\pmb{\Pi}}^\epsilon\<2>\right) \cdot R\left(  {\Gamma}_{q,p} - \id\right)Q_{\leq 0} j_p  K( \hat{\pmb{\Pi}}^\epsilon (\<3'>_{(e_p)}) ) \
\end{equation}

and set

\begin{align*}
\bigl( \hat{\Pi}^\epsilon_q  \hat{\Gamma}_{q,p}^\epsilon \<3'2>_{(e_p)}- \hat{\Pi}^\epsilon_p  \<3'2>_{(e_p)} \bigr)_{\hat{A}} &=
\hat{\Pi}^\epsilon_q  \hat{\Gamma}_{q,p}^\epsilon \<3'2>_{(e_p)}- \hat{\Pi}^\epsilon_p  \<3'2>_{(e_p)} \\
&-\bigl( \hat{\Pi}^\epsilon_q  \hat{\Gamma}_{q,p}^\epsilon \<3'2>_{(e_p)}- \hat{\Pi}^\epsilon_p  \<3'2>_{(e_p)} \bigr)_{\hat{B}} .
\end{align*}

\begin{align*}
\bigl( \Pi^\epsilon_\star  \Gamma_{\star,p} \<3'2>_{(e_p)}- \Pi^\epsilon_p \<3'2>_{(e_p)}\bigr)_A  (\phi_\star^\lambda) &=
\begin{tikzpicture}[scale=0.35,baseline=0.25cm]
\node at (0,-2)  [root] (root) {};
\node at (0,0)  [int] (int) {};
\node at (0,2.5)  [int] (middle) {};
\node at (-1.6,2)  [var] (left) {};
\node at (1.6,2)  [var] (right) {};
\node at (-1.5,4.5)  [var] (tl) {};
\node at (1.5,4.5)  [var] (tr) {};
%	\node at (0,5)  [var] (tm) {};
\node [above] at (0,4.8) {${\scriptscriptstyle \Gamma_{\star,p} e_p -e_p}$};
\draw[keps] (left) to  (int);
\draw[keps] (right) to (int);
\draw[keps] (tl) to  (middle);
\draw[keps] (tr) to (middle);
\draw[kernelBig] (0,5) to  (middle);
\draw[kernel2] (middle) to (int);
\draw[testfcn] (int) to  (root);
\end{tikzpicture}\; +
\begin{tikzpicture}[scale=0.35,baseline=0.25cm]
\node at (0,-2)  [root] (root) {};
\node at (0,-0)  [int] (int) {};
\node at (0,2.5)  [int] (middle) {};
\node at (-1.6,1.25)  [int] (left) {};
\node at (1.6,2)  [var] (right) {};
\node at (1.5,4.5)  [var] (tr) {};
%	\node at (0,6)  [var] (tm) {};
\node [above] at (0,4.8) {${\scriptscriptstyle \Gamma_{\star,p} e_p -e_p}$};
\draw[keps,bend right=40] (left) to  (int);
\draw[keps] (right) to (int);
\draw[keps,bend left=40] (left) to  (middle);
\draw[keps] (tr) to (middle);
\draw[kernelBig] (0,5) to  (middle);
\draw[kernel2] (middle) to (int);
\draw[testfcn] (int) to  (root);
\end{tikzpicture}\; +
\;
\begin{tikzpicture}[scale=0.35,baseline=0.25cm]
\node at (0,-2)  [root] (root) {};
\node at (0,0)  [int] (int) {};
\node at (0,2.5)  [int] (middle) {};
\node at (-1.6,2)  [var] (left) {};
\node at (1.6,2)  [var] (right) {};
\node at (-1,4.6)  [int] (tl) {};
%	\node at (1,4.6)  [var] (tr) {};
\node [above] at (1.1,4.4) {${\scriptscriptstyle \Gamma_{\star,p} e_p -e_p}$};
\draw[keps] (left) to  (int);
\draw[keps] (right) to (int);
\draw[keps,bend right = 60] (tl) to  (middle);
\draw[keps,bend left = 60] (tl) to (middle);
\draw[kernelBig,bend left = 60] (1,4.6)  to (middle);
\draw[kernel2] (middle) to (int);
\draw[testfcn] (int) to  (root);
\end{tikzpicture}
\\
&\qquad +
\;
\begin{tikzpicture}[scale=0.35,baseline=0.25cm]
\node at (0,-2)  [root] (root) {};
\node at (0,0)  [int] (int) {};
\node at (1,2.6)  [int] (middle) {};
\node at (-1,2.6)  [int] (left) {};
\node at (-0.5,4.5)  [var] (tl) {};
%	\node at (2.5,4.5)  [var] (tr) {};
\node at (1,5)  [var] (tm) {};
\node [above] at (3,4.2)  {${\scriptscriptstyle \Gamma_{\star,p} e_p -e_p}$};
\draw[keps,bend right = 60] (left) to  (int);
\draw[keps,bend left = 60] (left) to (int);
\draw[kernel2,bend left = 60] (middle) to (int);
\draw[keps] (tl) to  (middle);
\draw[kernelBig] (2.5,4.5)  to (middle);
\draw[keps] (tm) to  (middle);
\draw[testfcn] (int) to  (root);
\end{tikzpicture}\; +
\;
\begin{tikzpicture}[scale=0.35,baseline=0.25cm]
\node at (0,-2)  [root] (root) {};
\node at (0,0)  [int] (int) {};
\node at (1,2.6)  [int] (middle) {};
\node at (-1,2.6)  [int] (left) {};
\node at (0,4.6)  [int] (tl) {};
%	\node at (2,4.6)  [var] (tr) {};
\node [above] at (2.1,4.4) {${\scriptscriptstyle \Gamma_{\star,p} e_p -e_p}$};
\draw[keps,bend right = 60] (left) to  (int);
\draw[keps,bend left = 60] (left) to (int);
\draw[kernel2,bend left = 60] (middle) to (int);
\draw[keps,bend right = 60] (tl) to  (middle);
\draw[keps,bend left = 60] (tl) to (middle);
\draw[kernelBig,bend left = 60] (2,4.6) to (middle);
\draw[testfcn] (int) to  (root);
\end{tikzpicture}\; +  \;
2\; \begin{tikzpicture}[scale=0.35,baseline=0.25cm]
\node at (0,-2)  [root] (root) {};
\node at (0,0)  [int] (int) {};
\node at (0,2.5)  [int] (middle) {};
\node at (-1.6,1.25)  [int] (left) {};
\node at (1.6,1.25)  [int] (right) {};
%	\node at (0,5)  [var] (tm) {};
\node [above] at (0,4.8) {${\scriptscriptstyle \Gamma_{\star,p} e_p -e_p}$};
\draw[keps,bend right=40] (left) to  (int);
\draw[keps,bend left=40] (left) to  (middle);
\draw[keps,bend left=40] (right) to  (int);
\draw[keps,bend right=40] (right) to  (middle);
\draw[kernelBig] (0,5) to  (middle);
\draw[kernel2] (middle) to (int);
\draw[testfcn] (int) to  (root);
\end{tikzpicture}\;,
\end{align*}

we find that the latter term is given by

\begin{align}\label{eq:renormalised error bound 3'2 A}
\bigl( \hat{\Pi}^\epsilon_\star  \hat{\Gamma}_{\star,p} \<3'2>_{(e_p)}-& \hat{\Pi}^\epsilon_p \<3'2>_{(e_p)}\bigr)_{\hat{A}}  (\phi_\star^\lambda) =
\begin{tikzpicture}[scale=0.35,baseline=0.25cm]
\node at (0,-2)  [root] (root) {};
\node at (0,0)  [int] (int) {};
\node at (0,2.5)  [int] (middle) {};
\node at (-1.6,2)  [var] (left) {};
\node at (1.6,2)  [var] (right) {};
\node at (-1.5,4.5)  [var] (tl) {};
\node at (1.5,4.5)  [var] (tr) {};
\node [above] at (0,4.8) {${\scriptscriptstyle \Gamma_{\star,p} e_p -e_p}$};
%	\node at (0,5)  [var] (tm) {};
\draw[keps] (left) to  (int);
\draw[keps] (right) to (int);
\draw[keps] (tl) to  (middle);
\draw[keps] (tr) to (middle);
\draw[kernelBig] (0,5) to  (middle);
\draw[kernel2] (middle) to (int);
\draw[testfcn] (int) to  (root);
\end{tikzpicture}\; +
\begin{tikzpicture}[scale=0.35,baseline=0.25cm]
\node at (0,-2)  [root] (root) {};
\node at (0,-0)  [int] (int) {};
\node at (0,2.5)  [int] (middle) {};
\node at (-1.6,1.25)  [int] (left) {};
\node at (1.6,2)  [var] (right) {};
\node at (1.5,4.5)  [var] (tr) {};
%	\node at (0,6)  [var] (tm) {};
\node [above] at (0,4.8) {${\scriptscriptstyle \Gamma_{\star,p} e_p -e_p}$};
\draw[keps,bend right=40] (left) to  (int);
\draw[keps] (right) to (int);
\draw[keps,bend left=40] (left) to  (middle);
\draw[keps] (tr) to (middle);
\draw[kernelBig] (0,5) to  (middle);
\draw[kernel2] (middle) to (int);
\draw[testfcn] (int) to  (root);
\end{tikzpicture}\; +
\;
\begin{tikzpicture}[scale=0.35,baseline=0.25cm]
\node at (0,-2)  [root] (root) {};
\node at (0,0)  [int] (int) {};
\node at (0,2.5)  [int] (middle) {};
\node at (-1.6,2)  [var] (left) {};
\node at (1.6,2)  [var] (right) {};
\node at (-1,4.6)  [bluedot] (tl) {};
%	\node at (1,4.6)  [var] (tr) {};
\node [above] at (1.1,4.4) {${\scriptscriptstyle \Gamma_{\star,p} e_p -e_p}$};
\draw[keps] (left) to  (int);
\draw[keps] (right) to (int);
\draw[blue, keps,bend right = 60] (tl) to  (middle);
\draw[blue, keps,bend left = 60] (tl) to (middle);
\draw[kernelBig,bend left = 60] (1,4.6)  to (middle);
\draw[kernel2] (middle) to (int);
\draw[testfcn] (int) to  (root);
\end{tikzpicture}
\\
&\qquad + \;
\begin{tikzpicture}[scale=0.35,baseline=0.25cm]
\node at (0,-2)  [root] (root) {};
\node at (0,0)  [int] (int) {};
\node at (1,2.6)  [int] (middle) {};
\node at (-1,2.6)  [bluedot] (left) {};
\node at (-0.5,4.5)  [var] (tl) {};
%	\node at (2.5,4.5)  [var] (tr) {};
\node at (1,5)  [var] (tm) {};
\node [above] at (3,4.2)  {${\scriptscriptstyle \Gamma_{\star,p} e_p -e_p}$};
\draw[blue, keps,bend right = 60] (left) to  (int);
\draw[blue, keps,bend left = 60] (left) to (int);
\draw[kernel2,bend left = 60] (middle) to (int);
\draw[keps] (tl) to  (middle);
\draw[kernelBig] (2.5,4.5)  to (middle);
\draw[keps] (tm) to  (middle);
\draw[testfcn] (int) to  (root);
\end{tikzpicture} +
\;
\begin{tikzpicture}[scale=0.35,baseline=0.25cm]
\node at (0,-2)  [root] (root) {};
\node at (0,0)  [int] (int) {};
\node at (1,2.6)  [int] (middle) {};
\node at (-1,2.6)  [bluedot] (left) {};
\node at (0,4.6)  [bluedot] (tl) {};
%	\node at (2,4.6)  [var] (tr) {};
\node [above] at (2.1,4.4) {${\scriptscriptstyle \Gamma_{\star,p} e_p -e_p}$};
\draw[blue, keps,bend right = 60] (left) to  (int);
\draw[blue, keps,bend left = 60] (left) to (int);
\draw[kernel2,bend left = 60] (middle) to (int);
\draw[blue, keps,bend right = 60] (tl) to  (middle);
\draw[blue, keps,bend left = 60] (tl) to (middle);
\draw[kernelBig,bend left = 60] (2,4.6) to (middle);
\draw[testfcn] (int) to  (root);
\end{tikzpicture}\; +
\begin{tikzpicture}[scale=0.35,baseline=0.25cm]
\node at (0,-2)  [root] (root) {};
\node at (0,0)  [int] (int) {};
\node at (0,2.5)  [bluedot] (middle) {};
\node at (-1.6,1.25)  [bluedot] (left) {};
\node at (1.6,1.25)  [bluedot] (right) {};
%	\node at (0,5)  [var] (tm) {};
\node [above] at (0,4.8) {${\scriptscriptstyle \Gamma_{\star,p} e_p -e_p}$};
\draw[blue, keps,bend right=40] (left) to  (int);
\draw[blue, keps,bend left=40] (left) to  (middle);
\draw[blue, keps,bend left=40] (right) to  (int);
\draw[blue, keps,bend right=40] (right) to  (middle);
\draw[kernelBig] (0,5) to  (middle);
\draw[blue, kernel] (middle) to (int);
\draw[testfcn] (int) to  (root);
\end{tikzpicture}\; + \;
2\; \begin{tikzpicture}[scale=0.35,baseline=0.25cm]
\node at (0,-2)  [root] (root) {};
\node at (0,0)  [int] (int) {};
\node at (0,2.5)  [int] (middle) {};
\node at (-1.6,1.25)  [int] (left) {};
\node at (0,1.25)  [int] (right) {};
%	\node at (0,5)  [var] (tm) {};
\node [above] at (0,4.8) {${\scriptscriptstyle \Gamma_{\star,p} e_p -e_p}$};
\node[above] at (2,0) {${\scriptstyle 1}$};
\draw[keps,bend right=40] (left) to  (int);
\draw[keps,bend left=40] (left) to  (middle);
\draw[keps] (right) to  (int);
\draw[keps] (right) to  (middle);
\draw[kernelBig] (0,5) to  (middle);
\draw[kernel, bend left= 90] (middle) to (root);
%	\draw[kernel, bend left= 83] (middle) to (root);
\draw[testfcn] (int) to  (root);
\draw[gray, kernel, bend right= 70] (root) to (int);
\end{tikzpicture}\;. \nonumber
\end{align}

The conclusion of Proposition <ref> holds for $\tau_p\in \Func_W(\<3'2>)$.

The desired bounds on $\bigl( \hat \Pi_\star^{(\epsilon)}\<3'2>_{(e_\star)}\bigr)(\phi_\star^\lambda)$ follow from (<ref>) as in the proof of Lemma <ref>.
In order to establish the desired bounds on

$$\bigl( \hat{\Pi}^\epsilon_\star  \hat{\Gamma}_{\star,p} \<3'2>_{(e_p)}- \hat{\Pi}^\epsilon_p \<3'2>_{(e_p)}\bigr) (\phi_\star^\lambda) $$

we estimate (<ref>) and (<ref>) separately.
Indeed, the former is treated easily using the bounds already established in Lemma <ref> since it only involves a product of a smooth function with a distribution.
For the latter, note that for all terms in the Wiener Chaos decomposition except

\begin{equation}\label{eq:some more work}
\mathcal{P}^\epsilon (\phi^\lambda_\star):=\begin{tikzpicture}[scale=0.35,baseline=0.25cm]
\node at (0,-2)  [root] (root) {};
\node at (0,0)  [int] (int) {};
\node at (0,2.5)  [bluedot] (middle) {};
\node at (-1.6,1.25)  [bluedot] (left) {};
\node at (1.6,1.25)  [bluedot] (right) {};
%	\node at (0,5)  [var] (tm) {};
\node [above] at (0,4.8) {${\scriptscriptstyle \Gamma_{\star,p} e_p -e_p}$};
\draw[blue, keps,bend right=40] (left) to  (int);
\draw[blue, keps,bend left=40] (left) to  (middle);
\draw[blue, keps,bend left=40] (right) to  (int);
\draw[blue, keps,bend right=40] (right) to  (middle);
\draw[kernelBig] (0,5) to  (middle);
\draw[blue, kernel] (middle) to (int);
\draw[testfcn] (int) to  (root);
\end{tikzpicture}\; + \;
\begin{tikzpicture}[scale=0.35,baseline=0.25cm]
\node at (0,-2)  [root] (root) {};
\node at (0,0)  [int] (int) {};
\node at (0,2.5)  [int] (middle) {};
\node at (-1.6,1.25)  [int] (left) {};
\node at (0,1.25)  [int] (right) {};
%	\node at (0,5)  [var] (tm) {};
\node [above] at (0,4.8) {${\scriptscriptstyle \Gamma_{\star,p} e_p -e_p}$};
\node[above] at (2,0) {${\scriptstyle 1}$};
\draw[keps,bend right=40] (left) to  (int);
\draw[keps,bend left=40] (left) to  (middle);
\draw[keps] (right) to  (int);
\draw[keps] (right) to  (middle);
\draw[kernelBig] (0,5) to  (middle);
\draw[kernel, bend left= 90] (middle) to (root);
%	\draw[kernel, bend left= 83] (middle) to (root);
\draw[testfcn] (int) to  (root);
\draw[gray, kernel, bend right= 70] (root) to (int);
\end{tikzpicture}\;
\end{equation}

one can argue as for the terms in (<ref>).
In order to treat this last term, let us introduce the following notation:

* $f(z)= R\left(\Gamma_{\star,p} e_p -e_p\right)(z)$, which satisfies $|f(z)|\lesssim d_\fraks (z,\star)^{\delta_0}$

* $K^\epsilon(\star_1,\star_2)=
\tikzsetnextfilename{nondiv_kernel}
\begin{tikzpicture}[scale=0.35,baseline=-0.1cm]
\node at (4,0)  [root] (root) {};
\node at (0,0)  [root] (middle) {};

\node at (2,-1.5)  [int] (left) {};
\node at (2,1.5)  [int] (right) {};
\draw[keps,bend right=30] (left) to  (root);
\draw[keps,bend left=30] (left) to  (middle);
\draw[keps,bend left=30] (right) to  (root);
\draw[keps,bend right=30] (right) to  (middle);
\end{tikzpicture}\;,
which satisfies $K^\epsilon(\star_1,\star_2)\lesssim d_\fraks(\star_1,\star_2)^{-2}$,

and note that (<ref>) is given by

\begin{align*}
\mathcal{P}^\epsilon (\phi^\lambda_\star)=&\int_{\mathbb{R}\times M}\int_{\mathbb{R}\times M} K^\epsilon(x,z)K(x,z )f(z)\phi^\lambda(x)- K^\epsilon(x,z)K(x,z )\tau_{z,x} f(x)\phi^\lambda_\star(x)dx dz\\
&- \int_{\mathbb{R}\times M}\int_{\mathbb{R}\times M}K^\epsilon (x,z) R\left(Q_{\leq 1} j_\star K(\cdot, z)\right)(x)f(z)\phi_\star^\lambda(x)
dx dz \\
&+ \int_{\mathbb{R}\times M}\int_{\mathbb{R}\times M}  k_\epsilon (x)f(x)\phi^\lambda_\star (x) dx dz.
\end{align*}

Clearly $\int_{\mathbb{R}\times M}\int_{\mathbb{R}\times M} | k_\epsilon (x)f(x)\phi^\lambda_\star (x) |dx dz\lesssim \lambda^{\delta_0}$, so we turn to the other term. On the one hand, first integrating over $x\in \mathbb{R}\times M$ one finds uniformly in $\epsilon>0$.

\begin{align*}
&\Bigg| \int_{|x|<\frac{1}{2}d(x,z)} \Big( K^\epsilon(x,z)K(x,z )f(z)\phi^\lambda_\star (x) - K^\epsilon (x,z) R\left(Q_{\leq 1} j_\star K(\cdot, z)\right)(x)f(z)\phi^\lambda_\star(x) \Big)dx dz \\
&\qquad-\int_{|x|<\frac{1}{2}d(x,z)} K^\epsilon(x,z)K(x,z ) \tau_{z,x}f(x)\phi^\lambda_\star(x)
dx dz\Bigg| \\
&\lesssim  \int_{|x|<\frac{1}{2}d(x,z)} |K^\epsilon(x,z)|d(x,\star)^2  \sup_{|y|<\frac{1}{2}d(x,z)} |Q_{\leq 2} j_y K(\cdot, z)|  |f(z)| |\phi^\lambda_\star| dx dz \\
&\qquad +\int_{|x|<\frac{1}{2}d(x,z)} |K^\epsilon(x,z)K(x,z ) \tau_{z,x}f(x)\phi^\lambda_\star(x)| dx dz\\
%\lesssim&  \int_{|x|<\frac{1}{2}d(x,z)} d_\fraks(x,y)^{-2} d(x,\star)^2\sup_{|y|<\frac{1}{2}d(x,z)} |Q_{\leq 2} j_y K(\cdot, z)|  d_\fraks (z,\star)^{\delta_0} |\phi^\lambda(x)| dx dz \\
%&+\int_{|x|<\frac{1}{2}d(x,z)} d_\fraks(x,y)^{-2} |K(x,z )| d_\fraks (x,\star)^{\delta_0} |\phi^\lambda(x)| dx dz\\
%\lesssim&  \int_{|x|<\frac{1}{2}d(x,z)} d_\fraks(x,z)^{-2} d(x,\star)^2d(x,z)^{-5}  d_\fraks (z,\star)^{\delta_0} |\phi^\lambda(x)| dx dz \\
%&+\int_{|x|<\frac{1}{2}d(x,z)} d_\fraks(x,z)^{-2} d(x,z)^{-3}  d_\fraks (x,\star)^{\delta_0} |\phi^\lambda(x)| dx dz\\
%\lesssim&  \int_{|x|<\frac{1}{2}d(x,z)} d(x,z)^{-7}  d_\fraks (x,\star)^{\delta_0 +2} |\phi^\lambda(x)| dx dz \\
%&+\int_{|x|<\frac{1}{2}d(x,z)} d_\fraks(x,z)^{-5} d_\fraks (x,\star)^{\delta_0} |\phi^\lambda(x)| dx dz\\
%\lesssim&  \int d(\star,z)^{-7}  \left( d_\fraks(z,\star)^{\delta_0+2+|\fraks|} \lambda^{-|\fraks|} \wedge \lambda^{\delta_0+2} \right) dz  \\
%&+\int d_\fraks(\star ,z)^{-5} \left( d_\fraks(z,\star)^{\delta_0+|\fraks|} \lambda^{-|\fraks|} \wedge \lambda^{\delta_0} \right) dz\\
&\lesssim \lambda^{\delta_0-\kappa}
\end{align*}

Similarly, integrating first over $z$

\begin{align*}
&\Bigg| \int_{|x|<\frac{1}{2}d(x,z)} \Big( K^\epsilon(x,z)K(x,z )f(z)\phi^\lambda_\star (x) - K^\epsilon (x,z) R\left(Q_{\leq 1} j_\star K(\cdot, z)\right)(x)f(z)\phi^\lambda_\star(x) \Big) dx dz \\
&\qquad -\int_{|x|<\frac{1}{2}d(x,z)} K^\epsilon(x,z)K(x,z ) \tau_{z,x}f(x)\phi^\lambda_\star (x)
dx dz\Bigg| \\
&\leq \Bigg|\int_{|x|>\frac{1}{2}d(x,z)}  K^\epsilon(x,z)K(x,z )\left(f(z)-\tau_{z,x}f(x)\right)\phi^\lambda_\star(x)  dx dz\Bigg|\\
&\qquad + \int_{|x|>\frac{1}{2}d(x,z)}  | K^\epsilon (x,z) R\left(Q_{\leq 1} j_\star K(\cdot, z)\right)(x)f(z)\phi^\lambda_\star (x)
dx dz\\
&\lesssim \lambda^{\delta_0}
\end{align*}

Note that in order to estimate (<ref>), we needed to leverage positive and negative renormalisation for the set of edges connecting the same two vertices. Thus, estimating this diagram falls outside the scope of (naively) applying Theorem <ref>.

Note that the scalar $\phi^4_3$ equation can essentially be treated as in [28] by working with the choice of admissible realisation such that $R j_p 1= 1$ for all $p\in M$, since then $\hat{\Pi}^\epsilon_q  \hat{\Gamma}_{q,p}^\epsilon \<3'2>_{(j_p 1)}- \hat{\Pi}^\epsilon_p  \<3'2>_{(j_p 1)} =0 \ .$

§.§ The $\phi^3_4$ equation.

C^{ \<2>, 1}_\epsilon =\int_{\mathbb{R}^3}   \phi^\epsilon (t-s+r)\phi^\epsilon (t-s+r )
\bar{Z}_{s+s'+2\epsilon^2}(0)\
ds  ds' dr \ ,

$$C^{ \<2>,2}_\epsilon =\frac{1}{3}\int_{\mathbb{R}^3}  \phi^\epsilon (t-s+r)\phi^\epsilon (t-s+r )
(s+s'+2\epsilon)  \bar{Z}_{s+s'+2\epsilon^2}(0)\
ds  ds' dr  \ .

In order to describe the remaining counterterms, we use again graphical notation, with
$\tikzsetnextfilename{arrrr} \tikz[baseline=-0.1cm] \draw[red, kernel] (0,0) to (1,0) \enlarge;$
representing $\bar{Z}$,
$\tikzsetnextfilename{arr} \tikz[baseline=-0.1cm] \draw[red, keps] (0,0) to (1,0) \enlarge; $
representing the kernel
$\bar{Z}* \bar{\rho}_\epsilon$
for $\bar{\rho}_\epsilon(t,z) = \phi^\epsilon(t) \bar{Z}_{{\epsilon}^2}(x)$.
The special node $\tikzsetnextfilename{rootdotbg} \tikz[baseline=-3] \node[broot] {};$ represents $0\in \mathbb{R}^5$, while
$\tikzsetnextfilename{arbred} \tikz[baseline=-3] \node [reddot] {};$ represents a variable in $\mathbb{R}^5$ to be integrated out.

\begin{equation*}
C^{\!\!\<211>}_\epsilon = 2\;
\tikzsetnextfilename{phi34ct'brg}
\begin{tikzpicture}[scale=0.35,baseline=0.8cm]
\node at (-2,1)  [broot] (left) {};
\node at (0.0,3)  [reddot] (left1) {};
\node at (-2,5)  [reddot] (left2) {};
\node at (-2,3) [reddot] (variable3) {};
\node at (0,5.7) [reddot] (variable4) {};
\draw[red, kernel] (left1) to   (left);
\draw[red, kernel] (left2) to  (left1);
\draw[red, keps] (variable3) to  (left);
\draw[red, keps] (variable4) to   (left1);
\draw[red, keps] (variable3) to   (left2);
\draw[red, keps] (variable4) to   (left2);
\end{tikzpicture}\; ,
\qquad
C^{\<22j>}_\epsilon = 2\; \tikzsetnextfilename{phi34ct211''brg}
\begin{tikzpicture}[scale=0.35,baseline=.9cm]
%\node at (-1.5,1)  [dot] (left) {};
\node at (0.0,2)  [broot] (int) {};
\node at (-2,3)  [reddot] (left) {};
\node at (0,5.5)  [reddot] (ttop) {};
%	\node at (-3,5) [var] (variable4) {};
\node at (2,3)  [reddot] (right) {};
\node at (0,4) [reddot] (top) {};
%	\node at (3,5) [var] (variable2) {};
\draw[red, kernel] (left) to  (int);
\draw[red, kernel] (right) to  (int);
\draw[red, keps] (ttop) to  (right);
\draw[red, keps] (top) to  (right);
\draw[red, keps] (top) to  (left);
\draw[red, keps] (ttop) to (left);
\end{tikzpicture} \ ,
\qquad
C^{\<11> }_\epsilon =
\tikzsetnextfilename{phi34ct11'br'g}
\begin{tikzpicture}[scale=0.35,baseline=0.6cm]
\node at (-2,1)  [broot] (left) {};
\node at (-2,3)  [reddot] (left1) {};
\node at (0.0,2) [reddot] (variable) {};
\draw[red, kernel] (left1) to (left);
\draw[red, keps] (variable) to (left1);
\draw[red, keps] (left) to (variable);
\end{tikzpicture}\; .
\end{equation*}

Finally, set

\begin{equation}\label{countertermm}
\bar{C}_\epsilon= (C^{ \<2>, 1}_\epsilon,
C^{ \<2>,2}_\epsilon ,
C^{\!\!\<211>}_\epsilon ,
C^{\<22j>}_\epsilon ,
C^{\<11> }_\epsilon )\in \mathbb{R}^5 \ .
\end{equation}

We shall prove the following theorem.

Let $M$ be a compact four dimensional Riemannian manifold, $\xi$ be space time white noise on $M$
and $\xi_\epsilon$ its regularisation as described above. Let $u_0 \in \cC^{\alpha}(M) $ for $\alpha>-1$
and denote by $\Delta$ the Laplace–Beltrami operator and by $s:M\to \mathbb{R}$ the scalar curvature.
Then, for $\bar{C}_\epsilon$ as in (<ref>) and $u_\epsilon$ solving

\begin{equation}\label{eq:phi3_4}
\partial_t u_\epsilon + \Delta u_\epsilon = u^2_\epsilon - C^{\<2>, 1}_\epsilon  - C_\epsilon^{\<2>, 2} s -4 C^{\<11>}_\epsilon u_\epsilon  - 4C^{\!\!\<211>}_\epsilon - C^{\<22j>}_\epsilon +  \xi_\epsilon, \qquad u_\epsilon(0) =u_0 \ ,
\end{equation}

there exists a (random) $T>0$ and $u\in  \mathcal{D}'((0,T)\times M)$ such that $u_\epsilon \to u$ as $\epsilon\to 0$ in probability.

Note that for this equation one expects that the maximal existence time of the solutions is a.s. finite.

We proceed as by now standard and define edge types

* $\mathcal{E}_+= \{\mathcal{I} \}$,

* $\mathcal{E}_-= \{ \Xi \}$,

* $\mathcal{E}^+_0= \{\delta^{\mathcal{I}}\}$ with the injection $\iota:\mathcal{E}_+\to  \mathcal{E}^+_0$.

* Since the equations is scalar valued and the non-linearity parallel, we suppress the
$\mathcal{E}^{0}_{0}$ type edges.

We fix homogeneities $|\mathcal{I}|\in (2-\frac{1}{10}, 2)$, $|\Xi|\in (-3-\frac{1}{10}, -3)$
and the rule characterised by

$${R}(\mathcal{I})= \big\{ [\delta^{\mathcal{I}}]^2,\ [\delta^{\mathcal{I}} \mathcal{I}],\ [\mathcal{I}]^2, \ [\Xi]  \big\}\ .$$

Again suffices to solve the equation in $\mathcal{D}^\gamma$ for some $\gamma>1$ and thus, using again graphical notation where $\Xi= \<X>$, $\mathcal{I}$ is written as $\<I>$  and $\delta^{\mathcal{I}}$ as $\<1'>$,  the relevant trees are

$$\mathfrak{T}^\mathcal{J}:=\left\{ \<1>, \; \<20>, \;\<210>
\;\<1'>, \; \<2'0>\ \right\}$$

\left\{\ \<X>, \; \<2>, \; \<21>,  \; \<211>, \; \<22j>,
\; \<2'>, \; \<2''>, \; \<2'1> ,\; \<21'>\ \right\}\ . $$

Counterterms are only needed for negative subtrees with an even number of noises, namely
2>, 211>, 22j>, 11>.
We denote for
$$\bar{C}=(C^{( \<2>, 1)}, C^{ \<2>,2}, C^{\<211>}, C^{\<22j>}, C^{\<11>}) \in \mathbb{R}^5$$

by  $g_ {\bar C}\in \mathfrak{G}$ the element such that
$\mathfrak{g}_{\bar C}(\<2>)= C^{( \<2>, 1)} s + C^{ \<2>,2}, $
$\mathfrak{g}_{\bar C}(\tau)= C^\tau$ for $\tau\in \left\{\<211>, \ \<22j>,\ \<11> \right\} $ and $\mathfrak{g}_{\bar C}(\bar \tau)= e(\bar \tau)$ for any other negative subtree $\bar \tau\in \mathfrak{T}_-\setminus \left\{ \<2>,  \<211>, \<22j>,\<11> \right\}$ . We denote by $\mathfrak{G}_-'\subset \mathfrak{G}_-$ the subgroup generated by these $\mathfrak{g}_{\bar C} \in \mathfrak{G}_-$.

Denote by $Z^\epsilon=(\Pi^\epsilon, \Gamma^\epsilon)$ the canonical model for the smooth noise $\xi_\epsilon$, such that the abstract Integration map $\mathcal{I}$ realises $K_t(p,q):=G_t(p,q) \kappa(t)\kappa(d(p,q)/\bar{r})$, where
$G$ denotes the heat kernel, and such that the admissible realisation has the property that $R(j_\cdot 1)=1$.

We again write  $\hat{Z}^\epsilon:= (Z^\epsilon)^{\mathfrak{g}_{\bar C}}$ and
$\hat{Z}^\epsilon= (\hat{\Pi}^\epsilon, \hat{\Gamma}^\epsilon)$.

In the setting described above, let $\eta, \gamma\in \mathbb{R}$ such that
$-1<\eta<\alpha$ and $\big| |\Xi| + |\mathcal{I}|\big|<\gamma <\delta_0$.
Note that this time the homogeneity of $|\Xi|$ and $| \<2> |$ are
too low to apply the fixed point theorem <ref> directly. Fortunately, this can again be resolved in <cit.>.
One then finds that there exists a unique abstract (local) solution $\hat{U}^\epsilon\in \mathcal{D}^{\gamma, \eta}(\mathcal{T}^{\mathcal{I}})$
of the abstract lift of

$$\partial_t u-\Delta u= u^2 +\xi$$

for the renormalised models $\hat{Z}^\epsilon$.
Then, by Proposition <ref>
$\hat{u}= \mathcal{R} \hat{U}^{\epsilon}$ solves (<ref>).
The proof is thus complete, once the analogue of Proposition <ref> for the model $\hat{Z}^\epsilon$ of this section  is shown, which in turn follows Proposition <ref> below.

§.§.§ Stochastic estimates

In this section we prove the following proposition.

The conclusion of Proposition <ref> holds for $\tau_p \in {\mathcal{T}}^r_{\alpha, :}$ whenever $\alpha\leq 0$.

Throughout this section, we use (essentially) the same graphical as in Section <ref>.
Thus, we write

\begin{tikzpicture}[scale=0.35,baseline=0.3cm]
\node at (0,0.0)  [root] (root) {};
\node at (0,2.5) [reddot] (top) {};
\draw[red, keps] (top) to[bend left=60] (root);
\draw[red, keps] (top) to[bend right=60] (root);
\end{tikzpicture}\; =   C^{( \<2>, 1)}_\epsilon +
C^{ \<2>,2}_\epsilon s(\star)
\qquad \text{and } \qquad
\begin{tikzpicture}[scale=0.35,baseline=0.3cm]
\node at (0.0,0)  [root] (root) {};
\node at (0,2.5) [bluedot] (top) {};
\draw[blue,keps] (top) to[bend left=60] (root);
\draw[blue,keps] (top) to[bend right=60] (root);
\end{tikzpicture}\; = \;
\begin{tikzpicture}[scale=0.35,baseline=0.3cm]
\node at (0.0,0)  [root] (root) {};
\node at (0,2.5) [dot] (top) {};
\draw[keps] (top) to[bend left=60] (root);
\draw[keps] (top) to[bend right=60] (root);
\end{tikzpicture}\; - \;
\begin{tikzpicture}[scale=0.35,baseline=0.3cm]
\node at (0.0,0)  [root] (root) {};
\node at (0,2.5) [reddot] (top) {};
\draw[red, keps] (top) to[bend left=60] (root);
\draw[red, keps] (top) to[bend right=60] (root);
\end{tikzpicture}\;  .

The section $\begin{tikzpicture}[scale=0.35,baseline=0.3cm]
\node at (0.0,0)  [root] (int) {};
\node at (0,2.5) [bluedot] (top) {};
\draw[blue, keps] (top) to[bend left=60] (int);
\draw[blue, keps] (top) to[bend right=60] (int);
\end{tikzpicture}\;$ of $\mathcal{C}^\infty(\RR\times M)$
is uniformly bounded for $\epsilon\in (0,1]$ and converges uniformly.

We proceed as previously and find that

\begin{align*}
\pmb{\Pi}^{(\epsilon)}  \<1>  (t,p)&= \int_\mathbb{R}  \langle \xi , \phi^\epsilon (s- \cdot ) G_{t-s+\epsilon^2}(p, \cdot) \rangle \kappa (t-s) ds + r_\epsilon(t,p) .
\end{align*}

\begin{tikzpicture}[scale=0.35,baseline=0.3cm]
\node at (0.0,0)  [root] (root) {};
\node at (0,2.5) [dot] (top) {};
\draw[keps] (top) to[bend left=60] (root);
\draw[keps] (top) to[bend right=60] (root);
\end{tikzpicture}
=  \int_{\mathbb{R}^3} \int_M  \phi^\epsilon (t-s+r)\phi^\epsilon (t-s+r )
G_{s+s'+2\epsilon^2}(p, p) \kappa(s)\kappa (s')
ds  ds' dr + R_\epsilon \ ,

where $R_\epsilon$ satisfies the conclusion of the lemma.
Thus, the lemma follows from Theorem <ref>.

The Lemma follows from a simplified ad verbatim adaptation of the proofs of Lemma <ref> and <ref>.

The smooth functions

\tikzsetnextfilename{phi34ct'b}
\begin{tikzpicture}[scale=0.35,baseline=0.8cm]
\node at (-2,1)  [root] (left) {};
\node at (0.0,3)  [bluedot] (left1) {};
\node at (-2,5)  [bluedot] (left2) {};
\node at (-2,3) [bluedot] (variable3) {};
\node at (0,5.7) [bluedot] (variable4) {};
\draw[blue, kernel] (left1) to   (left);
\draw[blue, kernel] (left2) to  (left1);
\draw[blue, keps] (variable3) to  (left);
\draw[blue, keps] (variable4) to   (left1);
\draw[blue, keps] (variable3) to   (left2);
\draw[blue, keps] (variable4) to   (left2);
\end{tikzpicture}
\;
\tikzsetnextfilename{phi34ct'bl}
\begin{tikzpicture}[scale=0.35,baseline=0.8cm]
\node at (-2,1)  [root] (left) {};
\node at (0.0,3)  [dot] (left1) {};
\node at (-2,5)  [dot] (left2) {};
\node at (-2,3) [dot] (variable3) {};
\node at (0,5.7) [dot] (variable4) {};
\draw[kernel] (left1) to   (left);
\draw[kernel] (left2) to  (left1);
\draw[keps] (variable3) to  (left);
\draw[keps] (variable4) to   (left1);
\draw[keps] (variable3) to   (left2);
\draw[keps] (variable4) to   (left2);
\end{tikzpicture}\;
-C^{\<211>}_\epsilon\ ,
\qquad
\tikzsetnextfilename{phi34ct211''b}
\begin{tikzpicture}[scale=0.35,baseline=.9cm]
\node at (0,2)  [root] (int) {};
\node at (-2,3)  [bluedot] (left) {};
\node at (0.0,5.5)  [bluedot] (ttop) {};
\node at (2,3)  [bluedot] (right) {};
\node at (0,4) [bluedot] (top) {};
\draw[blue, kernel] (left) to  (int);
\draw[blue, kernel] (right) to  (int);
\draw[blue, keps] (ttop) to  (right);
\draw[blue, keps] (top) to  (right);
\draw[blue, keps] (top) to  (left);
\draw[blue, keps] (ttop) to (left);
\end{tikzpicture} \;
\tikzsetnextfilename{phi34ct211''bl}
\begin{tikzpicture}[scale=0.35,baseline=.9cm]
\node at (0.0,2)  [root] (int) {};
\node at (-2,3)  [dot] (left) {};
\node at (0,5.5)  [dot] (ttop) {};
\node at (2,3)  [dot] (right) {};
\node at (0,4) [dot] (top) {};
\draw[kernel] (left) to  (int);
\draw[kernel] (right) to  (int);
\draw[keps] (ttop) to  (right);
\draw[keps] (top) to  (right);
\draw[keps] (top) to  (left);
\draw[keps] (ttop) to (left);
\end{tikzpicture} \;

are uniformly bounded for $\epsilon\in (0,1]$ and converge uniformly as $\epsilon\to 0$. Furthermore, denoting the variable corresponding to the higher green dot
by $\star_1$ and the one corresponding to the lower one by $\star_2$,
one has

\tikzsetnextfilename{phi34ct11''bl}
\begin{tikzpicture}[scale=0.35,baseline=0.6cm]
\node at (-2,1)  [root] (left) {};
\node at (-2,3)  [root] (left1) {};
\node at (0.0,2) [bluedot] (variable) {};
\draw[blue, kernel] (left1) to (left);
\draw[blue, keps] (variable) to (left1);
\draw[blue, keps] (left) to (variable);
\end{tikzpicture}\;
\tikzsetnextfilename{phi34ct11''bldef}
\begin{tikzpicture}[scale=0.35,baseline=0.6cm]
\node at (-2,1)  [root] (left) {};
\node at (-2,3)  [root] (left1) {};
\node at (0.0,2) [dot] (variable) {};
\draw[kernel] (left1) to (left);
\draw[keps] (variable) to (left1);
\draw[keps] (left) to (variable);
\end{tikzpicture}\; - C^{\<11> }_\epsilon \delta_{\star_2} (\star_1) =
\mathcal{R}Q_\epsilon (\star_1, \star_2) -k_\epsilon(\star_2) \delta_{\star_2} (\star_1)

where the $k_\epsilon\in \mathcal{C}^\infty (\RR\times M)$ is uniformly bounded for $\epsilon \in (0,1]$ and converges uniformly as $\epsilon\to 0$.

We proceed exactly as in the proof of Propositions <ref> & <ref>. The term $\bigl(\hat \Pi_\star^{(\epsilon)}\<2>\bigr)(\phi_\star^\lambda)$ can be treated exactly as in the estimate of the analogue term in Proposition <ref>.
We next study the Wiener Chaos expansion of

$$\bigl( \hat{\Pi}_\star^\epsilon \<21>\bigr)(\phi_\star^\lambda), \qquad \bigl( \hat{\Pi}_\star^\epsilon \<22j>\bigr)(\phi_\star^\lambda), \qquad
\bigl( \hat{\Pi}^\epsilon_\star\<211>\bigr)(\phi_\star^\lambda) \ .$$

\bigl( \Pi_{\star}^\epsilon \<21>\bigr) (\phi_\star^\lambda)
\begin{tikzpicture}[scale=0.35,baseline=1.1cm]
\node at (0.8,1.5)  [root] (root) {};
%\node at (-1.5,1)  [dot] (left) {};
\node at (-1.3,3)  [dot] (left1) {};
\node at (-1.3,5)  [dot] (left2) {};
\node at (0.8,3) [var] (variable2) {};
\node at (0.8,4.3) [var] (variable3) {};
\node at (0.8,5.7) [var] (variable4) {};
\draw[testfcn] (left1) to (root);
\draw[kernel] (left2) to  (left1);
\draw[keps] (variable2) to  (left1);
\draw[keps] (variable3) to  (left2);
\draw[keps] (variable4) to (left2);
\end{tikzpicture}
\begin{tikzpicture}[scale=0.35,baseline=1.1cm]
\node at (0.8,1.5)  [root] (root) {};
%\node at (-1.5,1)  [dot] (left) {};
\node at (-1.3,3)  [dot] (left1) {};
\node at (-1.3,5)  [dot] (left2) {};
\node at (0.8,3) [var] (variable2) {};
%	\node at (0.8,4.3) [var] (variable3) {};
\node at (0.8,5) [dot] (variable4) {};
\draw[testfcn] (left1) to (root);
\draw[kernel] (left2) to  (left1);
\draw[keps] (variable2) to  (left1);
\draw[keps,bend left=40] (variable4) to  (left2);
\draw[keps,bend right=40] (variable4) to (left2);
\end{tikzpicture}
+ 2\;
\begin{tikzpicture}[scale=0.35,baseline=1.1cm]
\node at (0.8,1.5)  [root] (root) {};
%\node at (-1.5,1)  [dot] (left) {};
\node at (-1.3,3)  [dot] (left1) {};
\node at (-1.3,5)  [dot] (left2) {};
\node at (0.8,4) [dot] (variable2) {};
\node at (0.8,5.7) [var] (variable3) {};
\draw[testfcn] (left1) to (root);
\draw[kernel] (left2) to  (left1);
\draw[keps] (variable2) to  (left1);
\draw[keps] (variable3) to  (left2);
\draw[keps] (variable2) to (left2);
\end{tikzpicture}\;,

we find that

\bigl( \hat{\Pi}_\star^\epsilon \<21>\bigr)(\phi_\star^\lambda)
\begin{tikzpicture}[scale=0.35,baseline=1.1cm]
\node at (0.8,1.5)  [root] (root) {};
%\node at (-1.5,1)  [dot] (left) {};
\node at (-1.3,3)  [dot] (left1) {};
\node at (-1.3,5)  [dot] (left2) {};
\node at (0.8,3) [var] (variable2) {};
\node at (0.8,4.3) [var] (variable3) {};
\node at (0.8,5.7) [var] (variable4) {};
\draw[testfcn] (left1) to (root);
\draw[kernel] (left2) to  (left1);
\draw[keps] (variable2) to  (left1);
\draw[keps] (variable3) to  (left2);
\draw[keps] (variable4) to (left2);
\end{tikzpicture}
\begin{tikzpicture}[scale=0.35,baseline=1.1cm]
\node at (0.8,1.5)  [root] (root) {};
%\node at (-1.5,1)  [dot] (left) {};
\node at (-1.3,3)  [dot] (left1) {};
\node at (-1.3,5)  [bluedot] (left2) {};
\node at (0.8,3) [var] (variable2) {};
%	\node at (0.8,4.3) [var] (variable3) {};
\node at (0.8,5) [bluedot] (variable4) {};
\draw[testfcn] (left1) to (root);
\draw[kernel] (left2) to  (left1);
\draw[keps] (variable2) to  (left1);
\draw[blue, keps,bend left=40] (variable4) to  (left2);
\draw[blue, keps,bend right=40] (variable4) to (left2);
\end{tikzpicture}
+ 2\;
\begin{tikzpicture}[scale=0.35,baseline=1.1cm]
\node at (0.8,1.5)  [root] (root) {};
%\node at (-1.5,1)  [dot] (left) {};
\node at (-1.3,3)  [dot] (left1) {};
\node at (-1.3,5)  [dot] (left2) {};
\node at (0.8,4) [bluedot] (variable2) {};
\node at (0.8,5.7) [var] (variable3) {};
\draw[testfcn] (left1) to (root);
\draw[blue, kernel] (left2) to  (left1);
\draw[blue, keps] (variable2) to  (left1);
\draw[keps] (variable3) to  (left2);
\draw[blue, keps] (variable2) to (left2);
\end{tikzpicture}\;.

We observe that all three trees can by bounded straightforwardly using the estimates of Section <ref>.
Similarly one finds that

\begin{align*}
\bigl( \Pi_\star^\epsilon \<22j>\bigr)(\phi_\star^\lambda)
\begin{tikzpicture}[scale=0.35,baseline=.9cm]
\node at (0,0)  [root] (root) {};
%\node at (-1.5,1)  [dot] (left) {};
\node at (0,2)  [dot] (int) {};
\node at (-2,3)  [dot] (left) {};
\node at (-1,5) [var] (variable3) {};
\node at (-3,5) [var] (variable4) {};
\node at (2,3)  [dot] (right) {};
\node at (1,5) [var] (variable1) {};
\node at (3,5) [var] (variable2) {};
\draw[testfcn] (int) to (root);
\draw[kernel] (left) to  (int);
\draw[kernel] (right) to  (int);
\draw[keps] (variable2) to  (right);
\draw[keps] (variable1) to  (right);
\draw[keps] (variable3) to  (left);
\draw[keps] (variable4) to (left);
\end{tikzpicture}
+ 2
\begin{tikzpicture}[scale=0.35,baseline=.9cm]
\node at (0,0)  [root] (root) {};
%\node at (-1.5,1)  [dot] (left) {};
\node at (0,2)  [dot] (int) {};
\node at (-2,3)  [dot] (left) {};
\node at (-2,5) [dot] (variable3) {};
%	\node at (-3,5) [var] (variable4) {};
\node at (2,3)  [dot] (right) {};
\node at (1,5) [var] (variable1) {};
\node at (3,5) [var] (variable2) {};
\draw[testfcn] (int) to (root);
\draw[kernel] (left) to  (int);
\draw[kernel] (right) to  (int);
\draw[keps] (variable2) to  (right);
\draw[keps] (variable1) to  (right);
\draw[keps, bend left=40] (variable3) to  (left);
\draw[keps, bend right=40] (variable3) to (left);
\end{tikzpicture}\\
+ 4\;
\begin{tikzpicture}[scale=0.35,baseline=.9cm]
\node at (0,0)  [root] (root) {};
%\node at (-1.5,1)  [dot] (left) {};
\node at (0,2)  [dot] (int) {};
\node at (-2,3)  [dot] (left) {};
\node at (-3,5) [var] (variable4) {};
\node at (2,3)  [dot] (right) {};
\node at (0,4) [dot] (top) {};
\node at (3,5) [var] (variable2) {};
\draw[testfcn] (int) to (root);
\draw[kernel] (left) to  (int);
\draw[kernel] (right) to  (int);
\draw[keps] (variable2) to  (right);
\draw[keps] (top) to  (right);
\draw[keps] (top) to  (left);
\draw[keps] (variable4) to (left);
\end{tikzpicture}
+ \begin{tikzpicture}[scale=0.35,baseline=.9cm]
\node at (0,0)  [root] (root) {};
%\node at (-1.5,1)  [dot] (left) {};
\node at (0,2)  [dot] (int) {};
\node at (-2,3)  [dot] (left) {};
\node at (-2,5) [dot] (variable3) {};
%	\node at (-3,5) [var] (variable4) {};
\node at (2,3)  [dot] (right) {};
\node at (2,5) [dot] (variable1) {};
%	\node at (3,5) [var] (variable2) {};
\draw[testfcn] (int) to (root);
\draw[kernel] (left) to  (int);
\draw[kernel] (right) to  (int);
\draw[keps, bend left=40] (variable1) to  (right);
\draw[keps, bend right=40] (variable1) to  (right);
\draw[keps, bend left=40] (variable3) to  (left);
\draw[keps, bend right=40] (variable3) to (left);
\end{tikzpicture}
\begin{tikzpicture}[scale=0.35,baseline=.9cm]
\node at (0,0)  [root] (root) {};
%\node at (-1.5,1)  [dot] (left) {};
\node at (0,2)  [dot] (int) {};
\node at (-2,3)  [dot] (left) {};
\node at (0,5.5)  [dot] (ttop) {};
%	\node at (-3,5) [var] (variable4) {};
\node at (2,3)  [dot] (right) {};
\node at (0,4) [dot] (top) {};
%	\node at (3,5) [var] (variable2) {};
\draw[testfcn] (int) to (root);
\draw[kernel] (left) to  (int);
\draw[kernel] (right) to  (int);
\draw[keps] (ttop) to  (right);
\draw[keps] (top) to  (right);
\draw[keps] (top) to  (left);
\draw[keps] (ttop) to (left);
\end{tikzpicture}
\end{align*}

and thus

\begin{align*}
\bigl( \hat{\Pi}_\star^\epsilon \<22j>\bigr)(\phi_\star^\lambda)
= &
\begin{tikzpicture}[scale=0.35,baseline=.9cm]
\node at (0,0.0)  [root] (root) {};
%\node at (-1.5,1)  [dot] (left) {};
\node at (0,2)  [dot] (int) {};
\node at (-2,3)  [dot] (left) {};
\node at (-1,5) [var] (variable3) {};
\node at (-3,5) [var] (variable4) {};
\node at (2,3)  [dot] (right) {};
\node at (1,5) [var] (variable1) {};
\node at (3,5) [var] (variable2) {};
\draw[testfcn] (int) to (root);
\draw[kernel] (left) to  (int);
\draw[kernel] (right) to  (int);
\draw[keps] (variable2) to  (right);
\draw[keps] (variable1) to  (right);
\draw[keps] (variable3) to  (left);
\draw[keps] (variable4) to (left);
\end{tikzpicture}
+ 2
\begin{tikzpicture}[scale=0.35,baseline=.9cm]
\node at (0,0.0)  [root] (root) {};
%\node at (-1.5,1)  [dot] (left) {};
\node at (0,2)  [dot] (int) {};
\node at (-2,3)  [dot] (left) {};
\node at (-2,5) [bluedot] (variable3) {};
%	\node at (-3,5) [var] (variable4) {};
\node at (2,3)  [dot] (right) {};
\node at (1,5) [var] (variable1) {};
\node at (3,5) [var] (variable2) {};
\draw[testfcn] (int) to (root);
\draw[kernel] (left) to  (int);
\draw[kernel] (right) to  (int);
\draw[keps] (variable2) to  (right);
\draw[keps] (variable1) to  (right);
\draw[blue, keps, bend left=40] (variable3) to  (left);
\draw[blue, keps, bend right=40] (variable3) to (left);
\end{tikzpicture}\\
+ 4\;
\begin{tikzpicture}[scale=0.35,baseline=.9cm]
\node at (0,0.0)  [root] (root) {};
%\node at (-1.5,1)  [dot] (left) {};
\node at (0,2)  [dot] (int) {};
\node at (-2,3)  [dot] (left) {};
\node at (-3,5) [var] (variable4) {};
\node at (2,3)  [dot] (right) {};
\node at (0,4) [dot] (top) {};
\node at (3,5) [var] (variable2) {};
\draw[testfcn] (int) to (root);
\draw[kernel] (left) to  (int);
\draw[kernel] (right) to  (int);
\draw[keps] (variable2) to  (right);
\draw[keps] (top) to  (right);
\draw[keps] (top) to  (left);
\draw[keps] (variable4) to (left);
\end{tikzpicture}
+ \begin{tikzpicture}[scale=0.35,baseline=.9cm]
\node at (0,0.0)  [root] (root) {};
%\node at (-1.5,1)  [dot] (left) {};
\node at (0,2)  [dot] (int) {};
\node at (-2,3)  [dot] (left) {};
\node at (-2,5) [bluedot] (variable3) {};
%	\node at (-3,5) [var] (variable4) {};
\node at (2,3)  [dot] (right) {};
\node at (2,5) [bluedot] (variable1) {};
%	\node at (3,5) [var] (variable2) {};
\draw[testfcn] (int) to (root);
\draw[kernel] (left) to  (int);
\draw[kernel] (right) to  (int);
\draw[blue, keps, bend left=40] (variable1) to  (right);
\draw[blue, keps, bend right=40] (variable1) to  (right);
\draw[blue, keps, bend left=40] (variable3) to  (left);
\draw[blue, keps, bend right=40] (variable3) to (left);
\end{tikzpicture}
+2 \;
\begin{tikzpicture}[scale=0.35,baseline=.9cm]
\node at (0.0,0)  [root] (root) {};
%\node at (-1.5,1)  [dot] (left) {};
\node at (0,2)  [bluedot] (int) {};
\node at (-2,3)  [bluedot] (left) {};
\node at (0,5.5)  [bluedot] (ttop) {};
%	\node at (-3,5) [var] (variable4) {};
\node at (2,3)  [bluedot] (right) {};
\node at (0,4) [bluedot] (top) {};
%	\node at (3,5) [var] (variable2) {};
\draw[testfcn] (int) to (root);
\draw[blue, kernel] (left) to  (int);
\draw[blue, kernel] (right) to  (int);
\draw[blue, keps] (ttop) to  (right);
\draw[blue, keps] (top) to  (right);
\draw[blue, keps] (top) to  (left);
\draw[blue, keps] (ttop) to (left);
\end{tikzpicture} \ .
\end{align*}

In order to estimate the term

\begin{equation}\label{eq:for HQ-1}
\tikzsetnextfilename{phi34a1}
\begin{tikzpicture}[scale=0.35,baseline=.9cm]
\node at (0,0)  [root] (root) {};
%\node at (-1.5,1)  [dot] (left) {};
\node at (0,2)  [dot] (int) {};
\node at (-2,3)  [dot] (left) {};
\node at (-3,5) [var] (variable4) {};
\node at (2,3)  [dot] (right) {};
\node at (0,4) [dot] (top) {};
\node at (3,5) [var] (variable2) {};
\draw[testfcn] (int) to (root);
\draw[kernel] (left) to  (int);
\draw[kernel] (right) to  (int);
\draw[keps] (variable2) to  (right);
\draw[keps] (top) to  (right);
\draw[keps] (top) to  (left);
\draw[keps] (variable4) to (left);
\end{tikzpicture}
\end{equation}

we apply Theorem <ref>, while the other terms are easily bounded using the estimates of Section <ref>. Next note that

\bigl( \Pi_\star^\epsilon\<211>\bigr)(\phi_\star^\lambda)
\begin{tikzpicture}[scale=0.35,baseline=0.8cm]
\node at (0.0,-0.8)  [root] (root) {};
\node at (-2,1)  [dot] (left) {};
\node at (-2,3)  [dot] (left1) {};
\node at (-2,5)  [dot] (left2) {};
\node at (0,1) [var] (variable1) {};
\node at (0,3) [var] (variable2) {};
\node at (0,4.3) [var] (variable3) {};
\node at (0,5.7) [var] (variable4) {};

\draw[testfcn] (left) to (root);

\draw[kernel1] (left1) to   (left);
\draw[kernel] (left2) to  (left1);
\draw[keps] (variable2) to  (left1);
\draw[keps] (variable1) to   (left);
\draw[keps] (variable3) to   (left2);
\draw[keps] (variable4) to   (left2);
\end{tikzpicture}
\begin{tikzpicture}[scale=0.35,baseline=0.8cm]
\node at (0.0,-0.8)  [root] (root) {};
\node at (-2,1)  [dot] (left) {};
\node at (-2,3)  [dot] (left1) {};
\node at (-2,5)  [dot] (left2) {};
\node at (0,1) [var] (variable1) {};
\node at (0,3) [var] (variable2) {};
\node at (0,5) [dot] (variable3) {};
%	\node at (0,5.7) [var] (variable4) {};

\draw[testfcn] (left) to (root);

\draw[kernel1] (left1) to   (left);
\draw[kernel] (left2) to  (left1);
\draw[keps] (variable2) to  (left1);
\draw[keps] (variable1) to   (left);
\draw[keps, bend left=40] (variable3) to   (left2);
\draw[keps, bend right=40] (variable3) to   (left2);
\end{tikzpicture}
\;+\;
\begin{tikzpicture}[scale=0.35,baseline=0.8cm]
\node at (0.0,-0.8)  [root] (root) {};
\node at (-2,1)  [dot] (left) {};
\node at (-2,3)  [dot] (left1) {};
\node at (-2,5)  [dot] (left2) {};
\node at (0,2) [dot] (variable1) {};
\node at (0,4.3) [var] (variable3) {};
\node at (0,5.7) [var] (variable4) {};

\draw[testfcn] (left) to (root);

\draw[kernel1] (left1) to   (left);
\draw[kernel] (left2) to  (left1);
\draw[keps] (variable1) to  (left1);
\draw[keps] (variable1) to   (left);
\draw[keps] (variable3) to   (left2);
\draw[keps] (variable4) to   (left2);
\end{tikzpicture}
\;+2\;
\begin{tikzpicture}[scale=0.35,baseline=0.8cm]
\node at (0.0,-0.8)  [root] (root) {};
\node at (-2,1)  [dot] (left) {};
\node at (-2,3)  [dot] (left1) {};
\node at (-2,5)  [dot] (left2) {};
\node at (0,1) [var] (variable1) {};
\node at (0,4) [dot] (variable3) {};
\node at (0,5.7) [var] (variable4) {};

\draw[testfcn] (left) to (root);

\draw[kernel1] (left1) to   (left);
\draw[kernel] (left2) to  (left1);
\draw[keps] (variable3) to  (left1);
\draw[keps] (variable1) to   (left);
\draw[keps] (variable3) to   (left2);
\draw[keps] (variable4) to   (left2);
\end{tikzpicture}
\;+2\;
\begin{tikzpicture}[scale=0.35,baseline=0.8cm]
\node at (0.0,-0.8)  [root] (root) {};
\node at (-2,1)  [dot] (left) {};
\node at (0,3)  [dot] (left1) {};
\node at (-2,5)  [dot] (left2) {};
\node at (0,5) [var] (variable1) {};
\node at (-2,3) [dot] (variable3) {};
\node at (-0.5,6) [var] (variable4) {};

\draw[testfcn] (left) to (root);

\draw[kernel1] (left1) to   (left);
\draw[kernel] (left2) to  (left1);
\draw[keps] (variable3) to  (left);
\draw[keps] (variable1) to   (left1);
\draw[keps] (variable3) to   (left2);
\draw[keps] (variable4) to   (left2);
\end{tikzpicture}
\;+2\;
\begin{tikzpicture}[scale=0.35,baseline=0.8cm]
\node at (0.0,-0.8)  [root] (root) {};
\node at (-2,1)  [dot] (left) {};
\node at (0,3)  [dot] (left1) {};
\node at (-2,5)  [dot] (left2) {};
\node at (-2,3) [dot] (variable3) {};
\node at (0,5.7) [dot] (variable4) {};

\draw[testfcn] (left) to (root);

\draw[kernel1] (left1) to   (left);
\draw[kernel] (left2) to  (left1);
\draw[keps] (variable3) to  (left);
\draw[keps] (variable4) to   (left1);
\draw[keps] (variable3) to   (left2);
\draw[keps] (variable4) to   (left2);
\end{tikzpicture}
\begin{tikzpicture}[scale=0.35,baseline=0.8cm]
\node at (0.0,-0.8)  [root] (root) {};
\node at (-2,1)  [dot] (left) {};
\node at (-2,3)  [dot] (left1) {};
\node at (-2,5)  [dot] (left2) {};
\node at (0,2) [dot] (variable1) {};
\node at (0,5) [dot] (variable3) {};
%	\node at (0,5.7) [var] (variable4) {};

\draw[testfcn] (left) to (root);

\draw[kernel1] (left1) to   (left);
\draw[kernel] (left2) to  (left1);
\draw[keps] (variable1) to  (left1);
\draw[keps] (variable1) to   (left);
\draw[keps, bend left=40] (variable3) to   (left2);
\draw[keps, bend right=40] (variable3) to   (left2);
\end{tikzpicture}

and thus

\begin{align*}
\bigl( \hat{\Pi}^\epsilon_\star\<211>\bigr)(\phi_\star^\lambda)
\begin{tikzpicture}[scale=0.35,baseline=0.8cm]
\node at (0.0,-0.8)  [root] (root) {};
\node at (-2,1)  [dot] (left) {};
\node at (-2,3)  [dot] (left1) {};
\node at (-2,5)  [dot] (left2) {};
\node at (0,1) [var] (variable1) {};
\node at (0,3) [var] (variable2) {};
\node at (0,4.3) [var] (variable3) {};
\node at (0,5.7) [var] (variable4) {};
\draw[testfcn] (left) to (root);
\draw[kernel1] (left1) to   (left);
\draw[kernel] (left2) to  (left1);
\draw[keps] (variable2) to  (left1);
\draw[keps] (variable1) to   (left);
\draw[keps] (variable3) to   (left2);
\draw[keps] (variable4) to   (left2);
\end{tikzpicture}
\begin{tikzpicture}[scale=0.35,baseline=0.8cm]
\node at (0.0,-0.8)  [root] (root) {};
\node at (-2,1)  [dot] (left) {};
\node at (-2,3)  [dot] (left1) {};
\node at (-2,5)  [dot] (left2) {};
\node at (0,1) [var] (variable1) {};
\node at (0,3) [var] (variable2) {};
\node at (0,5) [bluedot] (variable3) {};
%	\node at (0,5.7) [var] (variable4) {};
\draw[testfcn] (left) to (root);
\draw[kernel1] (left1) to   (left);
\draw[kernel] (left2) to  (left1);
\draw[keps] (variable2) to  (left1);
\draw[keps] (variable1) to   (left);
\draw[blue, keps, bend left=40] (variable3) to   (left2);
\draw[blue, keps, bend right=40] (variable3) to   (left2);
\end{tikzpicture}
\;+\;
\begin{tikzpicture}[scale=0.35,baseline=0.8cm]
\node at (0.0,-0.8)  [root] (root) {};
\node at (-2,1)  [dot] (left) {};
\node at (-2,3)  [dot] (left1) {};
\node at (-2,5)  [dot] (left2) {};
\node at (0,2) [bluedot] (variable1) {};
\node at (0,4.3) [var] (variable3) {};
\node at (0,5.7) [var] (variable4) {};
\draw[testfcn] (left) to (root);
\draw[blue, kernel] (left1) to   (left);
\draw[kernel] (left2) to  (left1);
\draw[blue, keps] (variable1) to  (left1);
\draw[blue, keps] (variable1) to   (left);
\draw[keps] (variable3) to   (left2);
\draw[keps] (variable4) to   (left2);
\end{tikzpicture}
\;+\;
\begin{tikzpicture}[scale=0.35,baseline=0.8cm]
\node at (0.0,-0.8)  [root] (root) {};
\node at (-2,1)  [dot] (left) {};
\node at (-2,3)  [dot] (left1) {};
\node at (-2,5)  [dot] (left2) {};
\node at (0,2) [dot] (variable1) {};
\node at (0,4.3) [var] (variable3) {};
\node at (0,5.7) [var] (variable4) {};
\draw[testfcn] (left) to (root);
\draw[kernel, bend right=60] (left1) to   (root);
\draw[kernel] (left2) to  (left1);
\draw[keps] (variable1) to  (left1);
\draw[keps] (variable1) to   (left);
\draw[keps] (variable3) to   (left2);
\draw[keps] (variable4) to   (left2);
\end{tikzpicture}
\;+2\;
\begin{tikzpicture}[scale=0.35,baseline=0.8cm]
\node at (0.0,-0.8)  [root] (root) {};
\node at (-2,1)  [dot] (left) {};
\node at (-2,3)  [dot] (left1) {};
\node at (-2,5)  [bluedot] (left2) {};
\node at (0,1) [var] (variable1) {};
\node at (0,4) [bluedot] (variable3) {};
\node at (0,5.7) [var] (variable4) {};
\draw[testfcn] (left) to (root);
\draw[kernel1] (left1) to   (left);
\draw[kernel] (left2) to  (left1);
\draw[blue, keps] (variable3) to  (left1);
\draw[keps] (variable1) to   (left);
\draw[blue, keps] (variable3) to   (left2);
\draw[keps] (variable4) to   (left2);
\end{tikzpicture}
\;+2\;
\begin{tikzpicture}[scale=0.35,baseline=0.8cm]
\node at (0.0,-0.8)  [root] (root) {};
\node at (-2,1)  [dot] (left) {};
\node at (0,3)  [dot] (left1) {};
\node at (-2,5)  [dot] (left2) {};
\node at (0,5) [var] (variable1) {};
\node at (-2,3) [dot] (variable3) {};
\node at (-0.5,6) [var] (variable4) {};
\draw[testfcn] (left) to (root);
\draw[kernel1] (left1) to   (left);
\draw[kernel] (left2) to  (left1);
\draw[keps] (variable3) to  (left);
\draw[keps] (variable1) to   (left1);
\draw[keps] (variable3) to   (left2);
\draw[keps] (variable4) to   (left2);
\end{tikzpicture}
\;\\
\begin{tikzpicture}[scale=0.35,baseline=0.8cm]
\node at (0.0,-0.8)  [root] (root) {};
\node at (-2,1)  [dot] (left) {};
\node at (0,3)  [bluedot] (left1) {};
\node at (-2,5)  [bluedot] (left2) {};
\node at (-2,3) [bluedot] (variable3) {};
\node at (0,5.7) [bluedot] (variable4) {};
\draw[testfcn] (left) to (root);
\draw[blue, kernel] (left1) to   (left);
\draw[blue, kernel] (left2) to  (left1);
\draw[blue, keps] (variable3) to  (left);
\draw[blue, keps] (variable4) to   (left1);
\draw[blue, keps] (variable3) to   (left2);
\draw[blue, keps] (variable4) to   (left2);
\end{tikzpicture}
\;+2\;
\begin{tikzpicture}[scale=0.35,baseline=0.8cm]
\node at (0.0,-0.8)  [root] (root) {};
\node at (-2,1)  [dot] (left) {};
\node at (0,3)  [dot] (left1) {};
\node at (-2,5)  [dot] (left2) {};
\node at (-2,3) [dot] (variable3) {};
\node at (0,5.7) [dot] (variable4) {};
\draw[testfcn] (left) to (root);
\draw[kernel] (left1) to   (root);
\draw[kernel] (left2) to  (left1);
\draw[keps] (variable3) to  (left);
\draw[keps] (variable4) to   (left1);
\draw[keps] (variable3) to   (left2);
\draw[keps] (variable4) to   (left2);
\end{tikzpicture}
\begin{tikzpicture}[scale=0.35,baseline=0.8cm]
\node at (0.0,-0.8)  [root] (root) {};
\node at (-2,1)  [dot] (left) {};
\node at (-2,3)  [bluedot] (left1) {};
\node at (-2,5)  [dot] (left2) {};
\node at (0,2) [bluedot] (variable1) {};
\node at (0,5) [bluedot] (variable3) {};
%	\node at (0,5.7) [var] (variable4) {};
\draw[testfcn] (left) to (root);
\draw[blue, kernel] (left1) to   (left);
\draw[kernel] (left2) to  (left1);
\draw[blue, keps] (variable1) to  (left1);
\draw[blue, keps] (variable1) to   (left);
\draw[blue, keps, bend left=40] (variable3) to   (left2);
\draw[blue, keps, bend right=40] (variable3) to   (left2);
\end{tikzpicture} \;
\begin{tikzpicture}[scale=0.35,baseline=0.8cm]
\node at (0.0,-0.8)  [root] (root) {};
\node at (-2,1)  [dot] (left) {};
\node at (-2,3)  [dot] (left1) {};
\node at (-2,5)  [dot] (left2) {};
\node at (0,2) [dot] (variable1) {};
\node at (0,5) [bluedot] (variable3) {};
%	\node at (0,5.7) [var] (variable4) {};
\draw[testfcn] (left) to (root);
\draw[kernel , bend right=60] (left1) to   (root);
\draw[kernel] (left2) to  (left1);
\draw[keps] (variable1) to  (left1);
\draw[keps] (variable1) to   (left);
\draw[blue, keps, bend left=40] (variable3) to   (left2);
\draw[blue, keps, bend right=40] (variable3) to   (left2);
\end{tikzpicture} \;.
\end{align*}

Here we apply Theorem <ref> to bound the terms

\begin{equation}\label{eq:for HQ-2}
\tikzsetnextfilename{phi34a2}
\begin{tikzpicture}[scale=0.35,baseline=0.8cm]
\node at (0.0,-0.8)  [root] (root) {};
\node at (-2,1)  [dot] (left) {};
\node at (-2,3)  [dot] (left1) {};
\node at (-2,5)  [dot] (left2) {};
\node at (0,2) [dot] (variable1) {};
\node at (0,4.3) [var] (variable3) {};
\node at (0,5.7) [var] (variable4) {};
\draw[testfcn] (left) to (root);
\draw[kernel, bend right=60] (left1) to   (root);
\draw[kernel] (left2) to  (left1);
\draw[keps] (variable1) to  (left1);
\draw[keps] (variable1) to   (left);
\draw[keps] (variable3) to   (left2);
\draw[keps] (variable4) to   (left2);
\end{tikzpicture}
\; \qquad \text{and} \qquad
\tikzsetnextfilename{phi34a3}
\begin{tikzpicture}[scale=0.35,baseline=0.8cm]
\node at (0.0,-0.8)  [root] (root) {};
\node at (-2,1)  [dot] (left) {};
\node at (0,3)  [dot] (left1) {};
\node at (-2,5)  [dot] (left2) {};
\node at (0,5) [var] (variable1) {};
\node at (-2,3) [dot] (variable3) {};
\node at (-0.5,6) [var] (variable4) {};
\draw[testfcn] (left) to (root);
\draw[kernel1] (left1) to   (left);
\draw[kernel] (left2) to  (left1);
\draw[keps] (variable3) to  (left);
\draw[keps] (variable1) to   (left1);
\draw[keps] (variable3) to   (left2);
\draw[keps] (variable4) to   (left2);
\end{tikzpicture} \; ,
\end{equation}

while the remaining terms can again be bounded using the estimates in Section <ref>
The bounds on
$$\bigl( \hat{\Pi}_\star^\epsilon \<2'>\bigr)(\phi_\star^\lambda), \qquad \bigl( \hat{\Pi}_\star^\epsilon \<21'>\bigr)(\phi_\star^\lambda) , \qquad \bigl( \hat{\Pi}_\star^\epsilon \<2'1>\bigr)(\phi_\star^\lambda),
follow straightforwardly, where for the last term we use that admissible realisation satisfies $R(j_\cdot 1)=1$.

Let us observe that the trees appearing for the $\phi^3_4$ equation are formally identical to those for the KPZ equation and the power-counting is also similar.

§ KERNELS WITH PRESCRIBED SINGULARITIES

Throughout this section we fix vector bundles  $(E^{i},\nabla^E, \langle\cdot, \cdot\rangle_E)$ and $(F^{i},\nabla^F, \langle\cdot, \cdot\rangle_F)$ over $M$ for $i\in \mathbb{N}$.
A smooth kernel $K\in \cC^\infty(F\hotimes E^* \setminus \triangle)$, where $\triangle\subset M\times M$ denotes the diagonal[
Throughout this section, we shall use the lighter notation
$F\hotimes E^* \setminus \triangle$ to denote the restricted bundle with base space $M\times M\setminus \triangle$.

is called of order $\zeta$ if for any for $m \in \mathbb{N}$

$$\vertiii{K}_{\zeta; m} := \sup_{k\leq m} \sup_{p,q\in M} d_\fraks (p,q)^{k-\zeta} |j_{p,q} K|_k <+\infty \ .$$

§.§ Elementary estimates

In this section, we collect the analogue estimates to <cit.>, the proofs of which adapt mutatis mutandis.

For $i=1,2$ let $K_{i}\in \cC^\infty(F^{(i)}\hotimes (E^{(i)})^* \setminus\triangle)$ be supported on $\{(p,q)\in (M\times M)\setminus \triangle\   | \ d_\fraks(p,q)<1\}$
and of order $\zeta_i$.
Then, up to the canonical isomorphism $\left(F^{(1)}\hotimes (E^{(1)})^*\right) \otimes\left( F^{(1)}\hotimes (E^{(1)})^*\right)\simeq (F^{(1)}\otimes F^{(2)})\hotimes (E^{(1)}\otimes E^{(2)} )^* \ , $

$$ K_1\otimes K_{2}\in \cC^\infty\left( (F^{(1)}\otimes F^{(2)})\hotimes (E^{(1)}\otimes E^{(2)} )^* \setminus \triangle\right)\ .$$
Furthermore,  there exists $C>0$ such that

$$\vertiii{K_1\otimes K_2}_{\zeta_1+\zeta_2; m}\leq C\vertiii{K_1}_{\zeta_1; m}\vertiii{K_1}_{\zeta_2; m}\ .$$