\section{The reductions modulo 5}\label{sec:reductions}

In the notation of \cref{sec:notation5}, this section carries out step~(1) of \cref{sec:intro-uncond}: the monodromy at $\p$ of every member above $5$ of a compatible system as in \cref{sec:S} is not zero (\cref{thm:cases}), and so $\rhob(I_\p)$ is not contained in the scalar matrices (\cref{prop:notscalar}).

\subsection{Constituents}
We use Fontaine--Laffaille theory \cite{FontaineLaffaille} in the normalization of \cite[\S1.4]{BLGGT14}, for the unramified extension $\Qtf/\Q_5$ with ring of integers $\Z_{25}$.  Let $\mathcal{MF}$ be the category of finite $\Z_{25}\otimes_{\Z_5}\OO_E$-modules $\DFL$ with a decreasing filtration by submodules $\Fil^i\DFL$ that are direct summands over $\Z_{25}$, with $\Fil^0\DFL=\DFL$ and $\Fil^4\DFL=0$, and with Frobenius-semilinear maps $\varphi_i\colon\Fil^i\DFL\to\DFL$ such that $\varphi_i|_{\Fil^{i+1}\DFL}=5\varphi_{i+1}$ and $\sum_i\varphi_i(\Fil^i\DFL)=\DFL$.  There is an exact, fully faithful, covariant, $\OO_E$-linear functor $\TFL$ from $\mathcal{MF}$ to the category of finite $\OO_E$-modules with a continuous action of $G_{\Qtf}$, whose essential image is closed under subobjects and quotients.  For an object $\DFL$ killed by $\mE$ and an embedding $\tau\colon\Qtf\to E$, the \emph{jumps} of $\DFL$ at $\tau$ are the integers $i$ such that the $\tau$-component of $\Fil^i\DFL/\Fil^{i+1}\DFL$ is nonzero, counted with its dimension over $\FE$.  If $\DFL$ is torsion-free, then $\TFL(\DFL)\otimes\Q_5$ is crystalline and its Hodge--Tate weights at $\tau$ are the jumps of $\DFL/\mE\DFL$ at $\tau$; and every $G_{\Qtf}$-stable lattice in a crystalline representation with Hodge--Tate weights in $[0,3]$ is in the image of $\TFL$.  An object killed by $\mE$ that has a filtration by subobjects with graded pieces of rank one over $\F_{25}\otimes\FE$ is the reduction of a torsion-free object with a filtration lifting the given one \cite[Lemma~1.4.2]{BLGGT14}.  Let $\tau,\tau'$ be the two embeddings of $\Qtf$ in $E$.

Recall the set of types $\mathcal T=\{(0,0),(1,1),(2,2),(0,2),(2,0)\}$ of \cref{sec:irreducibility}.

\begin{lemma}\label{lem:constituents}
Let $\Lambda$ be a stable lattice.
\begin{enumerate}[(i)]
\item $\Rb_\Lambda|_{G_{\Qtf}}=\TFL(\bar\DFL)$ for an object $\bar\DFL$ whose jumps are $\{0,1,2\}$ at $\tau$ and at $\tau'$.  Every subquotient of $\Rb_\Lambda|_{G_{\Qtf}}$ is the image under $\TFL$ of a subquotient of $\bar\DFL$, and jumps are additive in exact sequences.
\item Let $\theta\colon G_K\to\FE^\times$ be a one-dimensional constituent of $\Rb^{\mathrm{ss}}$, and let $(k,k')$ be its jumps at $(\tau,\tau')$.  Then $(k,k')\in\mathcal T$.  In the unramified case, $k=k'$ and $\theta=\epsf^{-k}$.  In the ramified case, either $(k,k')=(0,2)$ for all such $\theta$, or $(k,k')=(2,0)$ for all such $\theta$.
\item $\Rb^{\mathrm{ss}}$ is a sum of three characters only in the unramified case, and then $\Rb^{\mathrm{ss}}=1\oplus\epsf^{-1}\oplus\epsf^{-2}$, three distinct characters.
\item Let $\Theta$ be a subquotient of $\Rb_\Lambda$ that is an extension of $\epsf^{-j}$, $j\in\{0,1\}$, by a subrepresentation isomorphic to $\epsf^{-2}$.  Then $\Theta$ is split as a representation of $G_{\Qtf}$.
\end{enumerate}
\end{lemma}

\begin{proof}
(i) By (M2), $\Lambda$ is a lattice in a crystalline representation with Hodge--Tate weights $\{0,1,2\}$ at both embeddings, so $\Lambda=\TFL(\DFL)$; by exactness and full faithfulness $\DFL$ is torsion-free, $\Rb_\Lambda=\TFL(\DFL/\mE\DFL)$, and the jumps of $\bar\DFL=\DFL/\mE\DFL$ are the Hodge--Tate weights.  The rest follows from the properties of $\TFL$ and the strictness of morphisms in $\mathcal{MF}$ with respect to the filtrations.

(ii) By (i), $\theta|_{G_{\Qtf}}$ is the image of an object of rank one with jumps $k,k'\in\{0,1,2\}$, which lifts to a torsion-free object, so $\theta|_{G_{\Qtf}}$ is the reduction of a crystalline character with Hodge--Tate weights $(k,k')$.  On the inertia group, identified with $\Z_{25}^\times$ by local class field theory, such a character is $x\mapsto\tau(x)^{\pm k}\tau'(x)^{\pm k'}$, so $\theta$ is $x\mapsto\bar\tau(x)^{\pm(k+5k')}$ there, where $\bar\tau(x)\in\FE^\times$ is the residue class of $\tau(x)$, and the sign is fixed by the normalizations.  By \cref{prop:ram}, $\theta$ is unramified outside $\{\p,5\}$, and $\theta|_{I_\p}=\bar\kappa_\lambda$.  Let $\theta_\p$ be the character of $\OO_\p^\times$ attached to $\theta|_{I_\p}$ by class field theory.  Its order divides $3$.  The idele class character of $\theta$ is trivial on $\mu_6\subset K^\times$, so $\theta_\p(\zeta)\,\bar\tau(\zeta)^{\pm(k+5k')}=1$ for $\zeta\in\mu_6$.  For $\zeta=-1$ this gives $(-1)^{k+k'}=1$, so $(k,k')\in\mathcal T$.  For $\zeta=\omega$ it gives $\theta_\p(\omega)=\bar\tau(\omega)^{\mp(k-k')}$, because $\omega^5=\omega^{-1}$.  In the unramified case $\theta_\p$ is trivial, so $k\equiv k'$ modulo $3$ and modulo $2$, hence $k=k'$; then $\theta\epsf^{k}$ is unramified at $5$ (all crystalline characters with the same Hodge--Tate weights agree on inertia), at $\p$ and everywhere else, so it is trivial, because $K$ has class number one.  In the ramified case $\theta_\p$ is a character of order $3$ of $\OO_\p^\times/(1+\p^2)\isom\mu_6$ (by (F3) and (M3)), so $\theta_\p(\omega)\ne1$ and $k\not\equiv k'\pmod 3$: $(k,k')$ is $(0,2)$ or $(2,0)$, and which of the two is determined by $\theta_\p(\omega)=\bar\kappa_\lambda(\Art_\p(\omega))$.

(iii) By (i) the jumps of the three characters at $\tau$ are $0,1,2$ in some order.  By (ii) this is impossible in the ramified case, and in the unramified case the characters are $\epsf^{-k}$ with $k=0,1,2$.  They are distinct because $\epsf$ has order $4$ on $G_K$.

(iv) $\Theta|_{G_{\Qtf}}=\TFL(\DFL_\Theta)$ with $\DFL_\Theta$ an extension of an object $\DFL_2$ of rank one with jumps $(j,j)$ by a subobject $\DFL_1$ of rank one with jumps $(2,2)$.  By strictness, $\Fil^i\DFL_\Theta=\DFL_\Theta$ for $i\le j$ and $\Fil^i\DFL_\Theta=\DFL_1$ for $j<i\le2$.  Let $e_\tau,e_{\tau'}$ generate the two components of $\DFL_1$ and let $f_\tau,f_{\tau'}\in\DFL_\Theta$ lift generators of the two components of $\DFL_2$.  Since $e_\tau\in\Fil^{j+1}\DFL_\Theta$, $\varphi_j(e_\tau)=5\varphi_{j+1}(e_\tau)=0$.  Write $\varphi_j(f_\tau)=\beta_\tau f_{\tau''}+y_\tau e_{\tau''}$, where $\tau''$ is the other embedding and $\beta_\tau\in\FE^\times$.  Replacing $f_{\tau''}$ by $f_{\tau''}+(y_\tau/\beta_\tau)e_{\tau''}$, for both $\tau$, does not change $\varphi_j(f_\tau)$ and makes $y_\tau=0$.  Then $f_\tau,f_{\tau'}$ span a subobject mapping isomorphically to $\DFL_2$: the extension of filtered modules splits, and so does $\Theta|_{G_{\Qtf}}$.
\end{proof}

With Hodge--Tate weights normalized as in \cref{sec:conventions}, (iv) says that an extension, in the image of $\TFL$, of a character by a character of larger Hodge--Tate weight splits.  For weights $0$ and $1$ this is the splitting of an extension of a multiplicative group scheme by an \'etale one, whereas extensions in the other direction (the Kummer ones) can be nonsplit.

\subsection{A lattice with a prescribed socle}

\begin{lemma}\label{lem:lattice}
Let $\Theta$ be a constituent of $\Rb^{\mathrm{ss}}$ of multiplicity one.  Then there is a stable lattice $\Lambda'$ such that $\Theta$ is the unique irreducible subrepresentation of $\Rb_{\Lambda'}$.
\end{lemma}

\begin{proof}
Passing to the dual of $R$ and to dual lattices, it suffices to find a stable lattice whose reduction has $\Theta$ as its unique irreducible quotient.  Fix a stable lattice $\Lambda_0$, and let $\mathfrak L$ be the set of stable lattices $\Lambda\subset\Lambda_0$ such that $\Theta$ is not a constituent of $\Lambda_0/\Lambda$.  It is closed under finite intersections.  Let $X_0$ be the smallest subrepresentation of $\Lambda_0/\mE\Lambda_0$ whose quotient does not have $\Theta$ as a constituent.  It is nonzero, and the image of every $\Lambda\in\mathfrak L$ in $\Lambda_0/\mE\Lambda_0$ contains it.  So the sets $\Lambda\smallsetminus\mE\Lambda_0$, $\Lambda\in\mathfrak L$, are nonempty and compact, and any finite number of them have a common point.  Hence $\Lambda_\infty=\bigcap_{\Lambda\in\mathfrak L}\Lambda$ is nonzero.  It is a $G_K$-stable $\OO_E$-submodule of $\Lambda_0$, so it is a lattice, because $R$ is irreducible (\cref{prop:member-irreducible}).  Then $\mE^m\Lambda_0\subset\Lambda_\infty$ for some $m$, and as $\Lambda_0/\mE^m\Lambda_0$ is finite, $\Lambda_\infty$ is a finite intersection of members of $\mathfrak L$; so $\Lambda_\infty$ is the smallest element of $\mathfrak L$.  If $\Rb_{\Lambda_\infty}$ had an irreducible quotient not isomorphic to $\Theta$, the kernel of the map from $\Lambda_\infty$ to it would be a smaller element of $\mathfrak L$.  So every irreducible quotient of $\Rb_{\Lambda_\infty}$ is isomorphic to $\Theta$, and there is only one, because $\Theta$ has multiplicity one.
\end{proof}

\needspace{5\baselineskip}\subsection{The main case analysis}

\begin{theorem}\label{thm:cases}
Let $\Rsys$ be a compatible system as in \cref{sec:S}, $\lambda$ a prime of its coefficient field above $5$, and $\Lambda$ a stable lattice of $R=R_\lambda$.  Then one of the following holds.
\begin{enumerate}[(A)]
\item $\Rb_\Lambda$ is irreducible; put $\Theta=\Rb_\Lambda$.
\item $\Rb^{\mathrm{ss}}=\theta\oplus\Theta$ with $\theta$ a character and $\Theta$ irreducible of dimension $2$.
\end{enumerate}
In both cases let $\Delta$ be the projective image of $\Theta$ and $L$ the field it cuts out.  Then:
\begin{enumerate}[(1)]
\item $L$ is a Galois extension of $\Q$ of degree $2|\Delta|$, it contains $K$, and it is unramified outside $\{3,5\}$;
\item the inertia groups of $L/K$ at $\p$ have order $5$; in particular $\bar U_\Theta\ne1$, and the matrix $N$ of \textup{(P3)} is not zero.
\end{enumerate}
\end{theorem}

\begin{proof}
After enlarging $E$, $\Rb^{\mathrm{ss}}$ is irreducible, or the sum of a character and an irreducible representation of dimension $2$, or a sum of three characters.  The polarization of $R$ gives $(\Rb^{\mathrm{ss}})^\cc\isom(\Rb^{\mathrm{ss}})^\vee\otimes\epsf^{-2}$.  In the first two cases this gives $\Theta^\cc\isom\Theta^\vee\otimes\epsf^{-2}$, so the kernel of the projective representation of $\Theta$ is stable under $\tilde\cc$ and $L/\Q$ is Galois; $L$ contains $K$, it is totally complex of degree $2|\Delta|$, and it is unramified outside $\{3,5\}$ by \cref{prop:ram}~(c).  Let $e_\p\in\{1,5\}$ be the order of the inertia groups of $L/K$ at $\p$ (\cref{prop:ram}~(b)).

\emph{Three characters.}  By \cref{lem:constituents}~(iii) we are in the unramified case and $\Rb^{\mathrm{ss}}=1\oplus\epsf^{-1}\oplus\epsf^{-2}$.  By \cref{lem:lattice} there is a stable lattice $\Lambda'$ such that $\epsf^{-2}$ is the unique irreducible subrepresentation of $\Rb_{\Lambda'}$.  An irreducible subrepresentation of $\Rb_{\Lambda'}/\epsf^{-2}$ is $\epsf^{-j}$ with $j\in\{0,1\}$, and its preimage in $\Rb_{\Lambda'}$ is a nonsplit extension of $\epsf^{-j}$ by $\epsf^{-2}$.  Its class is a nonzero element of $H^1(G_K,\FE(\eta))$ with $\eta=\epsf^{\,j-2}\in\{\epsf^{-1},\epsf^{-2}\}$.  This class is unramified outside $\{\p,5\}$ (\cref{prop:ram}~(c)) and trivial on $G_{\Qtf}$ (\cref{lem:constituents}~(iv)).  It is also trivial on $G_{K_\p}$: by the local Euler characteristic formula and local duality, $H^1(G_{K_\p},\FE(\eta))=0$, because $\eta$ and $\eta^{-1}\epsf$ are nontrivial on $G_{K_\p}$ (they are unramified at $\p$, and $\epsf(\Frob_\p)=3^{\pm1}$ has order $4$).  Let $K_\eta$ be the field cut out by $\eta$: it is $K(\zeta_5)$ or $K(\sqrt5)$, of degree $4$ or $2$ over $K$.  This degree is prime to $5$, so the restriction of the class to $G_{K_\eta}$ is a nonzero homomorphism $G_{K_\eta}\to\FE$, and it is unramified at every place.  It defines a cyclic unramified extension of $K_\eta$ of degree $5$, which contradicts \cref{lem:classfields}~(a).  So $\Rb^{\mathrm{ss}}$ is not a sum of three characters.

\emph{Cases (A) and (B) with $e_\p=1$.}  Then $L/K$ is unramified outside $5$, and \cref{prop:ram}~(d) gives $\rd(L)<3^{1/2}\cdot5^{3/2}$, $\rd(L)<3^{1/2}\cdot5^{7/5}$ if $25\nmid|\Delta|$, and $\rd(L)<3^{1/2}\cdot5$ if $5\nmid|\Delta|$.  We compare with \cref{prop:odlyzko}.
\begin{itemize}
\item If $5\nmid|\Delta|$ in case (A): $|\Delta|\ge9$ (\cref{lem:groups}~(b)), so $[L:\Q]\ge18$ and ${\rd(L)>9.2709>8.6603}$.
\item If $5\nmid|\Delta|$ in case (B): $\Delta$ is dihedral of order $2m\ge4$, or $A_4$, or $S_4$ (\cref{lem:groups}~(c)).  Each of these groups has a quotient that \cref{lem:classfields} forbids for the Galois group of an extension of $K$ unramified outside $5$ of degree prime to $5$: $(\Z/2)^2$ if $\Delta$ is dihedral with $m$ even, and $\Z/3$ if $\Delta\isom A_4$, by \cref{lem:classfields}~(b); and a dihedral group of order $2m'$ with $m'\ge3$ odd if $\Delta$ is dihedral with $m$ odd ($m'=m$) or $\Delta\isom S_4$ ($m'=3$, through its quotient $S_3$).  For a dihedral quotient of order $2m'$ the quadratic subextension is $K(\sqrt5)$ by \cref{lem:classfields}~(b), and it has a cyclic extension of degree $m'$ unramified outside $5$, which \cref{lem:classfields}~(c) forbids.
\item If $5\mid|\Delta|$ and $25\nmid|\Delta|$: $|\Delta|\ge60$ (\cref{lem:no-abelian}), so $[L:\Q]\ge120$ and ${\rd(L)>16.9886>16.4862}$.
\item If $25\mid|\Delta|$: $|\Delta|\ge3600$ (\cref{lem:no-abelian} and \cref{lem:groups}~(a)), so $[L:\Q]\ge7200\ge2002$ and $\rd(L)>21.4136>19.3650$.
\end{itemize}
Each case is contradictory, so $e_\p=5$.

As $e_\p=5$, the image of $\bar U_\Theta$ in the projective linear group of $\Theta$ is not trivial (\cref{prop:ram}~(b)), so $\bar U_\Theta\ne1$.  Then $U\ne1$, so the monodromy operator of $R$ at $\p$ is not zero, and neither is $N$ (\cref{prop:members}).
\end{proof}

\begin{remark}\label{rem:precedents}
The case $e_\p=1$ of this proof concerns a representation of $G_K$ modulo $5$ of Fontaine--Laffaille type with weights in $[0,2]$ whose projective image cuts out an extension of $K$ unramified outside $5$ (in the unramified case the representation itself is unramified outside $5$): the situation of the theorems of Abrashkin \cite{Abrashkin89} and Fontaine \cite{Fontaine93} on smooth proper schemes over rings of integers, which combine the ramification bound of \cref{thm:FA} with unconditional lower bounds for discriminants (see \cite[\S2.4]{Hattori19}).  Abrashkin asserts that $K$ satisfies his condition SH$(5,2)$ \cite[\S7, 5.2~a)]{Abrashkin89} (p.~516 of the English translation) and writes that the proof is analogous to his proof for $\Q$; no proof is printed.  That condition, with the upper ramification number $2/(5-1)=1/2$ itself allowed, as in his proof of 5.1 there, implies the case $e_\p=1$.  The proof above does not rely on this assertion.
\end{remark}

\subsection{Inertia at \texorpdfstring{$\p$}{p} for the Frey representation}

\begin{proposition}\label{prop:notscalar}
For every solution, the group $\rhob(I_\p)$ is not contained in the group of scalar matrices.
\end{proposition}

\begin{proof}
Suppose that it is.  We construct a compatible system as in \cref{sec:S} for which the matrix $N$ of (P3) is zero, which contradicts \cref{thm:cases}~(2).

By \eqref{eq:F3bar}, $\rhob|_{I_\p}=\bar\kappa$.  Let $I_1$ be the kernel of $\kappa$ on $I_\p$.  It is normal in $G_{\Q_3}$, because $\kappa^\cc=\kappa^{-1}$, and the prolongation $\rhob_{\mathcal G}$ (\cref{sec:systems}) is trivial on it, because $\epsb^{-2}\bar\delta_{K/\Q}$ is trivial on $I_\p$.  The group $G_{\Q_3}/I_1$ is the semidirect product of the finite group $\bar I=I_{\Q_3}/I_1$, of order $2$ or $6$, and the closure of the group generated by a Frobenius lift $\sigma_3$, which is isomorphic to $\hat\Z$.  Let $\OO'$ be the ring of integers of a finite extension of $\Q_p$ with residue field $\F'$ such that $\rhob_{\mathcal G}$ has values in $\Gthree(\F')$.

(1) As $|\bar I|$ is prime to $p$ and $\Gthree$ is smooth, $\rhob_{\mathcal G}|_{\bar I}$ lifts to a homomorphism $h\colon\bar I\to\Gthree(\OO')$.  The characters $\mult\circ h$ and $\varepsilon_p^{-2}\dK|_{\bar I}$ have order prime to $p$ and the same reduction, so they are equal.

(2) The scheme of elements $g$ of $\Gthree$ with $g\,h(x)\,g^{-1}=h(\sigma_3x\sigma_3^{-1})$ for all $x\in\bar I$ is the scheme of fixed points of an action of $\bar I$ on $\Gthree$, so it is smooth over $\OO'$ \cite[Proposition~3.4]{Edixhoven92}; it has the point $\rhob_{\mathcal G}(\sigma_3)$ over $\F'$, which by Hensel's lemma lifts to a point $g_0$ over $\OO'$.  The torus $\{(t\cdot1_3,t^2)\}$ is central in $\Gthree$, and $\mult(t\cdot1_3,t^2)=t^2$.  Multiplying $g_0$ by the element of this torus with $t^2=(\varepsilon_p^{-2}\dK)(\sigma_3)/\mult(g_0)$ and $t\equiv1$, we get a lift $g_1$ of $\rhob_{\mathcal G}(\sigma_3)$ with the same property and with $\mult(g_1)=(\varepsilon_p^{-2}\dK)(\sigma_3)$.

(3) The map $x\mapsto h(x)$, $\sigma_3\mapsto g_1$ extends to a continuous homomorphism $\rho'_{\mathcal G}\colon G_{\Q_3}\to\Gthree(\OO')$, trivial on $I_1$, which lifts $\rhob_{\mathcal G}|_{G_{\Q_3}}$ and has multiplier $\varepsilon_p^{-2}\dK$.  On $I_\p$ its $\GL_3$-component has order dividing $3$ and scalar reduction, so it is the scalar $\kappa$, as in the proof of \cref{lem:lifts-at-3}.  So $\rho'_{\mathcal G}$ has finite image on inertia, and its monodromy operator is zero.

(4) Let $\tau_3$ be the inertial type of $\rho'_{\mathcal G}$, and let $X$ be the generic fiber of the ring of framed liftings of $\rhob_{\mathcal G}|_{G_{\Q_3}}$ to $\Gthree$ of type $\tau_3$ with multiplier $\varepsilon_p^{-2}\dK$.  The locus of $X$ where the monodromy operator vanishes is a union of irreducible components of $X$, and it is regular.  This is the case $l\ne p$ of \cite[Theorem~3.3.7 and the paragraph before it]{BellovinGee19}; there $p$ is the residue characteristic of the local field and $l$ that of the coefficients, so here the local field is $\Q_3$ and $l$ is our $p$.  It applies to $\Gthree$: \cite[\S\S1.3.2, 3]{BellovinGee19} allow any finite extension of their $\Q_p$ (here $\Q_3$) and any smooth affine group scheme whose geometric fibers are reductive, not necessarily connected, and whose center is smooth (the center of $\Gthree$ is $\{(t\cdot1_3,t^2)\}\isom\GL_1$), and the condition on the multiplier is the condition $\mathrm{ab}\circ\rho=\psi$ of \cite[\S3]{BellovinGee19} for $\Gthree$ \cite[\S5.1]{BellovinGee19}, where $\Gthree$-valued representations at places of $F^+$ that are inert or ramified in $F$ are treated explicitly.

That every point of $X$ with vanishing monodromy lies on a component on which it vanishes identically is the statement of the paragraph before \cite[Theorem~3.3.7]{BellovinGee19}.  We recall the argument.  Let $x$ be a closed point of this locus (it suffices to treat these), with residue field $E_x$, and let $L_3/\Q_3$ be a finite Galois extension with $\tau_3|_{I_{L_3}}$ trivial.  The completed local ring at $x$ of the locus $N=0$ pro-represents the framed liftings of $\rho_x$ with multiplier $\varepsilon_p^{-2}\dK$ to artinian local $E_x$-algebras $B$ with residue field $E_x$ that factor through $G_{\Q_3}/I_{L_3}$: for a lifting $\rho_B$ of type $\tau_3$ with Weil--Deligne representation $(\phi_B,N_B)$ we have $\rho_B(\sigma)=\phi_B(\sigma)\exp(t_p(\sigma)N_B)$ for $\sigma\in I_{\Q_3}$ (\cref{sec:conventions}), and $\phi_B$ is trivial on $I_{L_3}$, since $\phi_B(I_{L_3})$ is a finite subgroup of the kernel of $\Gthree(B)\to\Gthree(E_x)$, which is torsion-free; so $N_B=0$ if and only if $\rho_B$ is trivial on $I_{L_3}$, and a lifting trivial on $I_{L_3}$ has type $\tau_3$, as representations of the finite group $I_{\Q_3}/I_{L_3}$ in characteristic $0$ do not deform.  As $I_{\Q_3}/I_{L_3}$ is finite and the quotient of $G_{\Q_3}/I_{L_3}$ by it is $\hat\Z$, these liftings are unobstructed, and their tangent space has dimension $\dim\Gthree^{\mathrm{der}}=9$, where $\Gthree^{\mathrm{der}}=\GL_3\times1$.  So the locus $N=0$ is regular of dimension $9$ at $x$; as $x$ is a closed point of the generic fiber of a complete local noetherian ring, the irreducible component of the locus through $x$ has dimension $9$, so it is an irreducible component of $X$, which is equidimensional of dimension $9$ \cite[Theorem~3.3.2]{BellovinGee19}.

Let $\Cthree'$ be the irreducible component of this locus through $\rho'_{\mathcal G}$ (it is unique, as the locus is regular): it is a polarized component at $3$ (these are defined through \cite[\S3]{BellovinGee19}, \cite[\S1.5.19]{CEG22}), and the monodromy operator vanishes at all its points.

(5) \Cref{thm:CEG} with the components $\Cthree'$ and $\Cp$ gives a compatible system as in \cref{sec:S} whose $p$-adic member has $N=0$ in (P3).
\end{proof}

If the unipotent part of $\rhob|_{I_\p}$ is not trivial, then every lift of $\rhob|_{G_{K_\p}}$ has $N\ne0$ (\cref{lem:lifts-at-3}), and no choice of component can force $N=0$.  At this point the argument leaves the prime $5$.