\section{The member at 3}\label{sec:three} Part~III works with the members of $\Rsys$ at the primes above $3$, which were of no use in Part~II: their Hodge--Tate weights $\{0,1,2\}$ do not lie in the Fontaine--Laffaille range $[0,\ell-2]$ for $\ell=3$. For them, ordinarity at $\p$ takes the place of Fontaine--Laffaille theory in bounding the ramification of the reductions, and weak admissibility forces it once the monodromy at $\p$ is a single Jordan block. We prove \cref{thm:three-main} by steps~(2)--(5) of \cref{sec:intro-uncond}: steps~(2) and~(3) in this section, steps~(4) and~(5) in \cref{sec:three-reductions}. Throughout this part, $p\ge17$ is a prime with $p\equiv1\pmod 3$, $\rho$ and $\rhob$ are the Frey representation of a solution of $x^3+y^3=z^p$ in coprime nonzero integers and its reduction (\cref{def:frey}), with the properties (F1)--(F6) of \cref{sec:F}, and $\Rsys=\{R_\lambda\}$ is a compatible system as in \cref{sec:S}, with coefficient field $M$; the polarized component at $3$ used to construct $\Rsys$ may be any, not necessarily $\Cthree$. As in Part~II, representations over finite fields and over $\ell$-adic fields are considered over an algebraic closure where convenient, and \emph{irreducible} means absolutely irreducible. The completion $K_\p=\Q_3(\sqrt{-3})$ has uniformizer $1-\omega$, residue field $\F_3$ and maximal unramified subfield $\Q_3$, and it contains the cube roots of unity. Let $G_{K_\p}\subset G_K$ be a decomposition group at $\p$ and $I_\p$ its inertia subgroup. Upper ramification groups $G^y$ ($y\ge-1$ real) are numbered as in \cref{sec:bounds} \cite[Chapitre~IV]{SerreCL}. For a finite Galois extension of $K_\p$, $G^0$ is its inertia group, and $G^y$ for $y>0$ is contained in its wild inertia group. \subsection{Regular monodromy at \texorpdfstring{$\p$}{p}}\label{sec:three-regular} For a square matrix $A$ we write ${}^tA$ for its transpose and ${}^tA^{-1}$ for the inverse of the transpose. A unipotent matrix $U\in\GL_3$ is \emph{regular} if $(U-1)^2\ne0$, that is, if it has a single Jordan block. A nilpotent $3\times3$ matrix $N$ is regular if $N^2\ne0$. \begin{lemma}\label{lem:cosquare} Let $k$ be a field, $A\in\GL_3(k)$, and $B=A\,{}^tA^{-1}$. If $B=\beta U$ with $\beta\in\bar k^\times$ and $U$ unipotent, then $B=1$ or $B$ is a regular unipotent matrix. \end{lemma} \begin{proof} As $\det B=1$, $\beta^3=1$. We have ${}^tB^{-1}={}^tA^{-1}A=A^{-1}BA$, so $B$ is conjugate to ${}^tB^{-1}$, whose eigenvalues are the inverses of those of $B$. All eigenvalues of $B$ equal $\beta$, so $\beta=\beta^{-1}$, and with $\beta^3=1$ this gives $\beta=1$: $B$ is unipotent. From $B\,{}^tA=A$ we get $A-{}^tA=(B-1)\,{}^tA$, so $B-1$ and $A-{}^tA$ have the same rank. The matrix $A-{}^tA$ is alternating (antisymmetric with zero diagonal), so its rank is even, in every characteristic. A unipotent $B\ne1$ with $(B-1)^2=0$ has $\rk(B-1)=1$; so this case does not occur. \end{proof} \begin{proposition}\label{prop:regular} Let $\ell\notin\{2,3\}$ be a prime and $r\colon G_K\to\GL_3(\Fbar_\ell)$ a continuous irreducible representation. Suppose that $r^\cc\isom r^\vee\otimes\nu|_{G_K}$ for a character $\nu$ of $G_\Q$, and that $r(\sigma)=\theta(\sigma)u(\sigma)$ for $\sigma\in I_\p$, with $\theta$ a character of $I_\p$ and $u(\sigma)$ unipotent. Then $u(I_\p)$ is trivial or generated by a regular unipotent matrix. \end{proposition} \begin{proof} The map $u=\theta^{-1}r|_{I_\p}$ is a homomorphism with finite image, consisting of unipotent matrices in characteristic $\ell$; so its image is an $\ell$-group. The isomorphism $r^\cc\isom r^\vee\otimes\nu|_{G_K}$ gives $P\in\GL_3(\Fbar_\ell)$ with $r(\tilde\cc h\tilde\cc^{-1})=\nu(h)\,P\,{}^tr(h)^{-1}P^{-1}$ for $h\in G_K$. Let $\sigma\in G_\Q\smallsetminus G_K$, and write $\sigma=\tilde\cc g$ with $g\in G_K$. Since $\nu$ takes values in an abelian group, $\nu(ghg^{-1})=\nu(h)$, and $r(\sigma h\sigma^{-1})=\nu(h)\,P_\sigma\,{}^tr(h)^{-1}P_\sigma^{-1}$ with $P_\sigma=P\,{}^tr(g)^{-1}$. Applying this to $\sigma h\sigma^{-1}\in G_K$ in place of $h$, and using $\nu(\sigma h\sigma^{-1})=\nu(h)$ again, we find \[ r(\sigma^2h\sigma^{-2})=(P_\sigma{}^tP_\sigma^{-1})\,r(h)\,(P_\sigma{}^tP_\sigma^{-1})^{-1}\qquad(h\in G_K). \] Since also $r(\sigma^2h\sigma^{-2})=r(\sigma^2)r(h)r(\sigma^2)^{-1}$ and $r$ is irreducible, Schur's lemma gives $r(\sigma^2)=\kappa_0\,P_\sigma{}^tP_\sigma^{-1}$ with $\kappa_0\in\Fbar_\ell^\times$. Now choose $\sigma$ as follows. Let $I_3\subset G_{\Q_3}$ be the inertia group containing $I_\p$, so that $I_\p=I_3\cap G_K$, and $P_3\subset I_3$ its wild inertia group, the unique pro-$3$ Sylow subgroup of $I_3$; it lies in $I_\p$, which has index $2$ in $I_3$. The quotient $I_3/P_3$ is procyclic, and the surjection $I_3\to\Gal(K_\p/\Q_3)$ factors through it, because $K_\p/\Q_3$ is tamely ramified. So every lift $\sigma\in I_3$ of a topological generator of $I_3/P_3$ lies outside $G_K$, and $\sigma^2\in I_\p$ maps to a topological generator of the tame quotient $I_\p/P_3$. As $\ell\ne3$, $u$ is trivial on $P_3$, and $u(I_\p)$ is generated by $u(\sigma^2)$. Now \[ P_\sigma{}^tP_\sigma^{-1}=\kappa_0^{-1}r(\sigma^2)=\kappa_0^{-1}\theta(\sigma^2)\,u(\sigma^2) \] is a scalar times a unipotent matrix. By \cref{lem:cosquare} (with $A=P_\sigma$) it is $1$ or a regular unipotent matrix. A scalar times a unipotent matrix determines the scalar, its only eigenvalue; so $\kappa_0^{-1}\theta(\sigma^2)=1$, and $u(\sigma^2)$ is $1$ or regular. \end{proof} \begin{corollary}\label{cor:regular} By \eqref{eq:F3bar}, $\rhob(\sigma)=\bar\kappa(\sigma)\bar u(\sigma)$ for $\sigma\in I_\p$, with $\bar u(\sigma)$ unipotent. The group $\bar u(I_\p)$ is generated by a regular unipotent matrix. For every compatible system as in \cref{sec:S}, the matrix $N$ of \textup{(P3)} is regular. \end{corollary} \begin{proof} Apply \cref{prop:regular} with $\ell=p\ge17$, $r=\rhob$ (irreducible by (F6)), $\nu=\epsb^{-2}$ (by (F4), $\rhob^\cc\isom\rhob^\vee\otimes\epsb^{-2}$; see \cref{lem:polarized}) and $\theta=\bar\kappa|_{I_\p}$. By \cref{prop:notscalar}, $\rhob(I_\p)$ is not contained in the scalar matrices, so $\bar u(I_\p)\ne1$, and it is generated by a regular unipotent matrix. Let $\Rsys$ be as in \cref{sec:S}. By (P1) and (F6), the reduction of every $G_K$-stable lattice in $R_p$ is $\rhob$, and the lattice is unique up to homothety. In a basis of such a lattice whose reduction is $\rhob$, the matrix $U(\sigma)=\kappa(\sigma)^{-1}R_p(\sigma)$ is integral for $\sigma\in I_\p$ (as $\kappa(\sigma)\in\mu_3$ is a unit), it equals $\exp(t_p(\sigma)N')$ by (P3), with $N'$ conjugate to $N$, and its reduction is $\bar\kappa(\sigma)^{-1}\rhob(\sigma)=\bar u(\sigma)$. If $N^2=0$, then $U(\sigma)-1=t_p(\sigma)N'$ and $(U(\sigma)-1)^2=0$, hence $(\bar u(\sigma)-1)^2=0$ for all $\sigma\in I_\p$, which the first part excludes. So $N^2\ne0$. \end{proof} \subsection{The member at a prime above 3}\label{sec:three-member} From now on in this part, $\lambda$ is a prime of $M$ above $3$ and $R=R_\lambda\colon G_K\to\GL_3(\overline M_\lambda)$. We use the notation of \cref{sec:notation5} with $3$ in place of $5$: $E\subset\overline M_\lambda$ is a finite extension of $\Q_3$ over which $R$ is defined, with ring of integers $\OO_E$, maximal ideal $\mE$ and residue field $\FE$; a \emph{stable lattice} is a $G_K$-stable $\OO_E$-lattice $\Lambda$ in the space of $R$; its reduction is $\Rb_\Lambda=\Lambda/\mE\Lambda$; and the semisimplification $\Rb^{\mathrm{ss}}$ of $\Rb_\Lambda$ does not depend on $\Lambda$. We enlarge $E$ when needed, and in particular assume that $E$ contains the images of the two embeddings of $K_\p$ in $\overline M_\lambda$. \begin{proposition}\label{prop:three-unram} The representation $R$ is unramified at every prime of $K$ other than $\p$. \end{proposition} \begin{proof} Let $\q\ne\p$ be a prime of $K$. Then $\q\nmid3$. By (P1) and (P2), $R_p$ is unramified at $\q$ if $\q\nmid p$ and crystalline at $\q$ if $\q\mid p$. By \cref{lem:lgc-members}~(a), $R$ is unramified at $\q$. \end{proof} \begin{proposition}\label{prop:three-wd} The restriction $R|_{G_{K_\p}}$ is de Rham, with Hodge--Tate weights $\{0,1,2\}$ for both embeddings of $K_\p$ in $\overline M_\lambda$, and $\WD(R|_{G_{K_\p}})^{\Fss}$ corresponds to $\WD(R_p|_{G_{K_\p}})^{\Fss}$ under an isomorphism of their coefficient fields that is the identity on $M$. So $I_\p$ acts on $\WD(R|_{G_{K_\p}})$ through a character with values in $\mu_3(\overline M_\lambda)$, which is $\kappa|_{I_\p}$ after an identification of the groups $\mu_3$ of the two coefficient fields (as in \textup{(M3)}), and the monodromy operator of $\WD(R|_{G_{K_\p}})$ is regular. \end{proposition} \begin{proof} The first assertion is part of the definition of a compatible system (\cref{sec:systems}), with (P2). By \cref{lem:lgc-members} with $\q=\p$, $\WD(R|_{G_{K_\p}})^{\Fss}$ corresponds to $\WD(R_p|_{G_{K_\p}})^{\Fss}$ (compose with the inverse of an isomorphism $j'\colon\overline\Q_p\to\C$ restricting to $\imath_\p$ on $M$). Here $\lambda$ lies above $3$, the residue characteristic of $\p$, and in this case \cref{lem:lgc-members} gives the monodromy operator of $\WD(R|_{G_{K_\p}})$ as well as its semisimplification (the Shin-regular case of \cite[Theorem~1.1]{BLGGTlgc2}; see the proof of \cref{lem:lgc-members}), and the regularity of the monodromy operator, shown next, depends on this. By (P3), $I_\p$ acts on $\WD(R_p|_{G_{K_\p}})$ through the scalar character $\kappa$, and its monodromy operator is conjugate to $N$, which is regular by \cref{cor:regular}; by \cref{lem:lgc-members}~(b), the same holds for $\WD(R|_{G_{K_\p}})$, after identifying the two groups $\mu_3$ as in the statement. \end{proof} Let $\kappa_\lambda\colon G_K\to\mu_3(\overline M_\lambda)$ be the character obtained from $\kappa$ through an identification as in \cref{prop:three-wd}, so that $I_\p$ acts on $\WD(R|_{G_{K_\p}})$ through $\kappa_\lambda|_{I_\p}$, the analogue for $\lambda\mid3$ of the character $\kappa_\lambda$ of \textup{(M3)}; in the unramified case $\kappa|_{I_\p}=1$ and either identification may be used. Twisting by $\kappa_\lambda$ does not change the subspaces stable under a subgroup of $G_K$, and $\kappa_\lambda\equiv1\pmod{\mE}$, because $\mu_3\subset1+\mE$. \begin{proposition}\label{prop:three-ordinary} Let $V=(R\otimes\kappa_\lambda^{-1})|_{G_{K_\p}}$. Then $V$ is semistable, and it has $G_{K_\p}$-stable subspaces $0\subset V_1\subset V_2\subset V$ of dimensions $1$ and $2$ whose graded pieces $V_1$, $V_2/V_1$, $V/V_2$ are the characters $\psi$, $\psi\varepsilon_3^{-1}$, $\psi\varepsilon_3^{-2}$, for one unramified character $\psi\colon G_{K_\p}\to E^\times$. In particular $R|_{G_{K_\p}}$ stabilizes the line $V_1$. For every stable lattice $\Lambda$ there is a basis of $\Lambda$ in which \[ \Rb_\Lambda(g)=\bar\psi(g)\begin{pmatrix}1&a(g)&c(g)\\0&1&b(g)\\0&0&1\end{pmatrix}\qquad(g\in G_{K_\p}), \] where $\bar\psi$ is the reduction of $\psi$, $a,b\in\Hom(G_{K_\p},\FE)$, and $c\colon G_{K_\p}\to\FE$ is continuous with $c(gh)=c(g)+c(h)+a(g)b(h)$. \end{proposition} \begin{proof} By \cref{prop:three-wd}, $R|_{G_{K_\p}}$ is de Rham, hence potentially semistable \cite{Berger02}, and so is $V$; and $I_\p$ acts trivially on $\WD(V)=\WD(R|_{G_{K_\p}})\otimes\WD(\kappa_\lambda^{-1}|_{G_{K_\p}})$ ($\WD$ is compatible with tensor products), by the choice of $\kappa_\lambda$. So $V$ is semistable \cite{Fontaine94,Fontaine94pst}. Let $D=D_{\mathrm{st}}(V)$ be its filtered $(\varphi,N)$-module (covariant), with coefficients in $E$. The maximal unramified subfield of $K_\p$ is $\Q_3$, so $D$ is a free $E$-module of rank $3$ with $E$-linear $\varphi$ and $N$, and $N\varphi=3\varphi N$; by \cref{prop:three-wd}, $N$ is regular. The $N$-stable subspaces of $D$ are $0\subset\ker N\subset\ker N^2\subset D$, and they are $\varphi$-stable. If $\varphi$ acts on $D/\ker N^2$ by $\alpha\in E^\times$, then, as $N$ induces isomorphisms $D/\ker N^2\to\ker N^2/\ker N\to\ker N$ and $\varphi N=3^{-1}N\varphi$, it acts on $\ker N^2/\ker N$ and on $\ker N$ by $\alpha/3$ and $\alpha/9$. The Hodge filtration lives on $D\otimes_{\Q_3}K_\p=\bigoplus_\tau D_\tau$, the sum over the two embeddings $\tau$ of $K_\p$ in $E$, and its jumps on each $D_\tau$ are $0,1,2$. For a $(\varphi,N)$-stable subspace $D'$ put $t_N(D')=v_3(\det(\varphi|D'))$ and $t_H(D')=\frac12\sum_\tau\sum_ii\dim\gr^i(D'_\tau)$, for the induced filtrations; these are the usual invariants of $D'$ divided by $[E:\Q_3]$ (for $\varepsilon_3$ both are $-1$). Weak admissibility says $t_H(D')\le t_N(D')$, with equality for $D'=D$. For $D$: $t_H(D)=\frac12(3+3)=3=t_N(D)=3v_3(\alpha)-3$, so $v_3(\alpha)=2$, and $\ker N$, $\ker N^2/\ker N$, $D/\ker N^2$ have slopes $0,1,2$. For $\ker N$: $t_N=0$ and $t_H\ge0$, so $t_H=0$ by weak admissibility, that is, the jump of the line $(\ker N)_\tau$ is $0$ for both $\tau$. For $\ker N^2$: $t_N=0+1=1$, while the two jumps of each plane $(\ker N^2)_\tau$ are distinct elements of $\{0,1,2\}$, so $t_H\ge\frac12(1+1)=1$; so $t_H=1$, and the jumps are $\{0,1\}$ for both $\tau$. Hence $\ker N$ and $\ker N^2$, with the induced structures, have $t_H=t_N$; they are weakly admissible, hence admissible \cite{ColmezFontaine00}, and they are $D_{\mathrm{st}}$ of $G_{K_\p}$-stable subspaces $V_1\subset V_2$ of $V$ of dimensions $1$ and $2$. The graded pieces $\ker N$, $\ker N^2/\ker N$, $D/\ker N^2$ have $N=0$, jumps $h=0,1,2$ at both embeddings, and $\varphi$-eigenvalues $\alpha/9$, $\alpha/3$, $\alpha$; so they are $D_{\mathrm{st}}$ of crystalline characters of Hodge--Tate weights $(h,h)$, that is, of characters $\psi_h\varepsilon_3^{-h}$ with $\psi_h$ unramified (a crystalline character with all Hodge--Tate weights $0$ is unramified). The eigenvalue of $\varphi$ on $D_{\mathrm{cris}}(\psi_h\varepsilon_3^{-h})$ is $3^h$ times that on $D_{\mathrm{cris}}(\psi_h)$, so the three unramified characters $\psi_h$ have the same eigenvalue $\alpha/9$, and they are equal; call them $\psi$. Let $\Lambda$ be a stable lattice. The subspaces $V_1\subset V_2$ are $G_{K_\p}$-stable subspaces of the space of $R$, so $\Lambda\cap V_1\subset\Lambda\cap V_2\subset\Lambda$ are $G_{K_\p}$-stable direct summands, and the graded pieces are lattices in the characters $\psi\varepsilon_3^{-h}\kappa_\lambda$, $h=0,1,2$. The reductions of these characters are all $\bar\psi$: $\bar\varepsilon_3$ is trivial on $G_{K_\p}$ because $K_\p$ contains the cube roots of unity, and $\bar\kappa_\lambda=1$. So in a basis of $\Lambda$ adapted to this flag, $\bar\psi^{-1}\Rb_\Lambda|_{G_{K_\p}}$ is upper unitriangular. Its entries $a$, $b$ next to the diagonal are then homomorphisms, and its corner entry $c$ satisfies $c(gh)=c(g)+c(h)+a(g)b(h)$. \end{proof} \begin{proposition}\label{prop:three-irreducible} The representation $R$ is irreducible. \end{proposition} \begin{proof} $R$ is semisimple, as is every member of a compatible system (\cref{sec:systems}). If $R=R_1\oplus R_2$ with $R_1,R_2\ne0$, then $\WD(R|_{G_{K_\p}})=\WD(R_1|_{G_{K_\p}})\oplus\WD(R_2|_{G_{K_\p}})$, and the monodromy operator is the sum of nilpotent operators on spaces of dimension at most $2$, so its square is zero. This contradicts \cref{prop:three-wd}. \end{proof} \section{The reductions of the member at 3, and the proof of Theorem~A}\label{sec:three-reductions} It remains to show that every reduction of the member $R$ at a prime above $3$ has trivial semisimplification (\cref{cor:three-trivial}). Then, by \cref{cor:three-inertia}, $R$ has the same image as its inertia group at $\p$, which stabilizes the line $V_1$ of \cref{prop:three-ordinary}; so $R$ is reducible, against \cref{prop:three-irreducible}. This proves \cref{thm:three-main}, which completes the proof of \cref{thm:main-uncond}. The notation is that of \cref{sec:three-member}. For a stable lattice $\Lambda$ let $L_\Lambda$ be the field cut out by $\Rb_\Lambda$. By \cref{prop:three-unram}, $L_\Lambda/K$ is a finite Galois extension unramified at every place other than $\p$, including the infinite places, as $K$ has no real place. On $G_{K_\p}$, $\Rb_\Lambda$ has the shape of \cref{prop:three-ordinary}. \subsection{Characters} \begin{proposition}\label{prop:three-characters} Every continuous character $\theta\colon G_K\to\Fbar_3^\times$ unramified outside $\p$ is trivial. In particular every one-dimensional constituent of $\Rb^{\mathrm{ss}}$ is trivial. \end{proposition} \begin{proof} The character $\theta$ has finite order prime to $3$, since $\Fbar_3^\times$ has no element of order $3$. If $\p^n$ is its conductor, it factors through the ray class group $\Cl_{\p^n}(K)$, which is $(\OO/\p^n)^\times/\mu_6$ because $K$ has class number one and no real place. The group $(\OO/\p^n)^\times$ has order $2\cdot3^{n-1}$, and $\mu_6$ injects into it for $n\ge2$ (\cref{sec:conventions}); so $\Cl_{\p^n}(K)$ is a $3$-group (of order $1$ for $n\le2$ and $3^{n-2}$ for $n\ge2$), and $\theta=1$. The constituents of $\Rb^{\mathrm{ss}}$ are unramified outside $\p$ by \cref{prop:three-unram}. \end{proof} \subsection{Cubic characters of \texorpdfstring{$K_\p$}{the completion}}\label{sec:three-kummer} As $\omega\in K_\p$, Kummer theory identifies $\Hom(G_{K_\p},\mu_3)$ with $K_\p^\times/K_\p^{\times3}$. We identify $\mu_3$ with $\F_3$ by $\omega\mapsto1$, and put \[ \mathcal H=\Hom(G_{K_\p},\F_3)\isom K_\p^\times/K_\p^{\times3}. \] Let $U^{(j)}=1+\p^j$ be the groups of principal units of $K_\p$, $\mathcal U\subset\mathcal H$ the image of $U^{(2)}$, and $e_\pi,e_2,e_1,e_0\in\mathcal H$ the classes of \[ 1-\omega,\qquad \omega,\qquad 4,\qquad 1+(1-\omega)^3 . \] We fix an isomorphism $H^2(G_{K_\p},\F_3)\isom\F_3$ (local class field theory). The cup product $\mathcal H\times\mathcal H\to H^2(G_{K_\p},\F_3)=\F_3$ is alternating. We call a plane $W\subset\mathcal H$ \emph{isotropic} if the cup product vanishes on $W\times W$. Under these identifications the cup product of the classes of $x,y\in K_\p^\times$ is, up to a sign, the cubic Hilbert symbol $(x,y)_\p$ \cite[Chapitre~XIV, \S\S1--2]{SerreCL}. \needspace{4\baselineskip}\begin{lemma}\label{lem:three-kummer}\leavevmode \begin{enumerate}[(a)] \item $\mathcal H$ has dimension $4$ over $\F_3$, and $(e_\pi,e_2,e_1,e_0)$ is a basis; $\mathcal U=\F_3e_1\oplus\F_3e_0$. \item Let $\alpha\in\mathcal H$, $\alpha\ne0$. The cubic extension of $K_\p$ cut out by $\alpha$ has upper ramification break $3$ if $\alpha\notin\langle e_2,e_1,e_0\rangle$, $2$ if $\alpha\in\langle e_2,e_1,e_0\rangle\smallsetminus\mathcal U$, and $1$ if $\alpha\in\mathcal U\smallsetminus\F_3e_0$, and it is unramified if $\alpha\in\F_3e_0$. In particular every $\alpha\in\mathcal H$ vanishes on $G^y_{K_\p}$ for $y>3$. \item In the basis $(e_\pi,e_2,e_1,e_0)$ the cup product has, up to a sign, the matrix \[ \begin{pmatrix}0&0&2&1\\0&0&1&0\\1&2&0&0\\2&0&0&0\end{pmatrix}. \] Exactly $40$ of the $130$ planes of $\mathcal H$ are isotropic, and $13$ of these meet $\mathcal U$ nontrivially. \end{enumerate} \end{lemma} \begin{proof} (a) As $\mu_3\subset K_\p$ and $[K_\p:\Q_3]=2$, $K_\p^\times/K_\p^{\times3}$ has order $3^{2+2}$. Write $\pi=1-\omega$, so that $3=-\omega^2\pi^2$. For $j\ge2$ and $v_\p(x)=j$ we have $(1+x)^3\equiv1+3x\pmod{\p^{j+3}}$, so cubing induces isomorphisms $U^{(j)}/U^{(j+1)}\to U^{(j+2)}/U^{(j+3)}$, and $U^{(4)}=(U^{(2)})^3$ by completeness. For a unit $z$, $(1+\pi z)^3=1+\pi^3(z^3-\omega^2z)+3\pi^2z^2\in U^{(4)}$, as $z^3\equiv z$ and $\omega\equiv1$ modulo $\p$. So the cubes of units are $\pm U^{(4)}$, and $K_\p^\times/K_\p^{\times3}$ is the product of the group generated by the class of $\pi$, of order $3$, and of $\OO_\p^\times/\{\pm1\}U^{(4)}=(\OO/9)^\times/\{\pm1\}$, of order $27$. The latter group is generated by the classes of $\omega$, $4$ and $1+\pi^3=-2(1+3\omega)$, which lie in $U^{(1)}$, $U^{(2)}$, $U^{(3)}$ with $v_\p(\omega-1)=1$, $v_\p(4-1)=2$, $v_\p(\pi^3)=3$; and $\mathcal U$ is the image of $U^{(2)}/U^{(4)}$, of order $9$, generated by $4$ and $1+\pi^3$. (b) Let $x$ represent $\alpha$; replacing $\alpha$ by $-\alpha$ does not change the extension. We use the valuation $v_\p$ of $K_\p$, extended to an algebraic closure. For a cyclic extension of degree $3$ of $K_\p$, totally ramified with break $t$ (the same in both numberings), the different has $v_\p$-valuation $2(t+1)/3$; and if the ring of integers is generated by a root $\Pi$ of a polynomial $f$, the different is generated by $f'(\Pi)$ \cite[Chapitre~III, \S6, and Chapitre~IV, \S1]{SerreCL}. If $\alpha\notin\langle e_2,e_1,e_0\rangle$ we may take $x=\pi z$ with $z$ a unit; $X^3-x$ is an Eisenstein polynomial, and $v_\p(3X^2)=2+\frac23$ at a root, so $t=3$. If $\alpha\in\langle e_2,e_1,e_0\rangle\smallsetminus\mathcal U$ we may take $x=1+y$ with $v_\p(y)=1$; then $T=x^{1/3}-1$ is a root of the Eisenstein polynomial $T^3+3T^2+3T-y$, whose derivative $3(T+1)^2$ has valuation $2$, so $t=2$. If $\alpha\in\mathcal U\smallsetminus\F_3e_0$ we may take $x=1+y$ with $v_\p(y)=2$; then $\Pi=\pi/T$ is a root of the Eisenstein polynomial $\Pi^3-(3\pi/y)\Pi^2-(3\pi^2/y)\Pi-\pi^3/y$, whose derivative at $\Pi$ has valuation $\min(2+\frac23,\,1+\frac13,\,2)=\frac43$, attained once, so $t=1$. If $\alpha\in\F_3e_0$ we may take $x=1+\pi^3z$ with $z$ a unit; then $S=T/\pi$ satisfies $S^3+(3/\pi)S^2+(3/\pi^2)S-z=0$, whose reduction $S^3-S-\bar z$ is separable ($3/\pi^2=-\omega^2\equiv-1$), so the extension is unramified. Finally, an element of $\mathcal H$ vanishes on $G^y_{K_\p}$ for every $y$ larger than the break of the extension it cuts out, because the upper numbering passes to quotients. (c) Besides $\p$, the only primes that divide the representatives $1-\omega$, $\omega$, $4$ and $1+\pi^3=-2(1+3\omega)$ are $(2)$ and $\lp=(2-\omega)=(1+3\omega)$, with residue fields $\F_4$ and $\F_7$, and $K$ has no real place. So, by the product formula, $(x,y)_\p$ is the inverse of the product of the tame symbols $\bigl((-1)^{v_\q(x)v_\q(y)}x^{v_\q(y)}y^{-v_\q(x)}\bmod\q\bigr)^{(\Nm\q-1)/3}$ at $\q=(2)$ and $\q=\lp$. This gives $(\pi,\omega)_\p=(\omega,1+\pi^3)_\p=(4,1+\pi^3)_\p=1$, $(\pi,4)_\p=\omega^2$ and $(\pi,1+\pi^3)_\p=(\omega,4)_\p=\omega$; with $(x,x)_\p=1$ and antisymmetry this is the matrix stated. Its Pfaffian is $1$, so the form is nondegenerate, and $\mathcal H$ has $(3+1)(3^2+1)=40$ isotropic planes among its $(3^4-1)(3^3-1)/((3^2-1)(3-1))=130$ planes (a symplectic space of dimension $4$ over $\F_q$ has $(q+1)(q^2+1)$ Lagrangian planes). The plane $\mathcal U$ is isotropic, and the isotropic planes meeting it only in $0$ are the graphs of the $3^3=27$ self-adjoint maps from a fixed isotropic complement of $\mathcal U$ to $\mathcal U$; so $13$ meet $\mathcal U$ nontrivially. \end{proof} \subsection{Heisenberg extensions of \texorpdfstring{$K_\p$}{the completion}}\label{sec:three-heis} The reductions in \cref{prop:three-ordinary}, divided by their unramified character, cut out, on the decomposition group at $\p$, extensions of $K_\p$ whose group is a subgroup of the group of upper unitriangular matrices over the residue field. The aim of this subsection is to prove \cref{prop:three-allk}, which bounds the upper ramification groups of these extensions; \cref{prop:three-rd} turns the bound into one for root discriminants. Over $\F_3$ the unitriangular group is the Heisenberg group of order $27$. A \emph{Heisenberg extension} of $K_\p$ is a finite Galois extension $L/K_\p$ whose group is isomorphic to it, and its \emph{plane} is the plane of $\mathcal H$ formed by the homomorphisms $G_{K_\p}\to\F_3$ that factor through $\Gal(L/K_\p)$. Throughout, $F$ is a finite extension of $K_\p$ in a fixed algebraic closure, $v_F$ its normalized valuation ($v_F(F^\times)=\Z$), $\OO_F$ its ring of integers, $\varpi_F$ a uniformizer, $e_F=v_F(3)$, and $U_F^{(j)}=1+\varpi_F^j\OO_F$ for $j\ge1$. As $\omega\in F$ and $(1-\omega)^2=-3\omega$, $e_F$ is even and $v_F(\omega-1)=e_F/2$. For a continuous homomorphism $\chi$ from $G_F$ to an abelian group and a real number $T$, we say that $\chi$ \emph{has break at most $T$} if $\chi(G_F^y)=0$ for every $y\ge-1$ with $y>T$; these $\chi$ form a group. For $u\in F^\times$ we say that $u$ has break at most $T$ if the character $g\mapsto g(u^{1/3})/u^{1/3}$ of $G_F$ does; this depends only on the class of $u$ in $F^\times/F^{\times3}$. Unramified characters have break at most $T$ for every $T\ge-1$. If $F'/F$ is cyclic of degree $3$ and totally ramified, then $\Gal(F'/F)^y=\Gal(F'/F)$ for $y\le t$ and $\Gal(F'/F)^y=1$ for $y>t$, for one integer $t\ge1$, the \emph{break} of $F'/F$ (the same in both numberings). When $F(u^{1/3})/F$ is cyclic of degree $3$ and totally ramified, its break is also called the \emph{break} of $u$. For a generator $s$ of $\Gal(F'/F)$ and every uniformizer $\Pi$ of $F'$ we have $v_{F'}(s(a)-a)\ge t+1$ for all $a\in\OO_{F'}$, with equality for $a=\Pi$ \cite[Chapitre~IV, \S1, Lemme~1]{SerreCL} (as $\OO_{F'}=\OO_F[\Pi]$, \cite[Chapitre~I, \S6, Propositions~17 and~18]{SerreCL}). Herbrand's functions are \[ \varphi_{F'/F}(u)=\begin{cases}u,&u\le t,\\ t+\frac{u-t}3,&u\ge t,\end{cases}\qquad \psi_{F'/F}(y)=\begin{cases}y,&y\le t,\\ t+3(y-t),&y\ge t.\end{cases} \] If $F'/F$ is unramified, $\varphi_{F'/F}=\psi_{F'/F}$ is the identity. We use three facts from \cite[Chapitre~IV]{SerreCL}. \begin{enumerate}[(R1)] \item The upper numbering passes to quotients (\S3, Proposition~14). \item If $N/F$ is finite Galois with group $G$, $F\subset F'\subset N$ with $F'/F$ Galois, and $H=\Gal(N/F')$, then $G^y\cap H=H^{\psi_{F'/F}(y)}$ for $y\ge-1$. (By \S1, Proposition~2, $H_u=G_u\cap H$; and $\psi_{N/F}=\psi_{N/F'}\circ\psi_{F'/F}$ by \S3.) \item If a character $\chi$ of $G_F$ has break at most $T$, then $\chi|_{G_{F'}}$ has break at most $\psi_{F'/F}(T)$, for $F'/F$ finite Galois. (Apply (R2) to the compositum $N$ of $F'$ and the field cut out by $\chi$, and (R1) twice.) \end{enumerate} \begin{lemma}\label{lem:kummer-E} Let $u\in F^\times$ and $e=e_F$. \begin{enumerate}[(a)] \item If $3\nmid v_F(u)$, then $F(u^{1/3})/F$ is cyclic of degree $3$, totally ramified, with break $3e/2$. \item If $u\in U_F^{(r)}$ with $r\ge1$, then $u$ has break at most $3e/2-r$, and at most $3e/2-r-1$ if moreover $3\mid r$ and $r<3e/2$. \item If $u\in U_F^{(r)}\smallsetminus U_F^{(r+1)}$ with $3\nmid r$ and $1\le r<3e/2$, then $F(u^{1/3})/F$ is cyclic of degree $3$, totally ramified, with break exactly $3e/2-r$. \end{enumerate} \end{lemma} For $F=K_\p$ ($e=2$) this is the computation in the proof of \cref{lem:three-kummer}~(b). We use it for $F$ cubic and totally ramified over $K_\p$ ($e=6$, breaks $9-r$) and for $F$ the unramified cubic extension of $K_\p$ ($e=2$). \begin{proof} (a) Let $\vartheta$ be a cube root of $u$ and $F'=F(\vartheta)$. As $v_F(\vartheta)=v_F(u)/3\notin\Z$, $F'/F$ is totally ramified of degree $3$. Choose integers $a,b$ with $a\,v_F(u)+3b=1$; then $\Pi=\vartheta^a\varpi_F^b$ is a uniformizer of $F'$. A generator $s$ of $\Gal(F'/F)$ maps $\vartheta$ to $\omega^{\pm1}\vartheta$, so $s(\Pi)/\Pi=\omega^{\pm a}$ with $3\nmid a$, and $v_{F'}(s(\Pi)-\Pi)=1+3v_F(\omega^{\pm a}-1)=1+3e/2$. The break is $3e/2$. (c) Put $z=u-1$, so $v_F(z)=r$, and let $T$ be a root of $(1+T)^3=u$, that is, $T^3+3T^2+3T=z$. If $v_F(T)\ge e/2$, all three terms on the left have valuation at least $3e/2>r$; so $\tau=v_F(T)0$. Then $s(\Pi)/\Pi=(1+\eta)^a$, and as $3\nmid a$, $v_F\big((1+\eta)^a-1\big)=v_F(\eta)$. So $v_{F'}(s(\Pi)-\Pi)=1+3(e/2-r/3)=1+3e/2-r$, and the break is $3e/2-r$. (b) We argue by descending induction on $r$. Let $\pi_0=\omega-1$, so $v_F(\pi_0)=e/2$ and $3/\pi_0^2=-\omega^2$. If $r\ge3e/2$, put $T=\pi_0S$: the equation $(1+T)^3=u$ becomes $S^3+(3/\pi_0)S^2+(3/\pi_0^2)S=(u-1)/\pi_0^3$, with $v_F(3/\pi_0)=e/2>0$ and $(u-1)/\pi_0^3$ integral, so its reduction is $S^3-S-\bar c$ for some $\bar c$ in the residue field, which is separable. If $r>3e/2$ then $\bar c=0$, the root $S=0$ of the reduction is simple, and by Hensel's lemma $u$ is a cube; if $r=3e/2$, $F(u^{1/3})/F$ is unramified. In both cases the claim holds. Let $r<3e/2$. If $3\mid r$, write $u=1+\varpi_F^r\varepsilon$ with $\varepsilon\in\OO_F$ and choose $a\in\OO_F$ with $a^3\equiv\varepsilon\pmod{\varpi_F}$ (the residue field is perfect); as $v_F(3a\varpi_F^{r/3})\ge e+r/3>r$, $u(1+a\varpi_F^{r/3})^{-3}\in U_F^{(r+1)}$, and $u$ has the class of an element of $U_F^{(r+1)}$, of break at most $3e/2-r-1$ by induction (this is the second assertion). If $3\nmid r$, either $u\in U_F^{(r+1)}$ and we conclude by induction, or (c) applies. \end{proof} \begin{lemma}\label{lem:sigma-quotients} Let $K_1/K_\p$ be cyclic of degree $3$ and totally ramified, with break $t$, let $\sigma$ generate $\Gal(K_1/K_\p)$, and let $\theta_1$ be a uniformizer of $K_1$. Write $U^{(j)}=U^{(j)}_{K_1}$. \begin{enumerate}[(a)] \item $\sigma(u)/u\in U^{(t+1)}$ for every unit $u$ of $K_1$, and $\sigma(\theta_1)/\theta_1\in U^{(t)}\smallsetminus U^{(t+1)}$. Hence $\sigma(w)/w\in U^{(t)}$ for every $w\in K_1^\times$. \item $\sigma(u)/u\in U^{(m+t)}$ for every $u\in U^{(m)}$, $m\ge1$. \end{enumerate} \end{lemma} \begin{proof} We have $v_{K_1}(\sigma(a)-a)\ge t+1$ for all $a\in\OO_{K_1}$, with equality for $a=\theta_1$. (a) follows from $\sigma(u)/u-1=(\sigma(u)-u)/u$ and $\sigma(\theta_1)/\theta_1-1=(\sigma(\theta_1)-\theta_1)/\theta_1$. (b) Write $u=1+\theta_1^ma$ with $a\in\OO_{K_1}$. Then $\sigma(u)-u=\theta_1^m\big((\sigma(\theta_1)/\theta_1)^m\sigma(a)-a\big)$, and by (a) $(\sigma(\theta_1)/\theta_1)^m\sigma(a)\equiv\sigma(a)\equiv a\pmod{\theta_1^t}$; so $v_{K_1}(\sigma(u)-u)\ge m+t$, and $\sigma(u)/u-1=(\sigma(u)-u)/u$. \end{proof} The next lemma, on the Kummer generators of a Heisenberg extension over a cubic subfield, is in substance due to Massy \cite{Massy87} and Sharifi \cite[Theorem~2.5]{Sharifi99}: every Heisenberg extension with a given plane is obtained by adjoining a cube root of an explicit element of the form below, unique up to multiplication by an element of the base field and a cube in the compositum of the two cubic extensions. Our short proof is by a different argument. That this construction does give Heisenberg extensions is also in \cite[\S2.2, Lemma~2.1 and Proposition~2.2]{MinacTan16}. \begin{lemma}\label{lem:heis-kummer} Let $L/K_\p$ be a finite Galois extension whose group $G$ is isomorphic to the Heisenberg group of order $27$, let $K_1\subset L$ be an extension of degree $3$ of $K_\p$, cut out by $\alpha\in\Hom(G,\F_3)$, $\alpha\neq0$, and let $\sigma$ generate $\Gal(K_1/K_\p)$. Then $H=\Gal(L/K_1)$ is elementary abelian of order $9$, and there are $w\in K_1^\times$ and $f\in K_\p^\times$ such that the Kummer group $D\subset K_1^\times/K_1^{\times3}$ of $L/K_1$ is generated by the classes of \[ z=\Nm_{K_1/K_\p}(w)\in K_\p^\times\qquad\text{and}\qquad f\,\sigma(w)^2\sigma^2(w)=f\,z\,\frac{\sigma(w)}{w}. \] Moreover the class of $z$ in $\mathcal H$ lies in $\Hom(G,\F_3)\smallsetminus\F_3\alpha$. In particular every element of $D$ is the class of $a\cdot(\sigma(w)/w)^k$ with $a\in K_\p^\times$ and $k\in\Z$. \end{lemma} \begin{proof} The center $Z$ of $G$ has order $3$ and equals $[G,G]$ and the Frattini subgroup; $H=\ker\alpha$ is a maximal subgroup, so it contains $Z$, and it is abelian of exponent $3$ (every maximal subgroup of the Heisenberg group is), of order $9$. As $\mu_3\subset K_1$, Kummer theory gives $D$, of order $9$, with $L=K_1(D^{1/3})$, and the pairing $H\times D\to\mu_3$, $\langle h,\delta\rangle=h(\delta^{1/3})/\delta^{1/3}$, is perfect. Let $M_0=L^Z$, the maximal abelian subextension; $M_0/K_\p$ has group $(\Z/3)^2$, so $M_0=K_1(z_0^{1/3})$ for some $z_0\in K_\p^\times$ whose class in $\mathcal H$ lies in $\Hom(G,\F_3)$ and not in $\F_3\alpha$ (the kernel of $\mathcal H\to K_1^\times/K_1^{\times3}$ is $\F_3\alpha$), and the class $[z_0]$ of $z_0$ in $K_1^\times/K_1^{\times3}$ is a nonzero element~of~$D$. \emph{$\sigma$ preserves $D$ and acts on it nontrivially.} Let $g\in G$ map to $\sigma$. For $\delta\in K_1^\times$ with $[\delta]\in D$, $g(\delta^{1/3})\in L$ is a cube root of $\sigma(\delta)$, so $\sigma$ preserves $D$, and for $h\in H$, $ghg^{-1}\big(g(\delta^{1/3})\big)=g\big(\langle h,\delta\rangle\delta^{1/3}\big)=\langle h,\delta\rangle\, g(\delta^{1/3})$, that is, $\langle ghg^{-1},\sigma(\delta)\rangle=\langle h,\delta\rangle$. If $\sigma$ acted trivially on $D$, perfectness would give $ghg^{-1}=h$ for all $h\in H$, and $G=\langle g,H\rangle$ would be abelian. So $\sigma-1\ne0$ on the plane $D$; as $(\sigma-1)^3=\sigma^3-1=0$ in characteristic $3$, $(\sigma-1)D=\ker(\sigma-1)$ is a line, and it contains $[z_0]\ne0$. Choose $\gamma\in K_1^\times$ with $D=\langle[z_0],[\gamma]\rangle$. Then $\sigma(\gamma)/\gamma\in z_0^{\,j}K_1^{\times3}$ with $j\in\{1,2\}$; put $z_1=z_0^{\,j}$ and write $\sigma(\gamma)=\gamma z_1c^3$ with $c\in K_1^\times$. \emph{Exponent $3$.} Fix cube roots $\gamma^{1/3},z_1^{1/3}\in L$. The lifts of $\sigma$ to $M_0=K_1(z_1^{1/3})$ form a coset of $\Gal(M_0/K_1)$, which permutes the three cube roots of $z_1$ simply transitively; so exactly one lift fixes $z_1^{1/3}$. Let $s\in G$ lift it. Then $s(\gamma^{1/3})$ is a cube root of $\sigma(\gamma)=\gamma z_1c^3$, so $s(\gamma^{1/3})=\zeta\gamma^{1/3}z_1^{1/3}c$ with $\zeta\in\mu_3$; replacing $s$ by $sh$ with $h\in Z=\Gal(L/M_0)$, which acts on $\gamma^{1/3}$ through a nontrivial character (as $L=M_0(\gamma^{1/3})$), we may take $\zeta=1$. Using $s(z_1^{1/3})=z_1^{1/3}$ and $s|_{K_1}=\sigma$, \[ s^3(\gamma^{1/3})=\gamma^{1/3}\cdot z_1^{1/3}c\cdot z_1^{1/3}\sigma(c)\cdot z_1^{1/3}\sigma^2(c)=\gamma^{1/3}\,z_1\Nm_{K_1/K_\p}(c). \] Every element of $G$ has order dividing $3$, so $s^3=1$ and $z_1\Nm_{K_1/K_\p}(c)=1$. \emph{Conclusion.} Put $w=\sigma^2(c)^{-1}$. Then $\Nm(w)=\Nm(c)^{-1}=z_1$ and $\sigma(w)=c^{-1}$. The element $\gamma_0=z_1\sigma(w)/w=\sigma(w)^2\sigma^2(w)$ satisfies $\sigma(\gamma_0)/\gamma_0=\sigma^2(w)\,w/\sigma(w)^2=\Nm(w)/\sigma(w)^3=z_1c^3=\sigma(\gamma)/\gamma$, so $f=\gamma/\gamma_0$ is fixed by $\sigma$, that is, $f\in K_\p^\times$. Hence $D=\langle[z_1],[f\gamma_0]\rangle$ with $z=z_1=\Nm(w)$, and the class of $z$ in $\mathcal H$ is $j$ times that of $z_0$. \end{proof} The exponent $3$ is used once, in $s^3=1$, and the lemma fails without it: the field $K_\p(\zeta_9,(1-\omega)^{1/9})$, whose group over $K_\p$ is non-abelian of order $27$ and exponent $9$, contains $K_1=K_\p((1-\omega)^{1/3})$, and its Kummer group over $K_1$ contains the class of $(1-\omega)^{1/3}$, which has valuation $1$ and break $9$ over $K_1$, whereas every element $a\,(\sigma(w)/w)^k$ with $a\in K_\p^\times$ has $v_{K_1}$-valuation divisible by $3$, as $K_1/K_\p$ is totally ramified. \begin{proposition}\label{prop:three-type} Let $L/K_\p$ be a finite Galois extension whose group is isomorphic to the Heisenberg group of order $27$, let $G^y$ \textup{(}$y\ge-1$\textup{)} be the upper ramification groups of $\Gal(L/K_\p)$, and let $W\subset\mathcal H$ be the plane of homomorphisms $G_{K_\p}\to\F_3$ that factor through $\Gal(L/K_\p)$. Then $W$ is isotropic, and: \begin{enumerate}[(i)] \item if $W\cap\mathcal U=0$, then $G^y\ne1$ for $y\le11/3$ and $G^y=1$ for $y>11/3$; \item if $W\cap\mathcal U$ is a line other than $\F_3e_0$, then $G^y\ne1$ for $y\le10/3$ and $G^y=1$ for $y>10/3$; \item if $e_0\in W$, then $G^y=1$ for $y>3$. \end{enumerate} Every isotropic plane falls under exactly one of \textup{(i)}, \textup{(ii)}, \textup{(iii)}. In particular every Heisenberg extension of $K_\p$ has largest upper ramification break at most $11/3$. \end{proposition} \begin{proof} Choose an isomorphism of $\Gal(L/K_\p)$ with the unitriangular group, and let $a,b,c$ be its entries in positions $(1,2)$, $(2,3)$, $(1,3)$. Then $a,b$ is a basis of $W$ and $c(gh)=c(g)+c(h)+a(g)b(h)$, so the cup product $a\cup b$ is the class of the coboundary of $c$, and $W$ is isotropic. \emph{Exhaustion.} Let $\mathcal H_0=\langle e_2,e_1,e_0\rangle$, the classes of units, which contains $\mathcal U$. By \cref{lem:three-kummer}~(c) the cup product on $\mathcal H_0$ has radical $\F_3e_0$, and it is nondegenerate on the plane $\mathcal H_0/\F_3e_0$, whose isotropic subspaces have dimension at most $1$. So an isotropic plane contained in $\mathcal H_0$ contains $e_0$. If $W\cap\mathcal U\ne0$ and $W\cap\mathcal U$ is not a line other than $\F_3e_0$, then $W\cap\mathcal U$ is $\F_3e_0$ or $\mathcal U$, and $e_0\in W$. If $W\cap\mathcal U$ is a line other than $\F_3e_0$, then $e_0\notin W$. So (i), (ii), (iii) are exhaustive and mutually exclusive. Moreover, in case (i), $W\not\subset\mathcal H_0$ (otherwise $\dim W\cap\mathcal U\ge2+2-3$), and in case (ii) also $W\not\subset\mathcal H_0$ (as $e_0\notin W$): in cases (i) and (ii), $W$ contains a class of nonzero valuation modulo $3$. \emph{Reduction to a cubic subfield.} Let $\alpha\in W$, $\alpha\ne0$, and let $K_1\subset L$ be the cubic extension of $K_\p$ cut out by $\alpha$, $H=\Gal(L/K_1)$, $D$ the Kummer group of $L/K_1$, $\sigma$ a generator of $\Gal(K_1/K_\p)$, and $\psi=\psi_{K_1/K_\p}$, $\varphi=\varphi_{K_1/K_\p}$. By \cref{lem:three-kummer}~(b), $K_1/K_\p$ is totally ramified with break $t\in\{1,2,3\}$, or unramified (put $t=0$). For $y>t$, $\alpha$ vanishes on $G^y$ (by (R1); if $t=0$, because $G^y\subset G^0$), so $G^y\subset H$ and, by (R2), \begin{equation}\label{eq:heis-restrict} G^y=G^y\cap H=H^{\psi(y)}\qquad(y>t). \end{equation} The characters of $H$ are the characters $h\mapsto\langle h,\delta\rangle$, $[\delta]\in D$, and by (R1) the character attached to $\delta$ vanishes on $H^u$ if and only if the character of $G_{K_1}$ cut out by $\delta$ vanishes on $G_{K_1}^u$. Hence, for $T\ge t$: \begin{itemize} \item if every element of $D$ has break at most $T$, then $H^u=1$ for $u>T$, and by \eqref{eq:heis-restrict} $G^y=1$ for $y>\varphi(T)$; \item if some $\delta$ with $[\delta]\in D$ cuts out a totally ramified extension of $K_1$ with break $T>t$, then $H^T\ne1$, and by \eqref{eq:heis-restrict} $G^y\supset H^{\psi(y)}\supset H^T\ne1$ for $t\varphi(3)=3$. \emph{Case \textup{(ii)}.} Take $\alpha\in W\cap\mathcal U$, $\alpha\notin\F_3e_0$, of break $1$, represented by $x\in U^{(2)}$ (\cref{lem:three-kummer}). So $t=1$, $K_1=K_\p(x^{1/3})$, $e_{K_1}=6$, and \cref{lem:kummer-E} applies over $K_1$ with $3e_{K_1}/2=9$. By \cref{lem:sigma-quotients}~(a), $\sigma(w)/w\in U^{(1)}_{K_1}$ has break at most $8$; elements of $K_\p^\times$ have break at most $\psi(3)=7$. So every element of $D$ has break at most $8$, and $G^y=1$ for $y>\varphi(8)=1+\frac73=\frac{10}3$. For the lower bound: the class of $z$ lies in $W\smallsetminus\F_3\alpha$ (\cref{lem:heis-kummer}), so $W=\F_3\alpha+\F_3[z]$; the valuation modulo $3$ is a linear form on $\mathcal H$ that vanishes at $\alpha$ and not on $W$, so $3\nmid v_\p(z)$. As $K_1/K_\p$ is totally ramified, $v_{K_1}(w)=v_\p(\Nm(w))=v_\p(z)$, so $3\nmid v_{K_1}(w)$. Write $w=\theta_1^{\,j}u$ with $\theta_1$ a uniformizer of $K_1$, $j=v_{K_1}(w)$ and $u$ a unit. By \cref{lem:sigma-quotients}~(a), $\sigma(\theta_1)/\theta_1\in U^{(1)}_{K_1}\smallsetminus U^{(2)}_{K_1}$, hence $(\sigma(\theta_1)/\theta_1)^j\in U^{(1)}_{K_1}\smallsetminus U^{(2)}_{K_1}$ as $3\nmid j$, and $\sigma(u)/u\in U^{(2)}_{K_1}$; so $\sigma(w)/w\in U^{(1)}_{K_1}\smallsetminus U^{(2)}_{K_1}$, which by \cref{lem:kummer-E}~(c) cuts out a totally ramified extension with break exactly $8$. As $\delta_0$ differs from $\sigma(w)/w$ by an element of $K_\p^\times$, of break at most $7$, $\delta_0$ also has break exactly $8$. So $G^y\ne1$ for $y\le\varphi(8)=10/3$. \emph{Case \textup{(i)}.} $W$ contains a class of nonzero valuation; replacing it by its negative, it has valuation $\equiv1\pmod3$, and it is represented by some $x$ with $v_\p(x)=1$. Take $\alpha=[x]$: $t=3$, $K_1=K_\p(\theta)$ with $\theta^3=x$, a uniformizer of $K_1$, $\OO_{K_1}=\OO_\p[\theta]$, and $\sigma(\theta)=\omega\theta$ for a suitable generator $\sigma$. Write $w=\theta^ju$ with $u$ a unit. Then $\sigma(w)/w=\omega^j\sigma(u)/u$ with $\sigma(u)/u\in U^{(4)}_{K_1}$ (\cref{lem:sigma-quotients}~(a)), of break at most $9-4=5$ (\cref{lem:kummer-E}~(b)), and $\omega^j\in K_\p^\times$ has break at most $\psi(3)=3$. So every element of $D$ has break at most $5$, and $G^y=1$ for $y>\varphi(5)=3+\frac23=\frac{11}3$. For the lower bound, write $u=\pm u_1$ with $u_1\in U^{(1)}_{K_1}$. Then $z=\Nm(w)=x^j\cdot(\pm1)\cdot\Nm(u_1)$, as $\Nm(\theta)=\theta^3=x$. Suppose that $u_1\in U^{(2)}_{K_1}$. Then $\Nm(u_1)\in U^{(2)}$: write $u_1=1+\zeta$ with $\zeta\in\theta^2\OO_{K_1}$, that is, $\zeta=a_0+a_1\theta+a_2\theta^2$ with $a_i\in\OO_\p$ and $v_\p(a_0),v_\p(a_1)\ge1$; then $\Nm(1+\zeta)=1+s_1+s_2+s_3$ with $s_1=\operatorname{Tr}(\zeta)=3a_0$, $s_2=\frac12(\operatorname{Tr}(\zeta)^2-\operatorname{Tr}(\zeta^2))$, $s_3=\Nm(\zeta)$, and as $\operatorname{Tr}(\theta)=\operatorname{Tr}(\theta^2)=0$ and the constant coefficient of $\zeta^2$ is $a_0^2+2a_1a_2x$, we get $v_\p(s_1)\ge3$, $v_\p(s_2)\ge4$, $v_\p(s_3)=v_{K_1}(\zeta)\ge2$. So the class of $\pm z\,x^{-j}$, which is $[z]-j\alpha$, lies in $W\cap\mathcal U=0$, and $[z]\in\F_3\alpha$, against \cref{lem:heis-kummer}. Hence $u_1\in U^{(1)}_{K_1}\smallsetminus U^{(2)}_{K_1}$, and $u_1=(1+\epsilon\theta)u_2$ with $\epsilon\in\{1,-1\}$ and $u_2\in U^{(2)}_{K_1}$. Now \[ \frac{\sigma(w)}w=\omega^j\cdot\frac{1+\epsilon\omega\theta}{1+\epsilon\theta}\cdot\frac{\sigma(u_2)}{u_2},\qquad \frac{1+\epsilon\omega\theta}{1+\epsilon\theta}=1+\frac{\epsilon(\omega-1)\theta}{1+\epsilon\theta}\in U^{(4)}_{K_1}\smallsetminus U^{(5)}_{K_1}, \] as $v_{K_1}(\omega-1)=3$; the middle factor cuts out a totally ramified extension of break exactly $9-4=5$ (\cref{lem:kummer-E}~(c)), $\sigma(u_2)/u_2\in U^{(5)}_{K_1}$ (\cref{lem:sigma-quotients}~(b)) has break at most $4$, and $\omega^j$ has break at most $3$. So $\sigma(w)/w$, and $\delta_0$, which differs from it by an element of $K_\p^\times$, have break exactly $5$, and $G^y\ne1$ for $y\le\varphi(5)=11/3$. \end{proof} \begin{corollary}\label{cor:three-type-LW} Let $W=\langle\alpha,\beta\rangle$ be an isotropic plane, $c\colon G_{K_\p}\to\F_3$ continuous with $c(gh)=c(g)+c(h)+\alpha(g)\beta(h)$, and $L_W$ the field cut out by \[ g\mapsto\begin{pmatrix}1&\alpha(g)&c(g)\\0&1&\beta(g)\\0&0&1\end{pmatrix}. \] Then the largest upper ramification break of $L_W/K_\p$ is $11/3$ if $W\cap\mathcal U=0$, it is $10/3$ if $W\cap\mathcal U$ is a line other than $\F_3e_0$, and it is at most $3$ if $e_0\in W$. In particular it is at most $10/3$ if $W\cap\mathcal U\neq0$. \end{corollary} \begin{proof} Such a $c$ exists because $W$ is isotropic. The image of the representation maps onto the unitriangular group modulo its center, because $\alpha,\beta$ are linearly independent, and the center is the Frattini subgroup; so the image is the whole unitriangular group, $\Gal(L_W/K_\p)$ is isomorphic to it, and $W$ is the plane of \cref{prop:three-type}. \end{proof} \needspace{4\baselineskip}\begin{remark}\label{rem:three-type-lit} By \cref{cor:three-type-LW}, each of the two irreducible representations $\rho_W$ of dimension $3$ of $\Gal(L_W/K_\p)$ has Artin conductor $f(\rho_W)=3(1+y_{\max})$, where $y_{\max}$ is the largest upper break. Indeed $f(\rho)=\int_{-1}^\infty\operatorname{codim}\rho^{G^y}\,dy$, by the definition in \cite[Chapitre~VI, \S2]{SerreCL} and a change of variable by Herbrand's function \cite[Chapitre~IV, \S3]{SerreCL}, and $\rho_W$ has no nonzero vector fixed by $G^y$ for $y\le y_{\max}$, because every nontrivial normal subgroup contains the center, on which $\rho_W$ is a nontrivial scalar. So $f(\rho_W)=14$ in case~(i) and $13$ in case~(ii). The bound $11/3$ also follows from a theorem of Abrashkin: for a finite extension $F_0$ of $\Q_\ell$, $\ell>2$, containing a primitive $\ell$-th root of unity, with ramification index $e_{F_0}$, and for $1\le s<\ell$, the maximal $\ell$-extension $F_0[s]$ of $F_0$ whose Galois group has exponent $\ell$ and nilpotency class at most $s$ satisfies $\Gal(F_0[s]/F_0)^y=1$ if and only if $y>e_{F_0}\big(1+\frac s{\ell-1}\big)-\frac{1-\delta_{1s}}\ell$, where $\delta_{1s}$ is $1$ for $s=1$ and $0$ otherwise \cite[Theorem~0.5 and the remark after it]{Abrashkin17}, \cite[Theorem~4.5]{Abrashkin17AIF}. Abrashkin's upper numbering is that of \cite{SerreCL}, with the valuation of the base field normalized to have value group $\Z$ \cite[\S4.1]{Abrashkin17}; as a check, for $s=1$ his value $e_{F_0}\ell/(\ell-1)$ is $3$ for $F_0=K_\p$, the largest break in \cref{lem:three-kummer}~(b). For $F_0=K_\p$ ($e_{F_0}=2$, $\ell=3$) and $s=2$ the value is $2\cdot2-\frac13=\frac{11}3$, which \cite[\S6.7]{Abrashkin17} obtains as $\varphi_{K_1/K_\p}(5)$ with $K_1=K_\p\big((1-\omega)^{1/3}\big)$, as in case~(i) of the proof of \cref{prop:three-type}. Every Heisenberg extension of $K_\p$, and every field cut out by a representation as in \cref{prop:three-allk}, lies in $K_\p[2]$, so its largest upper break is at most $11/3$ by (R1). \Cref{prop:three-type} shows in addition which planes attain the bound, and gives the values $10/3$ and at most $3$ for the others. Single planes were treated before: Sharifi's computation of conductors for the plane $\langle[\ell],[1-\ell]\rangle$ over $\Q_\ell(\zeta_\ell)$ \cite[Proposition~4.18]{Sharifi99} gives, for $\ell=3$, the plane $\langle[3],[-2]\rangle=\langle[3],[2]\rangle$ ($W_3$ of \cref{rem:three-heis-examples}), of type~(ii), with largest upper break $10/3$; and for the plane $\langle[\omega],[1-\omega]\rangle=\langle e_2,e_\pi\rangle$, of type~(i), compare the conductor of the top step computed by Davis and Pries \cite[Lemma~4.15]{DavisPries20}. We have not found in the literature the values of \cref{prop:three-type} for every isotropic plane, the conductors $14$ and $13$, or the bound $|G^y|\le27$ for $3\frac{11}3$. So $L/K_\p$ is totally ramified, its lower breaks are $2,5,11$, and the top step $L/K_\p(\zeta_9,\theta)$, whose group is the center (contained in every nontrivial $G_u$), has break $11$ and conductor $12$, the value of \cite[Lemma~4.15]{DavisPries20} at the prime $3$; and $d=3\cdot\frac{26}{27}+\frac89+\frac23\cdot\frac23=\frac{38}9$, so $v_L(\mathfrak D_{L/K_\p})=27d=114$ and the discriminant of $L/K_\p$ is $\p^{114}$. This agrees with the conductor--discriminant formula: the six characters of order $3$ of $\Gal(L/K_\p)$ outside $\F_3e_2$ have conductor $4$, the two in $\F_3e_2$ conductor $3$, and the two irreducible characters of degree $3$ conductor $14$ (\cref{rem:three-type-lit}), and $6\cdot4+2\cdot3+2\cdot3\cdot14=114$. \emph{Case~\textup{(ii)}: $W_3=\langle e_\pi+e_2,e_1\rangle=\langle[3],[2]\rangle$}, the plane of \cite[Proposition~4.18]{Sharifi99} at the prime $3$. Take $x=2$, of class $-e_1$, so that $t=1$ and $\theta=2^{1/3}$ is a unit of $K_1$, and $w=1+\theta$; then $z=3$, of class $-(e_\pi+e_2)$, and \[ L=K_\p\big(2^{1/3},3^{1/3},\eta\big),\qquad \eta^3=\frac{1+\omega\,2^{1/3}}{1+2^{1/3}}=1+\frac{(\omega-1)\,2^{1/3}}{1+2^{1/3}}\in U^{(1)}_{K_1}\smallsetminus U^{(2)}_{K_1}, \] as $v_{K_1}(1+\theta)=v_\p(3)=2$ and $v_{K_1}(\omega-1)=3$. Over $K_1$, $\eta^3$ has break $8$, that is, conductor $9=\ell^2-\ell+3$ for $\ell=3$, as in \cite[Proposition~4.18]{Sharifi99}, and $z$ has break $\psi_{K_1/K_\p}(3)=7$ (by (R2) for $K_\p(2^{1/3},3^{1/3})$). Hence $|G^y|$ is $27$ for $-111/3$, $|G^y|\le3$ for $10/33$ the restriction of $c$ to $G^y_{K_\p}$ depends linearly on the Plücker vector of $a\wedge b$, which lies in a space of dimension $5$. So it is enough to control it for five suitably chosen isotropic planes whose Plücker vectors form a basis. \Cref{cor:three-type-LW} says, for each isotropic plane, in which range of $y$ it vanishes, according to how the plane meets $\mathcal U$. The bounds stated come from a basis with one plane meeting $\mathcal U$ only in $0$, two meeting it in a line other than $\F_3e_0$, and two containing $e_0$, as in the proof. \begin{proof} Write $\mathcal H_k=\Hom(G_{K_\p},k)=\mathcal H\otimes_{\F_3}k$, and $a=\sum_ia_ie_i$, $b=\sum_ib_ie_i$ with $a_i,b_i\in k$, where $i$ runs over $\{\pi,2,1,0\}$ in this order. For $y>3$, every element of $\mathcal H_k$ vanishes on $G^y_{K_\p}$ (\cref{lem:three-kummer}~(b)); so $a$ and $b$ vanish there, and $\Upsilon(g)$, for $g\in G^y_{K_\p}$, is determined by $c(g)$. We show that $c$ vanishes on $G^y_{K_\p}$ for $y>11/3$, and that its values on $G^y_{K_\p}$ lie in an $\F_3$-line of $k$ for $10/33$, both $c'-c_{a\wedge b}$ and $Q$ vanish on $G^y_{K_\p}$, so \[ c|_{G^y_{K_\p}}=c_{a\wedge b}|_{G^y_{K_\p}}\qquad(y>3). \] For $y>3$ the map $\Phi_y\colon\mathcal P\to\Hom(G^y_{K_\p},\F_3)$, $\xi\mapsto c_\xi|_{G^y_{K_\p}}$, is $\F_3$-linear, and it does not depend on the choice of the $c_\xi$, since two choices differ by an element of $\mathcal H$. Let $W=\langle\alpha,\beta\rangle$ be an isotropic plane, and $c_W$ as in \cref{cor:three-type-LW}. Applying the identity just proved to $(\alpha,\beta,c_W)$ in place of $(a,b,c)$, we get $c_W=c_{\alpha\wedge\beta}$ on $G^y_{K_\p}$ for $y>3$. By \cref{cor:three-type-LW}, $c_W$ vanishes on $G^y_{K_\p}$ for $y>11/3$, for $y>10/3$ if $W\cap\mathcal U$ is a line other than $\F_3e_0$, and for $y>3$ if $e_0\in W$. So $\Phi_y(\alpha\wedge\beta)=0$ in these ranges. Let $\mathcal P_0\subset\mathcal P$ be the kernel of the map $\mathcal P\to\bigwedge^2(\mathcal H/\mathcal U)\isom\F_3$; in coordinates, $\xi\mapsto\xi_{\pi2}$. Consider the planes \begin{gather*} W_1=\langle e_\pi,e_2\rangle,\quad W_2=\langle e_\pi,e_1+e_0\rangle,\quad W_3=\langle e_\pi+e_2,e_1\rangle,\\ W_4=\langle e_2,e_0\rangle,\quad W_5=\mathcal U=\langle e_1,e_0\rangle, \end{gather*} which are isotropic by \cref{lem:three-kummer}~(c). (\Cref{rem:three-heis-examples} gives explicit Heisenberg extensions for $W_1$, $W_3$, $W_4$ and $W_5$, and the Supplement, item~S8, one for each of the five planes.) In the basis $(e_\pi\wedge e_2,e_\pi\wedge e_1,e_\pi\wedge e_0,e_2\wedge e_1,e_2\wedge e_0,e_1\wedge e_0)$ of $\bigwedge^2\mathcal H$ their Plücker vectors are \begin{gather*} \xi_1=(1,0,0,0,0,0),\quad \xi_2=(0,1,1,0,0,0),\quad \xi_3=(0,1,0,1,0,0),\\ \xi_4=(0,0,0,0,1,0),\quad \xi_5=(0,0,0,0,0,1). \end{gather*} They are linearly independent, so they form a basis of $\mathcal P$, which has dimension $5$; and $\xi_2,\dots,\xi_5$ form a basis of $\mathcal P_0$, which has dimension $4$ since $\xi_1\notin\mathcal P_0$. Now $W_1\cap\mathcal U=0$; $W_2\cap\mathcal U=\F_3(e_1+e_0)$ and $W_3\cap\mathcal U=\F_3e_1$ are lines other than $\F_3e_0$; and $e_0\in W_4\cap W_5$. So $\Phi_y(\xi_1)=0$ for $y>11/3$, $\Phi_y(\xi_2)=\Phi_y(\xi_3)=0$ for $y>10/3$, and $\Phi_y(\xi_4)=\Phi_y(\xi_5)=0$ for $y>3$. Hence $\Phi_y=0$ for $y>11/3$, and $\Phi_y(\mathcal P_0)=0$ for $y>10/3$. For $y>11/3$ this gives $c=0$ on $G^y_{K_\p}$, so $G^y=1$. For $10/30$. By \cref{prop:three-allk}, applied with $k=\FE$, $G^y=1$ for $y>11/3$, $|G^y|\le3$ for $10/311/3$, the proof gives $d<14/3$ and $\rd(L)<3^{17/6}$, and $3^{17/6}>22.48$ lies above the supremum $4\pi e^\gamma<22.39$ of the bounds of \cref{sec:odlyzko}; with them, $3^{449/162}<21.01$ lies below it. \subsection{Lower bounds} \begin{proposition}\label{prop:three-odlyzko} Let $L$ be a totally complex number field of degree $n$. Then \begin{center} \begin{tabular}{@{}lrrrrrr@{}} \toprule $n\ge$ & $1378$ & $674$ & $118$ & $80$ & $24$ & $20$\\ $\rd(L)>$ & $21.1515$ & $20.4476$ & $16.9179$ & $15.6281$ & $10.5939$ & $9.7691$\\ \bottomrule \end{tabular} \end{center} and $\rd(L)>3.2422$ if $n\ge4$. On the other hand $3^{449/162}<21.0084$, $T_3<19.3665$, $T_2<16.2359$, $3^{5/2}<15.5885$, $3^{11/6}<7.4942$, $T_1<9.5665$ and $\sqrt3<1.7321$. \end{proposition} \begin{proof} The right side of \eqref{eq:rdbound}, with $r_1=0$ and the test function $\Phi(x)=\Phi_0(x/b)/\cosh(x/2)$ of \cref{sec:odlyzko}, was evaluated with certified error bounds at $b=\frac{63}2,25,12,10,6,6,2$ for the seven degrees $n=1378,674,118,80,24,20,4$ (\cref{app:computations}, item~C5). For fixed $b$ the bound increases with $n$. The smallest margin is $2.54\cdot10^{-3}$ in $\log\rd$, at $n=80$, against $3^{5/2}$ (item~C5). The comparison at $n=674$ with $T_3$ is used in the case $s=3$ of the proof of \cref{prop:three-noirred}, where $\Delta$ is the projective image of an irreducible reduction $\Rb_\Lambda$ and $3^s$ the $3$-part of $|\Delta|$: with $3^{449/162}$ in place of $T_3$, step~(c) there would give only $|\Delta|\le688$, while for non-solvable $\Delta$ the count of step~(d) gives only $|\Delta|\ge540$. \end{proof} \subsection{The reductions} \begin{proposition}\label{prop:three-noirred} For every stable lattice $\Lambda$, $\Rb_\Lambda$ is reducible. \end{proposition} \begin{proof} Suppose that $\Rb_\Lambda$ is irreducible. Let $\Delta\subset\PGL_3(\FE)$ be its projective image, $L\subset L_\Lambda$ the field cut out by the projective representation, which is Galois over $K$ with group $\Delta$ and totally complex of degree $2|\Delta|$, and $P\subset\Delta$ an inertia group at a prime above $\p$. (a) \emph{$\Delta$ is generated by the conjugates of $P$, which is a $3$-group.} The subgroup generated by the inertia groups at the primes above $\p$ is normal, and its fixed field is unramified over $K$ at every place. A nontrivial extension of $K$ unramified at every place would be totally complex of degree at least $4$ with root discriminant $\rd(K)=\sqrt3$, which \cref{prop:three-odlyzko} excludes; so that subgroup is $\Delta$. And $P$ is a $3$-group, by the proof of \cref{prop:three-rd}. (b) \emph{$\Delta$ has no nontrivial normal $3$-subgroup.} Let $Q\subset\Delta$ be one, and $\tilde Q$ its preimage in $\Rb_\Lambda(G_K)$. The kernel of $\Rb_\Lambda(G_K)\to\Delta$ consists of scalar matrices, whose orders are prime to $3$; so $\tilde Q$ is the direct product of this kernel and of its unique Sylow $3$-subgroup $Q_3$, which is therefore normal in $\Rb_\Lambda(G_K)$. A $3$-subgroup of $\GL_3(\FE)$ is unipotent, so the vectors fixed by $Q_3$ form a nonzero subspace, stable under $\Rb_\Lambda(G_K)$. As $\Rb_\Lambda$ is irreducible, $Q_3$ acts trivially, so $Q_3=1$ and $Q=1$. (c) \emph{Orders.} $\Delta\ne1$, since otherwise $\Rb_\Lambda(G_K)$ would consist of scalars; by (a), $3$ divides $|\Delta|$. Let $3^s$ be the $3$-part of $|\Delta|$, so $s\ge1$ and $|P|\le3^s$. By \cref{prop:three-rd}, $\rd(L)<3^{449/162}$ and $\rd(L)\le T_s$; by \cref{prop:three-odlyzko}, $2|\Delta|<20$, $118$, $674$, $1378$ if $s=1$, $2$, $3$, $s\ge4$ respectively. So $|\Delta|\le9$, $58$, $336$, $688$ respectively. (d) \emph{Groups.} If $s=1$, then $|\Delta|\in\{3,6\}$, and the Sylow $3$-subgroup is normal, against (b). Let $s\ge2$; then $s\le5$, since $3^6>688$. Suppose first that $\Delta$ is solvable, and let $\Delta_{\mathrm F}$ be its Fitting subgroup. By (b), $\Delta_{\mathrm F}$ has order prime to $3$, and it contains its own centralizer in $\Delta$; so a Sylow $3$-subgroup of $\Delta$, of order $3^s$, acts faithfully on $\Delta_{\mathrm F}$ by conjugation, and $3^s$ divides $\lvert\Aut(\Delta_{\mathrm F})\rvert$. The order of $\Delta_{\mathrm F}$ divides the part $m$ of $|\Delta|$ prime to $3$, and by (c) $m\le6$, $12$, $8$, $2$ for $s=2,3,4,5$. But for every group $Y$ of order at most $12$ and prime to $3$, the $3$-part of $|\Aut(Y)|$ is at most $3$: these groups are $1$, $C_2$, $C_4$, $C_2^2$, $C_5$, $C_7$, $C_8$, $C_4\times C_2$, $D_4$, $Q_8$, $C_2^3$, $C_{10}$, $D_5$ and $C_{11}$, whose automorphism groups have orders $1$, $1$, $2$, $6$, $4$, $6$, $4$, $8$, $8$, $24$, $168$, $4$, $20$ and $10$. This contradicts $3^s\ge9$. If $\Delta$ is not solvable, it has a non-abelian simple composition factor $S$ with $|S|\le688$, so $S$ is one of $A_5$, $\mathrm{PSL}_2(\F_7)$, $A_6$, $\mathrm{PSL}_2(\F_8)$, $\mathrm{PSL}_2(\F_{11})$, of orders $60$, $168$, $360$, $504$, $660$ and with $3$-parts $3$, $3$, $9$, $9$, $3$ \cite{ATLAS}. The other composition factors contribute the rest of the $3$-part of $|\Delta|$. So $|\Delta|\ge60>58$ if $s=2$; $|\Delta|\ge\min(9\cdot60,\,3\cdot360)=540>336$ if $s=3$; and $|\Delta|\ge\min(27\cdot60,\,9\cdot360)=1620>688$ if $s\ge4$. In each case this is a contradiction. \end{proof} \begin{proposition}\label{prop:three-nodim2} $\Rb^{\mathrm{ss}}$ is not of the form $\theta\oplus\Theta$ with $\theta$ a character and $\Theta$ irreducible of dimension $2$. \end{proposition} \begin{proof} Suppose it is. For a suitable stable lattice $\Lambda$, $\Theta$ is a subquotient of $\Rb_\Lambda$, so by \cref{prop:three-ordinary} its restriction to $G_{K_\p}$ has, in a suitable basis, the form $g\mapsto\bar\psi(g)\bigl(\begin{smallmatrix}1&\eta(g)\\0&1\end{smallmatrix}\bigr)$ with $\eta\in\Hom(G_{K_\p},\FE)$. Let $\Delta$ be the projective image of $\Theta$ and $L$ the field it cuts out, Galois over $K$ with group $\Delta$, totally complex of degree $2|\Delta|$, unramified outside $\p$. The completions of $L$ at the primes above $\p$ are cut out by $\eta$: they are abelian of exponent $3$, without tame part, and with upper breaks at most $3$ by \cref{lem:three-kummer}~(b). As in the proof of \cref{prop:three-rd}, $d<1+3$ and $\rd(L)<3^{5/2}$, so $2|\Delta|<80$ by \cref{prop:three-odlyzko}, and $|\Delta|\le39$. As in (a) and (b) of the proof of \cref{prop:three-noirred}, $\Delta$ is generated by its elements of order a power of $3$ and has no nontrivial normal $3$-subgroup. Also $\Delta\ne1$, since $\Theta$ is irreducible, so $3$ divides $|\Delta|$. By Dickson's classification of the finite subgroups of $\PGL_2(\Fbar_3)$ \cite{Dickson01} (see \cite[Theorem~B and Remark~2.3]{Faber23}), a finite subgroup of $\PGL_2(\Fbar_3)$ of order divisible by $3$ without nontrivial normal $3$-subgroup is isomorphic to $A_4\isom\mathrm{PSL}_2(\F_3)$, $S_4\isom\PGL_2(\F_3)$, $A_5$, $\mathrm{PSL}_2(\F_{3^m})$ or $\PGL_2(\F_{3^m})$ with $m\ge2$. Of these only $A_4$ and $S_4$ have order at most $39$, and $S_4$ is not generated by its elements of order $3$. So $\Delta\isom A_4$, whose $3$-subgroups have order at most $3$: the inertia groups have order at most $3$, so $d\le\frac23\cdot4$ and $\rd(L)\le3^{11/6}<7.4942$, while $[L:\Q]=24$ gives $\rd(L)>10.5939$ (\cref{prop:three-odlyzko}). This is a contradiction. \end{proof} \begin{corollary}\label{cor:three-trivial} For every stable lattice $\Lambda$, $\Rb^{\mathrm{ss}}$ is the sum of three copies of the trivial character. \end{corollary} \begin{proof} After enlarging $E$, every constituent of $\Rb^{\mathrm{ss}}$ is absolutely irreducible, and $\Rb^{\mathrm{ss}}$ is irreducible, or the sum of a character and an irreducible representation of dimension $2$, or a sum of three characters. The first two cases are excluded by \cref{prop:three-noirred,prop:three-nodim2}, and in the third every character is trivial by \cref{prop:three-characters}. \end{proof} \subsection{Local generation} The following lemma is elementary and well known. It goes back to Shafarevich and Koch \cite{Shafarevich63,Koch02}, and Greenberg states it in the form ``$\p$ is totally ramified in $M/K$'' \cite[Remark~6.1.4]{Greenberg16}. Freitas, Kraus and Siksek prove that an $\ell$-extension of a number field unramified away from one prime $\q$ and the infinite places is totally ramified at $\q$ when $\ell$ does not divide the narrow class number \cite[Theorem~9(b)]{FreitasKrausSiksek20}. We include the proof. \needspace{4\baselineskip}\begin{lemma}\label{lem:local-generation} Let $\ell$ be an odd prime and $K'$ a number field with only one prime $\q$ above $\ell$ and with class number prime to $\ell$. Let $\Omega$ be the maximal pro-$\ell$ extension of $K'$ unramified outside $\q$, and $G=\Gal(\Omega/K')$. Then the inertia group in $G$ of a prime of $\Omega$ above $\q$ is $G$. \end{lemma} \begin{proof} Let $T\subset G$ be such an inertia group; it is compact, hence closed. As $G$ is a pro-$\ell$ group, a closed subgroup that maps onto $G/\Phi(G)$, where $\Phi(G)$ is the Frattini subgroup (the closure of $G^\ell[G,G]$), is $G$ itself. So it suffices to show that every continuous homomorphism $\chi\colon G\to\F_\ell$ with $\chi(T)=0$ is zero. Such a $\chi$ vanishes on all the inertia groups above $\q$, which are conjugate to $T$; so it cuts out an extension of $K'$ of degree $1$ or $\ell$ that is unramified at every finite place, and also at the infinite places, as $\ell$ is odd. As $\ell$ does not divide the class number, the extension is trivial and $\chi=0$. \end{proof} \begin{corollary}\label{cor:three-inertia} Let $R'\colon G_K\to\GL_3(\overline\Q_3)$ be a continuous representation, unramified outside $\p$, whose reduction has trivial semisimplification. Then $R'(G_K)=R'(I_\p)$. \end{corollary} \begin{proof} The image of $R'$ is compact, so $R'$ is defined over the ring of integers $\OO'$ of a finite extension of $\Q_3$, with uniformizer $\varpi$ and residue field $k'$, and stabilizes a lattice. The reduction of the lattice has trivial semisimplification, so in a suitable basis of the lattice it is upper unitriangular, and $R'(G_K)$ lies in the preimage in $\GL_3(\OO')$ of the group of upper unitriangular matrices of $\GL_3(k')$, a pro-$3$ group (an extension of a finite $3$-group by $1+\varpi M_3(\OO')$). So $R'$ factors through $\Gal(\Omega/K)$, where $\Omega$ is the maximal pro-$3$ extension of $K$ unramified outside $\p$. The field $K$ has one prime above $3$ and class number one, so by \cref{lem:local-generation} the image of $I_\p$ in $\Gal(\Omega/K)$ is the whole group. Hence $R'(G_K)=R'(I_\p)$. \end{proof} \subsection{The theorem} \begin{theorem}\label{thm:three-main} Let $p\ge17$ be a prime with $p\equiv1\pmod3$. Then $x^3+y^3=z^p$ has no solution in coprime nonzero integers. \end{theorem} \begin{proof} Suppose that $(a,b,c)$ is a solution. By \cref{prop:kraus}, $3\mid c$, $p\nmid c$ and $7\nmid abc$, so the Frey representation $\rho$ of the solution is defined and has the properties (F1)--(F6) of \cref{sec:F}. By \cref{prop:reasonable,lem:realizable,thm:CEG} there is a compatible system $\Rsys$ as in \cref{sec:S}. Let $\lambda$ be a prime of its coefficient field above $3$ and $R=R_\lambda$. By \cref{prop:three-unram}, $R$ is unramified outside $\p$; by \cref{prop:three-ordinary}, $R|_{G_{K_\p}}$ stabilizes a line $V_1$; by \cref{cor:three-trivial}, the reduction of $R$ has trivial semisimplification. By \cref{cor:three-inertia}, $R(G_K)=R(I_\p)\subset R(G_{K_\p})$, so $V_1$ is stable under $G_K$, and $R$ is reducible. This contradicts \cref{prop:three-irreducible}. \end{proof} \begin{corollary}\label{cor:three-all} For every integer $n\ge3$ the equation $x^3+y^3=z^n$ has no solution in coprime nonzero integers. \end{corollary} \begin{proof} An integer $n\ge3$ has an odd prime factor $q$, or is divisible by $4$; a solution $(a,b,c)$ for the exponent $n$ gives the solution $(a,b,c^{n/q})$ for the exponent $q$, or $(a,b,c^{n/4})$ for the exponent $4$. For the exponents $m$ with $3\le m\le10^9$, in particular $m=4$ and the primes $q\le13$, there is no solution by \cite[Theorem~2]{ChenSiksek09}; for the primes $q\ge17$ with $q\equiv2\pmod3$ by \cite[Theorem~3]{Freitas16}; and for the primes $q\ge17$ with $q\equiv1\pmod 3$ by \cref{thm:three-main}. \end{proof} \begin{remark}\label{rem:three-context} The two-dimensional analogue of \cref{prop:three-noirred,prop:three-nodim2} is a theorem of \c{S}eng\"un: there is no irreducible continuous representation of $G_K$ in $\GL_2(\Fbar_3)$ unramified outside $\p$ \cite[Theorem~C]{Sengun09}. The method, a ramification bound played against lower bounds for discriminants, is Tate's \cite{Tate94}; it was refined by Moon and Taguchi \cite{MoonTaguchi03} and extended to representations of higher dimension by Moon \cite{Moon00}, and Ghitza and Yamauchi prove, under the generalized Riemann hypothesis, that $G_\Q$ has no irreducible totally real representation modulo $3$ of dimension $2$, $3$ or $4$ unramified outside $3$ \cite[Theorem~A~(b)]{GhitzaYamauchi25}. Dieulefait and Golfieri Madriaga \cite[\S2]{DieulefaitGolfieri25} also argue at a prime above $3$, for Fermat's equation over totally real fields when $3\mid xyz$, but differently: the Frey curve has multiplicative reduction there, and level lowering with the Hasse--Weil bound excludes large exponents. \end{remark} \begin{proposition}\label{prop:three-scope} There is no continuous representation $R\colon G_K\to\GL_3(\overline\Q_3)$ that is semisimple, unramified at every prime $\q\ne\p$, and de Rham at $\p$ with Hodge--Tate weights $\{0,1,2\}$ for both embeddings of $K_\p$, and such that the monodromy operator of $\WD(R|_{G_{K_\p}})$ is regular. \end{proposition} \begin{proof} The proofs of \cref{prop:three-ordinary,prop:three-irreducible} and of \cref{prop:three-characters} to \cref{cor:three-inertia}, and the last three sentences of the proof of \cref{thm:three-main}, use only these four properties of $R$, together with a character of $G_{K_\p}$ of $3$-power order through which $I_\p$ acts on $\WD(R|_{G_{K_\p}})$, which replaces $\kappa_\lambda|_{G_{K_\p}}$. Such a character exists. The action of $I_\p$ on $\WD(R|_{G_{K_\p}})$ has finite image and commutes with the regular monodromy operator, so it is through a scalar character $\chi$. The cube of $\chi$ is the restriction to $I_\p$ of $\varepsilon_3^{3}\det R$, a character of $G_K$ that is unramified outside $\p$ and de Rham at $\p$ with Hodge--Tate weights $0$, hence of finite order on $I_\p$, and so of finite order, as $K$ has class number one; it is a character of some ray class group $\Cl_{\p^n}(K)$, which is a $3$-group by the proof of \cref{prop:three-characters}. So $\chi$ has $3$-power order. It is invariant under conjugation by the Weil group of $K_\p$, so it factors through the image of $I_\p$ in $G_{K_\p}^{\mathrm{ab}}$, which is $\OO_\p^\times$ by local class field theory, and it extends to a character of $G_{K_\p}$ of the same order. \end{proof} \begin{remark}\label{rem:three-scope} Each of the last three properties of \cref{prop:three-scope} fails for an irreducible representation with the other three. \begin{enumerate}[(a)] \item \emph{Ramified outside $\p$}: the restriction to $G_K$ of the symmetric square of $H^1$ of the elliptic curve 15a1, of conductor $15$, is ramified at the inert prime $5\OO$, where the curve has multiplicative reduction. It is de Rham at $\p$ with Hodge--Tate weights $\{0,1,2\}$, and its monodromy operator at $\p$ is regular, as the curve has multiplicative reduction at $3$. \item \emph{Other Hodge--Tate weights}: for the newform $f=\eta(z)^6\eta(3z)^6$ of weight $6$ on $\Gamma_0(3)$, the restriction to $G_K$ of the symmetric square of the $3$-adic representation $\rho_f$ of $f$ has Hodge--Tate weights $\{0,5,10\}$. It is unramified outside $\p$, and its monodromy operator at $\p$ is regular: $f$ is new of level $3$, with $a_3(f)=9=3^{(6-2)/2}$, so $\rho_f$ is semistable at $3$ with nonzero monodromy operator \cite{Saito97}. \item \emph{Monodromy operator $0$}: the curve $y^2-3xy+\omega y=x^3$ over $K$ has discriminant $27\omega^2$ and $j$-invariant $27(-216-703\omega)$, so it has good reduction outside $\p$ and potentially good reduction at $\p$, and the symmetric square of its $H^1$ has monodromy operator $0$ at $\p$. It is unramified outside $\p$, and de Rham at $\p$ with Hodge--Tate weights $\{0,1,2\}$. \end{enumerate} These representations are irreducible. In (a) and (c) the curve has no complex multiplication (15a1 has non-integral $j$-invariant, and the $j$-invariant in (c) has no real conjugate, unlike every $j$-invariant with complex multiplication), so the image of $G_K$ acting on $H^1$ is open in $\GL_2(\Z_3)$ \cite[Chapter~IV, \S2.2]{SerreAbelian}, hence Zariski dense in $\GL_2$, and the symmetric square is irreducible. In (b), $\rho_f$ is irreducible: otherwise its semisimplification would be a sum of two characters unramified outside $3$ with Hodge--Tate weights $0$ and $5$, whose values at $\Frob_2$ have absolute values $1$ and $2^5$, while $a_2(f)=-6$. So the symmetric square of $\rho_f|_{G_K}$ is semisimple, with regular monodromy operator at $\p$, hence irreducible, as in the proof of \cref{prop:three-irreducible}. The following kinds of objects over $K$ fail one of the properties: induced and CM ones have $N=0$; endoscopic and CAP ones (for $\mathrm{U}(2,1)$) have reducible Galois representations \cite[\S3.2.3]{BellaicheChenevier04}, hence $N^2=0$; and symmetric squares of elliptic curves over $K$ with $N^2\neq0$, that is, of curves with potentially multiplicative reduction at $\p$, ramify outside $\p$ (by \cite[Theorem~C]{Sengun09} and \cite[Theorem~1]{FreitasKrausSiksek20}, after a quadratic twist). Moreover, there is no nonzero scalar-valued Picard modular cusp form of weight~$3$ (the weight of holomorphic $2$-forms) for the principal congruence subgroup of level~$\sqrt{-3}$ of $\SU(2,1)(\Z[\omega])$, nor, equivalently, for that of $\mathrm{U}(2,1)(\Z[\omega])$ with a power of $\det$ as character \cite[\S2.1 and Proposition~2.2]{BergstromvdGeer22}. \end{remark}