\section{The energy identities at a place}\label{sec:energy} In this section we treat diagram~$9$ and diagrams $14$ and $15$. Since the pieces of diagrams $14$ and $15$ carry no mass condition from \cite{AK12} (\S\ref{ss:Kgating}) and are present for every label set (\S\ref{ss:certs}), no choice of labels removes them. We therefore need an argument that does not go through the labels. By Theorem~\ref{thm:Kgating}, places of diagrams $15$ and $14$ carry solutions of $\mathrm{K15}$ and $\mathrm{K14}$, which exist on hypersurfaces of the space of cluster masses. We exclude the places of diagram $15$ by expanding at the place the potential $V=U+\frac12I$, which is constant along $C$ (Lemma~\ref{lem:U}) and whose gradient vanishes there (Lemma~\ref{lem:potential}). For diagram $14$ we derive in \S\ref{ss:d14} a condition on the leading cluster. With the same expansion we exclude the pieces $\pc{d14b}$ (Theorem~\ref{thm:D14}). In \S\ref{ss:d9} the expansion, in coordinates adapted to the diagram, excludes diagram~$9$ at every positive mass (Theorem~\ref{thm:D9}). Identities marked $(\dagger)$ were verified by computer in exact arithmetic; the programs are in the supplementary material. \subsection{Expansions at a place}\label{ss:expansion} Let $P$ be a singular place of $C$ and $o$ a body. We use the positions relative to $o$, $u_i=z_i-z_o$ and $v_i=w_i-w_o$ for $i\ne o$, as coordinates. By relabelling, Lemma~\ref{lem:potential} holds in them. So $\nabla_{(u,v)}V=0$ at all points of $X_m$. After the rotation of Lemma~\ref{lem:place}, each of these coordinates is $q=\vartheta^{d\,e(q)}y_q(\vartheta)$. Put $\tau=\vartheta^{d/2}$, so that $q=\tau^{2e(q)}y_q$, where $y_q$ is a convergent Puiseux series in $\tau$ with $y_q(0)\neq0$. We write $\bar y=y(0)$ for the \emph{leading point}. For a Puiseux series $g$ and $r\in\Q$, $[g]_r$ denotes its coefficient of $\tau^r$. Replacing $\tau$ by $\tau\varphi(\tau)$ and rotating by $\mu(\tau)$, with convergent Puiseux series $\varphi$, $\mu$ and $\varphi(0)=\mu(0)=1$, changes neither the leading point nor the identities below. We use this freedom in \S\ref{ss:order5}. \begin{lemma}\label{lem:identities} Suppose that $F(\tau,y):=V(\tau^{2e}y)=\sum_{n\ge0}\tau^{n-2}V_n(y)$, with $V_n$ holomorphic near $\bar y$ and the series convergent for $y$ near $\bar y$ and small $\tau\ne0$. Then $\nabla_yF(\tau,y(\tau))=0$, and for every $q\in\Q$ with $q\ne2$, \begin{equation}\label{eq:qid} \sum_{n\ge1}n\,\big[V_n(y(\tau))\big]_{q-n}=0 . \end{equation} \end{lemma} \begin{proof} Since $y=\tau^{-2e}x$ coordinatewise, $\nabla_yF=\tau^{2e}\nabla_xV=0$ along the place. As $V$ is constant on $C$, $\sum_n\tau^{n-2}V_n(y(\tau))$ is a constant $E$. The function $F$ satisfies $F(\lambda\tau,\lambda^{-2e}y)=F(\tau,y)$, both sides being $V$ at the same point. Differentiating at $\lambda=1$ and using $\nabla_yF=0$ at $y=y(\tau)$, we obtain $\sum_n(n-2)\tau^{n-2}V_n(y(\tau))=0$. Adding twice the first identity gives $\sum_nn\,\tau^{n-2}V_n(y(\tau))=2E$. Its coefficient of $\tau^{q-2}$ is \eqref{eq:qid}. Each coefficient involves finitely many $n$, since $\tau^{n-2}V_n(y(\tau))$ has valuation at least $n-2$. \end{proof} \subsection{Diagram 15 at order three}\label{ss:d15} Let $P$ have type $\pc b_k(j)$. By Theorem~\ref{thm:Kgating}(c), $P$ has diagram $15$ with $zw$-cluster $Q=Q_j$ and, after exchanging the colours if necessary, $w$-edge $kj$. We take $o=k$ and write $Q'=Q\setminus\{k\}=\{a,b,c\}$ for the three \emph{outer} bodies. By Lemma~\ref{lem:zwK4} and Lemma~\ref{lem:orders}, $e(u_i)=e(v_i)=1$ for $i\in Q'$, $e(u_j)=2$ and $e(v_j)=-2$. Write $\hat z_i=\bar y_{u_i}$, $\hat w_i=\bar y_{v_i}$ for $i\in Q'$, $\hat z_j=\bar y_{u_j}$, $\hat w_j=\bar y_{v_j}$, and $\hat p_{il}$ for the leading coefficients of the $p_{il}$. \begin{lemma}\label{lem:d15lead} \begin{enumerate}[label=\textup{(\alph*)}] \item The cluster coefficients $(\hat z_i,\hat w_i)_{i\in Q'}$, with $\hat z_k=\hat w_k=0$, and $(\hat p_{il})_{i,l\in Q}$ solve $\mathrm{K15}(m_Q,k)$ with $c=-m_j\hat w_j/M$. \item $\sum_{i\in Q'}m_i\hat z_i=0$. \item $\hat p_{ij}^2\,\hat z_i\,(-\hat w_j)=1$ for $i\in Q'$. \end{enumerate} \end{lemma} \begin{proof} (a) By Theorem~\ref{thm:Kgating}(b) the cluster coefficients solve $\mathrm{K15}(m_Q,k)$. The constant $c$ is the coefficient of $w_k$ at exponent $-2$, which by $Mw_k=-\sum_{i\ne k}m_iv_i$ is $-m_j\hat w_j/M$. (b) For a solution of K15, $\sum_im_i\hat z_iF^w_i=\sum_{i0$ for real $t\ne0$ and $Q(0)=m_a^2m_b>0$. \end{proof} \begin{corollary}\label{cor:equilateral} If $m_a=m_b=m_c$, then $Q=m_a^3(t^2+t+1)^2$, and the outer bodies of a cluster with a bad shape form an equilateral triangle centred at body $k$: $\{\hat z_a,\hat z_b,\hat z_c\}=\lambda\{1,\jj,\jj^2\}$ with $\jj=e^{2\pi i/3}$. \end{corollary} \begin{proof} For equal masses, $(1+t)^2(1+t^2)+t^2=(t^2+t+1)^2$. Hence $t\in\{\jj,\jj^2\}$, $x$ is proportional to $(1,\jj,\jj^2)$ up to order, and $\hat z_i=x_i^{-2}$. \end{proof} \begin{proposition}\label{prop:eqfull} Let $\nu\in\C\setminus\{0\}$. No solution of $\mathrm{K15}((1,1,1,\nu),4)$ has $z_1-z_4$, $z_2-z_4$, $z_3-z_4$ proportional to $(1,\jj,\jj^2)$ or to $(1,\jj^2,\jj)$. \end{proposition} \begin{proof} After a normalization we reduce the equations to four polynomials in $\nu^2$ and two further unknowns, eliminate $\nu^2$, take resultants in one of the two unknowns, and show that the resulting polynomials in the other have no common root except $0$ and $1$. Exchanging bodies $2$ and $3$ reduces the second case to the first. We normalize $z=(1,\jj,\jj^2,0)$ by the symmetries of K15. Momentum gives $3c+\nu(c+d)=0$, so that $c\ne0$. We take $c=1$ by the remaining scaling. Let $s_j$ be the coordinate $p$ of the side opposite body $j\le3$ and $r_j=p_{j4}$, write $s_j=\sigma_0+\sigma_1\jj^{j-1}+\sigma_2\jj^{1-j}$, and put $k=3\nu^2$, $x=\sigma_1$ and $P=\sigma_0\sigma_2$. With $w_{il}=1/(p_{il}^2z_{il})$, the rows $F^w$ give $\nu r_j=-\sigma_0+\jj^{j-1}+\sigma_2\jj^{1-j}$. The rows $F^z$ then become three polynomial equations $Z_0,Z_1,Z_2$ in $(\sigma_0,\sigma_1,\sigma_2,k)$. The closure of the triangles in $w$ gives three more, $C_0,C_1,C_2$. We use $Z_0$, $Z_1$, $Z_2$, $C_0$ and $C_1$. Since $\prod_js_j\ne0$, they force $P\ne0$. \begin{enumerate}[label=(\arabic*)] \item With $t=\sigma_2$, $Z_2=(P^2-P+t^3)/t$. Hence $t^3=P(1-P)$ and $P\ne1$. \item Modulo $t^3=P(1-P)$, the equations $Z_1$, $C_1$, $Z_0$ and $C_0$ are, up to sign, products of powers of $t$ and $P$ with \[ Q-kK,\quad (1-P)(x^4+3P^2)-2PQx,\quad 1-6P+3P^2+P^3+k\big(PQ+(1-P)x^3+6P(1-P)x\big) \] and with a polynomial $C_0'$ of degree $1$ in $k$ $(\dagger)$, where $Q=P^2-P+1$ and $K=(1-P)x^2+Px+P(1-P)$. We call the three displayed polynomials (I), (II) and (III). A solution therefore gives $(P,x,k)$ with $P\notin\{0,1\}$ at which (I), (II), (III) and $C_0'$ vanish. \item Write (III) $=A_3+kB_3$ and $C_0'=A_0+kB_0$ with $A_3,B_3,A_0,B_0$ free of $k$. Put $G_3=KA_3+QB_3$ and $G_0=KA_0+QB_0$. Since (I) is $Q-kK$, we have $G_3=K\cdot\mathrm{(III)}+B_3\cdot\mathrm{(I)}$ and $G_0=K\,C_0'+B_0\cdot\mathrm{(I)}$ $(\dagger)$, so that $G_3$ and $G_0$ vanish at $(P,x)$. The resultants with (II) factor as \[ \mathrm{Res}_x(\mathrm{II},G_3)=P^3(P-1)^2H_{18}(P),\qquad\mathrm{Res}_x(\mathrm{II},G_0)=27P^6(P-1)^5(P^2-3P+1)^4H_{32}(P), \] where $H_{18}$ and $H_{32}$ are irreducible over $\Q$, of degrees $18$ and $32$, with leading coefficients $5$ and $100$ $(\dagger)$. Since $P\notin\{0,1\}$, $H_{18}(P)=0$ and $(P^2-3P+1)H_{32}(P)=0$. But $H_{18}\equiv1735830-4544451P$ modulo $P^2-3P+1$, a nonzero polynomial with a rational root, while the roots of $P^2-3P+1$ are irrational. Thus $H_{18}$ and $P^2-3P+1$ have no common root. Since $7$ divides neither leading coefficient and the Euclidean algorithm for $H_{32}$ and $H_{18}$ modulo $7$ ends in a nonzero constant, $H_{18}$ and $H_{32}$ have no common root either. Hence no such $P$ exists. \end{enumerate} The polynomials and the table of the Euclidean algorithm, which has $19$ rows, are in the supplementary material. \end{proof} \begin{corollary}\label{cor:d15equal} Let the masses be positive. If the three outer masses $m_i$, $i\in Q_j\setminus\{k\}$, are equal, no singular place of $C$ has type $\pc b_k(j)$. \end{corollary} \begin{proof} By Corollary~\ref{cor:equilateral} the leading cluster would be an equilateral solution of $\mathrm{K15}((1,1,1,\nu),4)$ after relabelling and scaling, with $\nu=m_k/m_a>0$. Proposition~\ref{prop:eqfull} excludes it. \end{proof} \subsection{Diagram 15 at order five}\label{ss:order5} On the six coordinates $(u_i,v_i)_{i\in Q'}$ of the outer bodies of a cluster with body $k$ at the origin, put \[ \Psi_c(u,v)=U_Q(u,v)+\frac c2\sum_{i\in Q'}m_iu_i . \] Let $m_i\ne0$ for $i\in Q$. A solution of $\mathrm{K15}(m_Q,k)$ then has $c\ne0$ and $M_Q\ne0$, by $cM_Q+dm_k=0$ with $d\ne0$. Since $\partial U_Q/\partial u_i=-\frac12m_iF^w_i$ and $\partial U_Q/\partial v_i=-\frac12m_iF^z_i$, for $c\ne0$ and $M_Q\ne0$ the critical points of $\Psi_c$ are the solutions of $\mathrm{K15}(m_Q,k)$ with body $k$ at the origin and the given $c$ (the equations of body $k$ follow from $\sum_im_iF_i=0$, with $d=-cM_Q/m_k$). Let $H$ be the Hessian of $\Psi_c$ at such a solution, and put $X=(\hat z,-3\hat w)$, $E=(\hat z,\hat w)$ and $\tilde m=(m_a,m_b,m_c)$. \begin{lemma}\label{lem:Hstruct} Let $m_i\ne0$ for $i\in Q$. At a solution of $\mathrm{K15}(m_Q,k)$ with body $k$ at the origin, $HX=0$, $HE=c\,(\tilde m,0)$, and $\sum_{i\in Q'}m_i\xi_{u_i}=0$ for every $\xi\in\ker H$. \end{lemma} \begin{proof} Since $U_Q$ is homogeneous of degree $-\frac12$ in $u$ and in $v$ separately, $\Psi_c(\lambda u,\lambda^{-3}v)=\lambda\Psi_c(u,v)$. The orbit of a critical point under this scaling therefore consists of critical points. So its tangent vector $X$ lies in $\ker H$. Next, $\sum(u\partial_u+v\partial_v)\Psi_c=-U_Q+\frac c2\sum m_iu_i$. Differentiating at a critical point, where $\nabla U_Q=-\frac c2(\tilde m,0)$, gives $HE=c(\tilde m,0)$. Finally $\sum u\partial_u\Psi_c-\sum v\partial_v\Psi_c=\frac c2\sum m_iu_i$. Differentiating along $\xi\in\ker H$ at a critical point, we obtain $\frac c2\sum m_i\xi_{u_i}=0$. Here $c\ne0$ as $m_k\ne0$. \end{proof} The configuration is \emph{nondegenerate} if $\ker H=\C X$, and has \emph{corank} $r$ if $\dim\ker H=r+1$. For a place of type $\pc b_k(j)$ we take $c=-m_j\hat w_j/M$ as in Lemma~\ref{lem:d15lead}, so that $V_0=\Psi_c$ when $y_{v_j}=\hat w_j$. For $\xi\in\C^6$ put $\kappa_3(\xi)=\sum_{i\in Q'}\frac{\partial V_3}{\partial y_{u_i}}(\bar y)\,\xi_{u_i}$. By Lemma~\ref{lem:d15expansion}, \[ \kappa_3(\xi)=-\tfrac12m_j(-\hat w_j)^{-1/2}\sum_{i\in Q'}m_ix_i^3\,\xi_{u_i},\qquad V_5(\bar y)=\tfrac12m_j\hat z_j(-\hat w_j)^{-1/2}\sum_{i\in Q'}m_ix_i^3 , \] and $\kappa_3(X)$ is proportional to $\sum m_ix_i^3\hat z_i=\sum m_ix_i=0$. \begin{theorem}\label{thm:order5} Let the masses satisfy $m_i\ne0$, $M\ne0$ and $M_Q\ne0$, and let $P$ be a singular place of type $\pc b_k(j)$. \begin{enumerate}[label=\textup{(\alph*)}] \item If the leading cluster configuration is nondegenerate, then $\sum_{i\in Q'}m_ix_i^3=0$. \item If it has corank one, then $\kappa_3$ vanishes on $\ker H$. \end{enumerate} \end{theorem} \begin{proof} We fix an outer body $a$. By the freedom described in \S\ref{ss:expansion} we may assume $y_{v_j}(\tau)\equiv\hat w_j$ and $y_{u_a}(\tau)\equiv\hat z_a$: the conditions are $\mu^{-1}\varphi^4y_{v_j}=\hat w_j$ and $\mu\varphi^{-2}y_{u_a}=\hat z_a$, which determine $\varphi^2$ and then $\mu$. Write $y(\tau)-\bar y=(\delta(\tau),\delta_{u_j}(\tau))$ with $\delta$ the cluster part, so that $\delta_{u_a}=0$. Let $S=\{\xi\in\C^6:\xi_{u_a}=0\}$. Since $X\in\ker H$ and $X_{u_a}=\hat z_a\ne0$, the kernel of the restriction of the form $H$ to $S$ is $K_S=\ker H\cap S$, of dimension equal to the corank. Choose a complement $W$ of $K_S$ in $S$ and write $\delta=\xi(\tau)+\eta(\tau)$ with $\xi(\tau)\in K_S$, $\eta(\tau)\in W$. Let $\alpha$ be the valuation of $\xi(\tau)$ ($\alpha=\infty$ if $\xi\equiv0$). By Lemmas~\ref{lem:identities} and \ref{lem:d15expansion}, the cluster components of $\tau^2\nabla_yF=0$ are $\nabla\Psi_c(\bar y+\delta)+\tau^3\nabla V_3(y)+O(\tau^5)=0$, because $V_2$ does not involve the cluster and $V_0=\Psi_c$ along the normalized place. Taylor expansion gives $H\delta+R(\delta)+\tau^3\nabla V_3(y)+O(\tau^5)=0$ with $R(\delta)=O(|\delta|^2)$. Taking the components in $S^*$ and using that $H|_S$ maps $W$ isomorphically onto its image, the valuation of $\eta$ is at least the minimum of $3$, $2\alpha$, $\alpha+\val\eta$ and $2\val\eta$. Hence \begin{equation}\label{eq:etaval} \val\eta\ge\min(3,2\alpha). \end{equation} Pairing the same equation with $\xi_0\in K_S$ and using $\xi_0^{\mathsf T}H=0$ gives \begin{equation}\label{eq:bif} \xi_0\cdot R(\delta)+\tau^3\,\xi_0\cdot\nabla V_3(y(\tau))+O(\tau^5)=0 . \end{equation} Since $y_{v_j}$ is constant, $V_3(y(\tau))=V_3(\bar y)+\kappa_3(\delta)+O(|\delta|^2)$ with $V_3(\bar y)=0$. \eqref{eq:qid} with $q=5$ reads $[V_1]_4+2[V_2]_3+3[V_3]_2+4[V_4]_1+5V_5(\bar y)=0$. By \eqref{eq:V2flat} this becomes \begin{equation}\label{eq:q5} 3\big[V_3(y(\tau))\big]_2+5V_5(\bar y)=0 . \end{equation} (a) Since $K_S=0$ in this case, $\delta=\eta$ has valuation at least $3$ by \eqref{eq:etaval}. Hence $[V_3(y(\tau))]_2=0$. Then \eqref{eq:q5} gives $V_5(\bar y)=0$, i.e.\ $\sum m_ix_i^3=0$. (b) Here $K_S=\C\xi_0$ and $\xi(\tau)=a(\tau)\xi_0$ with $a(\tau)=a_\alpha\tau^\alpha+\dots$, $a_\alpha\ne0$. Suppose $\kappa_3(\xi_0)\ne0$. If $\alpha<2$, we apply \eqref{eq:qid} with $q=3+\alpha$: the terms $n=1,4$ vanish, the term $n=2$ is $2[V_2]_{1+\alpha}=0$ by \eqref{eq:V2flat}, and the terms $n\ge5$ have negative index. This leaves $[V_3(y(\tau))]_\alpha=0$. But by \eqref{eq:etaval} the valuations of $\eta$ and of $|\delta|^2$ exceed $\alpha$. Therefore $[V_3(y(\tau))]_\alpha=a_\alpha\kappa_3(\xi_0)\ne0$, a contradiction. Hence $\alpha\ge2$. Then $\val\eta\ge3$ and $\val R(\delta)\ge4$. The coefficient of $\tau^3$ in \eqref{eq:bif} is then $\xi_0\cdot\nabla V_3(\bar y)=\kappa_3(\xi_0)$, which must vanish, again a contradiction. Thus $\kappa_3(\xi_0)=0$. As $\kappa_3(X)=0$ and $\ker H=\C X\oplus\C\xi_0$, the form $\kappa_3$ vanishes on $\ker H$. \end{proof} The next statement says that three nonzero complex numbers with positive weights cannot have vanishing weighted sums of $x$, $x^3$ and $x^{-2}$; for four numbers this fails (remark (c) after the proof). Theorems~\ref{thm:order3} and~\ref{thm:order5} need only $m_i,M,M_Q\ne0$, but this statement needs positive masses. Corollary~\ref{cor:degenerate} and Proposition~\ref{prop:D14N} use it. \begin{proposition}\label{prop:P11} Let $m_a,m_b,m_c>0$. There is no $x\in(\C^*)^3$ with $\sum m_ix_i=\sum m_ix_i^3=\sum m_ix_i^{-2}=0$. \end{proposition} \begin{proof} We say that three directions in $\R/2\pi\Z$ \emph{surround} $0$ if they do not lie in a closed half-circle, that is, if each of the three cyclic gaps between them is less than $\pi$. We use two facts. (A) Let $w_1,w_2,w_3\in\C^*$ and $\mu_1,\mu_2,\mu_3>0$ with $\sum\mu_iw_i=0$. Then the arguments of the $w_i$ surround $0$, or all $w_i$ lie on one line through $0$. Indeed, if the arguments do not surround $0$, all $w_i$ lie in a closed half-plane $\{\mathrm{Re}(\bar uz)\ge0\}$ with $|u|=1$. Then $0=\sum\mu_i\mathrm{Re}(\bar uw_i)$ is a sum of nonnegative terms, so that all $w_i$ lie on the line $iu\R$. (B) There are no $\theta_1,\theta_2,\theta_3$ such that each of the sets $\{\theta_i\}$, $\{2\theta_i\}$ and $\{3\theta_i\}$ surrounds $0$. Indeed, the three conditions are unchanged by a common shift of the $\theta_i$ and by relabelling. If $\{\theta_i\}$ surrounds $0$, we may therefore take $\theta=(0,\pi-s,\pi+t)$ with $s,t>0$ and $s+t<\pi$, so that the gaps are $\pi-s$, $s+t$ and $\pi-t$. Then $2\theta\equiv(0,-2s,2t)$, with gaps $2t$, $2\pi-2s-2t$ and $2s$. The set $\{2\theta_i\}$ surrounds $0$ exactly when $s<\frac\pi2$, $t<\frac\pi2$ and $s+t>\frac\pi2$. Assume this. Then $3s,3t\in(0,\frac{3\pi}2)$. After a rotation by $\pi$ the set $\{3\theta_i\}$ is $\{\pi,-3s,3t\}$. If $3s<\pi$ and $3t<\pi$, its gaps are $\pi-3t$, $\pi-3s$ and $3s+3t>\frac{3\pi}2$. If $3t\ge\pi>3s$, it lies in $[\pi,2\pi]$. If $3s\ge\pi>3t$, it lies in $[0,\pi]$. If $3s\ge\pi$ and $3t\ge\pi$, it lies in $[\frac\pi2,\frac{3\pi}2]$. In no case does it surround $0$. Now let $x$ be a solution, and put $\theta_i=\arg x_i$. \begin{enumerate}[label=(\arabic*)] \item $\{\theta_i\}$ surrounds $0$. This is (A) with $w=x$: if all $x_i$ lay on a line through $0$, then $x_i=\lambda r_i$ with $r_i\in\R^*$, and $\sum m_ix_i^{-2}=\lambda^{-2}\sum m_ir_i^{-2}\ne0$. \item $\{2\theta_i\}$ surrounds $0$. This is (A) with $w=x^{-2}$, whose arguments are the $-2\theta_i$ (a set surrounds $0$ if and only if its mirror image does): if all $x_i^{-2}$ lay on a line through $0$, the $\theta_i$ would lie among four directions at right angles, and three of these lie in a closed half-circle, against (1). \item $\{3\theta_i\}$ surrounds $0$. This is (A) with $w=x^3$: if all $x_i^3$ lay on a line through $0$, then $\theta_i=\psi+n_i\frac\pi3$ with $n_i\in\Z/6$. By (1) the cyclic gaps between the $n_i$ are less than $3$. As they are also positive with sum $6$, they all equal $2$. So the $n_i$ have the same parity, all $x_i^3$ lie on one ray, and $\sum m_ix_i^3\ne0$. \end{enumerate} Steps (1)--(3) contradict (B). \end{proof} \emph{Remarks.} (a) A second proof eliminates $x$. With $x=(1,t,-(m_a+m_bt)/m_c)$, the conditions are $Q(t)=0$, with $Q$ as in Lemma~\ref{lem:badshape}, and $C(t)=m_c^2(m_a+m_bt^3)-(m_a+m_bt)^3=0$. A solution therefore gives a common root of $Q$ and $C$, so that their resultant \[ \Res_t(Q,C)=m_a^2m_b^2m_c^4(m_a+m_b+m_c)^2\,P_{11}(m_a,m_b,m_c) \] vanishes. But $P_{11}$ is a symmetric polynomial of degree $11$ that is positive on the positive orthant (Appendix~\ref{app:P11}). (b) The positivity of all three masses is used in (A) and in steps (1) and (3). The statement has no uniform version on the closed simplex: for $m=(\frac12,\frac12,\varepsilon)$ and $x=(1,-1,i\sqrt\varepsilon)$ we have $\sum m_ix_i^{-2}=0$, while $\sum m_ix_i=i\varepsilon^{3/2}$ and $\sum m_ix_i^3=-i\varepsilon^{5/2}$ tend to $0$ with $\varepsilon$. Correspondingly, $P_{11}(\frac12,\frac12,0)=0$. (c) For four points the analogous statement fails: equal weights at $1,i,-1,-i$ satisfy all three conditions. In the proof above, (B) is where the number three enters. Note that for equal outer masses the bad shape is $(1,\jj,\jj^2)$ and $\sum m_ix_i^3=3m_a\ne0$, in accordance with Proposition~\ref{prop:P11}. \begin{corollary}\label{cor:degenerate} Let $m\in(\R_{>0})^5$. At a singular place of type $\pc b_k(j)$ the leading cluster configuration is degenerate, and there is $\xi\in\ker H\setminus\C X$ with $\kappa_3(\xi)=0$. \end{corollary} \begin{proof} A nondegenerate configuration would give a bad shape with $\sum m_ix_i^3=0$ (Theorems~\ref{thm:order3} and \ref{thm:order5}(a)), which is excluded by Proposition~\ref{prop:P11}. At corank one, $\kappa_3$ vanishes on $\ker H$ by Theorem~\ref{thm:order5}(b). At corank at least two, $\kappa_3$ is a linear form on $\ker H/\C X$, a space of dimension at least two, and has a nonzero kernel there. \end{proof} \subsection{Degenerate leading configurations}\label{ss:residual} For a function $\lambda$ on the six pairs of $Q$, let $L_\lambda$ be the grounded Laplacian of the complete graph on $Q$ with edge weights $\lambda_{il}$, grounded at $k$: the $3\times3$ matrix with $(L_\lambda\xi)_i=\sum_{l\ne i}\lambda_{il}(\xi_i-\xi_l)$ for $i\in Q'$, where $\xi_k=0$. At a solution of K15 put $q_{il}=m_im_lp_{il}^3$ and $\varrho_{il}=z_{il}/w_{il}$. \begin{lemma}\label{lem:laplacian} At a solution of $\mathrm{K15}(m_Q,k)$ with body $k$ at the origin, \[ H=\begin{pmatrix}L_{3q/(4\varrho)}&L_{q/4}\\ L_{q/4}&L_{3q\varrho/4}\end{pmatrix}, \] and the system reads $L_qz=0$, $L_{q\varrho}w=0$ and $L_qw=c\,\tilde m$. \end{lemma} \begin{proof} For one pair, $p=(zw)^{-1/2}$ and, with $\alpha=\zeta/z$, $\beta=\eta/w$ for a displacement $(\zeta,\eta)$, $p(z+s\zeta,w+s\eta)=p(1+s\alpha)^{-1/2}(1+s\beta)^{-1/2}$, whose second derivative at $s=0$ is $\frac p4(3\alpha^2+2\alpha\beta+3\beta^2)$. With $p^2zw=1$ the three coefficients are $\frac34q/\varrho$, $\frac14q$ and $\frac34q\varrho$ times $\zeta^2$, $2\zeta\eta$, $\eta^2$ (divided by $m_im_l$). Summing over the pairs, we obtain $H$. The linear term of $\Psi_c$ does not contribute. The system is $(L_qz)_i=m_iF^z_i$ and $(L_qw)_i=m_iF^w_i$, and $q_{il}\varrho_{il}w_{il}=q_{il}z_{il}$. \end{proof} \begin{theorem}\label{thm:residual} Let the outer masses be positive and $m_k\ne0$, let $(\hat z,\hat w)$ solve $\mathrm{K15}(m_Q,k)$ with body $k$ at the origin, suppose $\sum_{i\in Q'}m_ix_i=0$ with $x_i^2=1/\hat z_i$, and let $\xi\in\ker H\setminus\C X$ satisfy $\sum_{i\in Q'}m_ix_i^3\xi_{u_i}=0$. Then one of the following holds: \begin{enumerate}[label=\textup{(\roman*)}] \item the outer masses are equal and $\{\hat z_i\}=\lambda\{1,\jj,\jj^2\}$; \item \textup{(G2)} $\ker L_q=\C\hat z$ and $L_{q\varrho}\hat z=0$; \item \textup{(G1)} $L_q=\sigma\,\tilde m\tilde m^{\mathsf T}$ for some $\sigma$, and $\rank L_{q\varrho}=1$. \end{enumerate} \end{theorem} \begin{proof} We write $\xi=(\zeta,\eta)$. The linear forms $\ell_1(\zeta)=\sum m_i\zeta_i$ and $\ell_2(\zeta)=\sum m_ix_i^3\zeta_i$ vanish on $\zeta$ (Lemma~\ref{lem:Hstruct} and the hypothesis on $\xi$) and on $\hat z$ ($\sum m_i\hat z_i=0$ and $\sum m_ix_i^3\hat z_i=\sum m_ix_i=0$). If they are proportional, then $x_a^3=x_b^3=x_c^3$. In that case, since the $\hat z_i=x_i^{-2}$ are distinct, $x$ is proportional to $(1,\jj,\jj^2)$ up to order, and $\sum m_ix_i=0$ with real positive $m_i$ forces equal masses: case (i). If they are not proportional, their common kernel is $\C\hat z$, which gives $\zeta=\gamma\hat z$, and $\xi-\gamma X=(0,\eta')$ with $\eta'=\eta+3\gamma\hat w\neq0$ lies in $\ker H$. By Lemma~\ref{lem:laplacian}, $(0,\eta')\in\ker H$ means $L_q\eta'=L_{q\varrho}\eta'=0$. Now $\hat z\in\ker L_q$ and $\hat w\in\ker L_{q\varrho}$. Hence $\rank L_q\le2$. If $\rank L_q=2$, then $\eta'\in\C\hat z$ and $L_{q\varrho}\hat z=0$: case (ii). If $\rank L_q\le1$, we write $L_q=\sigma vv^{\mathsf T}$. Then $L_q\hat w=\sigma v(v\cdot\hat w)=c\tilde m\ne0$ forces $v\in\C\tilde m$ and $\ker L_q=\tilde m^\perp$. If $\rank L_{q\varrho}=2$, then $\eta'\in\C\hat w$ and $\tilde m\cdot\hat w=0$, contradicting $\sigma(\tilde m\cdot\hat w)\tilde m=c\tilde m$ with $c\ne0$. As $L_{q\varrho}\ne0$, $\rank L_{q\varrho}=1$: case (iii). \end{proof} Case (iii) is excluded by the following statement, which does not use the bad shape. \begin{proposition}\label{prop:G1} Let $(\hat z,\hat w)$ solve $\mathrm{K15}(m_Q,k)$ with body $k$ at the origin, where $m_a$, $m_b$, $m_c$, $m_k$ and $M_{Q'}=m_a+m_b+m_c$ are nonzero, and suppose that $\rank L_q\le1$ and $\rank L_{q\varrho}\le1$. Then $(m_k/M_{Q'})^2=-1/s^3$ for a number $s$ with $s^3=\pm3\sqrt{-3}$. In particular $m_k/M_{Q'}$ is not real, and the masses are not all positive. \end{proposition} \begin{proof} We use the rank conditions to write $q_{il}$ and $\varrho_{il}$ as products of factors attached to~$i$ and to~$l$, deduce from the rows $D_{il}$ that the three points $1/\hat z_i$ form an equilateral triangle, and compute $s$ and $(m_k/M_{Q'})^2$ from this triangle. (1) Since $L_q\hat w=c\tilde m\ne0$, $\rank L_q=1$ and $L_q=\sigma vv^{\mathsf T}$. Then $\sigma v(v\cdot\hat w)=c\tilde m$ gives $L_q=\sigma\tilde m\tilde m^{\mathsf T}$ with $\sigma\ne0$. Reading off the entries, we obtain $q_{il}=-\sigma m'_im'_l$ on all six pairs, where $m'=(m_a,m_b,m_c,-M_{Q'})$. Thus $p_{il}^3=-\sigma$ on the outer pairs and $p_{ik}^3=\sigma M_{Q'}/m_k$. The off-diagonal entries $-q_{il}\varrho_{il}$ of $L_{q\varrho}$ are nonzero. Hence $L_{q\varrho}=\sigma'v'v'^{\mathsf T}$ with $\sigma'\ne0$, and in the same way $q_{il}\varrho_{il}=-\sigma'\tilde v_i\tilde v_l$ with $\tilde v=(v'_a,v'_b,v'_c,-v'_a-v'_b-v'_c)$, all of whose entries are nonzero. Hence $\varrho_{il}=\gamma\beta_i\beta_l$ on all six pairs, with $\gamma=\sigma'/\sigma$ and $\beta_i=\tilde v_i/m'_i\ne0$. (2) We put $\psi_i=1/\hat z_i$. On the pairs through $k$, $\hat z_i=\gamma\beta_i\beta_k\hat w_i$. Inserting this into $\varrho_{il}=\gamma\beta_i\beta_l$ on the outer pairs gives $(1-\beta_i/\beta_k)\psi_i=(1-\beta_l/\beta_k)\psi_l$. If $\beta_i=\beta_k$ for one $i$, all the $\beta$ are equal, $\varrho$ is constant and $\hat w$ is a multiple of $\hat z$, which is excluded. So the common value $\psi_0$ is nonzero, $\beta_i=\beta_k(\psi_i-\psi_0)/\psi_i$, and $1/\hat w_i=\gamma\beta_i\beta_k\psi_i=A(\psi_i-\psi_0)$ with $A=\gamma\beta_k^2$. (3) Since the $p_{ik}^3$ are equal, $p_{ik}=\varepsilon_i\pi_1$ with $\varepsilon_i^3=1$. The rows $D_{ik}$, $p_{ik}^2\hat z_i\hat w_i=1$, then read $\psi_i(\psi_i-\psi_0)=\varkappa\varepsilon_i^2$ with $\varkappa=\pi_1^2/A$. (4) Likewise $p_{il}=\varepsilon_{il}\pi_0$ on the outer pairs. As $\hat z_{il}\hat w_{il}=(\psi_i-\psi_l)^2/\big(A\psi_i\psi_l(\psi_i-\psi_0)(\psi_l-\psi_0)\big)$, the rows $D_{il}$ read $(\psi_i-\psi_l)^2=\varepsilon_{il}\varepsilon_i^2\varepsilon_l^2\,\varkappa/\lambda$ with $\lambda=\pi_0^2/\pi_1^2$. So the squared sides of the triangle $\psi_a,\psi_b,\psi_c$ agree up to cube roots of unity: $\theta=(\psi_c-\psi_a)/(\psi_b-\psi_a)$ satisfies $\theta^6=(\theta-1)^6=1$ with $\theta\ne0,1$, whence $|\theta|=|\theta-1|=1$ and $\theta=e^{\pm i\pi/3}$. The triangle is equilateral: $\psi_i=h+R\varpi_i$ with $\{\varpi_a,\varpi_b,\varpi_c\}=\{1,\jj,\jj^2\}$ and $R\ne0$. (5) The polynomial $\Phi(\varpi)=(h+R\varpi)(h+R\varpi-\psi_0)=R^2\varpi^2+R(2h-\psi_0)\varpi+h(h-\psi_0)$ takes the values $\Phi(\varpi_i)=\varkappa\varepsilon_i^2$. Hence $R^2=\frac13\sum_i\Phi(\varpi_i)\varpi_i=\varkappa s/3$ with $s=\sum_i\varepsilon_i^2\varpi_i$, a sum of three cube roots of unity. If the three terms are distinct, $s=0$ and $R=0$, which is excluded. If they are all equal to some $\alpha$, then $\Phi(\varpi)=\varkappa\alpha\varpi^2$ at three points and hence identically, so $R(2h-\psi_0)=h(h-\psi_0)=0$, which forces $\psi_0=0$, excluded. So exactly two terms are equal and $s^3=\pm3\sqrt{-3}$. (6) By (4), $\lambda^3=\varkappa^3/(\psi_b-\psi_a)^6$, and $(\psi_b-\psi_a)^2=R^2(\varpi_b-\varpi_a)^2=-3R^2\varpi_a\varpi_b$, so that $(\psi_b-\psi_a)^6=-27R^6=-\varkappa^3s^3$ and $\lambda^3=-1/s^3$. (7) We choose $\pi_0=p_{ab}$ and $\pi_1=p_{ak}$. Then (1) gives $(\pi_0/\pi_1)^3=-m_k/M_{Q'}$. So $(m_k/M_{Q'})^2=\lambda^3=-1/s^3$, a nonzero imaginary number. \end{proof} In case (ii), $\hat z$ and $\hat w$ lie in the kernel of $L_{q\varrho}$. They are independent, since $\hat w=\lambda\hat z$ would give $L_q\hat w=0$. Hence $L_{q\varrho}=\sigma'uu^{\mathsf T}$ with $\sigma'\ne0$ and $u=\hat z\times\hat w$. That is, $q_{il}\varrho_{il}=-\sigma'\tilde u_i\tilde u_l$ on the six pairs, where $\tilde u=(u_a,u_b,u_c,-u_a-u_b-u_c)$ is the affine dependence of the four points $(\hat z_i,\hat w_i)\in\C^2$. Then $q_{il}=-\sigma'\tilde u_i\tilde u_lw_{il}/z_{il}$. With this, $L_q\hat w=c\tilde m$ reads \begin{equation}\label{eq:G2mass} c\,m_i=-\sigma'\tilde u_iS_i,\qquad S_i=\sum_{l\ne i}\tilde u_l\,w_{il}^2/z_{il}\qquad(i\in Q'). \end{equation} \begin{lemma}\label{lem:G2mass} For any four points $(z_i,w_i)$, $i\in Q$, with $(z_k,w_k)=(0,0)$ and distinct $z_i$, and with $\tilde u$ and $S_i$ formed from them as above, $\sum_{i\in Q}\tilde u_iS_iz_i^j=0$ for $j=0,1,2$. In case (ii), therefore, $\sum_{i\in Q'}m_i\hat z_i^2=0$. \end{lemma} \begin{proof} When we group the double sum by pairs, the pair $\{i,l\}$ contributes $\tilde u_i\tilde u_lw_{il}^2(z_i^j-z_l^j)/(z_i-z_l)$, that is $0$, $\tilde u_i\tilde u_lw_{il}^2$ and $\tilde u_i\tilde u_lw_{il}^2(z_i+z_l)$ for $j=0,1,2$. Since $\sum_i\tilde u_i=0$, any two functions $f$, $g$ on $Q$ satisfy $\sum_{\{i,l\}}\tilde u_i\tilde u_l(f_i-f_l)(g_i-g_l)=-\big(\sum_i\tilde u_if_i\big)\big(\sum_i\tilde u_ig_i\big)$. For $j=1$ we take $f=g=w$. The result vanishes because $\sum_i\tilde u_iw_i=0$. For $j=2$ we write $(w_i-w_l)(z_i+z_l)=\big((wz)_i-(wz)_l\big)+(w_iz_l-w_lz_i)$. The first part gives $-\big(\sum\tilde u_iw_i\big)\big(\sum\tilde u_iw_iz_i\big)=0$. In the second part the pairs through $k$ contribute nothing. On the outer pairs $w_bz_c-w_cz_b=-u_a$ and cyclically, so that the second part equals $-\tilde u_a\tilde u_b\tilde u_c\big((w_b-w_c)+(w_c-w_a)+(w_a-w_b)\big)=0$. The last statement follows from \eqref{eq:G2mass} and $\hat z_k=0$. \end{proof} Since a vector orthogonal to $\hat z$ and to $(\hat z_i^2)$ is proportional to $(1/P'(\hat z_i))_{i\in Q'}$ with $P(t)=t\prod_{i\in Q'}(t-\hat z_i)$, by partial fractions, Lemma~\ref{lem:G2mass} says that in case (ii) the outer masses are proportional to $(1/P'(\hat z_i))_{i\in Q'}$. With the bad shape, this leaves only the equilateral configuration. \begin{theorem}\label{thm:wtriangle} Let $m_a,m_b,m_c>0$, and let $z_a,z_b,z_c$ be distinct nonzero complex numbers with $\sum_im_iz_i=\sum_im_iz_i^2=0$ and $\sum_im_ix_i=0$ for some $x_i$ with $x_i^2z_i=1$. Then $m_a=m_b=m_c$ and $\{z_a,z_b,z_c\}=\lambda\{1,\jj,\jj^2\}$ for some $\lambda\in\C^*$. \end{theorem} \begin{proof} The plane $m^\perp=\{v\in\C^3:\sum_im_iv_i=0\}$ contains $z$ and $z^2=(z_i^2)$, which are independent because the $z_i$ are distinct and nonzero. It is closed under complex conjugation because the $m_i$ are real. Hence $\bar z=\alpha z+\beta z^2$. If $\beta=0$, then $|\alpha|=1$ and the $z_i$ lie on a line through $0$, so that $\sum m_iz_i^2\ne0$. Thus $\beta\ne0$. The scaling $z\mapsto\lambda z$, $x\mapsto\lambda^{-1/2}x$ preserves the hypotheses and multiplies $\beta$ by $\bar\lambda/\lambda^2$. So we may take $\beta=1$: $\bar z_i=\alpha z_i+z_i^2$. Conjugating and substituting gives $z_i=|\alpha|^2z_i+(\alpha^2+\bar\alpha)z_i^2+2\alpha z_i^3+z_i^4$. Hence the $z_i$ are the three roots of \[ g_\alpha(t)=t^3+2\alpha t^2+(\alpha^2+\bar\alpha)t+|\alpha|^2-1. \] Thus $z_a+z_b+z_c=-2\alpha$ and $z_az_bz_c=1-|\alpha|^2$. For $i\ne l$ this gives $z_i\bar z_l+\bar z_iz_l=z_iz_l(2\alpha+z_i+z_l)=-z_az_bz_c=|\alpha|^2-1$. Therefore \[ 0=\Big|\sum_im_iz_i\Big|^2=\sum_im_i^2|z_i|^2-(1-|\alpha|^2)\sum_{i0$. Thus $\sin3\varphi=0$. For $e^{3i\varphi}=1$, $E=-\rho^7+6\rho^6+5\rho^5+12\rho^4+25\rho^3+6\rho^2+3\rho+8>0$. For $e^{3i\varphi}=-1$, $E=(1-\rho)^2(\rho^5+8\rho^4+10\rho^3+24\rho^2+13\rho+8)>0$. This contradicts $E=0$. Therefore $\alpha=0$. Then $g_0(t)=t^3-1$, the $z_i$ are the cube roots of unity, and $\sum m_iz_i=0$ with real $m_i$ forces $m_a=m_b=m_c$. \end{proof} The proof of Theorem~\ref{thm:wtriangle} uses three small computations: the reduction of $t^3(\gamma+\delta t)^2-1$ modulo the cubic $g_\alpha$, the resultant of the two binary quadratic forms $q_2$ and $q_1$, and the polar form of $N$. \begin{proposition}\label{prop:G2} A configuration of case \textup{(ii)} of Theorem~\ref{thm:residual} with a bad shape and positive outer masses, as in that theorem, has equal outer masses and $\{\hat z_i\}=\lambda\{1,\jj,\jj^2\}$. \end{proposition} \begin{proof} The outer masses are positive and the $\hat z_i$ distinct and nonzero. Lemma~\ref{lem:G2mass} gives $\sum m_i\hat z_i=\sum m_i\hat z_i^2=0$. The bad shape gives $\sum m_ix_i=0$ with $x_i^2\hat z_i=1$. Theorem~\ref{thm:wtriangle} applies. \end{proof} Note that the bad shape is needed in Proposition~\ref{prop:G2}: for all positive outer masses there are distinct nonzero $z_i$ with $\sum m_iz_i=\sum m_iz_i^2=0$. So the conclusion of Lemma~\ref{lem:G2mass} alone does not exclude them. \begin{corollary}\label{cor:nod15} For every $m\in(\R_{>0})^5$, no singular place of an irreducible curve $C\subseteq Y_m$ with $U\not\equiv0$ on $C$ has type $\pc b_k(j)$. Hence the pieces $\pc{d15}(j,k)$ may be removed from $\mathrm{pieces}(S(m))$ in Theorem~\ref{thm:types}, for every $m$, $j$ and $k$. \end{corollary} \begin{proof} By Theorem~\ref{thm:order3} the leading cluster has a bad shape. By Corollary~\ref{cor:degenerate} it is degenerate with a vector $\xi$ as in Theorem~\ref{thm:residual}, since $\kappa_3(\xi)$ is a nonzero multiple of $\sum_{i\in Q'}m_ix_i^3\xi_{u_i}$. Case (iii) is excluded by Proposition~\ref{prop:G1}. In case (ii), Proposition~\ref{prop:G2} gives the configuration of case (i). Corollary~\ref{cor:d15equal}, which applies as all masses are positive, excludes case (i). \end{proof} \subsection{Diagram 14}\label{ss:d14} Let $P$ have type $\sigma_k(j)(t)$ with $0-\sigma$ for $i\ne k$, and $p_{kj}^3w_{kj}$ has exponent $-\sigma$ with coefficient $-\hat p_{kj}^3\hat w_j$. Thus $C^w_i$ has the coefficient $m_j\hat w_j\big(-\frac1M+[i=k]\hat p_{kj}^3\big)$ at exponent $-\sigma$. Since $\sum_im_iC^w_i=0$, we obtain $\hat p_{kj}^3=M_Q/(m_kM)$. On the right side the term $i=k$ vanishes. The rest has the coefficient $-\frac{m_j}M\hat w_j\sum_{i\in Q'}m_i\hat z_i$ at exponent $1-\sigma<2$, which therefore vanishes. \end{proof} \input{sec/s5_d14} \begin{remark}\label{rem:d14a} The identities of \S\ref{ss:expansion} also apply at a place of type $\pc u_0(j)$, the piece $\pc{d14a}(j)$, where the fifth body stays at distance of order one from the cluster while the cluster shrinks. There the pairings of $\nabla_yV$ with the $z$- and $w$-scalings of the cluster, together with the constancy of $V$, give at first order a signed dipole condition $\sum_ls_lm_l(\hat z_l-\hat z_G)=0$ when the leading coefficients $\hat p_{jl}$ have unequal signs $s_l$, a centred quadrupole condition $\sum_lm_l(\hat z_l-\hat z_G)^2=0$ when they have equal signs, the same conditions in $w$ at the next order, and in the boundary case a single scalar $g^{\mathsf T}H^+g$. These are the first-order conditions of the analysis of $\pc{d14a}$ in \S\ref{sec:curves}. The identities of \S\ref{ss:expansion} therefore do not exclude $\pc{d14a}$ at every positive mass: unlike the conditions of Theorems~\ref{thm:order3} and \ref{thm:order5}, these conditions hold at some symmetric K14 configurations with positive masses, for instance centrally symmetric ones. We treat the pieces $\pc{d14a}$ instead by the point certificates of \S\ref{sec:points}. \end{remark} \input{sec/s5_d9}