%
%

\subsection{Diagram 9}\label{ss:d9}

We now use the expansion of \S\ref{ss:expansion} to exclude diagram~$9$ at all positive masses.

\begin{theorem}\label{thm:D9}
Let $m\in(\R_{>0})^5$, and let $C\subseteq Y_m$ be an irreducible curve on which some $r_{kl}^2$ is not constant and $U\not\equiv0$. No singular place of $C$ has diagram~$9$. Hence the pieces $\pc{d9}(T;ax,by)$ may be removed from $\mathrm{pieces}(S(m))$ in Theorem~\ref{thm:types}, for all $m$, $T$, $ax$ and $by$.
\end{theorem}

Let $P$ be a singular place of diagram~$9$. After a relabelling of the bodies and, if necessary, an exchange of the colours, the triangle is $T=\{1,2,3\}$, the $w$-edges are $14$ and $25$, and $x(P)=\pc n(\{1,2,3\};14,25)$ (Table~\ref{tab:types}). By Appendix~\ref{app:types}, $(\zeta_{kl},\omega_{kl})$ is $(1,1)$ on $12$, $13$, $23$ and $(2,-2)$ on $14$, $25$. On the other five pairs $\zeta_{kl}=1$, by $z_{45}=z_{41}+z_{12}+z_{25}$, $z_{15}=z_{12}+z_{25}$ and the like, and $\omega_{kl}=2x_{kl}-1$, which is $-1$ on $45$ and $-2$ on $15$, $24$, $34$, $35$.

For $h\in\C^5$ let $h^*=\frac{m_4h_4+m_5h_5}{m_4+m_5}-\frac{m_1h_1+m_2h_2+m_3h_3}{M_T}$. We use the positions relative to body~$3$, $u_i=z_i-z_3$ and $v_i=w_i-w_3$, in the following combinations: the \emph{coarse} coordinates $u_1$, $u_2$, $v_1$, $v_2$, $\Omega=w^*$, and the \emph{fine} coordinates $u_{41}$, $u_{52}$, $v_{45}$, where $u_{kl}=u_k-u_l$ and $v_{kl}=v_k-v_l$. They form an invertible linear function of $(u_1,u_2,u_4,u_5)$ and one of $(v_1,v_2,v_4,v_5)$. Their exponents are $1$, $1$, $1$, $1$, $-2$ and $2$, $2$, $-1$. For $\Omega$ this follows from $w_{14}=-\Omega-\frac{m_5}{m_4+m_5}v_{45}+\frac{m_2v_{12}+m_3v_1}{M_T}$, since $w_{14}$ has exponent $-2$ and the last two terms have exponents $-1$ and $1$. The coordinates of \S\ref{ss:expansion} do not serve here. For each body $o$ one of the pairs $14$, $25$ avoids $o$, the positions of its two bodies relative to $o$ have exponent $1$ and the same leading coefficient, and so the corresponding $p_{kl}$ is not holomorphic in $y$ near $\bar y$. We therefore state Lemma~\ref{lem:identities} for linear coordinates of the kind above.

\begin{lemma}\label{lem:adapted}
Let $q_1,\dots,q_8$ be coordinates such that $(q_1,\dots,q_4)$ is an invertible linear function of $(u_1,\dots,u_4)$ and $(q_5,\dots,q_8)$ one of $(v_1,\dots,v_4)$, and suppose that no $q_i$ vanishes identically on $C$. Then, after the rotation of Lemma~\ref{lem:place}, $q_i=\tau^{2e(q_i)}y_{q_i}$ with $y_{q_i}(0)\ne0$. Lemma~\ref{lem:identities} holds in the coordinates $q$, and the reparametrizations and rotations of \S\ref{ss:expansion} change neither the leading point nor the identities of that lemma.
\end{lemma}

\begin{proof}
Since each $q_i$ is of $z$-type or of $w$-type, Lemma~\ref{lem:place} gives the first statement and a rotation by $\mu(\tau)$ multiplies $y_{q_i}$ by $\mu(\tau)^{\pm1}$. The proof of Lemma~\ref{lem:identities} uses three facts: $\nabla V=0$ at the points of $X_m$ in the coordinates used, $V$ is constant on $C$, and $F(\lambda\tau,\lambda^{-2e}y)=F(\tau,y)$. The last two do not depend on the coordinates. The first holds in the coordinates $q$, because $\nabla_{(u,v)}V=0$ on $X_m$ (\S\ref{ss:expansion}) and $q$ is an invertible linear function of $(u,v)$.
\end{proof}

We write the coefficients $V_n$ of $F(\tau,y)=V(\tau^{2e}y)$ as functions of the coordinates, with $u_1$ standing for $y_{u_1}$ and so on, and put $u_{12}=u_1-u_2$, $v_{12}=v_1-v_2$, $\nu=M_T(m_4+m_5)/M$, $\nu_{45}=m_4m_5/(m_4+m_5)$,
\begin{gather*}
U_T=m_1m_2(u_{12}v_{12})^{-1/2}+m_1m_3(u_1v_1)^{-1/2}+m_2m_3(u_2v_2)^{-1/2},\\
L(u)=\frac{m_4u_1+m_5u_2}{m_4+m_5}-\frac{m_1u_1+m_2u_2}{M_T}.
\end{gather*}

\begin{lemma}\label{lem:d9exp}
At a place of type $\pc n(\{1,2,3\};14,25)$, the function $F(\tau,y)=\sum_{n\ge0}\tau^{n-2}V_n(y)$ satisfies the hypothesis of Lemma~\ref{lem:identities} in the coordinates above, with $V_1=0$ and
\begin{align*}
V_0&=U_T+\tfrac\nu2\,\Omega\,L(u),\\
V_2&=m_1m_4(u_{41}\Omega)^{-1/2}+m_2m_5(u_{52}\Omega)^{-1/2}+m_4m_5(u_{12}v_{45})^{-1/2}\\
&\qquad+\tfrac{\nu_{45}}2\,u_{12}v_{45}+\tfrac\nu2\,\Omega\,\frac{m_4u_{41}+m_5u_{52}}{m_4+m_5},\\
V_3&=m_1m_5(-u_{12}\Omega)^{-1/2}+m_2m_4(u_{12}\Omega)^{-1/2}+m_3m_4(u_1\Omega)^{-1/2}+m_3m_5(u_2\Omega)^{-1/2},
\end{align*}
where the square roots are the branches of the $p_{kl}$ at $P$. The coefficients $V_0$ and $V_3$ do not depend on the fine coordinates.
\end{lemma}

\begin{proof}
The coordinates have the $\tau$-exponents $2,2,2,2,-4$ and $4,4,-2$, all even. So each $z_{kl}$ is $\tau^{2\zeta_{kl}}(\bar z_{kl}+\tau^2c_{kl})$. Here $\bar z_{kl}$ is a linear form in the coordinates whose value at $\bar y$ is the leading coefficient $\hat z_{kl}\ne0$. The function $c_{kl}$ is a polynomial in $\tau^2$ and the coordinates. Likewise $w_{kl}=\tau^{2\omega_{kl}}(\bar w_{kl}+\tau^2c'_{kl})$. For instance $z_{15}=\tau^2(u_{12}-\tau^2u_{52})$ and $w_{15}=\tau^{-4}\big(-\Omega+\tau^2\frac{m_4v_{45}}{m_4+m_5}+\tau^6\frac{m_2v_{12}+m_3v_1}{M_T}\big)$. Hence
\[
p_{kl}=\tau^{-2x_{kl}}(\bar z_{kl}\bar w_{kl})^{-1/2}\big(1+\tau^2c_{kl}/\bar z_{kl}\big)^{-1/2}\big(1+\tau^2c'_{kl}/\bar w_{kl}\big)^{-1/2}
\]
is $\tau^{-2x_{kl}}$ times a convergent series in $\tau^2$ whose coefficients are holomorphic near $\bar y$. The exponent $-2x_{kl}$ is $-2$ on the pairs of $T$, $0$ on $14$, $25$, $45$ and $1$ on $15$, $24$, $34$, $35$. On the pairs of $T$ the series is $1$. The products $\bar z_{kl}\bar w_{kl}$ are $u_{12}v_{12}$, $u_1v_1$, $u_2v_2$ on $12$, $13$, $23$, then $u_{41}\Omega$, $u_{52}\Omega$, $u_{12}v_{45}$ on $14$, $25$, $45$, and $-u_{12}\Omega$, $u_{12}\Omega$, $u_1\Omega$, $u_2\Omega$ on $15$, $24$, $34$, $35$. For the moment of inertia we use $I=I_T+\nu z^*w^*+\nu_{45}z_{45}w_{45}$ with $I_T=M_T^{-1}\sum_{k<l\in T}m_km_lz_{kl}w_{kl}$. Here $I_T$ contributes only at $\tau^4$; $z^*=L(u)+(m_4u_{41}+m_5u_{52})/(m_4+m_5)$ and $w^*=\Omega$, so $\nu z^*w^*$ contributes at $\tau^{-2}$ and $\tau^0$; and $z_{45}w_{45}=\big(u_{12}+\tau^2(u_{41}-u_{52})\big)v_{45}$ contributes at $\tau^0$ and $\tau^2$. Collecting terms gives the lemma.
\end{proof}

\begin{lemma}\label{lem:d9lead}
At a place of type $\pc n(\{1,2,3\};14,25)$, let $\bar y_c$ be the coarse part of the leading point, and $\nabla_f$ the gradient in the fine coordinates.
\begin{enumerate}[label=\textup{(\alph*)}]
\item $\nabla V_0(\bar y_c)=0$ and $\nabla_fV_2(\bar y)=0$.
\item $(m_4+m_5)\hat z_{3l}=\rho_{3l}\hat z_{12}$ for $l=1,2$, where
\[
\rho_{31}=\frac{m_2m_4-m_1m_5-m_3m_5}{m_3},\qquad\rho_{32}=\frac{m_2m_4-m_1m_5+m_3m_4}{m_3}
\]
are nonzero real numbers with $\rho_{32}-\rho_{31}=m_4+m_5$.
\item $\hat Z_{12}/m_3=\hat Z_{23}/m_1=\hat Z_{31}/m_2=:\kappa$, where $\hat Z_{kl}=\hat p_{kl}^3\hat z_{kl}$, and $\kappa\ne0$.
\item $U_T(\bar y_c)=m_1m_2\hat p_{12}+m_1m_3\hat p_{13}+m_2m_3\hat p_{23}=0$.
\end{enumerate}
\end{lemma}

\begin{proof}
(a) By Lemmas~\ref{lem:adapted} and~\ref{lem:d9exp}, $\nabla_yF(\tau,y(\tau))=0$. Its coarse components, multiplied by $\tau^2$, are $\nabla V_0(y_c(\tau))+O(\tau^2)$. Its fine components are $\nabla_fV_2(y(\tau))+O(\tau^2)$.
(b) The $\Omega$-component of $\nabla V_0$ is $\frac\nu2L(u)$. So $\bar u=(\bar u_1,\bar u_2)=(-\hat z_{31},-\hat z_{32})$ satisfies $L(\bar u)=0$ and $\bar u_1-\bar u_2=\hat z_{12}$. These two linear conditions determine $\bar u$, since the coefficients of $L$ have the sum $m_3/M_T\ne0$. The vector $-\hat z_{12}(\rho_{31},\rho_{32})/(m_4+m_5)$ satisfies both. The $\hat z_{3l}$ are nonzero, as $\zeta_{3l}=1$. (These are the positions of \cite[\S8.9.2]{AK12}.)
(c) Since $p_{kl}^2z_{kl}w_{kl}=1$, $\partial p_{kl}/\partial w_{kl}=-\frac12p_{kl}^3z_{kl}$. As $w_{12}=v_1-v_2$, $w_{13}=v_1$ and $w_{23}=v_2$, the $v$-components of $\nabla V_0(\bar y_c)=0$ are $-\frac12m_1(m_2\hat Z_{12}+m_3\hat Z_{13})=0$ and $-\frac12m_2(m_1\hat Z_{21}+m_3\hat Z_{23})=0$.
(d) By (c) and $\hat p_{kl}=\hat p_{kl}^3\hat z_{kl}\hat w_{kl}=\hat Z_{kl}\hat w_{kl}$, we get $U_T(\bar y_c)=\kappa m_1m_2m_3(\hat w_{12}+\hat w_{23}+\hat w_{31})=0$.
\end{proof}

\begin{lemma}\label{lem:d9kernel}
Let $H$ be the Hessian of $V_0$ at $\bar y_c$ in the coordinates $(u_1,u_2,v_1,v_2,\Omega)$, and let $\bar u=(\bar u_1,\bar u_2)$ and $\bar v=(\bar v_1,\bar v_2)$. Then
\[
\ker H=\big\{\big(a\bar u,\,b\bar v,\,-\tfrac{3a+b}2\,\bar\Omega\big):a,b\in\C\big\}.
\]
\end{lemma}

\begin{proof}
Since $U_T$ is homogeneous of degree $-\frac12$ in $u$ and in $v$, $u\cdot\nabla_uU_T=v\cdot\nabla_vU_T=-\frac12U_T$. By Lemma~\ref{lem:d9lead}(a), $\nabla_vU_T=0$ and $\nabla_uU_T=-\frac\nu2\bar\Omega\nabla L$ at $\bar y_c$. The term $\frac\nu2\Omega L(u)$ of $V_0$ is linear in $u$ and does not involve $v$. Differentiating the two Euler identities, we obtain at $\bar y_c$
\[
H_{vu}\bar u=-\tfrac12\nabla_vU_T=0,\qquad H_{vv}\bar v=-\tfrac32\nabla_vU_T=0,\qquad H_{uu}\bar u=-\tfrac32\nabla_uU_T,\qquad H_{uv}\bar v=-\tfrac12\nabla_uU_T,
\]
where $H_{uu},\dots$ are the blocks of $H$. Moreover the $\Omega$-row of $H$ is $(\frac\nu2\nabla L,0,0)$.

Let $\delta=(\delta u,\delta v,\delta\Omega)\in\ker H$. The $\Omega$-row gives $L(\delta u)=0$. So $\delta u=a\bar u$ for some $a\in\C$, as $\ker L=\C\bar u$. The $v$-rows then give $H_{vv}\delta v=0$. We show that $H_{vv}$ has rank $1$. As $\partial^2p_{kl}/\partial w_{kl}^2=\frac34p_{kl}/w_{kl}^2$ and, by Lemma~\ref{lem:d9lead}(c), $m_km_l\hat p_{kl}/\hat w_{kl}=m_km_l\hat Z_{kl}=\kappa m_1m_2m_3$ for $kl=12,23,31$, the quadratic form of $H_{vv}$ is
\[
\tfrac34\kappa m_1m_2m_3\Big(\frac{s_{12}^2}{\hat w_{12}}+\frac{s_{23}^2}{\hat w_{23}}+\frac{s_{31}^2}{\hat w_{31}}\Big),\qquad s_{12}=\delta v_1-\delta v_2,\quad s_{23}=\delta v_2,\quad s_{31}=-\delta v_1,
\]
on the plane $s_{12}+s_{23}+s_{31}=0$. Its polar form pairs the point $(\hat w_{12},\hat w_{23},\hat w_{31})$, given by $\delta v=\bar v$, with $s$ to $\frac34\kappa m_1m_2m_3(s_{12}+s_{23}+s_{31})$, which vanishes on the plane. At $s=(1,-1,0)$ the form is $\frac34\kappa m_1m_2m_3(\hat w_{12}^{-1}+\hat w_{23}^{-1})=-\frac34\kappa m_1m_2m_3\hat w_{31}/(\hat w_{12}\hat w_{23})\ne0$. So $\ker H_{vv}=\C\bar v$ and $\delta v=b\bar v$ for some $b\in\C$. Finally the $u$-rows give $aH_{uu}\bar u+bH_{uv}\bar v+\frac\nu2\delta\Omega\,\nabla L=\frac\nu2\big(\frac{3a+b}2\bar\Omega+\delta\Omega\big)\nabla L=0$ with $\nabla L\ne0$. Conversely, one checks by the same computation that each vector of the stated form lies in $\ker H$. The vectors with $(a,b)=(1,-1)$ and $(-2,-2)$ are the tangent vectors of the rotation and of the scaling $(u,v,\Omega)\mapsto(\lambda^{-2}u,\lambda^{-2}v,\lambda^4\Omega)$.
\end{proof}

The next statement is the analogue of Theorem~\ref{thm:order3}: the leading potential between $T$ and $\{4,5\}$ vanishes.

\begin{proposition}\label{prop:d9cross}
At a place of type $\pc n(\{1,2,3\};14,25)$, $V_3(\bar y)=m_1m_5\hat p_{15}+m_2m_4\hat p_{24}+m_3m_4\hat p_{34}+m_3m_5\hat p_{35}=0$.
\end{proposition}

\begin{proof}
By Lemma~\ref{lem:adapted} we may normalize the place as in the proof of Theorem~\ref{thm:order5}, so that $y_{u_1}(\tau)\equiv\bar u_1$ and $y_\Omega(\tau)\equiv\bar\Omega$: the conditions $\mu\varphi^{-2}y_{u_1}=\bar u_1$ and $\mu^{-1}\varphi^4y_\Omega=\bar\Omega$ determine $\varphi^2$ and then $\mu$. We write $y(\tau)=\bar y+(\delta_c(\tau),\delta_f(\tau))$ with the coarse part $\delta_c$ and the fine part $\delta_f$. Then $\delta_c$ lies in $\mathcal S=\{\xi\in\C^5:\xi_{u_1}=\xi_\Omega=0\}$. By Lemma~\ref{lem:d9kernel}, $\mathcal S\cap\ker H=0$, since $a\bar u_1=0$ and $(3a+b)\bar\Omega=0$ force $a=b=0$. As $\dim\mathcal S+\dim\ker H=5$, the restriction of the bilinear form $H$ to $\mathcal S$ is nondegenerate: a vector of $\mathcal S$ that is $H$-orthogonal to $\mathcal S$ is $H$-orthogonal to $\mathcal S+\ker H=\C^5$. By Lemma~\ref{lem:d9lead}(a), the coarse components of $\tau^2\nabla_yF(\tau,y(\tau))=0$ read $H\delta_c+O(|\delta_c|^2)+O(\tau^2)=0$. Their components along $\mathcal S$ give $\val\delta_c\ge\min(2\val\delta_c,2)$. Hence $\val\delta_c\ge2$. The fine components of $\nabla_yF(\tau,y(\tau))=0$ read $\nabla_fV_2(\bar y+\delta)+O(\tau^2)=0$. By Lemma~\ref{lem:d9exp}, the matrix of second derivatives of $V_2$ in the fine coordinates is diagonal at $\bar y$, with the nonzero entries $\frac34m_1m_4\hat p_{14}/\bar u_{41}^2$, $\frac34m_2m_5\hat p_{25}/\bar u_{52}^2$ and $\frac34m_4m_5\hat p_{45}/\bar v_{45}^2$. With $\nabla_fV_2(\bar y)=0$ and $\val\delta_c\ge2$ this gives $\val\delta_f\ge\min(2\val\delta_f,2)$, hence $\val\delta_f\ge2$. Therefore $V_2(y(\tau))=V_2(\bar y)+O(\tau^2)$. Now \eqref{eq:qid} with $q=3$ reads $[V_1]_2+2[V_2]_1+3V_3(\bar y)=0$, as in the proof of Theorem~\ref{thm:order3}. Here $V_1=0$ and $[V_2(y(\tau))]_1=0$. Thus $V_3(\bar y)=0$.
\end{proof}

\begin{proof}[Proof of Theorem~\textup{\ref{thm:D9}}]
Suppose that $P$ has diagram~$9$, with the labelling above. On the pairs $kl=15,24,34,35$, Lemma~\ref{lem:d9exp} gives $\hat p_{kl}=(\bar z_{kl}\bar w_{kl})^{-1/2}$ at $\bar y$, with $\bar w_{kl}=-\bar\Omega$. By Lemma~\ref{lem:d9lead}(b), $(m_4+m_5)\bar z_{kl}=\tilde\rho_{kl}\hat z_{12}$ at $\bar y$, where $\tilde\rho_{15}=m_4+m_5$, $\tilde\rho_{24}=-(m_4+m_5)$, $\tilde\rho_{34}=\rho_{31}$ and $\tilde\rho_{35}=\rho_{32}$. Let $K$ be a square root of $-(m_4+m_5)/(\hat z_{12}\bar\Omega)$. Then $\hat p_{kl}=\pm K\tilde\rho_{kl}^{-1/2}$ if $\tilde\rho_{kl}>0$, and $\hat p_{kl}=\pm iK|\tilde\rho_{kl}|^{-1/2}$ if $\tilde\rho_{kl}<0$. So Proposition~\ref{prop:d9cross} says that two sums of real numbers $\pm m_km_l|\tilde\rho_{kl}|^{-1/2}$ vanish, one over the pairs with $\tilde\rho_{kl}>0$ and one over the pairs with $\tilde\rho_{kl}<0$. Each sum therefore has at least two terms. The pair $15$ is in the first sum, $24$ is in the second, and $\rho_{31}<\rho_{32}$. Hence $\rho_{31}<0<\rho_{32}$, the first sum is over $15$, $35$ and the second over $24$, $34$, and they vanish only if
\begin{equation}\label{eq:C12}
m_1^2\rho_{32}=m_3^2(m_4+m_5),\qquad m_2^2\rho_{31}=-m_3^2(m_4+m_5).
\end{equation}
Let $\alpha=\hat p_{12}/\hat p_{23}$ and $\beta=\hat p_{12}/\hat p_{31}$. By Lemma~\ref{lem:d9lead}(c), $\hat p_{kl}^3\hat z_{kl}$ is $\kappa m_3$, $\kappa m_1$, $\kappa m_2$ for $kl=12,23,31$. By Lemma~\ref{lem:d9lead}(b), $(m_4+m_5)\hat z_{23}=-\rho_{32}\hat z_{12}$ and $(m_4+m_5)\hat z_{31}=\rho_{31}\hat z_{12}$. With Lemma~\ref{lem:d9lead}(d) these relations give
\begin{gather*}
m_1(m_4+m_5)\alpha^3=-m_3\rho_{32}=m_1m_5-m_2m_4-m_3m_4,\\
m_2(m_4+m_5)\beta^3=m_3\rho_{31}=m_2m_4-m_1m_5-m_3m_5,\\
m_1m_3\alpha+m_2m_3\beta+m_1m_2\alpha\beta=\alpha\beta\,U_T(\bar y_c)/\hat p_{12}=0.
\end{gather*}
By \eqref{eq:C12}, $\alpha^3=-(m_3/m_1)^3$ and $\beta^3=-(m_3/m_2)^3$. So $\alpha=-\eta m_3/m_1$ and $\beta=-\eta'm_3/m_2$ with $\eta^3=\eta'^3=1$. The third relation becomes $m_3^2(\eta\eta'-\eta-\eta')=0$, that is, $(\eta-1)(\eta'-1)=1$. This is impossible, because $|\eta-1|$ and $|\eta'-1|$ are $0$ or $\sqrt3$.

The last statement follows, since Theorem~\ref{thm:types} places the type of a singular place in a piece of its own diagram.
\end{proof}

\begin{remark}\label{rem:akprop8}
The statement of \cite[Prop.~8]{AK12} covers diagram~$9$: along a singular sequence that approaches one of diagrams $9$--$16$, every 4-product $\cP_{ij}$ is bounded. Its proof does not give this bound for diagram~$9$. For this diagram the exponents of the distances given in the proof form the type $\pc n(\{1,2,3\};14,25)$ above, the case of \cite[\S8.9.2]{AK12}. At a place of this type $e(\cP_{15})=2(x_{15}+x_{23}+x_{34}+x_{24})=-1$, and likewise $e(\cP_{24})=-1$, so these two 4-products are unbounded. By Theorem~\ref{thm:D9}, diagram~$9$ does not occur on a nontrivial curve at positive masses. We do not use \cite[Prop.~8]{AK12}.
\end{remark}

\begin{remark}\label{rem:torque}
At a place of diagram~$1$ or~$4$ with triangle $T$, the torque identity of \cite[\S5.0]{AK12} shows that the leading potential between $T$ and the two other bodies vanishes as well; at positive masses this confines such places to finitely many explicit curves of masses \cite[Thm.~9.2]{Tom26}. Chen and Tong state the stronger result that no such place exists at positive masses \cite[Thms.~5.13 and~8.2]{CT26}.
\end{remark}