\subsection{The restricted problem and the base value}\label{sec:restricted} We now compute the preimages of the points with coordinates $(A,P,Q)=(A_0,0,0)$ for $A_0$ large. Since $A=\log\bigl((m_1m_2m_3)^{1/3}/m_4\bigr)$, body~$4$ is almost massless in this limit. The preimages lie near the face $K=0$ of $\Om$, where they correspond to the relative equilibria of the restricted four-body problem inside the triangle of the primaries. At $G=0$ the three primaries have equal masses. These equilibria are then a minimum and three saddles of the potential; the passage from the restricted problem to $F$ reverses the signs. The value $(A_0,0,0)$ therefore has one preimage with $J<0$ and three with $J>0$ (Proposition~\ref{prop:base}). A base value at equal masses would need Albouy's count of the CCs of four equal masses and the signs of Proposition~\ref{prop:cert}(g), but the limit of large $A$ needs neither of these. Theorem~\ref{thm:N}(g) then recovers Albouy's count. Of the earlier results the proofs of Lemmas~\ref{lem:face}--\ref{lem:collarsign} and Propositions~\ref{prop:collar} and~\ref{prop:base} use only Lemmas~\ref{lem:dziobek}, \ref{lem:angles} and~\ref{lem:axis}(a). In particular they use none of the certificates of Section~\ref{sec:cert} and none of Sections~\ref{sec:proper}--\ref{sec:injective}. The supplementary file \texttt{deg\_restricted.py} checks identities of this subsection by exact symbolic computation. We cite its checks by their labels. \emph{The barycentric form.} Let $\mathcal A\subset(0,1)^3$ be the set of points at which the solution of Lemma~\ref{lem:angles} exists with $\theta\in(0,\pi)^3$, so that $\Om=\{a\in\mathcal A:K(a)<0\}$. Lemma~\ref{lem:angles} shows that $\mathcal A$ is open and that $K$ is analytic on it. On $\mathcal A$ we have $w_i>0$ and $S_i>0$. We write $\beta_i=S_i/S$ for the barycentric coordinates of $x_4$ in the outer triangle, and \[ \ell_i=\log\frac{\beta_i}{w_i},\qquad \bar\ell=\tfrac13(\ell_1+\ell_2+\ell_3). \] Then $\hat m_i=\ell_i+\log(w_1w_2w_3)$ in \eqref{eq:F}. For the vectors $E_1$ and $E_2$ of Section~\ref{sec:coords}, let $\varpi:\R^3\to\R^2$ be the linear map $\varpi(\xi)=(\xi\cdot E_1,\,\xi\cdot E_2)$; its kernel is spanned by $(1,1,1)$. Since $E_1$ and $E_2$ form an orthonormal basis of the plane $\xi_1+\xi_2+\xi_3=0$, the restriction of $\varpi$ to this plane is an isometry onto $\R^2$. With $G=\varpi(\mu)$ we obtain \begin{equation}\label{eq:bary} G=\varpi(\ell),\qquad A=\hA-\log(-K),\qquad \hA=\bar\ell+\log(w_1w_2w_3), \end{equation} where $\hA$ is as in Section~\ref{sec:quantities}. The functions $\varpi(\ell)$ and $\bar\ell+\log(w_1w_2w_3)$ are analytic on all of $\mathcal A$. Through \eqref{eq:bary} we regard $G$ and $\hA$ as analytic functions on $\mathcal A$. In this subsection $\nabla f$ denotes the Euclidean gradient of a function $f$ on $\mathcal A$ with respect to $a=(a_1,a_2,a_3)$, and of a function on the plane with respect to $p$, not the gradient in the frame $(e_s,e_r,e_\varphi)$ of Section~\ref{sec:quantities}. Dot products and determinants of vectors of $\R^3$ are taken in the coordinates $a$. \emph{The face $K=0$.} Let $\mathcal K=\{a\in\mathcal A:K(a)=0\}$. We fix the unit equilateral triangle with the positively oriented vertices \[ v_1=\Bigl(\frac12,\frac1{2\sqrt3}\Bigr),\qquad v_2=\Bigl(-\frac12,\frac1{2\sqrt3}\Bigr),\qquad v_3=\Bigl(0,-\frac1{\sqrt3}\Bigr), \] that is, $v_i=\varpi(e_i)/\sqrt2$, where $e_1,e_2,e_3$ is the standard basis of $\R^3$. Let $\mathcal T$ be its open interior. For $p\in\mathcal T$ let $a(p)=(|p-v_1|,|p-v_2|,|p-v_3|)$ and $w_i(p)=|p-v_i|^{-3}-1$. The barycentric coordinates of $p$ are the affine functions $\beta_i(p)=\frac13+2\,p\cdot v_i$. The \emph{restricted map} is \[ g:\mathcal T\to\R^2,\qquad g(p)=\varpi(\ell(p)),\qquad \ell_i(p)=\log\frac{\beta_i(p)}{w_i(p)} . \] \begin{lemma}\label{lem:face} \begin{enumerate}[label=\textup{(\alph*)},leftmargin=2em] \item The map $p\mapsto a(p)$ is a bijection from $\mathcal T$ onto $\mathcal K$, and its differential has rank $2$ at every point. At $a(p)$ the configuration is $(v_1,v_2,v_3,p)$ up to an isometry, and $\theta_i=\alpha_i=\arccos\zeta_i$ with $\zeta_i$ as in Lemma~\ref{lem:alpha}. \item At every point of $\mathcal K$, for $l=1,2,3$, \[ \frac{\partial}{\partial a_l}(\alpha_1+\alpha_2+\alpha_3)=-\frac2{\sqrt3}\,\frac{a_l\beta_l}{\beta_1\beta_2\beta_3}<0\qquad\text{and}\qquad\frac{\partial K}{\partial a_l}<0 . \] In particular $\mathcal K$ is an analytic surface with normal vector $\nabla K$, and $\nabla K\cdot(1,1,1)<0$ on $\mathcal K$. \item $G(a(p))=g(p)$ for every $p\in\mathcal T$. \end{enumerate} \end{lemma} \begin{proof} (a) Let $a\in\mathcal K$. From $k_i=w_i(q_i-1)=0$ and $w_i>0$ we get $q_i=1$, that is $R_i=1$: the outer triangle is equilateral with unit sides. Since $\theta\in(0,\pi)^3$ and $\sum_i\theta_i=2\pi$, body~$4$ lies strictly inside it. An isometry that maps $x_i$ to $v_i$ for $i\le3$ maps $x_4$ to a point $p\in\mathcal T$; then $a=a(p)$. Conversely, let $p\in\mathcal T$. Then $|p-v_i|<1$: an interior point of a triangle is closer to each vertex than the farther of the two other vertices (proof of Lemma~\ref{lem:dziobek}). The angles $\alpha_i$ at $p$ between $v_j-p$ and $v_k-p$ lie in $(0,\pi)$ and add up to $2\pi$, and the law of cosines gives $\cos\alpha_i=\zeta_i$, since $|v_j-v_k|=1$. With $R_i=1$, these angles and $K=0$ solve the system of Lemma~\ref{lem:angles} at $a(p)$. They are its solution by the uniqueness in that lemma, so $a(p)\in\mathcal K$. The map $p\mapsto a(p)$ is injective: a point of the plane is determined by its distances to three non-collinear points. The differential of $p\mapsto a(p)$ has the unit vectors $(p-v_i)/|p-v_i|$ as rows. Any two of them are independent, since $p$ does not lie on a line through two vertices. (b) Let $\theta_i(a,K)=\arccos\bigl((a_j^2+a_k^2-R_i)/(2a_ja_k)\bigr)$ with $R_i=(1+K/w_i)^{-2/3}$, as in the proof of Lemma~\ref{lem:angles}. Differentiating $\sum_i\theta_i(a,K(a))=2\pi$ on $\mathcal A$ gives \[ \nabla K=-\frac{\nabla_a\sum_i\theta_i}{\partial_K\sum_i\theta_i}, \] where $\partial_K\theta_i<0$ as in that proof. At $K=0$ we have $R_i=1$ and $\partial R_i/\partial a=0$, so $\nabla_a\theta_i=\nabla\alpha_i$ on $\mathcal K$. We compute $\nabla\alpha_i$ with the configuration of (a). The triangle $pv_jv_k$ has area $S_i=\frac{\sqrt3}4\beta_i$. For $\{l,j,k\}=\{1,2,3\}$ we have $p-v_l=\beta_j(v_j-v_l)+\beta_k(v_k-v_l)$. It follows that $a_l^2=\beta_j^2+\beta_j\beta_k+\beta_k^2$ and, since $\sum_i\beta_i=1$, \[ a_l^2-a_k^2+1=\beta_j+2\beta_k . \] Differentiating $\cos\alpha_i=(a_j^2+a_k^2-1)/(2a_ja_k)$ and using $2S_i=a_ja_k\sin\alpha_i$ gives $\partial\alpha_i/\partial a_j=-(a_j^2-a_k^2+1)/(4a_jS_i)$. As $\alpha_l$ does not depend on $a_l$, \[ \frac{\partial}{\partial a_l}\sum_i\alpha_i=-\frac{\beta_j+2\beta_k}{\sqrt3\,a_l\beta_j}-\frac{\beta_k+2\beta_j}{\sqrt3\,a_l\beta_k}=-\frac{2(\beta_j^2+\beta_j\beta_k+\beta_k^2)}{\sqrt3\,a_l\beta_j\beta_k}=-\frac{2a_l}{\sqrt3\,\beta_j\beta_k} \] (check E6). With $\partial_K\sum_i\theta_i<0$, the formula for $\nabla K$ shows that all the partial derivatives $\partial K/\partial a_l$ are negative. In particular $\nabla K\ne0$ on $\mathcal K=\mathcal A\cap\{K=0\}$, which is thus an analytic surface with normal vector $\nabla K$. (c) By (a), $\beta_i(a(p))=\beta_i(p)$ and $w_i(a(p))=w_i(p)$. Statement (c) then follows from \eqref{eq:bary}. \end{proof} \emph{The restricted problem.} For $m=(m_1,m_2,m_3)\in(0,\infty)^3$ and $p\in\R^2\setminus\{v_1,v_2,v_3\}$ let \[ V^{\mathrm r}_m(p)=\sum_{i=1}^3m_i\Bigl(\frac1{|p-v_i|}+\frac{|p-v_i|^2}2\Bigr). \] Since the gradient of $|p-v_i|^{-1}+\frac12|p-v_i|^2$ is $-w_i(p)(p-v_i)$, \begin{equation}\label{eq:gradV} \nabla V^{\mathrm r}_m(p)=-\sum_{i=1}^3m_iw_i(p)\,(p-v_i). \end{equation} Thus $\nabla V^{\mathrm r}_m(p)=0$ is the equation of body~$4$ in \eqref{eq:ccnorm} for the configuration $(v_1,v_2,v_3,p)$, an equation that does not involve $m_4$; the critical points of $V^{\mathrm r}_m$ are therefore the relative equilibria of the restricted problem with primaries of masses $m_i$ at the points $v_i$. We write $\log m=(\log m_1,\log m_2,\log m_3)$. \begin{lemma}\label{lem:restricted} Let $m\in(0,\infty)^3$. A point $p\in\mathcal T$ is a critical point of $V^{\mathrm r}_m$ if and only if $g(p)=\varpi(\log m)$, and then $m$ is a positive multiple of $m(p)=(\beta_i(p)/w_i(p))_i$. \end{lemma} \begin{proof} The vectors $\xi\in\R^3$ with $\sum_i\xi_i(p-v_i)=0$ form a line, since $p-v_1$ and $p-v_2$ are independent for $p\in\mathcal T$. It contains $(\beta_i(p))_i$, since $\sum_i\beta_i(p)(p-v_i)=p-p=0$. Equation \eqref{eq:gradV} shows that $p$ is a critical point of $V^{\mathrm r}_m$ if and only if $(m_iw_i(p))_i$ is a multiple of $(\beta_i(p))_i$, that is, if and only if $m$ is a positive multiple of $m(p)$. As the kernel of $\varpi$ is spanned by $(1,1,1)$, this holds if and only if $\varpi(\log m)=\varpi(\log m(p))=g(p)$. \end{proof} \emph{Equal primaries.} We write $V=V^{\mathrm r}_{(1,1,1)}$; its critical points in $\mathcal T$ are the relative equilibria of the restricted problem with three equal primaries inside their triangle. That there are exactly four of them is known \cite[Teor.~45]{Bar08} (see the references there). We include a short proof, which also gives the signs of the Hessians. For $p=(x,y)$ we put $p^\perp=(-y,x)$. \begin{lemma}[equal primaries]\label{lem:eqprim} The function $V$ has exactly four critical points in $\mathcal T$: the centre $0$, where $\operatorname{Hess}V(0)=(3+\frac92\sqrt3)\,I$, and three points $p_1$, $p_2$, $p_3$, which the rotation through $120^\circ$ about $0$ permutes and at which $\det\operatorname{Hess}V(p_k)<0$. The point $p_3$ lies on the open segment from $0$ to the midpoint of $v_1v_2$. \end{lemma} \begin{proof} We write $u_i=|p-v_i|^2$, $f_0(u)=u^{-3/2}$, $w_i=f_0(u_i)-1$ and, as in \eqref{eq:gradV}, $Y=-\nabla V=\sum_iw_i\,(p-v_i)$. Differentiating once more and using $\nabla w_i=-3u_i^{-5/2}(p-v_i)$, we get \[ \operatorname{Hess}V=\sum_{i=1}^3\Bigl(-w_i\,I+3u_i^{-5/2}\,(p-v_i)(p-v_i)^{\mathsf T}\Bigr). \] At the centre $u_i=\frac13$. With $\sum_iv_iv_i^{\mathsf T}=\frac12I$ this yields $\operatorname{Hess}V(0)=-3(3\sqrt3-1)\,I+\frac{27\sqrt3}2\,I=(3+\frac92\sqrt3)\,I$ (check E2). \emph{Step 1: the critical points lie on the medians.} Let $(i,j,k)$ be a cyclic permutation of $(1,2,3)$ and $t_i=p\cdot v_i$. As $|v_i|^2=\frac13$, we have $u_i=|p|^2+\frac13-2t_i$. Moreover $v_j-v_k=\sqrt3\,v_i^\perp$, so that $\det[v_i,p]=p\cdot v_i^\perp=(t_j-t_k)/\sqrt3=(u_k-u_j)/(2\sqrt3)$. Since $(p-v_i)\cdot p^\perp=\det[v_i,p]$, we obtain \[ Y\cdot p^\perp=\frac{w_1(u_3-u_2)+w_2(u_1-u_3)+w_3(u_2-u_1)}{2\sqrt3}=\frac1{2\sqrt3}\det\begin{pmatrix}1&1&1\\u_1&u_2&u_3\\f_0(u_1)&f_0(u_2)&f_0(u_3)\end{pmatrix}. \] The constants $-1$ in the $w_i$ drop out. The determinant is alternating in its three columns. If $u_11304$, while the first has absolute value at most $\sqrt3\cdot\frac94\cdot(\frac85)^{7/2}<1.75\cdot2.25\cdot5.2<21$. Next, $\psi(0)=\frac43-\frac{\sqrt3}2>0$. At the centre, where $d=d_0=\frac1{2\sqrt3}$, we have $\psi(d_0)=0$ and $\psi'(d_0)=\partial_y^2V(0)=3+\frac92\sqrt3>0$. Thus $\psi<0$ just below $d_0$; hence $\psi$ has a zero $d_s\in(0,d_0)$ by the intermediate value theorem. A strictly convex function has at most two zeros. The zeros of $\psi$ on $(0,\frac{\sqrt3}2)$ are therefore $d_s$ and $d_0$. By convexity, $\psi'(d_s)<(\psi(d_0)-\psi(d_s))/(d_0-d_s)=0$. Thus the critical points of $V$ on this median are $0$ and $p_3=(0,y_s)$, where $y_s=d_0-d_s>0$. The rotations through $\pm120^\circ$ about $0$ preserve $V$ and $\mathcal T$ and map this median onto the other two. This yields $p_1$ and $p_2$, and with Step~1 the four critical points. \emph{Step 3: the Hessian at $p_3$.} By the reflection symmetry, $\operatorname{Hess}V(p_3)$ is diagonal. Moreover $\partial_y^2V(p_3)=\psi'(d_s)<0$. On the median, $p-v_1=(-\frac12,\,y-\frac1{2\sqrt3})$, $p-v_2=(\frac12,\,y-\frac1{2\sqrt3})$ and $p-v_3=(0,\,y+\frac1{\sqrt3})$, so the formula for $\operatorname{Hess}V$ gives \[ \partial_x^2V(0,y)=-2f_0(u_1)-f_0(u_3)+3-f_0'(u_1),\qquad u_3-u_1=\sqrt3\,y . \] The equation $Y_y(0,y)=0$ reads $E(y)=0$, where $E(y)=\sqrt3\,Y_y(0,y)=(1+\sqrt3y)\bigl(f_0(u_3)-1\bigr)-(1-2\sqrt3y)\bigl(f_0(u_1)-1\bigr)$. Expanding both sides, we find \[ \sqrt3\,y\,\partial_x^2V(0,y)-\bigl[f_0(u_3)-f_0(u_1)-f_0'(u_1)(u_3-u_1)\bigr]=-E(y). \] At a critical point $(0,y)$ the right side vanishes. At $p_3$ we have $y_s>0$ and $u_3>u_1$, so the bracket is positive, since a strictly convex function lies above its tangent lines. Therefore $\partial_x^2V(p_3)>0$ and $\det\operatorname{Hess}V(p_3)=\partial_x^2V(p_3)\,\partial_y^2V(p_3)<0$. The rotations carry this to $p_1$ and $p_2$. \end{proof} Numerically $d_s=0.0497168\ldots$, $p_3=(0,0.238958\ldots)$, $\partial_x^2V(p_3)=32.2297\ldots$ and $\partial_y^2V(p_3)=-8.6257\ldots$ The signs of the critical points are the signs of the Jacobian of $g$, by the following identity. \begin{lemma}[sign identity]\label{lem:signid} For $p\in\mathcal T$ let $C(p)$ be the linear map of $\R^2$ with $C(p)\varpi(e_l)=\beta_l(p)(v_l-p)$ for $l=1,2,3$. Then \[ \operatorname{Hess}V^{\mathrm r}_{m(p)}(p)=-C(p)\,Dg(p),\qquad \det C(p)=\tfrac32\,\beta_1(p)\beta_2(p)\beta_3(p)>0 . \] In particular $\operatorname{sign}\det Dg(p)=\operatorname{sign}\det\operatorname{Hess}V^{\mathrm r}_{m(p)}(p)$. \end{lemma} \begin{proof} The map $C(p)$ is well defined: $\varpi(e_1)$ and $\varpi(e_2)$ form a basis of $\R^2$, $\sum_l\varpi(e_l)=0$ and $\sum_l\beta_l(p)(v_l-p)=0$. We omit the argument $p$ and write $a_l=|p-v_l|$. Differentiating \eqref{eq:gradV} and using $\nabla w_l=-3a_l^{-5}(p-v_l)$ and $m_lw_l=\beta_l$ for $m=m(p)$, we get \[ \operatorname{Hess}V^{\mathrm r}_{m(p)}=\sum_l\Bigl(-\beta_lI+\frac{3\beta_l}{a_l^5w_l}\,(p-v_l)(p-v_l)^{\mathsf T}\Bigr). \] From $\beta_l=\frac13+2\,p\cdot v_l$ we get $\nabla\ell_l=2v_l/\beta_l+3a_l^{-5}w_l^{-1}(p-v_l)$. With $Dg=\sum_l\varpi(e_l)\nabla\ell_l^{\mathsf T}$ this gives \[ C\,Dg=\sum_l\Bigl(2(v_l-p)v_l^{\mathsf T}-\frac{3\beta_l}{a_l^5w_l}\,(p-v_l)(p-v_l)^{\mathsf T}\Bigr). \] With $\sum_l\beta_l=1$, $\sum_lv_lv_l^{\mathsf T}=\frac12I$ and $\sum_lv_l=0$, the sum of the two matrices is $-I+2\sum_lv_lv_l^{\mathsf T}-2p\bigl(\sum_lv_l\bigr)^{\mathsf T}=0$. This identity does not use the relation $a_l=|p-v_l|$ (check E8b). Finally, $C$ maps the basis $\varpi(e_1),\varpi(e_2)$, of determinant $1/\sqrt3$, to the vectors $\beta_1(v_1-p),\beta_2(v_2-p)$, whose determinant is $\beta_1\beta_2$ times twice the area $\frac{\sqrt3}4\beta_3$ of the positively oriented triangle $pv_1v_2$. So $\det C=\frac32\beta_1\beta_2\beta_3$ (check E8c). The last assertion holds as $\det(-C\,Dg)=\det C\det Dg$ for $2\times2$ matrices. \end{proof} \emph{The collar.} For $A_0\in\R$ and $G_0\in\R^2$ we write $F^{-1}(A_0,G_0)$ for the set of $a\in\Om$ with $A(a)=A_0$ and $G(a)=G_0$, which is the fibre of $F$ over the point with coordinates $(A,P,Q)=(A_0,G_0)$. We show that for $A_0$ large this fibre corresponds to $g^{-1}(G_0)$. The first step locates it near $\mathcal K$. \begin{lemma}\label{lem:edge} Let $B\ge0$, and let $a^{(n)}\in\Om$ satisfy $|G(a^{(n)})|\le B$ and $A(a^{(n)})\to+\infty$. Then a subsequence of $(a^{(n)})$ converges to a point of $\mathcal K$. \end{lemma} \begin{proof} We scale the masses of $a^{(n)}$ so that $m_1+m_2+m_3=1$. Since \eqref{eq:ccnorm} is linear in the masses, this does not change the CC. The vector $\mu-A(1,1,1)$ lies in the plane $\xi_1+\xi_2+\xi_3=0$, on which $\varpi$ is an isometry. Therefore $|\mu_i-A|\le|G|\le B$ for each $i$. It follows that $m_i/m_j\le e^{2B}$ and $m_i\ge e^{-2B}/3$ for $i,j\le3$. Moreover $m_4=(m_1m_2m_3)^{1/3}e^{-A}\le e^{-A}\to0$. With $x_4$ at the origin, $|x_i|=a_i<1$; the positions are therefore bounded. First, all spokes are bounded below. Since $R_i>1$ by Lemma~\ref{lem:dziobek} and $R_i=r_{jk}^2\le(a_j+a_k)^2$, we have $a_j+a_k>1$ for all pairs: at most one spoke is shorter than $1/2$. The equation of body~$4$ in \eqref{eq:ccnorm} reads $\sum_{i\le3}m_iw_i(x_i-x_4)=0$. The $i$-th term of this sum has length $m_i(a_i^{-2}-a_i)$, which is at most $7m_i/2$ if $a_i\ge1/2$. Let $a_i$ be the shortest spoke. The two other terms, whose spokes are at least $1/2$, have total length at most $7/2$. Then $m_i(a_i^{-2}-a_i)\le7/2$ and $a_i^{-2}\le1+\frac{21}2e^{2B}$. So $c_B=(1+\frac{21}2e^{2B})^{-1/2}$ is a lower bound for all spokes. We pass to a subsequence along which the masses converge to $\bar m$ and the positions to $\bar x$. Then $\bar m_4=0$, $\bar m_i\ge e^{-2B}/3$ for $i\le3$, $a^{(n)}\to\bar a$ and $\bar a_i\ge c_B$. In the equation of an outer body $j$, the term of body~$4$ has length $m_4(a_j^{-2}-a_j)\le m_4c_B^{-2}\to0$, while the other terms depend continuously on the positions, since $r_{ij}>1$. In the limit, \[ \sum_{i\le3,\ i\ne j}\bar m_i\,(\bar x_i-\bar x_j)\,(\bar r_{ij}^{-3}-1)=0,\qquad j=1,2,3,\qquad\text{with }\bar r_{ij}\ge1 . \] If $\bar x_1,\bar x_2,\bar x_3$ were collinear, then at an end body $j$ the two vectors $\bar x_i-\bar x_j$ would have the same direction and coefficients $\bar m_i(\bar r_{ij}^{-3}-1)\le0$. Both coefficients would vanish, and the two other bodies would coincide, a contradiction. For each $j$ the two vectors $\bar x_i-\bar x_j$ are then independent, so all coefficients vanish. Hence the limit outer triangle is equilateral with unit sides. It contains $\bar x_4$, a limit of interior points, which is not a vertex, since all $\bar a_l$ are positive. As the only points of a closed unit equilateral triangle at distance $1$ from a vertex are the two other vertices, all $\bar a_i$ are less than $1$; hence $w_i\to\bar a_i^{-3}-1>0$. The equation of body~$4$ shows that the barycentric coordinates of $x_4$ in the outer triangle are $\beta_i=m_iw_i/\sum_lm_lw_l$. These converge to positive limits. Hence $\bar x_4$ lies inside the limit triangle, so $\bar a=a(p)$ for some $p\in\mathcal T$. Lemma~\ref{lem:face}(a) then gives $\bar a\in\mathcal K$. \end{proof} \begin{lemma}\label{lem:orient} For $p\in\mathcal T$ let $\tau_1(p)$ and $\tau_2(p)$ be the partial derivatives of $a$ at $p$ with respect to the two Cartesian coordinates of the plane. Then \[ \det[\nabla K(a(p)),\tau_1(p),\tau_2(p)]<0\qquad\text{for every }p\in\mathcal T . \] \end{lemma} \begin{proof} Lemma~\ref{lem:face} shows that $\tau_1(p)$ and $\tau_2(p)$ span the tangent plane of $\mathcal K$ at $a(p)$ and that $\nabla K(a(p))$ is a nonzero normal vector. The determinant is then a continuous function on $\mathcal T$ without zeros. Since $\mathcal T$ is connected, its sign is constant. We may evaluate it at $p=0$, where $a(0)=3^{-1/2}(1,1,1)$ and $\beta_l=\frac13$. Lemma~\ref{lem:face}(b) shows that the three partial derivatives of $K$ are equal there: $\nabla K=\frac13K_s(1,1,1)$, where $K_s$ is the derivative of $K$ along $s\mapsto s(1,1,1)$ at $s=1/\sqrt3$. By Lemma~\ref{lem:angles}, the symmetric solution $\theta_i=2\pi/3$, $R_i=3s^2$ yields $K=(s^{-3}-1)\bigl((\sqrt3\,s)^{-3}-1\bigr)$ on the axis (Lemma~\ref{lem:axis}(a)). Differentiating at $s=1/\sqrt3$ gives $K_s=-3\sqrt3\,(3\sqrt3-1)$. Also $\tau_k(0)=-\sqrt3\,(v_{1,k},v_{2,k},v_{3,k})$, where $v_{l,k}$ is the $k$-th coordinate of $v_l$. Hence $\det[(1,1,1),\tau_1(0),\tau_2(0)]$ is $3$ times twice the area $\frac{\sqrt3}4$ of the positively oriented triangle $v_1v_2v_3$, that is $\frac{3\sqrt3}2$. At $p=0$ we obtain \[ \det[\nabla K,\tau_1,\tau_2]=\frac{K_s}3\cdot\frac{3\sqrt3}2=-\frac92\,(3\sqrt3-1)<0 \] (check E7). \end{proof} \emph{Collar coordinates.} Let $p_0\in\mathcal T$. Since $\nabla K\cdot(1,1,1)<0$ on $\mathcal K$, by the implicit function theorem there is an analytic function $t(K,p)$ near $(0,p_0)$ with $t(0,p)=0$ and $K\bigl(a(p)+t(K,p)(1,1,1)\bigr)=K$. Let $\Xi(K,p)=a(p)+t(K,p)(1,1,1)$; its differential at $(0,p)$ has the columns $(1,1,1)/(\nabla K\cdot(1,1,1))$, $\tau_1(p)$ and $\tau_2(p)$. It is invertible: $\tau_1(p)$ and $\tau_2(p)$ span the tangent plane of $\mathcal K$ at $a(p)$ (Lemma~\ref{lem:face}) and $(1,1,1)$ is not tangent to $\mathcal K$. By the inverse function theorem, $\Xi$ maps a neighbourhood $(-\eta,\eta)\times D$ of $(0,p_0)$ diffeomorphically onto a neighbourhood of $a(p_0)$ in $\mathcal A$. It maps $(-\eta,0)\times D$ into $\Om$ and $\{0\}\times D$ into $\mathcal K$. In the coordinates $(K,p)$ the functions $G$ and $\hA$ are analytic, and $G(0,p)=g(p)$ by Lemma~\ref{lem:face}(c). \begin{lemma}\label{lem:collarsign} Let $p_0\in\mathcal T$ with $\det Dg(p_0)\ne0$. As $a\to a(p_0)$ in $\Om$, \[ (-K)\,J\ \to\ \sqrt3\,\frac{|\nabla K|^2\,\det Dg(p_0)}{\det[\nabla K,\tau_1,\tau_2]}, \] where $\nabla K$, $\tau_1$ and $\tau_2$ are taken at $a(p_0)$ and $p_0$. In particular $J\ne0$ and $\operatorname{sign}J=-\operatorname{sign}\det Dg(p_0)$ at the points of $\Om$ close to $a(p_0)$. \end{lemma} \begin{proof} We use the collar coordinates at $p_0$ and put $\omega=-\log(-K)$ for $K<0$. Then $A=\hA+\omega$ and $\partial_\omega=(-K)\,\partial_K$. As $(K,p)\to(0,p_0)$ with $K<0$, \[ D_{(\omega,p)}(A,P,Q)=\begin{pmatrix}1+(-K)\,\partial_K\hA&\partial_p\hA\\(-K)\,\partial_KG&D_pG\end{pmatrix}\ \to\ \begin{pmatrix}1&\partial_p\hA\\0&Dg(p_0)\end{pmatrix}, \] because $\hA$ and $G$ are analytic in $(K,p)$ and $G(0,p)=g(p)$. With $\det\partial(A,P,Q)/\partial\mu=1/\sqrt3$ and $d\omega/dK=1/(-K)$, \[ (-K)\,J=\sqrt3\,\frac{\det D_{(\omega,p)}(A,P,Q)}{\det D\Xi(K,p)}\ \to\ \sqrt3\,\frac{\det Dg(p_0)}{\det D\Xi(0,p_0)} . \] Differentiating $K(\Xi(K,p))=K$ gives $\nabla K\cdot\partial_K\Xi=1$. At $K=0$ the column $\partial_K\Xi$ is then $\nabla K/|\nabla K|^2$ plus a vector tangent to $\mathcal K$, that is, plus a combination of $\tau_1$ and $\tau_2$. Therefore $\det D\Xi(0,p_0)=\det[\nabla K,\tau_1,\tau_2]/|\nabla K|^2$, which is negative by Lemma~\ref{lem:orient}. \end{proof} \begin{proposition}[collar]\label{prop:collar} Let $G_0\in\R^2$, and suppose that $g^{-1}(G_0)=\{q_1,\dots,q_n\}$ is finite and that $\det Dg(q_l)\ne0$ for $l=1,\dots,n$. For every $A_0$ large enough, $F^{-1}(A_0,G_0)$ consists of exactly $n$ points, one near each point $a(q_l)$. At the point near $a(q_l)$ we have $J\ne0$ and $\operatorname{sign}J=-\operatorname{sign}\det Dg(q_l)$. \end{proposition} \begin{proof} For each $l$ we take collar coordinates $(K,p)$ at $q_l$ on $(-\eta,\eta)\times D_l$ and choose the discs $D_l$ so small that the sets $\Xi((-\eta,\eta)\times D_l)$ are disjoint. Since $D_pG(0,q_l)=Dg(q_l)$ is invertible, the implicit function theorem gives, after shrinking $\eta$ and $D_l$, a unique solution $p=q_l(K)\in D_l$ of $G(K,p)=G_0$ for each $K\in(-\eta,\eta)$. At $K=0$ the solution is $q_l(0)=q_l$. For $K\in(-\eta,0)$ let $\chi_l(K)=\hA(K,q_l(K))-\log(-K)$, the value of $A$ at this solution. Since the derivative of $\chi_l$ is a bounded function minus $1/K$, there is $\eta'\in(0,\eta)$ such that $\chi_l$ increases strictly on $(-\eta',0)$. It tends to $+\infty$ as $K\to0$. A point of $F^{-1}(A_0,G_0)$ in $\Xi((-\eta',\eta')\times D_l)$ has $K<0$, $p=q_l(K)$ and $\chi_l(K)=A_0$. Conversely, such a $K$ determines the point $\Xi(K,q_l(K))$ of $F^{-1}(A_0,G_0)$. For $A_0>\max_l\chi_l(-\eta'/2)$ the equation $\chi_l(K)=A_0$ has exactly one solution in $(-\eta',0)$, by monotonicity and the intermediate value theorem. Each of the $n$ sets $\Xi((-\eta',\eta')\times D_l)$ then contains exactly one point of $F^{-1}(A_0,G_0)$. This point tends to $a(q_l)$ as $A_0\to+\infty$. It remains to show that, for $A_0$ large, $F^{-1}(A_0,G_0)$ lies in the union $\mathcal V$ of these sets. Otherwise there are $A^{(n)}\to+\infty$ and points $a^{(n)}\in F^{-1}(A^{(n)},G_0)\setminus\mathcal V$. By Lemma~\ref{lem:edge} with $B=|G_0|$, a subsequence converges to a point $\bar a=a(\bar p)$ of $\mathcal K$. Since $G$ is continuous on $\mathcal A$, $g(\bar p)=G(\bar a)=G_0$, so $\bar a$ is one of the points $a(q_l)$. This is impossible because $\mathcal V$ is a neighbourhood of these points. The signs follow from Lemma~\ref{lem:collarsign}. \end{proof} We now take $G_0=0$. \begin{proposition}[base value]\label{prop:base} For every sufficiently large $A_0$, the point with coordinates $(A,P,Q)=(A_0,0,0)$ is a regular value of $F$ with exactly four preimages. At one of them $J<0$, and at the other three $J>0$. \end{proposition} \begin{proof} Let $m=(1,1,1)$, so that $\varpi(\log m)=0$. Lemma~\ref{lem:restricted} shows that $g^{-1}(0)$ is the set of critical points of $V=V^{\mathrm r}_m$ in $\mathcal T$ and that at each of them $m(p)$ is a positive multiple of $m$. This set is $\{0,p_1,p_2,p_3\}$ by Lemma~\ref{lem:eqprim}. As $\operatorname{Hess}V^{\mathrm r}_{tm}=t\operatorname{Hess}V^{\mathrm r}_m$ for $t>0$, Lemma~\ref{lem:signid} implies $\operatorname{sign}\det Dg(p)=\operatorname{sign}\det\operatorname{Hess}V(p)$ at these points. This sign is $+$ at the centre and $-$ at $p_1$, $p_2$ and $p_3$ (Lemma~\ref{lem:eqprim}). Therefore $g^{-1}(0)$ is finite and $\det Dg\ne0$ at each of its four points. Proposition~\ref{prop:collar} with $G_0=0$ shows that for $A_0$ large the set $F^{-1}(A_0,0)$ consists of exactly four points, at which $J\ne0$ and $\operatorname{sign}J=-\operatorname{sign}\det Dg$, that is, $-$, $+$, $+$, $+$. \end{proof} The preimage with $J<0$ is the axis point at which $A=A_0$: this point has $G=0$ and $J<0$ by Lemma~\ref{lem:axis}(c),(d) and Proposition~\ref{prop:ax}.