\section{Weighted leading forms}\label{sec:weights} Fix $f\in\C[x_1,x_2,x_3,x_4]$ with nonzero constant Hessian determinant. All Hessian ranks are taken over the rational function field in the source variables unless a specialization is specified. For rational weights $w_1,\ldots,w_4$, the weight of $\prod_{i=1}^4 x_i^{a_i}$ is $\sum_{i=1}^4 a_i w_i$. The leading form is the sum of the terms of maximal weight. Let $E$ be this maximal weight for $f$ and put \[ W=4E-2\sum_{i=1}^4 w_i. \] In position $(i,j)$ of the Hessian, differentiating twice lowers the weight of every surviving monomial by $w_i+w_j$. Hence each product in the determinant expansion has weight at most $W$, and its component of weight $W$ comes from the leading form. Since the full determinant is a nonzero constant, $W\ge0$. For $W>0$ the leading form has zero Hessian determinant, and for $W=0$ it has the same Hessian determinant as the full polynomial. Along an affine path of weights, each monomial weight is affine in the path parameter. We choose paths on which the monomials of the current leading form stay tied, and call the first parameter at which another monomial joins them a \emph{wall}. The finite monomial support gives a strict gap until that parameter. At a wall, a small further step in a specified direction selects the terms of greatest slope from the tied terms, since all others have a strict gap. \begin{lemma}\label{lem:endpoint} If all weights are positive and the leading form has nonzero constant Hessian determinant, the gradient of the full polynomial is a polynomial automorphism. \end{lemma} \begin{proof} Let $P$ be the leading form and $A=\Hess P(0)$. This matrix is invertible. Both $A$ and $A^{-1}$ have entries supported on pairs $(i,j)$ with $w_i+w_j=E$, because $A$ identifies each weight block with its complementary block. Thus component $i$ of $A^{-1}\grad f$ has weight at most $w_i$. The part of its linear terms in variables of weight $w_i$ is $x_i$: the leading form contributes $A^{-1}A X=X$, where $X=(x_1,\ldots,x_4)^{\mathsf T}$, and the lower-weight terms contribute only weights less than $w_i$. Any nonlinear monomial of weight at most $w_i$ uses only variables of weight less than $w_i$, since every factor has positive weight. Successive substitution in increasing order of the weights gives the polynomial inverse. \end{proof} An \emph{apex direction} is a vector $d\in\C^4$ such that the Zariski closure of the gradient image is invariant under translation by $\lambda d$ for all $\lambda\in\C$. The apex directions form a vector space. A nonzero such direction is a projective image apex in the terminology of \cite[Theorem~7.2 and Corollary~7.3]{debondt}: the condition that each polynomial relation on the gradient still vanishes after translation by $\lambda d$ is equivalent to invariance of the closure. The \emph{complementary source space} of an apex space is its annihilator under the source-target pairing $(d,v)\mapsto d^{\mathsf T}v$. For Hessian rank three, de Bondt's classification gives the following possibilities. If the apex space is one-dimensional, align it with a target coordinate axis. Over the rational function field of the corresponding source variable, the potential is a polynomial in a linear form of the other three variables plus an affine polynomial in them. The Hessian restricted to the complementary source space has rank one. Rank zero would give full Hessian rank at most two, since only one row and column are added. If the apex space has dimension two, the potential is jointly affine in the two complementary source variables. If it has dimension three, there is a one-dimensional constant source kernel. For Hessian rank two, either there is a two-dimensional constant source kernel or the apex space is one-dimensional and the potential is jointly affine in the three complementary source variables. The gradient of a form of weight $E$ is equivariant for source scaling with weights $w_i$ and target scaling with weights $E-w_i$. Here and below, rational weights may be cleared of denominators to define the diagonal action of $\C^\times$. The equal-weight blocks on source and target coincide. The apex space is invariant under the target scaling, and the constant Hessian kernel under source scaling. Both spaces therefore split into their intersections with the equal-weight blocks. Subtracting an affine polynomial translates the gradient and does not change its Hessian. Let $P$ be a wall form and $H$ the leading form just before the wall. Use the old weights from a point on the preceding open segment, with $E$ the old weight of $H$. Then $P-H$ consists of terms of weight less than $E$. Write $\adj$ for the classical adjugate. Each product in entry $(i,j)$ of $\adj(\Hess P)$ has old weight at most $3E-2\sum_k w_k+w_i+w_j$, because the deleted row and column omit weights $w_j,w_i$. Its component at that bound is the corresponding entry for $H$. For constant vectors $u,v$, the Hessian minor on $u,v$ is the determinant of the two-by-two matrix of the Hessian bilinear form in those vectors. If $u,v$ are homogeneous for the old weights, the same comparison shows that the old minor survives as the maximal-weight part of the new one. If a constant vector $d$ has a unique nonzero coordinate of greatest old weight and the corresponding principal cofactor of $\Hess H$ is nonzero, then $d^{\mathsf T}\adj(\Hess P)d\ne0$: its term of greatest old weight comes from that diagonal entry alone. When $\Hess P$ has rank three, the gradient image closure is an irreducible hypersurface. Its reduced equation has nonzero gradient on the smooth locus. The image of $\grad P$ is dense, so this normal, evaluated at $\grad P$, is nonzero over the rational function field. Differentiating the equation shows that the normal spans $\ker\Hess P$. The adjugate is a nonzero scalar multiple of the outer product of this normal with itself. Since a translation direction is tangent to the hypersurface, its adjugate contraction vanishes. Under a constant invertible change from the column $X$ of source coordinates to $X'=SX$, the gradient at $X'$ is $S^{-\mathsf T}\grad f(S^{-1}X')$. Thus an apex $d$ becomes $S^{-\mathsf T}d$. Suppose the source and target coordinates are split into corresponding base and fiber blocks, and the apex plane is the graph of a matrix $U$ from the target base block to its fiber block. The source shear $\mathrm{base}'=\mathrm{base}+U^{\mathsf T}\mathrm{fiber}$ sends this graph to the target base block. At a wall it preserves weights because the graph is invariant under diagonal target scaling. In the applications a variable added to another at a crossing has strictly smaller old weight. Substituting the inverse shear preserves the old leading form, since every term in the expansion involving a replacement variable has smaller old weight than the original term. Let $A,B$ be nonzero weighted homogeneous polynomials in two variables of positive weights, of respective weighted degrees $p,q>0$. If $dA\wedge dB=0$, contracting this equality with the weighted Euler field gives $pA\,dB=qB\,dA$. Clear the weight denominators to take $p,q$ integral. Then $d(A^q/B^p)=0$, so $A^q/B^p\in\C^\times$. For each irreducible factor, its multiplicities in $A,B$ are in the ratio $p:q$. Dividing that ratio by $\gcd(p,q)$ gives $A=aS^r$, $B=bS^s$ for nonzero constants $a,b$, positive integers $r,s$, and a common polynomial $S$. The equality $A=aS^r$ forces $S$ to be weighted homogeneous by comparing its smallest and largest weights. Call $S$ genuinely binary if it is not a polynomial in a nonzero constant linear form. Equivalently, $D_vS$ is nonzero for every nonzero constant base vector $v$, where $D_v$ denotes differentiation in the constant direction $v$. In fact a vanishing derivative makes $S$ independent of the coordinate in direction $v$. If $S$ is a polynomial in a linear form and the base weights differ, that form is proportional to a coordinate. Otherwise its highest ordinary power contains monomials of different weights and the same ordinary degree. In the one-apex form, let $x$ be the source variable corresponding to the apex, of nonzero weight $w_x$. Call it the active variable, and call the three remaining variables $Y=(Y_1,Y_2,Y_3)^{\mathsf T}$ passive. Passive degree means the total degree in $Y_1,Y_2,Y_3$. The part of passive degree at least two is a linear combination over $\C(x)$ of the powers $L^k$ for $k\ge2$, where $L=\ell(x)^{\mathsf T}Y$ for a coefficient column $\ell(x)\in\C(x)^3$. We call $L$ the phase. A phase variable is a passive variable with nonzero coefficient in $L$. When the passive Hessian has rank one, normalize a nonzero entry $\ell_i$ to $1$. The passive Hessian is the second derivative in $L$ times $\ell\ell^{\mathsf T}$, so the ratio of its $(i,j)$ to its $(i,i)$ entry is $\ell_j$. This ratio is homogeneous of weight $w_{Y_i}-w_{Y_j}$, where $w_{Y_k}$ denotes the weight of $Y_k$. A nonzero homogeneous rational function of $x$ has no zero or pole in $\C^\times$, since scaling would give infinitely many. Thus $\ell_j$ is zero or a constant times $x^{(w_{Y_i}-w_{Y_j})/w_x}$ with integral exponent. For $k\ge2$, the coefficient of $L^k$ is the coefficient of $Y_i^k$ in the restriction obtained by setting the other passive variables to zero, so it is polynomial in $x$. \subsection{Two fiber variables with a common binary polynomial} \begin{lemma}\label{lem:cotangent-propagation} Suppose the leading form is $R(x,y)+zA(x,y)+wB(x,y)$ with $A=aS^r$ and $B=bS^s$ for nonzero constants $a,b$, positive integers $r,s$, and a genuinely binary polynomial $S$. The weights of $x,y$ are positive, the weights of $z,w$ are distinct, and $W>0$. Then after a constant linear change $f$ is jointly affine in two variables, with a Keller map as its pair of coefficients. \end{lemma} \begin{proof} Raise both fiber weights at unit speed, fixing the base weights. After a small step the leading form is $H=zA+wB$. The value of $W$ is constant. Define \[ U=S_y^2 S_{xx}-2S_xS_yS_{xy}+S_x^2S_{yy}. \] If $U=0$, the tangent direction is constant on each irreducible component of a regular generic level of $S$. On the chart $S_y\ne0$, the derivative of $S_x/S_y$ along $S_y\partial_x-S_x\partial_y$ equals $U/S_y^2$. A rational function on a complex curve with zero differential is constant. On a component with $S_y=0$, the chart $S_x\ne0$ gives the same conclusion with the roles of $x,y$ reversed. A constant tangent direction forces a linear form annihilating that direction to be constant on the component. Its closure is then a line contained in the level. Lines in different levels are disjoint and hence parallel. Comparing each component with a different level gives a fixed direction along all generic components. The derivative of $S$ in that direction vanishes on a dense set, so $S$ is a polynomial in a constant linear form. This contradiction gives $U\ne0$. Write $A',A'',B',B''$ for the first and second derivatives of the one-variable polynomials $aT^r,bT^s$ at $T=S$. The base Hessian of $H$ is \[ (zA'+wB')\Hess S+(zA''+wB'')(\grad S)(\grad S)^{\mathsf T}. \] The cofactor complementary to $z$ is the determinant of this two-by-two block bordered by $B'\grad S$ and a zero diagonal entry. Expanding along the last row and column evaluates it as the negative of the base Hessian on $B'(S_y,-S_x)^{\mathsf T}$. The outer-product term vanishes on this vector, so the cofactor is $-(B')^2(zA'+wB')U$. The cofactor complementary to $w$ is $-(A')^2(zA'+wB')U$. Both are nonzero. The minors on $v$ and one of the fiber axes are $-(D_v A)^2$ and $-(D_v B)^2$, respectively, for any nonzero constant base vector $v$. Since $D_v A=A'D_vS$ and $D_v B=B'D_vS$, both minors are nonzero. At a first wall the two cofactors survive, so the new form has rank three. Its apex space exists by Theorem~\ref{thm:r4}. It has zero intersection with the fiber plane: this intersection is weight invariant, and the two distinct fiber weights and nonzero cofactors exclude both fiber axes. If the apex space were a line, its weight block would contain at most one fiber axis. Choose the other fiber and an old-weight-homogeneous base direction annihilated by the apex. Such a base direction exists because the two base weights are either equal, or the apex has at most one base component. The surviving mixed minor in these directions contradicts complementary rank at most one. The apex space is therefore a graph over the base plane. The transpose graph shear puts the new form into jointly affine form in the two fibers. At a wall a distance $\tau>0$ along this segment, a monomial of total fiber degree $j$ tied at the wall has old weight $E+(1-j)\tau$. In particular its fiber-free part vanishes and its fiber-linear part in the old coordinates is exactly $zA+wB$. The inverse shear fixes the two fibers and is the identity on the base at $z=w=0$. Substituting it into a form with zero fiber-free part leaves the fiber-linear part unchanged. Thus the shear restores $H$ exactly, and the weights can continue increasing at the same speeds. Every genuine wall requires a base/fiber weight equality. The base weights stay fixed, so the four base/fiber pairs yield at most four walls. Beyond the last wall, a term of total fiber degree at least two would overtake $H$. The full polynomial is therefore jointly affine in the fibers. Lemma~\ref{lem:lift} gives the required planar coefficient map. \end{proof} \subsection{A phase involving three variables} \begin{lemma}\label{lem:three-phase} Under the standing hypothesis on $f$, its leading form cannot be \[ a(x)(Y_1+b x^rY_2+c x^sY_3)^m \] with a nonzero polynomial $a(x)$, $m\ge2$, $bc\ne0$, distinct exponents $0,r,s\in\mathbb Z$, active weight $w_x=1$, three distinct passive weights greater than one, and $W>0$. The expression is understood to be polynomial even if some phase coefficients are Laurent monomials. \end{lemma} \begin{proof} For a polynomial $F(x,L)$ over $\C(x)$ in the passive linear form $L=\ell(x)^{\mathsf T}Y$, the passive Hessian is $F_{LL}\ell\ell^{\mathsf T}$. The mixed Hessian row is $(F_{xL}+F_{LL}\partial_x L)\ell^{\mathsf T}+F_L(\ell')^{\mathsf T}$, where a prime denotes differentiation in $x$. To compute the principal cofactor deleting a passive variable, keep the other two passive indices $i,j$. Expansion along the active row and column yields $-F_{LL}$ times the square of the mixed row evaluated on $(\ell_j,-\ell_i)^{\mathsf T}$. Thus this cofactor equals $-F_{LL}F_L^2(\ell_i'\ell_j-\ell_j'\ell_i)^2$. For coefficients proportional to $1,x^r,x^s$ with distinct exponents all three cofactors are nonzero. Raise all passive weights at unit speed. The leading weight has slope $m$ and $W$ has positive slope $4m-6$. At the first new wall the cofactors survive. The wall form has rank three, and its apex space is the target $x$ axis, since the distinct weights and surviving cofactors exclude all other axes and Theorem~\ref{thm:r4} makes the apex space nonzero. A term of passive degree $j$ tied at the wall a distance $\tau>0$ from the preceding weights has old weight $E+(m-j)\tau$. Thus degrees below $m$ are absent and the degree-$m$ part is the old form. In the complementary rank-one representation its pure $Y_1^m$ coefficient is nonzero. With the $Y_1$ coefficient of the linear form normalized to one, the ratios of the $Y_1^{m-1}Y_j$ coefficients to $m$ times the $Y_1^m$ coefficient recover the other two phase coefficients. They are exactly $b x^r,c x^s$. The highest power at a genuine wall therefore has degree greater than $m$. A small common raise retains that power with the same phase. The passive degree of the full polynomial bounds this strictly increasing sequence. The final degree equals the maximal passive degree $D$ of the full polynomial, since a term of larger passive degree would overtake the final form. Let $N_D$ be the sum of the terms of passive degree $D$. Its Hessian determinant is zero by weights $(0,1,1,1)$, for which $W=4D-6>0$. At any finite point on the final segment, the leading form of $N_D$ is the same as that of $f$. Its three nonzero passive cofactors survive the addition of the lower-weight terms of $N_D$. Each nonzero passive vector has a unique nonzero coordinate of greatest weight on that segment, so its adjugate contraction is nonzero. Under the weights $(0,1,1,1)$, the apex space splits into its active and passive intersections. The passive intersection vanishes, and Theorem~\ref{thm:r4} gives the active axis as the apex space. It follows that $N_D=a(x)(l(x)^{\mathsf T}Y)^D$ for a rational coefficient vector $l$. Its components are linearly independent over $\C$: otherwise a constant passive kernel would make the rank-three gradient image a hyperplane with three independent translations. The Wronskian, the determinant with rows $l^{\mathsf T},(l')^{\mathsf T},(l'')^{\mathsf T}$ with primes denoting derivatives in $x$, is nonzero. To see this, at any point where they are regular, use constant row operations on the three independent Taylor series to obtain distinct leading orders $n_1,n_2,n_3$. The coefficient of order $n_1+n_2+n_3-3$ in their Wronskian is the product of their nonzero leading coefficients and the determinant with rows $1$, $n_i$, and $n_i(n_i-1)$ indexed by columns $i=1,2,3$. This Vandermonde determinant is nonzero. Choose $x_0$ where the original coefficients have no poles, $a(x_0)\ne0$, and $l(x_0),l'(x_0),l''(x_0)$ are independent. Translate $x=x_0+\xi$ and take the three passive coordinates \[ u=l(x_0)^{\mathsf T}Y,\qquad v=l'(x_0)^{\mathsf T}Y,\qquad t=\tfrac12 l''(x_0)^{\mathsf T}Y. \] In these coordinates the Taylor series of $l(x_0+\xi)^{\mathsf T}Y$ has initial terms $u+\xi v+\xi^2t$, and every remaining coefficient is of order at least three in $\xi$. Thus a monomial $\xi^ku^iv^jt^h$ of $N_D$ satisfies $k\ge j+2h$. Its terms with equality give \[ P_0=a(x_0)(u+\xi v+\xi^2t)^D. \] Subtract the affine part of the full polynomial and assign weights $(-\eta,1,1+\eta,1+2\eta)$ to $(\xi,u,v,t)$, with $\eta>0$. The weight of $\xi^ku^iv^jt^h$ is $i+j+h+\eta(j+2h-k)$. For passive degree $d=i+j+h$ with $2\le d0$. Then $\grad f$ is a polynomial automorphism and $p=1$. \end{lemma} \begin{proof} Raise the weights of $y,w$ at unit speed and that of $z$ at speed $m$. The leading weight has slope $m$, so $W$ has slope $2m-4\ge0$. For a form $F(x,L)+B(x)z$ with $L=w+r(x)y$, primes denoting derivatives in $x$, the vector \[ n=(0,-B',F_Lr',rB')^{\mathsf T} \] is in the Hessian kernel in coordinate order $(x,y,z,w)$. The passive rows annihilate it because their passive contributions involve only the phase direction $(r,0,1)$ and the $x$ component of $n$ is zero. In the active row the difference between the $xy$ entry and $r$ times the $xw$ entry is $F_Lr'$, while the $xz$ entry is $B'$. These terms cancel on $n$. The cofactor deleting $y$ is $-F_{LL}(B')^2$ by expansion in its $z$ row. Since it is nonzero, the Hessian has rank three, and its adjugate is $-F_{LL}nn^{\mathsf T}$ by comparison of the $yy$ entry. With $r=x^A$, $B=b x^p$, and $F=a x^\ell L^m$, all three passive cofactors are nonzero. Before a wall the old passive weights can be taken distinct, since both differences of the $z$ weight from a phase weight have slope $m-1>0$. The old contraction criterion excludes every passive apex even at a repeated new weight. The active weight is strictly smaller than each passive weight, so weight invariance separates its axis from the passive block. Theorem~\ref{thm:r4} therefore makes the apex space the target $x$ axis. Let $\gamma,\delta$ be the weights of $z,w$ at the wall, and let $E$ be its leading weight. The pure $w^m$ term allows normalization of the coefficient of $w$ in the passive linear form. For a monomial in $x,y,w$ of passive degree $j$ at the wall a distance $\tau>0$ from the old weights, the old weight is $E_{\mathrm{old}}+(m-j)\tau$. Consequently the restriction to $z=0$ has no terms of passive degree below $m$, and its degree-$m$ part agrees with $a x^\ell(w+x^A y)^m$. Comparing the pure $w^m$ and the $yw^{m-1}$ coefficients fixes the coefficient of $y$ in the phase. By the homogeneous ratio formula, the resulting phase is \[ w+x^A y+kx^{\delta-\gamma}z. \] Every phase power of degree $j\ge2$ has coefficient proportional to $x^{E-j\delta}$, by its pure $w^j$ term, and only powers $j\ge m$ occur. The coefficient of a term linear in $z$ alone has old weight $E_{\mathrm{old}}$ when multiplied by $z$, because $z$ has slope $m$. Thus the affine part is exactly $b x^p z$. If the new $z$ coefficient is nonzero and its exponent differs from $0,A$, take the highest power, with degree $M$, by raising all three phase variables at unit speed for a small step. The affine $z$ term then drops, and Lemma~\ref{lem:three-phase} gives a contradiction. If the exponent is $0$ or $A$, remove it by $w'=w+kz$ or $y'=y+kz$, respectively. Each shear preserves the old leading form, since the weight of $z$ crosses the weight of the affected phase variable from below. Restart the path with the highest phase power $M$ and with $z$ speed $M$. At every wall either $M>m$ or one of the two shears occurs. The degree in the passive variables bounds $M$, and each difference of the $z$ weight from a phase weight strictly increases on all segments. There are at most two shears. After the last wall the path can be extended indefinitely. Each monomial then satisfies \[ \deg_y+\deg_w+m\deg_z\le m, \] where $\deg_y,\deg_w,\deg_z$ in this inequality are the individual monomial exponents. Thus the full potential is $F(x,y,w)+zB(x)$, and the coefficient $b x^p$ survives in $B$. Expanding its Hessian determinant along the $z$ row and column gives $-(B')^2\det\Hess_{y,w}F$. It follows that $B'$ and $\det\Hess_{y,w}F$ are nonzero constants, and $p=1$. The output $f_z$ determines $x$ by an affine inverse. The two outputs $F_y,F_w$ have a polynomial inverse in $y,w$ over $\C(x)$ by the Hessian theorem in two variables over a field of characteristic zero \cite{hess3}. Subtract their values at $(y,w)=(0,0)$ to obtain a map with zero constant term. Its linear coefficient matrix is invertible over $\C[x]$ because its determinant is the constant binary Hessian determinant. To construct its formal inverse, at each total degree in the two output variables solve for the unknown homogeneous part by applying this inverse matrix to the error determined by lower degrees. This constructs a unique formal inverse over $\C[x]$. Over $\C(x)$ it equals the Taylor expansion at the origin of the polynomial inverse of the shifted map. It therefore truncates, and restoring the output shift gives a polynomial inverse over $\C[x]$. Finally $f_x=F_x+B'z$ determines $z$ polynomially. \end{proof} \begin{lemma}\label{lem:unary} Suppose the leading form is $a x^\ell y^m+b x^p z+c x^q w$ with nonzero coefficients and integer exponents $\ell\ge0$, $m\ge2$, $p>q\ge1$. Assume $w_x=1$, the other weights exceed one, and $W>0$. Then there is an inverse or a two-fiber reduction. \end{lemma} \begin{proof} Write $\beta$ for the weight of $y$ and put $E=\ell+m\beta$. The fiber weights are $E-p,E-q$. Raise $\beta$ at unit speed and both fiber weights at speed $m$. Then $W=2(p+q-1-\beta)$, with zero at $\beta=p+q-1$. For a wall at distance $\tau$ along the segment, a tied monomial $x^i y^j z^k w^\ell$ has old weight $E_{\mathrm{old}}+[m-j-m(k+\ell)]\tau$, where $E_{\mathrm{old}}$ is the leading weight at the start of the segment. At a wall before that value, the nonzero old principal cofactors deleting $z,w$ survive. Deleting $z$ gives the determinant in $(x,y,w)$, which expands along the $w$ row to $-H_{yy}(c q x^{q-1})^2$, where $H$ denotes the old leading form. Deleting $w$ similarly gives $-H_{yy}(b p x^{p-1})^2$. Thus the wall form has rank three and has a nonzero apex space by Theorem~\ref{thm:r4}. This space has zero intersection with the fiber plane, whose weights are distinct. If it is a line other than the target $x$ axis, its weight block contains $y$ and at most one fiber axis. The minor in $x$ and the other fiber survives and is nonzero, contrary to complementary rank at most one. If the apex space has dimension two, it is a graph over $(x,y)$. Only $y$ can mix with a fiber at a weight equality. After the transpose shear, the new form is $R(x,y)+zG(x,y)+wK(x,y)$, with $\det D(G,K)=0$. The coefficients have distinct degrees $p,q$. On $y=0$ they equal $b x^p,c x^q$, since the corresponding terms have old weight $E_{\mathrm{old}}$ and the shear preserves the old leading form. If the common polynomial is genuinely binary, Lemma~\ref{lem:cotangent-propagation} applies. Otherwise it is a power of $x$, since the base weights differ and the old $x$ powers are nonzero. This fixes both coefficients exactly. The old-weight formula with $k=\ell=0$ excludes degrees in $y$ below $m$ and fixes the degree-$m$ term. Retain the largest $y$-degree $M$ and both fiber terms by taking slopes $1,M,M$ for $y,z,w$. If $M=m$, there is a genuine wall only when the graph shear is nontrivial. If the apex space is the target $x$ axis, the complementary Hessian has rank one and the phase is \[ y+d x^{\beta-(E-p)}z+e x^{\beta-(E-q)}w \] at the wall. The $y^m$ term permits normalization of the coefficient of $y$ to one. Setting $z=w=0$ and comparing old weights excludes all powers below $m$, including affine terms in $y$ and the active-only term. The monomials linear in a single fiber alone have old weight $E_{\mathrm{old}}$ since each fiber has slope $m$. This retains exactly the affine partners $b x^p z,c x^q w$. If a fiber occurs in the phase with exponent zero, add the corresponding constant multiple of that fiber to $y$ as a new coordinate. This shear preserves the old leading form since the fiber weight crosses the weight of $y$ from below. Three remaining nonzero phase entries have distinct exponents, and choosing their highest power by a small common raise contradicts Lemma~\ref{lem:three-phase}. With two entries, raise the two phase weights at speed one and the absent variable at the speed of the highest power. That selects the largest power and the absent variable's affine partner, to which Lemma~\ref{lem:two-phase} applies after naming the phase variables by their increasing weights. With one phase entry, restart at the slopes $1,M,M$. All restarts use finite support and a sufficiently small positive step. Both fiber weights minus $\beta$ strictly increase on every segment, so at most two shears occur. Every other continuation increases $m$ to a higher passive degree. The finite sequence therefore exits by a previous lemma or reaches $W=0$ with positive weights. Lemma~\ref{lem:endpoint} gives the inverse there. \end{proof} \begin{lemma}\label{lem:unique-min} Suppose a leading form has Hessian rank three, $W>0$, and a one-dimensional apex space. Assume the weights are positive and the apex lies in a smallest-weight block. Then there is an inverse or a two-fiber reduction. \end{lemma} \begin{proof} A constant change within that block makes the apex the active coordinate $x$, and we scale the weights to make its weight one. The complementary rank-one form has a phase linear in the passive variables over $\C(x)$. Normalize one nonzero phase coefficient to one. The others are constant multiples of Laurent monomials in $x$, by weight invariance of the passive Hessian image. Constant changes within the passive weight blocks leave at most one phase entry per block. A variable absent from the phase has a nonconstant affine coefficient. Otherwise it gives a constant source kernel and a hyperplane image, contradicting uniqueness of the apex. Two absent variables cannot have equal weights because their affine coefficients would be proportional monomials. Let $m\ge2$ be the degree in the phase. Raise its variable weights at speed one and each absent variable at speed $m$. For a small step the leading form is the highest phase power together with the affine terms in the absent variables. All passive weights now exceed one. With three phase variables Lemma~\ref{lem:three-phase} gives a contradiction. With two phase variables, normalize the coefficient of the larger-weight variable to one. The other coefficient is a nonzero constant times a positive integral power of $x$, since its exponent is the difference of their weights. Absorb the constant by rescaling that variable and apply Lemma~\ref{lem:two-phase}. With one phase variable, the other two variables have nonconstant affine coefficients of distinct positive integral exponents, so Lemma~\ref{lem:unary} applies. In each case the coefficient of the pure power of the greatest-weight phase variable is a polynomial in $x$, which gives the nonnegative exponent in the required normal form. \end{proof} \begin{lemma}\label{lem:three-partners} Suppose the leading form is $\sum_{i=1}^3 b_i x^{p_i}Y_i$, with $b_i\ne0$, integers $p_1>p_2>p_3\ge1$, $w_x=1$, passive weights greater than one, and $W>0$. Then there is an inverse or a two-fiber reduction. \end{lemma} \begin{proof} Raise the three passive weights at unit speed. The leading weight has slope one and $W$ has slope $-2$. Before $W=0$ the old minors in $(x,Y_i)$ survive, since each is $-(b_i p_i x^{p_i-1})^2\ne0$. The passive weights have the form $E-p_i$ and stay distinct. A constant source kernel cannot occur, since it would contain a weight axis and each axis is excluded by an old mixed minor. A rank-two wall therefore has one apex. It must be $x$, because the complementary variables otherwise include an old mixed minor. The rank-two affine form then fixes all three partners by weight, and an active-only term would have exceeded the old leading weight. This cannot give a genuine wall. At a rank-three wall the apex space has dimension one or two by Theorem~\ref{thm:r4} and the exclusion of a constant kernel. A single apex again has to be $x$, so Lemma~\ref{lem:unique-min} applies. A two-apex plane must contain $x$ because the complementary pair otherwise has a surviving mixed minor. The potential then is jointly affine in two fibers with coefficient degrees two of the $p_i$. Their common polynomial either is genuinely binary, giving Lemma~\ref{lem:cotangent-propagation}, or is a power of $x$ by the old nonzero $x$ powers. In the latter case both coefficients are unchanged. A genuinely new term must be nonlinear in the remaining passive variable, since an active-only term would have exceeded the old degree and the old form fixes all passive-linear terms. Retaining its highest power and the two affine partners by the paired raise gives Lemma~\ref{lem:unary}. If the first wall occurs at $W=0$, or if no wall occurs by then, the leading form has nonzero constant Hessian determinant and the weights are positive. Lemma~\ref{lem:endpoint} applies. \end{proof} \subsection{A pure power of a lower-weight variable} \begin{lemma}\label{lem:low-base} Suppose the leading form is $a y^M+b x^p z+c x^q w$, with $abc\ne0$ and integers $00$. It is greater than $1$ after any positive step because $M\beta-q\ge p-q\ge1$ initially. The two fiber weights differ by $p-q>0$, so the only possible equalities are \[ M\beta-p=\beta,\qquad M\beta-p=1,\qquad \beta=1. \] The corresponding values of $\beta$ are $p/(M-1)$, $(p+1)/M$, and $1$. They are in increasing order if $pM\beta_{\mathrm{old}}$. The affine terms in $z,w$ are fixed by the old comparison. The form after the shear is therefore the original leading form. Its gradient image closure is the product of $\C^2$ with the closure of the curve $(b x^p,c x^q)$: for each $x\ne0$, the $x$-output varies freely with the fibers and the $y$-output varies freely with $y$. Hence there are at least two independent apex directions, contradicting the one-dimensional apex space. If the apex space is a plane, it is transverse to the fiber plane. The graph shear puts the form into $R(x,y)+zG(x,y)+wK(x,y)$. It preserves the old leading form since it only mixes a fiber of smaller old weight into a base variable. The coefficients of degrees $p,q$ have vanishing Jacobian by Lemma~\ref{lem:lift} and $W>0$. They therefore have a common weighted homogeneous polynomial. Their restrictions to $y=0$ are $b x^p,c x^q$ by the old-weight formula for the monomials linear in one fiber and independent of $y$. If it is genuinely binary, apply Lemma~\ref{lem:cotangent-propagation}. Otherwise the old nonzero powers of $x$ make it a power of $x$ for unequal base weights. At $\beta=1$ it may be a power of $x+\lambda y$, rectified by an additional old-weight-preserving shear. This recovers both old affine partners. The same comparison of $y$ powers fixes $R=a y^M$: higher powers exceed the new weight, and lower powers exceed the old weight. Hence every continuing wall is a constant shear at one of the three displayed equalities, with the same leading form afterward. There are finitely many such walls. If there is no earlier exit, the path reaches $\beta=p+q-1$ with positive weights and $W=0$. At that parameter the full leading form has the original nonzero constant Hessian determinant, including if another monomial ties there. Lemma~\ref{lem:endpoint} gives a polynomial inverse. \end{proof}