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\title{A singular normal affine surface\\with free tangent sheaf}
\author{OpenAI}
\date{October 29, 2026}
\begin{document}
\maketitle
\begin{abstract}
We construct a finitely generated normal integral $\CC$-algebra $A$
of Krull dimension two whose module of derivations
$\Der_{\CC}(A)$ is free of rank two, though $A$ has a regular
localization at a nonregular maximal ideal. This disproves the
Lipman--Zariski conjecture in characteristic zero. The surface
constructed can be chosen smooth away from a single point. Its
smooth locus has logarithmic Kodaira dimension two, and iterative
blow-up at the module of derivations does not resolve the singular
point.
\end{abstract}
\section{Introduction}\label{sec:intro}
\subsection*{The question}
Let $A$ be a finitely generated integral $\CC$-algebra. On the
affine scheme $\Spec A$ the sheaf of vector fields is
\[
\Der_{\CC}(A)
=\Hom_A(\Omega^1_{A/\CC},A),
\]
where $\Omega^1_{A/\CC}$ is the module of K\"ahler differentials.
If $\Spec A$ is smooth of dimension $d$, this module is free of
rank $d$. The Lipman--Zariski conjecture asserts the converse
over fields of characteristic zero: free derivation module of rank
$d$ should force regularity, i.e.\ smoothness.
We prove this conjecture false over $\CC$.
\begin{theorem}\label{thm:main}
There exist a finitely generated normal integral $\CC$-algebra $A$
of Krull dimension two and a maximal ideal $\m\subset A$ such that
\[
\Der_{\CC}(A)\simeq A^{\oplus2}
\qquad\text{and}\qquad
A_\m\text{ is not regular}.
\]
\end{theorem}
The surface $\Spec A$ can be chosen to be smooth away from $\m$.
The singular local ring $A_\m$ has geometric genus at least $33$
and canonical discrepancy $-5$; its resolution has a unique
exceptional curve of genus $21$ and self-intersection $-10$.
\subsection*{Why it matters, and what was known}\label{subsec:context}
Lipman's 1965 paper on free derivation modules shows that local
freeness of the tangent sheaf forces normality~\cite{Lipman1965}.
The conjecture asks whether that hypothesis excludes the remaining
normal singularities.
Much is known in the positive direction. In positive characteristic
the failure is easy to produce: over a perfect field of
characteristic $p$, the surface $xy-z^p=0$ has tangent sheaf freely
generated by $x\partial_x-y\partial_y$ and $\partial_z$
\cite{GrafPositive2022}. Over $\CC$ the situation is subtler. The
classes of singularities for which the conjecture was previously
known to hold include:
\begin{itemize}
\item positively graded cones with isolated singularity and
$\m$-adically free derivation module (Hochster, 1977
\cite{Hochster1977});
\item hypersurfaces (Scheja--Storch 1972
\cite{SchejaStorch1972});
\item locally complete intersections in characteristic zero
(K\"allstr\"om, corrected 2018, \cite{Kallstrom2011});
\item isolated singularities in dimension at least three
(van Straten--Steenbrink 1985
\cite{VanStratenSteenbrink1985});
\item log canonical spaces (Greb--Kebekus--Kov\'acs--Peternell
2011 \cite{GKKP2011}, Druel 2014 \cite{Druel2014},
Graf--Kov\'acs 2014 \cite{GrafKovacs2014});
\item low-genus surfaces, including $p_g\le 1$
(Graf 2021 \cite{GrafLowGenus2021}); more generally $p_g-g-b\le 2$
where $g$ is the total genus of the exceptional curves and $b$ is
the first Betti number of the dual graph (Bergner--Graf 2020
\cite{BergnerGraf2020});
\item normal affine surfaces whose tangent bundle on the smooth
locus is trivial and whose smooth locus has logarithmic Kodaira
dimension at most one (Biswas--Gurjar--Kolte 2014
\cite{BiswasGurjarKolte2014}).
\end{itemize}
The analytic germ we construct is Gorenstein (so log canonical if
discrepancies were at least $-1$), but its canonical discrepancy is
$-5$, its exceptional curve has genus $21$, and
$p_g-g-b=33-21-0=12>2$. So it lies beyond every positive result
just listed. In particular the example also shows
$\bar\kappa(X_{\text{smooth}})=2$ in Corollary~\ref{cor:log-kodaira},
answering the analogous global question implicitly posed by the
Biswas--Gurjar--Kolte result.
Two further obstructions are exact. J\"order's theorem
\cite{Jorder2014} says that if the tangent sheaf admits a local
frame of pairwise commuting vector fields at a point, that point is
smooth. Our local frame therefore cannot be made commuting near the
singular point. Second, the Barajas--Ch\'avez-Mart\'inez--
Romano-Vel\'azquez question \cite{BarajasChavezRomano2026} asks
whether iterating the blow-up at the module of derivations resolves
any singularity. Here every such blow-up is already the identity,
so the answer is negative in characteristic zero
(Corollary~\ref{cor:derivation-blowups}).
\subsection*{Proof structure}\label{subsec:structure}
The proof proceeds in two halves: first an analytic construction,
then an algebraization.
\paragraph{Analytic half.}
Fix a smooth complex curve $C$ and a line bundle $L$ with
$L^{\otimes4}\simeq K_C$. In Section~\ref{sec:curve} we construct
$C$ so that:
\begin{itemize}
\item $C$ lies on $\PP^1\times\PP^1$, has genus $21$, and has
fourth-order contact with an auxiliary elliptic curve $\Gamma$
at six marked points $q_1,\ldots,q_6$;
\item there is a two-dimensional space $H^0(C,L)$ spanned by two
sections $r_1,r_2$ whose common zero divisor is the reduced
divisor $Z=q_1+\cdots+q_6$;
\item the trace of a degree-four map $f:C\to\PP^1$ identifies a
certain two-dimensional quotient $W$ of $H^0(C,\Lambda)$ with
$H^0(\Gamma,\Phi)^*$, up to a specific scalar and sign, and
supplies normalized residue data at $Z$.
\end{itemize}
Sections~\ref{sec:trace} and~\ref{sec:jets} then show that the
analytic neighborhood of $C$ of total space of $L^{-1}$ can be
modified by finitely many holomorphic changes of gluing so that
two meromorphic vector fields with well-chosen jets exist. The
leading coefficients of those fields are $r_1,t$ etc., and the
determinant of the two fields is required to be $t^5(1+O(t))$.
Section~\ref{sec:analytic} realizes the neighborhoods on an
actual Hausdorff surface $Y$ and uses cohomology vanishing plus
Grauert's criterion to contract $C$ to a singular point $x$ of a
normal surface $X$ with $\cT_X\simeq\OO_X^2$. Neighboring
singularities are excluded by adjunction/adjunction-discrepancy.
\paragraph{Algebraic half.}
Section~\ref{sec:algebra} uses Artin approximation of a system of
equations together with a finite relation matrix. A finite
determinacy statement (Lemma~\ref{lem:finite-determinacy}) shows
that sufficiently close approximation retains the complete local
ring. Faithful flatness of completion then descends both tangent
freeness and the local ring properties from the completion to a
finitely generated $\CC$-algebra $A$, and an affine neighborhood
shrink produces the globally free module in
Theorem~\ref{thm:main}.
\subsection*{Basic notation}
Line bundles are written additively: $aL+bH$ is
$L^{\otimes a}\otimes H^{\otimes b}$, and $H^q(L)$ is $H^q(C,L)$.
For two rows $X=(X_1,X_2)$, $Y=(Y_1,Y_2)$ put
$X\wedge Y=X_1Y_2-X_2Y_1$. On a quadratic space with form $Q$ we
write $\langle x,x\rangle=Q(x)$ and take the associated bilinear
form to satisfy $(Q(x+y)-Q(x)-Q(y))/2$. A residue on an annular
overlap means the contour integral divided by $2\pi i$. In all
adjunction and residue formulas the normal differential comes first:
locally $\Res_{y=0}((dy/y)\wedge dz)=dz$. All varieties and germs
are over $\CC$. Standard facts about curve cohomology and adjunction
are cited to \cite{Hartshorne1977}; facts about coherent analytic
sheaves and normal spaces to \cite{GrauertRemmert1984}. The main
objects of the paper appear in Table~\ref{tab:main-objects}.
\begin{table}[htbp]
\centering\small
\begin{tabular}{@{}p{0.20\textwidth}p{0.72\textwidth}@{}}
\toprule
Object & Meaning / normalization \\
\midrule
$C$, $Z$ & Smooth genus-$21$ curve; six marked points $q_j$. \\
$H$, $L$ & $\deg H = 4$, $L=H+Z$, $\deg L=10$; $K_C=4L$ (fixed isomorphism).\\
$\Gamma$ & Auxiliary elliptic curve in $\PP^1\times\PP^1$; auxiliary support.\\
$Y$, $\cI$, $\cF$ & Neighborhood of $C$; $\cI=\OO_Y(-C)$, $\cF=\cT_Y(-\log C)$.\\
$\pi:Y\to X$ & Contraction of $C$ to a point $x$; $C^2=-10$. \\
$A$, $\m$ & Affine algebra and marked singular point. \\
\bottomrule
\end{tabular}
\caption{Main objects, in order of use.}
\label{tab:main-objects}
\end{table}
\section{A prescribed curve and its residue data}\label{sec:curve}
We need a smooth projective curve $C$ with a fourth root $L$ of its
canonical bundle, and with six marked points where a specific
quadratic-space structure is available. The plan uses an auxiliary
elliptic curve $\Gamma$: its points supply the six marked points, and
a trace pairing on $C$ later transports the quadratic data from
$\Gamma$ to $C$. This section produces that data. Proposition
\ref{prop:curve-data} is the endpoint; everything later depends only
on it.
\subsection{The elliptic curve}\label{subsec:elliptic}
Let $\Gamma$ be a smooth elliptic curve. Choose
\begin{itemize}
\item a degree-two line bundle $F_0$ on $\Gamma$, and
\item a line bundle $D_0$ on $\Gamma$ of exact order four in
$\Pic(\Gamma)$ (equivalently, a point of $\Gamma[4]\setminus\Gamma[2]$),
\end{itemize}
and set $S_0=F_0+D_0$. Both $F_0$ and $S_0$ have two-dimensional
spaces of sections, and for a suitable choice of origin on $\Gamma$ the
two associated involutions are $p\mapsto a-p$ and $p\mapsto a+d_0-p$
with $d_0\neq 0$ representing $D_0$. Two distinct points cannot be
interchanged by both involutions, and no point is fixed by both because
their difference is a nonzero translation. The product map
\[
\Gamma\;\longrightarrow\;\PP^1\times\PP^1
\]
is therefore injective and immersive, hence a closed embedding; its
image has bidegree $(2,2)$. From now on $\Gamma\subset\Sigma:=\PP^1
\times\PP^1$ denotes this image. Also write
\smash{$F=\OO_\Sigma(1,0)$}, \smash{$S=\OO_\Sigma(0,1)$} for the two
ruling line bundles on $\Sigma$; on restricting to $\Gamma$ they
coincide with $F_0$ and $S_0$ respectively.
On $\Gamma$ put
\[
\Delta=2D_0,\qquad \Phi=F_0+\Delta,\qquad M=3\Phi.
\]
Then $\Delta$ is nontrivial of exact order two and $2\Delta=0$. The
intersection form on $\Sigma$ gives
\[
\Gamma|_\Gamma=2F_0+2S_0=4F_0+\Delta,
\qquad 2M=6F_0,
\]
where the second equality uses $6\Delta=0$. Consequently
\begin{equation}\label{eq:bundle-identities}
\begin{split}
4M&=12F_0=(2\Gamma+4F)|_\Gamma,\\
2M-(\Gamma+F)|_\Gamma&=2\Phi-(F_0+D_0+S_0)=\Phi,\\
(\Gamma+2F)|_\Gamma-2M&=(F_0+2D_0+S_0)+2F_0-6F_0=\Delta.
\end{split}
\end{equation}
Fix isomorphisms realizing these identities once and for all.
Because $K_\Sigma+\Gamma=(0,0)$ on $\Sigma$ and $\Gamma$ is reduced,
there is a logarithmic two-form $\Omega$ on $\Sigma$ with simple pole
along $\Gamma$ and no other poles or zeros; its residue
$\omega=\Res_\Gamma\Omega$ is a nowhere-vanishing holomorphic
differential on $\Gamma$. Here and throughout, the residue places the
normal differential first. Choose a basis $u=(u_1,u_2)$ of
$H^0(\Sigma,F)$ and put
\[
\eta_\Gamma=u_1\,du_2-u_2\,du_1\in H^0(\Gamma,K_\Gamma+2F_0).
\]
This is independent of the local frame of $F_0$.
\subsection{The auxiliary quadratic form}\label{subsec:quadratic}
On $\Gamma$ the section $\eta_\Gamma/\omega$ of $2F_0=2\Phi$ is where
the auxiliary quadratic space enters.
\begin{lemma}\label{lem:aux-quadratic}
There is a unique $Q\in\Sym^2 H^0(\Gamma,\Phi)$ whose image under
multiplication is $-\eta_\Gamma/\omega\in H^0(\Gamma,2\Phi)$. It is
nondegenerate, and for every $q\in\Gamma$, the projective direction
$[\mathrm{ev}_q]$ is isotropic for $Q$ if and only if $q$ is a
ramification point of the degree-two map $F_0:\Gamma\to\PP^1$.
\end{lemma}
\begin{proof}
Let $\iota_F$ and $\iota_\Phi$ be the two degree-two involutions on
$\Gamma$; they satisfy $\iota_\Phi=t_\Delta\iota_F$ where $t_\Delta$
is translation by the nonzero two-torsion element of $\Delta$. The
fixed points of $\iota_F$ are the four ramification points of
$F_0$. They are paired without fixed points by $\iota_\Phi$ (a
two-torsion translation does not fix them, because $\Delta\neq 0$).
Their total divisor is the pullback under $\Phi$ of a divisor of
degree $2+2=4$ on $\PP^1$; equivalently, the ramification divisor of
$F_0$ on $\Gamma$ belongs to the pullback under $\Phi$ of a zero
divisor of a binary quadratic with two distinct roots on the target.
The corresponding element of $\Sym^2 H^0(\Gamma,\Phi)$ multiplies,
after fixing the identification $2\Phi=2F_0$, to a nonzero multiple of
$\eta_\Gamma/\omega$; scaling it fixes the constant. Uniqueness
follows because pullback $\Phi^*:\Sym^2H^0(\PP^1,\OO(1))\to
H^0(\Gamma,2\Phi)$ is injective ($\Phi$ is surjective of degree two,
not constant). Nondegeneracy and the isotropic-direction statement
are immediate from the identification: the associated bilinear form is
the pullback of a binary quadratic with two distinct roots, which is
nondegenerate by inspection of its matrix.
For any $q\in\Gamma$, evaluation in a local frame of $\Phi$ gives a
projective direction $[\mathrm{ev}_q]\in\PP(H^0(\Gamma,\Phi)^*)$.
By construction, $[\mathrm{ev}_q]$ is isotropic for $Q$ exactly when
$q$ is a ramification point of $F_0$.
\end{proof}
For a quadratic form $Q$ on a two-dimensional complex vector space,
we use the polar form
\[
x\cdot y=\tfrac12\bigl(Q(x+y)-Q(x)-Q(y)\bigr),
\qquad x^2=Q(x).
\]
\subsection{Six normalized vectors in the quadratic space}\label{subsec:six-vectors}
The data we need comes in six copies: a pair of projective directions
in the quadratic space, each direction hit twice, whose sum has
controlled square and whose evaluation normalization is fixed at all
six marked points.
\begin{lemma}\label{lem:six-vectors}
In a two-dimensional nondegenerate complex quadratic space there are
three distinct nonisotropic projective directions, avoiding any
prescribed finite set of directions, and nonzero vectors
$w_1,\ldots,w_6$ using each of these directions twice, with the two
vectors in each pair equal, and satisfying
\begin{equation}\label{eq:six-vector-conditions}
w=\sum_{j=1}^6 w_j,\qquad
w^2=90,\qquad
w_j^2+w_j\cdot w=1\quad(1\le j\le 6).
\end{equation}
\end{lemma}
\begin{proof}
Choose coordinates $Q(x,y)=xy$, so that the polar form is
$(x,y)\cdot(x',y')=(xy'+yx')/2$. Set
\[
h=\sqrt{90}>0,\quad R=\sqrt{1+h^2/4}>0,\quad b=\frac{2h}{R},\quad
d(z)=\Bigl(Rz-\frac h2,\;\frac Rz-\frac h2\Bigr).
\]
A direct expansion gives
\[
d(z)^2+d(z)\cdot(h,h)=R^2-\frac{h^2}{4}=1
\qquad\text{for all }z\in\CC^*.
\]
Let $\lambda\in\CC^*$ be small and general, and let $z_1,z_2,z_3$
be the roots of
\[
X^3-bX^2+b\lambda X-\lambda=0.
\]
Vieta's formulas give
\[
\sum_{i=1}^3 z_i=b,\qquad
\sum_{i=1}^3\frac1{z_i}=b,\qquad
2\sum_{i=1}^3 d(z_i)=(h,h).
\]
Thus $w_j=d(z_{\lceil j/2\rceil})$ for $j=1,\ldots,6$ satisfies
\eqref{eq:six-vector-conditions}.
Distinctness of the three directions $[d(z_1)],[d(z_2)],[d(z_3)]$
follows from the analysis of $\lambda\to 0$: the discriminant of the
cubic is $-4b^3\lambda+O(\lambda^2)$, nonzero for general $t$, so the
roots are distinct. As $\lambda\to 0$ two roots go to zero and one
goes to $b$; the small-root direction is
$(Rz-h/2,R/z-h/2)$ up to scale, whose projective ratio
$z(Rz-h/2)/(R-hz/2)$ has nonzero derivative $-h/(2R)$ at $z=0$, so
the two small roots give distinct directions; the third giving
$(3h/2,(R^2-h^2)/(2h))$ is distinct because the entries are both
nonzero and would otherwise force $h^2=R^2+h^2/4$, impossible.
Finally apply the isometry $(x,y)\mapsto(\nu x,\nu^{-1}y)$ to all six
vectors simultaneously. On nonisotropic projective directions this
multiplies the ratio by $\nu^2$, so finitely many values of $\nu\in
\CC^*$ are avoided by any prescribed finite set. All equations in
\eqref{eq:six-vector-conditions} are preserved.
\end{proof}
Apply Lemma~\ref{lem:six-vectors} to $Q$ from
Lemma~\ref{lem:aux-quadratic}, avoiding the branch values of the
$\Phi$-map. Each of the three directions is $[\mathrm{ev}_q]$ for two
distinct points of $\Gamma$. Label the six points
$q_1,\ldots,q_6$ so that $[w_j]=[\mathrm{ev}_{q_j}]$, and set
\[
Z=q_1+\cdots+q_6\in |3\Phi|=|M|.
\]
Let $\beta$ be a section of $M$ with divisor $Z$. Nonisotropy guarantees
$F_0$ is unramified at all six $q_j$; the six $F_0$-values are not all
equal, since a two-point fiber of $\Gamma\to\PP^1$ cannot contain six
distinct points.
\subsection{Curve with fourth-order contact}\label{subsec:contact}
We want $C\subset\Sigma$ passing through the six marked points with
fourth-order contact with $\Gamma$. That way, jet conditions on $\Gamma$
of length at most four (which are easy: $\Gamma$ is an elliptic curve)
transfer to $C$.
Write $\gamma$ for the section of $\OO_\Sigma(\Gamma)$ defining
$\Gamma$. By \eqref{eq:bundle-identities} the fourth power of $\beta$
in $H^0(\Gamma,4M)$ agrees with the restriction of a section of
$\OO_\Sigma(2\Gamma+4F)$, so we may choose
\[
\kappa\in H^0(\Sigma,\,2\Gamma+4F),\qquad
\kappa|_\Gamma=\beta^4.
\]
Existence holds because the restriction kernel
$\OO_\Sigma(\Gamma+4F)=\OO_\Sigma(6,2)$ has $H^1=0$, so restriction on
sections is surjective.
A general such $\kappa$ defines a smooth connected curve $C$: the
linear system is base-point-free off $\Gamma$ (Bertini), nonzero at
$\Gamma\setminus Z$ for any choice, and at each $q_j$ the normal
derivative can be varied by global sections of $\OO_\Sigma(6,2)$, so
finitely many hyperplane conditions suffice. In particular there is
a choice with the normal derivative at each $q_j$ nonzero.
Concretely, near $q_j$ choose a local equation $y=0$ for $\Gamma$, a
coordinate $z$ on $\Gamma$ vanishing at $q_j$, and compatible frames
in which $\kappa$ has coefficient $k(y,z)$ and $\beta$ has coefficient
$b(z)$, with $k(0,z)=b(z)^4$, $b'(0)\neq 0$, $k_y(0,0)\neq 0$. The
implicit function theorem writes $C$ locally as $y=\varphi(z)$ with
\begin{equation}\label{eq:contact-expansion}
\varphi(z)=-\frac{b'(0)^4}{k_y(0,0)}\,z^4+O(z^5).
\end{equation}
Thus the contact order is exactly four:
$\mathrm{div}_C(\gamma|_C)=4Z$, and the six marked points are read
both on $\Gamma$ and on $C$.
The bidegree of $C$ is $2\Gamma+4F=(8,4)$. The restriction sequence
for $C$ together with $H^1(\Sigma,\OO(-8,-4))=0$ gives
$H^0(C,\OO_C)=\CC$, so $C$ is connected and irreducible. Set
\begin{equation}\label{eq:curve-bundles}
H=F|_C,\qquad L=H+Z,\qquad \theta=2L.
\end{equation}
Let $\xi\in H^0(C,\OO_C(Z))$ be the canonical section. The divisor
equality $\mathrm{div}_C(\gamma)=4Z$ determines an isomorphism
$4\OO_C(Z)\simeq\OO_C(\Gamma)$ by $\xi^4\mapsto\gamma|_C$. Combined
with adjunction $K_C=(\Gamma+4F)|_C$ this gives
\begin{equation}\label{eq:canonical-root}
4L=4H+4Z\simeq(\Gamma+4F)|_C\simeq K_C.
\end{equation}
We fix the last isomorphism by requiring, for a local section $s$ of
$\OO_\Sigma(\Gamma+4F)$,
\begin{equation}\label{eq:adjunction-normalization}
s|_C\longmapsto \Res_C\!\left(\frac{s\gamma\Omega}{\kappa}\right),
\end{equation}
with the normal differential first. This pins down the scalar in
\eqref{eq:canonical-root}.
\subsection{Section counts and evaluation maps}\label{subsec:section-counts}
The key computation concerns a family of adjoint line bundles on
$\Sigma$ whose restrictions give both $K_C-jH$ on $C$ and a line
bundle on $\Gamma$ whose cohomology is elementary.
\begin{lemma}\label{lem:adjoint-jets}
For $0\le j\le 3$ put
\[
\mathcal A_j=\Gamma+(4-j)F=(6-j,2),\qquad
N_j=\mathcal A_j|_\Gamma=(8-j)F_0+\Delta.
\]
Restriction to $C$ identifies $H^0(\Sigma,\mathcal A_j)$ with
$H^0(C,K_C-jH)$ and maps it onto $H^0(\Gamma,N_j)$. For
$1\le \ell\le 4$, evaluation on the length-$\ell$ subschemes at the
six marked points has the same rank on $C$ as on $\Gamma$. The
ranks are:
\begin{equation}\label{eq:rank-table}
\begin{array}{c|c|c|c|c|c}
j & \ell & h^0(\Gamma,N_j) & N_j-\ell Z
& h^0(\Gamma,N_j-\ell Z) & \text{evaluation rank}\\ \hline
0 & 1 & 16 & 5F_0 & 10 & 6\\
1 & 2 & 14 & \Phi & 2 & 12\\
2 & 2 & 12 & \Delta & 0 & 12\\
3 & 2 & 10 & -F_0+\Delta & 0 & 10
\end{array}
\end{equation}
\end{lemma}
\begin{proof}
Restriction to $C$ has kernel $\mathcal A_j-C=(-2-j,-2)$, whose
$H^0$ and $H^1$ vanish by product cohomology; and restriction to
$\Gamma$ has kernel $\mathcal A_j-\Gamma=(4-j,0)$, whose $H^1$
vanishes since $4-j\ge 0$. Adjunction identifies the two images.
By \eqref{eq:contact-expansion} the ideal of the length-$\ell$
subscheme of $C$ at $q_j$, in the local surface $(\Sigma,q_j)$ with
coordinates $(y,z)$, is $(y-\varphi(z),z^\ell)=(y,z^\ell)$ for
$\ell\le 4$. Thus the two length-$\ell$ schemes coincide as embedded
subschemes, hence as $C$- and $\Gamma$-subschemes. Consequently the
evaluation ranks agree.
On $\Gamma$ one has $\deg F_0=2$, $\deg\Delta=0$, so
$N_j-\ell Z=(8-j-3\ell)F_0+(1-\ell)\Delta$ has degree
$(16-2j)-6\ell$. In every row of \eqref{eq:rank-table} this is either
positive (then $h^0$ equals the degree because $\Gamma$ is elliptic)
or zero, in which case the relevant bundle is a nonzero translation
bundle with $h^0=0$, or negative (then $h^0=0$). This gives the
displayed values; subtracting from $h^0(\Gamma,N_j)$ gives the
evaluation ranks.
In particular, $H^0(\Sigma,\mathcal A_3)$ is $12$-dimensional
(equal to $h^0(\Sigma,(3,2))=4\cdot 3$), the kernel in restriction to
$\Gamma$ is two-dimensional, and the resulting $10$-dimensional
system injects into its length-$12$ evaluation space.
\end{proof}
\begin{proposition}[Curve data]\label{prop:curve-data}
The curve constructed above satisfies the following.
\begin{enumerate}
\item Numerical and canonical data.
\[
g(C)=21,\quad \deg H=4,\quad \deg L=10,\quad
L=H+Z,\quad \theta=2L,\quad K_C=4L,
\]
with the last isomorphism given by \eqref{eq:adjunction-normalization}
and $\xi^4\mapsto\gamma|_C$. The curve $C$ has fourth-order contact
with $\Gamma$ at each $q_j$.
\item The map $f:C\to\PP^1$ given by $u$ is finite of degree four,
unramified at each $q_j$, and its values $f(q_j)$ are not all equal.
The zero divisor of $u_1$ can be chosen as a reduced fiber disjoint
from $Z$.
\item Section dimensions.
\begin{equation}\label{eq:section-dims}
\begin{gathered}
h^0(C,jH)=j+1\quad(0\le j\le 3),\\
h^0(C,L)=h^0(C,H+2Z)=2,\qquad
h^0(C,\theta)=3,\qquad
h^0(C,\OO_C(Z))=1,\\
h^0(C,\theta+H)=6.
\end{gathered}
\end{equation}
Moreover
\begin{equation}\label{eq:section-multiplication}
\begin{split}
H^0(C,L)&=\xi H^0(C,H),\\
H^0(C,H+2Z)&=\xi^2H^0(C,H),\\
H^0(C,\theta)&=\xi^2H^0(C,2H).
\end{split}
\end{equation}
In particular $r=\xi u$ is a basis of $H^0(C,L)$ with reduced
common zero divisor exactly $Z$, and every section of $\theta$
vanishes twice along $Z$.
\item For every $k\ge 2$ the evaluation map
\begin{equation}\label{eq:evaluation-surjectivity}
H^0(C,(k+1)L)\longrightarrow H^0(Z,((k+1)L)|_Z)
\end{equation}
is surjective.
\item Auxiliary data on $\Gamma$ is preserved:
\eqref{eq:bundle-identities} and Lemma~\ref{lem:aux-quadratic} hold;
the divisor $Z$ belongs to $|M|$; the six vectors
$w_j\in H^0(\Gamma,\Phi)^*$ have projective evaluation directions at
$q_j$, use three distinct directions, and satisfy
\eqref{eq:six-vector-conditions} with $w^2=90=9\deg L$.
\end{enumerate}
\end{proposition}
\begin{proof}
Contact and connectedness were shown above. From $C=(8,4)$ and
$K_C=(6,2)|_C$, intersection on the quadric gives
\[
\deg H=(1,0)\cdot(8,4)=4,\quad
\deg K_C=(6,2)\cdot(8,4)=40,\quad
g(C)=21.
\]
Since $\deg Z=6$ we have $\deg L=10$, and $4\deg L=40=\deg K_C$, so
$K_C=4L$ as line bundles (with the fixed scalar above).
The map $f:C\to\PP^1$ given by $u$ is the first projection on
$\Sigma$ restricted to the irreducible $C$; it is finite of degree
four. At $q_j$ the contact of order four makes the differential of
$f$ agree with that of $F_0|_\Gamma$; the latter is nonzero because
$F_0$ is unramified at $q_j$. The values $f(q_j)$ are distinct
projective directions equal to the $F_0$-values at $q_j$, which are
not all equal. Choose a regular fiber of $f$ away from these finitely
many values; a determinant-one change of basis of $u$ makes it the zero
divisor of $u_1$. Such a change preserves $\eta_\Gamma$ and $\eta_C$
and thereby the form $Q$.
Section dimensions without $Z$. For $0\le j\le 3$ the sequence
$0\to\OO_\Sigma(j-8,-4)\to\OO_\Sigma(j,0)\to\OO_C(jH)\to 0$ has
kernel with $H^0=H^1=0$, hence $h^0(C,jH)=h^0(\Sigma,\OO(j,0))=j+1$.
The divisor dimension formula. For a line bundle $B$ on $C$ and an
effective divisor $D$,
\begin{equation}\label{eq:dimension-formula}
h^0(C,B+D)-h^0(C,B)
=\deg D-\mathrm{rank}\bigl[H^0(C,K_C-B)\to H^0(D,(K_C-B)|_D)\bigr].
\end{equation}
This is Riemann--Roch followed by Serre duality: tensor the exact
sequence $0\to\OO_C(-D)\to\OO_C\to\OO_D\to 0$ by $B$, use
$h^1(C,B)=h^0(C,K_C-B)^*$, and dualize the restriction map.
Apply \eqref{eq:dimension-formula} with $B=0,Z,2Z$. Specifically:
\begin{itemize}
\item $B=0$, $D=Z$ with $K_C-B=K_C$: kernel is $K_C=C_0+C_1+\cdots+C_5$
sum of $6$ points on $C$; rank of evaluation is row $(0,1)$ of
\eqref{eq:rank-table}, i.e.\ $6$. Then
$h^0(C,\OO_C(Z))-h^0(C,\OO_C)=6-6=0$, so
$h^0(C,\OO_C(Z))=1$.
\item $B=H$, $D=2Z$ with $K_C-H=3H$: rank is row $(1,2)$, i.e.\
$12$. Then $h^0(C,H+2Z)-h^0(C,H)=12-12=0$, so
$h^0(C,H+2Z)=2$.
\item $B=2H$, $D=2Z$ with $K_C-2H=2H$: rank is row $(2,2)$, i.e.\
$12$. Then $h^0(C,2H+2Z)-h^0(C,2H)=12-12=0$, so
$h^0(C,2H+2Z)=3$.
\item $B=3H$, $D=2Z$ with $K_C-3H=H$: rank is row $(3,2)$, i.e.\
$10$. Then $h^0(C,3H+2Z)-h^0(C,3H)=12-10=2$, so
$h^0(C,3H+2Z)=6$.
\end{itemize}
Inclusions $H^0(C,H)\hookrightarrow H^0(C,L)\hookrightarrow
H^0(C,H+2Z)$ between spaces of dimensions $2$ and $2$ force equality
$h^0(C,L)=2$, and $H^0(C,\theta)=H^0(C,2L)=H^0(C,2H+2Z)=3$.
Inclusion $H^0(C,L)=\xi H^0(C,H)\subset H^0(C,L)$: indeed, $r=\xi u$
spans $H^0(C,L)$, giving \eqref{eq:section-multiplication} for the
first line, and the second and third follow by the evident signs in
adjunction and the adjoint-jet identification. The assertion about
the common zero divisor of $r$ follows because $u$ has no base point
and $Z$ is reduced.
Finally, \eqref{eq:evaluation-surjectivity} follows from the divisor
sequence for $N=(k+1)L$, Serre duality, and the containment
$H^0(C,(3-k)L)\hookrightarrow H^0(C,(3-k)L+Z)$: for $k=2$ this is
$L\hookrightarrow H+2Z$ between $2$-dimensional spaces, an iso; for
$k=3$ it is $\OO_C\hookrightarrow\OO_C(Z)$ between $1$-dimensional
spaces, an iso; for $k\ge 4$ the target bundle has degree
$10(3-k)+6\le -4$, so both vanish. Item~5 follows from
Lemma~\ref{lem:six-vectors} and the choice of $Z$.
\end{proof}
Two further facts from Riemann--Roch and Serre duality, used later:
\begin{equation}\label{eq:boundary-cohomology}
h^1(C,L)=h^0(C,3L)=12,\qquad h^0(C,4L)=21.
\end{equation}
Indeed, $h^0(C,L)-h^1(C,L)=10+1-21=-10$ and $K_C-L=3L$, so
$h^1(C,L)=h^0(C,3L)$; and $h^0(C,4L)=h^0(C,K_C)=g=21$.
Finally, in compatible frames $r=zu$ at a marked point, so
$r'(0)=u(0)$. The six rows $r'(q_j)$ span $\CC^2$: their projective
directions are the values $f(q_j)$, which are not all equal, and
changing frames or local parameters rescales each row separately.
\section{The trace form and normalized residues}\label{sec:trace}
Section~\ref{sec:curve} gives us the curve $C$ and a quadratic space
structure on the auxiliary elliptic curve $\Gamma$. To use it on $C$,
we need to transport the quadratic data to $C$. This is the job of the
trace pairing of a finite morphism. The endpoint is
Proposition~\ref{prop:residue-data}: a meromorphic pair of sections
$P\in H^0(C,\theta+Z)^2$ with residues at $Z$ pinned so that the
low-order obstruction in Section~\ref{sec:jets} vanishes.
Throughout we retain the data and normalizations of
Proposition~\ref{prop:curve-data}. Put
\[
\theta=2L,\qquad
\Lambda=\theta+H=3H+2Z,\qquad
f:C\to\PP^1
\]
where $f$ is the degree-four morphism defined by the pencil
$u=(u_1,u_2)\in H^0(C,H)^{\oplus 2}$. Note
$2\Lambda\simeq K_C+2H$.
\subsection{A perfect trace form}\label{subsec:trace-form}
\begin{lemma}\label{lem:trace-pairing}
The rank-four vector bundle $E=f_*\Lambda$ on $\PP^1$ carries a perfect
symmetric pairing
\begin{equation}\label{eq:trace-form}
E\otimes E\longrightarrow\OO_{\PP^1},\qquad
(T_1,T_2)\longmapsto
\Tr_{C/\PP^1}\!\left(\frac{T_1T_2}{\eta_C}\right),
\end{equation}
with $\eta_C=u_1du_2-u_2du_1 \in H^0(C,K_C+2H)$. Its splitting type
is $(3,0,0,-3)$. Multiplication by $\xi^2$ identifies
$\OO_{\PP^1}(3)$ with a positive line subbundle $\mathcal P\to E$. The
form on global sections has radical $\xi^2H^0(C,3H)$, and the induced
form
\[
\langle\ ,\ \rangle_W\quad\text{on}\quad
W=H^0(C,\Lambda)/\xi^2H^0(C,3H),\qquad \dim W=2,
\]
is nondegenerate. For every $x\in\PP^1$, evaluation at $x$ identifies
$W$ with the fiber $(\mathcal P^\perp/\mathcal P)_x$, with the induced
form.
For $q\in Z$ choose local frames $\mathfrak f$ of $H$ and $\tau$ of
$\Lambda$ near $q$ and a parameter $z$ at $q$. Write, using the fixed
identification $2\Lambda=K_C+2H$,
\[
\tau^2=s(z)\,dz\,\mathfrak f^2,\qquad
\eta_C=\mathcal E(z)\,dz\,\mathfrak f^2.
\]
Let $\ell_q\in W^*$ be coefficient evaluation in $\tau$. Then in the
dual form on $W^*$,
\begin{equation}\label{eq:dual-evaluation}
\ell_q^2=\frac{\mathcal E(q)}{s(q)}\neq 0.
\end{equation}
\end{lemma}
\begin{proof}
The map $f$ is finite and flat of degree four. The identity
$2\Lambda=K_C+2H$ identifies $\Lambda^2$ with the relative dualizing
bundle. We check locality and perfection directly.
At a ramification point choose a parameter $w$ with $x=w^e$ on the
target. A frame of $H$ pulled back from the target can be chosen so
that $\eta_C=dx\,\mathfrak f^2$. Write $\tau^2=a(w)dw\,\mathfrak f^2$
with $a$ a unit. The local trace pairing is
$(b,c)\mapsto\Tr(a(w)bc/(e w^{e-1}))$. On the basis $1,w,\ldots,
w^{e-1}$ the pairing without the unit $a$ has matrix
\[
\Tr\!\left(\frac{w^{i+j}}{e w^{e-1}}\right)
=\begin{cases}1&i+j=e-1,\\0&i+j\neq e-1,\end{cases}
\quad 0\le i,j<e.
\]
This is holomorphic and perfect; multiplication by the unit $a$ is an
automorphism. Direct sum over points above a target disk gives
perfection of the pairing.
On an unramified fiber the trace is the sum of its four contributions,
with no division by the degree.
The splitting type: $E$ is a vector bundle on $\PP^1$, so a direct sum
of line bundles. It is self-dual under \eqref{eq:trace-form}, so its
splitting degrees pair as $(a,b,-b,-a)$ with $a\geq b\geq 0$.
Proposition~\ref{prop:curve-data} gives
\[
h^0(E(-1))=h^0(C,\theta)=3,\qquad
h^0(E(-2))=h^0(C,H+2Z)=2.
\]
For $\OO(\lambda)=E$-summand of degree $\lambda$:
$h^0(\OO(\lambda-1))=\lambda$ if $\lambda>0$ and $0$ otherwise;
$h^0(\OO(\lambda-2))=(\lambda-1)$ if $\lambda\ge 2$ and $0$ otherwise.
The solutions of $a^+ +b^+=3$ and $(a-1)^+(b-1)^+=2$ for
$a\ge b\ge 0$ give $a=3$, $b=0$. Hence
$E\simeq\OO(3)\oplus\OO^{\oplus2}\oplus\OO(-3)$.
The map $\OO(3)\to E$ from multiplication by $\xi^2$ has zero
projection to the $\OO$, $\OO$ and $\OO(-3)$ summands (by degree), so
its image $\mathcal P$ is a subbundle equal to the positive summand
and is defined everywhere. Injectivity of $H^0(\PP^1,\OO(3))\to
H^0(C,3H)$ (both 4-dimensional) gives $H^0(\mathcal P)=\xi^2H^0(C,3H)$.
The positive summand $\mathcal P$ is isotropic because
$\Hom(\OO(6),\OO)=0$, and it pairs trivially with the two degree-zero
summands. Perfection gives
$\mathcal P^\perp\simeq\OO(3)\oplus\OO^{\oplus2}$ and
$\mathcal P^\perp/\mathcal P\simeq\OO^{\oplus 2}$ with a nondegenerate
constant form. All global sections of $E$ lie in $\mathcal P^\perp$.
Taking sections gives $W$ is 2-dimensional and the induced form is
nondegenerate.
For \eqref{eq:dual-evaluation}, at $q\in Z$ the fiber algebra is
one-dimensional because $f$ is unramified. The trace pairing restricted
to the $q$-factor has form coefficient $s(q)/\mathcal E(q)$; evaluation
at $q$ vanishes on every other factor and on $\mathcal P$. The metric
dual of coefficient evaluation has, on the factor at $q$, square
$\mathcal E(q)/s(q)$, and it lies in $\mathcal P_{f(q)}^\perp$.
Passing to the quotient preserves the square. Finally $\mathcal E(q)
\neq 0$ by unramifiedness.
\end{proof}
\subsection{Identification by principal parts}\label{subsec:principal-parts}
The trace form gives a quadratic space on $C$. We now identify it with
$H^0(\Gamma,\Phi)^*$ via residue duality.
\begin{lemma}\label{lem:principal-parts}
There is an isomorphism $\iota:W\to H^0(\Gamma,\Phi)$ that identifies
the full principal parts of $T/\xi^2$ on $C$ and of $\iota(T)/\beta^2$
on $\Gamma$ via residue duality, and identifies the projective
evaluations at corresponding points of $Z$. Its dual
$\iota^*:H^0(\Gamma,\Phi)^*\to W^*$ is an isometry for $Q$
(Lemma~\ref{lem:aux-quadratic}) and the dual trace form
(Lemma~\ref{lem:trace-pairing}).
\end{lemma}
\begin{proof}
Set $N_\Sigma=\Gamma+F$, $N_\Gamma=F_0+\Delta+F_0+D_0=5F_0+\Delta$
and $D=2Z$. Fourth-order contact identifies $D$ on $C$ and on $\Gamma$
as the same embedded length-twelve subscheme. The isomorphism
$N_\Sigma|_C\simeq K_C-3H$ comes from adjunction against the fixed
volume (\eqref{eq:adjunction-normalization}). The differential $\omega$
trivializes $K_\Gamma$.
For a line bundle $B$ on a smooth curve, the principal parts of
$B(D)$ at $D$ and the jets of $K-B$ are perfect dual by
\[
(a,s)\longmapsto\sum_{q\in Z}\Res_q(as).
\]
In a parameter and mutually dual frames, coefficients
$a_{-2}z^{-2}+a_{-1}z^{-1}$ pair with $s_0+s_1z$ as $a_{-2}s_1+a_{-1}
s_0$. Thus the two full principal-part spaces
\[
H^0(C,\OO_C(3H+2Z)/\OO_C(3H)),\qquad
H^0(\Gamma,\OO_\Gamma(-N_\Gamma+2Z)/\OO_\Gamma(-N_\Gamma))
\]
are each identified by residues with
$H^0(D,N_\Sigma|_D)^*$.
Both complete adjoint systems are restrictions of the same ambient
system $H^0(\Sigma,N_\Sigma)$. Restriction kernels to $C$ and to
$\Gamma$ are
\[
N_\Sigma-C=(-5,-2),\qquad N_\Sigma-\Gamma=(1,0),
\]
with $H^1=0$ by product cohomology (both indices nonnegative or
appropriate for $H^1=0$). Their images on the common $D$ therefore
coincide. Equivalently, an ambient section vanishing on $\Gamma$
vanishes to order at least four on $C$ at $Z$ and contributes nothing
to these residue tests.
The principal-part exact sequence and Serre duality say the image of
$H^0(C,\Lambda)$ by $T\mapsto\operatorname{pp}(T/\xi^2)$ is precisely
the annihilator of the common adjoint jet space, and its kernel is
$\xi^2H^0(C,3H)$. On $\Gamma$ the same is true for
\[
H^0(\Gamma,\Phi)\longrightarrow
H^0(\Gamma,\OO_\Gamma(-N_\Gamma+2Z)/\OO_\Gamma(-N_\Gamma)),
\quad \widetilde T\longmapsto\operatorname{pp}(\widetilde T/\beta^2),
\]
since $\Phi=2M-N_\Gamma$ and $(\beta)=Z$. This map is injective since
its kernel would be a section of $-N_\Gamma$ (degree $-10$). The two
identifications with the same annihilator yield
$\iota:W\to H^0(\Gamma,\Phi)$, identifying full principal parts. As a
sanity check, $h^0(N_\Gamma)=\deg N_\Gamma=10$ and restriction to $D$
is injective since $\deg(N_\Gamma-2Z)=-2$, so the annihilator of the
image has $12-10=2$ dimensions.
For the metric, near $q\in Z$ choose $(y,z)$ and frames $\mathfrak g$
of $\Gamma$ and $\mathfrak f$ of $F$ with $\gamma=y\mathfrak g$,
$\Omega=e(y,z)(dy/y)\wedge dz$, $e(0,z)=1$. Write
\[
\kappa=k(y,z)\mathfrak g^2\mathfrak f^4,\qquad
\beta=b(z)\mathfrak m,\qquad
\mathfrak m^4=\mathfrak g^2\mathfrak f^4|_\Gamma,
\]
so that $k(0,z)=b(z)^4$ and $b'(0)k_y(0,0)\neq 0$. Put
$K(z)=k_y(0,z)$; then the equation of $C$ reads
\[
y|_C=-\frac{b(z)^4}{K(z)}+O(z^8),\quad
e|_C=1+O(z^4),\quad k_y|_C=K(z)+O(z^4).
\]
The adjunction convention sends the frame
$\mathfrak g\mathfrak f^4$ to
\begin{equation}\label{eq:adjunction-frame}
\Res_C\!\left(\frac{e(y,z)dy\wedge dz}{k(y,z)}\right)
=\frac{e|_C}{k_y|_C}\,dz.
\end{equation}
Sign is positive because $dy\wedge dz=(dk\wedge dz)/k_y$.
Let $a=\xi/z$ be the local frame of $\OO_C(Z)$ and $\tau=\mathfrak
f^3a^2$ the resulting frame of $\Lambda$. The fourth-power
identification sends $a^4$ to $(y|_C/z^4)\mathfrak g$, so by
\eqref{eq:adjunction-frame},
\[
\tau^2=s(z)\,dz\,\mathfrak f^2,\qquad
s(z)=\frac{e|_C y|_C}{z^4 k_y|_C},\qquad
s(0)=-\frac{b'(0)^4}{k_y(0,0)^2}.
\]
The minus sign comes from solving for $y|_C$.
Write $T=\pi(z)\tau$ and $\iota(T)=\widetilde T=\widetilde\pi(z)
\widetilde\tau$ in the frame
$\widetilde\tau=\mathfrak m^2/(\mathfrak g\mathfrak f)$ of $\Phi$.
For an ambient test section $A(y,z)\mathfrak g\mathfrak f$ the two
residue tests are
\[
\Res_0\!\left(\frac{A|_C\,\pi\,e|_C}{k_y|_C\,z^2}\,dz\right),\qquad
\Res_0\!\left(\frac{A(0,z)\widetilde\pi}{b(z)^2}\,dz\right).
\]
The restrictions of $A$ to the two curves agree modulo $z^4$; the
equality of the functionals for all test jets is exactly
\[
\operatorname{pp}\!\left(\frac{\pi}{K(z)z^2}\right)
=\operatorname{pp}\!\left(\frac{\widetilde\pi}{b(z)^2}\right),
\qquad
\widetilde\pi\equiv\frac{(b(z)/z)^2}{K(z)}\,\pi\pmod{z^2}.
\]
Writing $b=b_1z+b_2z^2+\cdots$, $K=K_0+K_1z+\cdots$,
$\pi=\pi_0+\pi_1z+\cdots$, both principal part functionals yield the
two-coefficient triangular transformation
\[
\widetilde\pi_0=\frac{b_1^2}{K_0}\pi_0,\qquad
\widetilde\pi_1=\frac{b_1^2}{K_0}\pi_1
+\left(\frac{2b_1b_2}{K_0}-\frac{b_1^2K_1}{K_0^2}\right)\pi_0.
\]
This is invertible and coordinate-independent.
For the metric, write $\eta_C=\mathcal E_C\,dz\,\mathfrak f^2$ and
$\eta_\Gamma=\mathcal E_\Gamma\,dz\,\mathfrak f^2$. Tangency gives
$\mathcal E_C(0)=\mathcal E_\Gamma(0)$. If $\ell$ and $\widetilde\ell$
are coefficient evaluations in $\tau$ and $\widetilde\tau$, then
$\iota^*\widetilde\ell=\alpha\ell$ with $\alpha=b_1^2/K_0$. By
\eqref{eq:dual-evaluation} and the formula for $s(0)$,
\[
(\iota^*\widetilde\ell)^2
=\alpha^2\frac{\mathcal E_C(0)}{s(0)}
=-\mathcal E_\Gamma(0).
\]
The compatible bundle identification gives $\widetilde\tau^2=
\mathfrak f^2$, so the last quantity is exactly the value assigned by
$Q$ to $\widetilde\ell\in H^0(\Gamma,\Phi)^*$, since the section
defining $Q$ multiplies to $-\eta_\Gamma/\omega$. The evaluations at
$Z$ have three distinct projective directions; a binary quadratic
that matches $Q$ at all three is identically equal. Hence $\iota^*$
is an isometry, with sign and scalar fixed.
\end{proof}
\subsection{The meromorphic section with prescribed residues}\label{subsec:residue-data}
We now convert the abstract isometry into concrete pointwise data at
the six marked points.
On a disk centered at $q_j$ choose a parameter $z_j$ and a frame
$\ell_j^{\mathrm{fr}}$ of $L$ whose fourth power maps to $dz_j$ under
the fixed isomorphism $L^4=K_C$. Set
\[
a_j=\xi/z_j,\quad
\mathfrak f_j=\ell_j^{\mathrm{fr}}/a_j,\quad
\tau_j=(\ell_j^{\mathrm{fr}})^2\mathfrak f_j;
\]
these frame $\OO_C(Z)$, $H$ and $\Lambda$ respectively. If
$r=\xi u$, then $r=z_j u$ in these frames. The fiber-linear function
$t_j$ on $L^{-1}$ dual to $\ell_j^{\mathrm{fr}}$ gives compatible
split coordinates $(t_j,z_j)$.
Since $\tau_j^2=dz_j\,\mathfrak f_j^2$, Lemma~\ref{lem:trace-pairing}
says that if $\ell_j\in W^*$ denotes coefficient evaluation in
$\tau_j$, then
\begin{equation}\label{eq:compatible-norm}
\ell_j^2=(u_1u_2'-u_2u_1')(0).
\end{equation}
Here and below primes denote $z_j$-derivatives.
\begin{proposition}[Residue data]\label{prop:residue-data}
After possibly a determinant-one change of basis $u$ (preserving
$Q$, $L$, and the normalization), with the zero divisor of $u_1$ a
reduced fiber of $f$ disjoint from $Z$, the following hold. Set
$r=\xi u$, $v_j^*=\iota^*w_j\in W^*$ and $V=\sum_{j=1}^6 v_j^*$. For
any mutually disjoint compatible coordinate disks as above, there are
nonzero constants $c_j\in\CC$ and a pair
$P\in H^0(C,\theta+Z)^{\oplus 2}$ such that:
\begin{enumerate}
\item $c_j\ell_j=v_j^*$ for each $j$;
$V^2=90=9\deg L$; and $(v_j^*)^2+v_j^*\cdot V=1$ for each $j$.
\item On disk $j$, in the compatible frames, put
$m_j=c_j/z_j$, $R_j=m_jr$, $W_j=R_j(0)\wedge R_j'(0)$, and
$p_j=P-m_jr'$. Then $P$ has principal part $R_j(0)/z_j$
in the frame $(\ell_j^{\mathrm{fr}})^2$ of $\theta$; the rows
$R_j,p_j$ are holomorphic near $0$; $R_j(0)\neq 0$.
\item Residue identities at each $j$:
\[
W_j=(v_j^*)^2,\qquad p_j(0)\wedge R_j(0)=1,\qquad
\sum_{j=1}^6(W_j-1)=-9\deg L.
\]
\item Global residue pairing. With $J=u\wedge P\in H^0(C,\Lambda)$,
\[
\langle[J],[T]\rangle_W=-V([T])
\quad\text{for every }T\in H^0(C,\Lambda).
\]
\item Changing $P$ by a holomorphic pair of sections of $\theta$ does
not change $[J]$. The pole prescription is invariant under changes
of compatible coordinates.
\end{enumerate}
\end{proposition}
\begin{proof}
A determinant-one change of basis of $u$ preserves $\eta_C$ and
$\eta_\Gamma$ and hence the form $Q$. Avoiding the finitely many
branch values and the image of $Z$, we can make $u_1$'s zero divisor
a suitable reduced fiber. The isometry $\iota^*$ transports the
equations for $w_j$ to
$c_j\ell_j=v_j^*$; since $\ell_j$ is nonzero and $v_j^*\neq 0$ there
is a unique such $c_j$. Locally $R_j=c_ju$ is holomorphic and nonzero
at the origin, and \eqref{eq:compatible-norm} gives
$R_j(0)\wedge R_j'(0)=c_j^2\ell_j^2=(v_j^*)^2$.
Every prescribed set of simple principal parts of $\theta$ at $Z$ is
realisable by a section of $\theta+Z$: the obstruction lies in
$H^1(C,\theta)$, which by Serre duality pairs with
$H^0(C,K_C-\theta)=H^0(C,\theta)$. Every section of $\theta$ vanishes
twice along $Z$ (Proposition~\ref{prop:curve-data}(3)), so all such
residues vanish. Applying this separately to both entries produces
$P$ with the prescribed principal part. Then $m_jr'$ has the same
simple pole part, so $p_j=P-m_jr'$ is holomorphic.
The simple pole of $P$ at $q_j$ is proportional to $u(q_j)$; hence
$u\wedge P$ is holomorphic there. Away from $Z$ it is already
holomorphic. So $J=u\wedge P\in H^0(C,\Lambda)$. To compute its
class, fix $T\in H^0(C,\Lambda)$ and consider the meromorphic
canonical differential $\nu=P_1T/u_1$; its degree is
$\theta+(\theta+H)-H=2\theta=K_C$. Its only possible poles are at $Z$
and at the simple zero fiber of $u_1$. At $q_j$ the compatible frames
make the canonical product equal to the coefficient product times
$dz_j$, so $\Res_{q_j}\nu=c_j\ell_j(T)$. At a zero $q$ of $u_1$, the
section $u_2$ is nonzero, $J(q)=-u_2(q)P_1(q)$ and
$\eta_C(q)=-u_2(q)du_1(q)$, so the residue is
$(JT/\eta_C)(q)$. Summing residues on the zero fiber gives the value
of $\Tr(JT/\eta_C)$ there; this trace is a global holomorphic function
on $\PP^1$ (Lemma~\ref{lem:trace-pairing}), hence constant. The
residue theorem gives
\[
\langle[J],[T]\rangle_W
=-\sum_j c_j\ell_j(T)=-V([T]),
\]
which is the stated sign.
Locally, since $r=zu$,
\[
P=p+\frac{c}{z}u+cu',\qquad cJ=R\wedge p+R\wedge R'.
\]
So $c_j\ell_j(J)=R_j(0)\wedge p_j(0)+W_j$ on disk $j$. On the other
hand, \eqref{eq:trace-form} applied to $\iota^*$ gives
\[
c_j\ell_j(J)=-v_j^*\cdot V=W_j-1,
\]
because $v_j^*\cdot V=1-W_j$. Thus $R_j(0)\wedge p_j(0)=-1$, i.e.\
$p_j(0)\wedge R_j(0)=1$. Summing $W_j-1=-v_j^*\cdot V=-v_j^*\cdot V$
over $j$ yields $\sum_j(W_j-1)=-V\cdot V=-V^2=-90=-9\deg L$.
The useful local expansion, used below, is
\begin{equation}\label{eq:P-local-expansion}
P=\frac{R_j(0)}{z_j}+p_j(0)+2R_j'(0)+O(z_j).
\end{equation}
If $P$ is replaced by $P+S$ with $S\in H^0(C,\theta)^{\oplus 2}$, then
$u\wedge S\in\xi^2H^0(C,3H)$ by Proposition~\ref{prop:curve-data}(3),
so $[J]$ is unchanged.
Finally, invariance under compatible coordinate changes: let
$\zeta=\psi(z)$ be another centered parameter and $g(z)\ell^
{\mathrm{fr}}$ the corresponding frame, so $g^4=\psi'$. The frame
transforms have
\[
a_{\rm new}=\frac{z}{\psi(z)}a,\qquad
\mathfrak f_{\rm new}=g\frac{\psi(z)}{z}\mathfrak f,\qquad
\tau_{\rm new}=g^3\frac{\psi(z)}{z}\tau.
\]
Writing $g_0=g(0)$, one gets $\ell_{\rm new}=g_0^{-7}\ell$, so the
same $v^*$ prescribes $c_{\rm new}=g_0^7c$. Also $r_{\rm new}=g^{-1}r$
and $R_{\rm new}(0)=g_0^2R(0)$. The coefficient row of the global
$\theta$-section transforms as $P_{\rm new}=g^{-2}P$; its
$\zeta^{-1}$-coefficient is $\psi'(0)g_0^{-2}R(0)=g_0^2R(0)$, exactly
the new prescribed $R_{\rm new}(0)$, so the same global $P$ realizes
the new prescription. Differentiating $R_{\rm new}=g_0^2R$ at $0$ in
the new coordinate gives $W_{\rm new}=W$; and
$p_{\rm new}(0)-g_0^{-2}p(0)$ is proportional to $R(0)$, so
$p_{\rm new}(0)\wedge R_{\rm new}(0)=p(0)\wedge R(0)$.
\end{proof}
\section{Gluing the vector-field jets}\label{sec:jets}
We now do the actual construction on the total space of $L^{-1}$. The
task is to pick finite-order jets of two vector fields whose determinant
is a unit times a prescribed factor, so that the next section can lift
them holomorphically. This is the only place where substantial
bookkeeping happens.
\subsection{Coordinates, transitions, and obstruction sheaves}\label{subsec:coords}
On the total space of $L^{-1}$ choose a base coordinate $z$ and a
fiber-linear coordinate $t$ measuring the equation of the zero section.
We work with the local frame $(E_t,\partial_z)$ where $E_t=t\partial_t$.
Choose coordinates adapted to $L^4=K_C$: under $z\mapsto\widetilde z
=\psi(z)$ we require $\widetilde t=g(z)t$ with $g^4=\psi'$. The
meromorphic two-form
\begin{equation}\label{eq:volume-form}
\sigma=t^{-5}\,dt\wedge dz
\end{equation}
is invariant under such split changes. We seek rows
$A=(A_1,A_2)$, $B=(B_1,B_2)$ of functions of $(t,z)$ defining
\begin{equation}\label{eq:vector-fields}
v_i=t(A_i(t,z)E_t+B_i(t,z)\partial_z)\quad(i=1,2),
\end{equation}
with Taylor expansions $A=\sum_{k\ge 0}A_k t^k$,
$B=\sum_{k\ge 0}B_k t^k$ where $A_k=(A_{k1},A_{k2})$ and
$B_k=(B_{k1},B_{k2})$. At leading order,
\begin{equation}\label{eq:leading-coefficients}
A_0=r,\qquad B_0=0.
\end{equation}
Then
\begin{equation}\label{eq:determinant-form}
\sigma(v_1,v_2)=t^{-2}(A\wedge B),\qquad
v_1\wedge v_2=t^2(A\wedge B)E_t\wedge\partial_z.
\end{equation}
Under a split change $(\widetilde z,\widetilde t)=(\psi(z),g(z)t)$
of the above type, put $a=(\log g)'$. A term
$t^{k+1}(A_kE_t+B_k\partial_z)$ transforms by
\begin{equation}\label{eq:split-transform}
\widetilde A_k=g^{-(k+1)}(A_k+aB_k),\qquad
\widetilde B_k=g^{3-k}B_k.
\end{equation}
Thus $A_k$ has \emph{type} $L^{k+1}$ (scales by $g^{-(k+1)}$) and
$B_k$ has \emph{type} $L^{k-3}$ (scales by $g^{3-k}$). The
homogeneous-term sheaf at order $k$ fits into the extension
\begin{equation}\label{eq:homogeneous-extension}
0\to L^{k+1}\to\mathcal E_k\to L^{k-3}\to 0.
\end{equation}
For the cover: take $U_0=C\setminus Z$ and six mutually disjoint disks
$U_j$ about the $q_j$. A positively large multiple of $Z$ gives an
ample line bundle; the complement of a general hyperplane section in
its very ample embedding is affine, hence $U_0$'s analytification is
Stein. The disks and annular intersections are Stein as well. By
\eqref{eq:homogeneous-extension} and Cartan's Theorem B on Stein
spaces~\cite{Cartan1952}, horizontal sections admit holomorphic
vertical lifts on each chart. Obstruction computations below use
contour residues $\frac{1}{2\pi i}\int\omega$ around omitted points,
which is a form of Serre duality for overlap functions holomorphic on
annuli; see~\cite{SerreDuality1955}.
\subsection{Transition generators and the bracket}\label{subsec:generators}
The changes of gluing are compositions of time-one flows of holomorphic
vector fields
\begin{equation}\label{eq:generator}
D=t^k(bE_t+h\partial_z),\qquad k>0,
\end{equation}
where $b,h$ are holomorphic on the annular overlap in the disk
coordinate. Use the flow of $-D$, so that the action on other vector
fields is $\exp(\mathrm{ad}\,D)$. Each generator vanishes on the zero
section and changes its equation only by $t\mapsto t+O(t^2)$, so the
zero section and its conormal bundle are preserved.
\begin{lemma}\label{lem:bracket}
For $D$ as in \eqref{eq:generator} and
$V=t^{\ell+1}(AE_t+B\partial_z)$,
\begin{align}
[D,V]=t^{k+\ell+1}\bigl(&((\ell+1-k)bA+hA'-Bb')E_t\notag\\
&+((\ell+1)bB+hB'-khA-Bh')\partial_z\bigr).
\label{eq:general-bracket}
\end{align}
Moreover, the logarithmic divergence
\begin{equation}\label{eq:divergence}
\mathcal L_D\sigma=t^k\bigl((k-4)b+h'\bigr)\sigma.
\end{equation}
For $k\neq 4$ the choice $b=h'/(4-k)$ preserves $\sigma$ exactly.
The first effect on the leading row $(A,B)$ of~\eqref{eq:vector-fields}
is
\begin{equation}\label{eq:leading-action}
\Delta A_k=(1-k)br+hr',\qquad
\Delta B_k=-khr.
\end{equation}
\end{lemma}
\begin{proof}
Compute $[t^k bE_t,t^{\ell+1}AE_t]$ etc.\ in the two components of
the bracket. For the divergence, the coefficient of $dt\wedge dz$ in
$\sigma$ is $t^{-5}$, so
\[
\mathrm{div}_\sigma D
=t^5\partial_t(t^{k-4}b)+t^5\partial_z(t^{k-5}h)
=t^k((k-4)b+h').
\]
Setting $\ell=0$, $A=r$, $B=0$ in the bracket gives
\eqref{eq:leading-action}.
\end{proof}
On a compact subannulus every $D$ in \eqref{eq:generator} is
holomorphic near the zero section and vanishes there; so its time-one
flow and inverse exist after shrinking the fiber disc. A finite
composition has the computed Taylor series. There are no triple
overlaps, so no additional cocycle relation is imposed. The resulting
transition germs will be realized on an actual surface in
Section~\ref{sec:analytic}.
\subsection{The first two orders}\label{subsec:first-two}
Order one. On the $j$-th overlap, take $h=m=m_j$ and $b=m'/3$. Choose
the inside and outside values
\begin{equation}\label{eq:order-one}
\begin{array}{c|cc}
&A_1&B_1\\\hline
U_j&p&mr\\
U_0&P&0
\end{array}
\end{equation}
By \eqref{eq:leading-action} the first transition adds
$(mr',-mr)$ to the inside row, giving exactly $(P,0)$ on the outside.
The degree-one coefficient of $A\wedge B$ vanishes on both sides.
Order two. Write $U=B_2$. On each chart solve
\begin{equation}\label{eq:order-two-horizontal}
r\wedge U=1-A_1\wedge B_1.
\end{equation}
On the disk write $r=z\rho$ with $\rho$ nowhere zero after shrinking.
The right-hand side vanishes at $z=0$ by \eqref{eq:P-local-expansion}
in Proposition~\ref{prop:residue-data}, so division by $z$ reduces
\eqref{eq:order-two-horizontal} to a surjective wedge with $\rho$.
On $U_0$, $r$ has no common zero and the corresponding wedge map of
coherent sheaves is surjective on sections by Steinness. So holomorphic
$U$ exists. After the order-one transition, the difference of the two
chosen $U$'s on an overlap wedges to zero with $r$; it is therefore a
unique holomorphic multiple of $r$. An order-two transition with
$b=h'/2$ removes it (its horizontal effect is $-2hr$). Fix the
outside $U$ throughout this matching.
It remains to match vertical lifts $A_2$. The obstruction lies in
$H^1(C,L^3)^2$, tested against $H^0(C,K_C\otimes L^{-3})=H^0(C,L)$
with the two components $r_1,r_2$ as Serre-dual sections.
\begin{lemma}\label{lem:order-two-obstruction}
The order-two vertical obstruction vanishes for all four entries.
\end{lemma}
\begin{proof}
Write $D^{(2)}=A_2$ as the vertical row (not a transition generator).
For $i,j\in\{1,2\}$ put
\begin{equation}\label{eq:psi-order-two}
\Psi_{ij}
=\left(r_iD^{(2)}_j+\tfrac14 r_i'U_j+\tfrac34 r_j'U_i\right)dz.
\end{equation}
Once horizontal rows match, the difference of $\Psi_{ij}$ for inside
and outside is exactly $r_i$ times the $j$-th vertical difference,
multiplied by $dz$.
Put $\epsilon_{12}=1$, $\epsilon_{21}=-1$, $\epsilon_{11}=\epsilon_{22}
=0$. Under a split change with $a=(\log g)'$,
\[
\widetilde r_i=g^{-1}r_i,\quad
\partial_{\widetilde z}\widetilde r_i=g^{-5}(r_i'-ar_i),\quad
\widetilde U_i=gU_i,\quad
\widetilde D^{(2)}_i=g^{-3}(D^{(2)}_i+aU_i).
\]
Substituting in \eqref{eq:psi-order-two},
\[
\widetilde\Psi_{ij}-\Psi_{ij}
=\tfrac34\epsilon_{ij}\,d\log g
\quad\text{on }U_0,
\]
using $r\wedge U=1$ on the outside. The section $r_1=\xi u_1$ of $L$
has $\widetilde r_1=g^{-1}r_1$, so
$\Psi_{ij}+\tfrac34\epsilon_{ij}d\log r_1$ is intrinsic on $U_0$ and
has only simple poles at the four zeros of $u_1$, each of residue
$\tfrac34\epsilon_{ij}$. The residue theorem on the complement of
small disks gives total contour residue $-3\epsilon_{ij}$ for this
intrinsic differential, and subtracting the six simple logarithmic
residues at $Z$ yields
\begin{equation}\label{eq:outside-total}
\sum_{q\in Z}\Res_q\Psi_{ij}^{\mathrm{out}}
=-\tfrac34\epsilon_{ij}\deg L.
\end{equation}
This argument uses only contour integrals at $Z$ and makes no
meromorphic extension assumption.
Inside. Before transport, $\Psi_{ij}$ is holomorphic on the disk.
The order-two transition changes
\[
\delta D^{(2)}=hr'-\tfrac12h'r,\qquad\delta U=-2hr,
\]
so $\delta\Psi_{ij}=-\tfrac12d(hr_ir_j)$, an exact differential with
zero contour residue. It remains to compute the effect of the order-one
transition, including its second exponential term.
The bracket of $t(m'E_t/3+m\partial_z)$ with $t^2(pE_t+mr\partial_z)$
gives
\[
\left(\tfrac13m'p+mp'-\tfrac13mm''r,\quad
\tfrac23mm'r+m^2r'-mp\right)
\]
in $(A_2,U)$, and half of the second bracket with $trE_t$ gives
\[
\left(\tfrac23mm'r'+\tfrac12m^2r''+\tfrac16mm''r,\quad
-\tfrac13mm'r-m^2r'\right).
\]
Summing:
\begin{align}
\Delta A_2&=\tfrac13m'p+mp'+\tfrac23mm'r'
+\tfrac12m^2r''-\tfrac16mm''r,\notag\\
\Delta U&=-mp+\tfrac13mm'r.
\label{eq:second-order-expansion}
\end{align}
Abbreviate $R=mr$, $m=c/z$, so $r=zR/c$. Substituting in
\eqref{eq:second-order-expansion}:
\[
\Delta A_2=-\frac{cR}{z^3}+\frac{cR'}{3z^2}
+\frac{cR''}{2z}-\frac{cp}{3z^2}+\frac{cp'}{z},\qquad
\Delta U=-\frac{cp}{z}-\frac{cR}{3z^2}.
\]
Equating the Laurent residues of $r_i\Delta A_{2,j}$ and
$r_i'\Delta U_j$ with coefficients at $0$:
\begin{align}
F_{ij}=\Res_0(r_i\Delta A_{2,j}\,dz)
&=-\tfrac13R_ip_j-\tfrac23R_iR_j'-R_i'R_j,\notag\\
H_{ij}=\Res_0(r_i'\Delta U_j\,dz)
&=-R_ip_j-\tfrac13R_iR_j'-\tfrac23R_i'R_j.
\label{eq:laurent-residues}
\end{align}
The total inside residue is thus
\begin{equation}\label{eq:total-residue-matrix}
T_{ij}=F_{ij}+\tfrac14H_{ij}+\tfrac34H_{ji}
=-\tfrac7{12}R_ip_j-\tfrac9{12}R_jp_i
-\tfrac54R_iR_j'-\tfrac{17}{12}R_i'R_j.
\end{equation}
Decompose $T$ into symmetric and antisymmetric parts. The symmetric
part is
\[
T_{ij}^{\mathrm{sym}}
=-\tfrac23(R_ip_j+R_jp_i+2R_iR_j'+2R_i'R_j),
\]
and the antisymmetric part is
\[
T_{ij}^{\mathrm{alt}}
=\tfrac1{12}\epsilon_{ij}
\bigl((R\wedge p)(0)+(R\wedge R')(0)\bigr).
\]
To identify the symmetric part intrinsically, observe that
$P=p+mr'=\frac{R(0)}{z}+p(0)+2R'(0)+O(z)$ and $P_iP_j$ is a
meromorphic section of $L^4=K_C$ with local expression $P_iP_j\,dz$.
Then $\Res_0(P_iP_j\,dz)=R_ip_j+R_jp_i+2R_iR_j'+2R_i'R_j$, so the sum
of every symmetric entry over $Z$ vanishes by the residue theorem on
$C$. For the antisymmetric part, $R(0)\wedge p(0)=-1$ and
\[
\sum_{q\in Z}T_{ij}^{\mathrm{alt}}
=\tfrac1{12}\epsilon_{ij}\sum_{\nu=1}^6(W_\nu-1)
=-\tfrac34\epsilon_{ij}\deg L.
\]
So the total inside sums are
$(0,-\tfrac34\deg L\cdot\epsilon_{12},\tfrac34\deg L\cdot
\epsilon_{21},0)$, which equal the four outside totals in
\eqref{eq:outside-total}. Every Serre-duality pairing of the vertical
difference vanishes, so the corresponding class in $H^1(C,L^3)^2$ is
zero and holomorphic vertical corrections complete the order-two
matching.
\end{proof}
\subsection{Higher orders and determinant targets}\label{subsec:higher-orders}
From order three on, fields and determinant are extended together.
Write the determinant equation as
\begin{equation}\label{eq:target}
A\wedge B=t^2G,\qquad G=1+G_1t+G_2t^2+\cdots.
\end{equation}
The field equation in degree $k$ uses $G_{k-2}$; to compare horizontal
rows we need target compatibility already established. The target is
the local bivector $G\sigma^{-1}$, living in the line bundle
\[
\mathcal M=\cI^4\det\cF=\cI^5\det\cT_Y
\]
with $\mathcal M|_C=L^4\otimes\cT_C\simeq\OO_C$ (using $L^4=K_C$).
The local frame of $\mathcal M$ is $\sigma^{-1}$; the degree-$k$
associated graded is $L^k$. Once transitions stop preserving $\sigma$,
$G$ is not an invariant scalar. The transformation law is: if
$q=\mathrm{div}_\sigma D$, then
\begin{equation}\label{eq:target-transform}
[D,G\sigma^{-1}]=(D(G)-qG)\sigma^{-1}.
\end{equation}
An order-$k$ transition changes a target with constant term $1$ first
in degree $k$ of $G$, with contribution $-((k-4)b+h')$.
The factor $t^2$ in \eqref{eq:target} and $B_0=0$ determine the order
in which the choices can be made: the field coefficient $A_a$ appears
in $A\wedge B$ at degrees $\ge a+1$, and $B_a$ at degrees $\ge a$.
Concretely for step $k$:
\begin{itemize}
\item a global addition to $A_{k-1}$ affects degree $k$ of $A\wedge B$
(i.e.\ $G_{k-2}$);
\item $B_k$ affects degree $k$ (also $G_{k-2}$);
\item $A_k$ affects degree $k+1$ (i.e.\ $G_{k-1}$).
\end{itemize}
All targets referenced here are already chosen and fixed. So we can
freely change $A_{k-1}$ to solve the current equation without breaking
earlier ones, and $G_k$ will be picked only after fields match through
index $k$.
We maintain the following invariant after step $n\geq 2$: field rows
through index $n$ match; determinant equations are imposed through
degree $n$ in $A\wedge B$; coefficients $G_0=1,\ldots,G_n$ are chosen
and transform compatibly. This is true for $n=2$ with $G=1$ because
both transitions of Section~\ref{subsec:first-two} preserve $\sigma$
exactly.
Assume it holds through $k-1$ for $k\ge 3$. On each chart we need
\begin{equation}\label{eq:horizontal-induction}
r\wedge B_k=G_{k-2}-\sum_{a=1}^{k-1}A_a\wedge B_{k-a}.
\end{equation}
This is an equation in $L^{k-2}$. Indeed, under
\eqref{eq:split-transform} the extra terms on the right are multiples
of $\sum_{a=1}^{k-1}B_a\wedge B_{k-a}=0$.
\paragraph{Making \eqref{eq:horizontal-induction} divisible at $Z$.}
Add a row $a\in H^0(C,L^k)^2$ to $A_{k-1}$. Such a change preserves
matching through index $k-1$ and all determinant equations already
imposed (because $B_0=0$: the new term $t^kaE_t$ first contributes
$a\wedge B_1$ in degree $k$, and the equation for degree $k$ has not
yet been imposed). At each $q_j$, the row $B_1(q_j)=R_j(0)$ is
nonzero, and \eqref{eq:evaluation-surjectivity} (
Proposition~\ref{prop:curve-data}(4)) gives
$H^0(C,L^k)\to H^0(Z,L^k|_Z)$ surjective for $k\ge 3$. The scalar
maps $a(q_j)\mapsto a(q_j)\wedge B_1(q_j)$ are surjective, so we can
make the right side of \eqref{eq:horizontal-induction} vanish at each
$q_j$. Since $r=z\rho$ has only a simple common zero, no derivative
condition is needed. Dividing by $z$ on each disk and using Steinness
on $U_0$ gives holomorphic solutions $B_k$.
The previous transitions act on $a$ in the current degree. Only the
first transition contributes at row index $k$:
\begin{equation}\label{eq:old-transition-effect}
\delta A_k=\tfrac{k-1}{3}m'a+ma',\qquad\delta B_k=-ma.
\end{equation}
All later generators and repeated brackets affect strictly higher
indices. The horizontal contribution satisfies $r\wedge(-ma)=a\wedge
(mr)$, exactly the corresponding determinant change on the disk. We
incorporate these in the matching below.
\paragraph{Matching horizontal rows.}
Choose holomorphic vertical lifts of the new $B_k$. After transport
by previous transitions, lower degree-$k$ rows already agree. The
degree-$k$ difference of determinants is $r\wedge\delta B_k$ (because
$\delta A_k\wedge B_0=0$). Target compatibility through degree $k-2$
of $G$ makes this difference zero. So the horizontal discrepancy is a
holomorphic multiple of $r$ on each overlap, and \eqref{eq:leading-action}
removes it by a choice of $h$ in an order-$k$ transition. Use $b=h'$
when $k=3$ and $b=0$ when $k\ge 4$. The new transition first affects
$G$ in degree $k$, so it does not alter the target comparison in
degree $k-2$ just used.
\paragraph{Matching vertical rows.}
After horizontal matching, the remaining class lies in
$H^1(C,L^{k+1})^2$.
For $k\ge 4$ it vanishes because
\begin{equation}\label{eq:vertical-vanishing}
H^1(C,L^{k+1})\simeq H^0(C,L^{3-k})^*=0.
\end{equation}
Holomorphic vertical corrections on the charts solve the matching.
For $k=3$ the obstruction lies in $H^1(C,K_C)^2$, tested by the
constants in the two entries. Additional overlap freedom kills it: on
the $j$-th overlap, replace $h$ by $h+\lambda_j/z$ and compensate by
$B_3\mapsto B_3+3\lambda_jr/z$. This is holomorphic, and preserves
both the current determinant equation and horizontal matching (the
transition adds horizontal row $-3\lambda_jr/z$, its negation). Keep
$A_3$ unchanged in the disk coordinate. The change in the transported
vertical difference at index three is
\begin{equation}\label{eq:order-three-residue}
\lambda_j\!\left(\frac{r'}z+\frac{2r}{z^2}\right),\qquad
\Res_0\!\left[\lambda_j\!\left(\frac{r'}z+\frac{2r}{z^2}\right)dz\right]
=3\lambda_jr'(0).
\end{equation}
The six rows $r'(q_j)$ span $\CC^2$ (Section~\ref{sec:curve}), so
$(\lambda_1,\ldots,\lambda_6)\mapsto3\sum_j\lambda_jr'(q_j)$ has
rank two and cancels both obstruction entries. Order three is complete.
The correction \eqref{eq:order-three-residue} can change the degree-four
coefficient of $A\wedge B$; that is harmless because $G_2=0$ has been
chosen but its determinant equation is first imposed at step four.
The modified order-three transition still has $b=h'$ and preserves
$\sigma$ exactly.
\paragraph{Extending the determinant target.}
Choose $G_k$ compatibly after the order-$k$ transition.
For $k=3$, all transitions preserve $\sigma$, so $G=1$.
For $k=4$: the first target discrepancy from the order-four transition
with $b=0$ is $-h'$ in degree four. The horizontal coefficient of an
order-four generator has type $L^4\otimes\cT_C=\OO_C$; more explicitly
$\widetilde h=g^{4-k}h=h$, so $h'dz=dh$ is an intrinsic exact
differential on the overlap. Its contour residue vanishes; its
pairing with the constant section of $\OO_C$ (the only Serre-dual test)
also vanishes because the residue is zero. So the cocycle splits and
gives holomorphic chart functions $G_4$ with $G_1=G_2=G_3=0$.
For $k>4$, once lower coefficients are compatible, the new leading
discrepancy lies in $L^k$, and
\begin{equation}\label{eq:target-vanishing}
H^1(C,L^k)\simeq H^0(C,L^{4-k})^*=0,
\end{equation}
so it always splits and defines $G_k$. In every case the extension
leaves all earlier equations alone (since $G_k$ is used only in degree
$k+2$ of $A\wedge B$). This completes the induction invariant.
We stop at index seven. A row of index $k$ contributes degree $k+1$ in
the logarithmic frame $(E_t,\partial_z)$, so matching through index
seven prescribes a section modulo $\cI^9\cF$. This threshold is
needed for the vanishing argument of Section~\ref{sec:analytic}: the
unprescribed tail decomposes at each step into extensions of $L^{n-4}$
by $L^n$, and both have vanishing first cohomology for $n\ge 9$.
\begin{proposition}[Finite compatible jets]\label{prop:jets}
For the data of Proposition~\ref{prop:curve-data}, with the
normalization of Proposition~\ref{prop:residue-data}, there exist
finitely many holomorphic transition germs (each a composition of
flows \eqref{eq:generator} of orders $1,\dots,7$) and holomorphic
local rows $A_k,B_k$ for $0\le k\le 7$ such that:
\begin{itemize}
\item the transitions fix $C$ and preserve its conormal bundle $L$;
\item the vector fields \eqref{eq:vector-fields} match through row
index seven;
\item $A\wedge B=t^2(1+O(t))$ on every chart.
\end{itemize}
On any surface neighborhood $Y$ realizing these transition germs, with
$\cI=\OO_Y(-C)$ and $\cF=\cT_Y(-\log C)$, the rows give two sections
of $\cF/\cI^9\cF$, both in $\cI\cF/\cI^9\cF$. Their determinant has
leading term $t^4E_t\wedge\partial_z$, equivalently
$t^5\partial_t\wedge\partial_z$, with coefficient $1$.
\end{proposition}
\begin{proof}
Perform the first two orders and the induction through $k=7$. Only
finitely many flows occur, each an actual holomorphic germ on an
annular neighborhood of the zero section. Terms with row indices
$0,\ldots,7$ have frame coefficients of degrees $1,\ldots,8$, and
matching those terms is matching modulo $\cI^9\cF$. Finally
\eqref{eq:determinant-form} and the degree-two determinant equation
give
\[
v_1\wedge v_2=t^4(1+O(t))E_t\wedge\partial_z
=t^5(1+O(t))\partial_t\wedge\partial_z.
\]
This statement uses only finite compatible jets.
\end{proof}
\section{Analytic realization and contraction}\label{sec:analytic}
The finite jets of Section~\ref{sec:jets} live in overlapping charts.
We now realize them on a single Hausdorff complex surface $Y$, lift
them to holomorphic vector fields by cohomology vanishing, and contract
$C$ to a singular point.
\subsection{A Hausdorff surface neighborhood}\label{subsec:realization}
\begin{lemma}\label{lem:realization}
The overlap maps of Proposition~\ref{prop:jets} define, after shrinking
their domains, a smooth Hausdorff complex surface $Y$ containing $C$.
The conormal bundle is $L$, hence $C^2=-\deg L=-10$. The prescribed
rows give two global sections
$\bar v_1,\bar v_2\in H^0(Y,\cF/\cI^9\cF)$ with local logarithmic and
ordinary determinants $t^4(1+O(t))$ and $t^5(1+O(t))$ in compatible
coordinates.
\end{lemma}
\begin{proof}
Let $N=L^{-1}$ be the split normal bundle. All changes to its
disk-to-outside identifications are finite compositions of flows of
holomorphic vector fields
\[
D=t^k\bigl(b(z)t\partial_t+h(z)\partial_z\bigr),\qquad k\ge 1,
\]
on punctured disk neighborhoods. On a compact annulus every $D$
vanishes along the zero section; the local existence theorem for
holomorphic differential equations, applied uniformly on the annulus,
gives the time-one map and inverse after shrinking the fiber radius.
There are finitely many such flows; successive shrinking produces
their compositions on annular bands. Denote the resulting
inside-to-outside maps by $\Phi_j$, $1\le j\le 6$.
\emph{Quotient construction.} Choose mutually disjoint coordinate
disks $D_j$ about the six $q_j$, and in each radii
$0<a_j<a'_j<b'_j<b_j$ so that $\Phi_j$ and $\Phi_j^{-1}$ are defined
on neighborhoods of the closed annulus $a_j\le |z_j|\le b_j$ in
their split bundles. Retain a small open neighborhood $V_j$ of the
zero section over $|z_j|<b'_j$ and a small open neighborhood $V_0$ of
the zero section over $C\setminus\cup_j\{|z_j|\le a'_j\}$. Choose the
transverse thicknesses small enough that $\Phi_j$ and $\Phi_j^{-1}$
are defined throughout the retained parts of the closed annuli and
\begin{equation}\label{eq:collar-bound}
|z_j(\Phi_j(p))-z_j(p)|<\tfrac12(a'_j-a_j)
\end{equation}
there. This is possible because $\Phi_j$ restricts to the identity on
the zero section. Make the retained annular images land over disjoint
$D_j$.
Set $A_j=\{p\in V_j:\,a_j<|z_j(p)|<b_j,\;\Phi_j(p)\in V_0\}$ and
identify $p\in A_j$ with $\Phi_j(p)\in V_0$. These are biholomorphic
identifications between open subsets. Each point of $V_0$ lies in at
most one image because the retained annuli lie over disjoint $D_j$.
Thus every equivalence class in $V_0\sqcup V_1\sqcup\cdots\sqcup V_6$
has at most two elements: the equivalence relation consists exactly
of the diagonal, these six graphs, and their inverses. No triple
overlaps occur.
\emph{Separation.} Each graph is closed relative to $V_j\times V_0$.
Suppose $p_\nu\in A_j$, $p_\nu\to p\in V_j$, $\Phi_j(p_\nu)\to q\in V_0$.
The base radii of the sources have limit in $[a_j,b'_j]$ with
$b'_j<b_j$; the flows are defined across the closed annulus, so
continuity gives $q=\Phi_j(p)$. If the source radius limit were $a_j$,
inequality~\eqref{eq:collar-bound} would give $|z_j(q)|<a'_j$,
contradicting $q\in V_0$. So $a_j<|z_j(p)|<b_j$, hence $p\in A_j$.
Fiber boundary issues do not arise, since both limit points lie in
the retained open sets.
The equivalence relation is closed. The quotient map is open (gluing
open sets by biholomorphisms), so the quotient is Hausdorff. Each
$V_i$ maps injectively and openly into $Y$, giving $Y$ a smooth
complex surface structure. The image of the zero section is $C$ with
its original structure.
\emph{Conormal bundle.} Each flow fixes $C$ and induces the identity
on its conormal bundle. In the split identification the coordinate has
$t'=t+O(t^2)$, so composing with the original split changes leaves
$\cI/\cI^2\simeq L$. Hence $C^2=-\deg L=-10$.
\emph{Jet sections.} In the logarithmic frame $(t\partial_t,\partial_z)$
the rows $0,\ldots,7$ prescribe coefficients up to degree $8$ in $t$.
The matching of Proposition~\ref{prop:jets} is matching modulo
$\cI^9\cF$ on the actual overlaps. So the rows descend to global
sections $\bar v_1,\bar v_2$ of $\cF/\cI^9\cF$. Their logarithmic
determinant is $t^2(A\wedge B)=t^4(1+O(t))$. Replacing $t\partial_t$
by $\partial_t$ multiplies by $t$, giving ordinary determinant
$t^5(1+O(t))$.
\end{proof}
\subsection{Cohomology vanishing and lifting}\label{subsec:lifting}
We now lift the jets to holomorphic vector fields.
\begin{lemma}\label{lem:graded-vanishing}
For every $n\ge 9$,
\[
H^1(C,\cF|_C\otimes L^n)=0.
\]
Consequently, with $\mathcal E=\cI^9\cF$ and every $r\ge 1$,
\[
H^1(Y,\mathcal E/\cI^r\mathcal E)=0.
\]
\end{lemma}
\begin{proof}
The logarithmic tangent sheaf is locally free of rank two and fits
into the log restriction exact sequence
\begin{equation}\label{eq:log-restriction}
0\to\OO_C\to\cF|_C\to\cT_C\to 0,
\end{equation}
with the first map sending $1$ to the normal Euler field $t\partial_t$
and the second restricting a vector field to $C$. Since $K_C=L^4$,
tensoring by $L^n$ gives
\[
0\to L^n\to\cF|_C\otimes L^n\to L^{n-4}\to 0.
\]
Serre duality identifies the duals of the first cohomologies with
\[
H^0(C,L^{4-n})\quad\text{and}\quad H^0(C,L^{8-n}),
\]
both zero for $n\ge 9$ because $\deg L>0$. The cohomology sequence
gives $H^1(C,\cF|_C\otimes L^n)=0$.
Since $C$ is Cartier and $\cF$ is locally free,
\[
\cI^n\cF/\cI^{n+1}\cF\simeq\cF|_C\otimes L^n.
\]
The quotient $\mathcal E/\cI^r\mathcal E=\cI^9\cF/\cI^{9+r}\cF$ has
a finite filtration with graded terms for $9\le n<9+r$, each with
vanishing $H^1$ by the first part. Successive cohomology sequences
give the second part.
\end{proof}
\begin{lemma}\label{lem:actual-lifting}
After shrinking $Y$ about $C$, the sections $\bar v_1,\bar v_2$ lift
to holomorphic sections $v_1,v_2$ of $\cF$ framing
$\cT_{Y\setminus C}$.
\end{lemma}
\begin{proof}
The intersection matrix of $C$ is the negative-definite $(-10)$. So
Grauert's contraction~\cite[\S4.8(e), pp.~366--367]{Grauert1962}
contracts $C$ to a point $x$ of a normal surface $X$ with
\[
\pi:Y\to X,\quad
\pi^{-1}(x)=C\quad(\text{set}),
\quad Y\setminus C\simeq X\setminus\{x\},
\]
and $\pi_*\OO_Y=\OO_X$. Choose a sufficiently small Stein representative
$X$ about $x$ and replace $Y$ by its inverse image.
\emph{$R^1\pi_*\mathcal E=0$.} Let $\mathfrak m_x$ be the ideal sheaf
of $x$ in $X$ and $\mathcal J=\mathfrak m_x\OO_Y$. Both $\mathcal J$
and $\cI$ vanish exactly on $C$; $\cI$ is radical. The analytic
Nullstellensatz~\cite[Ch.~II, Thm.~4.22]{Demailly2012} and
Noetherianity give $\cI^a\subseteq\mathcal J$ locally at each point
of $C$; compactness of $C$ gives a single exponent $a$. After
shrinking,
\begin{equation}\label{eq:cofinal}
\cI^a\subseteq\mathcal J\subseteq\cI,\qquad
\cI^{ar}\subseteq\mathcal J^r\subseteq\cI^r
\quad(r\ge 1).
\end{equation}
The proper analytic direct-image theorem makes
$R^1\pi_*\mathcal E$ coherent. The Grauert theorem on formal
functions~\cite[\S6, Haupts\"atze~I, IIa, p.~59]{Grauert1960}
identifies the completed stalk with
\begin{equation}\label{eq:formal-functions}
\widehat{(R^1\pi_*\mathcal E)_x}
\simeq\varprojlim_{r\geq 1}H^1(Y,\mathcal E/\mathcal J^r\mathcal E).
\end{equation}
More precisely, with the two ideal filtrations compared by
\eqref{eq:cofinal}: the transition map from index $ar$ to index $r$ in
$\varprojlim_r H^1(Y,\mathcal E/\mathcal J^r\mathcal E)$ factors
through
\[
H^1(Y,\mathcal E/\cI^{ar}\mathcal E)=0
\]
by Lemma~\ref{lem:graded-vanishing}. The inverse system is pro-zero,
so its inverse limit is zero. The finite $\OO_{X,x}$-module
$(R^1\pi_*\mathcal E)_x$ has zero completion, hence is zero by
Nakayama. Away from $x$ the map $\pi$ is an isomorphism, so
$R^1\pi_*\mathcal E=0$ globally.
\emph{Lifting.} The sheaf $\pi_*\mathcal E$ is coherent. Cartan's
Theorem B on the Stein space $X$, together with the Leray spectral
sequence for $\pi$, gives $H^1(Y,\mathcal E)=0$. The cohomology
sequence of
\[
0\to\mathcal E\to\cF\to\cF/\cI^9\cF\to 0
\]
lifts both $\bar v_i$ to holomorphic sections $v_i$ of $\cF$.
\emph{Frame property.} In each divisor chart let
$v_i^{(8)}$ be the polynomial representative of the prescribed jet.
Then $v_i^{(8)}\in\cI\cF$ and $v_i-v_i^{(8)}\in\cI^9\cF$. Hence the
logarithmic determinant of the pair differs from $t^4(1+O(t))$ by an
element of $\cI^{10}\det\cF$; its leading coefficient is still $1$
even at the common zeros of $r$. Ordinary determinant:
\begin{equation}\label{eq:ordinary-determinant}
v_1\wedge v_2=t^5u(t,z)\,\partial_t\wedge\partial_z,\qquad u(0,z)=1.
\end{equation}
The unit condition holds locally around each point of $C$; by
compactness and properness, after further shrinking $X$ it holds on
all of $Y$. So $v_1,v_2$ frame $\cT_{Y\setminus C}$.
\end{proof}
\subsection{The analytic singular point}\label{subsec:counterexample}
\begin{proposition}\label{prop:analytic-counterexample}
There is a normal complex surface germ $(X,x)$ with an isolated
singularity and free analytic derivation sheaf of rank two:
\[
\cT_X=\mathcal H\mathrm{om}_{\OO_X}(\Omega^1_X,\OO_X)\simeq
\OO_X^{\oplus 2}.
\]
Its resolution $Y\to X$ has exceptional set $C$, a smooth curve of
genus $21$ and self-intersection $-10$.
\end{proposition}
\begin{proof}
Take the contraction of Lemma~\ref{lem:actual-lifting}. Suppose $x$
were smooth. A small neighborhood of $x$ would admit a
nowhere-vanishing holomorphic two-form $\omega$; its pullback is
holomorphic on $Y$ and non-vanishing off $C$. So
\[
\mathrm{div}(\pi^*\omega)=aC\quad(a\ge 0),\qquad
K_Y\cdot C=aC^2=a\cdot(-10)\le 0.
\]
But adjunction gives
\[
K_Y\cdot C=2g(C)-2-C^2=40+10=50>0,
\]
a contradiction. So $x$ is singular, and it is the unique singular
point after shrinking.
Let $X^\circ=X\setminus\{x\}$ and $j:X^\circ\hookrightarrow X$. The
frame on $X^\circ$ from Lemma~\ref{lem:actual-lifting} gives two
holomorphic vector fields $w_1,w_2$ there. Normality in dimension two
gives the Hartogs extension
$\OO_X\simeq j_*\OO_{X^\circ}$~\cite[Ch.~II,
Cor.~7.8]{Demailly2012}. This extends the $w_i$: embed $(X,x)$ in a
smooth analytic germ with coordinates $x_1,\ldots,x_e$, extend the
functions $w_i(x_\alpha)$ to functions $a_{i\alpha}$ on $X$, verify
that
$\sum_\alpha a_{i\alpha}\partial f/\partial x_\alpha=0$ for every
local defining equation $f$ (by density and reducedness), and get
sections of $\cT_X$. For any local derivation $w$ on $X$, the frame on
$X^\circ$ writes $w=h_1w_1+h_2w_2$ there with holomorphic coefficients;
both coefficients extend by Hartogs, giving the equality on $X$ by
applying the difference to embedding coordinates. So
$\OO_X^{\oplus 2}\to\cT_X$ is surjective. Injectivity: if
$h_1w_1+h_2w_2=0$ on $X$, then both $h_i=0$ on $X^\circ$, hence on $X$.
\end{proof}
\begin{remark}\label{rem:invariants}
The canonical discrepancy and a lower bound on $p_g$ follow from the
construction. The two-form $\omega^\circ$ dual to the frame generates
the canonical sheaf on $X^\circ$; reflexivizing across the normal
point gives $\omega_X\simeq\OO_X$, so (choosing $K_X=0$) $X$ is
Gorenstein. By \eqref{eq:ordinary-determinant},
\[
\pi^*\omega^\circ=t^{-5}u(t,z)^{-1}dt\wedge dz,
\]
which is meromorphic on $Y$ with divisor exactly $-5C$. So
$K_Y=\pi^*K_X-5C$ and the discrepancy of $C$ is $-5$.
For $p_g$, put $\OO_{nC}=\OO_Y/\cI^n$. From
\eqref{eq:boundary-cohomology} and \eqref{eq:dimension-formula},
$h^1(C,L)=12$. Then
\[
0\to L\to\OO_{2C}\to\OO_C\to 0,
\]
with $H^2(C,L)=0$, gives
$h^1(\OO_{2C})=h^1(C,L)+h^1(C,\OO_C)=12+21=33$. For every $n$ the
sequence $0\to L^n\to\OO_{(n+1)C}\to\OO_{nC}\to 0$ shows
$H^1(\OO_{(n+1)C})\to H^1(\OO_{nC})$ is surjective. Formal functions
identifies their inverse limit with the completion of
$(R^1\pi_*\OO_Y)_x$, which is a finite module supported at $x$;
$p_g(X,x)\ge 33$. With $g=21$, $b=0$, we get $p_g-g-b\ge 12$.
\end{remark}
\section{Algebraization}\label{sec:algebra}
The analytic germ $(X,x)$ from Section~\ref{sec:analytic} is what we
want, but we need a finitely generated $\CC$-algebra. Approximating
defining equations alone is insufficient: an approximation $\widetilde f$
of $f$ need not have $P/(\widetilde f)$ isomorphic to $P/(f)$ if the
relations between the $f_i$ are lost. The key is to approximate the
equations and a finite relation matrix \emph{together}. This follows
Hironaka's equation-and-relation determinacy~\cite[\S3,
Thm.~3.3]{Hironaka1969}, used in Artin's algebraization of isolated
singularities~\cite[Thms.~3.8--3.9]{Artin1969}. We prove the finite
determinacy statement we need, including fixed estimates that avoid
any complete-intersection hypothesis.
\subsection{Finite determinacy of equations with relations}\label{subsec:finite-determinacy}
Put
\[
P=\CC[[x_1,\ldots,x_e]],\quad
\mathfrak n=(x_1,\ldots,x_e),\quad
I=(f_1,\ldots,f_s),\quad R=P/I,
\]
with $f=(f_1,\ldots,f_s)\in\mathfrak n P^s$. Write
$\mathfrak m_R=\mathfrak n R$, and use an overbar for reduction modulo
$I$. Choose an $s\times t$ matrix $D$ whose columns generate all
relations on $f$; thus
\begin{equation}\label{eq:ideal-presentation}
P^t\xrightarrow{D}P^s\xrightarrow{f}I\to 0
\end{equation}
is exact. Define
\[
\psi:R^s\to R^t,\quad v\mapsto v\overline D,\quad
K=\ker\psi,\quad
J=\sum_{j=1}^e R\overline{\partial_j f}.
\]
Differentiating $fD=0$ gives $J\subseteq K$. Right-exact reduction of
\eqref{eq:ideal-presentation} gives a presentation of $I/I^2$, whose
dual identifies
\begin{equation}\label{eq:conormal-dual}
K=\Hom_R(I/I^2,R),\qquad T^1=K/J.
\end{equation}
\begin{lemma}[Finite determinacy]\label{lem:finite-determinacy}
Suppose $T^1$ in \eqref{eq:conormal-dual} has finite length. Then
there exist integers $N,n_0\ge 0$ depending only on $f,D$ such that
the following holds. Suppose a row $g\in P^s$ and an $s\times t$
matrix $B$ satisfy
\[
gB=0,\quad g-f\in\mathfrak n^{n_0}P^s,\quad B-D\in\mathfrak n^N
P^{s\times t}.
\]
Then there is a continuous $\CC$-algebra automorphism $\Phi$ of $P$
and an invertible $U\in\mathrm{GL}_s(P)$ with
\[
f=\Phi(g)U,
\]
$\Phi(x_j)\equiv x_j\pmod{\mathfrak n^2}$ and $U\equiv 1\pmod{\mathfrak n}$.
Consequently $P/(g)\simeq P/(f)$ as complete local $\CC$-algebras.
\end{lemma}
\begin{proof}
We improve equations one order at a time. The exact relation $gB=0$
makes the equation error, modulo fixed $I$, nearly lie in $K$; Artin--Rees
lets us correct it into $K$; finite length of $K/J$ expresses the
high-order corrected class by coordinate derivatives; a change of
generators removes the remaining $I$-valued error. Corrections converge.
\emph{Fixed bounds.} For a map $u:M\to M'$ of finite modules over a
Noetherian local ring with maximal ideal $\mathfrak a$, the
Artin--Rees lemma~\cite[Tag~00IN]{StacksApproximation} applied to
$\mathrm{im}\,u\subset M'$ gives $a\ge 0$ with
\[
\mathrm{im}\,u\cap\mathfrak a^rM'\subset u(\mathfrak a^{r-a}M)
\qquad(r\ge a).
\]
Choose nonnegative $\alpha,\beta,d$ such that
\begin{align}
\mathrm{im}\,\psi\cap\mathfrak m_R^rR^t&
\subset\psi(\mathfrak m_R^{r-\alpha}R^s),
\label{eq:ar-image}\\
K\cap\mathfrak m_R^rR^s&\subset\mathfrak m_R^{r-\beta}K,
\label{eq:ar-kernel}\\
I^{\oplus s}\cap\mathfrak n^rP^s&
\subset\{fC:C\in\mathfrak n^{r-d}P^{s\times s}\}
\label{eq:ar-ideal}
\end{align}
for the relevant exponents. In \eqref{eq:ar-ideal}, Artin-Rees is
applied to $P^{s\times s}\to P^s$, $C\mapsto fC$, with image $I^{\oplus s}$;
the filtration on $I^{\oplus s}$ comes from $P^s$.
Choose $\ell\ge 0$ with $\mathfrak m_R^\ell K\subset J$ (possible since
$K/J=T^1$ has finite length). Put
\begin{equation}\label{eq:constants}
c_0=\beta+\ell,\quad c=c_0+d,\quad N\ge\alpha+1,\quad
n_0\ge\max\{c+N,c+2,2c+1\}.
\end{equation}
\emph{One improvement step.} Suppose $n\ge n_0$ and
$h=g-f\in\mathfrak n^nP^s$, $B-D\in\mathfrak n^NP^{s\times t}$, $gB=0$.
The exact relation, reduced modulo fixed $I$, gives
\begin{equation}\label{eq:high-order-relation}
\bar h\bar D=\bar g\bar D=-\bar g(\bar B-\bar D)
=-\bar h(\bar B-\bar D)\in\mathfrak m_R^{n+N}R^t.
\end{equation}
The factor $\bar h$ on the right gives order $n+N$.
By \eqref{eq:ar-image} there is
\[
\bar\epsilon\in\mathfrak m_R^{n+N-\alpha}R^s
\subset\mathfrak m_R^{n+1}R^s\quad\text{with}\quad
\psi(\bar\epsilon)=\psi(\bar h).
\]
So $\kappa=\bar h-\bar\epsilon\in K\cap\mathfrak m_R^nR^s\subset
\mathfrak m_R^{n-\beta}K\subset\mathfrak m_R^{n-c_0}J$.
Lift coefficients of $\kappa$ in generators of $J$ to
$a_j\in\mathfrak n^{n-c_0}$; lift $\bar\epsilon$ to
$\epsilon\in\mathfrak n^{n+1}P^s$. Then
\[
w=h-\sum_{j=1}^e a_j\partial_jf-\epsilon
\in I^{\oplus s}\cap\mathfrak n^{n-c_0}P^s.
\]
Using \eqref{eq:ar-ideal}, $w=fC$ with $C\in\mathfrak n^{n-c}P^{s\times s}$.
Thus
\begin{equation}\label{eq:linear-correction}
h=\sum_{j=1}^e a_j\partial_jf+fC+\epsilon,\quad
a_j,C\in\mathfrak n^{n-c},\quad\epsilon\in\mathfrak n^{n+1}P^s.
\end{equation}
Put $q=n-c$ and define
\begin{equation}\label{eq:one-step}
\phi(x_j)=x_j-a_j,\quad M=(1+C)^{-1},\quad
g_{\mathrm{new}}=\phi(g)M,\quad B_{\mathrm{new}}=(1+C)\phi(B).
\end{equation}
The bounds give $q\ge\max\{N,2\}$, $2q\ge n+1$, $n+q-1\ge n+1$.
In particular $\phi$ is a formal automorphism with identity linear
part and $M$ is invertible. Taylor expansion gives
\[
\phi(f)-f+\sum a_j\partial_jf\in\mathfrak n^{2q}P^s,\qquad
\phi(h)-h\in\mathfrak n^{n+q-1}P^s.
\]
Combining with \eqref{eq:linear-correction}:
\[
\phi(g)\equiv f(1+C)\pmod{\mathfrak n^{n+1}P^s},\quad
g_{\mathrm{new}}-f\in\mathfrak n^{n+1}P^s.
\]
And
$g_{\mathrm{new}}B_{\mathrm{new}}=\phi(g)\phi(B)=\phi(gB)=0$.
For $B$: substitution $\phi$ changes any power series by an element
of $\mathfrak n^q$, and $C\in\mathfrak n^q$, so
\[
B_{\mathrm{new}}-D=C\phi(B)+(\phi(B)-B)+(B-D)\in\mathfrak n^NP^{s\times t}.
\]
\emph{Iteration.} Starting from $(g_{n_0},B_{n_0})=(g,B)$, repeat for
$n=n_0,n_0+1,\dots$. Set
$\Phi_{n_0}=\id$, $U_{n_0}=1$,
$\Phi_{n+1}=\phi_n\circ\Phi_n$,
$U_{n+1}=\phi_n(U_n)M_n$, so that $g_n=\Phi_n(g)U_n$.
Since $(\phi_n-\id)(P)\subset\mathfrak n^{n-c}$ and
$M_n-1\in\mathfrak n^{n-c}P^{s\times s}$, consecutive images
$\Phi_n(x_j)$ and matrices $U_n$ differ in orders tending to $\infty$;
they converge in $P$ to $\Phi(x_j)$ and $U$. Then $\Phi$ has identity
linear part (so is an automorphism), $U\equiv 1\pmod{\mathfrak n}$ (so
is invertible), and passing to the limit in $g_n=\Phi_n(g)U_n$ gives
$f=\Phi(g)U$.
\end{proof}
\begin{remark}\label{rem:finite-determinacy-note}
The original $D$ generates all relations, so \eqref{eq:conormal-dual}
holds. The approximating $B$ need only satisfy the exact zero product
and be close to $D$; no relation generation is required. So
Lemma~\ref{lem:finite-determinacy} concerns solutions of $gB=0$, not
arbitrary truncations of the defining equations.
\end{remark}
\subsection{The algebraization proposition}\label{subsec:algebraization}
\begin{proposition}[Algebraization]\label{prop:algebraization}
Let $(X,x)$ be a normal complex analytic germ of dimension $d\ge 1$,
smooth away from $x$, with holomorphic tangent module free of rank $d$.
There exist a finitely generated normal integral $\CC$-algebra $A$
and a maximal ideal $\mathfrak p\subset A$ such that
\[
\dim A=d,\quad A/\mathfrak p=\CC,\quad
\widehat{A_{\mathfrak p}}\simeq\widehat{\OO_{X,x}},\quad
\Der_\CC(A)\simeq A^d.
\]
The variety $\Spec A$ can be chosen smooth away from $\mathfrak p$.
If $x$ is singular, $A_{\mathfrak p}$ is nonregular.
\end{proposition}
\begin{proof}
\emph{Setup.} Write $S=\OO_{X,x}=\CC\{x_1,\ldots,x_e\}/I_{\mathrm{an}}$.
Choose a finite row $f$ generating $I_{\mathrm{an}}$ and a finite
matrix $D$ generating all its relations over the convergent ring.
Completion is exact on finite modules, so they give
\eqref{eq:ideal-presentation} over $P=\CC[[x_1,\ldots,x_e]]$, with
$R=\widehat S=P/(f)$. Complex analytic local rings are
excellent~\cite[Prop.~2.1(1)]{Cutkosky2017}, so the normal local
ring $S$ completes to a normal local domain $R$; also
\begin{equation}\label{eq:completion-invariants}
\dim R=\dim S=d,\quad
R\text{ regular }\Leftrightarrow S\text{ regular}.
\end{equation}
\emph{$T^1$ has finite length.} On the smooth locus, the
tangent-to-normal map is surjective (a smooth subspace is locally
regular); its cokernel is a coherent module supported at the isolated
$x$, hence finite length. Completion preserves this finite-length
support, yielding the analogue of \eqref{eq:conormal-dual} for $f,D$
with finite-length $T^1$. So Lemma~\ref{lem:finite-determinacy}
applies.
\emph{Tangent modules.} The holomorphic differentials of $S$ have a
finite Jacobian presentation; its dual is the tangent module, which is
free of rank $d$ by hypothesis. Flat completion of the kernel gives
\begin{equation}\label{eq:continuous-tangent}
\Der_\CC^{\mathrm{cont}}(R,R)
=\ker\bigl(R^e\to R^s,\,(v_j)\mapsto\sum_j v_j\overline{\partial_jf}
\bigr)\simeq R^d.
\end{equation}
A continuous derivation of a power-series quotient is determined by
values on $x_j$, and the displayed relations are exactly the descent
conditions.
\emph{Approximation.} Let
$H=(\CC[x_1,\ldots,x_e]_{(x_1,\ldots,x_e)})^h$ be the henselization;
its completion is $P$. Apply Artin's approximation theorem over $H$
to the system of $t$ equations
$\sum_i G_iB_{i\nu}=0$ ($1\le\nu\le t$) in the $s+st$ entries of
$G\in H^s$, $B\in H^{s\times t}$, with formal solution $(f,D)$
\begin{equation}\label{eq:artin-system}
\sum_{i=1}^s G_iB_{i\nu}=0\quad(1\le\nu\le t).
\end{equation}
Get $g,B$ with $gB=0$ exactly and $g\equiv f$, $B\equiv D$ to any
order~\cite[Thm.~1.10]{Artin1969}, \cite[Tags~07QY, 07QZ]{StacksApproximation}.
Take that order $\ge\max\{n_0,N\}$ with $n_0,N$ from
Lemma~\ref{lem:finite-determinacy}; the lemma then gives
\begin{equation}\label{eq:formal-equivalence}
P/(g)\simeq R.
\end{equation}
\emph{Spreading to a finite-type neighborhood.} Henzelization is the
filtered colimit of pointed \'etale neighborhoods. So the finitely
many entries of $g,B$ are represented on a single such neighborhood,
and after passing to a later one the identities
\eqref{eq:artin-system} hold there as well. Expressing this stage
over the local polynomial ring, finite presentation and clearing
denominators spread it to an affine \'etale neighborhood of the origin
in $\mathbb A^e_\CC$.
The result is an affine \'etale map
$\mathrm{Spec}\,E\to\mathbb A^e_\CC$, a marked point $\mathfrak q$
over the origin with residue field $\CC$, and a row $g$ of $E$ with
$g(\mathfrak q)=0$. The completed ambient local ring is $P$. Put
$A'=E/(g)$, $\mathfrak p'=\mathfrak q/(g)$, $T=A'_{\mathfrak p'}$.
Then $A'$ is finitely generated over $\CC$, and exactness of
completion plus \eqref{eq:formal-equivalence} gives
\begin{equation}\label{eq:completed-local-ring}
\widehat T=P/(g)\simeq R.
\end{equation}
\emph{Descent of tangent freeness.} Because $E$ is \'etale over
$\mathbb A^e_\CC$, its differentials have basis $dx_1,\ldots,dx_e$
near $\mathfrak q$. Completing the finite Jacobian presentation of
$\Omega^1_{T/\CC}$ produces the continuous presentation of $P/(g)$.
For a finitely presented module, flat base change commutes with its
dual (express the dual as a kernel of finite-free maps). Chasing
through \eqref{eq:completed-local-ring} and
\eqref{eq:continuous-tangent}:
\begin{equation}\label{eq:tangent-descent}
\Der_\CC(T)\otimes_T\widehat T
\simeq\Der_\CC^{\mathrm{cont}}(\widehat T,\widehat T)
\simeq\widehat T^d.
\end{equation}
Let $M=\Der_\CC(T)$. From \eqref{eq:tangent-descent}, the residue
vector space of $M$ is $d$-dimensional. Lift a basis to $d$ elements
of $M$: Nakayama gives a surjection $T^d\to M$, which after
completion is an isomorphism of free rank-$d$ modules and hence
iso; faithful flatness of completion makes the kernel zero, so
$M\simeq T^d$.
\emph{Local ring properties.} $T\to\widehat T$ is injective and
$\widehat T$ is a domain; so $T$ is a domain. For normality: if
$a/b\in\mathrm{Frac}(T)$ is integral over $T$, it is integral over
the normal domain $\widehat T$, hence lies in $\widehat T$. Then
$a\in b\widehat T\cap T=bT$ (last equality by faithful flatness),
so $a/b\in T$. Completion also preserves Krull dimension and
$1+\dim_{\CC(\mathfrak p)}\mathfrak p/\mathfrak p^2$; by
\eqref{eq:completion-invariants} and \eqref{eq:completed-local-ring},
$T$ has dimension $d$ and is nonregular whenever $S$ is.
\emph{Isolated singularity.} Let $\mathfrak j_{\mathrm{an}}$ be the
$d$th Fitting ideal of the holomorphic differential module of $S$.
The analytic Jacobian criterion identifies its zero set with the
nonsmooth locus, so some power of $\mathfrak m_S$ lies in
$\mathfrak j_{\mathrm{an}}$. Fitting ideals commute with base change.
The Jacobian presentations above and \eqref{eq:completed-local-ring}
give, by downward contraction from faithful completion,
\begin{equation}\label{eq:isolated-descent}
\mathfrak m_T^b\subset\mathrm{Fitt}_d(\Omega^1_{T/\CC})
\quad\text{for some }b\ge 0.
\end{equation}
On the reduced pure-$d$-dimensional neighborhood constructed next,
the algebraic Jacobian criterion identifies this containment as
nonsmoothness supported only at $\mathfrak p$.
\emph{Affine neighborhood.} $\Der_\CC(A')$ localizes at $\mathfrak p'$
to $\Der_\CC(T)$ because $\Omega^1_{A'/\CC}$ is finitely presented.
The chosen basis spreads after inverting an element outside
$\mathfrak p'$: it defines a map free-to-derivation whose kernel and
cokernel vanish at $\mathfrak p'$, so a further principal shrinking
makes the map an isomorphism.
Since $T$ is a domain, exactly one irreducible component of
$\Spec A'$ passes through $\mathfrak p'$, and the support of the
nilradical avoids $\mathfrak p'$. Remove the other components and
that support. The remaining neighborhood is reduced and irreducible.
Its normal locus is open (finite type over $\CC$ is excellent) and
contains $\mathfrak p'$ by normality of $T$; restrict there. The
irreducible neighborhood has dimension $d$. Its nonsmooth locus is
closed; by \eqref{eq:isolated-descent} it has no positive-dimensional
component through $\mathfrak p'$; remove the components that do not
contain $\mathfrak p'$.
All the restrictions above are open neighborhoods of $\mathfrak p'$.
Choose a principal affine neighborhood contained in their
intersection, write it $\Spec A$ with marked maximal
$\mathfrak p$. Then $A$ is finitely generated, normal, integral,
dimension $d$; $A_{\mathfrak p}=T$; $\Der_\CC(A)\simeq A^d$; residue
field and completed local ring are unchanged; the variety is smooth
away from $\mathfrak p$. This proves all the assertions.
\end{proof}
\begin{proof}[Proof of Theorem~\ref{thm:main}]
Proposition~\ref{prop:analytic-counterexample} gives a singular
normal complex analytic surface germ, smooth away from its marked
point, with free holomorphic tangent module of rank two. Apply
Proposition~\ref{prop:algebraization}; its coordinate ring $A$ is
finitely generated, normal, integral over $\CC$, dimension $2$, with
$\Der_\CC(A)\simeq A^2$ and $A_{\mathfrak p}$ nonregular. So the
derivation module is free while $A$ is not regular. This is the
required counterexample over $\CC$.
\end{proof}
\section{Geometric consequences}\label{sec:consequences}
The counterexample carries secondary structure worth recording.
\subsection*{Logarithmic Kodaira dimension}
For a smooth quasiprojective surface $U$ with smooth projective
compactification $\overline U$ and simple normal-crossing boundary
$D=\overline U\setminus U$, the logarithmic Kodaira dimension is
\[
\overline\kappa(U)=\kappa(K_{\overline U}+D).
\]
This value does not depend on the compactification.
\begin{corollary}\label{cor:log-kodaira}
The smooth locus $U=X_{\mathrm{reg}}$ of the affine surface
$X=\Spec A$ in Theorem~\ref{thm:main} has $\overline\kappa(U)=2$.
\end{corollary}
\begin{proof}
The free tangent sheaf on $X$ restricts to a trivial tangent bundle on
$U$; the top exterior power of that bundle gives a nowhere-vanishing
bivector on $U$, whose dual is a nowhere-vanishing two-form on $U$.
If $\overline\kappa(U)\le 1$, the theorem of Biswas, Gurjar, and
Kolte~\cite[Theorem~0.1(1)]{BiswasGurjarKolte2014} makes the normal
affine surface $X$ smooth, contrary to Theorem~\ref{thm:main}. The
only remaining possible value for a surface is two.
\end{proof}
\subsection*{Failure of derivation blowups to resolve}
For a finite $A$-module $M$ of generic rank $r$, viewed as a coherent
sheaf on $X$, the blowup at $M$ is universal among proper birational
morphisms from integral varieties to $X$ for which the pullback of $M$
modulo torsion is locally free of rank $r$. Every such morphism factors
uniquely through the blowup; see~\cite[\S1.6, Thm.~1.17]{BarajasChavezRomano2026}.
\begin{corollary}\label{cor:derivation-blowups}
Starting with the singular surface $X=\Spec A$ of
Theorem~\ref{thm:main}, every iteration of blowing up at the
derivation module is isomorphic to $X$. The same holds if each
blowup is followed by normalization. Neither procedure resolves the
singular point.
\end{corollary}
\begin{proof}
The derivation module is free. Its blowup is the identity
morphism. The same holds after composition with normalization, since
$X$ is already normal.
\end{proof}
\section*{Conclusion}
The surface $X$ of Theorem~\ref{thm:main} is normal, singular at
exactly one point, and its module of algebraic vector fields is free
of rank two. This is a counterexample to the Lipman--Zariski
conjecture in characteristic zero. The method combines six features
that seem to be forced by the problem:
\begin{enumerate}
\item Algebraic construction of a curve with fourth root of $K_C$ and
prescribed residue data (Section~\ref{sec:curve}).
\item A trace pairing that transports quadratic structures from an
auxiliary elliptic curve to the plane curve (Section~\ref{sec:trace}).
\item Successive gluing of vector-field jets in which the leading
determinant coefficient is arranged globally while the vertical
obstruction is measured by Serre duality and cancelled by residue
identities (Section~\ref{sec:jets}).
\item Realization of finite jets on a Hausdorff neighborhood and
lifting to holomorphic fields by cohomology vanishing
(Section~\ref{sec:analytic}).
\item Contraction to a normal surface point at which the exceptional
curve has self-intersection $-10$ (Lemma~\ref{lem:actual-lifting}).
\item Algebraization via simultaneous approximation of equations and a
finite relation matrix, faithful descent of the tangent module,
and affine shrinking (Section~\ref{sec:algebra}).
\end{enumerate}
The construction also shows that:
\begin{itemize}
\item The frame cannot be chosen commuting (J\"order).
\item Iterated blowup at the derivation module does not resolve the
singular point.
\item The smooth locus has logarithmic Kodaira dimension two.
\item The canonical discrepancy of the exceptional curve is $-5$;
the geometric genus at the singular point satisfies $p_g\ge 33$
(the bound $p_g-g-b\ge 12$ for a normal singularity resolved on one
side by a curve of genus $g=21$ and $b=0$).
\end{itemize}
\section*{Acknowledgements}
The author thanks the open literature for the many positive results
that focused attention on the missing case of surfaces in
characteristic zero.
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\end{document}