\section{Finite determinants for the cubic construction}
\label{sec:cubic-determinants}

The four determinants used in the cubic construction can be specified
without listing their entries.  We give the point coordinates by an
integer recurrence, construct the matrices by polynomial evaluation,
and record the elimination results.  All the arithmetic in the first
three constructions takes place in $\mathbb F_{1009}$.

\begin{proposition}\label{prop:cubic-certificates}
The cubic construction has the following nonvanishing properties.
\begin{enumerate}
\item For $m=10$, the middle multiplication map is an isomorphism.
\item The $160$ root rows in the passage from $m=10$ to $m=12$
      are independent on the new source.
\item The $26$-dimensional pair block is invertible.
\item The final four-dimensional block is invertible.
\end{enumerate}
Each assertion holds on a nonempty open subset of the corresponding
space of cone and trace parameters over every characteristic-zero field.
\end{proposition}

\subsection{Conventions for the computations}

For $h\in\mathbb Z/2^{64}\mathbb Z$, replace $h$ by
\[
 h\longleftarrow
 6364136223846793005h+1442695040888963407\pmod {2^{64}}.
\]
After this replacement define a draw from $[a,b)$ by
\begin{equation}\label{eq:cubic-draw}
 a+\bigl(\lfloor h/2^{32}\rfloor\bmod(b-a)\bigr).
\end{equation}
Every draw advances the state once, including a draw whose value is
subsequently discarded.  Residues in $\mathbb F_{1009}$ are represented
by $0,\ldots,1008$.

The degree-$e$ monomials in $z_1,\ldots,z_v$ are ordered
lexicographically by decreasing exponent: first those with exponent
$e$ on $z_1$, then those with exponent $e-1$, and so on, applying the
same rule recursively to the remaining variables.  A subcollection
inherits this order.  Polynomial coefficients are column vectors.
For a matrix $A$, its canonical kernel matrix $K(A)$ is obtained as
follows.  Scan columns from left to right; in each column use the first
available nonzero row as pivot, normalize it, and eliminate below it.
The free columns, in increasing order, determine the kernel basis:
set the corresponding free coordinate to one, the other free coordinates
to zero, and solve backwards for the pivot coordinates.  Thus
$AK(A)=0$.

The determinant recorded below is that of the submatrix on the selected
pivot columns and the original pivot rows, the latter listed in their
selection order.  It is the product of the unnormalized pivots.  These
conventions specify both the intermediate kernel bases and each reported
minor uniquely.

\subsection{The initial middle map and the root rows}

Begin with state $h=20261108$.  Use trace indices $0,\ldots,9$ and
single-edge set $\{\{1,7\},\{2,8\}\}$.  A cone datum is a pair
$(\sigma,\mu)$, with $\sigma=(1,t)$ and $\mu\ne0$.
For a request for $c\in\{1,2\}$ cone data, first draw $c$ distinct
slopes from $[0,1009)$, rejecting repeated draws, and then draw their
$c$ heights from $[1,1009)$ in the same order.

For each vertex, in increasing order, request two cone data.  Then visit
all pairs $i<j$ in lexicographic order.  For each pair draw
$\lambda_{ij}$ from $[1,1009)$, then request one cone datum on a single
edge and two otherwise.  The trace supports are $e_i$ and
$e_i+\lambda_{ij}e_j$.  A datum $(\sigma,\mu)$ over support $y$ means
the actual point $(\sigma,\mu y)$ in coordinates
$(s,t,y_0,\ldots,y_9)$.
Finally draw $110$ free trace points.  Each has first coordinate one
and its other nine coordinates are successive draws from $[0,1009)$.
List the trace points as vertices, edge supports, and free points,
in these respective orders.

Let $E_P$ evaluate the monomials of $(s,t)_3$ at the $108$ cone points,
and let $E_T$ evaluate the degree-three trace monomials at the $165$
trace points.  Put
\[
 K=K(E_P),\qquad V=K(E_T).
\]
Form $M$ by multiplying the columns of $K$ successively by
$s,t,y_0,\ldots,y_9$, and then the columns of $V$ by $s,t$.
For each multiplier retain the order of the kernel basis.  The rows
are the ordered monomials of $(s,t)_4$.  This is the actual middle
multiplication matrix.

Continue the same sequence of draws.  Add trace indices $10,11$ and
single edges
\[
 \{i,10\}\ (1\le i\le4),\qquad
 \{i,11\}\ (7\le i\le9).
\]
Keep all old data.  First generate the two new vertices' cone data,
then visit all pairs $i<j$ in lexicographic order and generate data
only for pairs with a new endpoint, using the preceding rule.
Let $E'_P$ be the new cone evaluation matrix and $K'=K(E'_P)$.
Let $R$ be the rows of $K'$ belonging to monomials not involving
$y_{10},y_{11}$, and set $K^0=K'K(R)$.

Form a matrix $C$ whose columns are the products of $K^0$ by the twelve
old variables, followed by the products of $K'$ by $y_{10},y_{11}$.
It has $254$ columns.  Let $C_{\ge2}$ comprise its rows of total
$(s,t)$-degree at least two, and put
\[
 \Phi=C K(C_{\ge2}).
\]
Its $177$ columns have $(s,t)$-degree one and restrict to zero when
$y_{10}=y_{11}=0$.  For $b=1,2$, extract their coefficients of
$y_{9+b}$; these are $\langle s,t\rangle$-valued quadrics in the ten
old trace variables.

For the first direction put $N_1=\{1,2,3,4\}$, and for the second put
$N_2=\{7,8,9\}$.  The ordered face lists are:
\begin{enumerate}
\item the vertices in $N_b$, in increasing order;
\item the pairs $i<j$, in lexicographic order, for which
$\mathbf1_{\{i,j\}\in H}+|\{i,j\}\cap N_b|\ge1$;
\item for $b=1$, the faces
$\{5\},\{6\},\{7\},\{8\},\{9\},\{0\},
\{7,8\},\{7,9\},\{8,9\}$; for $b=2$, the faces
$\{0\},\{1\},\{2\},\{3\},\{4\},\{5\},\{6\},
\{1,2\},\{1,3\},\{2,3\}$.
\end{enumerate}
For each face draw $\eta$ from $[1,1009)$, using support $e_i$ for a
vertex and $e_i+\eta e_j$ for a pair $i<j$.  The draw is made even
for a vertex.  Complete the first direction before the second.
Evaluate the corresponding vector-valued quadrics, listing the
$s$-component before the $t$-component at each point.  Stacking these
rows gives the root matrix.

The resulting elimination is as follows:
\begin{center}
\begin{tabular}{lrrr}
\hline
Matrix & Size & Rank & Selected determinant\\
\hline
$E_P$ & $108\times144$ & $108$ & $1008$\\
$E_T$ & $165\times220$ & $165$ & $274$\\
$M$ & $650\times542$ & $542$ & $995$\\
$E'_P$ & $147\times196$ & $147$ & $903$\\
$R$ & $144\times49$ & $36$ & $697$\\
$C_{\ge2}$ & $287\times254$ & $77$ & $859$\\
First root direction & $86\times177$ & $86$ & $217$\\
Second root direction & $74\times177$ & $74$ & $588$\\
Root matrix & $160\times177$ & $160$ & $103$\\
\hline
\end{tabular}
\end{center}
All entries in the last column are in $\mathbb F_{1009}$.
The first two ranks give $\dim K=36$ and $\dim V=55$.
Every column of $M$ vanishes at the cone points, and its target has
dimension $650-108=542$.  The nonzero determinant therefore proves
assertion~(1).

For assertion~(2), restriction of $K'$ lands in the old
$36$-dimensional kernel.  Moreover, the high-normal image of $C$ has
at most eight coefficients of normal degree three and $69$ of normal
degree two: these are precisely the coefficients involving a new trace
variable.  Its rank is consequently at most $77$ before specialization.
The table proves equality.  The $17$ higher-order columns described in
the cubic construction lie in the kernel of the root rows.  The root
rank $160$ is therefore the required rank on their quotient.

\subsection{The pair block}

Restart the recurrence at $h=851$ and use ordered indices
\[
 0<5<6<10<11<12<13,
 \qquad b=10,\ a=11,\ w_1=12,\ w_2=13.
\]
Every draw in this paragraph is from $[1,1009)$.  At each vertex draw
two distinct slopes, rejecting repeats, then their two heights.
Visit all edges lexicographically.  On an edge, first draw its slope
or distinct slopes, then its $\lambda$, then its heights.  The only
single edges are $\{11,12\}$ and $\{10,13\}$.

Construct the linear lifts $L_c(Q)$ from the vertex and edge equations
\eqref{eq:cubic-local-lifts}.  At a single edge solve its one equation
by setting the $t$-component of the coefficient vector to zero.
Let $Q$ run in the order $s^2,st,t^2$.  In direction $w_1$, list the
three motion columns for $i=a$, then the two extra columns for $a$,
then the three motion columns for $i=b$, then the five half-face
columns for $a$.  In direction $w_2$, list the three motion columns for
$i=a$, then the three motion columns for $i=b$, then the two extra
columns for $b$, then the five half-face columns for $b$.
Explicitly, these columns are
\[
 \begin{array}{ll}
 \text{motion:}& y_i(L_{w_j}(Q)-L_i(Q)),\\
 \text{extra:}& y_iL_{w_j}(s\ell),\quad y_iL_{w_j}(t\ell),\\
 \text{half-face:}& \ell y_i y_c,
 \end{array}
\]
where $i=a$ in direction one and $i=b$ in direction two for the last
two types, $\ell=t_{iw_j}s-t$, and $c$ runs over
$i$, the other member of $\{a,b\}$, $0,5,6$, in that order.
All columns are restricted to coordinates $0,5,6,10,11$.

For each of thirteen samples, draw five point coordinates in this
order and then two direction coordinates $(v_1,v_2)$.  Multiply a
column from direction $j$ by $v_j$, evaluate it, and list its two
components as consecutive rows.  The resulting $26\times26$ matrix
has determinant
\[
 527\pmod {1009}.
\]
This proves assertion~(3).  In particular the computation uses the
same cone coefficients in every column; the entries are not chosen
independently.

\subsection{The four-dimensional block}

Here it is simpler to give rational coordinates.  Write $x,u,v$ for
the leftover old coordinate and the two new coordinates.  All edge
parameters $\lambda$ equal one.  Use the following cone data:
\begin{center}
\begin{tabular}{ll}
\hline
Trace support & $(\sigma,\mu)$\\
\hline
$e_u$ & $((1,0),1),\ ((0,1),2)$\\
$e_v$ & $((1,0),3),\ ((0,1),4)$\\
$e_x+e_u$ & $((1,2),2)$\\
$e_x+e_v$ & $((1,1),3),\ ((2,1),5)$\\
$e_u+e_v$ & $((1,1),2),\ ((1,-1),3)$\\
\hline
\end{tabular}
\end{center}
At the single edge $xu$, set the $s$-component of the lift coefficient
to zero.  The four columns are
\[
 (2,-1)xuv,\qquad uv(L_v(Q)-L_u(Q)),\quad Q=s^2,st,t^2.
\]
Their evaluation at $(x,u,v)=(1,1,1),(1,2,3)$, in component order
$s,t$, is
\[
 \begin{pmatrix}
 2&-2/15&-1/15&2/15\\
 -1&-7/60&7/30&17/60\\
 12&-63/10&1/10&33/10\\
 -6&-1/5&39/10&-23/10
 \end{pmatrix},
 \qquad \det=-\frac{603}{40}.
\]
\subsection{Passage to characteristic zero}

Lift the primitive parameters in the first three constructions to
their integer representatives, and form the cone points
$(\sigma,\mu y)$ over $\mathbb Q$. The selected pivot minors are
units at $1009$, so the same kernel charts are defined over
$\mathbb Z_{(1009)}$ and reduce to the displayed matrices.
The structural rank bounds and the nonzero pivot minors preserve
all intermediate dimensions. Thus each final nonzero minor lifts
to a nonzero rational minor. Clearing denominators gives a nonzero
polynomial in the parameters; the rational fourth determinant
does so directly. These polynomials remain nonzero over any
characteristic-zero field, proving
Proposition~\ref{prop:cubic-certificates}.