\section{Two normal directions and an integral trace}
\label{sec:integral-two-normal}

The preceding hyperplane construction has a numerical limit.  Beyond
that limit we use a subspace of codimension two and move its points in
the two independent normal directions.  We first prove the transfer
when the evaluation map on the trace is square.  The same construction,
with one additional free projection, will treat the nonsquare case.

\subsection{Three interpolation facts}

We record the linear algebra used in both constructions.  A finite
dimensional space of scalar polynomials has maximal-rank evaluation at
general points of an irreducible affine open set: choose the points
successively, observing that a nonzero polynomial cannot vanish on that
open set.  The statement also holds for homogeneous polynomials after
choosing an affine chart.

\begin{lemma}[Two groups of scalar evaluations]
\label{lem:two-group-evaluation}
Let \(V\) be a finite dimensional subspace of \(V_A\oplus V_B\), where
\(V_A,V_B\) are spaces of scalar polynomials on irreducible varieties.
Write
\[
 a=\dim V,\qquad
 h_A=\dim\bigl(V\cap(V_A\oplus0)\bigr),\qquad
 h_B=\dim\bigl(V\cap(0\oplus V_B)\bigr).
\]
If \(h_A,h_B\le b\), evaluation of the first component at \(b\)
general points and of the second component at \(b\) independently
general points has kernel of dimension \(\max(0,a-2b)\).
Both evaluations can simultaneously be chosen injective on their
respective pure subspaces.
\end{lemma}

\begin{proof}
The first projection has dimension \(a-h_B\).  Its evaluation at \(b\)
general points has maximal rank; hence the kernel remaining in \(V\)
has dimension \(\max(a-b,h_B)\).  At the same time the evaluation is
injective on the pure \(A\)-subspace.  Consequently the remaining
kernel projects injectively into \(V_B\).  Evaluation at the second
group of points cuts this scalar polynomial space with maximal rank.
Its final dimension is
\[
 \max\bigl(0,\max(a-b,h_B)-b\bigr)=\max(0,a-2b).
\]
Injectivity on the pure \(B\)-subspace is another nonempty open
condition on the second group, and can be imposed at the same time.
\end{proof}

Fix \(T=k[y_1,\ldots,y_m]\), set \(S=T[s,t]\), and identify
\(\Lambda=\{s=t=0\}\) with \(\mathbf P^{m-1}\).

\begin{lemma}[Restriction of a point ideal]
\label{lem:height-degeneration}
Let \(e\ge1\), \(C=\dim S_e\), \(M_e=\dim T_e\), and \(q=C-g\),
where \(0<g\le M_e\).  Suppose that the degree-\(e\) ideal component
of \(M_e-g\) general points of \(\Lambda\) has dimension \(g\) and
injective multiplication by \(T_1\).
For \(q\) general points \(X\) with \(s,t\ne0\), the space
\(G=I_X(e)\) has dimension \(g\); restriction \(G\to T_e\) is
injective, and its image \(\overline G\) has injective multiplication
\[
 \overline G\otimes T_1\longrightarrow T_{e+1}.
\]
\end{lemma}

\begin{proof}
Put \(r_0=C-M_e=\dim(s,t)_e\).  Retain \(r_0\) points with nonzero
normal coordinates at which evaluation on \((s,t)_e\) is an
isomorphism.  Put the remaining
\(q-r_0=M_e-g\) points on \(\Lambda\), in general position.
Given \(f\in T_e\) vanishing at the latter points, a unique element
of \((s,t)_e\) corrects its values at the retained points.
Thus restriction identifies the degree-\(e\) ideal of this
configuration with the ordinary point-ideal component specified in
the hypothesis.  This also proves independence of all \(q\) scalar
conditions in degree \(e\).

On the open set where those scalar conditions are independent, their
kernels form a vector bundle.  Injectivity of restriction and of its
product map are open rank conditions on this bundle.  They hold at
the configuration just constructed, and therefore on a nonempty
open set of configurations with both normal coordinates nonzero.
\end{proof}

Here is the interpolation statement needed for the attached
projections.  In particular, the projections of the points on a
slice need not be independent of the ideal defined by those points.

\begin{lemma}[Interpolation at attached projections]
\label{lem:attached-code}
Let \(H_A=\{t=0\}\), let \(D\ge2\), and put
\[
 L=N_{D-1}(m+1),\qquad M=N_{D-1}(m).
\]
Let \(A'\) consist of \(b\) general points of \(H_A\setminus\Lambda\),
where \(b\le M\) and \(b<L\).  Suppose multiplication on
\(W_A=I_{A'}(D-1)\) by \(k[H_A]_1\) is injective.  Set
\[
 U_A=\sum_{i=1}^m y_i\bigl(W_A|_\Lambda\bigr)\subset T_D.
\]
Then \(\dim U_A=m(L-b)\).  Let \(Z_2\subset\Lambda\) be a general
ordered set containing the projections of \(A'\), with all its
other points independently general relative to \(A'\).  Provided
\(\lvert Z_2\rvert\ge\dim U_A\), every subset of \(Z_2\) of
cardinality \(\dim U_A\) is unisolvent for \(U_A\).
Consequently, in
\[
 Q_A=k^{Z_2}/\operatorname{ev}_{Z_2}(U_A),
\]
every \(\dim Q_A\) coordinate classes form a basis.
\end{lemma}

\begin{proof}
The kernel of restriction from \(W_A\,k[H_A]_1\) to \(\Lambda\)
is exactly \(sW_A\).  Indeed a polynomial in this product space
which is divisible by \(s\) vanishes at \(A'\), and division by
\(s\), which does not vanish there, gives an element of \(W_A\).
The assumed product injection therefore gives
\(\dim U_A=(m+1)(L-b)-(L-b)=m(L-b)\).

We first show that \(U_A\) interpolates arbitrary values at the
projections of \(A'\).  Choose representatives in one affine chart.
For \(v\in k[H_A]_{D-2}\), there exists \(w_0\in T_{D-1}\) such that
\[
 w_0(y(a'))=-s(a')v(a')\qquad(a'\in A'),
\]
since \(b\le M\).  Hence \(w_0+sv\in W_A\).
Using this separately for \(v_1,\ldots,v_m\), the values of \(U_A\)
contain those of
\(-s\sum_i y_i v_i\) at \(A'\).  The polynomials
\(\sum_i y_i v_i\) constitute the degree-\((D-1)\) part of
the ideal \((y_1,\ldots,y_m)\) in \(k[H_A]\), a space of dimension
\(L-1\ge b\).  Its evaluation at general \(A'\) is onto, and
multiplication by the nonzero values of \(s\) preserves this property.

For any chosen subset of \(Z_2\) of size \(\dim U_A\), first impose
the rows which belong to these attached projections.  They are
independent by the preceding paragraph.  The remaining labels are
independently general relative to \(A'\), and their evaluations
successively cut the remaining scalar polynomial space.  This
proves unisolvence for that subset.  There are finitely many subsets,
so all these open conditions hold simultaneously.  The assertion
about coordinate classes follows by taking complementary subsets.
\end{proof}

\begin{lemma}[A first response in two normal directions]
\label{lem:two-normal-response}
Let \(A(\epsilon)\) be a matrix of regular functions near
\(\epsilon=0\), with kernel \(K\) and cokernel \(Q\) at zero.
Suppose \(Q=Q_A\oplus Q_B\), and that its induced first derivative is
\[
 J=\sum_z v_z e_z^A\otimes\alpha_z+
              u_z e_z^B\otimes\beta_z.
\]
Assume that each of the families \((\alpha_z)\) and \((\beta_z)\)
spans \(K^*\), and that in each target every subset of at most
\(\dim Q_A\), respectively \(\dim Q_B\), of the indicated coordinate
classes is independent.  There are choices of \(u_z,v_z\) for which
\(J\) has rank \(\min(\dim K,\dim Q)\).  For these choices
\(A(\epsilon)\) has maximal rank for general \(\epsilon\).
\end{lemma}

\begin{proof}
Write \(h=\min(\dim K,\dim Q)\), and choose \(h_A,h_B\) with
\(h_A+h_B=h\), \(h_A\le\dim Q_A\), and \(h_B\le\dim Q_B\).
Choose \(h_A\) independent members of the first source family.
Extend them to \(h\) independent functionals using the second
family.  Set the corresponding velocities equal to one and all
others to zero.  The selected target vectors are independent,
within each group by hypothesis and between groups by the direct
sum.  Thus \(J\) factors as a surjection onto \(k^h\) followed by
an injection into \(Q\), and has rank \(h\).

Lemma~\ref{lem:first-variation} now gives maximal rank of
\(A(\epsilon)\) for general \(\epsilon\).
\end{proof}

\subsection{The integral-trace transfer}

\begin{proposition}
\label{prop:integral-two-normal}
Let \(D\ge4\), \(n=m+2\), and \(m\ge D(D-1)\).  Suppose
\(E(m,D)\) is an integer and the degree-\(D\) ideal generation
statement holds in \(m\) variables.  If the ideal generation
statements in degrees \(D-1\) and \(D-2\) hold in every number
of variables, then the degree-\(D\) statement holds in \(n\)
variables.
\end{proposition}

\begin{proof}
It suffices to construct the two critical evaluation maps.
Put \(e=D-2\), and use the following dimensions:
\[
 \begin{gathered}
 F=N_D(m),\quad M=N_{D-1}(m),\quad L=N_{D-1}(m+1),\\
 C=N_e(n),\quad R=N_e(m+1),\quad M_e=N_e(m).
 \end{gathered}
\]
Pascal's identity and the consecutive binomial ratios give
\begin{equation}
\label{eq:two-normal-identities}
 \begin{gathered}
 N_D(n)-F=2L+C,\qquad N_{D-1}(n)=L+C,\\
 DF=mL,\qquad (m+1)C=(D-1)L,\\
 \frac LM=\frac{m+D-1}{m},\qquad
 \frac LR=\frac{m+D-1}{D-1}.
 \end{gathered}
\end{equation}
Take \(r=\lfloor E(n,D)\rfloor\) or \(\lceil E(n,D)\rceil\).
Set
\begin{equation}
\label{eq:integral-two-normal-counts}
 \begin{aligned}
 \Delta&=r-E(n,D),& p&=N_D(n)-r,\\
 p_2&=\frac{DF}{D+1},&
 p_1&=\frac{D(2L+C)}{D+1}-\Delta,\\
 b&=\left\lceil\frac{DL}{D+1}\right\rceil,&
 \theta&=b-\frac{DL}{D+1},\\
 q&=p_1-2b=\frac{DC}{D+1}-\Delta-2\theta,&
 g&=C-q=\frac C{D+1}+\Delta+2\theta .
 \end{aligned}
\end{equation}
All point counts are integers, \(\lvert\Delta\rvert<1\), and
\(0\le\theta<1\).

\emph{The numerical margins.}
We need the degree-\((D-1)\) injection for \(b\) points in \(m+1\)
variables and the degree-\(e\) injection for \(M_e-g\) points in
\(m\) variables.  The inequalities
\[
 R\le b\le M<L,\qquad L-b<\frac LD
\]
follow from \(m\ge D(D-1)\) and
\eqref{eq:two-normal-identities}; indeed \(b\ge DR\).
They put the first ideal component in its initial-degree cell
and on the injective side.

For the second component, direct cancellation of the binomial
coefficients gives
\begin{equation}
\label{eq:two-normal-margin}
 \frac{M_e}{D-1}-\frac C{D+1}
 =
 \frac{\prod_{j=2}^{e-1}(m+j)}{e!}
 \left(\frac{m(m+1)}{D-1}
 -\frac{(m+D-2)(m+D-1)}{D+1}\right).
\end{equation}
The empty product is one.  The numerator of the expression in
parentheses, over \(D^2-1\), is
\[
 2m\bigl(m-(D-1)(D-2)+1\bigr)-(D-1)^2(D-2).
\]
It is increasing for \(m\ge D(D-1)\), and its value at that
endpoint is \((D^2-1)(3D-2)\).  The prefactor in
\eqref{eq:two-normal-margin} is \(M_e/(m(m+1))\ge1/2\).
Consequently
\[
 \frac{M_e}{D-1}-\frac C{D+1}\ge\frac{3D-2}{2}\ge5,
 \qquad 0<g<\frac C{D+1}+3<\frac{M_e}{D-1}.
\]
The consecutive binomial ratio then gives
\(M_e-g\ge N_{e-1}(m)\).  Thus the second injection is also
available in its initial-degree cell.  In addition,
\begin{equation}
\label{eq:two-normal-pure-margin}
 (m+1)g<b.
\end{equation}
In fact the difference is greater than
\(L/(D+1)-3(m+1)\), which is positive because
\[
 C\ge N_2(n)>\frac{m^2}{2}
 \ge\frac{D^2(D-1)^2}{2}>3(D^2-1).
\]
We also have \(0<q<C\), \(q>C-M_e\), and \(p_2>p_1\).
For the last assertion,
\[
 p_2-p_1=\frac{(m-2D)L-DC}{D+1}+\Delta>0,
\]
since \(m-2D\ge4\), \(DC/L=D(D-1)/(m+1)<1\), and \(L>D+1\).
Finally
\begin{equation}
\label{eq:two-normal-trace-large}
 p_2\ge M,\qquad
 \frac{p_2}{M}=\frac{m+D-1}{D+1}\ge D-1.
\end{equation}

\emph{The marked family.}
Choose \(b\) points \(A'\) on \(H_A=\{t=0\}\) with \(s\ne0\),
\(b\) points \(B'\) on \(H_B=\{s=0\}\) with \(t\ne0\), and
\(q\) points \(X\) with \(s,t\ne0\).  Their projections to
\(\Lambda\) are required to be distinct.  Let \(Z_2\) consist
of these \(p_1\) projections and \(p_2-p_1\) further general
points of \(\Lambda\).  Put
\[
 Z=A'\cup B'\cup X\cup Z_2 .
\]
The parameter space is irreducible and dominates the space of
ordered \(p_2\)-tuples on \(\Lambda\): prescribe the projections
first and choose the normal coordinates freely.  Every rank
condition below is a nonempty open condition on this same
parameter space.  They can therefore be imposed simultaneously.
In proving that an individual condition is nonempty, we are
free to fix one group of points and vary the other groups.

Scalar evaluation at \(Z\) is onto in degree \(D\).  Interpolate
first on \(Z_2\) using \(T_D\), then on \(A'\) using
\(s\,k[H_A]_{D-1}\), and on \(B'\) using
\(t\,k[H_B]_{D-1}\).  These corrections are possible since \(b<L\)
and vanish on the preceding groups.  Finally \(stS_e\) corrects
the values on \(X\), since \(q<C\).  A linear form nonzero at
all the points gives the same assertion in degree \(D+1\).

\emph{The two slices.}
Apply Lemma~\ref{lem:attached-code} to \(A'\) and \(B'\).
The spaces \(U_A,U_B\subset T_D\) both have dimension \(m(L-b)\),
and their evaluations on \(Z_2\) are injective.  Their quotient
codes
\[
 Q_A=k^{Z_2}/\operatorname{ev}(U_A),\qquad
 Q_B=k^{Z_2}/\operatorname{ev}(U_B)
\]
have the common dimension
\begin{equation}
\label{eq:two-normal-code-dimension}
 c=p_2-m(L-b)=m\theta,
\end{equation}
and every \(c\) coordinate classes form a basis.

The trace evaluation for \(\Omega_\Lambda(D+1)\) is an
isomorphism by the hypothesis and the integral count.  We claim
that the slice evaluation for \(\Omega_{H_A}(D+1)\) at
\(A'\cup Z_2\) is injective.  Write a section as
\((F_1,\ldots,F_m,F_s)\).  Its tangential trace is zero, so
\(F_i=sw_i\); vanishing at \(A'\) gives \(w_i\in W_A\).
Euler's relation gives \(F_s=-\sum y_iw_i\).
The normal trace values vanish on \(Z_2\), and injectivity of
evaluation on \(U_A\) implies \(\sum y_iw_i\in sW_A\).
The resulting linear relation on \(W_A\) is zero by its product
injection.  Thus the section is zero.

The cokernel of this slice evaluation is precisely \(Q_A\).
Indeed first remove arbitrary tangential trace values by a trace
Euler section.  Then remove arbitrary values at \(A'\) by sections
\((sw_1,\ldots,sw_m,-\sum y_iw_i)\), interpolating the \(w_i\)
there.  The remaining representatives are scalar normal values
on \(Z_2\); those arising from sections zero at \(A'\) are exactly
\(\operatorname{ev}(U_A)\), up to sign.  The same argument
identifies the other slice cokernel with \(Q_B\).

\emph{The ambient kernel.}
By Lemma~\ref{lem:height-degeneration}, the space \(G=I_X(e)\)
has dimension \(g\), restricts injectively to \(T_e\), and its
restricted image has injective multiplication by \(T_1\).
Scalar independence also gives
\(\dim I_X(D-1)=L+C-q=L+g\).

Let \(A_0\) be ambient Euler evaluation on \(Z\).
The restriction of any section in its kernel to either slice
vanishes tangentially, by the slice injections just proved.
Its components therefore have the form
\begin{equation}
\label{eq:two-normal-axis-kernel}
 F_i=st a_i,\qquad F_s=t a_s,\qquad F_t=s a_t,\qquad
 \sum_i y_i a_i+a_s+a_t=0.
\end{equation}
At \(X\), this means \(a_i\in G\) and \(a_s\in I_X(D-1)\);
the Euler relation then gives \(a_t\in I_X(D-1)\).
The remaining conditions are \(a_t|_{A'}=0\) and
\(a_s|_{B'}=0\).

Consider the space of pairs of scalar polynomials obtained from
\begin{equation}
\label{eq:two-normal-pair-map}
 G^m\oplus I_X(D-1)\longrightarrow
 k[H_A]_{D-1}\oplus k[H_B]_{D-1},\qquad
 (a_i,a_s)\longmapsto
 \left((a_s+\sum y_i a_i)|_{H_A},\,a_s|_{H_B}\right).
\end{equation}
This map is injective.  If both outputs vanish, restriction to
\(\Lambda\) and the product injection on \(\overline G\) give
\(a_i=0\) for every \(i\).  Then \(a_s=stc_0\), with \(c_0\)
vanishing at \(X\).  The polynomial \(sc_0\in G\) restricts
to zero, so it too is zero.

The pair space has dimension \(L+(m+1)g\).  Its pure subspaces
both have dimension \((m+1)g\): zero second component means
\(a_s=sf\), \(f\in G\), while zero first component means
\(a_s+\sum y_i a_i=tf\), \(f\in G\).
By \eqref{eq:two-normal-pure-margin} and
Lemma~\ref{lem:two-group-evaluation}, imposing the \(A'\) and
\(B'\) evaluations leaves a kernel of dimension
\begin{equation}
\label{eq:integral-two-normal-kernel-dimension}
 \dim\ker A_0=\max(0,d),\qquad
 d=L+(m+1)g-2b=2m\theta+(m+1)\Delta .
\end{equation}
We impose also the injectivity of the two pure evaluations.

\emph{The first response.}
The source dimension minus target dimension of \(A_0\) is
\((m+1)\Delta\).  If \(d\le0\), it is already injective.
If \(d>0\) and \(c=0\), it is already surjective.
It remains to consider \(d,c>0\).  In that case
\eqref{eq:integral-two-normal-kernel-dimension} gives
\(\dim\operatorname{coker}A_0=2c\).
The natural map
\[
 \operatorname{coker}A_0\longrightarrow Q_A\oplus Q_B
\]
is onto.  To lift a pair of slice representatives, use the
trace Euler isomorphism to make their tangential trace values
agree.  Compatible slice values lift to ambient fiber values;
the two normal trace values are independent, and the values at
\(X\) can be chosen freely.  Since both spaces have dimension
\(2c\), this map is an isomorphism.

Write \(K=\ker A_0\), and use a superscript \(0\) for restriction
to \(\Lambda\).  Each map
\[
 K\longrightarrow T_{D-1},\qquad
 F\longmapsto a_s^0,\quad F\longmapsto a_t^0
\]
is injective.  For if \(a_s^0=0\), then
\(a_s|_{H_B}=th\) with \(h\in k[H_B]_e\).
It vanishes on \(B'\), and \(b\ge R\), so generality of \(B'\)
gives \(h=0\).  The pair \eqref{eq:two-normal-pair-map} is
therefore pure \(A\), and injectivity of its \(A'\) evaluation
gives \(F=0\).  The other map is treated symmetrically.
Since \(p_2\ge M\), scalar degree-\((D-1)\) evaluation on
\(Z_2\) is injective.  Thus each family of functionals
\((a_s^0(z))_{z\in Z_2}\) and \((a_t^0(z))_{z\in Z_2}\)
spans \(K^*\).

Move a trace point \(z\) by
\(s(z)=\epsilon u_z\), \(t(z)=\epsilon v_z\), keeping its
trace coordinates fixed, and leave \(A',B',X\) fixed.
Formula~\eqref{eq:two-normal-axis-kernel} shows that the
first derivative on kernel and cokernel is
\[
 J=\sum_{z\in Z_2}v_z e_z^A\otimes
             [F\mapsto a_s^0(z)]
   +u_z e_z^B\otimes[F\mapsto a_t^0(z)].
\]
The tangential components start in order \(\epsilon^2\).
The coordinate-code property and
Lemma~\ref{lem:two-normal-response} give a motion for which
nearby Euler evaluation has maximal rank.  Distinctness and
scalar postulation persist on an open set.  The Euler
identification then gives the required maximal-rank
multiplication map at the chosen critical count.
Both choices of \(r\) are covered, so the point-count reduction
proves the proposition.
\end{proof}