Proof. From a point on a line, any positive distance has at most two realizations on that line. For a circle, use a sample point on the circle: the intersection with a circle of prescribed positive radius about that point has at most two points. These observations include the sample point itself after using \(M=1+|\Delta(P)|\).

The planar bound is Corollary 65. The other bounds follow by applying the lower-dimensional assertion to the points in the flat, whose pairwise distances are among the distances of \(P\). Since \(M\le B^2\), the planar bound is \(O(B^2\log B)\), while a proper flat of dimension \(u\ge3\) contains \(O(B^u)\) points. Thus, for example, \(c_0=1/2\) works after advancing along the sequence. A complex affine flat containing real points contains their real affine span, whose real dimension is at most its complex dimension. Applying the real bound to that span proves the last assertion. ◻

Proof. First discard pairs on lines containing more than \(H\) sample points, where \(H\) will be a sufficiently large fixed constant. A generic planar projection preserves the relevant distinct points, lines, and incidences. The rich-line consequence of Theorem 54 therefore bounds the number of physical lines with at least \(t\) points by \[C(N^2/t^3+N/t),\qquad t\ge2.\] If \(L_{\max}=o(N)\) is the maximum occupancy, a dyadic sum gives \[\#\{\text{discarded pairs}\} \le C N^2/H+C N L_{\max}=O(N^2/H)+o(N^2).\]

At each \(p\), independently of the other stars, greedily remove directions on any irreducible curve spanning at most a projective plane that currently contains more than \(K A\) times its degree many partners. Each removal contains at least one previously unremoved partner, so the process terminates. Record the curves chosen in that star. Their total degree is at most \(N/(KA)\), because their charged partner sets are disjoint. Now keep an unordered pair only when it survived both endpoint procedures.

There must be fixed \(H,K\) and a subsequence on which a positive fraction of pairs survives. Otherwise, a diagonal choice with \(H,K\to\infty\) sufficiently slowly would make the surviving fraction tend to zero, while the first, rich-line discard also has vanishing fraction. The recorded curves would then cover all but \(o(N^2)\) pairs from one endpoint or the other and have total degree at most \(N/(KA)=o(N/A)\) in every star. This contradicts Theorem 31. Fix such \(H,K\) and a positive-density subsequence.

We prove (2) by induction on \(a\). For \(a=0\) it is the light-line bound. Fix a variety \(Y\) of dimension \(a\ge1\), a center \(p\), and the selected partners counted on \(Y\). Let \(U\) be the real linear span of their displacements from \(p\), and put \(\ell=\dim U\). There is nothing to prove if this collection is empty.

Consider a full radius class about \(p\), meaning all points of \(P\) at one prescribed distance from \(p\). Suppose it has \(L\ge2M\) distinct orthogonal projections to \(U\), and choose physical points \(z_1,\ldots,z_L\in P\) in that class representing those projections. For each counted partner \(q\), the distances \(|q-z_i|\), including zero when it occurs, take at most \(M\) values. Consequently at least \[L^2/M-L\ge L^2/(2M)\] ordered pairs \(i\ne j\) have equal distance to \(q\). As the \(z_i\) lie in a common radius class, \[|q-z_i|=|q-z_j| \quad\Longrightarrow\quad (q-p)\cdot(z_i-z_j)=0.\] The projection of \(z_i-z_j\) to \(U\) is nonzero. The resulting hyperplane section is proper on \(Y\): if its linear equation vanished on \(Y\), it would vanish on all counted displacements and hence on \(U\), which is impossible for its nonzero normal in \(U\). Bézout’s theorem and the induction hypothesis bound the number of partners on each section by \(C\deg(Y)A^{a-1}\). Double counting gives \[\#\{q\text{ counted}\}\frac{L^2}{2M} \le C L^2\deg(Y)A^{a-1}.\] This is (2), since \(M\le A\).

In the remaining case, each of the at most \(M\) radius classes has fewer than \(2M\) projections to \(U\). Thus the full sample has \(O(M^2)\) fibers on which both the radius about \(p\) and the projection to \(U\) are fixed. A fiber lies on a sphere in an affine space of dimension \(d-\ell\). If \(\ell\ge4\) and \(d-\ell\ge3\), lower-dimensional induction bounds each fiber by \(O(M^{(d-\ell)/2})\). If \(d-\ell=2\), the circle bound is \(O(M)\); if \(d-\ell\le1\), the fiber has at most two points. In all cases the total is \(O(M^{d/2})\), contradicting \(N/M^{d/2}\to\infty\). Thus this alternative is impossible when \(\ell\ge4\).

It remains to consider \(\ell\le3\). For \(a\ge2\), Lemma 2 bounds the counted partners by \(O(A^2)\), which is sufficient. If the span is the entire space, then \(d=3\) and the trivial bound \(N=A^{3/2}\) is sufficient instead. For \(a=1\), either \(Y\subset\mathbb P(U_{\mathbb C})\), in which case the greedy curve bound applies, or its intersection with that projective subspace is proper and consists of at most \(\deg Y\) points. The light-line bound handles the latter case. This completes the induction. Fixed increases in the constants at the finitely many dimensions are harmless. ◻

Proof. First consider any subset \(S\) of the partners in one star, counting physical partners even when their projective directions agree. Form the projective envelope of those directions at degree \(D\). Assign each direction to an irreducible component containing it, and group these components by dimension. For a group of total degree \(a_j\) and dimension \(j\), Proposition 3 bounds the assigned physical partners by \(C a_j A^j\). The component Hilbert lower bound in Lemma 13, applied at the defining degree \(D\), gives \[H_S(D)=H_{\mathop{\mathrm{Env}}_D(S)}(D)\ge c_d a_jD^j.\] There are at most \(d\) groups, and \(D\asymp A\). Therefore \[ H_S(D)\ge c_1|S| \tag{3}\] with a fixed \(c_1>0\), uniformly for every subset \(S\).

Take a maximal undirected subset of the edges from Proposition 3 whose evaluations are independent at both endpoints. Each omitted edge is blocked at an endpoint: its evaluation vector lies in the span of the selected vectors there. Charge it to one such endpoint. If the selected degree at \(p\) is \(s_p\), all vectors charged to \(p\) together have rank at most \(s_p\). Equation (3) bounds their number by \(s_p/c_1\). Summing over vertices bounds the number of omitted edges by \(2/c_1\) times the number of selected edges. Positive density follows.

Independent evaluation vectors mean that the evaluation map onto the coordinates indexed by the star is surjective. Taking the coordinate basis in its image gives the asserted homogeneous separators. ◻

Proof. In Theorem 46, take \(e=d+1\), so \(r=d\). Physical separators from Lemma 8, one in \(p\) and one in \(q\), have a product of degree \(O(M)\). Express them in the coordinates \(v_0,y_0\) and homogenize with the last coordinate of \(v\), which is \(1\) on every parameter. They give positive-degree homogeneous separators of degree \(O(M)\le O(A)\).

Suppose parameters lie in one projective maximal isotropic plane. Their original real representatives in (7) belong to its underlying isotropic vector space. Polarization gives, for any two of them, \[(v_0-v'_0)\cdot(y_0-y'_0)=0,\qquad |p-p'|^2=|q-q'|^2.\] The first physical coordinate \(p\) is therefore injective on this set of parameters, because \(p=p'\) forces \(q=q'\). There are at most \(N\le A^{d-1}\) such parameters. The total number of parameters is at most \(N^2=A^d\). Thus Specification (a) of Theorem 46 applies, with one fixed constant absorbing the separator degree.

For \(h=0\), an irreducible variety containing a same-dimensional flat equals that flat. For \(h=d\), the total-count bound suffices. These give the two endpoints. ◻

Proof. The local skew-form model has \(e=d\), \(r=d-1\), and parameters \[[v:y]=[u+w:u-w].\] They lie on the split quadric, with nonzero \(v\). They are distinct. Indeed, proportional parameters give \((u',w')=\lambda(u,w)\). Since the vectors are real and nonzero, \(\lambda\) is real; uniqueness of a selected partner in its projective direction first gives \(u'=u\), then \(\lambda=1\) and \(w'=w\).

Multiply a homogeneous direction separator for \(u\) by one for \(w\) and substitute \(u=(v+y)/2\), \(w=(v-y)/2\). The product has degree \(O(A)\) and separates the corresponding neighbor parameter from all others. No physical scale ambiguity remains, because each selected star contains at most one physical partner in each projective direction.

For parameters in a common maximal isotropic plane, polarization on the original real representatives gives \[(u_i-w_i)\cdot(u_j+w_j) +(u_j-w_j)\cdot(u_i+w_i)=0,\] or equivalently \[u_i\cdot u_j=w_i\cdot w_j.\] The rule \(u_i\mapsto w_i\) is a well-defined real isometry on their spans: every linear relation is preserved by equality of the Gram matrices and positive definiteness of the real norm. Extend it to some \(S\in O(d)\). Every neighbor in this isotropic plane, together with the center, is a match for \(x\mapsto q+S(x-p)\).

For \(d\ge4\), the number of such neighbors is at most \(N=A^{d/2}\le A^{d-2}\). For \(d=3\), a collection with at least \(C_{\mathrm{rich}}A\) neighbors would, together with the center, violate the survivor property established when the rich match edges were deleted. The surviving cap is therefore \(O(A)\). This establishes precisely the isotropic hypothesis \(O(A^{r-1})\) in every dimension. In particular no restriction to orientation-preserving motions was made.

Factor the polynomial on \(F_{p,q}\) into irreducible hypersurfaces. Each component has excess dimension \(h=d-2=r-1\) above a neighbor flat, and lies in a hypersurface of degree at most \(A\). Specification (b) of Theorem 46 applies to it; that specification has no total-count hypothesis. Summing its bound \(C\deg(V)A^{d-2}\) over components of total degree at most \(A\) proves the result. ◻

Proof. Let \(\mathcal F\) be the family of retained global pair flats and \(m=|\mathcal F|\). At degree \(D=\lfloor A\rfloor\), form the envelope of their union. Each flat is contained in one of its irreducible components. Assign it to one such component and group components by their dimensions \(s_0+h\), \(0\le h\le d\). If a group has total degree \(a_h\), global concentration bounds its assigned flats by \(C a_h A^h\). The Hilbert lower bound at the defining degree gives \[H_{\bigcup\mathcal F}(D) \ge c a_hD^{s_0+h}.\] There are only \(d+1\) groups. Since \(D\asymp A\), \[ H_{\bigcup\mathcal F}(D)\ge c_4mA^{s_0}. \tag{9}\] This includes the case in which the envelope is the entire ambient space.

Choose each flat independently with a small fixed probability \(\theta>0\). With probability at least \(1/2\), the number selected is at most \(2\theta m\), by Markov’s inequality. On that event their restriction rank is at most \[2\theta m\binom{D+s_0}{s_0}\le C_4\theta m A^{s_0}.\] Choose \(\theta\) so small that this is less than the right side of (9).

At any vertex the sampled neighbors have binomial distribution of mean at least \(\theta\delta(\mathcal H)\). The probability of fewer than half that mean is at most \(\exp(-c\theta\delta(\mathcal H))\). There are at most \(N^2\) vertices, and \(\delta(\mathcal H)\gg A^{d-1}\), a fixed positive power of \(N\). A union bound shows that, with probability tending to one, every vertex has at least \(\theta\delta(\mathcal H)/2\gg A^{d-1}\) sampled neighbors. Both events thus hold for some choice of sample.

Its rank is strictly smaller than that of all flats. Hence a polynomial \(f\) of degree at most \(D\) vanishes on every sampled flat but does not vanish identically on some \(F\in\mathcal F\). For each sampled neighbor \(F'\), it vanishes on the neighbor flat \(F\cap F'\) inside \(F\). These neighbor flats are distinct by the local parameter check above. There are more than \(C A^{d-1}\) of them, contradicting Lemma 6 applied to the nonzero restriction \(f|_F\). ◻

Proof. For \(p\in P\) define \[s_p(x)=\prod_{\rho\in\Delta(P)} \frac{\sum_{j=1}^n(x_j-p_j)^2-\rho^2}{-\rho^2}.\] Every denominator is nonzero. The stated values on the real sample follow even though we subsequently regard \(s_p\) as a polynomial over \(\mathbb C\).

We prove (11) by induction on \(u\) and add over the irreducible components of \(V\). For \(u=0\) the number of points is at most \(a\). Suppose \(V\) is irreducible of positive dimension and contains a sample point \(p\). The polynomial \(s_p\) does not vanish identically on \(V\), because it is nonzero at \(p\), and it vanishes at every other sample point of \(V\). Its proper section has degree at most \(2Ma\). The induction hypothesis therefore gives \[|P\cap V|\le 1+C_n(2Ma)M^{u-1}.\] Increasing the dimension-dependent constant proves the claim. This argument requires no smoothness or general-position assumption on the sample points.

Finally use the polynomials \[ 1-\sum_{p\in P}s_p(x),\qquad (x_j-p_j)s_p(x)\quad(p\in P,\ 1\le j\le n). \tag{12}\] They vanish at every point of \(P\). At any common zero, the first equation forces some \(s_p\) to be nonzero, and the corresponding \(n\) coordinate equations then force \(x=p\). Thus there are no additional zeros. ◻

Proof. The coordinate parameters form a regular sequence. The regular local ring \(R\) is Cohen–Macaulay, so the \(c\) elements \(f_i\), being a system of parameters, also form a regular sequence (The Stacks Project Authors 2026, Tags 00OQ, 00NQ, and 02JN). Their Koszul complexes are therefore free resolutions of \(R/(f)\) and \(R/(Z)=F\), respectively (The Stacks Project Authors 2026, Tag 062F). The matrix \(Q\) induces a chain map from the first Koszul resolution to the second, lifting the quotient map \(R/(f)\longrightarrow F\); its map in top degree is multiplication by \(\det Q\).

For completeness, the vanishing needed to see that this top map is nonzero is especially simple. The coordinate Koszul resolution gives \[\operatorname{Ext}^i_R(F,R)=0\quad(i<c),\qquad \operatorname{Ext}^c_R(F,R)=F.\] Every finite-length \(R\)-module has a composition series with factors \(F\). The long exact sequence for \(\operatorname{Ext}\) therefore gives \(\operatorname{Ext}^i_R(K,R)=0\) for every finite-length \(K\) and \(i<c\). Apply this to the kernel \(K\) of \(R/(f)\longrightarrow F\). The long exact sequence supplies an injection \[\operatorname{Ext}^c_R(F,R)\lhook\joinrel\longrightarrow \operatorname{Ext}^c_R(R/(f),R).\] Under the identifications provided by the two Koszul resolutions, this map sends the class of \(1\) to the class of \(\det Q\), up to an irrelevant sign. Its image is nonzero. ◻

Proof. Fix an isolated point \(z\) of the original system. Its local defining ideal is primary to the maximal ideal at \(z\). We may choose \(c\) constant linear combinations \(f_1,\ldots,f_c\) of the given equations whose local ideal has finite colength. To see this, choose the combinations successively, avoiding the finitely many minimal primes of the preceding proper intersections other than the maximal ideal. The original ideal is not contained in any of those primes. Over an infinite field the forbidden coefficient vectors form a finite union of proper linear subspaces, so each choice is possible. Each \(f_i\) still has degree at most \(L\). The same avoidance can be imposed simultaneously at any finite list of isolated points. Thus a generic choice of \(c\) combinations is a local parameter sequence at every point of that list, although the argument below only needs this property at \(z\).

Introduce two sets of variables \(Y,Z\). Telescoping one coordinate at a time gives a divided-difference matrix \(Q(Y,Z)\) such that \[ f(Y)-f(Z)=Q(Y,Z)(Y-Z). \tag{13}\] Every entry has degree at most \(L-1\) in \(Y\), and hence \(\det Q(Y,Z)\) has degree at most \(c(L-1)\) in \(Y\). Take the coefficients in the local Artin algebra \[A_z=F[Z]_{(Z-z)}/(f_1(Z),\ldots,f_c(Z)).\] If \(y\ne z\) is any common zero of the original system, then \(f(y)=0\) and one coordinate of \(y-Z\) is a unit in \(A_z\). Multiplying \(Q(y,Z)(y-Z)=0\) by the adjugate of \(Q(y,Z)\) shows that \[\det Q(y,Z)=0\quad\hbox{in }A_z.\] This works for every such \(y\), whether or not it is isolated.

On setting \(Y=z\) in (13), we obtain \(f(Z)=Q(z,Z)(Z-z)\). Lemma 9, after translation, says that \(\det Q(z,Z)\) is nonzero in \(A_z\). Choose an \(F\)-linear functional \(\lambda:A_z\longrightarrow F\) which is nonzero on this element. Applying \(\lambda\) coefficientwise to \(\det Q(Y,Z)\) and normalizing produces a polynomial \(s_z\) of the required degree, with \(s_z(z)=1\) and \(s_z(y)=0\) at every other common zero.

Here are explicit first-order interpolants, which also show that the number of points does not enter the degree bound. The polynomial \[u_z=3s_z^2-2s_z^3\] has value \(1\) and all first derivatives zero at \(z\), and has value and all first derivatives zero at the other selected points. The polynomial \[v_{z,j}=(Y_j-z_j)s_z^2\] has zero value at every selected point, derivative \(1\) in coordinate \(j\) at \(z\), and all other prescribed first derivatives zero. Linear combinations of the \(u_z\) and \(v_{z,j}\) give arbitrary first-order data. Their degrees are at most \(\max\{3c(L-1),2c(L-1)+1\}\le C_cL\).

The exact-isolation Equations (12), with these normalized \(s_z\) and the set \(S\), now have degree at most \(c(L-1)+1\) and have precisely \(S\) as their common zero set. The empty selection is isolated by the equation \(1=0\). ◻

Proof. For a point \(z\in S\), treat each list separately. If \(z\in S_\nu\), take its normalized separator in that list. If \(z\notin S_\nu\), the exact-isolation part of Lemma 10 supplies an equation for \(S_\nu\) which does not vanish at \(z\); normalize its value there to \(1\). The product over the lists is \(1\) at \(z\) and zero at every other point of their union. Its degree is at most \(C_cL_*\). The first-order construction in the preceding proof costs a further fixed factor only. ◻

Proof. We first justify the upper bound with its precise leading coefficient. On a projective closure, polynomial restrictions inject into the global sections of \(\mathcal O_V(D)\). A general hyperplane avoids the associated components and has reduced pure-dimensional intersection with a reduced variety; this is the reduced hyperplane-section form of Bertini’s Theorem in characteristic zero. To see the reduced-scheme assertion, the universal hyperplane incidence is a projective-space bundle over the reduced variety, hence is reduced. Its generic fiber over the dual projective space is reduced by localization and is geometrically reduced in characteristic zero. The openness of reduced fibers (The Stacks Project Authors 2026, Tag 0578) gives the assertion for a general hyperplane. Avoiding the components gives pure codimension one, and the section has degree \(a\). The restriction exact sequence gives \[h^0(V,\mathcal O_V(D))\le h^0(V,\mathcal O_V(D-1))+ h^0(V\cap H,\mathcal O_{V\cap H}(D)).\] For dimension zero, the last space has dimension equal to the number of reduced points, namely the degree. In every dimension, \(h^0(V,\mathcal O_V)\le a\): it is at most the number of irreducible components, since a regular function on an irreducible projective variety is constant. Induction on dimension, followed by summation in \(D\), gives \[h^0(V,\mathcal O_V(D)) \le a\sum_{j=0}^D\binom{j+k-1}{k-1} =a\binom{D+k}{k}\] for \(k\ge1\). This proves (14); alternatively one may add the bound for the irreducible components. Affine homogenization gives exactly the same restriction rank.

For the lower bound, choose generic linear coordinates \[(t_1,\ldots,t_k,y_1,\ldots,y_{n-k})\] such that projection to the \(t\) coordinates is finite of degree \(\deg V_i\) on every irreducible component \(V_i\) of the family. This is the usual generic linear normalization: in the projective closure the projection center is disjoint from \(V_i\), and a general fiber is a complementary linear section. In characteristic zero the generic fibers are reduced. Put \(K=\mathbb C(t_1,\ldots,t_k)\) and let \(F\) be an algebraic closure of \(K\). The geometric generic fiber of the selected family consists of \(a\) distinct points of \(F^{n-k}\).

These points are isolated in the generic fiber of the entire given system. Indeed, if \(W\) is any different irreducible component of that system, then \(V_i\cap W\) is a proper subvariety of \(V_i\). Its projection to the \(k\)-dimensional base is not dominant. Thus no point of the generic fiber of \(V_i\) lies on \(W\). This argument also handles components \(W\) of dimension larger than \(k\); their generic fibers may be positive-dimensional, but they do not contain the selected sheets.

The equations on the generic fiber have degree at most \(L\) in \(y\). Lemma 10 says that evaluation on its \(a\) selected points is surjective from \(F[y]_{\le q}\) for some \(q\le C_nL\). Since ordinary monomials span this space, we can select \(a\) monomials \(m_1(y),\ldots,m_a(y)\) of degree at most \(q\) whose evaluation columns are linearly independent over \(F\). For each \(i\), multiply \(m_i\) by all base monomials \(t^\alpha\) with \(|\alpha|\le D-q\). The resulting global polynomials have total degree at most \(D\) and are linearly independent on \(V\). In fact a relation over \(\mathbb C\) would give \[\sum_i\left(\sum_\alpha c_{i,\alpha}t^\alpha\right) m_i\big|_{\text{generic fiber}}=0.\] Independence over \(F\) forces each coefficient polynomial in \(t\) to vanish, and algebraic independence of the \(t_j\) forces every \(c_{i,\alpha}\) to be zero. Counting these monomials proves (15). Notice that no rational coefficient has been cleared: only ordinary transverse monomials were selected.

We explain the passage to degree \(L\), since the lower bound above initially starts at a constant multiple of \(L\). In a monomial order refining total degree, the standard monomials of \(I(V)\) form a set closed under division, and their number of degree at most \(D\) is \(H_V(D)\). Fix an integer \(b\ge1\). The map \[x^\alpha\longmapsto x^{\lfloor\alpha/b\rfloor}\] sends standard monomials of degree at most \(bL\) to standard monomials of degree at most \(L\). Each image has at most \(b^n\) preimages, by the possible coordinatewise remainders. Hence \[ H_V(bL)\le b^n H_V(L). \tag{17}\] Take a fixed \(b\) large enough that \(bL-q\ge L\) and use (15). This proves (16).

For the union-of-systems assertion, take a generic base projection simultaneously for the finitely many systems and components. On its geometric generic fiber, Lemma 12 supplies the same surjectivity at degree \(C_nL_*\). The monomial argument and (17) are unchanged. This proof does not claim that a component of one system stays an isolated component of the union of all the original loci; it uses the selected finite lists on the generic fiber instead. ◻

Proof. For the lower bound, \(H_V(L)\) is at least the rank of each pure dimensional subfamily. Lemma 13 bounds each of those below by \(c_na_kL^k\); taking a maximum loses at most the fixed factor \(n+1\). For the upper bound, restriction to the entire union injects into the direct sum of its restrictions to the pure-dimensional subfamilies. Apply (14) to each. ◻

Proof. There is nothing to prove when \(k=n\): the only possible nonempty family is the full ambient space, with codimension zero. Otherwise choose the generic base and transverse coordinates used in Lemma 13, and keep the notation \(K=\mathbb C(t)\) and \(F=\overline K\). Let \(b\) be the total degree of the components on which the asserted generic Jacobian rank fails. Their geometric generic fibers contribute \(b\) of the \(a\) sheets.

We give the vector-space argument carefully. Write \[W_D=\mathbb C[t,y]_{\le D},\qquad E_D:W_D\longrightarrow F^a, \quad f\longmapsto (f(t,z_i))_{i=1}^a.\] These are a vector space and a linear map over \(\mathbb C\), and \[\ker_{\mathbb C}E_D=I(V)_{\le D},\qquad \mathop{\mathrm{rank}}_{\mathbb C}E_D=H_V(D).\] No linearity of this truncated kernel over \(K\) is asserted.

At a failing sheet \(z_i\), the common kernel of the ambient gradients of the equations in \(I(V)_{\le D}\) has dimension at least \(k+1\) over \(F\). Its intersection with the \((n-k)\)-dimensional vertical space therefore contains a nonzero vector \(v_i\). The tangent space of the component projects isomorphically to the base at a generic sheet, so this vertical vector is transverse to the component. Define a first derivative functional \[J_i(f)=v_i\mathbin{\cdot}\partial_yf(t,z_i).\] Its coefficients may be algebraic functions of \(t\). They are coefficients of this functional, not functions to be differentiated. By construction \(J_i\) annihilates \(\ker_{\mathbb C}E_D\). If \(J\) collects these \(b\) functionals, then \[ \mathop{\mathrm{rank}}_{\mathbb C}(E_D,J)=\mathop{\mathrm{rank}}_{\mathbb C}E_D=H_V(D). \tag{19}\]

First-order interpolation on the geometric generic fiber gives surjectivity onto these \(a\) values and \(b\) transverse derivatives from \(F[y]_{\le q}\), where \(q\le C_nL\). Select \(a+b\) ordinary \(y\)-monomials \(m_i\) of degree at most \(q\) with linearly independent augmented columns \(w_i\in F^{a+b}\). Vertical differentiation gives the identity \[(E_D,J)(t^\alpha m_i)=t^\alpha w_i \qquad(|\alpha|\le D-q).\] All these input polynomials lie in \(W_D\). A relation over \(\mathbb C\) among their columns becomes an \(F\)-linear relation among the \(w_i\), with coefficients that are polynomials in \(t\). Independence over \(F\), followed by algebraic independence of \(t\) over \(\mathbb C\), makes every coefficient zero. Thus \[ (a+b)\binom{D-q+k}{k}\le H_V(D) \le a\binom{D+k}{k}. \tag{20}\] For \(D\ge2q\), the ratio of the two binomial coefficients is at most \(1+C_nq/D\); this follows by multiplying the \(k\) factors \((D+i)/(D-q+i)\). Rearranging (20) proves (18). When \(k=0\) the binomial coefficients are both \(1\) and the argument gives \(b=0\).

Choose a basis \(f_1,\ldots,f_s\) of \(I(V)_{\le D}\). All \((n-k)\)-row Jacobian minors have degree at most \((n-k)(D-1)\). On each retained component at least one is nonzero. A generic constant linear combination \(g\) of all these minors is consequently nonzero on every retained component, since the excluded coefficient vectors are a finite union of proper linear subspaces. Wherever \(g\ne0\), some \(n-k\) of the \(f_i\) have independent differentials. Their zero locus is a smooth \(k\)-fold locally by the complex implicit function theorem. The pure \(k\)-dimensional set \(V\) is contained in it and has the same local dimension, so its reduced germ is that smooth germ. All remaining \(f_i\) vanish on this germ as well. Two distinct irreducible components cannot share this germ: an open common germ is Zariski dense in either component and would make them equal. ◻

Proof. For \(k=0\) all the cuts are empty, and take \(g=1\). Suppose \(k\ge1\) and let \(C\) be the union of their reduced supports. Bézout gives \(\deg C\le aL'\) in dimension \(k-1\), so \[H_C(D)\le aL'\binom{D+k-1}{k-1}.\] Let \(U_D=I(C)_{\le D}\), and call a component \(V_i\) forced if every polynomial in \(U_D\) vanishes identically on it. If their union has degree \(b\), the kernel inclusion \(I(C)_{\le D}\subset I(\bigcup_{i\text{ forced}}V_i)_{\le D}\) implies \[H_{\bigcup_{i\text{ forced}}V_i}(D)\le H_C(D).\] The lower bound of Lemma 13 applies to this subfamily with the same \(L\). For \(D\ge C_nL\) it bounds the left side below by \(c_n bD^k\), proving (21).

For each unforced component, the polynomials in \(U_D\) which vanish identically on it form a proper linear subspace of \(U_D\). Choose \(g\) outside the finite union of these subspaces. On increasing the fixed lower bound for \(D/L'\), the degree estimate makes \(b<a\) for a nonempty family, so such a \(g\) is a nonzero polynomial. It has precisely the required properties. ◻

Proof. An empty set \(S\) is handled by \(g=1\). Otherwise \(E\ge r\), because \(f\) is a nonzero polynomial vanishing to order at least \(r\) somewhere. First work over an algebraically closed field of characteristic \(p\), where \(p\ge r\). Set \[m=\left\lceil\frac{np}{r}\right\rceil.\] The Frobenius map is surjective on this field, so grouping monomials by their exponent vectors modulo \(p\) gives a unique expression \[ f(X)^m=\sum_{0\le i_1,\ldots,i_n<p}X^i h_i(X)^p. \tag{22}\] Every nonzero \(h_i\) has \[\deg h_i\le mE/p\le nE/r+E/p\le(n+1)E/r.\]

Fix \(z\in S\). The order of \(f^m\) at \(z\) is at least \(mr\ge np\). Every monomial in \(X-z\) of total degree at least \(np\) has some coordinate exponent at least \(p\). It follows that \[f^m\in J_z=((X_1-z_1)^p,\ldots,(X_n-z_n)^p).\] Modulo \(J_z\), the identity \(h_i(X)^p\equiv h_i(z)^p\) holds. The monomials \(X^i\), with every digit below \(p\), form a basis of the quotient by \(J_z\): translating the usual basis \((X-z)^i\) is an invertible triangular change of basis. Reducing (22) modulo \(J_z\) therefore gives \(h_i(z)=0\) for every \(i\). This applies at every point of \(S\). Evaluation at \(x_*\) shows that at least one \(h_i(x_*)\) is nonzero, because \(f(x_*)^m\ne0\). That \(h_i\) proves the assertion in characteristic \(p\).

We include a concrete specialization argument to transfer this fixed degree assertion to \(\mathbb C\). Put \(q=\lfloor(n+1)E/r\rfloor\) and suppose that no degree-\(q\) polynomial separates the given \(S=\{z_1,\ldots,z_s\}\) from \(x_*\). In the finite-dimensional space of polynomials of degree at most \(q\), this says that evaluation at \(x_*\) belongs to the linear span of the evaluations at the \(z_j\). Thus there are \(\lambda_1,\ldots,\lambda_s\in\mathbb C\) with \[ x_*^\alpha=\sum_{j=1}^s\lambda_j z_j^\alpha \qquad(|\alpha|\le q). \tag{23}\] Let \(R\subset\mathbb C\) be the finitely generated \(\mathbb Z\)-algebra generated by the coefficients of \(f\), all coordinates of these points, the \(\lambda_j\), and \(f(x_*)^{-1}\). All hypotheses are polynomial identities over \(R\): order at least \(r\) is expressed by the vanishing of the coefficients of total degree below \(r\) in \(f(z_j+Y)\). The inverse in \(R\) preserves \(f(x_*)\ne0\) under every specialization. The identities (23) also hold in \(R\).

Such an \(R\) has specializations to algebraically closed fields of arbitrarily large positive characteristic. Here is the required algebraic justification. Noether normalization for the finitely generated domain \(R\otimes_{\mathbb Z}\mathbb Q\) gives algebraically independent elements \(t_1,\ldots,t_b\) over which it is finite. After inverting one nonzero integer \(a\), these elements belong to \(R[1/a]\) and the finitely many generators of \(R[1/a]\) satisfy monic equations over \(\mathbb Z[1/a,t_1,\ldots,t_b]\). Hence \(R[1/a]\) is finite integral over that polynomial ring. For every prime \(p\nmid a\), the lying-over property supplies a maximal ideal above \((p,t_1,\ldots,t_b)\). Its residue field is a finite extension of \(\mathbb F_p\), and embeds in \(\overline{\mathbb F}_p\).

Choose such a prime \(p\ge r\). Specializing the coefficients, points, and identities gives characteristic-\(p\) data satisfying the hypotheses and (23). The characteristic-\(p\) construction produces a degree-\(q\) polynomial which vanishes at all the specialized \(z_j\) but not at the specialized \(x_*\). This contradicts (23). Repeated specialized points cause no problem: the evaluation identity and the compression construction remain valid. The contradiction proves the complex assertion. ◻

Proof. Expand the determinant using columns from \(A(p)\) and from \(A(x)-A(p)\). A term with fewer than \(h\) latter columns uses more than \(q-h\) constant columns and is zero, because those columns have rank at most \(q-h\). Each remaining term belongs to the \(h\)th power of the maximal ideal at \(p\). This is an assertion in the ambient polynomial ring; no regularity of a subvariety containing the points is needed. ◻

Proof. For (i), if \(p\notin X\), projection is a morphism and the pullback of a target hyperplane is a hyperplane section of \(X\). Intersecting \(k\) general target hyperplanes and counting the \(t\) points over a general image point gives \(t\deg Y=a\).

Suppose \(p\in X\). We give the intersection calculation that also justifies the formula when \(X\) is singular or not Cohen–Macaulay. For the blowup construction and the degree formula for a linear system with base locus, see (Fulton 1998, Appendix B.6 and Proposition 4.4). Let \(b:\widetilde X\longrightarrow X\) be the blowup of the ideal of \(p\), and let \(E\) be its exceptional Cartier divisor. The blowup is the relative projective spectrum of the Rees algebra \(\bigoplus_{j\ge0}\mathcal I_p^j\), where \(\mathcal I_p\) is the ideal sheaf of \(p\); it is the closure of the graph of projection from \(p\). Write \(H\) for the pullback of the hyperplane divisor of \(X\). The linear forms through \(p\) generate the ideal of \(p\) twisted by \(\mathcal O_X(1)\), so the resolved projection \(f:\widetilde X\longrightarrow Y\) satisfies \[f^*\mathcal O_Y(1)=\mathcal O_{\widetilde X}(H-E).\] Its top self-intersection is consequently \(t\deg Y\), by taking general hyperplanes on the target.

We recall each term in this self-intersection. The divisor \(H\) restricts trivially to \(E\), since \(E\) lies over the point \(p\). Therefore \[H^{k-j}E^j=0\qquad(1\le j<k).\] Moreover \[E=\operatorname{Proj}\!\left( \bigoplus_{j\ge0}\mathfrak m_p^j/\mathfrak m_p^{j+1}\right), \qquad \mathcal O(E)|_E=\mathcal O_E(-1).\] The Hilbert polynomial of this associated graded ring shows that the degree of \(E\) with respect to \(\mathcal O_E(1)\) is \(\mathop{\mathrm{mult}}_pX\). Thus \(E^k=(-1)^{k-1}\mathop{\mathrm{mult}}_pX\). Also \(H^k=a\), as can be computed using general hyperplanes avoiding \(p\). Expanding gives \[t\deg Y=(H-E)^k=a+(-1)^kE^k=a-\mathop{\mathrm{mult}}_pX.\] Only Cartier-divisor intersection numbers on the integral blowup are used here. In particular, the calculation does not identify multiplicity with the length of an arbitrary local linear section.

For (ii), the case \(n=k+1\) is the identity projection. Otherwise choose a linear center \(\Lambda\simeq\mathbb P^{n-k-2}\). For a fixed point \(q\), let \(J(q,X)\) be the closure of the union of the lines joining \(q\) to points of \(X\) different from \(q\). Its dimension is at most \(k+1\). Choose one smooth point \(q_0\in X\). A general \(\Lambda\) is disjoint from each of \[X,\quad J(p_i,X),\quad J(q_0,X),\quad T_{q_0}X, \quad\text{and, if specified, }J(x_*,X).\] There are finitely many conditions, and each is a nonempty open condition on the Grassmannian: the dimension of the relevant closed set plus \(n-k-2\) is strictly less than \(n\).

The resulting projection defines a morphism \(\pi:X\longrightarrow \mathbb P^{k+1}\). It is finite. Indeed, a fiber lies in a projective linear space in which \(\Lambda\) is a hyperplane; a positive-dimensional projective fiber would meet that hyperplane, contradicting \(\Lambda\cap X=\varnothing\). Avoidance of the joins makes the fibers over the specified \(p_i\), and over \(q_0\), singletons. Avoidance of \(T_{q_0}X\) makes the differential injective at \(q_0\), so the fiber at \(q_0\) is a reduced point. More explicitly, its Artinian local ring has zero cotangent space and hence zero maximal ideal by Nakayama’s Lemma. The generic rank of a finite module is at most the dimension of any fiber, again by Nakayama’s Lemma. The generic degree is therefore one. The image is an integral \(k\)-fold in \(\mathbb P^{k+1}\), hence a hypersurface, and its degree equals \(a\) by pulling back \(k\) general hyperplanes. The additional join condition ensures \(\pi(x_*)\notin\pi(X)\).

It remains to verify the local multiplicity assertion without a smoothness assumption at \(p_i\). Write \[A=\mathcal O_{\pi(X),\pi(p_i)},\qquad B=\mathcal O_{X,p_i},\] with maximal ideals \(\mathfrak m\) and \(\mathfrak n\). Because the fiber is a singleton, \(B\) is a finite local \(A\)-algebra; because the projection is birational, it has rank one and \(A\subset B\). The quotient \(B/A\) is annihilated by a nonzero element of \(A\) and therefore has dimension at most \(k-1\). The Artin–Rees property gives a fixed \(c\) such that \[\mathfrak m^jA\subseteq\mathfrak m^jB\cap A \subseteq\mathfrak m^{j-c}A\qquad(j\ge c).\] Using the exact sequence \[0\longrightarrow A/(\mathfrak m^jB\cap A) \longrightarrow B/\mathfrak m^jB \longrightarrow (B/A)/\mathfrak m^j(B/A)\longrightarrow0\] shows that \(A/\mathfrak m^jA\) and \(B/\mathfrak m^jB\) have the same leading coefficient of degree \(k\) in their lengths. Since \(\mathfrak mB\subseteq\mathfrak n\), \[\dim_{\mathbb C} B/\mathfrak m^jB\ge \dim_{\mathbb C}B/\mathfrak n^j.\] Comparison of the leading coefficients proves (25). ◻

Proof. For a zero-dimensional irreducible variety, \(X\) is one reduced point, and a linear polynomial suffices. In positive dimension apply Lemma 19(ii) to the projective closure, the finite set \(S\), and \(x_*\). Compose a homogeneous defining equation of the image with the linear forms of the projection and dehomogenize in the original affine chart. The resulting polynomial has degree at most \(a\) and is nonzero at \(x_*\). Choose the target chart to contain the finitely many specified images. Its dehomogenization differs locally from this polynomial by a unit, so its order at every \(p\in S\) is at least \(\mathop{\mathrm{mult}}_pX\): composition does not decrease order, and hypersurface multiplicity equals the order of its equation. Apply Lemma 17 with \(r=\lfloor a/2\rfloor+1\). ◻

Proof. Choose an affine hyperplane through \(p+T_pX\) which does not contain \(z\). Its linear equation \(\ell\) restricts to an element of \(\mathfrak m_{X,p}^2\), because its value and differential on \(X\) both vanish at \(p\). On the curve it therefore belongs to \(\mathfrak m_{C,p}^2\), and it is not identically zero.

The one-dimensional reduced local ring of the curve is Cohen–Macaulay. Thus, writing \(R=\mathcal O_{C,p}\), the local intersection number with the hyperplane is \(\operatorname{length}(R/(\ell))\), and \(\operatorname{length}(R/(\ell^j))= j\operatorname{length}(R/(\ell))\). Since \((\ell^j)\subseteq\mathfrak m_{C,p}^{2j}\), comparison with the Hilbert–Samuel leading term gives \[\operatorname{length}(R/(\ell))\ge2\mathop{\mathrm{mult}}_pC.\] Bézout bounds this local intersection number by \(\deg C\). The final assertion follows from (24). ◻

Proof. Write \(C=\{F=0\}\) with \(F\) homogeneous and squarefree. A general first polar \[G=\lambda_0\frac{\partial F}{\partial X_0} +\lambda_1\frac{\partial F}{\partial X_1} +\lambda_2\frac{\partial F}{\partial X_2}\] has degree \(a-1\) and has no irreducible factor in common with \(F\). Indeed, for every irreducible factor \(F_i\) some derivative of \(F_i\) is nonzero modulo \(F_i\) in characteristic zero. Modulo \(F_i\), the corresponding derivative of \(F\) is this derivative times the product of the other factors. A generic coefficient vector avoids vanishing identically on each of the finitely many components.

In an affine chart at \(q\), write \(f\) for the local equation of the curve. The three dehomogenized derivatives of \(F\) are, after renaming the coordinates, \[f_x,\qquad f_y,\qquad af-xf_x-yf_y.\] Each has order at least \(m_q-1\), so the local equation of \(G\) has order at least \(m_q-1\). The elementary local intersection inequality gives \[I_q(F,G)\ge m_q(m_q-1).\] One can see that inequality directly on the reduced local branches: a branch of multiplicity \(b\) has a primitive parametrization whose two coordinates have minimum order \(b\), and substituting into a polynomial of order \(c\) gives order at least \(bc\). Add over the branches to obtain the displayed bound. Since \(F\) and \(G\) have no common component, Bézout gives total intersection number \(a(a-1)\) and in particular makes the set of points with \(m_q\ge2\) finite. This proves the first inequality. The second follows from \(m^2\le2m(m-1)\) for every integer \(m\ge2\). ◻

Proof. Choose a direction chart by requiring one coordinate of the direction to equal one. Write \(y-x=a u\) and \(z-x=b u\) in that chart; the two nonzero longitudinal parameters \(a,b\) are distinct. For a source velocity \(\xi\in T_xX\), variation of \(x\) with the other endpoint fixed has direction derivative \(-\bar\xi/a\) at \(y\), and \(-\bar\xi/b\) at \(z\). The first coordinates of these two image tangent vectors are both \(\xi\). Their difference is therefore vertical and has direction component \((b^{-1}-a^{-1})\bar\xi\).

At a general pair the product image has dimension \(k+l\): its first coordinate records the \(k\) source directions, and injectivity of the target tangent modulo the connecting line supplies \(l\) vertical directions. If the further root exists generally, the two irreducible product images coincide. Indeed they have the same dimension \(k+l\), and the first is contained in the second on a dense set. At a general smooth image point their tangent spaces agree. The vertical kernel of the first product image is exactly \(\overline{T_yY}\). The preceding subtraction gives \(\overline{T_xX}\subseteq\overline{T_yY}\) and hence \(k\le l\). This calculation also proves the local assertion: the full-rank first product identifies its tangent image with \(T W\), whose vertical kernel is \(\overline{T_qY}\). Source variation with the second root held fixed still gives tangent vectors in \(T W\). No assertion that the second root belongs to the finite sample is used.

When \(k=l\), the projected tangents agree. Thus \[T_xX\subseteq T_yY+\mathbb C(x-y), \qquad T_yY\subseteq T_xX+\mathbb C(x-y).\] Fix a general \(y\). Project \(X\) projectively away from the affine tangent plane \(y+T_yY\). The first inclusion says that this projection has zero differential wherever it is defined. In characteristic zero a rational map with zero differential has zero-dimensional image: on function fields, every function in its image has zero differential and is constant. Its irreducible image is consequently one point. Thus \(X\) lies in an affine \((k+1)\)-flat containing \(y+T_yY\). The case in which \(X\) already lies in the projection center gives the same conclusion. Interchanging \(X,Y\) bounds both affine spans by \(k+1\). If one span has dimension \(k+1\), the preceding flat for each general point of the other variety must equal that span, so it contains both varieties. If both spans have dimension \(k\), both varieties are flats, and the displayed tangent inclusions say directly that their affine join has dimension at most \(k+1\). ◻

Proof. First discard pairs for which a general connecting line is tangent at one endpoint. For a fixed target component \(Y\), choose a general smooth \(y\in Y\). Every source component tangent in this sense is contained in the fixed proper affine flat \(y+T_yY\). Its total assigned source mass is at most \(NB^{-c_0}\). After division by \(n_x\), this is \(B^{-c_0+o(1)}\). Average with target weights. The argument with source and target interchanged removes tangencies at the other endpoint. The choices of tangent points can be made simultaneously for the finite lists.

For \(k=l\), also discard pairs lying in a common \((k+1)\)-flat when one member is nonlinear. That nonlinear member spans the flat: its span cannot have dimension \(k\) unless it is itself a flat. Fixing that member therefore pins down a proper flat containing the other member’s assigned points. The same flat bound and weighted averaging show that these pairs have product weight \(o(1)\).

Lemma 23 now rules out a third moving root except when \(k=l\) and the source and first target are flats in a common \((k+1)\)-flat \(H\). Every further listed target with the same projection must also lie in \(H\): for a general \(x\in X\), its affine join with \(x\) equals \(H\). It is a flat, since the nonlinear exception was removed. Each such flat contributes exactly one moving root. Thus it remains to bound the number of listed flats in \(H\) for most weighted pairs.

The flat members are already distinct within each family. In the slice, identify a flat occurring in both families and add its two weights; this changes the bounds by at most a factor two. Fix a small constant \(\eta>0\). Delete flats of normalized weight less than \(e^{-1}\exp(-\eta w)\); their total weight is \(\exp(-\eta w+o(w))\). Slice all remaining flats by a common generic real projective subspace of codimension \(k\). Each becomes a point, and each generated \((k+1)\)-flat becomes a line. For these finite families, genericity ensures that distinct generated flats yield distinct lines and that a sliced point belongs to such a line exactly when the original flat is contained in the corresponding \(H\). The number of sliced points is at most \(e\exp(o(w))\).

Let \(t_H\) be their number on the line for \(H\). The lower weight cutoff and the physical flat bound give \[t_H\le e\exp(\eta w+o(w))B^{-c_0}.\] Choose \(\eta<c_0/4\). Since \(w\le\log B\), the last bound is at most \(eB^{-c_0/2}\) for large \(B\). By the rich-line form of Szemerédi–Trotter, Theorem 54, the number of these lines at dyadic occupancy \(t\) is \[O\!\left(\frac{e^2\exp(o(w))}{t^3} +\frac{e\exp(o(w))}{t}\right).\] Each has pair weight at most \(t^2e^{-2}\exp(o(w))\). Summing for \(t\ge\exp(\epsilon w)\) gives \[\exp(-\epsilon w+o(w))+B^{-c_0/2}\exp(o(w))=o(1).\] Delete these pairs and, for each source, delete the corresponding targets from its list. A pair which remains belongs to an \(H\) with at most \(\exp(\epsilon w)\) remaining target flats. Removing targets cannot increase this number. This proves the assertion.

For the announced applications, the degree and per-degree mass bounds of Proposition 27 give the common-window hypotheses. In subpower windows all testing degrees are \(B^{1+o(1)}\), so the number of \(l\)-flat carriers is \(B^{d-l+o(1)}\) and their normalized weights are at most \(B^{-(d-l)+o(1)}\). These are the second set of hypotheses. Exceptional intersections of distinct projected images are not part of this statement: at a generic direction, only components having the same image contribute. ◻

Proof. Write \(n_i\) for the assigned mass of \(Y_i\) and \(d_i=\deg Y_i\). Lemma 8 gives \[\sum_i n_i\le N,\qquad n_i\le C M d_i. \tag{\ref*{geo:terminal}.1}\] We treat lines first, then nonlinear curves.

Proof. Choose \(s_*\to\infty\) sufficiently slowly that \[s_*\le\tfrac14\log B,\qquad \delta\exp(6s_*)\longrightarrow0.\] Start with width \(\log B\), halve the width at each step, and stop on entry to the first width in \([s_*,2s_*]\). On entry to a step of width \(s\), the unresolved set is isolated by equations of degree \[ B\exp((1+o(1))s). \tag{33}\] This is true initially by Lemma 8, since \(M=o(B^2)\). Except at the initial step, the unresolved set also has no nonzero polynomial of degree at most \(B\exp(-s)\) capturing as many as \(N\exp(-\sqrt{2s})\) points.

At a step which is not the final one, first remove greedily batches of at least \(N\exp(-\sqrt{s})\) points on zero sets of nonzero polynomials of degree at most \(B\exp(-s/2)\). These batches are sent to the regularization below. From the remaining points remove greedily batches of at least \(N\exp(-\sqrt{s})\) points which occur as isolated zero-dimensional components of some system of degree at most \(B\exp(s/2)\). Only the union of these latter batches proceeds to the next window. The final remainder is sent to regularization if its size is at least \(N\exp(-s^{1/3})\), and otherwise is discarded.

There are at most \(\exp(\sqrt{s})\) batches of either kind. Isolated point interpolation in Lemma 10 gives equations of degree \(O(B\exp(s/2))\) for precisely each batch of the second kind. Indeed, from normalized separators \(q_i\) for its points \(p_i\) use \(1-\sum_iq_i\) and all coordinates of \((x-p_i)q_i\). Their common zero set is precisely the batch. A union of finitely many such zero sets is cut out by products, taking one equation from each system; the degrees add. Thus the union proceeding to the next step has isolation degree at most \[B\exp(s/2+O(\sqrt{s})).\] This is (33) at width \(s/2\). The first greedy operation supplies exactly the asserted lower boundary there. Consequently the final core has isolation degree at most \(B\exp(3s_*)\) for large indices, and hence \(EL^2=o(N)\).

The shrinking windows have produced the core and the batches that will become non-core pieces. It remains to give each retained batch one path of proper cuts and to ensure that a proper polynomial test between cut times captures a vanishing fraction of the piece. We now enforce this condition while discarding only \(o(N)\) points. For each width \(s\) use a grid in normalized time from \(-1\) to \(1+o(1)\), containing the upper isolation degree, with mesh tending to zero and \(G=O(1+\sqrt{\log s})\) grid points. Index the potential steps by increasing grid time, and, within one grid time, by decreasing positive dimension. There are \(K=O_d(G)\) indices. Choose thresholds \(\lambda_i=\exp(-a_i)\) with \[ a_K=s^{1/10},\qquad a_i=10a_{i+1}. \tag{34}\] The same indexed thresholds are used in all trees of this width.

In a node of dimension \(k\) and original size \(m\), repeatedly extract at least \(\lambda_i m\) remaining points where a polynomial of the current testing degree vanishes properly on every local component. An extracted set becomes a child, with that polynomial imposed, and is processed in dimension \(k-1\) at the same time. Here \(m\) stays the original mass of this node during all its extractions. The unextracted residual advances to the next time, unless its size is less than \(\sqrt{\lambda_i}m\), in which case it is discarded. At the last time, at most \(d\) generic combinations of the isolation equations finish the path: on any positive-dimensional local component through a counted point, some isolation equation is nonidentically zero, so a generic combination is a proper cut simultaneously on all such components.

Every path visits an index at most once. Extraction decreases dimension; taking a residual increases time. Sibling extractions are alternative branches, not repeated steps on one path. Therefore \[\sum_i a_i=s^{1/10+o(1)}=o(s^{1/3}),\qquad \sum_{r>i}a_r<\frac{a_i}{9}.\] If a residual survives at step \(i\), every later terminal piece inside it has size at least \(m\exp(-a_i/2-\sum_{r>i}a_r)\). A proper cut available when extraction stopped captured fewer than \(m\exp(-a_i)\) residual points. Its captured fraction in that terminal piece is consequently at most \[ \exp\left(-\frac{a_i}{2}+\sum_{r>i}a_r\right) \le \exp(-7a_i/18)\longrightarrow0. \tag{35}\] This proves the required no-capture assertion on any limiting open plateau. To allow a fixed multiplicative factor in the testing degree, choose a slightly later grid time which is still inside that plateau.

At a fixed index the nodes in all these trees are disjoint. Their discarded residual mass is at most \(\exp(-a_i/2)\) times their total mass. Thus the regularization discards at most \[ C_d N(1+\sqrt{\log s})\exp(-s^{1/10}/2) \tag{36}\] points at width \(s\). This estimate is summable over windows even if \(s_*\) grows arbitrarily slowly. To see this, reverse their order and bound by widths \(2^m u\), \(m\ge0\), with \(u\ge s_*\). Put \(t=u^{1/10}\) and \(c=(\log2)/10\). The inequalities \[2^{m/10}\ge1+cm,\qquad \sqrt{\log(2^m u)}\le\sqrt{\log u}+\sqrt{m\log2}\] bound the sum of the fractions in (36) by \[C_d e^{-t/2}\sum_{m\ge0} (1+\sqrt{\log u}+\sqrt{m\log2})e^{-ctm/2} =O_d((1+\sqrt{\log u})e^{-u^{1/10}/2})=o(1).\] The separately discarded small remainders have fractions \(\exp(-s^{1/3})\) and are summable by the same argument. No factor equal to the number of windows is used.

The discarded mass is therefore negligible across all windows. We check the remaining size and cut-time conditions. Each retained terminal piece has size at least its initial batch size times \(\exp(-o(s^{1/3}))\), and hence is \(N\exp(-o(s))\). The inherited lower boundary is negligible even relative to the smallest such piece, since \[\frac{N\exp(-\sqrt{2s})} {N\exp(-\sqrt{s}-o(s^{1/3}))} =\exp(-(\sqrt2-1)\sqrt{s}+o(s^{1/3}))\to0.\] There are exactly \(d\) cuts on a path. Their times have subsequential limits \(b_1\le\cdots\le b_d\) in \([-1,1]\). For a first-type batch, some cut occurs by limiting time \(-1/2\): otherwise its defining polynomial would capture the entire piece in an initial full-dimensional no-capture interval. For a remainder piece the last cut cannot have limiting time less than \(1/2\). Otherwise its path would isolate more than \(N\exp(-\sqrt{s})\) points by equations of degree at most \(B\exp(s/2)\), contradicting termination of the second greedy operation. This proves the imbalance and all stated properties. The estimates were uniform over nodes and pieces, which also proves the final sequence quantifier. ◻

Proof. Apply Bézout successively, retaining the proper intersections through counted points. It gives the carrier degree bound and, after applying the later cuts to a specified carrier subcollection, (39). Applied to the entire path it gives \(n\le(\prod_iD_i)\exp(o(s))\). Since \(n=N\exp(-o(s))\) and \(N=B^d\), this proves \(\sum_i b_i\ge0\).

We prove (40) by propagation through the finitely many plateaux; the assertion is always for every fixed positive density subset, including subsets chosen during the proof. Before the first cut, no nonzero polynomial of the testing degree can vanish on \(S\). Its envelope is therefore the ambient space, whose rank is comparable to \(T_s(b)^d\).

On a plateau of dimension \(l\), let \(V\) be the testing-degree envelope of \(S\). At all but \(o(n)\) points of \(S\), at least one local carrier is contained in \(V\). Indeed, if no carrier through a point is contained, a generic linear combination of the envelope equations is proper on all those carriers and vanishes at the point. There are finitely many components and sample points, so the combination can be chosen simultaneously; no-capture then applies.

For this envelope, choose at each such point one contained local carrier, and choose one irreducible component of \(V\) containing each distinct carrier so chosen. Let the points inherit the component chosen for their carrier. Group all irreducible components of \(V\) by dimension, counting the points by these choices; components receiving no points remain in the degree of their group. Some dimension \(h\ge l\) group, of total degree \(a\), covers a fixed positive fraction of \(S\). Writing \(S'\) for that fraction and \(T=T_s(b)\), the component Hilbert bounds give \[H_S(T)=H_V(T)\gtrsim aT^h,\qquad H_{S'}(T')\lesssim a(T'+1)^h\quad(T'<T).\] The lower estimate uses that the group consists of components of the envelope at its defining degree \(T\); the upper estimate uses only that \(S'\) lies on that group. Eliminating \(a\) gives \[H_S(T)\gtrsim H_{S'}(T') \left(\frac{T}{T'+1}\right)^h.\] Take \(T'=T_s(b')\) at a time \(b'\) just before the cut starting the current plateau, and apply (40) there to \(S'\). The displayed inequality increases that lower bound by a factor at least \((T/T_s(b'))^h\), to exponential accuracy. Since \(h\ge l\), this is at least the growth required by \(\sigma\) on the plateau. Take \(b'\) arbitrarily close to the boundary; the logarithmic endpoint error can be made smaller than any prescribed fixed multiple of \(s\). This proves the next plateau, including after a group of simultaneous limiting cuts.

There is one possible boundary convention: if the first cut has testing degree only \(\exp(o(s))\), there may be no preceding testing time of degree at least one. Rank one at that boundary is already the required bound to exponential accuracy, even if several first cuts have that same time. Use it as the start of the induction.

The same envelope argument advances across the final cut, with required growth exponent zero. Since rank is at most \(N\), it gives \(\prod_iD_i\le N\exp(o(s))\), proving the reverse inequality \(\sum_i b_i\le0\). The profile is nonzero by the imbalance in Proposition 26. Its monotonicity and zero sum give \(b_1<0<b_d\). If \(b_1+b_2\ge0\), then \(b_2>0\) and the \(d-2\) later entries are at least \(b_2\), contradicting the zero sum because \(d>2\). This proves (38). ◻

Proof. Every carrier used here is an irreducible component of its fixed path locus. Indeed, at every point of the piece on that carrier, the successive proper cuts give the current local dimension. A larger irreducible component containing the carrier would pass through that point and contradict this local dimension. The path locus may have larger components elsewhere; they remain in its equations.

Let the plateau start at time \(\beta<b\), and choose \(b<b'<b''\) in the same plateau. The carrier components are therefore isolated components of their cut system, whose equations have degree at most \(B\exp((\beta+o(1))s)\). Lemma 16, used at degree \(T_s(b')\), pools proper component cuts of degree \(O(T_s(b))\) into one polynomial. It is nonidentically zero except on components of total degree at most \[e_j\exp(-(b'-b)s+o(s)).\] By (39) and \(\prod_iD_i=N\), all points of the piece touching those exceptional components have mass \(o(n)\). At every remaining counted point the pooled polynomial is proper on every local carrier. No-capture at \(b'\) proves the assertion.

Apply Lemma 15 at degree \(T_s(b')\). Since the cut-system equations have degree at most \(B\exp((\beta+o(1))s)\), the components on which the common equations fail to have generic full rank have total degree at most \[e_j\exp(-(b'-\beta)s+o(s)).\] The mass budget removes all piece points touching those components at a cost of \(o(n)\). On the other components a generic linear combination of maximal minors is nonidentically zero. Its zero set contains every rank failure, and its degree is \(O(T_s(b'))\). The preceding individual-cut conclusion applies to this linear combination using the remaining interior margin before \(b''\), and removes its zeros at another \(o(n)\) points. At a surviving point the implicit function theorem gives a smooth locus of the correct dimension; any two carrier components through that point would have the same germ and hence would coincide. The equations vanish on the whole family, including components whose generic regularity failed.

For pooled pieces, use the union-of-systems versions of Lemmas 16 and 15. Each carrier remains a component of its own path system. On a generic transverse fiber, Lemma 12 interpolates the selected isolated lists, including first jets, with an effective bound equal to the sum of the individual system bounds: separators for the other lists are multiplied, using their exact-isolation equations for a test point outside a list. If these bounds are \(L_\nu\), then for \(\exp(o(s))\) pieces with the common limiting profile, \[L_*=\sum_\nu L_\nu\le B\exp((\beta+o(1))s).\] The strict interior gaps above absorb this change in the effective bound and the \(\exp(o(s))\) changes in the degree budgets.

Mass accounting for such a union is performed separately in each original piece. Its own later cuts bound all of its points touching exceptional components of its own current system. We do not apply one piece’s later cuts to points assigned to another piece. Rank failures on every good assigned component are then discarded by that piece’s no-capture assertion. The equations still vanish on the whole union, so full rank at a surviving point also excludes intersections with a carrier from another piece. The sequence quantifier makes the discarded fractions uniform over these pieces; summing their masses creates no loss equal to the number of pieces.

Finally, the real points of a nonreal complex affine flat lie in a proper complex affine linear section of that flat. Choose one such degree-one proper cut for each nonreal flat, and no cut on the other carriers. The individual-cut assertion just proved makes their total assigned mass negligible. ◻

Proof. For an actual source point \(p\), apply the individual-cut conclusion of Proposition 28 on every target carrier \(Y\) on which the specialization is nonidentically zero. For the other case, fix \(Y\). If the polynomial is not identically zero on \(X\times Y\), choose a generic \(y\in Y\) for which its specialization is nonidentically zero on each such \(X\). The same individual-cut conclusion on the source bounds points \(p\) where the polynomial vanishes identically on \(Y\). Average this bound with the target masses of the \(Y\)’s. The fixed assigned sets are disjoint within each factor, so these weighted sums are unambiguous even when carriers meet. For pooled pieces, retain their original mass accounting. All bounds are uniform over the polynomial and the pieces by their sequence formulation.

For a system of equations, pull back each equation and clear its chart denominator separately. The resulting numerator polynomials vanish at the actual system points and have the common degree bound. On every component product carrying a pair on the chart whose image is not contained in the system, at least one numerator is nonidentically zero. A generic linear combination of the numerators is therefore nonidentically zero on every such product simultaneously. It has the same degree bound: a finite basis of their span suffices, regardless of the number of system equations. Apply the first assertion to this combination.

For rank tests use their cleared pullback numerators. When a rank condition is expressed by minors whose cleared numerators obey the common degree bound, a generic linear combination of those numerators is nonidentically zero on every relevant component product where the rank condition holds generically, and its zero set contains every rank failure. A bounded number of the resulting tests has a union of failure sets of size \(o(n_xn_y)\). A slightly later interior time absorbs the uniform constant in the degree bound. This proves the last assertion without assuming that sample points are generic on their carriers. ◻

Proof. Assign each point of every non-core piece to one of its carrier curves after the \((d-1)\)st cut. Omit assignments to nonreal lines. Such a line contains at most one real point, so the omitted fraction in a piece is at most \[\frac{e_{d-1}\exp(o(s))}{n} =D_d^{-1}\exp(o(s))=o(1),\] using (38). This is uniform over pieces. Every point is assigned at most once; pool repeated curves. Apply Proposition 25 once to this global assignment. It removes only \(o(N^2)\) ordered incidences, and hence leaves the required almost-complete unoriented coverage. Assignments retain their original meaning if pieces are later pooled.

If the core has positive relative size, the cone over a directional family has degree \(O(E)\) and dimension at most two. The core’s isolation equations, restricted to its components, and Bézout bound its intersections with the core by \(O(EL^2)\) per center. This is \(o(N)\), so neither orientation can cover a positive fraction of the core pairs. This resolves the first alternative. Otherwise discard the core.

Here are details of the remaining selection, including the order of the limits. First fix a precision \(1/h\), and partition the compact space of normalized drop lists and values of \(s/\log B\in[0,1]\) into finitely many bins of that diameter. Also bin the dyadic index modulo an integer tending to infinity with \(h\). Two independent random retained points fall in one common bin with probability bounded below by a positive constant depending only on \(h\). Thus the relative uncovered probability conditional on this agreement tends to zero for fixed \(h\). Select a pair of pieces with small relative uncovered probability, and pass to an orientation containing at least half the covered pairs. Now let \(h\) tend to infinity sufficiently slowly with the sample index. This standard diagonal choice retains conditional coverage while making the profiles and normalized widths converge to the same limits.

The selected dyadic indices are either equal on a subsequence, or their difference tends to infinity, by their congruence condition. In the latter case the width ratio tends to infinity. Their normalized widths have the same limit, so that limit must be zero. This proves the two window alternatives.

There is a useful refinement at the outermost window. If its total mass tends to zero, discard it before the selection. If its mass is bounded below on a subsequence, work only there. For each fixed precision choose a largest profile bin and use its pool as both factors. It has a fixed positive mass at that precision, and its relative uncovered fraction tends to zero. In the diagonal choice increase precision slowly enough that its mass is at least \(N\sqrt\delta\) and \(N\exp(-o(\log B))\). This is possible because \(\delta\to0\). Each constituent piece has size \(N\exp(-o(s))\) uniformly, so there are only \(\exp(o(s))\) of them. Their profiles differ by \(o(1)\), and the conclusions for pooled families follow from Proposition 28 and Lemma 29.

Finally, Szemerédi–Trotter, with a generic planar projection, makes the fraction of ordered pairs on lines of occupancy above a threshold \(H\) arbitrarily small as \(H\to\infty\): summing its dyadic rich-line estimate gives \(O(N^2/H+N h_{\max})\), and the proper-flat bound gives \(h_{\max}=o(N)\). At each fixed selection precision first choose \(H\) large enough for its conditioning probability. Increase the precision sufficiently slowly that \(H\le B^{o(1)}\) and all the preceding conditional errors still vanish. This is compatible with the outer-pool mass condition. All constants and precisions are fixed before taking each asymptotic limit; the final diagonal sequence merely combines these finitely specified requirements. ◻

Proof. Choose fixed times \(\tau_{\rm eq}<\tau_{\rm test}<\tau\) between the start of the plateau and \(\tau\). Use the earlier part of this interval for the equation choices in Proposition 28, leaving the strict gap up to \(\tau_{\rm test}\) for its minor and proper-cut tests at the sample points. Apply its union version to the two physical carrier families. It removes the negligible point sets excluded by its regularity conclusion and supplies common equations for the whole union. At each remaining point there is now a unique smooth carrier. Denote by \(g_{\rm ph}\) the common combination of maximal minors whose zeros were removed in this step. Discard every such point whose carrier is a nonreal flat, using the degree-one proper sections in the last paragraph of that proposition. This costs another \(o(n_x)+o(n_y)\) points. The source and target assignments are now unambiguous. Each assigned carrier belongs to the point’s own original piece path: that path already supplies a carrier through the point, and uniqueness in the full union identifies it with the assigned member. For two copies of the same pool take the union of the two negligible point deletions in both roles; the resulting assignment is the same. Extend the retained assignments by assigning each deleted point to a containing carrier of its original piece before any later product tests, using the same extension in both roles for a common pool. Product testing on the original sets then retains the present assignments under restriction, and its \(o(n_xn_y)\) pair loss is also \(o(m_xm_y)\). Pool equal retained carriers within each of the two lists and omit members with zero mass. All these choices have been made inside the interval ending at \(\tau\), which gives the stated degree bound.

The pooled budgets, with each original piece still supplying the bound for its own points, give \[ \begin{gathered} \max\left\{\sum_X\deg X,\sum_Y\deg Y\right\}\le e_j\exp(o(s)),\\ \frac{n_X}{m_x}\le\frac{\deg X}{e_j}\exp(o(s)),\qquad \frac{n_Y}{m_y}\le\frac{\deg Y}{e_j}\exp(o(s)). \end{gathered} \tag{44}\] Indeed, the mass of a carrier in one piece is at most \(\deg(X)(N/e_j)\exp(o(s))\), and pooling \(\exp(o(s))\) pieces changes this by only \(\exp(o(s))\). The remaining point counts are still \(N\exp(-o(s))\). A flat has degree one, so its number and its normalized weight satisfy the hypotheses of Lemma 24 with \(w=s\) and \(e=e_j\). The flat members are real by the preceding deletion.

Apply that lemma now with the fixed \(\varepsilon\). Let \(\mathcal Y_X\) be its target list for \(X\), and set \(\mathcal R=\{(X,Y):Y\in\mathcal Y_X\}\). The removed carrier pairs have normalized product weight \(o(1)\), hence account for \(o(m_xm_y)\) actual pairs. This deletion includes pairs generically tangent at either endpoint. Thus the target tangent projects injectively at a general pair, and the first coordinate of \(\psi\) records all source tangent directions; the product image has dimension \(2k\). The preimages of a general \(2k\)-fold linear section of that image are isolated away from the base locus of the bounded-degree map on \(X\times Y\). The isolated-intersection form of Bézout therefore gives \(\deg Z_{X,Y}\le C_d\deg X\deg Y\), and (42) follows from (44).

We prove the uniform subfamily estimate while keeping these same lists. Images belonging to different source carriers are distinct: the closure of the first projection of \(Z_{X,Y}\) is \(X\). Fix one source \(X\) and one of its distinct images \(Z\). To identify its source fibers, take its closure \(\widehat Z\subset X\times\mathbb P^{d-1}\) before restricting the direction coordinate to our affine chart. Localizing the ideal of the rational image at \(\mathbb C(X)\) identifies the generic fiber with the closure of the projection of \(Y\) from the generic source point. This fiber is geometrically integral, since it is the image closure of the geometrically integral variety \(Y_{\mathbb C(X)}\). After restricting \(X\) to a dense open, the projective family is flat with geometrically integral fibers of a constant degree \(d_Z\); this uses generic flatness, geometric integrality after shrinking, and constancy of the Hilbert polynomial in a flat projective family (The Stacks Project Authors 2026, Tags 052A, 0559, 0C0E, and 0B9T). We may also require that projection of \(Y\) from each \(x\) in this open set has dimension \(k\), by generic target transversality. Its closed image is then a \(k\)-dimensional irreducible subvariety of the \(k\)-dimensional integral fiber, so it equals the whole fiber. Make these restrictions simultaneously for the finite image and target lists. Thus all targets with \(Z_{X,Y}=Z\) have the same complete projected image at a common general \(x\in X\). Write \(t_x(Y)\) for their generic map degrees. The fixed list bound says \[\sum_{Y:\,(X,Y)\in\mathcal R,\ Z_{X,Y}=Z}t_x(Y) \le \exp(\varepsilon s).\] For a general \(x\), the point-projection formula gives \(\deg Y-\operatorname{mult}_xY=t_x(Y)d_Z\). The multiplicity is zero unless \(X=Y\). In that case \(x\) is smooth and the multiplicity is one; a coincident flat pair was deleted as tangential. Consequently \(\deg Y\le2t_x(Y)d_Z\) for every target in this sum.

Choose \(k\) general affine hyperplanes in the source coordinates. They meet \(X\) in \(\deg X\) points of the preceding open set. Then choose \(k\) general hyperplanes in the direction coordinates; on each complete fiber they give \(d_Z\) points, with multiplicity. The choices can be made so all these points lie in the selected label chart and avoid intersections between distinct images. The resulting mixed section of \(Z\) therefore has \(\deg X\,d_Z\) points. The \(2k\) blockwise hyperplanes are also affine linear equations in the full label space. Their homogenizations restrict to linear equations on the ordinary projective closure of \(Z\), where all the counted affine points are isolated intersections. The isolated-intersection form of Bézout bounds their total multiplicity by the ordinary degree, even if the same equations have additional components at infinity. Applying this to a reduced image subfamily and summing the preceding map-degree bound gives \[\sum_{\substack{(X,Y)\in\mathcal R\\Z_{X,Y}\in\mathcal J}} \deg X\deg Y \le C_d\exp(\varepsilon s) \deg\bigl(\bigcup_{Z\in\mathcal J}Z\bigr).\] Multiplying the two weight bounds in (44) proves (43). No further generic-depth deletion is made when \(\mathcal J\) is chosen. ◻

Proof. On the initial full-dimensional interval, Lemma 29, or its two successive full-dimensional no-capture arguments, forces the testing-degree envelope of a dense label graph to be the whole label space. Its rank is comparable to \(T^{2d-1}\). Since \(D_1D_2\ge T^2\) there, this is at least the right side of (45).

On a positive-dimensional plateau after \(j\ge1\) cuts put \(k=d-j\). Choose a physical regularity time between the start of this plateau and the current testing time \(b\). When \(k<d-1\), apply Lemma 32 with any fixed \(\varepsilon>0\). When \(k=d-1\), apply Proposition 28 to the union of the two physical families, discard its rank-failure and nonreal-flat points, and assign each survivor to its unique carrier from its original path. Use the same supported extension to the original point sets before product testing, and pool equal retained carriers with their assigned masses. The fixed tangent-flat argument and the proper-flat cap then discard carrier pairs generically tangent at either endpoint. In both cases the point and carrier-pair losses remove only a negligible fraction of the graph, and every remaining complete product image \(\overline{\psi(X\times Y)}\) has dimension \(2k\). Product testing forces these complete images into the testing-degree envelope at almost all remaining graph pairs.

For this fixed envelope, choose one irreducible envelope component containing each distinct complete image occurring at those surviving pairs. Let the pairs with that image inherit its choice. Group all envelope components by dimension and count each surviving pair in the group of its chosen component; components receiving no pairs remain in the degree of their group. Every chosen component has dimension at least \(2k\), so some dimension \(h\ge2k\) group carries a positive fraction of the graph. Write \(a\) for its degree and \(\Gamma'\) for the graph portion it carries.

If \(h\ge2k+1\), the Hilbert bounds give \[H(T)\gtrsim aT^h,\qquad H_{\psi(\Gamma')}(T')\lesssim a(T'+1)^h.\] Borrow (45) from a time just before the cut starting this plateau. The required right side grows with exponent at most \(2k+1\) within the plateau, so growth with exponent \(h\) suffices. Endpoint approximation costs an arbitrarily small fixed multiple of \(s\) in the logarithm.

It remains to bound \(a\) when \(h=2k\). Each image assigned to this group is an irreducible \(2k\)-fold contained in its chosen irreducible \(2k\)-dimensional envelope component, and therefore equals that component. If \(k=d-1\), necessarily \(j=1\), and the projection of a generic target hypersurface from a generic source point fills direction space. For each source carrier \(X\) the product image is thus \(X\times\mathbb P^{d-1}\) in these label coordinates. A positive mass of pairs requires a positive source mass. The carrier mass budget therefore gives \[a\gtrsim e_1\exp(-o(s))=D_1\exp(-o(s)).\] The resulting rank \(D_1T^{2d-2}\exp(-o(s))\) is sufficient: on this plateau \(T\le D_2\) to exponential accuracy and \[D_1^2T^{2d-2}\min(1,T/(D_1D_2)) \le D_1T^{2d-2}.\]

Now suppose \(k<d-1\). The images assigned to this group represent a positive amount of assigned product mass, since they carry \(\Gamma'\). Apply (43) to this subfamily. Its degree is at most \(a\), so \[a\gtrsim e_j^2\exp(-\varepsilon s-o(s)).\] Since \(\varepsilon\) is arbitrary, \(a\ge e_j^2\exp(-o(s))\). The Hilbert lower bound now gives \(H(T)\ge e_j^2T^{2k}\exp(-o(s))\), which is stronger than (45) on this plateau.

These alternatives propagate through the finitely many distinct cut times. If the first boundary has degree \(\exp(o(s))\) and there is no earlier testing time of degree at least one, use rank one there; the required boundary value is itself \(\exp(o(s))\). This also handles several simultaneous first cuts. Every borrowed graph portion has fixed positive density after passage to a subsequence, so the induction applies to it as asserted. ◻

Proof. Put \(T_i=T_s(t_i)\) and form \[V=\operatorname{Env}_{T_0}(U).\] For each distinct image \(Z\subset U\), choose once an irreducible component \(W(Z)\) of \(V\) containing the whole image. Every assigned pair represented by \(Z\) inherits that choice. Since \(\dim Z=2k\), a chosen component has dimension at least \(2k\); equality implies \(W(Z)=Z\).

We first show that the images assigned to larger components represent only \(o(n^2)\) product mass. Otherwise, after taking a subsequence, some dimension \(h\ge2k+1\) group represents a fixed positive normalized mass. Form an auxiliary graph \(\Lambda_h\) from all actual ordered pairs in those assigned products, omitting the diagonal where \(\psi\) is undefined. Disjointness of the assignments makes its size equal to that product mass before the omission, and the diagonal has at most \(n\) pairs. Thus \(\Lambda_h\) is dense in \(S\times S\). This graph is independent of the covered graph \(\Gamma\); the arbitrary-graph scope of Lemma 33 applies to it. Every label of \(\Lambda_h\) lies in the chosen dimension-\(h\) component containing its complete product image.

Write \(a_h\) for the total degree of the dimension-\(h\) component group of \(V\), including components receiving no images. Its upper Hilbert bound at an earlier degree \(T'\) controls \(H_{\psi(\Lambda_h)}(T')\), while its lower envelope bound gives \(H_V(T_0)\gtrsim a_hT_0^h\). Borrow the common-window label rank at a time just before \(\beta\), with the fixed endpoint tolerance chosen above. If \(\beta=-1\) and no earlier testing degree is at least one, use the rank-one boundary convention in Lemma 33; then all \(j\) earlier cut times equal \(-1\) and the required boundary rank is one to exponential accuracy. On the final plateau \(\sigma(b)=e_jT_s(b)^k\). Propagating with \(h\ge2k+1\) therefore gives \[H_V(T_0)\ge e_j^2T_0^{2k} \min\left\{\frac{T_0}{T_s(\beta)}, \frac{T_0}{D_1D_2}\right\} \exp(-\xi s-o(s)),\] where the fixed \(\xi>0\) can be smaller than both gaps specified after (50). The two factors have normalized logarithms \(t_0-\beta\) and \(t_0-(1+b_1+b_2)\), respectively, so the lower bound has a strict positive margin. But envelope equality and the upper Hilbert bound on the pure \(2k\)-dimensional family \(U\) give \[H_V(T_0)=H_U(T_0)\le C e_j^2T_0^{2k}\exp(o(s)).\] This contradiction proves the claimed product-mass deletion.

Let \(\mathcal I\) be the distinct images left after that deletion. Each is an actual irreducible \(2k\)-dimensional component of the single locus \(V\), defined by equations of degree at most \(T_0\); larger components elsewhere in \(V\) are kept in those equations. Also \(\deg\mathcal I\le\deg U\le e_j^2\exp(o(s))\). Apply Lemma 15 to this pure component family at degree \(T_1\). Its equations vanish on the whole family \(\mathcal I\). The exceptional image subfamily \(\mathcal E\) has total degree at most \[C\deg(\mathcal I)\frac{T_0}{T_1} \le e_j^2\exp(-(t_1-t_0)s+o(s)).\] Use (43) for this particular subfamily of the already fixed \(U\). Its represented normalized pair mass is at most \[\exp(-(t_1-t_0-\varepsilon)s+o(s))=o(1).\] Here the same \(\varepsilon\)-lists chosen before \(U\) are used; no new generic-depth deletion depends on \(\mathcal E\). Let \(\mathcal R_{\rm sm}\) consist of the carrier pairs represented by \(\mathcal I\setminus\mathcal E\). The two image deletions have omitted only \(o(n^2)\) assigned product mass.

The regularity lemma supplies a common polynomial \(g\) of degree \(O(T_1)\), a linear combination of maximal Jacobian minors, which is nonzero on each image in \(\mathcal I\setminus\mathcal E\). The single rational pullback \(g\circ\psi\) has a cleared numerator of degree \(O(T_1)\) on the chosen charts, nonidentically zero on every carrier pair in \(\mathcal R_{\rm sm}\). Product testing, with its internal times in \((t_1,t_2)\), removes its actual zeros at a cost of \(o(n^2)\) pairs. At every surviving label the common equations are locally a smooth \(2k\)-fold. They still vanish on all of \(\mathcal I\), including its exceptional components.

It remains to test the two tangent injectivities at actual pairs. Write \(f_a\) for the common physical equations chosen before \(\tau\), and \(v=q-p\). At a point \(r\in S\), their Jacobian has kernel \(T_r\), the tangent space of its unique carrier. Thus \(v\notin T_r\) exactly when at least one polynomial \(\nabla f_a(r)\mathbin{\cdot}v\) is nonzero. These tests have degree \(O(T_s(\tau))\) in \((p,q)\). On every retained carrier pair the source collection and the target collection each contain a generically nonzero test, by the fixed generic-depth deletion. For each endpoint, choose a generic constant linear combination nonzero on every retained product. Product testing removes its actual zeros at cost \(o(n^2)\). The target injectivity, together with the source coordinate of \(\psi\), gives differential rank \(2k\); the source injectivity is retained separately for the secant step.

If \(k\ge2\), every remaining physical carrier spans dimension at least \(k+2\) by the preceding carrier deletion. Equality of the two projected tangent spaces at a general retained pair would put both carriers in a common \((k+1)\)-flat by Lemma 23, which is impossible. This inequality also has a bounded-degree test using the same physical equations. Put \(c'=d-k\) and write \(a_i(r)=\nabla f_i(r)\). On the full-rank transverse locus the row vectors \[(a_i(r)\mathbin{\cdot}v)a_j(r) -(a_j(r)\mathbin{\cdot}v)a_i(r)\] span the \((c'-1)\)-dimensional annihilator of \(T_r+\mathbb Cv\). Indeed, they lie in that annihilator, and choosing one row with nonzero value on \(v\) expresses every row combination annihilating \(v\) in their span. The two projected tangent spaces are unequal exactly when the combined row spans for \(r=p,q\) have rank at least \(c'\). Their \(c'\times c'\) minors are polynomials of degree \(O(T_s(\tau))\). Choose a generic combination nonzero on every retained product, and apply product testing once more to remove its actual zeros. All these tests fit in the interval reserved below \(t_2\). Since \(s=\log B\) and \(t_2<1\), their degrees and the common label-equation degrees are at most \(B^{2-c}\) for some fixed \(c>0\). This proves the lemma. ◻

Proof. We first explain the degree and mass bookkeeping common to all three cases, then prove their depth bounds.

Proof. On the initial interval target no-capture, applied to the pullback of a direction polynomial, gives rank comparable to \(T^{d-1}\). Its ratio to \(\sigma(b)T/A\) is \[\frac{A}{T^2}=\exp(-2bs).\] This has a strict positive margin near the first cut, since \(b_1<0\).

On a plateau of dimension \(l\), consider the testing-degree envelope of the directions to the current row subset. For each fixed assigned target carrier whose projection is not contained in this envelope, a generic linear combination of its pulled-back equations is a proper cut on that carrier. The combination can be chosen simultaneously for the finite carrier list. Use a slightly later interior time for the fixed pullback and pooling costs. Every counted target on such a carrier lies in the cut. The individual-cut conclusion of Proposition 28 therefore removes only \(o(|P_y|)\) such targets, uniformly in the center and the envelope. Hence the complete projection of the assigned carrier is contained in the envelope at almost all remaining counted targets.

For this fixed envelope, choose one irreducible component containing each distinct projected carrier occurring at those targets, and let the targets inherit the choice for their projected carrier. Group all envelope components by dimension, counting targets by these choices; components receiving no targets remain in the degree of their group. Some group of dimension \(h\ge l\) carries a fixed positive fraction of the target mass.

If \(h>l\), that group borrows the preceding bound with growth exponent \(h\ge l+1\). This preserves the required growth exponent of \(\sigma T\) on the plateau. Approximate the preceding endpoint closely enough that the lost logarithmic tolerance is a small fraction of the existing \(c\).

If \(h=l\), each projected carrier assigned to this group equals its chosen envelope component, since both are irreducible of dimension \(l\). The projected degree lower bound (61) and the envelope Hilbert bound give \[H(T)\gtrsim T^l e_j/R\,\exp(-o(s)).\] Relative to \(\sigma T/A=e_jT^{l+1}/A\), this supplies the ratio \(A/(RT)\), to exponential accuracy. It suffices to have a strict negative normalized logarithm for \(R D_{j+1}/A\), the worst point at the next endpoint. The three choices give, respectively, \[ b_1+b_{j+1},\qquad -\sum_{i>j+1}b_i,\qquad b_{j+1}-2. \tag{66}\] The first is strictly negative on the hypersurface plateau, by \(b_1+b_2<0\), and whenever \(b_{j+1}\le0\). For wider centers all remaining cases have \(b_{j+1}>0\) and \(l<d-1\), so the third choice applies and has margin at least one. For narrower centers and \(b_{j+1}>0\), the second choice has a strict margin when \(l\ge2\), since its sum includes a positive later entry. Only \(l=1\) gives zero rather than a strict margin.

There are at most \(d\) distinct cut times. Choose all endpoint approximations and fixed logarithmic tolerances smaller than a sufficiently small fraction of the finitely many positive margins just displayed. Decrease \(c\) when necessary. This completes the propagation. The degree bounds and prior rank statements quantify over every dense restriction, so choosing a different dimension group or dense subset for different centers does not change the argument. ◻

Proof of Theorem 31. The profile-pair selection exhausts the possibilities. A positive core is impossible by its isolation degree. The common-window rank bound contradicts coverage except at the outermost endpoint, where the lift argument gives the contradiction. For separated windows, strict row rank propagation contradicts coverage except on the final narrow-center curve plateau, which the preceding argument resolves by common regularity and terminal deletion. Thus the proposed almost-complete sparse-cone coverage cannot exist. ◻

Proof. A generic real planar projection preserves the distinct points and all noncollinear triples of a finite affine configuration. It therefore suffices to work in the plane. Theorem 54 gives \[ \#\{L:|L\cap X|\ge u\} \le C_0\left(\frac{n^2}{u^3}+\frac nu\right), \qquad u\ge2. \tag{72}\] For a large fixed integer \(u_0\), sum over dyadic occupancies \(u=2^j u_0\), up to the maximum occupancy \(\delta_0 n\). The number of ordered pairs of distinct points on lines in these buckets is at most \[C_1\sum_{u} \left(\frac{n^2}{u}+nu\right) \le C_2\left(\frac{n^2}{u_0}+\delta_0 n^2\right).\] Choose \(u_0\) large and then \(\delta_0\) small, both absolute, so this is less than one half of all ordered pairs. A positive fraction of the pairs consequently lie on lines containing fewer than \(u_0\) points. Each such line accounts for fewer than \(u_0^2\) ordered pairs, so there are at least \(c_0n^2\) spanned lines for an absolute \(c_0>0\). The finitely many small values of \(n\) cause no difficulty: when \(\delta_0n<2\), the hypothesis is impossible for \(n\ge2\).

The sum, over the points, of the numbers of incident spanned lines is at least \(2c_0n^2\), whereas each point is incident to at most \(n-1\) such lines. It follows that at least \(c_0n\) points are incident to at least \(c_0n\) spanned lines, after decreasing \(c_0\) if necessary. Take \(\gamma_0\le c_0\). A generic projective chart puts a finite projective configuration in the affine plane without changing its incidences. ◻

Proof. We may assume \(H\le m\). First consider a plane for which a richest line contains at least half of \(X\cap\Pi\). That line contains at least \(H\) points, and there are at least \(H\) points of \(X\cap\Pi\) off the line. By (72), applied after a generic projection, the number of \(H\)-rich lines of \(X\) is \[O\left(\frac{m^2}{H^3}+\frac mH\right)=O(m/H).\] Here \(H\ge\sqrt m\), after increasing \(C_2\). For any fixed line, distinct planes containing it have disjoint sets of points off that line. There are therefore at most \(m/H\) planes of the present kind through each such rich line. This part contributes \(O(m^2/H^2)\).

For the remaining planes no line contains half their points. Bucket their occupancies in \([K,2K)\), where \(K=2^jH\), and let \(f\) denote the number in one bucket. Any fixed spatial line belongs to at most \(2m/K\) of these planes, since each such plane has at least \(K/2\) points off the line and those off-line sets are disjoint.

For \(x\in X\), let \(r_x\) be the number of bucket planes containing \(x\), and put \(I=\sum_xr_x\). Thus \(Kf\le I<2Kf\). The sum \(J\), over ordered distinct pairs of bucket planes, of their common sample-point counts satisfies \[ J=\sum_xr_x(r_x-1) \ge \frac{K^2f^2}{m}-2Kf. \tag{73}\]

To bound \(J\) from above, fix one bucket plane \(\Pi\). Separate its intersection lines with the other planes according as their occupancy is less than \(C_3\sqrt K\) or at least \(C_3\sqrt K\), where \(C_3\ge2\) is a fixed constant. The first class contributes at most \(C_3\sqrt K\,f\). Within \(\Pi\), the number \(a\) of distinct lines in the second class obeys, again by (72), \[a\le C_4\left(C_3^{-3}+C_3^{-1}\right)\sqrt K =O(\sqrt K).\] Their total summed occupancy is \(O(K)\). Indeed, if a point is on \(s\ge1\) of these lines, then \(s\le1+\binom{s}{2}\); and two distinct lines have at most one common point. Hence the total occupancy is at most \[|X\cap\Pi|+\binom a2=O(K).\] Each such intersection line belongs to at most \(2m/K\) bucket planes. Their contribution for the fixed \(\Pi\) is therefore \(O(m)\). Summing over \(\Pi\) gives \[ J\le C_5 f(m+\sqrt K\,f). \tag{74}\] For sufficiently large \(C_2\), the assumption \(K\ge C_2m^{2/3}\) ensures \(K^2/m\ge2C_5\sqrt K\). Combining (73) and (74), and using \(K\le m\), now gives \(f=O(m^2/K^2)\). The sum over dyadic \(K\ge H\) is geometric, which proves the lemma. ◻

Proof of Lemma 41. We first prove the general-type case by a polynomial argument. Every non-line-type match set contains a noncollinear triple. The images of an ordered noncollinear triple determine a spatial isometry up to two choices, and determine it uniquely if its orientation is specified. Thus all motion lists considered in the proof of Lemma 41 are finite.

Separate the two orientations. Composing the second cloud with a fixed reflection when necessary reduces either list to reversing motions \(g(x)=Sx+h\), with \(\det S=-1\). Next rotate the second cloud generically, applying the same rotation to every motion in the finite list. We may and do arrange that:

1. \(\ker(I+S)\) is a line for each counted motion;

2. projection perpendicular to that line preserves noncollinearity for every noncollinear triple in its match set.

To justify these simultaneous choices, for any one fixed \(S\), left multiplication by a rotation ranges over the entire negative component of \(O(3)\). Outside the proper locus \(S=-I\), its \(-1\) eigenspace is a line with arbitrary generic direction. For a fixed noncollinear triple, the projection fails precisely when that line direction belongs to the two-dimensional direction space of the triple’s plane. This is a proper algebraic condition. There are finitely many motions and triples, so a rotation avoiding all these proper conditions exists. Counts, distances, and line/plane occupancies are preserved by this change of coordinates.

Proof. Put \(\mu=N/r\ge1\). Choose a containing half-open cube and subdivide a dyadic cube whenever it contains more than \(\mu\) sample points. Call these subdivided cubes heavy. The procedure terminates: distinct points in a finite cloud have positive minimum separation, so sufficiently small cubes contain at most one point. If the root cube is light, it alone is the required partition.

Otherwise the heavy cubes form a finite rooted tree, with edges to heavy children. Let \(L\) be the number of its leaves and \(B\) the number of its vertices with at least two heavy children. The heavy leaves are disjoint and each contains more than \(\mu\) points, so \(L<r\). The elementary rooted-tree identity gives \(B\le L-1\). Vertices with exactly one heavy child break into maximal chains. Each chain ends at a different branching vertex or heavy leaf, so there are at most \(B+L\) such chains.

For a chain \(Q_0\supset Q_1\supset\cdots\supset Q_a\), where each \(Q_{i+1}\) is the unique heavy child of \(Q_i\), consider the differences \[D_i=Q_i\setminus Q_{i+1}\qquad(0\le i<a).\] Each is a union of at most seven light children and contains at most \(7\mu\) points. In chain order, close a run of consecutive differences as soon as its mass reaches \(\mu\). Every completed run has mass in \([\mu,8\mu)\), and at most one residual run per chain has smaller mass. A residual run of zero mass is retained when necessary to cover the containing cube. Every run is a difference of a cube and one strict dyadic descendant, since \[D_i\cup\cdots\cup D_j=Q_i\setminus Q_{j+1}.\] At each branching vertex and each heavy leaf, take its light children as separate cube regions.

These regions partition the original root cube: they group the terminal cubes of the full subdivision without changing their union. The half-open convention makes the assertion exact also on boundaries. Completed runs are disjoint and have mass at least \(\mu\), so there are at most \(r\) of them. There are at most \(B+L\) residual runs and at most \(8(B+L)\) separate light children. Consequently \[|\mathcal R_X|\le r+9(B+L)\le19r,\] and each region has mass at most \(8\mu\).

We prove a thickness property for these regions. If the outer cube of a region \(R\) has side \(S\), then for every \(x\in R\) and \(0<s\le S\), there is a measurable set \(E\subset R\) with \[ E\subset B(x,\sqrt3\,s),\qquad \operatorname{vol}(E)\ge(s/4)^3. \tag{82}\] Choose the unique dyadic subcube \(A_0\) of the outer cube containing \(x\), with side \(a\) satisfying \(s/2<a\le s\). For a cube region, take \(E=A_0\). If \(R=Q\setminus Q'\) and \(A_0\cap Q'=\varnothing\), the same choice works. In the remaining case dyadic nesting, together with \(x\in A_0\setminus Q'\), says that \(Q'\) is a strict descendant of \(A_0\). A child of \(A_0\) other than the child leading to \(Q'\) lies in \(R\), has side \(a/2>s/4\), and lies within distance \(\sqrt3\,s\) of \(x\). Use this child as \(E\). This proves (82), including at all included half-open boundary points. The statement is unchanged by a rigid motion.

Consider the two disjoint region families \(g(\mathcal R_X)\) and \(\mathcal R_Y\). Charge each intersecting pair to the member with smaller outer side, breaking ties by a fixed preference between the families. Fix a charged region \(R\) with outer side \(s\), and let \(c_R\) be the center of its outer cube. For each partner \(R'\), choose \(x_{R'}\in R\cap R'\). The outer side of \(R'\) is at least \(s\), so (82) gives a set \[E_{R'}\subset R'\cap B(x_{R'},\sqrt3\,s),\qquad \operatorname{vol}(E_{R'})\ge(s/4)^3.\] All these sets lie in \(B(c_R,3\sqrt3\,s/2)\). They are pairwise disjoint because the partners belong to one partition. Comparing their volumes with that ball bounds their number by an absolute constant. Summing the charges over at most \(38r\) regions proves the intersection estimate. A matched point lies in a unique region of each partition, so the assertion about matches follows as well. ◻

Proof. For nonparallel lines choose unit directions \(a_1,a_2\) and \(b_1,b_2\). The orthogonal part \(S\) of a motion must satisfy \(Sa_i=\varepsilon_i b_i\), with \(\varepsilon_i\in\{1,-1\}\). There are at most four sign choices. Each compatible choice determines the isometry on the two-dimensional span of \(a_1,a_2\), and there are at most two choices on its perpendicular line. For each resulting \(S\), the condition on its translation \(h\) imposed by each line is an affine line in translation space with direction \(b_i\). Those two affine lines are nonparallel, so their intersection has at most one point. This proves the bound of eight, whether the original lines intersect or are skew.

For parallel lines choose unit directions \(a,b\). Let \(d\) be the nonzero perpendicular separation vector from \(L_1\) to \(L_2\), and let \(e\) be the analogous vector for the ordered target pair. Orthogonality forces \[Sa=\pm b,\qquad Sd=e.\] These equations determine \(S\) on a two-dimensional space, with at most two choices on its perpendicular line. There are therefore at most four orthogonal parts in total. For any fixed one, mapping the first source line onto the first target line fixes the translation modulo a scalar multiple of \(b\); the second-line condition is then either impossible or automatically satisfied. This proves the geometric assertion.

Parametrize one such translation family by a real number \(\tau\). Each pair of a source sample point on \(L_1\) and a target sample point on \(L'_1\) determines a unique \(\tau\). Thus \[\sum_\tau w_1(\tau) \le |X\cap L_1|\,|Y\cap L'_1|\le H^2, \qquad w_2(\tau)\le H.\] Only finitely many \(\tau\) have positive first weight. The product sum for this linear part is at most \(H^3\); summing over at most four parts proves (83). This argument sums the actual translations and introduces no discretization of the continuous family. ◻

Proof. In one parallelism class of size \(q\), intersect the lines with the plane through the origin perpendicular to their common direction. Their intersection points are distinct, and line separation equals the Euclidean distance of these points. Form the graph whose edges join points at distance \(\rho\). Two different vertices have at most two common neighbors, because two distinct circles of radius \(\rho\) have at most two common points. If the graph has \(e\) edges and degrees \(d_z\), then \[\sum_z\binom{d_z}{2}\le2\binom q2, \qquad (2e)^2\le q\sum_zd_z^2\le3q^3.\] In the last inequality we used the preceding bound and \(2e\le q(q-1)\). Hence there are \(O(q^{3/2})\) ordered pairs in this class. Sum over the classes, using \(\sum_jq_j^{3/2}\le(\sum_jq_j)^{3/2}\le L^{3/2}\). ◻

Proof of Theorem 38. We prove the stronger assertion stated at the start of the section. Let \(X,Y\) be clouds of at most \(N\) points, put \(A=N^{2/3}\), and suppose that each line contains at most \(2A\) points of either cloud. Write \(k_g=k_g(X,Y)\) throughout this proof. We prove (70) with this notation. This implies the asserted result by (71). Alternatively, the claim for a congruent copy follows by composing the motions with the inverse of the congruence, which preserves their match counts and bijects the two orientation classes as appropriate.

Fix the absolute constants \[\alpha=\frac1{14},\qquad \beta=\frac1{4K_{\mathrm{part}}},\qquad \nu=\frac{\eta_1}{4}.\] Choose \(C_{\mathrm{rich}}\) sufficiently large that \[ C_{\mathrm{rich}}>4,\qquad \frac\beta4 C_{\mathrm{rich}}^{9/14} \ge C_{\mathrm{loc}},\qquad \eta_1\beta C_{\mathrm{rich}}^{3/7}\ge1,\qquad \nu C_{\mathrm{rich}}\ge4. \tag{84}\] All constants on the right have already been fixed. These choices are independent of \(N\) and of the clouds.

Every motion with at least \(C_{\mathrm{rich}}A\) matches has a noncollinear matched triple, because the line cap is \(2A\). The finite choice of source and target ordered triples and the at-most-two extensions per triple prove finiteness of the motion set in the sum. In particular, possible continuous families of motions fixing a line are below the threshold unless they have additional noncollinear matches, in which case this same finite triple argument applies.

Bucket the high motions by \[k\le k_g(X,Y)<2k,\qquad k=tA,\qquad t=2^jC_{\mathrm{rich}}\le N^{1/3}.\] If no such \(t\) exists, the sum is empty. In a nonempty bucket set \[ r=t^{1+\alpha},\qquad m=8N/r,\qquad b=\beta k/r=\beta A t^{-\alpha}. \tag{85}\] Since \(t\le N^{1/3}\) and \(\alpha<2\), we have \(1\le r\le N\), so Lemma 42 applies to both clouds. It supplies at most \(19r\) regions in each partition, each with at most \(m\) points.

For the partitions just chosen, put \(X_R=X\cap R\) and \(Y_{R'}=Y\cap R'\). The local count is \(n_{R,R'}(g)=k_g(X_R,Y_{R'})\). These counts sum to \(k_g\), and at most \(K_{\mathrm{part}}r\) are nonzero. By (81), discarding those below \(b\) removes at most \(K_{\mathrm{part}}rb=k/4\) matches. At least \(3k/4\) matches remain. Bucket each remaining local count into \(\ell\le n_{R,R'}(g)<2\ell\), with \(\ell=2^i b\), and use the line/planar/general classification of Lemma 41 for the clouds \(X_R,Y_{R'}\) in that local bucket. A single global motion may contribute different local types in different region pairs.

The local lower threshold is valid, since \[ \frac{b}{m^{2/3}} =\frac\beta4\,t^{(2-\alpha)/3} =\frac\beta4\,t^{9/14} \ge C_{\mathrm{loc}}. \tag{86}\] We shall also need the full-cloud rich-line threshold \(q=\eta_1b\) to be at least \(\sqrt N\). The inequality \(t\le N^{1/3}\) gives \(N^{1/6}\ge t^{1/2}\), so \[ \frac{q}{\sqrt N} =\eta_1\beta N^{1/6}t^{-\alpha} \ge\eta_1\beta t^{1/2-\alpha} =\eta_1\beta t^{3/7}\ge1. \tag{87}\] In particular these rich lines contain at least two points in every nonempty bucket. One may also record the power margin \(b\ge\beta N^{9/14}\), although only (87) is needed below.

Proof. Suppose \(f_1,\ldots,f_{i-1}\) have been chosen, and list the finitely many components \(W\) of their zero set which contain \(V\). Each has codimension \(i-1\). None is contained in \(Z(I_{\le\delta_i}(V))\): otherwise a component of that latter set would contain \(W\), hence contain \(V\) and have codimension at most \(i-1\), contrary to the definition of \(\delta_i\). Consequently the members of \(I_{\le\delta_i}(V)\) that vanish on any fixed \(W\) form a proper linear subspace. Choose \(f_i\) outside their finite union. It cuts every such \(W\) properly. The principal ideal theorem then gives codimension exactly \(i\) for every new component containing \(V\). This includes \(i=1\), when the preceding zero set is the ambient space, and proves the construction.

For the second assertion, suppose no component \(W\) in the indicated prefix has \(I_{\le m}(V)\subset I(W)\). The same finite-union argument provides \(g\in I_{\le m}(V)\) which vanishes identically on none of them. All components of \(Z(f_1,\ldots,f_j,g)\) containing \(V\) then have codimension \(j+1\). Since \(m\ge\delta_j\), every one of these equations belongs to \(I_{\le m}(V)\). Thus every component of \(Z(I_{\le m}(V))\) containing \(V\) has codimension at least \(j+1\). This contradicts \(m<\delta_{j+1}\). For a component \(W\) where the inclusion does hold, the reverse inclusion follows from \(V\subset W\), giving (98). ◻

Proof of Theorem 47. Fix the polynomials from Lemma 48. We will prove, for \(0\le s\le c\), that every component \(U\) of the first \(s\) cuts containing \(V\) satisfies \[ \deg U\ge\gamma_n^{\,s}\prod_{i=1}^s\delta_i \tag{99}\] for one sufficiently small \(0<\gamma_n\le1\) depending only on \(n\). For \(s=0\) both sides are \(1\).

We record the precise consequence of Lemma 13 used here. There are constants \(a_n>0\) and \(b_n\ge1\) such that an irreducible \(k\)-dimensional variety \(X\) satisfies \[ H_X(m)\le b_n\deg(X)m^k\qquad(m\ge1), \tag{100}\] and, if \(X\) is a component of a system whose equations have degrees at most \(m\), then \[ H_X(m)\ge a_n\deg(X)m^k\qquad(m\ge1). \tag{101}\] The second assertion is the bound at the equation degree in (16), applied with \(L=m\). It permits other components of the system to have larger dimension. The lower bound also holds for \(X=\mathbb A^n\), after decreasing \(a_n\), directly from \(H_{\mathbb A^n}(m)=\binom{m+n}{n}\).

Suppose (99) is known for smaller indices, and let \(U\) be a component at index \(s\). If \(\delta_s=1\), the entire product in (99) is \(1\), and \(\deg U\ge1\) proves the claim. Otherwise set \[m=\delta_s-1,\qquad j=\max\{i:0\le i<s,\ \delta_i<\delta_s\}.\] Thus \[ \delta_j\le m<\delta_{j+1} =\cdots=\delta_s\le2m. \tag{102}\] Lemma 48 supplies a component \(W\) at index \(j\) with \(H_W(m)=H_V(m)\). Its dimension is \(n-j\), and the degrees of its defining prefix equations are at most \(\delta_j\le m\). The induction hypothesis and (101) therefore give \[H_V(m)=H_W(m) \ge a_n\gamma_n^{\,j} \left(\prod_{i=1}^j\delta_i\right)m^{n-j}.\] This formula also applies when \(j=0\). Since \(V\subset U\), restriction from \(U\) to \(V\) is surjective on polynomial restriction spaces, so \(H_U(m)\ge H_V(m)\). On the other hand, \(\dim U=n-s\) and (100) bounds \(H_U(m)\) above. Combining the bounds, then using (102), yields \[\begin{split} \deg U &\ge \frac{a_n}{b_n}\gamma_n^{\,j} \left(\prod_{i=1}^j\delta_i\right)m^{s-j}\\ &\ge \frac{a_n}{2^n b_n}\gamma_n^{\,j} \prod_{i=1}^s\delta_i. \end{split}\] Choose \(\gamma_n\le\min\{1,a_n/(2^n b_n)\}\). Since \(j<s\), the last expression is at least the right side of (99), completing the induction. Notice that no containment relation between \(U\) and the chosen \(W\) was needed; both contain \(V\).

At \(s=c\), every prefix component containing \(V\) has the same dimension as \(V\), so \(V\) itself is such a component. Applying (99) to it gives \[\prod_{i=1}^c\deg f_i \le\prod_{i=1}^c\delta_i \le\gamma_n^{-c}\deg V \le\gamma_n^{-n}\deg V.\] This proves the theorem with \(C_n=\gamma_n^{-n}\). ◻

Proof. Induct on \(a\). The assertion for \(a=0\) says that a point contains at most one sample parameter. Otherwise, if \(Y\cap\mathcal P\ne\varnothing\), choose \(\lambda\in Y\cap\mathcal P\) and a separator \(q\) for it. Since \(q(\lambda)\ne0\), the section \(Y\cap Z(q)\) is proper, and it contains all the other counted parameters. Its components have dimension \(a-1\) and total degree at most \(C_0A\deg Y\). Applying the induction hypothesis to those components gives \[|Y\cap\mathcal P| \leq 1+C_{a-1}C_0\deg(Y)A^a.\] The initial \(1\) is absorbed because \(A\geq1\) and \(\deg Y\geq1\). There are only boundedly many possible dimensions. ◻

Proof. Put \(P=I(V)\) and consider the regular local ring \[R=\mathbb C[x_1,\ldots,x_n]_P,\qquad \mathfrak m=PR,\qquad K=R/\mathfrak m=\mathbb C(V).\] Its dimension is \(c\). Because \(V\) is an isolated component of the common zero set at its generic point, \(J=(f_1,\ldots,f_c)R\) is \(\mathfrak m\)-primary. Its generators form a parameter sequence, which is regular because \(R\) is Cohen–Macaulay (The Stacks Project Authors 2026, Tags 00OQ, 00NQ, and 02JN). We first bound the length of \(R/J\) by a dimensional constant.

Write \(k=\dim V\). A general projective linear projection of \(\overline V\) to \(\mathbf P^k\), with center in the hyperplane at infinity disjoint from \(\overline V\), is finite and has degree \(\deg V\). Finiteness follows because a positive-dimensional projective fiber would meet its hyperplane lying in the projection center. Restricting to the affine chart and changing affine coordinates gives base coordinates \(t_1,\ldots,t_k\) and complementary coordinates \(w_1,\ldots,w_c\). Over \(F=\mathbb C(t_1,\ldots,t_k)\), the generic fiber of \(V\) is a closed point of \(\mathbb A^c_F\) with residue field \(K\), where \([K:F]=\deg V\). Localizing \(F[w_1,\ldots,w_c]\) at that point gives the same ring \(R\).

After extending \(F\) to an algebraic closure, this point splits into \([K:F]\) points, since the extension is separable. The sum of their local intersection multiplicities for the equations \(f_i\) is \([K:F]\operatorname{length}_R(R/J)\). They are isolated common zeros; other components of the system do not affect their isolated Bézout bound (Fulton 1998, Theorem 12.3 and Example 12.3.1). The equations \(f_i\) define hypersurfaces in regular local ambient rings, so these factors are Cohen–Macaulay and the isolated complete-intersection multiplicity equals the quotient length; see (Fulton 1998, Example 12.3.7(ii)–(iii)) for this identification and its Cohen–Macaulay hypothesis. Hence \[ \operatorname{length}_R(R/J) \leq\frac{\prod_i\deg f_i}{\deg V}\leq C_n. \tag{103}\] Choose an integer \(L=O_n(1)\) at least this length. The maximal ideal of an Artinian local ring of length at most \(L\) has \(L\)th power zero, so \[ \mathfrak m^L\subset J. \tag{104}\]

We construct the \(g_i\) one at a time. Suppose \(g_1,\ldots,g_u\) have independent classes in \(\mathfrak m/\mathfrak m^2\), where \(u<c\). They are part of a regular parameter system, and therefore \[R_u=R/(g_1,\ldots,g_u),\qquad \mathfrak m_u=\mathfrak mR_u\] is a regular local ring of positive dimension. The images of the \(f_i\) generate an ideal containing \(\mathfrak m_u^L\). At least one image has finite \(\mathfrak m_u\)-adic order \(b\) with \(1\leq b\leq L\). Otherwise all the images would lie in \(\mathfrak m_u^{L+1}\), whence \(\mathfrak m_u^L=\mathfrak m_u^{L+1}\) and Nakayama’s lemma would give \(\mathfrak m_u^L=0\), impossible in this positive-dimensional regular local ring.

We need derivatives which do not disturb the already chosen equations. For \(u=0\) use coordinate derivations. For \(u>0\), choose a \(u\)-square Jacobian block \(J_0\) of \((g_1,\ldots,g_u)\) whose determinant \(\Delta\) is nonzero in \(K\). If \(J_t\) is any remaining column, form a polynomial vector field having coefficient \(\Delta\) in coordinate \(t\) and coefficients \(-\operatorname{adj}(J_0)J_t\) in the chosen coordinates. These \(n-u\) fields, denoted by \(\delta_\alpha\), satisfy \[\delta_\alpha g_i=0\quad(1\leq i\leq u)\] as polynomial identities. Thus they extend to \(R\) and descend to \(R_u\). If the \(g_i\) have maximum degree \(E_u\), the coefficient degrees of these fields are \(O_n(E_u)\).

At the generic point, the fields span the common kernel of the \(d g_i\). The conormal map \(\mathfrak m/\mathfrak m^2\to K^n\), \(f\mapsto df\), is injective there, as \(V\) is generically smooth in characteristic zero. It follows that the residue evaluations of the \(\delta_\alpha\) span the dual of \(\mathfrak m_u/\mathfrak m_u^2\). These residue evaluations are \(K\)-linear: the extra Leibniz term from differentiating a coefficient has a factor in \(\mathfrak m_u\).

Take an \(f_i\) of order \(b\) as above. A derivation lowers adic order by at most one, so fewer than \(b\) of these derivations still give zero residue. Some composition of \(b\) of them has nonzero residue. Indeed, the leading term of \(f_i\) belongs to the degree-\(b\) piece of the graded ring of \(R_u\) (The Stacks Project Authors 2026, Tag 00NO), namely \[\mathfrak m_u^b/\mathfrak m_u^{b+1} \simeq\operatorname{Sym}^b_K(\mathfrak m_u/\mathfrak m_u^2).\] Its polarization is nonzero in characteristic zero. Since the residue functionals span the dual, some \(b\)-tuple detects it. In computing that residue, terms in which a derivation hits a coefficient retain a positive-order factor and vanish. This gives exactly the asserted iterated derivative.

Stop immediately before the last differentiation and call the resulting polynomial \(g_{u+1}\). Its residue in \(R_u\) is zero, so it belongs to \(P\). Its derivative has nonzero residue, so its class in \(\mathfrak m_u/\mathfrak m_u^2\) is nonzero. It adds an independent conormal. This procedure uses at most \(L-1\) derivatives per stage and at most \(c\) stages. Applying a polynomial field of coefficient degree \(T\) to a degree-\(B\) polynomial gives degree at most \(B+T-1\); the bounds on \(L\) and \(c\) therefore keep every \(g_i\) of degree \(O_n(D)\).

Finally choose a \(c\)-square minor \(H_0\) of the Jacobian of the \(g_i\) which is nonzero on \(V\). Its degree is \(O_n(D)\). Where it is nonzero, the common zero set of the \(g_i\) is locally smooth of dimension \(n-c\). The contained variety \(V\) has the same dimension, so its germ equals that smooth germ. This proves the assertion about tangent spaces. ◻

Proof. Use Lemma 50 to choose \(g_1,\ldots,g_c\) and an initial smooth open set in \(V\). In the construction below we remove zeros of a bounded number of polynomials, each of degree \(O_e(D)\) and nonzero on \(V\). Their product, together with the initial Jacobian minor, will be \(H\). We continue to call the current open set \(\Omega\).

First consider tangent directions. Under \(S^*=\bigwedge^2E\), the annihilator of \(K_v\) is \(v\wedge E\). A bivector belongs to this space if and only if its wedge with \(v\) is zero. Thus the necessary condition \(K_v\subset T_zV\) is the linear system \[ v\wedge dg_i(z)=0\qquad(1\leq i\leq c). \tag{106}\] Its vector solution space has dimension at most two. To see this, let a nonzero bivector \(w\) satisfy \(v\wedge w=0\) for a nonzero \(v\). Splitting off \(\mathbb Cv\) shows that \(w=v\wedge a\) for some \(a\); the vectors \(v'\) satisfying \(v'\wedge w=0\) are precisely \(\operatorname{span}(v,a)\). If two independent conormals had a common two-dimensional solution space, both would lie in its one-dimensional exterior square, a contradiction. A two-dimensional solution space therefore forces \(c=1\).

Work separately in each of the \(e\) normalized charts \(v_a=1\). Gaussian elimination over \(\mathbb C(V)\), followed by removal of nonzero rank minors, puts the system throughout \(\Omega\) into one of three forms: no solution, a graph \(v=b(z)\), or \[ v=b(z)+\tau a_1(z). \tag{107}\] In the last case choose a free coordinate of \(v\) as \(\tau\), so that its coordinate in \(b\) is zero and in \(a_1\) is one. All coefficients are regular on \(\Omega\) and admit ambient rational representatives with numerator and denominator degrees \(O_e(D)\), since the matrices have bounded size. Rank conditions which vanish generically vanish on all of \(V\); excluding the nonzero minors consequently controls all remaining fibers. Denote the resulting solution variety by \(\Gamma_0\). It is a smooth graph over \(\Omega\), or a smooth graph over \(\Omega\times\mathbb A^1\).

We now test whether the tangent directions integrate to a contained flat. Write \(\varepsilon_1,\ldots,\varepsilon_e\) for a basis of \(E^*\). In the chart \(v_a=1\), exterior products of the covectors \(\varepsilon_i-v_i\varepsilon_a\), \(i\ne a\), form a basis \(B_1(v),\ldots,B_{s_0}(v)\) of \(K_v\). Their coefficients have degree at most two in \(v\). Consider the horizontal fields \[X_\alpha(z,v)=(B_\alpha(v),0).\] Their \(z\)-components are tangent to \(V\), by (106). They are tangent to \(\Gamma_0\) if and only if the derivatives of all coordinates of its graph formula vanish in direction \(B_\alpha(v)\), with \(\tau\) fixed in (107). Holding \(\tau\) fixed is required because it is one of the coordinates of the normalized vector \(v\). Call these derivative expressions \(T_\beta\).

With no free coordinate the \(T_\beta\) are rational functions of \(z\). With a free coordinate they are polynomials in \(\tau\) of bounded degree, whose coefficients have ambient rational degree \(O_e(D)\). This follows by substituting (107) in the quadratic expressions \(B_\alpha\) and differentiating its rational coefficients. Differentiation along \(T_zV\) is independent of the ambient representatives chosen, since it annihilates functions vanishing on \(V\).

Every contained flat passes this first test: on a small open part of the flat, \(v\) is fixed and the direction equations continue to hold. If every \(T_\beta\) is identically zero, retain \(\Gamma_0\); its horizontal fields are already tangent everywhere. If there is no free coordinate but one test is nonzero, remove the zero set of its numerator and discard this chart. No contained flat in that chart can meet the remaining open set.

It remains to treat a free coordinate and a nonidentically zero test. Let \(P(\tau)\) be the monic gcd of the tests over \(\mathbb C(V)\), and let \(G(\tau)\) be its monic squarefree part. The Euclidean algorithm gives identities \[T_\beta=P Q_\beta,\qquad P=\sum_\beta U_\beta T_\beta, \qquad G\mid P,\quad P\mid G^q\] for a bounded integer \(q\). Exclude the poles of all coefficients used in these identities. They then show in every remaining fiber that the common roots of the tests are exactly the roots of \(G\). If \(G=1\), the chart contributes nothing. Otherwise exclude also the zero set of its discriminant. The root variety \[\Sigma_G=\{(z,\tau)\in\Omega\times\mathbb A^1:G(z,\tau)=0\}\] is now finite étale over \(\Omega\), with bounded degree. In particular it is smooth, and its roots form local holomorphic sheets.

All these exclusions have total degree \(O_e(D)\). Indeed, the number of tests and their degrees in \(\tau\) are bounded. Gcd, squarefree part, quotient, and discriminant computations therefore use a bounded number of field operations. A bounded number of operations on ambient fractions of degree \(O_e(D)\) preserves that degree bound. Every excluded denominator or pivot has nonzero restriction to \(V\). The same observation applies to the following second test.

Along a contained flat, \(G(z,\tau)=0\) persists with \(\tau\) fixed. Thus a second necessary condition is \[R_\alpha(z,\tau) =\sum_{m=1}^{n}(B_\alpha(v(z,\tau)))_m \frac{\partial G}{\partial z_m}(z,\tau)=0.\] These expressions again have bounded degree in \(\tau\) and coefficient degrees \(O_e(D)\). Form the monic polynomial \[P_2=\gcd(G,R_1,\ldots,R_{s_0})\] and exclude the poles needed to specialize its quotient and Bézout identities. Its root variety \(\Sigma_{P_2}\) is empty or finite étale over \(\Omega\). Because \(G\) remains squarefree and \(P_2\) divides it, the factorization \(G=P_2(G/P_2)\) has disjoint root sets in every remaining fiber. Thus \(\Sigma_{P_2}\) is locally a union of whole sheets of \(\Sigma_G\).

At a point of \(\Sigma_{P_2}\), the first test makes \(X_\alpha\) tangent to \(\Gamma_0\), where its coordinates are \((B_\alpha(v),0)\) in \((z,\tau)\). The second test makes it tangent to \(G=0\). Hence it is tangent to the selected sheet of \(\Sigma_{P_2}\). No further sequence of tangency tests is needed. Retain its irreducible components as the incidences. A smooth variety has disjoint irreducible components. Each component of this finite étale cover maps openly and closedly to the irreducible base, hence surjectively; their number and fiber sizes are bounded by \(\deg G\).

We verify containment rather than merely tangency. On every retained incidence the fields \(X_\alpha\) are tangent. Their ambient flows are \[(z,v)\longmapsto(z+tB_\alpha(v),v),\] since \(v\) is constant. Uniqueness of local holomorphic flows keeps these translations in the incidence for small \(t\). Composing them produces an analytic open subset of \(z+K_v\), because the \(B_\alpha(v)\) form a basis of \(K_v\). Every polynomial vanishing on \(V\) vanishes on that open subset, hence on the whole affine flat. Thus every retained point really represents a contained flat.

After processing all normalized charts, intersect their open sets and restrict their incidences to this common \(\Omega\). The product of the excluded polynomials is nonzero on irreducible \(V\) and has degree \(O_e(D)\). Every nonempty restricted incidence remains irreducible and dominant. Its fibers are finite of bounded size, except for the retained affine-line graphs; the latter arose only in codimension one.

If a contained flat meets \(\Omega\), choose a chart for its \(v\) and a point of \(F_\lambda\cap\Omega\). Both tests are necessary along that flat, and the specialized gcd identities therefore place the point in a retained incidence. This proves coverage.

Finally fix one incidence and a nonempty fiber of (105). Its normalized vector \(v\) is fixed, and the fiber is closed in \(F_\lambda\cap\Omega\). The preceding flows preserve both \(v\) and \(zv\) and give an analytic open subset of that flat in the fiber. Such an open subset is Zariski dense in the irreducible variety \(F_\lambda\cap\Omega\), so the fiber equals this whole intersection. It has dimension \(s_0\). The fiber-dimension theorem now gives \[\dim\Lambda=\dim\Gamma-s_0=(k+j)-s_0=h+j.\] ◻

Proof. The fibers of the parameter map are the nonempty sets \(F_\lambda\cap\Omega\), of dimension \(s_0\), whereas \(\dim\Gamma=k+j\). The fiber dimension theorem therefore gives \(\dim\Lambda=h+j\). We prove the degree estimate and the assertion for \(j=1\) in turn.

Proof of Theorem 46. Fix the sample, \(A\), and one of Specifications (a) and (b). For \(0\leq h<r\), let \(R_h\) be the supremum of \[\frac{\#\{\lambda\in\mathcal P:F_\lambda\subset V\}} {\deg(V)A^h}\] over the allowed irreducible varieties of dimension \(s_0+h\). An empty supremum is zero. These suprema are finite for the fixed sample: each numerator is at most \(|\mathcal P|\), and each denominator is at least one. As noted after the theorem, \(R_0\leq1\). In (a) also put \[R_r=|\mathcal P|/A^r\leq C_0.\] In (b) the index \(r\) is unavailable and will never be used.

We first refine the choice of approximate equations. Take the \(c=n-\dim V\) equations from Theorem [conc:walsh], ordering their degrees as \(d_1\leq\cdots\leq d_c\). They form a parameter sequence in the regular local ring \(\mathbb C[S]_{I(V)}\), so every initial segment of length \(i\) has height \(i\) (The Stacks Project Authors 2026, Tags 00NQ, 02JN, and 00NA). Choose a polynomial \(q\) of smallest positive degree in \(I(V)\). We may use \(q\) as the first equation and preserve the upper bounds \(d_i\) on the remaining ones. Indeed, suppose a replacement prefix of length \(i-1\) has height \(i-1\). Its minimal primes have that height, because a partial parameter sequence in this regular local ring is a regular sequence. None can contain the ideal generated by the first \(i\) original equations, whose height is \(i\). A generic constant linear combination of those \(i\) equations avoids all these minimal primes and extends the replacement prefix to height \(i\). Its degree is at most \(d_i\). Starting with \(q\) and continuing gives a parameter system again. Consequently \(V\) remains a component of its global common zero set. Replace \(d_1\) by \(\deg q\). The nondecreasing upper bounds still satisfy \[ \prod_{i=1}^c d_i\leq C_n\deg V. \tag{125}\] The later actual degrees may be smaller than their upper bounds; this only strengthens the relevant Bézout estimates.

Choose constants \[K_1>K_2>\cdots>K_{r-1}>C_0,\] to be fixed below in terms of \(e,C_0\). Consider an allowed carrier \(V\) of dimension \(s_0+h\), with \(1\leq h<r\).

Suppose first that every description degree \(d_i\) is at most \(K_hA\). Lemma 51 supplies a proper exceptional section \(V\cap Z(H)\) with \[\deg H\leq C_eK_hA.\] Outside it, there are boundedly many ruling incidences. For one with \(j=0\), Proposition 52 gives a parameter variety of dimension \(h\) and degree at most \(C_e\deg V\); the separator bound then counts at most \(C_{e,C_0}\deg(V)A^h\) parameters on it. For one with \(j=1\), that proposition says \(h=r-1\) and the parameter variety is a maximal isotropic \(\mathbf P^r\). Assumption (ii) counts at most \(C_0A^h\) parameters on it. Thus all represented flats together cost at most \(C_{e,C_0}\deg(V)A^h\).

The remaining flats are in \(V\cap Z(H)\). Its components, if any, have dimension \(s_0+h-1\) and total degree at most \(C_eK_hA\deg V\). They remain allowed in (b), since they are subsets of the original allowed carrier. The definition of \(R_{h-1}\) gives \[ \frac{\#\{\lambda:F_\lambda\subset V\}}{\deg(V)A^h} \leq C(1+K_hR_{h-1}). \tag{126}\] Here and below \(C\) depends only on \(e,C_0\), not on the \(K_i\).

Suppose instead that some \(d_i>K_hA\). Let \(p<c\) be the length of the prefix with \(d_i\leq K_hA\). The local prefix is a regular sequence of height \(p\). A minimal prime of it in \(\mathbb C[S]_{I(V)}\) contracts to a global component \(W\) of that prefix, containing \(V\) and of codimension \(p\). Isolated-component Bézout gives \[\deg W\leq\prod_{i\leq p}d_i.\] For \(p=0\), take \(W=S\) and the empty product to be one. Put \(i=r-p>h\), so \(\dim W=s_0+i\). From (125), \[ \deg V\geq C_n^{-1}\deg W(K_hA)^{i-h}. \tag{127}\] In Case (b), the minimal first equation has degree at most \(C_0A\). Since \(K_h>C_0\), it belongs to the prefix. Therefore \(p\geq1\), \(i<r\), and \(W\) lies in that same hypersurface of degree at most \(C_0A\). Thus \(W\) is allowed in (b). In (a) the ambient possibility \(i=r\) is already controlled. In either case, counting first on \(W\) yields \[ \frac{\#\{\lambda:F_\lambda\subset V\}}{\deg(V)A^h} \leq \frac{C R_i}{K_h^{i-h}}. \tag{128}\]

Taking suprema in the two alternatives gives the finite system \[ R_h\leq\max\left\{ C(1+K_hR_{h-1}),\quad \max_{\substack{i>h\\i\text{ available}}} \frac{CR_i}{K_h^{i-h}}\right\}, \qquad1\leq h<r. \tag{129}\] An empty inner maximum is omitted. We close this system explicitly, so that no constant depends on the description degrees or on the size of the sample.

Enlarge \(C\) to at least one. Choose \[B_0\geq\max(1,C_0),\qquad c_1>4C+4,\qquad \kappa>4Cc_1.\] Take \(K_{r-1}\geq\max(2,C_0+1)\), and successively set \(K_h=\kappa K_{h+1}\) for \(h=r-2,\ldots,1\). Define barriers \[B_h=c_1K_hB_{h-1}\quad(1\leq h<r), \qquad B_r=B_0\quad\hbox{in Case (a)}.\] For interior indices \(h<i<r\), \[\frac{CB_i}{B_hK_h^{i-h}} =C c_1^{i-h}\prod_{j=h+1}^{i}\frac{K_j}{K_h} \leq C(c_1/\kappa)^{i-h}<\frac14.\] If the ambient index is available, \(B_h\geq c_1K_hB_0\) gives \[\frac{CB_r}{B_hK_h^{r-h}} \leq\frac{C}{c_1K_h^{r-h+1}}<\frac12.\] The other transition satisfies \[\frac{C(1+K_hB_{h-1})}{B_h}\leq\frac{2C}{c_1}<\frac12.\] All choices so far depend only on \(e,C_0\).

If the maximum \(T=\max_h R_h/B_h\) were greater than one, a maximizing index would be interior: \(R_0\leq B_0\), and the ambient endpoint, when present, obeys the same bound. Substitute \(R_j\leq TB_j\) into (129). Since \(T>1\), each term in its right side is strictly less than \(TB_h\), contradicting maximality. Hence \(R_h\leq B_h\) for every available index. This is (96), with a constant depending only on \(e,C_0\). ◻

Proof. Deleting at most one edge for each intersecting pair produces a planar simple graph. Euler’s formula therefore gives \(e\le3v+X\). Select each vertex independently with probability \(p\) and keep the induced drawing. An edge survives with probability \(p^2\) and an intersecting pair with probability \(p^4\), because its four endpoints are different. Taking expectations in the same inequality gives \(p^2e\le3pv+p^4X\). Set \(p=4v/e\) and rearrange. ◻

Proof. The assertion is immediate if \(n=0\) or \(L=0\), so assume both are positive. Join consecutive points of \(Q\) on each line of \(\mathcal L\). The resulting graph is simple: a pair of distinct points determines just one line. If \(e\) is its number of edges, then \(e\ge I-L\). Edges on the same line have disjoint interiors. Edges on two different lines cross at most once, so their crossing count is at most \(\binom L2\). Lemma 53 gives either \(e<4n\) or \(e^3\le32n^2L^2\), proving the incidence bound. Concurrent crossings cause no problem: count pairs of intersecting edges, or separate the crossings in a sufficiently small disk without changing the bound.

For a set of \(s\) points each incident to at least \(k\) lines, the bound gives \(ks\le C(s^{2/3}L^{2/3}+s+L)\). If \(k\) exceeds a fixed constant, absorb the term \(Cs\) and compare the two remaining terms, obtaining \(s\le C'(L^2/k^3+L/k)\). For bounded \(k\ge2\), count pairs of lines: \(s\le\binom L2\), which gives the same assertion after increasing the constant. Interchanging points and lines by planar projective duality proves the rich-line statement. One can instead apply the incidence bound directly to the finite collection of rich lines, using that two points determine at most one line for bounded \(k\).

For a finite configuration in higher-dimensional affine space, choose a linear projection to \(\mathbb R^2\) which is injective on its points and sends its lines to distinct lines. Each forbidden coincidence is a proper algebraic condition on the projection, and finitely many such conditions can be avoided. Existing incidences survive, which is all that an upper bound requires. For projective data first choose an affine chart containing every specified point and in which no specified line is contained in the hyperplane at infinity. Apply the affine argument. ◻

Proof. For \(m<\binom{D+3}{3}\) finite-volume bounded measurable sets in \(\mathbb R^3\), normalize the coefficients of a degree-at-most-\(D\) polynomial to lie on their unit sphere. Its signed-volume imbalances on the \(m\) sets form a continuous odd map to \(\mathbb R^m\). Continuity follows by dominated convergence, since the zero set of a nonzero polynomial has Lebesgue measure zero. The latter fact follows by induction on the number of variables and Fubini’s Theorem. Restrict the coefficient sphere to an \((m+1)\)-dimensional linear subspace and apply Borsuk–Ulam. We obtain a polynomial bisecting all \(m\) sets.

Replace each point in each of \(m\) finite sets by a disjoint small ball of the same radius within that set, apply this assertion, and let the radius tend to zero. A convergent subsequence of normalized coefficient vectors gives a nonzero polynomial for which each strict sign contains at most half the points of each finite set. Indeed, a strict positive value at a point persists on its small ball and for nearby coefficient vectors; more than half of the points of either sign would contradict the volume bisection.

Iterate this finite bisection \(J\) times, bisecting all the at most \(2^{j-1}\) sign classes at stage \(j\). A degree \(C2^{j/3}\) suffices at that stage. The product \(f\) of these \(J\) polynomials has degree at most \(C2^{J/3}\), and every strict sign class has at most \(2^{-J}|S|\) points. Choose \(2^J\) comparable to \(r^3\). A line outside \(Z(f)\) is cut into at most \(\deg f+1\) intervals by its intersections with \(Z(f)\); all the factors have constant sign on each interval. This proves the last assertion as well. ◻

Proof. At a noncritical point all contained lines are tangent to the smooth surface. The second fundamental form vanishes in their directions. A homogeneous quadratic on a two-dimensional vector space which vanishes in three distinct projective directions vanishes identically. This proves the first assertion.

Square-freeness in characteristic zero implies that a generic constant linear combination of the first derivatives of \(f\) shares no irreducible factor with \(f\). All critical lines are common line components of their zero sets, so Bézout gives the second assertion. Restriction of \(f\) and its derivatives to a line proves the assertion about many critical points.

Here is an explicit polynomial test for flatness. With \(H_f\) denoting the Hessian matrix and \(e_1,e_2,e_3\) the coordinate vectors, put \[T_j=\bigl(H_f(e_j\mathbin\times\nabla f)\bigr) \mathbin\times\nabla f,\qquad j=1,2,3.\] The nine scalar components have degree at most \(3D\). At a regular point the vectors \(e_j\times\nabla f\) span the tangent plane. Thus all \(T_j\) vanish exactly when \(H_f\) maps every tangent vector into the normal line, which is exactly vanishing of the second fundamental form. If a line contains more than \(3D\) flat points, \(f\) and all these components vanish identically on it. It has a regular point, so it is a flat line.

It remains to bound flat lines on nonplanar components. Fix an irreducible real factor \(g\) of \(f\) which contains a flat line and is not linear. That line has a point regular for \(f\), and consequently \(g\) has a smooth real point. If every component of every \(T_j\) were divisible by \(g\), the second fundamental form would vanish on a neighborhood of that real point in \(Z(g)\). The differential of the unit normal would vanish there. The normal would be constant on a smaller connected neighborhood, and that neighborhood would be contained in its tangent plane. The implicit function theorem then makes it an open subset of that plane. Restricting \(g\) to the plane shows that its plane equation divides \(g\), contrary to irreducibility and \(\deg g>1\). Therefore some component \(T\) is not divisible by \(g\), and Bézout bounds their common lines by \(3D\deg g\).

Every flat line belongs to exactly one irreducible factor of \(f\): membership in two factors would make the entire line critical. Summing the preceding bound over the nonplanar factors gives at most \(3D^2\) flat lines. Factors having no smooth real point contain no flat line and do not need a separate argument. ◻

Proof. We first prove a uniform version. Let \(S\) be a set of \(s\) points each incident to between \(k\) and \(2k\) lines. Suppose at least \(L/100\) lines each contain at least \(sk/(100L)\) points of \(S\). We will prove the stated bound for \(s\).

Fix a sufficiently small absolute \(\varepsilon>0\). We may assume \[ s>A\left(L^{3/2}k^{-2}+Lk^{-1}\right), \tag{130}\] where \(A\) will be chosen sufficiently large. By Theorem 54, this assumption implies \(s\le C L^2k^{-3}\). Choose \[r=\left\lfloor\theta L^2s^{-1}k^{-3}\right\rfloor,\] where the constant \(\theta\) is sufficiently large. Thus \(r\ge1\) and \(r\) is comparable to the expression inside the floor. Lemma 55 gives a polynomial \(f\) of degree \(D\le Cr\). We claim that at most \(\varepsilon s\) points of \(S\) lie outside \(Z(f)\).

Otherwise, among its \(O(r^3)\) sign cells, at least \(c_\varepsilon r^3\) contain at least \(c_\varepsilon s/r^3\) points of \(S\). To verify this, discard cells with fewer than a small fixed multiple of \(\varepsilon s/r^3\) points and use the uniform upper bound \(Cs/r^3\) for the others. Each line meets at most \(D+1=O(r)\) cells. One of the retained cells is therefore met by at most \(C_\varepsilon L/r^2\) lines. Apply Theorem 54 inside it: \[\frac{s}{r^3} \le C_\varepsilon\left(\frac{L^2}{r^4k^3} +\frac L{r^2k}\right).\] Substitution of \(r\) gives \[s\le C_\varepsilon\left(\theta^{-1}s +\theta L^3s^{-1}k^{-4}\right).\] Choose \(\theta\) first so that the first term can be absorbed. The resulting inequality \(s\le C_\varepsilon\sqrt\theta L^{3/2}k^{-2}\) contradicts (130) when \(A\) is sufficiently large. This proves the claim.

Replacing \(f\) by its square-free part does not change its zero set and can only lower its degree. The same estimates and (130) give \[ D\le C\theta L^2s^{-1}k^{-3},\qquad \frac D{sk/L}\le\frac{C\theta}{A^2},\qquad \frac D{\sqrt L}\le\frac{C\theta}{Ak}. \tag{131}\] Write \(\mathcal L_Z\) for the lines contained in \(Z(f)\), and \(S_*\) for the points of \(S\) lying on at least three lines of \(\mathcal L_Z\). A point of \((S\cap Z(f))\setminus S_*\) lies on at least \(k-2\) lines outside \(\mathcal L_Z\). Each such line has at most \(D\) points on \(Z(f)\). Hence \[|S\setminus S_*|\le\varepsilon s+\frac{LD}{k-2} \le\left(\varepsilon+\frac{3C\theta}{A^2}\right)s.\] Choose \(A\) so large that this is at most \(2\varepsilon s\).

Let \(\mathcal L_0\) be the at least \(L/100\) lines in the uniformity assumption. If such a line contains fewer than \(sk/(200L)\) points of \(S_*\), it contains at least \(sk/(200L)\) points of \(S\setminus S_*\). But this latter set has at most \(2k|S\setminus S_*|\le4\varepsilon sk\) incidences. Thus at most \(800\varepsilon L\) lines of \(\mathcal L_0\) fail to contain \(sk/(200L)\) points of \(S_*\). With \(\varepsilon\) small enough, at least \(L/200\) lines succeed.

All points of \(S_*\) are critical or flat by Lemma 56. On each successful line one of these two kinds occurs at least \(sk/(400L)\) times. By (131) this is more than \(3D\), on enlarging \(A\). Every successful line is consequently a critical line or a flat line. Lemma 56 and the last bound in (131) show that fewer than \(L/400\) of them are critical or flat on nonplanar components. At least \(L/400\) therefore lie in plane components of \(Z(f)\). There are at most \(D\) such components, so one contains at least \(L/(400D)\) lines. It follows that \[B_0\ge\frac L{400D}\ge c\frac{sk^3}{L}, \qquad s\le C LB_0k^{-3}.\] Together with the alternative to (130), this proves the uniform version with an absolute constant \(C_0\).

We remove the uniformity assumption by induction on \(L\), keeping this same constant \(C_0\). Let \(\mathcal L_1\) consist of the lines containing at least \(sk/(100L)\) points of \(S\). If \(|\mathcal L_1|\ge L/100\), we have already finished. Otherwise the other lines supply fewer than \(sk/100\) incidences, and at least \(9s/10\) points of \(S\) lie on at least \(9k/10\) lines of \(\mathcal L_1\). Partition these points according as their multiplicity in \(\mathcal L_1\) is at least \(k\) or less than \(k\). One class has at least \(9s/20\) points. Its multiplicities lie between \(k_1\) and \(2k_1\), where either \(k_1=k\) or \(k_1=\lceil9k/10\rceil\ge3\). The induction hypothesis applies with \(L_1=|\mathcal L_1|<L/100\) and the unchanged plane cap \(B_0\). Consequently \[s\le\frac{20}{9}C_0 \left(\frac{L_1^{3/2}}{k_1^2} +\frac{L_1B_0}{k_1^3}+\frac{L_1}{k_1}\right) \le C_0\left(\frac{L^{3/2}}{k^2} +\frac{LB_0}{k^3}+\frac Lk\right),\] because \((20/9)(1/100)(10/9)^3<1\). Empty configurations start the induction. Finally divide all \(k\)-rich points into multiplicity intervals \([2^jk,2^{j+1}k)\) and sum this uniform-multiplicity bound. The three resulting geometric series converge. ◻

Proof. There are finitely many isometries with at least two matches: two distinct ordered matches determine at most two isometries. Separate the two orientation classes and reflect one cloud for the reversing class. A generic rotation of \(V\) makes \(I+R\) invertible for the linear part \(R\) of every counted direct isometry. Write \(J(x_1,x_2)=(-x_2,x_1)\). In its planar Cayley chart a direct isometry is represented by a point \((t,c)\in\mathbb R\times\mathbb R^2\), and a match \(x\mapsto y\) is represented by the line \[ c=\frac{x-y}{2}-tJ\frac{x+y}{2}. \tag{132}\] Indeed, putting \(Z=tJ\), the match equation is equivalent to \((I+Z)y=(I-Z)x-2c\), and \((I+Z)^{-1}(I-Z)\) is orthogonal of determinant one. Conversely every direct isometry with \(I+R\) invertible has this representation.

The \(uv\) lines are distinct: their intercept and slope determine \(x-y\) and \(x+y\), hence both endpoints. A plane in \((t,c)\)-space has an equation \(a\cdot c+bt=d\). If \(a=0\) it contains no such line. Otherwise, for a line (132) contained in the plane, \[a\cdot(x-y)=2d,\qquad a\cdot J(x+y)=2b.\] With \(x\) fixed these determine \(y\) uniquely, because \(a\) and \(J^Ta\) are independent; the same holds with \(y\) fixed. Thus the plane cap is \(B_0\le\min(u,v)\). Apply Theorem 57. Its middle term is bounded by its first because \(B_0\le\sqrt{uv}\le k\sqrt{uv}\), and its last is bounded by its first because \(k\le\min(u,v)\le\sqrt{uv}\). The bounded orientation split completes the proof. ◻

Proof. Planes are ruled. An irreducible complex projective quadric has rank three or four: rank three gives a cone over a conic, and in rank four a linear coordinate change gives \(X_0X_3=X_1X_2\), whose two rulings follow by fixing the ratio in either pair of factors. Hence we may assume \(d\ge3\). Write its irreducible defining polynomial as \(F\). After a linear coordinate change, \(A=F_z\) is not identically zero. Put \(B=F_x\), \(C=F_y\) and form the tangent vector \[V(u,v)=(Au,Av,-Bu-Cv).\] The binary forms \[q(u,v)=D^2F[V,V],\qquad c(u,v)=D^3F[V,V,V]\] are obtained by differentiating \(F\) only, with the displayed vector held fixed. Their coefficient degrees are at most \(3d-4\) and \(4d-6\). Let \(R\) be their homogeneous binary resultant. The Sylvester determinant is homogeneous of degree three in the quadratic coefficients and degree two in the cubic coefficients; therefore \[ \deg R\le3(3d-4)+2(4d-6)=17d-24. \tag{133}\] The resultant vanishes precisely when the two binary forms have a common projective root; this includes identically zero forms.

Every affine line in \(Z(F)\) lies in \(Z(R)\). At a point where \(A\ne0\), use its tangent direction, for which all directional derivatives vanish. At a point where \(A=0\), the two forms become \[q=F_{zz}(Bu+Cv)^2,\qquad c=-F_{zzz}(Bu+Cv)^3,\] so they again have a common projective root. This also handles a line lying entirely in \(Z(F_z)\).

We next prove that \(F\mid R\) implies ruledness. Let \(K=\mathbb C(Z(F))\). This field is a finite separable extension of \(\mathbb C(x,y)\). Extend \(\partial_x,\partial_y\) to an algebraic closure of \(K\), where they commute, and regard \(z\) as an algebraic function of \(x,y\). Write \[r=z_{xx},\qquad s=z_{xy},\qquad t=z_{yy}.\] If this Hessian is zero, \(D=\partial_x\) satisfies \(D^2x=D^2y=D^2z=0\), and the formal argument below already gives a line. Otherwise a constant invertible linear change of \(x,y\) ensures \(t\ne0\). Thus the binary quadratic of second derivatives does not vanish in the infinite-slope direction.

Since \(F\mid R\), there is a finite common tangent slope \(a\) in the algebraic closure of \(K\). Put \(w=z_x+az_y\). Implicit differentiation, keeping \(a\) fixed in the directional derivatives, gives \[\begin{align*} D^2F[(1,a,w),(1,a,w)]&=-F_z Q(a),\\ D^3F[(1,a,w),(1,a,w),(1,a,w)] &=-F_z P(a)-3Q(a)(F_{xz}+aF_{yz}+wF_{zz}), \end{align*}\] where \[Q(a)=r+2sa+ta^2,\qquad P(a)=z_{xxx}+3az_{xxy}+3a^2z_{xyy}+a^3z_{yyy}.\] Consequently \(Q(a)=P(a)=0\).

Set \(D=\partial_x+a\partial_y\). If \(a\) is a simple root of \(Q\), differentiating \(Q(a)=0\) gives \[0=D(Q(a))=P(a)+2(s+ta)Da,\] so \(Da=0\). If it is a repeated root, then \(s=-ta\) and \(r=ta^2\). Use the mixed-partial identities \(r_y=s_x\), \(s_y=t_x\): \[t_x=-t_ya-ta_y,\qquad t_ya^2+2taa_y=-t_xa-ta_x =t_ya^2+taa_y-ta_x.\] Their difference gives \(t(a_x+aa_y)=0\), so \(Da=0\) in this case as well. It follows that \[Dx=1,\quad Dy=a,\quad Dz=w,\qquad D^2x=D^2y=D^2z=0.\] The formal Taylor map \(\exp(TD)=\sum_{j\ge0}T^jD^j/j!\) is an algebra homomorphism in characteristic zero: the product rule proves this coefficient by coefficient. Apply it to \(F(x,y,z)=0\) to obtain \[F(x+T,y+aT,z+wT)=0.\] Thus the geometric generic point of the surface lies on a contained line. The set of lines on its projective closure is a closed subset of the Grassmannian: substituting the line parameterization in the homogenized equation gives polynomial equations for it. The projective point–line incidence variety has closed image in the surface. Since its image contains the geometric generic point, it is the whole surface. This proves ruledness.

If the surface is not ruled, \(F\) and \(R\) are therefore relatively prime, and Bézout with (133) bounds their common lines by \(17d^2\). ◻

Proof. The algebraic set of lines on a nonplanar irreducible surface has dimension at most one. Otherwise its incidence variety would have dimension at least three, and through a generic smooth point of \(S\) there would be infinitely many contained lines. All would lie in the tangent plane. Infinitely many distinct lines through that point fill the tangent plane algebraically, making this plane a component of \(S\), a contradiction. Each curve component of the line variety covers \(S\): its union is closed by projectivity, and infinitely many distinct lines cannot be contained in a proper closed subset of an irreducible surface. Indeed that subset has finitely many curve components. All other components of the line variety are isolated points and thus represent finitely many lines.

Suppose a generic point had at least two generators. Fix a generic generator \(\ell\) from a component \(\Gamma_1\). A dense set of points of \(\ell\) then belongs to a second generator distinct from \(\ell\). These second generators are infinitely many, since each meets \(\ell\) in only one point. Some curve component \(\Gamma_j\) therefore has infinitely many members meeting \(\ell\), and closedness makes every member meet \(\ell\). For each \(j\), the set of members of \(\Gamma_1\) meeting every member of \(\Gamma_j\) is closed: it is the intersection of the closed meeting loci for the individual lines. These finitely many closed sets cover a dense subset of \(\Gamma_1\). Irreducibility makes one of them all of \(\Gamma_1\). Call the corresponding component \(\Gamma_2\). This includes the case \(\Gamma_2=\Gamma_1\); the diagonal alone was not counted as a second generator.

If three generic members of \(\Gamma_1\) are pairwise skew, they determine a unique smooth quadric, and their common transversals are exactly its opposite ruling. To see this directly, take the first two lines as complementary coordinate two-planes in \(\mathbb C^4\) and the third as the graph of an invertible linear map between them. A transversal is the projectivization of the span of \(u\) and its image, with \([u]\in\mathbb P^1\); their union is a nonsingular quadric. Hence \(\Gamma_2\) would cover that quadric and \(S\) would equal it.

If generic pairs in \(\Gamma_1\) are not skew, every pair meets, because the meeting condition is closed in \(\Gamma_1\times\Gamma_1\). A family of pairwise meeting lines is coplanar or concurrent. Indeed, fix two meeting at \(p\) and spanning a plane \(H\). Every line meeting both is contained in \(H\) or passes through \(p\); irreducibility of the family forces one alternative throughout. The first makes \(S\) a plane. The second makes it a cone with vertex \(p\). Such a nonplanar cone contains no line \(\ell\) avoiding \(p\): all the joins from \(p\) to the points of \(\ell\) would give the entire plane spanned by \(p\) and \(\ell\) in \(S\). It consequently has a unique generator through a generic point, contradicting the assumption of a second generator. This proves generic uniqueness. It also proves uniqueness of \(\Gamma\), since every curve component covers \(S\).

Call a point or line exceptional if it has the corresponding infinite-incidence property in the statement. If \(p\) is an exceptional point, the closed subfamily of lines through it contains the whole generator curve. Thus all generators pass through \(p\). There cannot be two such points: a generator through both would be their unique joining line. Hence there is at most one exceptional point.

An exceptional line meets every generator, since infinitely many generators meet it and \(\Gamma\) is irreducible. Three pairwise skew exceptional lines would force \(S\) into their quadric as above. Two exceptional lines cannot meet: all generators meeting both would lie in their plane or pass through their common point. The former makes \(S\) planar, while the latter makes the common point exceptional and prevents either line from meeting infinitely many generators away from it. Thus there are at most two exceptional lines. Finally, a contained line outside \(\Gamma\) has all its points covered by generators. Infinitely many generators meet it at distinct points, and removing the possible one exceptional point does not change this fact. It is therefore exceptional. ◻

Proof. First suppose the projective closure of every irreducible component is neither a plane nor a smooth projective quadric. Across all the components there are at most \(d\) exceptional points and at most \(2d\) exceptional lines by Lemma 60. Intersections involving an exceptional line cost at most \(2dM\) points. The exceptional points themselves cost at most \(d\).

Fix any remaining line \(\ell\). Choose a generic complex plane \(H\) through it and restrict the square-free equation of the entire union to \(H\). Divide out the entire power of the linear factor defining \(\ell\). The residual plane curve \(C_H\) has degree at most \(d-1\) and does not contain \(\ell\). Dividing out the entire power is necessary when \(\ell\) is a singular line of a component.

We claim that every remaining intersection \(q=\ell\cap\ell'\) belongs to \(C_H\). The nonexceptional line \(\ell'\) belongs to the generator curve of some component \(S_j\). Take \(H\) not containing \(\ell'\). If only finitely many generators of \(S_j\) meet \(\ell\), then the map sending a generator to its intersection with \(H\) is regular near \(\ell'\). On a dense open subset its values lie in \(C_H\). As \(C_H\) is closed, its value at \(\ell'\) is also in \(C_H\), which says \(q\in C_H\). This argument applies even at a singular point of the generator curve.

The alternative of infinitely many generators meeting \(\ell\) cannot occur at the present \(q\). If \(\ell\subset S_j\), either these meetings range through infinitely many nonexceptional points, making \(\ell\) exceptional, or all generators pass through a single exceptional point on \(\ell\). In the second case every generator other than \(\ell\) meets it only at that point. If \(\ell\not\subset S_j\), the finite set \(\ell\cap S_j\) contains all the meetings; irreducibility then forces all generators through one point of this finite set, which is exceptional. Both cases contradict the choice of \(q\). This proves the claim. Bézout in \(H\) now bounds the number of such \(q\) on \(\ell\) by \(d-1\). Summing and restoring the exceptions bounds the total by \((d-1)M+2dM+d\le4dM\).

For the general union, extract all lines in each plane component or component with smooth projective quadric closure, one component at a time. A smooth real projective quadric having real lines gives a regulus; a quadric without real lines contributes nothing. If an extracted group has \(a_j\) lines, then \(a_j\le s\) and its internal intersections number at most \(a_j^2/2\le s a_j/2\). Every line left after its extraction is not contained in that component, so it meets it in at most the degree of the component points. All such cross intersections cost at most \(dM\). The remaining components and lines are covered by the preceding \(4dM\) estimate. Since the extracted groups are disjoint, their internal contributions sum to at most \(sM/2\). ◻

Proof. Write \(H(m,s)=m^{3/2}+ms\) and induct on \(m\), keeping the same parameter \(s\) throughout. Choose \(K\) larger than all the absolute constants required below. The case \(m\le1\) is immediate. Suppose, towards a contradiction, that \(P_2>KH(m,s)\). Repeatedly delete a line having fewer than \[T=\frac K4(\sqrt m+s)\] distinct intersection points with other currently retained lines. Each deletion loses fewer than \(T\) two-rich points. The remaining set of \(M\le m\) lines therefore has more than \(3KH(m,s)/4\) two-rich points and is nonempty. Every remaining line has at least \(T\) distinct intersections with the other remaining lines.

Select each remaining line independently with probability \(64/K\). For each fixed remaining line choose one witness line at every distinct intersection point. These witnesses are distinct. The number \(X\) of selected witnesses is a binomial random variable with mean at least \(16\sqrt m\). The elementary binomial tail bound \(\Pr(X<\mathbb EX/2)\le\exp(-\mathbb EX/8)\) gives \[\Pr(X<8\sqrt m)\le e^{-2\sqrt m}.\] For completeness, this tail bound follows by applying Markov’s inequality to \(e^{-\lambda X}\) and the inequality \(1+p(e^{-\lambda}-1)\le\exp(p(e^{-\lambda}-1))\), then taking \(\lambda=\log2\); the resulting exponent is at most \(-\mathbb EX/8\). A union bound over all remaining lines costs at most \(me^{-2\sqrt m}\le e^{-2}<1/4\). Markov’s inequality also shows that with probability at least \(1/2\) the number \(q\) of selected lines is at most \(128M/K\). There is therefore a choice with both properties. It necessarily has \(q\ge1\).

Set \(D=\lceil\sqrt{6q}\rceil\). A polynomial of degree at most \(D\) has \(\binom{D+3}{3}\) coefficients, and vanishing on a line imposes at most \(D+1\) linear conditions. Since \((D+3)(D+2)>6q\), there is a nonzero real polynomial vanishing on every selected line. Moreover \[D<4\sqrt q,\qquad D^2\le2048m/K,\] and \(D<8\sqrt m\) for our choice of \(K\). Each remaining line contains at least \(8\sqrt m\) distinct points on the selected lines, so the polynomial vanishes identically on all \(M\) remaining lines. Replace it by its square-free part.

Factor the polynomial over \(\mathbb C\). Retain the real, absolutely irreducible, geometrically ruled factors. At most \(b\le18D^2\) of the \(M\) lines fail to lie on their union. Indeed, Lemma 59 bounds the lines on each nonruled factor of degree \(d_i\) by \(17d_i^2\). A real line on a factor which is not real up to a scalar also lies on its distinct conjugate; Bézout bounds such lines by \(d_i^2\). Summing these bounds and using \(\sum d_i\le D\) proves the assertion. Thus \[b\le36864m/K\le m/4.\]

Every one of these \(b\) lines meets the retained ruled union in at most \(D\) points, since it is not contained in that union. Lemma 61 bounds intersections among the ruled lines by \(5Dm+sm/2\). The induction hypothesis, with the original cap \(s\), bounds intersections among the \(b\) other lines by \(KH(b,s)\). Altogether the remaining configuration has at most \[6Dm+sm/2+KH(b,s) \le(49+K/4)H(m,s)\] two-rich points. Here \(D<8\sqrt m\) and \(H(b,s)\le H(m,s)/4\) were used. Taking \(K>98\), in addition to the preceding requirements, contradicts the lower bound \(3KH(m,s)/4\). This completes the induction. ◻

Proof. Distinctness and the plane bound were proved in Corollary 58. Fix \(x\in U\) and allow the other endpoint \(y\) to range over all of \(\mathbb R^2\). These lines partition \((t,c)\)-space: at every point their equation uniquely determines \[y=(I+tJ)^{-1}\bigl((I-tJ)x-2c\bigr).\] Their nonparallel directions and lack of intersections mean that they are pairwise skew even projectively. The polynomial vector field \[W_x(t,c)=\bigl(1+t^2,(tI+J)(c-x)\bigr)\] is nowhere zero over \(\mathbb R\) and tangent to this family. Indeed along its line through \((t,c)\) the derivative of \(c\) with respect to \(t\) is \((tI+J)(c-x)/(1+t^2)=-J(x+y)/2\).

Let a regulus have irreducible quadratic equation \(f=0\). Every line in this fixed-\(x\) family contained in the regulus is a common line of \(f\) and \(W_x\cdot\nabla f\). The latter has degree at most three. If there are at least seven such lines, Bézout forces \[W_x\cdot\nabla f=h f\] for a polynomial \(h\). Every member of the fixed-\(x\) family through a real point of the regulus is then entirely contained in it. To verify this without an appeal to invariance of flows, restrict to that line and write \(g(t)=f(t,c(t))\). The identity becomes \[(1+t^2)g'(t)=h(t,c(t))g(t).\] A nonzero polynomial \(g\) cannot have a real root: at a root its derivative has strictly smaller vanishing order, while \(1+t^2\) does not vanish. The chosen line has such a root, so \(g\) is identically zero.

Thus this fixed-\(x\) family supplies a full real affine ruling of the regulus. Its lines are pairwise projectively skew, so they all belong to the same ruling. Conversely, coverage of every real affine point implies that no real affine line of that ruling is missing. A ruling line contained in the plane at infinity is irrelevant here. Different fixed-\(x\) families share no line, because the slope and intercept recover both endpoints. There are only two rulings, so at most two values of \(x\) contribute seven or more lines. They contribute at most \(2v\) lines in all; the others contribute at most \(6u\). This proves the claimed cap. ◻

Proof. Two distinct ordered matches determine a unique orientation-preserving planar isometry whenever the matched segments have the same positive length. There are consequently finitely many direct isometries with at least two matches from \(Q\) to itself. Rotate a second copy of \(Q\) by a generic rotation \(R_0\) to form \(Q'=R_0Q\). We may ensure that none of the direct isometries with two matches from \(Q\) to \(Q'\) has linear part \(-I\): each is \(R_0g\) for one of the finitely many just described, and each excludes only one rotation \(R_0\). All these motions therefore lie in the Cayley chart used in (132). Translations lie in this chart as well.

Use the \(n^2\) lines for the pairs \((x,y)\in Q\times Q'\). Lemma 63 gives plane cap \(n\) and regulus cap \(8n\). Theorem 62 bounds the two-rich points by \(Cn^3\). For \(3\le k\le n\), Theorem 57 bounds the \(k\)-rich points by \[C\left(\frac{n^3}{k^2}+\frac{n^3}{k^3}+\frac{n^2}{k}\right) \le C'\frac{n^3}{k^2}.\] Multiplicity equals the number of matches, by the bijective Cayley representation, and it is at most \(n\).

For each positive distance \(\rho\) let \(a_\rho\) be the number of ordered pairs of distinct points of \(Q\) at distance \(\rho\). The corresponding number for \(Q'\) is also \(a_\rho\). Every ordered pair of equally long nonzero segments, one from each cloud, gives the unique direct isometry described above. A motion with \(k\) matches contributes exactly \(k(k-1)\) such pairs. Splitting multiplicities into intervals \([2^j,2^{j+1})\), \(j\ge1\), and using the preceding bounds yields \[\sum_\rho a_\rho^2 =\sum_g k_g(k_g-1) \le C n^3\log(2n).\] Since \(\sum_\rho a_\rho=n(n-1)\), Cauchy–Schwarz gives \[|\Delta(Q)|\ge \frac{n^2(n-1)^2}{Cn^3\log(2n)} \ge c\frac n{\log(2n)}.\] ◻

Proof. For an occupancy \(n\ge2\), Theorem 64 gives \(n\le CM\log(2n)\). The elementary inequality \(\log(2n)\le C\sqrt n\) first yields \(n\le C'M^2\); substitute this back into the logarithm to get \(n\le C''M\log(2M)\). Occupancies zero and one are harmless. If \(M=o(B^2)\), eventually \(M\le B^2\), so the uniform bound is \(O(B^2\log(2B))\), which is \(B^{2+o(1)}\). ◻