The higher-dimensional Erdős distinct-distances conjecture

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 B.13. 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 B.2 therefore bounds the number of physical lines with at least t points by

If L_{\max}=o(N) is the maximum occupancy, a dyadic sum gives

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 6.1. Fix such H,K and a positive-density subsequence.

We prove (2.2) 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

ordered pairs i\ne j have equal distance to q. As the z_i lie in a common radius class,

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

This is (2.2), 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 2.1 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 2.2 bounds the assigned physical partners by C a_j A^j. The component Hilbert lower bound in Lemma 3.6, applied at the defining degree D, gives

There are at most d groups, and D\asymp A. Therefore

with a fixed c_1>0, uniformly for every subset S.

Take a maximal undirected subset of the edges from Proposition 2.2 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 (2.3) 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 A.1, take e=d+1, so r=d. Physical separators from Lemma 3.1, 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 (2.7) belong to its underlying isotropic vector space. Polarization gives, for any two of them,

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 A.1 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

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

or equivalently

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 A.1 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

There are only d+1 groups. Since D\asymp A,

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

Choose \theta so small that this is less than the right side of (2.9).

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 2.5 applied to the nonzero restriction f|_F.□

Proof. For p\in P define

Every denominator is nonzero. The stated values on the real sample follow even though we subsequently regard s_p as a polynomial over \C.

We prove (3.2) 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

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

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 [25, Tags 00OQ, 00NQ, and 02JN]. Their Koszul complexes are therefore free resolutions of R/(f) and R/(Z)=F, respectively [25, 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

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

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

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

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

This works for every such y, whether or not it is isolated.

On setting Y=z in (3.4), we obtain f(Z)=Q(z,Z)(Z-z). Lemma 3.2, 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

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

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 (3.3), 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 3.3 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 [25, 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

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

for k\ge1. This proves (3.5); 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

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=\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 3.3 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 \C would give

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 (3.6). 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

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

Take a fixed b large enough that bL-q\ge L and use (3.6). This proves (3.7).

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 3.5 supplies the same surjectivity at degree C_nL_*. The monomial argument and (3.8) 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 3.6 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 (3.5) 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 3.6, and keep the notation K=\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

These are a vector space and a linear map over \C, and

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

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_{\C}E_D. If J collects these b functionals, then

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

All these input polynomials lie in W_D. A relation over \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 \C, makes every coefficient zero. Thus

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 (3.11) proves (3.9). 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

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

The lower bound of Lemma 3.6 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 (3.12).

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

The Frobenius map is surjective on this field, so grouping monomials by their exponent vectors modulo p gives a unique expression

Every nonzero h_i has

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

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 (3.13) 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 \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\C with

Let R\subset\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 (3.14) 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 (3.14). 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 (3.14). 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 hth 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 [14, 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

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

Moreover

The Hilbert polynomial of this associated graded ring shows that the degree of E with respect to \mathcal O_E(1) is \mult_pX. Thus E^k=(-1)^{k-1}\mult_pX. Also H^k=a, as can be computed using general hyperplanes avoiding p. Expanding gives

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\PP^{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

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 \PP^{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 \PP^{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

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

Using the exact sequence

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,

Comparison of the leading coefficients proves (3.16).□

Proof. For a zero-dimensional irreducible variety, X is one reduced point, and a linear polynomial suffices. In positive dimension apply Lemma 3.12(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 \mult_pX: composition does not decrease order, and hypersurface multiplicity equals the order of its equation. Apply Lemma 3.10 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

Bézout bounds this local intersection number by \deg C. The final assertion follows from (3.15).□

Proof. Write C=\{F=0\} with F homogeneous and squarefree. A general first polar

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,

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

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

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 4.1 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

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 B.2, the number of these lines at dyadic occupancy t is

Each has pair weight at most t^2e^{-2}\exp(o(w)). Summing for t\ge\exp(\epsilon w) gives

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 5.2 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 3.1 gives

We treat lines first, then nonlinear curves.

Assigned lines. Centers on their assigned target line contribute at most 2M\sum_i n_i=o(N^2). Otherwise the projected line is itself a line component of C_p, and corresponds to a real plane H containing p and the assigned line. Put

Planes with S_H<A account for at most NAE=\delta N^2 pairs, because a center has at most E line components in C_p. For the other planes it suffices to prove \sum_{S_H\ge A}h_HS_H=o(N^2). They form a finite family determined by pairs of distinct assigned lines: one line has weight at most 2M=o(A).

If an assigned line in H has weight b, then

To prove this, choose plane coordinates with that line as the first axis and map a point (x_1,x_2) to (x_1,x_1^2+x_2^2). Every fiber has size at most two. A distance from an axis point (a,0) becomes a line equation v-2au=r^2-a^2 in these coordinates. There are at most bM distinct such lines, with different slopes for different axis points, and the incidence count is at least bh_H/2. The point-line estimate gives

For b above a sufficiently large fixed constant, absorb the h_H term. If h_H=O(M), use b\le2M; otherwise absorb bM as well and cube the remaining inequality. Both cases give (4.1). For bounded b, the case h_H\le1 is immediate. Otherwise choose two distinct sample points in H. Every point of P\cap H is an intersection of two circles whose radii belong to the at most M distance values; two distinct-center circles have at most two intersections. Thus h_H\le2M^2, which proves the bounded-b case too.

Group the planes by dyadic values h\le h_H<2h and s\le S_H<2s, where h,s\gtrsim A. For a dyadic line weight b, let N_b be the total assigned mass of all lines with weight in [b,2b). Equation (4.1) permits only b\le L=CM^2/h. Let F be the number of planes in this (h,s) group. A generic hyperplane slice turns the finite line arrangement into points and the relevant planes into lines, preserving exactly the needed incidences. A further generic planar projection permits use of the real point-line estimate. Applying it to each weight class and multiplying by its weight yields

Since L/s\le CM^2/A^2=o(1), the last term is absorbed, giving

The contribution of N/s to \sum hsF is at most CN B^{2+o(1)}(1+\log N)^2=o(N^2), using the plane cap and d\ge3. For the other contribution, Hölder’s inequality gives

Consequently

For each fixed b, summing over dyadic h, equivalently over inverse dyadic L\ge b, costs a bounded geometric sum. Then \sum_bN_b^2\le N^2, and \sum_{s\gtrsim A}s^{-2}=O(A^{-2}). The result is O(N^2M^2/A^2)=o(N^2), completing the line case.

Nonlinear curves when d\ge4. If \overline{\pi_pY_i} spans at most a projective plane, the affine span of Y_i has dimension at most three. A curve with three-dimensional span requires the center to lie in that fixed three-flat; the flat cap and (126.1) in the form \sum n_i\le N show that these pairs cost o(N^2). For a planar curve, centers in its own plane also cost o(N^2). It remains to consider planar curves viewed from outside their planes. Projection is then a projective isomorphism of the plane onto its image and preserves the curve’s degree.

At a center p, group curves by their common nonlinear projection \gamma\subset C_p, writing e_\gamma=\deg\gamma and W_{p,\gamma}=\sum_{i\text{ in the group}}n_i. Each individual mass is at most CMe_\gamma. For two different curves in this group, their planes are distinct and lie, with p, in a projective three-space. Their intersection is a projective line L.

For a fixed ordered pair of such curves, there are at most C e_\gamma possible centers. Here is the required stabilizer argument. Each perspective between the two planes fixes L pointwise and maps one curve to the other. Comparing two perspectives gives a projectivity of the first plane preserving its nonlinear curve and fixing L pointwise. In the affine chart with L at infinity, it has the form u\mapsto\lambda u+b. A nonidentity translation cannot preserve an irreducible nonlinear curve: the infinite translation orbit of any point lies on a line, whose closure would be contained in that curve. Thus the scalar homomorphism on this stabilizer has trivial kernel; the stabilizer is abelian because every commutator is a translation. If it is nontrivial, a nonidentity element has a unique affine fixed point c, fixed also by every commuting element. The group consists of homotheties about c. The orbit of a generic point is free and lies on its line through c, so Bézout bounds the group order by the curve degree. This argument also rules out an infinite group. Finally a perspective between distinct planes determines its center uniquely: it is the intersection of the joining lines of corresponding points, two suitably chosen pairs being enough. The claimed bound on centers follows.

Discard groups with W_{p,\gamma}\le KMe_\gamma, for a sufficiently large fixed K. Their total contribution is at most KNME=o(N^2). In every remaining group the off-diagonal ordered weight satisfies

If the surviving pair mass were at least \varepsilon N^2, weighted Cauchy–Schwarz, first in \gamma and then in p, would give

For each fixed ordered pair of distinct curves, degree preservation and the center bound give \sum_p e_\gamma^{-1}\le C. The left side is therefore at most C\sum_{i,j}n_i n_j\le CN^2, a contradiction since E=o(N).

Nonlinear curves when d=3. The degree and multiplicity analysis of curves sharing a radial projection here is closely related to the curve–cone estimates of Tidor, Yu, and Zakharov [26, Sections 10.2–10.3].

We first delete centers where \operatorname{mult}_pY_i>d_i/2. There are only boundedly many such centers for each curve. To see this, take a generic birational plane projection, preserving or increasing multiplicities at this finite list of points, as in Lemma 3.12. For a reduced plane curve of degree D, intersection with a general polar derivative gives

Indeed the intersection multiplicity at a point of multiplicity m is at least m(m-1), and m^2\le2m(m-1) for m\ge2. A nonlinear irreducible degree-two curve has no point of multiplicity greater than one; for all other degrees this bounds the number with multiplicity greater than half the degree by an absolute constant. The deletion therefore costs O(\sum n_i)=O(N) pairs. Projection to a line requires the curve and center to lie in one fixed plane; these pairs cost o(N^2) by the plane cap.

Group the remaining curves at p by their common nonlinear projection \gamma. Let t_i be the generic degree of \pi_p:Y_i\dashrightarrow \gamma in this group. The point-projection formula, obtained from the linear-system degree formula with its base-point contribution [14, Proposition 4.4], and the multiplicity deletion give

An ordered triple (i,j,k) in a group is called valid if each index occurs at most its corresponding map degree. We claim that for each fixed ordered triple of curve indices

Choose v general on Y_i, simultaneously for the finite list of centers in this sum. Distinct centers give distinct directions from v, since v can avoid all the lines joining pairs of centers. The map degrees t_i,t_j,t_k still refer to projection from each original center p; projection from v will be used to count the directions of those centers.

We first claim that, viewed from v, each of Y_j,Y_k projects birationally to a plane curve, and that the two plane images are distinct unless Y_j=Y_k. A general line joining v\in Y_i to a point of either target curve is tangent at neither endpoint. Otherwise, after fixing the tangent endpoint, the other nonlinear curve would lie in its tangent line. Thus the non-tangency hypothesis of Lemma 4.1 holds.

Failure of birationality would give a further point on the same target curve over a general joining direction. Equality of the two plane images for Y_j\ne Y_k would give a further point on the other target curve. In either case Lemma 4.1 puts Y_i and the first target curve in a common plane. At any center p contributing to the sum, the curves belong to one original nonlinear group. If the two curves are distinct, this is impossible: projection from a point in their plane has a line image, whereas projection from outside that plane is injective. If the first target is Y_i itself, the conclusion makes Y_i planar. The repeated first index then makes validity at that center require t_i\ge2, whereas a planar curve in a nonlinear group is projected from outside its plane and has map degree one. This proves the claim.

For a valid center p, the line pv consequently gives at least

distinct preimages other than v on the respective curves. They can be chosen generic on those curves by the original choice of v. Birationality identifies the normalizations of the projective closures of the curve and its plane image. These distinct generic smooth preimages therefore give distinct normalization points, hence distinct branches at the corresponding point of the plane image. Validity and (4.4) imply

If the two plane images are distinct, Bézout bounds the sum of their local multiplicity products at these distinct projected centers by d_jd_k. If they agree, then j=k, and the residual multiplicity is at least two: it is t_j\ge2 when i\ne j, and t_i-1\ge2 when all indices agree. Equation (4.3) bounds the sum of squared residual multiplicities by 2d_j^2. Both cases give (4.5).

Discard once more groups of mass W_{p,\gamma}\le KMe_\gamma. Their contribution is O(NME)=o(N^2). In a larger group, an invalid ordered triple must repeat an index whose map degree is at most two. Such an index has mass at most 4CMe_\gamma by (126.1) and (4.4). The weight of invalid triples is at most

Choose K so this is at most half of W_{p,\gamma}^3. If the remaining pair mass were at least \varepsilon N^2, Hölder in each row and then in the centers would imply

Indeed \sum_\gamma W_{p,\gamma}^3/e_\gamma^2\ge (\sum_\gamma W_{p,\gamma})^3/E^2, and the sum of the row cubes is at least their total cubed divided by N^2. On the other hand, (4.5) bounds this expression by C\sum_{i,j,k}n_i n_j n_k\le CN^3. For d=3 we have E=\delta N^{1/3}, so N^4/E^2 exceeds N^3 by an unbounded factor. This contradiction proves the proposition.□

Proof. Choose s_*\to\infty sufficiently slowly that

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

This is true initially by Lemma 3.1, 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 3.3 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

This is (5.2) 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

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

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

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

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

bound the sum of the fractions in (5.5) by

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

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, (5.8). 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 (5.9) 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

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

Take T'=T_s(b') at a time b' just before the cut starting the current plateau, and apply (5.9) 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 5.1. 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 (5.7).□

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 3.9, 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

By (5.8) 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 3.8 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

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 3.9 and 3.8. Each carrier remains a component of its own path system. On a generic transverse fiber, Lemma 3.5 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,

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 5.3 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

using (5.7). This is uniform over pieces. Every point is assigned at most once; pool repeated curves. Apply Proposition 4.3 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 5.3 and Lemma 5.4.

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 5.3, 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

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 4.2 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 (6.2) follows from (6.4).

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\PP^{d-1} before restricting the direction coordinate to our affine chart. Localizing the ideal of the rational image at \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_{\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 [25, 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

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

Multiplying the two weight bounds in (6.4) proves (6.3). No further generic-depth deletion is made when \mathcal J is chosen.□

Proof. On the initial full-dimensional interval, Lemma 5.4, 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 (6.5).

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 6.2 with any fixed \varepsilon>0. When k=d-1, apply Proposition 5.3 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

Borrow (6.5) 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\PP^{d-1} in these label coordinates. A positive mass of pairs requires a positive source mass. The carrier mass budget therefore gives

The resulting rank D_1T^{2d-2}\exp(-o(s)) is sufficient: on this plateau T\le D_2 to exponential accuracy and

Now suppose k<d-1. The images assigned to this group represent a positive amount of assigned product mass, since they carry \Gamma'. Apply (6.3) to this subfamily. Its degree is at most a, so

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 (6.5) 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

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 6.3 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 6.3; 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

where the fixed \xi>0 can be smaller than both gaps specified after (6.10). 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

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 3.8 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

Use (6.3) for this particular subfamily of the already fixed U. Its represented normalized pair mass is at most

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 4.1, 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

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.

Components, multiplicity, and incidence deletions. Proposition 5.2 gives

for every subcollection \mathcal A. Since \sum_i b_i=0 and |P_y|=N\exp(-o(s)), every subcollection of fixed positive assigned mass has degree at least e_j\exp(-o(s)).

Every Y\in\mathcal Y is an irreducible component of the fixed target path locus, by the component argument at the start of the proof of Proposition 5.3. Larger components elsewhere remain in the target equations.

Projection from p can drop the dimension of Y only if its generic fiber is a line. For a fixed general smooth y\in Y, this puts p in the proper affine tangent flat y+T_yY. The flat cap makes the proportion of such centers at most B^{-c_0+o(1)}, uniformly for each Y. Also, Corollary 3.13 puts centers where \operatorname{mult}_pY>\deg(Y)/2 in a hypersurface of bounded degree. The initial full-dimensional no-capture property of the center piece makes their proportion o(1): its initial cutoff is B^{1+o(1)}, well above every bounded degree. Average both conclusions with weights n_Y/|P_y|. They remove only o(1) of the center–component incidence mass.

At every retained incidence, the point-projection formula gives

where t_p(Y) is the generic map degree. For a fixed p, group components having the same projected image Z and define their depth to be the sum of their map degrees. If all retained groups have depth at most R\exp(\eta s), then any subcollection of degree a has projected degree at least

Thus it suffices to bound depth after negligible weighted incidence deletions. Fixed constants and the errors in (6.17) are absorbed by choosing \eta>0 sufficiently smaller than the desired \epsilon.

Whenever the deleted incidence mass is \alpha=o(1), Markov’s inequality leaves all but \sqrt\alpha of the centers with deleted target mass at most \sqrt\alpha. Every later subcollection of fixed positive mass still has positive mass after this deletion. This is the uniformity mechanism in the proposition. No union bound over the possible subcollections will be needed.

The first-cut bound. Use the system at the first plateau after a drop. Its equations have degrees at most D_1\exp(o(s)), and every current descendant is contained in this first system. The regularity conclusion of Proposition 5.3 makes its relevant carrier unique and smooth at all but o(|P_y|) assigned target points. For each such point q, centers p in its affine tangent flat form a B^{-c_0+o(1)} fraction of the source piece.

If, for a center and a current descendant Y, the generic joining ray were contained in the first system, then the same containment holds for every point of Y where the ray is defined, by polynomial identity. At an assigned regular point q the line would lie locally in its unique first carrier, so its direction would be tangent there. The preceding averaging therefore discards all such center–descendant incidences at negligible mass. On every other generic ray at least one first-system equation restricts nontrivially. Its degree bounds the number of distinct intersections with all current target components by D_1\exp(o(s)). In characteristic zero, the generic fibers defining the map degrees consist of distinct points. This proves the depth bound for (6.14). Only regularity of the carrier is used here; the original path equations need not themselves have full Jacobian rank.

Centers in the narrower window. Suppose v/s\to0. The center piece can be simultaneously isolated as zero-dimensional components of equations of degree

where the second error is signed. Fix Y\in\mathcal Y, and choose y general on it, also general for the projections from all finitely many actual centers. For each center, count the additional moving roots z\ne y,p on target components whose images agree with \overline{\pi_pY}. These points are generic on their components. Here this includes avoiding their intersections with the other members of the same finite family \mathcal Y. Their number is the group depth minus the one fixed root y.

Parametrize such pairs by

Impose the center-isolating equations on this parametrization. At every counted solution the image point p is isolated in the center system. At its root z, the target union is locally just the one member of \mathcal Y containing that root, by the preceding genericity condition. The generic ray has finite intersection with that component, so the fiber is locally finite. Thus the solution is isolated on the variety parametrized by (z,t). This variety has dimension l+1 and degree at most a dimensional constant times \sum_Z\deg Z; the pulled-back equations have degree at most 2L. The isolated-intersection form of Bézout therefore bounds the total number, over all centers, by

Possible larger components of the center equations away from their isolated sample points do not affect this count.

Divide (6.20) by |P_x|. Its value, to the stated accuracy, is

The fixed root adds only one; R=B^{1+o(1)}\to\infty. For every fixed logarithmic slack \eta>0, Markov’s inequality bounds the proportion of centers whose depth at this Y exceeds R\exp(\eta s) by \exp(-\eta s+o(s)). Weighted averaging in Y, followed by the row deletion already described, proves (6.15).

Centers in the wider window: the generic lists. Suppose v/s\to\infty, b_{j+1}>0, and l<d-1. Choose an interior center plateau just to the right of zero, before the first strictly positive center drop. Its carrier dimension is

All componentwise proper cuts of degree B\exp(Cs), with C fixed, are negligible for the center piece. To verify the degree comparison, choose a fixed positive interior testing time \tau on this center plateau. Since s=o(v), B\exp(Cs)=o(B\exp(\tau v)), and leave a second interior time for the pooling step of Proposition 5.3.

For each p\in P_x, fix one carrier X(p) of this center plateau which contains it. The distinct values form a source family \mathcal X, with disjoint assigned sets S_X=\{p\in P_x:X(p)=X\}. Apply the nonreal-flat deletion in Proposition 5.3 to this source assignment and to the fixed target assignment on \mathcal Y. Use it to omit every nonreal flat member from both input families, and omit any member of zero assigned mass. The removed assignments have source mass r_x=o(|P_x|) and target mass r_y=o(|P_y|). Discard every incidence whose center belongs to the removed source set or whose target component is removed. Their total normalized weight is at most

Call the surviving distinct families \mathcal X_0,\mathcal Y_0, keeping their assigned sets, and put

Both are at least N\exp(-o(\log B)), and every flat member of these families is real. When k=l, the degree and assigned-mass budgets, using v,s=o(\log B), give at most B^{d-l+o(1)} flat members in each family and normalized weight at most B^{-(d-l)+o(1)} for each one. Thus Lemma 4.2 applies with error scale \log B and, when k=l, with e=B^{d-l}. Its removed product weight has normalized incidence weight

in the incidence normalization fixed above.

The lemma supplies a list \mathcal Y_X\subseteq\mathcal Y_0 for each X\in\mathcal X_0 such that for every Y\in\mathcal Y_X, a general line through x\in X,y\in Y has at most

moving roots on that list. Using a smaller fixed exponent in the lemma, if necessary, ensures this displayed bound for all sufficiently large B. This is a bound on all possible generic list meetings, not just on the roots visible in the finite sample. For a fixed general x, a proper subvariety of an l-dimensional listed target has projected dimension less than l, so cannot supply roots over the generic direction image of Y. For a retained actual center p, use only the list indexed by its fixed carrier X(p), further restricted by the earlier incidence deletions at p.

We prove that excessive depth at an actual center is detected by a proper cut on X of degree B\exp(O(s)). Put

Because s=o(\log B), we have q=o(r) and r=B^{1+o(1)}. Fix X and Y\in\mathcal Y_X for the following construction.

The generic list bound does not yet control actual centers. We encode intersections with a joining line by two binary forms. Excessive retained depth will force a large rank decrease in their Sylvester matrix. Multiplicity compression will then place these centers in a proper cut of degree fitting the center plateau.

A fixed generic target point and two binary forms. Write the current target path locus as

This is the fixed full path locus, including components omitted from the generic-depth input family. Every retained target is an irreducible component of Z, as proved above. Choose y\in Y general with two sets of conditions. First, the generic list-root bound and the non-tangency conditions must hold for generic x\in X. A dense open subset of X\times Y has a dense open fiber over general y, so this is possible. Second, for each of the finitely many retained actual centers p assigned to X and each listed target Y' with l-dimensional image equal to \overline{\pi_pY}, require all points of the generic fiber over \pi_py to be affine, distinct, different from p, and outside every other irreducible component of Z. The generically finite map from Y' has a dense open image locus with these properties: every excluded proper subset of Y' has image of dimension at most l-1. Pulling back these open sets to Y and intersecting the finite list preserves a nonempty open set. Also avoid the finitely many actual centers. This fixes one y for all the specializations below.

Choose an integer L\ge\max(L_0,B) with L=B\exp(O(s)). Homogenize each path equation to its own degree and take all its homogeneous monomial multiples to degree L. Their span has projective base locus exactly V(f_1^{\mathrm h},\ldots,f_m^{\mathrm h}): at every projective point some monomial of each prescribed degree is nonzero. This base locus may contain extra components at infinity; we retain all of them. The closure of each retained affine target is still a component, since any larger component containing it would meet the affine chart and contradict the component property.

Parametrize the line through x and y by

For x\ne y, this identifies \mathbb P^1 with the line. The parameters of y,x, and infinity are respectively [1:0],[1:1],[0:1], independently of x. Every restricted equation is a binary form of degree L, whose coefficients are polynomials of degree at most L in x. The general line is not contained in the base locus. At the chosen smooth point y, outside its other components, such containment would put its direction in T_yY, contradicting the retained non-tangency condition.

Over K=\mathbb C(X), let H be the greatest common divisor of the restricted system and put g=\deg H\le L. After division by H the binary linear system has no base point. Two generic members of it have no common zero: choose one nonzero member, then choose the second to avoid its finitely many zeros. Consequently two generic constant complex linear combinations of the original degree-L forms restrict to binary forms F_x,G_x whose generic gcd is exactly H. The good choices form a nonempty algebraic open set over K and admit constant complex choices: expanding the coefficients of a nonzero polynomial over K in a complex-linearly independent basis shows that it cannot vanish on every tuple of complex constants. This also covers g=L, when the residual forms are constants.

The Sylvester matrix and specialization of its divisor. Let S(x) be the 2L-square matrix of

Its entries have ambient degree at most L in x, and its generic rank on X is \rho=2L-g. Indeed, after writing F_x=Hf and G_x=Hh with f,h coprime, the kernel consists of (hQ,-fQ) with \deg Q=g-1, and has dimension g (zero if g=0).

Consider an actual center p assigned to X at which the retained depth over \overline{\pi_pY} exceeds r. By the choice of y, at least r-O(1) distinct common roots of F_p,G_p are affine points of retained components, different from p,y, and each lies on no other component of the fixed base locus. Call these distinguished roots.

The generic gcd divisor specializes with its full length, even when X is singular at p. To see this explicitly, choose an irreducible algebraic curve through p whose general point lies in the open set where the preceding generic properties, including the gcd degree, hold. To obtain it, take successive general linear sections through p until the local dimension is one, avoiding every positive-dimensional component of the excluded closed set at each step. That closed set then has local dimension at most zero. Choose an irreducible curve component through p; it meets the good open set. Normalize it and use the discrete valuation ring at a point above p. Over its fraction field choose the gcd form, and scale it so its coefficients are integral and at least one is a unit. Its reduction H_0 is a nonzero homogeneous form of degree g. It divides both specialized forms. In fact, the quotients are integral: the content law for polynomials over a discrete valuation ring says that a primitive factor times a quotient of negative coefficient valuation cannot have all coefficients integral. Reducing the integral factorizations proves divisibility. Thus H_0 defines a common limiting divisor D_0 of length g on \mathbb P^1.

At most q distinguished roots lie in the support of D_0. For this assertion, pass to a finite extension splitting the generic gcd and a valuation ring above the chosen one. Each distinct generic root has a unique projective specialization, obtained by scaling its two coordinates to be integral with one a unit. Each physical root belongs to some fixed irreducible component of the projective base locus. Closedness forces its specialization to remain in that component. A distinguished point lies only on its own retained component, so a branch specializing to it must already have come from that retained component. It cannot be either immobile endpoint, whose fixed parameters specialize to y and p, nor a root fixed at infinity. The generic list bound permits at most q remaining distinct roots. Their specializations have at most q support points, regardless of the multiplicities carried by the gcd divisor.

If F_p,G_p are not both zero, their common divisor therefore has degree at least

This includes the case that exactly one form is zero: its common divisor with the other is the other degree-L form, and the matrix has rank L. In all these cases

for large B. If both forms vanish, S(p)=0 and the rank decrease is \rho\ge L\ge r, giving the same conclusion. Working throughout with binary forms of degree L records any loss of an affine leading coefficient as a root at infinity; hence that phenomenon introduces no exception to the comparison.

We have shown that depth exceeding r forces a matrix rank decrease of at least r/2. It remains to turn this into a proper polynomial cut of degree B\exp(O(s)), to which center no-capture applies.

Multiplicity compression and the center no-capture bound. Choose a \rho-square minor Q(x) nonzero generically on X. It has degree at most 2L^2 and has ambient multiplicity at least r/2 at every excessive actual center. For clarity, at such a center the selected submatrix has rank at most \rho-r/2. Constant invertible row and column operations put its value into rank normal form. At least \lceil r/2\rceil rows then have every entry in the ambient maximal ideal at p, so every determinant term belongs to its \lceil r/2\rceil-th power. No smoothness of X, or agreement between ambient generic rank and generic rank on X, is needed for this conclusion.

Choose a test point of X at which Q is nonzero. The multiplicity compression lemma, Lemma 3.10, produces a polynomial which vanishes at all these excessive centers, is nonzero at the test point, and has degree

It is consequently a proper cut on X. For each fixed Y, these cuts can be chosen separately on the center components X with Y\in\mathcal Y_X. Their degree bound is uniform. The componentwise pooling and no-capture statement on the chosen center plateau makes the captured center mass o(|P_x|). This is uniform in the choice of Y: a contrary sequence of targets would contradict the same no-capture statement, which allows arbitrary polynomials of the specified degree. Averaging with the normalized weights n_Y/|P_y| therefore removes only o(1) of all incidences.

On the remaining incidences, every group in the retained part of \mathcal Y_{X(p)} at p has depth at most r. Deleting further targets cannot increase depth. Apply the row deletion and (6.19). Since rounding r costs a factor 1+o(1), this proves (6.16) and completes the proof of the proposition.□

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

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 5.3 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 (6.21) and the envelope Hilbert bound give

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,

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 6.1. 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 B.2 gives

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

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 (7.3), applied after a generic projection, the number of H-rich lines of X is

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

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 (7.3),

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

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

For sufficiently large C_2, the assumption K\ge C_2m^{2/3} ensures K^2/m\ge2C_5\sqrt K. Combining (7.4) and (7.5), 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 7.4. 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 7.4 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. 1.

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

2. 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.

7.2.1 The two families of null three-flats

Use coordinates (w,c)\in\mathbb R^3\oplus\mathbb R^3, with Zx=w\times x, and equip this six-space with the nondegenerate quadratic form

Its associated symmetric bilinear form is

For a possible point pair (x,y)\in X\times Y, set v=(x+y)/2, b=(x-y)/2, and define

This is an affine three-flat. Its directions obey \delta c=-\delta w\times v, so \mathfrak q(\delta w,\delta c)=0. Its direction space is therefore maximal totally null and equals its \mathcal B-orthogonal space.

Fix one reversing motion, and choose positively oriented orthonormal coordinates with its axis \ker(I+S) equal to \mathbb R e_3. The perpendicular block S_\perp belongs to SO(2), and I+S_\perp is invertible. On the full graph y=Sx+h, there are constants v_0,\omega\in\mathbb R and d=(d_1,d_2)\in\mathbb R^2 such that

Here v_0=h_3/2, and (I-S_\perp)(I+S_\perp)^{-1}=\omega J. The map from the physical point x to (v_1,v_2,b_3) is an invertible real affine map: its linear blocks are (I+S_\perp)/2 and 1.

Define the motion flat by

It is another affine three-flat. Its directions satisfy

with \delta c_3 free, so they too are maximal totally null. If y=g(x), its intersection with the pair flat is the plane

These are the usual graph-dual planes of the transformed physical points (v_1,v_2,b_3). The orthonormal coordinates used to describe a particular G_g are only a means of writing its equations: within the normalized orientation class under consideration, every flat is regarded in the same original parameter six-space. Simultaneous rotations (w,c)\mapsto(Rw,Rc) preserve \mathfrak q, so the null properties just checked hold there for the whole family.

We record the needed distinctness assertions explicitly. First, different matched physical points give different planes H_{g,x}, since their transformed coordinates are distinct. Second, in a fixed pair flat F_{x,y}, whose free coordinates are w, the incident plane from a motion with unit axis a has equation w\cdot a=\omega. This affine plane determines (a,\omega) up to simultaneous sign. Since the quarter-turn operator J_a:x\mapsto a\times x on a^\perp also changes sign with a, the operator \omega J_a, and hence S, is determined. The match g(x)=y then fixes h. Hence different incident motions give distinct planes in F_{x,y}.

Finally, the whole affine flat G_g determines g. Its direction space intersects the pure-translation subspace \{\delta w=0\} in precisely the axis line \{\delta c\parallel a\}, recovering that unoriented axis. Choose either orientation of it. Projection of G_g to w-space recovers \omega; its direction coupling in (7.8) recovers v_0; and the affine constants then recover d. Equation (7.7) gives

and Sa=-a. These reconstruct the physical motion, independently of the arbitrary oriented coordinates used in the reconstruction. Thus the G_g are distinct as well.

7.2.2 Polynomial construction and descent

Suppose that one orientation class contains at least

general-type motions, where the absolute constant K_3 will be chosen large. Restrict to exactly T of them. Form the bipartite incidence graph between these motions and their point-pair labels. It has at least T\ell edges and at most m^2 label vertices. Repeatedly remove a motion vertex of degree below \ell/4, or a label vertex of degree below T\ell/(4m^2). The total number of edges removed in the first way is less than T\ell/4, and in the second way less than T\ell/4. A nonempty graph remains, in which the two minimum degree bounds hold. Every retained motion has between \ell/4 and 2\ell retained matches. Its retained physical cloud still has fewer than \eta_1\ell points on any line and fewer than \eta_2\ell on any plane.

An ambient polynomial of degree at most D has \binom{D+6}{6} coefficients. Vanishing on any one affine three-flat imposes at most \binom{D+3}{3} linear conditions. Thus a sufficiently large absolute multiple of T^{1/3}, rounded upward, is a degree D for which there is a nonzero polynomial f vanishing on all retained G_g. We may take

We used \ell\le m here; if this fails the motion bucket is empty. Choose K_3 sufficiently large that

Every retained pair flat then contains more than D distinct incident planes on which f vanishes. A nonzero polynomial of degree at most D on an affine three-space cannot vanish on more than D distinct planes: their distinct affine linear equations would all be factors. Therefore f vanishes on every retained pair flat as well.

We next show that every first partial derivative of f vanishes on every retained motion flat. For a polynomial h, define its raised gradient by

If h vanishes on a maximal null affine flat, its raised gradient at a point of that flat belongs to its direction space, because that space equals its orthogonal. At an intersection of three transverse planes H_{g,x_i} in G_g, the raised gradient of f therefore belongs to the directions of G_g and of all three incident pair flats. Its direction in G_g belongs to all three plane directions, whose intersection is zero. Hence df=0 there, so all ordinary first partial derivatives vanish at that point.

There are enough such points to force a derivative to vanish on the whole of G_g. Work in its transformed physical cloud of n_g\in[\ell/4,2\ell) retained points. Affine invertibility preserves all line and plane occupancy bounds. Since \eta_1\ell\le\delta_0 n_g, Lemma 7.2 gives at least \gamma_0\ell/4 anchor points, each with at least \gamma_0\ell/4 distinct outgoing projective directions. For a fixed such anchor, a projective line of directions corresponds to a physical plane through the anchor and contains at most \eta_2\ell of these distinct directions. The choice of \eta_2 permits a second application of Lemma 7.2. It gives at least c_7\ell^2 distinct physical planes through the anchor and two other noncollinear points, for an absolute c_7>0.

All these spanned planes are nonvertical in the coordinates (v_1,v_2,b_3), by the generic projection chosen above. Each is the graph of a unique affine function of (v_1,v_2). Graph duality in (7.9) therefore converts them into c_7\ell^2 distinct transverse triple-intersection points in the anchor plane H_{g,x}. They lie on the arrangement of at most 2\ell distinct lines obtained by intersecting this anchor plane with the other incident planes. A line in this arrangement contains at most 2\ell distinct intersection points with the other lines. Since each of the designated points lies on at least two arrangement lines, counting these incidences shows that at least c_8\ell distinct lines each contain at least c_8\ell designated points, for an absolute c_8>0.

Choose a fixed \varepsilon>0 smaller than both c_8/2 and \gamma_0/8. After K_3 has been fixed, increase C_{\mathrm{loc}} so that \ell\ge C_{\mathrm{loc}}m^{2/3} and (7.10) imply D<\varepsilon\ell. Each first derivative of f, whose degree is less than D, now vanishes on all the rich lines just found: its univariate restriction has more zeros than its degree. It consequently vanishes on the entire anchor plane, by the same distinct-factor argument in dimension two. There are at least \gamma_0\ell/4>D distinct anchor planes, so the derivative vanishes on G_g.

Every first derivative therefore vanishes on all retained motion flats, and the earlier propagation through the retained label degrees makes it vanish on all retained pair flats. The same argument applies to it, with strictly smaller degree and the same geometric configuration. Iteration implies that a nonzero constant derivative of f vanishes on a nonempty flat, a contradiction in characteristic zero. This proves the general-type bound O(m^3/\ell^{3/2}) for each orientation class.

7.2.3 Planar type

For two finite planar clouds of sizes a,a', Corollary B.6 bounds the number of planar isometries giving at least u matches, for an integer u\ge3, by

The corollary counts both orientations. Its proof uses the higher-richness Guth–Katz line theorem [16, Theorem 4.5] with a plane cap; that theorem has no regulus hypothesis.

Now take a planar-type motion in the local bucket. Choose a plane \Pi with at least \eta_2\ell matched source points and put \Pi'=g\Pi. Remove a richest ambient line from X\cap\Pi, and a richest ambient line from Y\cap\Pi'. Each deletion loses fewer than \eta_1\ell matches, by the non-line-type assumption. At least

matches remain. In particular e_X(\Pi),e_Y(\Pi')\ge\eta_0\ell.

Bucket these two effective sizes in [h,2h) and [h',2h'), where h,h' run dyadically from \eta_0\ell. Choose C_{\mathrm{loc}} also large enough that \eta_0\ell\ge C_2m^{2/3} and \eta_0\ell\ge3. Lemma 7.3 supplies at most O(m^4/(h^2(h')^2)) ordered supporting plane pairs in these buckets. For each fixed pair, choose the two richest lines once and for all, and choose orthonormal affine coordinates on the two planes. Write a,a' for the sizes of the reduced planar clouds, so a<2h and a'<2h'. A motion being counted restricts to a planar isometry with at least k_0=\lceil\eta_0\ell\rceil\ge3 matches. If k_0>\min\{a,a'\}, there are no such restrictions; otherwise (7.11), applied with u=k_0, bounds them by O((hh')^{3/2}/\ell^2). Each planar restriction extends to a spatial isometry in at most two ways, according to the image of the unit normal. The constants therefore remain absolute.

Multiplying and summing the two geometric dyadic series gives

The last inequality uses \ell\ge C_{\mathrm{loc}}m^{2/3}. Together with the general-type proof, this completes Lemma 7.4, taking K_{\mathrm{loc}} larger than the sum of the constants obtained for the two orientations and the planar-type estimate. The order of choices was \eta_2, \eta_1, then K_3, then C_{\mathrm{loc}}, and finally K_{\mathrm{loc}}.□

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

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

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

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

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 (7.13), 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 (7.13) gives a set

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

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

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 (7.14). 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

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 7.1. 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 (7.1) with this notation. This implies the asserted result by (7.2). 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

Choose C_{\mathrm{rich}} sufficiently large that

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

If no such t exists, the sum is empty. In a nonempty bucket set

Since t\le N^{1/3} and \alpha<2, we have 1\le r\le N, so Lemma 7.5 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 (7.12), 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 7.4 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

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

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 (7.18) is needed below.

Motions with substantial non-line-type mass. Let \mathcal G_0 consist of the motions in the global bucket whose retained local non-line-type counts have total at least k/4. For a fixed region pair and a local richness bucket, Lemma 7.4 bounds the number of non-line-type motions by K_{\mathrm{loc}}m^3/\ell^{3/2}. Their local match mass is therefore at most 2K_{\mathrm{loc}}m^3/\sqrt\ell. Summing the geometric series over \ell=2^i b gives O(m^3/\sqrt b) for each fixed region pair. The bound remains valid after restriction to our global bucket. There are at most (19r)^2 possible region pairs, whence

The equality N^3\sqrt A=N^4/A uses A=N^{2/3}.

Motions whose retained mass is mostly line type. For every remaining motion, at least k/2 retained matches belong to line-type local buckets. In each such region pair, choose one source line containing at least \eta_1\ell of its local matches, and retain those matches as assigned line matches. Since n_{R,R'}(g)<2\ell, this retains at least \eta_1/2 of that local count. Thus the assigned line-match mass W_g satisfies

A physical source point participates in at most one region-pair count for a fixed motion. Hence no assigned match is repeated. Pool assignments belonging to the same full source line L, and write its weight as w_{g,L}. Its target line is necessarily gL, and

Every line appearing on either side contains at least \eta_1b=q ambient sample points, because it was chosen in a local bucket with \ell\ge b.

Let \mathcal L_X,\mathcal L_Y be the full lists of these q-rich lines of the two clouds. These are finite lists of spanned lines. Equations (7.3) and (7.18) imply

Denote a common upper bound in (7.21) by L_*=C'N/b.

The diagonal weights can be removed without losing a fixed fraction. Indeed, by (7.15) and (7.20), W_g\ge\nu k\ge4A. Consequently

Sum this quantity over the remaining motions. For nonparallel source line pairs, there are at most L_*^4 prescriptions of an ordered source pair and its ordered target pair. By Lemma 7.6, each prescription has at most eight motions, and each product of weights is at most (2A)^2. Their total contribution is thus O(A^2L_*^4).

For a parallel ordered source pair, fix its positive separation \rho. Lemma 7.7 gives at most O(L_*^{3/2}) ordered target pairs at that separation, including all target parallelism classes. There are at most L_*^2 source pairs. For every prescription, the pooled selected weights are bounded by the full line-match weights in Lemma 7.6; with H=2A, its bound is O(A^3). Therefore the total parallel contribution is O(A^3L_*^{7/2}).

If \mathcal G_1 denotes the remaining motions, the two cases and (7.22) give

This counts both orientations. In the parallel case it sums the whole continuous translation family through its finite positive match support, rather than assuming a bounded number of motions for the two prescribed lines.

Summing the global buckets. From b=\beta A t^{-\alpha}, with its fixed positive \beta, the two terms in (7.23) are

Summing these increasing geometric series up to t\le N^{1/3} gives respectively

Both are bounded by O(N^{10/3})=O(N^4/A); in fact they save the powers N^{-4/7} and N^{-1/12}, respectively. The decreasing geometric series in (7.19) gives

Finally k_g<2k in each global bucket, so replacing its cardinality times k^2 by the sum of the actual k_g^2 costs at most a factor of four. The two types exhaust every bucket, proving (7.1).□

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 (A.4).□

Proof of Theorem A.2. Fix the polynomials from Lemma A.3. We will prove, for 0\le s\le c, that every component U of the first s cuts containing V satisfies

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 3.6 used here. There are constants a_n>0 and b_n\ge1 such that an irreducible k-dimensional variety X satisfies

and, if X is a component of a system whose equations have degrees at most m, then

The second assertion is the bound at the equation degree in (3.7), 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 (A.5) is known for smaller indices, and let U be a component at index s. If \delta_s=1, the entire product in (A.5) is 1, and \deg U\ge1 proves the claim. Otherwise set

Thus

Lemma A.3 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 (A.7) therefore give

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 (A.6) bounds H_U(m) above. Combining the bounds, then using (A.8), yields

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 (A.5), 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 (A.5) to it gives

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

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

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 [25, 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 [14, 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 [14, Example 12.3.7(ii)–(iii)] for this identification and its Cohen–Macaulay hypothesis. Hence

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 Lth power zero, so

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

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

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 [25, Tag 00NO], namely

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 A.5 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

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

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

Their z-components are tangent to V, by (A.12). 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 (A.13). 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 (A.13) 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

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

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

These expressions again have bounded degree in \tau and coefficient degrees O_e(D). Form the monic polynomial

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

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 (A.11). 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

□

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.

Compactifying the parameter space. On \mathbf{P}(E), let L=\mathcal O_{\mathbf{P}(E)}(-1) be the tautological line bundle and put Q=(E\otimes\mathcal O)/L. Evaluation of a skew form on the line L gives the exact sequence

Indeed, the value of a skew form on L is a map from L to its annihilator L^\perp\subset E^*, every such map extends to a skew form on E, and the kernel consists of the forms on E/L. The ranks of these three bundles are s_0,n,r, respectively.

All projective bundles below parametrize lines. Consider

The second presentation is obtained by tensoring the first bundle with L. We use the line convention of [14, Section 3.1 and Example 3.1.1]. Since L\oplus Q^* is a subbundle of the trivial bundle with fiber E\oplus E^*, it induces a morphism \mu:B\to\mathcal Q. Explicitly, write a point in the first presentation as (L,[t:\psi]), with \psi:L\to L^\perp. For any nonzero v\in L,

Both rescaling (t,\psi) and replacing v by another generator of L leave this point unchanged. Its quadratic value is zero because \psi(v) annihilates v.

Over \mathcal Q^\circ=\{[v:y]\in\mathcal Q:v\ne0\}, the inverse to \mu recovers L=\mathbb{C}v and [1:\psi], where \psi(v)=y. Thus \mu is an isomorphism over this open set. A boundary point [0:y] has fiber \mathbf{P}(\ker y): its preimages have t=0, and L can be any line annihilated by y.

Let \mathcal T=\mathcal O_B(-1) be the tautological line in the first presentation of (A.15), and set

The tautological line in the second presentation is \mathcal T\otimes\rho^*L. Consequently

The classes a and \ell are globally generated. We do not need positivity of b.

The projective incidence and its boundary. Pull back (A.14) to B. Define \mathcal U as the inverse image of \mathcal T under the surjective bundle map

It is a vector bundle of rank s_0+1, with exact sequence

Its inclusion in the trivial bundle defines a morphism \mathbf{P}(\mathcal U)\to\mathbf{P}(\mathbb{C}\oplus S). The pullback of the hyperplane class under this morphism is

For a parameter represented by (L,[1:\psi]), choose v\in L\setminus\{0\} and write y=\psi(v). The projective fiber maps isomorphically to

Over (L,[t:\psi]), the defining condition is instead (x_0,z|_L)\in\mathbb{C}(t,\psi). In particular,

Thus no boundary parameter occurs over an affine point [1:z].

Lift the parameter image of \Gamma to B using the isomorphism over \mathcal Q^\circ, and denote its reduced irreducible closure by \widetilde\Lambda. The map \mu:\widetilde\Lambda\to\Lambda is birational. Put

The variety X is irreducible and has dimension s_0+h+j=k+j. The point coordinate and the lifted parameter give an embedding of \Gamma as a locally closed subset of X: in its normalized direction chart the inverse recovers the original pair (z,v). Since the dimensions agree, this subset is dense and contains a dense open subset of X. The projective point map on X therefore has image exactly \overline V. More explicitly, its image is closed by properness; density shows that it is contained in the closure of the point image of \Gamma, and dominance of \Gamma\to\Omega gives the opposite inclusion. We obtain a surjective morphism

Here and below we also use a,b,\ell for their restrictions and pullbacks to the varieties on which they occur.

The degree of a generic point fiber. Fix an affine z\in S. By (A.20), the full incidence over [1:z], before restriction to \widetilde\Lambda, has parameters [v:zv] with v\ne0, and is isomorphic to \mathbf{P}(E). On this projective space \ell is the ordinary hyperplane class: the map [v]\mapsto[v:zv] is induced by an injective linear map E\to E\oplus E^*.

Choose a dense open subset X^\circ\subset X lying in the lift of \Gamma. Every component of its complement has dimension at most k+j-1. A component which dominates V therefore has generic fiber dimension at most j-1; a component which does not dominate V is absent over a general point. The complement cannot supply a top-dimensional component of the generic fiber of f.

When j=0, the generic fiber thus consists of at most C_e points, as does the fiber of \Gamma\to\Omega. When j=1, the generic fiber of \Gamma\to\Omega is an affine line of normalized vectors. Its closure in \mathbf{P}(E) is a projective line, of degree one against \ell, and this is the only top-dimensional generic fiber component of f. These components have multiplicity one. Indeed, the fibers on the dense open ruling incidence are smooth: its projection is finite étale or a smooth affine-line family. This determines the generic multiplicity of each component, which cannot be changed by a complement of smaller fiber dimension.

If d_\Gamma denotes the degree of the geometric generic fiber against \ell^j, we have proved

Proper pushforward and the projection formula, applied to fundamental cycles, now give

The coefficient in the first identity is computed on the generic fiber. Any other component of a representative hyperplane intersection has image of dimension less than k and pushes forward to zero. These formulas do not require smoothness of X or \overline V; see [14, Sections 1.4, 2.3, 2.5].

A positive Segre-class formula. For projective bundles of lines, the projective bundle formula is

This formula applies equally to a vector bundle on a singular base [14, Sections 3.1–3.3]. We compute it before restriction to \widetilde\Lambda.

Dualizing the tautological sequence on \mathbf{P}(E) and tensoring by L^*=\mathcal O(1) gives

Together with (A.14), this implies

Since c_t(\mathcal T)=1-bt, Equations (A.18) and (A.17) yield

Expanding

and taking the coefficient of t^h gives

For integers p\ge u+1 and u\ge0, these scalar coefficients are nonnegative. Indeed, c(p,0)=1, and c(p,1)=p-2\ge0 in its asserted range. For u\ge2, the identity (1+z)^2=(1+2z)+z^2 gives

The second term lies in the same range, because p-2\ge(u-2)+1. Induction proves the claim, with a binomial coefficient equal to zero when its lower index exceeds its upper index. In (A.25), the required inequality is e-1-i\ge h-i+1, which follows from h\le r-1=e-2. The coefficient of \ell^h is c(e-1-h,0)=1.

All mixed top intersections of a and \ell on \widetilde\Lambda are nonnegative. They can be computed by successive general members of globally generated linear systems, whose intersections with the fundamental cycle are effective or empty. Birationality of \widetilde\Lambda\to\Lambda, followed by (A.25), (A.23), and (A.22), therefore gives

The tangent constraint in the affine-line case. Suppose now that j=1. Lemma A.6 gives \operatorname{codim}_S V=1, so h=r-1 and \dim\Lambda=r. We first show that the tangent space of \Lambda at a general smooth point is maximal isotropic for the tangent quadratic form of \mathcal Q.

Choose a general smooth parameter \lambda=[v:y]\in \Lambda\cap\mathcal Q^\circ, fix z_0\in F_\lambda, and let U=E/\mathbb{C}v, of dimension r. With a representative (v,y) fixed, tangent vectors to \mathcal Q are pairs (\dot v,\dot y) satisfying \dot y(v)+y(\dot v)=0, modulo \mathbb{C}(v,y). There is an isomorphism

The covector \beta annihilates v, by skew symmetry and z_0v=y. Adding a multiple of (v,y) changes neither coordinate, and the kernel before quotienting is exactly that line, proving the isomorphism. The tangent quadratic form, defined up to scale on \lambda^\perp/\lambda, becomes \beta(\alpha), because

Let \Theta\subset U\oplus U^* be the resulting r-dimensional image of T_\lambda\Lambda.

The affine incidence over the smooth part of \Lambda\cap\mathcal Q^\circ is an affine bundle with fibers F_\lambda, and its point image lies in V by the preceding compactification. At a point z=z_0+D\in F_\lambda, with D\in K_v=\bigwedge^2U^*, differentiating zv=y gives

The vertical tangent space maps onto K_v. Evaluation at v identifies S/K_v with U^*, so modulo these vertical directions the differential is

Every parameter tangent vector lifts because the incidence is an affine bundle. At general points its differential has rank at most \dim V=n-1. Hence (A.28) has rank less than r for general D and general \lambda. Its determinant is polynomial in the entries of D; for each general \lambda it therefore vanishes for every skew form D on U.

Linear algebra of the singular skew pencil. We prove that this determinant condition forces \Theta to be isotropic. Put

If p=\dim A_*, then \dim B_*=r-p and \dim H_*=p. The plane \Theta defines a linear map

The map in (A.28) is the identity on B_*. On taking the quotient by this subspace, the determinant condition is precisely

Here the restriction is regarded as a map A_*\to H_*^*. If p=0, the original map is the identity on U^*, a contradiction. Thus p>0.

We claim that A_*=H_*. Otherwise let q=\dim(A_*\cap H_*)<p, and choose bases

All vectors displayed here are jointly linearly independent when the common vectors are listed only once. A restriction matrix of an ambient skew form on A_*\times H_* may therefore be chosen arbitrarily except that its common q-by-q block must be skew. Each such prescription extends to a skew form on A_*+H_*, and then to U.

There is an invertible restriction matrix of this kind. For even q, use a nondegenerate skew block on the common indices and the identity between the remaining row and column indices. For odd q, use a nondegenerate skew block on c_1,\ldots,c_{q-1}, the block

with rows (c_q,a_1) and columns (c_q,h_1), and an identity matrix on the unused extra indices. Its two off-diagonal entries pair different independent vectors and so can both be chosen equal to one without violating skew symmetry. In either case we obtain an ambient skew form D_0 with invertible restriction. The polynomial \det(C-tD_0|_{A_*\times H_*}) then has a nonzero leading coefficient, contradicting (A.29). This proves A_*=H_*.

If p were even, a nondegenerate skew form on A_*, extended to U, would give the same contradiction. Hence p is odd. We may now regard C as a bilinear form on A_*. If it is not skew, polarization in characteristic zero gives x\in A_* with C(x,x)\ne0. Choose a complement to \mathbb{C}x, and choose a skew form D_0 which annihilates x and is nondegenerate on this even-dimensional complement. Extend it to U. In a basis beginning with x, the coefficient of t^{p-1} in \det(C-tD_0|_{A_*}) is

To obtain this power of t, the determinant must use all remaining rows and columns from D_0, leaving the entry C(x,x). This also covers p=1, with the empty determinant equal to one. Again we contradict (A.29); therefore C is skew.

Finally B_*=A_*^\perp. For every (\alpha,\beta)\in\Theta we consequently have \beta(\alpha)=C(\alpha,\alpha)=0. Polarization shows that \Theta is totally isotropic. Its dimension is r in the nondegenerate 2r-dimensional space U\oplus U^*, so it is maximal isotropic. We have proved this tangent property on a dense open subset of the smooth locus of \Lambda.

Local rigidity of maximal null tangent spaces. We finish by showing directly that this tangent property makes \Lambda projectively linear. In the vector space defining \mathcal Q, choose a hyperbolic pair e_0,f_0, with orthogonal complement W, and normalize the quadratic form as

Choose the pair so that a general smooth point of \Lambda lies in the chart

The tangent quadratic form in this chart is q_0(dx): the derivative of the displayed lift is dx-(dq_0)_x(dx)f_0, with quadratic value q_0(dx). Shrink to a complex analytic neighborhood in the smooth locus on which the maximal null tangent property holds everywhere.

Linear coordinates (u,w)\in\mathbb{C}^r\oplus\mathbb{C}^r on W can be chosen so that

Here the dot denotes the standard complex bilinear pairing. For justification, choose a basis of the null tangent plane and a dual basis in a complementary plane. Subtracting half the Gram matrix in the first set of directions makes the complementary plane null as well. Sums and differences of these bases, with rescaling over \mathbb{C}, give the displayed model. The projection to the u coordinates is invertible on the tangent space at x_*, so the holomorphic inverse function theorem writes the local submanifold as w=F(u).

Write F_i=\partial_iF and F_{ij}=\partial_i\partial_jF. Null tangency says

Differentiating these identities and commuting mixed partials gives

The vectors F_1,\ldots,F_r are a basis, since their Gram matrix is the identity. Thus F_{ij}=0 for all i,j, and F is affine on a smaller connected neighborhood. The local submanifold in W is an open subset of an affine plane x_*+L_0, where L_0 is an r-dimensional totally isotropic linear subspace.

Let B_0(x,w)=q_0(x+w)-q_0(x)-q_0(w) be the polarization. For w\in L_0, q_0(x_*+w)=q_0(x_*)+B_0(x_*,w), so the lift in (A.30) is affine-linear on this plane. Its projective closure is the projectivization of

This vector space has dimension r+1: the first summand has nonzero e_0 coordinate, and the second maps injectively to L_0. Its quadratic form vanishes identically. Indeed, it vanishes on the displayed affine lift, and homogeneity extends this identity to its linear span. We have obtained a projective linear \mathbf{P}^r\subset\mathcal Q containing a nonempty analytic open subset of the smooth irreducible variety \Lambda.

Such an open subset is Zariski dense, since a proper algebraic subset has smaller dimension and cannot contain a smooth neighborhood. Hence \Lambda is contained in this \mathbf{P}^r, and equality follows from \dim\Lambda=r. This proves the exceptional assertion and completes the proof.□

Proof of Theorem A.1. Fix the sample, A, and one of Specifications (a) and (b). For 0\leq h<r, let R_h be the supremum of

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

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 A.2, 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 [25, 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

The later actual degrees may be smaller than their upper bounds; this only strengthens the relevant Bézout estimates.

Choose constants

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 A.6 supplies a proper exceptional section V\cap Z(H) with

Outside it, there are boundedly many ruling incidences. For one with j=0, Proposition A.7 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

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

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 (A.31),

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

Taking suprema in the two alternatives gives the finite system

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

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

For interior indices h<i<r,

If the ambient index is available, B_h\geq c_1K_hB_0 gives

The other transition satisfies

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 (A.35). 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 (A.2), 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 B.1 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

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

where A will be chosen sufficiently large. By Theorem B.2, this assumption implies s\le C L^2k^{-3}. Choose

where the constant \theta is sufficiently large. Thus r\ge1 and r is comparable to the expression inside the floor. Lemma B.3 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 B.2 inside it:

Substitution of r gives

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 (B.1) 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 (B.1) give

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

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 B.4. On each successful line one of these two kinds occurs at least sk/(400L) times. By (B.2) this is more than 3D, on enlarging A. Every successful line is consequently a critical line or a flat line. Lemma B.4 and the last bound in (B.2) 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

Together with the alternative to (B.1), 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

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

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 (B.3) contained in the plane,

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 B.5. 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

The binary forms

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

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

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

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

where

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

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:

Their difference gives t(a_x+aa_y)=0, so Da=0 in this case as well. It follows that

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

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 (B.4) 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 B.8. 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

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

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

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 B.7 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

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 B.9 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

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 B.6. 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

Their nonparallel directions and lack of intersections mean that they are pairwise skew even projectively. The polynomial vector field

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

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

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 (B.3). Translations lie in this chart as well.

Use the n^2 lines for the pairs (x,y)\in Q\times Q'. Lemma B.11 gives plane cap n and regulus cap 8n. Theorem B.10 bounds the two-rich points by Cn^3. For 3\le k\le n, Theorem B.5 bounds the k-rich points by

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

Since \sum_\rho a_\rho=n(n-1), Cauchy–Schwarz gives

□

Proof. For an occupancy n\ge2, Theorem B.12 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)}.□