advanced36 min studyLesson 619 in Mathematics

Incidence geometry and crossing arguments

Convert point–line incidences into graph edges and use the crossing lemma only in its valid density range.

On this page

What you’ll understand

  • Define finite point–line incidences without multiplicity ambiguities.
  • Explain the random-subgraph proof mechanism of the crossing lemma.
  • Derive the sparse and dense branches of a Szemerédi–Trotter incidence bound.

Counting how often points lie on lines becomes unexpectedly constrained when one connects consecutive points along each line. Too many resulting edges would force many crossings, but the original lines can intersect only once per pair. This tension produces the Szemerédi–Trotter bound. The proof works because the constructed graph is simple and because a quantitative crossing estimate is invoked only after its edge-density hypothesis is checked. Ignoring either condition turns an elegant argument into an incorrect universal formula.

From planar sparsity to a crossing lower bound

A drawing of a finite simple graph places distinct vertices in the plane and draws edges as curves with no unrelated vertex in an edge interior. We may use a minimum-crossing drawing in general position, with adjacent edges not crossing and no triple crossing. Deleting one edge per crossing leaves a planar simple graph, yielding the rough lower bound crossing number at least E − 3n. This estimate becomes weak when the original graph is dense.

Select every vertex independently with probability p and retain induced edges. Expected vertex and edge counts are pn and p²E. A crossing between two independent edges survives with probability p⁴ because four distinct endpoints must survive. Taking expectations in the rough planar bound and choosing p = 4n/E gives the crossing lemma with constant 1/64, provided E is at least 4n so the chosen probability is at most one. This is a proof sketch with the drawing simplifications stated explicitly.

cr⁡(G)≥E364n2for a simple graph with n>0, E≥4n\operatorname{cr}(G)\ge\frac{E^3}{64n^2}\qquad\text{for a simple graph with }n>0,\ E\ge4n
The crossing number is minimized over appropriate planar drawings. Below the edge-density threshold this particular cubic lower bound is not asserted.↔ Scroll wide equations horizontally.

Incidences create a simple geometric graph

Let P be n distinct points and L be m distinct lines in the real plane. An incidence is a pair consisting of a point and a line containing it. If either set is empty, the incidence count is zero and the bound is immediate; henceforth assume n and m are positive. Ignore lines with no incident point and call the number of remaining lines m′. Along each such line, join consecutive points of P. A line with k incident points contributes k − 1 edges, so the total edge count is I − m′.

The graph is simple because two distinct points determine only one line, and consecutive-point edges have no other P-point in their interior. Two supporting lines cross at most once, so the displayed drawing has at most m choose two crossings; intersections at graph vertices do not count as edge crossings. If several edges cross concurrently at a point outside P, a local perturbation separates them into ordinary pair crossings without exceeding the same bound from pairs of supporting lines. These facts rely on distinct lines and the real planar setting. Repeated line copies or arbitrary curves can create multiplicities requiring different estimates.

E=I−m′,m′≤m,cr⁡(G)≤(m2)⟹I≤C(n2/3m2/3+n+m)E=I-m',\quad m'\le m,\qquad \operatorname{cr}(G)\le\binom m2\quad\Longrightarrow\quad I\le C\bigl(n^{2/3}m^{2/3}+n+m\bigr)
The last conclusion uses an absolute constant C and combines a sparse-graph case with the crossing lemma in the dense case. Here m′ counts only lines containing at least one input point.↔ Scroll wide equations horizontally.

Keep both branches of the argument

If E is less than 4n, then I = E + m′ is less than 4n + m, already an acceptable incidence bound. Otherwise combine the cubic crossing lower bound with the quadratic upper bound from pairs of supporting lines. Solving gives E at most 32 to the power one-third times n to the power two-thirds times m to the power two-thirds. Adding m′ gives the dense contribution; retaining both branches explains the n + m terms.

The big-O notation hides an absolute constant, not a promise that the leading mixed-power term dominates for all n and m. A single line can contain all n points, so an n term is indispensable. Likewise one point can lie on all m lines, requiring an m term. The bound concerns a worst-case maximum; it does not predict the exact count in one configuration or imply that every configuration achieves that growth. Small examples are useful for checking the construction even when constants make the upper bound loose.

Check the range before using the crossing lemma

A finite simple graph has n = 100 vertices and E = 800 edges. What crossing-number lower bound follows from the stated lemma, and why is the substitution allowed?

  1. The density threshold is 4n = 400, and 800 is at least 400. The required sampling probability in the proof sketch is p = 400/800 = 1/2, a legitimate probability.

  2. Compute the numerator E³ = 800³ = 512,000,000. The denominator is 64 × 100² = 640,000.

  3. Divide to obtain crossing number at least 800. This is a lower bound for every drawing of that graph, not a statement that one particular drawing has exactly 800 crossings.

The result

The crossing lemma guarantees at least 800 crossings. Its threshold check is part of the argument, not an optional preliminary.

Explore the model

Interactive lab

Interactive lab · matrix

Incidences survive an invertible shear

Pause and Reset, then set diagonal scale a = 0.5 and shear b = 1. Consider the four final corners and the four infinite lines containing the boundary edges. Predict eight point–line incidences, just as for the original square, even though lengths and angles change. Play to inspect the selected final parallelogram. This affine example does not enumerate arbitrary incidences or compute crossing numbers.

Incidences survive an invertible shearThe selected matrix is A = [[a,b],[0,1]]. Playback interpolates geometry between the identity and A; readouts always describe A. Its determinant is a and its singular values come from AᵀA. Positive a makes every supported matrix invertible. This two-dimensional family is not a general eigensolver or a learned network. Current readings: Area scale · det A: 1.4; Singular values · σmax, σmin: 1.597, 0.876; Condition number · κ₂(A): 1.823. All parameters are adjustable with the labeled controls below.Matrix A with first row 1.4, 0.6 and second row zero, one; display interpolates between identity and A while final determinant remains 1.4.−4−2024−2−1012x · equal geometric scaleyA = [ 1.4 0.6 ; 0 1 ]Display blend: 0% · final values below
Interpolated geometryOriginal unit squareSelected final transform
Area scale · det A
1.4
Singular values · σmax, σmin
1.597, 0.876
Condition number · κ₂(A)
1.823
t = 0 s

The selected matrix is A = [[a,b],[0,1]]. Playback interpolates geometry between the identity and A; readouts always describe A. Its determinant is a and its singular values come from AᵀA. Positive a makes every supported matrix invertible. This two-dimensional family is not a general eigensolver or a learned network. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forAn invertible affine map preserves incidence even when metric appearances change.

Check your understanding

Why can the point–line construction use the simple-graph crossing lemma?

Try it yourself

Take the nine points of a three-by-three integer grid and the six lines consisting of its three rows and three columns. Count incidences and the edges joining consecutive collinear points. Which branch of the proof applies?

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

Every chosen line contains three points, and it contributes two consecutive-point edges.

Compare with the solution

There are six lines with three points each, giving I = 18. Each contributes two edges, so E = 12 = I − m with m = 6. Since n = 9, the crossing threshold is 4n = 36 and E is below it. The sparse branch gives I ≤ 4n + m = 42, a valid but loose bound. The graph drawing has no edge crossings away from vertices; trying to apply the dense cubic estimate here would wrongly force a positive crossing lower bound.

Take it with you

  • Consecutive point connections convert incidences into a simple graph with E = I − m′.
  • The cubic crossing bound requires the dense regime E ≥ 4n.
  • Linear terms in the incidence theorem handle sparse or highly unbalanced configurations.

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