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The celebrated Szemerédi-Trotter theorem gives a bound for the set of incidences between a finite set of points
and a finite set of lines
in the Euclidean plane
. Specifically, the bound is
or
to denote the statement that
for some absolute constant
. In particular, the number of incidences between
points and
lines is
. This bound is sharp; consider for instance the discrete box
with
being the collection of lines
. One easily verifies that
,
, and
, showing that (1) is essentially sharp in the case
; one can concoct similar examples for other regimes of
and
.
On the other hand, if one replaces the Euclidean plane by a finite field geometry
, where
is a finite field, then the estimate (1) is false. For instance, if
is the entire plane
, and
is the set of all lines in
, then
are both comparable to
, but
is comparable to
, thus violating (1) when
is large. Thus any proof of the Szemerédi-Trotter theorem must use a special property of the Euclidean plane which is not enjoyed by finite field geometries. In particular, this strongly suggests that one cannot rely purely on algebra and combinatorics to prove (1); one must also use some Euclidean geometry or topology as well.
Nowadays, the slickest proof of the Szemerédi-Trotter theorem is via the crossing number inequality (as discussed in this previous post), which ultimately relies on Euler’s famous formula ; thus in this argument it is topology which is the feature of Euclidean space which one is exploiting, and which is not present in the finite field setting. Today, though, I would like to mention a different proof (closer in spirit to the original proof of Szemerédi-Trotter, and also a later argument of Clarkson et al.), based on the method of cell decomposition, which has proven to be a very flexible method in combinatorial incidence geometry. Here, the distinctive feature of Euclidean geometry one is exploiting is convexity, which again has no finite field analogue.
Roughly speaking, the idea is this. Using nothing more than the axiom that two points determine at most one line, one can obtain the bound
An inspection of the proof of (2) shows that it is only expected to be sharp when the bushes associated to each point
behave like “independent” subsets of
, so that there is no significant correlation between the bush
of one point and the bush of another point
.
However, in Euclidean space, we have the phenomenon that the bush of a point is influenced by the region of space that
lies in. Clearly, if
lies in a set
(e.g. a convex polygon), then the only lines
that can contribute to
are those lines which pass through
. If
is a small convex region of space, one expects only a fraction of the lines in
to actually pass through
. As such, if
and
both lie in
, then
and
are compressed inside a smaller subset of
, namely the set of lines passing through
, and so should be more likely to intersect than if they were independent. This should lead to an improvement to (2) (and indeed, as we shall see below, ultimately leads to (1)).
More formally, the argument proceeds by applying the following lemma:
Lemma 1 (Cell decomposition) Let
be a finite collection of lines in
, let
be a finite set of points, and let
. Then it is possible to find a set
of
lines in
, plus some additional open line segments not containing any point in
, which subdivide
into
convex regions (or cells), such that the interior of each such cell is incident to at most
lines.
The deduction of (1) from (2), (3) and Lemma 1 is very quick. Firstly we may assume we are in the range
Let be a parameter to be optimised later. We apply the cell decomposition to subdivide
into
open convex regions, plus a family
of
lines. Each of the
convex regions
has only
lines through it, and so by (2) contributes
incidences. Meanwhile, on each of the lines
in
used to perform this decomposition, there are at most
transverse incidences (because each line in
distinct from
can intersect
at most once), plus all the incidences along
itself. Putting all this together, one obtains
We optimise this by selecting ; from (4) we can ensure that
, so that
. One then obtains
We can iterate away the error (halving the number of lines each time) and sum the resulting geometric series to obtain (1).
It remains to prove (1). If one subdivides using
arbitrary lines, one creates at most
cells (because each new line intersects the existing lines at most once, and so can create at most
distinct cells), and for a similar reason, every line in
visits at most
of these regions, and so by double counting one expects
lines per cell “on the average”. The key difficulty is then to get
lines through every cell, not just on the average. It turns out that a probabilistic argument will almost work, but with a logarithmic loss (thus having
lines per cell rather than
); but with a little more work one can then iterate away this loss also. The arguments here are loosely based on those of Clarkson et al.; a related (deterministic) decomposition also appears in the original paper of Szemerédi and Trotter. But I wish to focus here on the probabilistic approach.)
It is also worth noting that the original (somewhat complicated) argument of Szemerédi-Trotter has been adapted to establish the analogue of (1) in the complex plane by Toth, while the other known proofs of Szemerédi-Trotter, so far, have not been able to be extended to this setting (the Euler characteristic argument clearly breaks down, as does any proof based on using lines to divide planes into half-spaces). So all three proofs have their advantages and disadvantages.

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