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Jozsef Solymosi and I have just uploaded to the arXiv our paper “An incidence theorem in higher dimensions“, submitted to Discrete and Computational Geometry. In this paper we use the polynomial Ham Sandwich method of Guth and Katz (as discussed previously on this blog) to establish higher-dimensional versions of the Szemerédi-Trotter theorem, albeit at the cost of an epsilon loss in exponents.

Recall that the Szemerédi-Trotter theorem asserts that given any finite set of points ${P}$ and lines ${L}$ in the plane ${{\bf R}^2}$, the number of incidences ${I(P,L) := \{ (p,\ell) \in P \times L: p \in \ell \}}$ has the upper bound

$\displaystyle |I(P,L)| \ll |P|^{2/3} |L|^{2/3} + |P| + |L|.$

Apart from the constant factor, this bound is sharp. As discussed in this previous blog post, this theorem can be proven by the polynomial method, and the strategy can be rapidly summarised as follows. Select a parameter ${D \geq 1}$. By the polynomial Ham Sandwich theorem, one can divide ${{\bf R}^2}$ into ${O(D^2)}$ cell interiors, each with ${O(|P|/D^2)}$ points, and incident to ${O(|L|/D)}$ lines on the average, plus a boundary set which is an algebraic curve of degree ${O(D)}$. To handle the contribution of each cell interior, one uses a more elementary incidence bound (such as the bound ${|I(P,L)| \ll |P| |L|^{1/2} + |L|}$ coming from the fact that two points determine at most one line); to handle the contribution on the cell boundary, one uses algebraic geometry tools such as Bezout’s theorem. One then combines all the bounds and optimises in ${D}$ to obtain the result.

As a general rule, the contribution of the cell interiors becomes easier to handle as ${D}$ increases, while the contribution of the cell boundaries become easier as ${D}$ decreases. As such, the optimal value of ${D}$ is often an intermediate one (in the case of Szemerédi-Trotter, the choice ${D = |P|^{2/3} |L|^{-1/3}}$ is typical). Among other things, this requires some control of moderately high degree algebraic sets, though in this planar case ${{\bf R}^2}$, one only needs to control algebraic curves in the plane, which are very well understood.

In higher dimensions, though, the complexity of the algebraic geometry required to control medium degree algebraic sets increases sharply; compare for instance the algebraic geometry of ruled surfaces appearing in the three-dimensional work of Guth and Katz as discussed here, compared with the algebraic geometry of curves in the two-dimensional Szemerédi-Trotter theorem discussed here.

However, Jozsef and I discovered that it is also possible to use the polynomial method with a non-optimised value of ${D}$, and in particular with a bounded value ${D=O(1)}$ of ${D}$, which makes the algebraic geometry treatment of the boundary significantly easier. The drawback to this is that the cell interiors can no longer be adequately controlled by trivial incidence estimates. However, if instead one controls the cell interiors by an induction hypothesis, then it turns out that in many cases one can recover a good estimate. For instance, let us consider the two-dimensional Szemerédi-Trotter theorem in the most interesting regime, namely when the ${|P|^{2/3} |L|^{2/3}}$ term on the RHS is dominant. If we perform the cell decomposition with some small parameter ${D}$, then we obtain ${O(D^2)}$ cells with ${O(|P|/D^2)}$ points and ${O(|L|/D)}$ lines on the average; applying the Szemerédi-Trotter theorem inductively to each of these cells, we end up with a total contribution of

$\displaystyle O( D^2 ) \times O( |P|/D^2)^{2/3} \times O( |L|/D)^{2/3} = O( |P|^{2/3} |L|^{2/3} )$

for the cell interiors, and so provided that one can also control incidences on the (low degree) cell boundary, we see that we have closed the induction (up to changes in the implied constants in the ${O()}$ notation).

Unfortunately, as is well known, the fact that the implied constants in the ${O()}$ notation degrade when we do this prevents this induction argument from being rigorous. However, it turns out this method does work nicely to give the weaker incidence bound

$\displaystyle |I(P,L)| \ll_\epsilon |P|^{2/3+\epsilon} |L|^{2/3} + |P| + |L|$

for any fixed ${\epsilon > 0}$; the point being that this extra epsilon of room in the exponents means that the induction hypothesis gains a factor of ${D^{-\epsilon}}$ or so over the desired conclusion, allowing one to close the induction with no loss of implied constants if all the parameters are chosen properly. While this bound is weaker than the full Szemerédi-Trotter theorem, the argument is simpler in the sense that one only needs to understand bounded degree algebraic sets, rather than medium degree algebraic sets. As such, this argument is easier to extend to higher dimensions.

Indeed, as a consequence of this strategy (and also a generic projection argument to cut down the ambient dimension, as used in this previous paper with Ellenberg and Oberlin, and an induction on dimension), we obtain in our paper a higher-dimensional incidence theorem:

Theorem 1 (Higher-dimensional Szemerédi-Trotter theorem) Let ${d \geq 2k}$, and let ${P}$ and ${L}$ be finite sets of points and ${k}$-planes in ${{\bf R}^d}$, such that any two ${k}$-planes in ${L}$ intersect in at most one point. Then we have ${|I(P,L)| \ll_{k,\epsilon} |P|^{2/3+\epsilon} |L|^{2/3} + |P| + |L|}$ for any ${\epsilon > 0}$.

Something like the transversality hypothesis that any two ${k}$-planes in ${L}$ intersect in at most one point is necessary, as can be seen by considering the example in which the points ${P}$ are all collinear, and the ${k}$-planes in ${L}$ all contain the common line. As one particular consequence of this result, we recover (except for an epsilon loss in exponents) the complex Szemerédi-Trotter theorem of Toth, whose proof was significantly more complicated; it also gives some matrix and quaternionic analogues, in particular giving new sum-product theorems in these rings. Note that the condition ${d \geq 2k}$ is natural, because when the ambient dimension is less than ${2k}$, it is not possible for two ${k}$-planes to intersect in just one point. Using the generic projection trick, one can easily reduce to the critical case ${d=2k}$.

Actually, for inductive reasons, we prove a more general result than Theorem 1, in which the ${k}$-planes are replaced instead by ${k}$-dimensional real algebraic varieties. The transversality condition is now that whenever a point ${p}$ is incident to two such varieties ${\ell, \ell'}$, that the tangent cones of ${\ell}$ and ${\ell'}$ at ${p}$ only intersect at ${p}$. (Also for technical reasons, it is convenient to consider a partial subset ${{\mathcal I}}$ of incidences in ${I(P,L)}$, rather than the full set of incidences. Indeed, it seems most natural to consider the triplet

$\displaystyle \begin{array}{ccccc} \\ && {\mathcal I} && \\ & \swarrow & & \searrow & \\ P & & & & L \end{array}$

as a single diagram, rather than to consider just the sets ${P}$ and ${L}$.)

The reason for working in this greater level of generality is that it becomes much easier to use an induction hypothesis to deal with the cell boundary; one simply intersects each variety ${\ell}$ in ${L}$ with the cell boundary, which usually lowers the dimension of the variety by one. (There are some exceptional cases in which ${\ell}$ is completely trapped inside the boundary, but due to the transversality hypothesis, this cannot contribute many incidences if the ambient dimension is ${2k}$ (so the cell boundary is only ${2k-1}$ dimensional).)

As one application of this more general incidence result, we almost extend a classic result of Spencer, Szemerédi, and Trotter asserting that ${n}$ points in the plane ${{\bf R}^2}$ determine ${O(n^{4/3})}$ unit distances from ${{\bf R}^2}$ to ${{\bf C}^2}$, at the cost of degrading ${O(n^{4/3})}$ to ${O_\epsilon(n^{4/3+\epsilon})}$.

The ham sandwich theorem asserts that, given ${d}$ bounded open sets ${U_1,\ldots,U_d}$ in ${{\bf R}^d}$, there exists a hyperplane ${\{ x \in {\bf R}^d: x \cdot v = c \}}$ that bisects each of these sets ${U_i}$, in the sense that each of the two half-spaces ${\{ x \in {\bf R}^d: x \cdot v < c \}, \{ x \in {\bf R}^d: x \cdot v > c \}}$ on either side of the hyperplane captures exactly half of the volume of ${U_i}$. The shortest proof of this result proceeds by invoking the Borsuk-Ulam theorem.

A useful generalisation of the ham sandwich theorem is the polynomial ham sandwich theorem, which asserts that given ${m}$ bounded open sets ${U_1,\ldots,U_m}$ in ${{\bf R}^d}$, there exists a hypersurface ${\{ x \in {\bf R}^d: Q(x)=0\}}$ of degree ${O_d( m^{1/d} )}$ (thus ${P: {\bf R}^d \rightarrow {\bf R}}$ is a polynomial of degree ${O(m^{1/n})}$ such that the two semi-algebraic sets ${\{ Q > 0 \}}$ and ${\{ Q < 0\}}$ capture half the volume of each of the ${U_i}$. (More precisely, the degree will be at most ${D}$, where ${D}$ is the first positive integer for which ${\binom{D+d}{d}}$ exceeds ${m}$.) This theorem can be deduced from the Borsuk-Ulam theorem in the same manner that the ordinary ham sandwich theorem is (and can also be deduced directly from the ordinary ham sandwich theorem via the Veronese embedding).

The polynomial ham sandwich theorem is a theorem about continuous bodies (bounded open sets), but a simple limiting argument leads one to the following discrete analogue: given ${m}$ finite sets ${S_1,\ldots,S_m}$ in ${{\bf R}^d}$, there exists a hypersurface ${\{ x \in {\bf R}^d: Q(x)=0\}}$ of degree ${O_d( m^{1/d} )}$, such that each of the two semi-algebraic sets ${\{ Q > 0 \}}$ and ${\{ Q < 0\}}$ contain at most half of the points on ${S_i}$ (note that some of the points of ${S_i}$ can certainly lie on the boundary ${\{Q=0\}}$). This can be iterated to give a useful cell decomposition:

Proposition 1 (Cell decomposition) Let ${P}$ be a finite set of points in ${{\bf R}^d}$, and let ${D}$ be a positive integer. Then there exists a polynomial ${Q}$ of degree at most ${D}$, and a decomposition

$\displaystyle {\bf R}^d = \{ Q = 0\} \cup C_1 \cup \ldots \cup C_m$

into the hypersurface ${\{Q=0\}}$ and a collection ${C_1,\ldots,C_m}$ of cells bounded by ${\{P=0\}}$, such that ${m = O_d(D^d)}$, and such that each cell ${C_i}$ contains at most ${O_d( |P|/D^d )}$ points.

A proof is sketched in this previous blog post. The cells in the argument are not necessarily connected (being instead formed by intersecting together a number of semi-algebraic sets such as ${\{ Q > 0\}}$ and ${\{Q<0\}}$), but it is a classical result (established independently by Oleinik-Petrovskii, Milnor, and Thom) that any degree ${D}$ hypersurface ${\{Q=0\}}$ divides ${{\bf R}^d}$ into ${O_d(D^d)}$ connected components, so one can easily assume that the cells are connected if desired. (Incidentally, one does not need the full machinery of the results in the above cited papers – which control not just the number of components, but all the Betti numbers of the complement of ${\{Q=0\}}$ – to get the bound on connected components; one can instead observe that every bounded connected component has a critical point where ${\nabla Q = 0}$, and one can control the number of these points by Bezout’s theorem, after perturbing ${Q}$ slightly to enforce genericity, and then count the unbounded components by an induction on dimension.)

Remark 1 By setting ${D}$ as large as ${O_d(|P|^{1/m})}$, we obtain as a limiting case of the cell decomposition the fact that any finite set ${P}$ of points in ${{\bf R}^d}$ can be captured by a hypersurface of degree ${O_d(|P|^{1/m})}$. This fact is in fact true over arbitrary fields (not just over ${{\bf R}}$), and can be proven by a simple linear algebra argument (see e.g. this previous blog post). However, the cell decomposition is more flexible than this algebraic fact due to the ability to arbitrarily select the degree parameter ${D}$.

The cell decomposition can be viewed as a structural theorem for arbitrary large configurations of points in space, much as the Szemerédi regularity lemma can be viewed as a structural theorem for arbitrary large dense graphs. Indeed, just as many problems in the theory of large dense graphs can be profitably attacked by first applying the regularity lemma and then inspecting the outcome, it now seems that many problems in combinatorial incidence geometry can be attacked by applying the cell decomposition (or a similar such decomposition), with a parameter ${D}$ to be optimised later, to a relevant set of points, and seeing how the cells interact with each other and with the other objects in the configuration (lines, planes, circles, etc.). This strategy was spectacularly illustrated recently with Guth and Katz‘s use of the cell decomposition to resolve the Erdös distinct distance problem (up to logarithmic factors), as discussed in this blog post.

In this post, I wanted to record a simpler (but still illustrative) version of this method (that I learned from Nets Katz), namely to provide yet another proof of the Szemerédi-Trotter theorem in incidence geometry:

Theorem 2 (Szemerédi-Trotter theorem) Given a finite set of points ${P}$ and a finite set of lines ${L}$ in ${{\bf R}^2}$, the set of incidences ${I(P,L):= \{ (p,\ell) \in P \times L: p \in \ell \}}$ has cardinality

$\displaystyle |I(P,L)| \ll |P|^{2/3} |L|^{2/3} + |P| + |L|.$

This theorem has many short existing proofs, including one via crossing number inequalities (as discussed in this previous post) or via a slightly different type of cell decomposition (as discussed here). The proof given below is not that different, in particular, from the latter proof, but I believe it still serves as a good introduction to the polynomial method in combinatorial incidence geometry.

The celebrated Szemerédi-Trotter theorem gives a bound for the set of incidences ${I(P,L) := \{ (p,\ell) \in P \times L: p \in \ell \}}$ between a finite set of points ${P}$ and a finite set of lines ${L}$ in the Euclidean plane ${{\bf R}^2}$. Specifically, the bound is

$\displaystyle |I(P,L)| \ll |P|^{2/3} |L|^{2/3} + |P| + |L| \ \ \ \ \ (1)$

where we use the asymptotic notation ${X \ll Y}$ or ${X=O(Y)}$ to denote the statement that ${X \leq CY}$ for some absolute constant ${C}$. In particular, the number of incidences between ${n}$ points and ${n}$ lines is ${O(n^{4/3})}$. This bound is sharp; consider for instance the discrete box ${P := \{ (a,b) \in {\bf Z}^2: 1 \leq a \leq N; 1 \leq b \leq 2N^2 \}}$ with ${L}$ being the collection of lines ${\{ (x,mx+b): m, b \in {\bf Z}, 1 \leq m \leq N, 1 \leq b \leq N^2 \}}$. One easily verifies that ${|P|=2N^3}$, ${|L| = N^3}$, and ${|I(P,L)| = N^4}$, showing that (1) is essentially sharp in the case ${|P| \sim |L|}$; one can concoct similar examples for other regimes of ${|P|}$ and ${|L|}$.

On the other hand, if one replaces the Euclidean plane ${{\bf R}^2}$ by a finite field geometry ${F^2}$, where ${F}$ is a finite field, then the estimate (1) is false. For instance, if ${P}$ is the entire plane ${F^2}$, and ${L}$ is the set of all lines in ${F^2}$, then ${|P|, |L|}$ are both comparable to ${|F|^2}$, but ${|I(P,L)|}$ is comparable to ${|F|^3}$, thus violating (1) when ${|F|}$ 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 ${|V|-|E|+|F|=2}$; 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

$\displaystyle |I(P,L)| \ll |P| |L|^{1/2} + |L|, \ \ \ \ \ (2)$

which is inferior to (1). (On the other hand, this estimate works in both Euclidean and finite field geometries, and is sharp in the latter case, as shown by the example given earlier.) Dually, the axiom that two lines determine at most one point gives the bound

$\displaystyle |I(P,L)| \ll |L| |P|^{1/2} + |P| \ \ \ \ \ (3)$

(or alternatively, one can use projective duality to interchange points and lines and deduce (3) from (2)).

An inspection of the proof of (2) shows that it is only expected to be sharp when the bushes ${L_p := \{ \ell \in L: \ell \ni p \}}$ associated to each point ${p \in P}$ behave like “independent” subsets of ${L}$, so that there is no significant correlation between the bush ${L_p}$ of one point and the bush of another point ${L_q}$.

However, in Euclidean space, we have the phenomenon that the bush of a point ${L_p}$ is influenced by the region of space that ${p}$ lies in. Clearly, if ${p}$ lies in a set ${\Omega}$ (e.g. a convex polygon), then the only lines ${\ell \in L}$ that can contribute to ${L_p}$ are those lines which pass through ${\Omega}$. If ${\Omega}$ is a small convex region of space, one expects only a fraction of the lines in ${L}$ to actually pass through ${\Omega}$. As such, if ${p}$ and ${q}$ both lie in ${\Omega}$, then ${L_p}$ and ${L_q}$ are compressed inside a smaller subset of ${L}$, namely the set of lines passing through ${\Omega}$, 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 ${L}$ be a finite collection of lines in ${{\bf R}^2}$, let ${P}$ be a finite set of points, and let ${r \geq 1}$. Then it is possible to find a set ${R}$ of ${O(r)}$ lines in ${L}$, plus some additional open line segments not containing any point in ${P}$, which subdivide ${{\bf R}^2}$ into ${O(r^2)}$ convex regions (or cells), such that the interior of each such cell is incident to at most ${O(|L|/r)}$ lines.

The deduction of (1) from (2), (3) and Lemma 1 is very quick. Firstly we may assume we are in the range

$\displaystyle |L|^{1/2} \ll |P| \ll |L|^2 \ \ \ \ \ (4)$

otherwise the bound (1) follows already from either (2) or (3) and some high-school algebra.

Let ${r \geq 1}$ be a parameter to be optimised later. We apply the cell decomposition to subdivide ${{\bf R}^2}$ into ${O(r^2)}$ open convex regions, plus a family ${R}$ of ${O(r)}$ lines. Each of the ${O(r^2)}$ convex regions ${\Omega}$ has only ${O(|L|/r)}$ lines through it, and so by (2) contributes ${O( |P \cap \Omega| |L|^{1/2}/r^{1/2} + |L| / r )}$ incidences. Meanwhile, on each of the lines ${\ell}$ in ${R}$ used to perform this decomposition, there are at most ${|L|}$ transverse incidences (because each line in ${L}$ distinct from ${\ell}$ can intersect ${\ell}$ at most once), plus all the incidences along ${\ell}$ itself. Putting all this together, one obtains

$\displaystyle |I(P,L)| \leq |I(P,L \cap R)| + O( |P| |L|^{1/2}/r^{1/2} + |L| r).$

We optimise this by selecting ${r \sim |P|^{2/3} / |L|^{1/3}}$; from (4) we can ensure that ${r \leq |L|/2}$, so that ${|L \cap R| \leq |L|/2}$. One then obtains

$\displaystyle |I(P,L)| \leq |I(P,L \cap R)| + O( |P|^{2/3} |L|^{2/3} ).$

We can iterate away the ${L \cap R}$ 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 ${{\bf R}^2}$ using ${r}$ arbitrary lines, one creates at most ${O(r^2)}$ cells (because each new line intersects the existing lines at most once, and so can create at most ${O(r)}$ distinct cells), and for a similar reason, every line in ${L}$ visits at most ${r}$ of these regions, and so by double counting one expects ${O(|L|/r)}$ lines per cell “on the average”. The key difficulty is then to get ${O(|L|/r)}$ 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 ${O( |L| \log |L| / r )}$ lines per cell rather than ${O(|L|/r)}$); 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 ${{\bf C}^2}$ 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.