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If $n \geq 3$ and $K$ is a field of characteristic not dividing $n$, containing a primitive $n$-th root of unity $\zeta$, then the $3n$ points of the form $(1:-\zeta^a:0)$, $(0:1:-\zeta^b)$, $(-\zeta^c:0:1)$ are a Sylvester-Gallai configuration. In particular, taking $n=p-1$, this gives an SG configuration over $\mathbb{Q}_p$ for $p \geq 5$.

answered Dec 31 '16 at 16:30

David E Speyer

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13

I believe that a finite geometric proof is given by Jean Dieudonné here:
Dieudonné, Jean, Les isomorphismes exceptionnels entre les groupes classiques finis, Can. J. Math. 6, 305-315 (1954). ZBL0055.01904.
If you go to Section 7 of the given paper you will see the proof. Dieudonné displays the required isomorphism using a 1-1 correspondence between the 40 ...

13

The proof for this appeared over a series of papers. The final one was
Jan Saxl, `On Finite Linear Spaces with Almost Simple Flag-Transitive Automorphism Groups' Journal of Combinatorial Theory, Series A 100, 322–348 (2002).
which includes references for all the papers.

12

The answer is yes, by an argument using the axiom of choice.
There are exactly continuum many lines in the plane, and so by the
axiom of choice, we may enumerate them in a well-ordered sequence
of length continuum.
Let's build the set $X$ in stages, so that by stage $\alpha$ we've
included two points from all the lines in the enumeration up to
$\alpha$ and ...

answered Dec 29 '16 at 5:11

Joel David Hamkins

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12

Six years later it's hard to know if this is still a question of interest. For the record (since I stumbled across the question), the answer is yes and it's a consequence of a more general phenomenon.
The main theorem of the paper of Stevens and Warren (Thm 5 in https://arxiv.org/abs/2102.05446) states that for any convex (or concave) functions $f, g:\mathbb{...

10

Here is a bit more symmetric version of the picture in the accepted answer:

10

Over any field $K$ of characteristic $\neq 2$ and containing $i := \sqrt{-1}$, there exists in $\mathbb{P}^2(K)$ a Sylvester-Gallai configuration with $12$ points given in affine coordinates by $(0,0)$, $(0,1)$, $(1,0)$, $(1,1)$, $(a,a)$, $(a,b)$, $(b,a)$, $(b,b)$ where $a := \frac{1+i}{2}$ and $b := \frac{1-i}{2}$, $\infty\cdot(0,1)$, $\infty\cdot(1,0)$, $\...

8

That is true, since this is so for $\mathbb RP^n$ - take $n+2$ vertices of a "regular simplex" in it -- i.e. take the regular simplex in $S^n$ and project its vertices to $\mathbb RP^n$.
In coordinates, take points $(1,0,\ldots,0)$, ... $(0,\ldots,0,1)$ in $\mathbb R^{n+2}$, take the hyperplane $\sum_i x_i=1$ and take the $S^n$ in it that passes through ...

8

As Will Sawin points out, this represents only a partial answer.
Theorem 5 of this paper
Laison, Joshua D., and Yulan Qing. "Subspace intersection graphs." Discrete Mathematics 310, no. 23 (2010): 3413-3416.
Journal link
proves that this graph
cannot be realized as the intersection graph ...

answered Aug 17 '15 at 13:35

Joseph O'Rourke

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8

Building on previous work by Paul Libois, and related to work by Libois'
student Jean van Buggenhaut from 1969, Francis Buekenhout considered and
solved this question in "Foundations of one Dimensional Projective Geometry based on Perspectivities" Abhandlungen aus dem Mathematischen Seminar der Universität Hamburg 43 (1975) 21-29. Note that his approach also ...

7

I don't entirely follow your proposed approach, but let me make a few remarks which might help you think about this:
(1) Clearly this can't work, since this argument purports to show that a Kakeya set must have positive Lebesgue measure which, in contrast to the finite field setting, we know is false in Euclidean space.
(2) Note that the discretization of ...

6

Following up on Matthias Wendt's comment, the language of Moufang sets is indeed a suitable axiomatic approach to (generalizations of) projective lines.
Formally speaking, a Moufang set is a set $X$ together with a collection of groups $U_x \leq \operatorname{Sym}(X)$ (one group for each $x \in X$), such that:
each $U_x$ fixes $x$ and acts sharply ...

6

Here is a slightly better lower bound. If there are fewer than $2q^2$ points then there is some line that hits the set at most once. Consider the $q+1$ planes containing a line containing one point. There must be at least $2q-2$ other points in each of those planes to meet all lines in the plane, for a total of $1+(q+1)(2q-2) = 2q^2-1$ points. So, $2q^2-1 \...

6

A more general result than what you want appears as Theorem 1 in http://arxiv.org/pdf/1002.2554. (A slightly weaker result had appeared before as Theorem 3.1 in http://arxiv.org/pdf/0909.5471).
Curiously, it is open if at least one of $A^2+A^2$ and $A^3+A^3$ is necessarily large.

6

We can get a lower bound on the order of $n \log n$.
I'll describe how to arrange $4^n$ points in general position to get $n 4^{n-1}$ squares.
The arrangement is described recursively. For the base case $n=1$, we have $4^1 = 4$ points, and you can probably guess how we should arrange them to get $1 \cdot 4^{1-1} = 1$ squares. Now suppose we have an ...

5

One defining feature of $\mathbb P^1(k)$ is that it provides a sharply 3-transitive permutation representation for $\operatorname{PGL}_2(k)$. I believe that the abstraction of projective line to "sharply 3-transitive permutation group" is the most studied one.
The characterization of sharply 3-transitive groups as groups of projectivities over KT-fields ...

answered Oct 16 '15 at 23:43

Gjergji Zaimi

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5

You might benefit from reading Section 5.3 of John Faulkner's book The role of nonassociative algebra in projective geometry AMS 2014. The results there may have been what you had in mind by 'axiomatizing the "quadrangular hexad" relation'. These are (almost) all very old theorems, appearing in (and mostly predating) Pickert's Projektive Ebenen book from ...

5

I believe this is correct (assuming $\lambda\gt0$).
If $\lambda$ is infinite then each connected component of $G$ has $\lambda$ vertices. Since the components can be handled independently, the problem reduces to the $\kappa=\lambda$ case, which can be done by a straightforward transfinite recursion.
If $\lambda$ is a positive integer, this is a classical ...

5

The current best Szemeredi-Trotter bound over finite fields also holds over arbitrary fields. It is in the wonderful paper "An Improved Point-Line Incidence Bound Over Arbitrary Fields" by Sophie Stevens and Frank de Zeeuw.
Unfortunately, this current best bound of $m^{11/15}n^{11/25}$ is closer to the trivial $m^{3/4}n^{3/4}$ than to the ...

4

To answer your questions:
1) A projective plane admits a circulant incidence matrix if and only if the automorphism group contains a cyclic group acting regularly on points and regularly on blocks. Equivalently, the projective plane comes from a difference set. The automorphism group of the projective plane coming from the Dickson near-field of order 9 is ...

4

The Szemerédi–Trotter bound is known to be false for circles (it is true for circles with the same radii). There is a construction that gives $N^{2/3}|C|^{2/3}\log^{1/3}N$ incidences. The current best upper bound is about $N^{6/11}|C|^{9/11}$, and this is conjectured to be far from tight (see for example this recent result of Sharir and Zahl). You can find a ...

4

Take a 2-$(v,4,1)$ design on $v$ points and delete one block, along with the four points on it. In the original system each point is on exactly $(v-1)/3$ blocks, so if we assume $v\ge25$ the geometry we get by deleting one block is hyperbolic, all blocks have at least three points, and there are blocks of size three and size four.

4

I became interested in Kakeya sets because they have the interesting property that a Kakeya set in a projective plane cannot be a subset of a blocking set, and with the exception of the full plane with two lines removed, Kakeya sets are the only sets with this property.
As far as state of the art in higher dimensions, not much seems known. The article http:/...

4

It seems that the matrix
$$
A = \left(\begin{array}{cccccc}
1 & 1 & 1 & 1 & 1 & 1 \\
1 & 1 & 1 & 1 & 1 & 1 \\
1 & 1 & 1 & 1 & 1 & 1 \\
1 & 1 & 1 & 0 & 1 & 1 \\
1 & 1 & 1 & 1 & 0 & 1 \\
1 & 1 & 1 & 1 & 1 & 0 \\
\end{array}\right)
...

answered Aug 20 '15 at 22:08

Noam D. Elkies

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3

I always remember this diagram that seems impossible to find online. I have no source for it (except that I remember that it is/was engraved in the sidewalk near the Weber Building on the Colorado State University campus in Fort Collins, CO).
Does this one count as nondegenerate or is it still bad because of the outer circle only touching the red, blue and ...

3

There are continuum-many pairwise non-isomorphic perfectly dense hypergraphs. Below is a sketch of a proof.
Given a countably infinite field $\mathbb{K}$, the projective plane $\mathbb{KP}_2$ over $\mathbb{K}$ can be seen as a perfectly dense hypergraph, where vertices are points and edges are lines. I will show that from the hypergraph structure of $\...

3

Is the cross-ratio well-defined even when several of the four arguments coincide?
No, it is not. Modeled as homogeneous coordinates over $k^2$ you get the null vector in cases where three or more points coincide, but that's not an element of your line any more.
If I understand you correctly, it would be sufficient to have a set of axioms for the projective ...

3

Here is an improvement of the upper bound which I found in ``The polynomial method in Galois geometries'' by Simeon Ball. See page number 4.
The known constructions are somewhat crude. For example, let $S$ be a
set of points of $AG(3,q)$ with the property that every line is
incident with a point of $S$. For $q$ square, the smallest known
example ...

3

This is false. I don't see how to get a counterexample from Andreas Blass's comment (the only uses for the existence of $\sqrt2$ which are obvious to me require a more flexible notion of incidence statement), so I am posting this as an answer although it is probably more complicated than necessary.
For fixed prime $p$, one can take a point and line ...

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