Most LaTeX-ers know about Tikz, which allows the user to create images in LaTeX without having to embed images created from an external program. The main advantages are that

1. The ambient LaTeX fonts are used in the image, so labels and such conform to the ambient style of the document.
2. The size of the .tex file is kept small, since it is only text you are creating.
3. It yields a picture that is smooth and that looks good upon zooming in (i.e., the resolution of the picture is good).
4. It is functional code so that you can automate the drawing of many pictures by giving commands such as “draw a line between these two points”.

The main disadvantage, is that there is a steep learning curve. The best way to learn is through examples, and even though I’m still a hack, my tikz code has improved via my copying segments of other people’s code. For geometry, there isn’t much out there, so I thought that I would dump some images here. Below are some Tikz pictures of configurations in finite geometry that I’ve collected and think should be available for everyone else to use. A big thanks to Stephen Glasby who went to a lot of trouble to make the two generalised hexagons of order 2. If you have suggestions on how I can simplify my code, please let me know.

### Desargues’ configuration, two ways

\begin{tikzpicture}
\tikzstyle{point1}=[ball color=cyan, circle, draw=black, inner sep=0.1cm]
\tikzstyle{point2}=[ball color=green, circle, draw=black, inner sep=0.1cm]
\tikzstyle{point3}=[ball color=red, circle, draw=black, inner sep=0.1cm]
\node (v1) at (0,8) [ball color=blue, circle, draw=black, inner sep=0.1cm] {};
\node (v2) at (0,6) [point1] {};
\node (v3) at (2,5.5) [point1] {};
\node (v4) at (1.5,4) [point1] {};
\node (v5) at (0,0) [point2] {};
\node (v6) at (2.75*2,8-2.75*2.5) [point2] {};
\node (v7) at (1.5*1.5,8-1.5*4) [point2] {};
\draw (v1) -- (v2) -- (v5);
\draw (v1) -- (v3) -- (v6);
\draw (v1) -- (v4) -- (v7);
\draw (v2) -- (v3) -- (v4) -- (v2);
\draw (v5) -- (v6) -- (v7) -- (v5);
\node (v8) at (intersection of v2--v3 and v5--v6) [point3] {};
\node (v9) at (intersection of v2--v4 and v5--v7) [point3] {};
\node (v10) at (intersection of v3--v4 and v6--v7) [point3] {};
\draw (v3) -- (v8) -- (v6);
\draw (v4) -- (v9) -- (v7);
\draw (v4) -- (v10) -- (v7);
\draw (v8) -- (v9) -- (v10);
\end{tikzpicture}


… and the second one:

\begin{tikzpicture}
\tikzstyle{point} = [ball color=black, circle,  draw=black, inner sep=0.1cm]
\foreach\x in {0, 72, 144, 216, 288}{
\begin{scope}[rotate=\x]
\coordinate (o1) at (-0.588, -0.809);
\coordinate (o2) at (0.588, -0.809);
\coordinate (c1) at (-1.1, 4.6);
\coordinate (c2) at (1.1, 4.6);
\coordinate (o3) at (0, 3.236);
\draw[color=black] (o3) -- (3.236*-0.588, 3.236*-0.809);
\draw[color=blue] (o1) ..  controls (c1) and (c2) ..  (o2);
\end{scope}
}
\foreach\x in {0, 72, 144, 216, 288}{
\begin{scope}[rotate=\x]
\coordinate (o2) at (0.588, -0.809);
\coordinate (o3) at (0, 3.236);
\fill[point] (o2) circle (2pt);
\fill[point] (o3) circle (2pt);
\end{scope}
}
\end{tikzpicture}


### The generalised quadrangle of order 2

I think Gordon gave me the original tikz code for this and then I tweaked it.

\begin{tikzpicture}
\tikzstyle{point}=[ball color=magenta, circle, draw=black, inner sep=0.1cm]
\foreach \x in {18,90,...,306}{
\node [point] (t\x) at (\x:2.65){};
}
\foreach \x in {54,126,...,342}{
\draw [color=blue, double=green](\x:1cm) circle (1.17557cm);
}
\fill [white] (0,0) circle (1cm);
\foreach \x in {54,126,...,342}{
\node[point] (i\x) at (\x:1cm) {};
\node[point] (o\x) at (\x:2.17557cm) {};
}

\draw [color=blue,double=green] (t90)--(o126)--(t162)--(o198)--(t234)--(o270)--(t306)--(o342)--(t18)--(o54)--(t90);
\draw (t90)--(i270)--(o270);
\draw (t162)--(i342)--(o342);
\draw (t234)--(i54)--(o54);
\draw (t306)--(i126)--(o126);
\draw (t18)--(i198)--(o198);
\end{tikzpicture}


### The two generalised hexagons of order 2

These pictures were originally drawn by Schroth in his 1999 paper, and then appeared in Burkard Polster’s book “A geometrical picture book”.

\begin{tikzpicture}
\foreach\n in {0, 1,..., 6}{
\begin{scope}[rotate=\n*51.4286]
\coordinate (a0) at (10,0);
\coordinate (b0) at (7,0);
\coordinate (c0) at (1.45,0);
\coordinate (d0) at (4.878,-0.4878);
\coordinate (e0) at (2.1729,0.37976);
\coordinate (f0) at (1.45,0.612);
\coordinate (g0) at (2.78,-0.585);
\coordinate (h0) at (4.074,0.7846);
\coordinate (i0) at (6.0976,2.9268);
\foreach\k in {1, 2,..., 6}{
\foreach\p in {a,b,c,d,e,f,g,h,i}{
\coordinate (\p\k) at ($(0,0)!1! \k*51.4286:(\p0)$);
}
}
\draw[thick,blue] (a0)--(b0)--(c0);
\draw[thick,blue] (d0)--(e0)--(f0);
\draw[thick,black] (g0)--(h0)--(i0);
\draw[thick,black] (a0)--(g1)--(a3);
\draw[thick,green] (i0)--(d1)--(i2);
\draw[thick,black] (c0)--(g3)--(d3);
\draw[thick,color=purple] (b0) .. controls (5.7,-1.8) and (4.6,-2.2)
.. (h6) .. controls (1.5,-3.2) and (-1,-2.4) .. (e4);
\draw[thick,black] (b0)--(h0)--(e2);
\draw[thick,purple] (f6)--(c0)--(f0);
\foreach\k in {1, 2,..., 6}{
\foreach\p in {a,b,c,d,e,f,g,h,i}{
}
}
\end{scope}
}
\end{tikzpicture}


\begin{tikzpicture}
\foreach\n in {0, 1,..., 6}{
\begin{scope}[rotate=\n*51.4286]
\coordinate (a0) at (85,0);
\coordinate (b0) at (55,0);
\coordinate (c0) at (12.5,0);
\coordinate (d0) at (8.5,1.3);
\coordinate (e0) at (16.2,9);
\coordinate (f0) at (30,14.3);
\coordinate (g0) at (26.6,17.0);
\coordinate (h0) at (26.3,22.4);
\coordinate (i0) at (29.5,28);
\foreach\k in {1, 2,..., 6}{
\foreach\p in {a,b,c,d,e,f,g,h,i}{
\coordinate (\p\k) at ($(0,0)!1! \k*51.4286:(\p0)$);
}
}
\draw[thick,black] (a0)--(e1)--(a3);
\draw[thick,green] (b0)--(h0)--(b2);
\draw[thick,purple] (f0)--(g0)--(f1);
\draw[thick,blue] (h0)--(i0)--(a1);
\draw[thick,purple] (h6)--(c0)--(d0);
\draw[thick,purple] (f0)--(e0)--(i3);
\draw[thick,black] (b0)--(i6)--(g6);
\draw[thick,color=blue] (g0) .. controls (27,2) and (17,-5)
.. (d6) .. controls (-7,-5) and (-5,-5) .. (c3);
\draw[thick,color=blue] (c0) .. controls (16,3) and (17,5)
.. (e0) .. controls (15,13) and (8,15) .. (d1);
\foreach\k in {1, 2,..., 6}{
\foreach\p in {a,b,c,d,e,f,g,h,i}{
}
}
\end{scope}
}
\end{tikzpicture}


The first is the Split Cayley hexagon as it is usually given, whilst the second is its dual.

### The projective plane of order 2

\begin{tikzpicture}
\tikzstyle{point}=[ball color=cyan, circle, draw=black, inner sep=0.1cm]
\node (v7) at (0,0) [point] {};
\draw (0,0) circle (1cm);
\node (v1) at (90:2cm) [point] {};
\node (v2) at (210:2cm) [point] {};
\node (v4) at (330:2cm) [point] {};
\node (v3) at (150:1cm) [point] {};
\node (v6) at (270:1cm) [point] {};
\node (v5) at (30:1cm) [point] {};
\draw (v1) -- (v3) -- (v2);
\draw (v2) -- (v6) -- (v4);
\draw (v4) -- (v5) -- (v1);
\draw (v3) -- (v7) -- (v4);
\draw (v5) -- (v7) -- (v2);
\draw (v6) -- (v7) -- (v1);
\end{tikzpicture}


Tomorrow the winner(s) of the Fields medal(s) for 2014 will be announced at the Seoul ICM and we’ll be eagerly awaiting the outcome.

Although he is only an outside chance (at least according to http://poll.pollcode.com/p6es9_result) one of the mathematicians “in the mix” is the UWA graduate Akshay Venkatesh. Actually Akshay went through Uni at the same time as Michael, although he’s 5 or 6 years younger than Michael. I remember first meeting them when I was chairing a seminar by Jack Koolen, and Akshay and Michael turned up. I must admit that on seeing this fresh-faced youngster (Michael) with someone barely of high-school age (Akshay), I assumed they were strays from some high-school visit and enquired whether they knew what sort of talk they were coming to!

Anyway, while I wish Akshay all the best, this post is really about the consequences of a possible award to him. Our university is obsessed by rankings and while the official goal “Top 50 by 2050″ doesn’t specify which ranking system it is referring to, it is widely understood to be the Shanghai Jiao-Tong ranking of world universities. Personally, I don’t find this goal at all motivating – firstly it is totally unachievable without either a truly massive increase in resources or by such a severe distortion of university activities to focus only on the ranking criteria that it would no longer be recognisable as a university.

The problem is that the SJT criteria are incredibly narrowly focussed on extreme events such as Fields Medals and Nobel Prizes. For each university, 10% of the score is based on alumni with  Nobel prizes/Fields Medals, while a further 20% is based on Faculty with Nobel prizes/Fields Medals.  Another 20% is based on highly-cited researchers, of which UWA has a handful (including Cheryl). So 50% of the ranking is based on a tiny number of individuals. Of course universities devious enough about the rankings can appoint a few Nobel prize winners on reasearch-only positions and move up a few slots, whether or not the Nobel prize winner ever meets a student, or indeed ever steps on campus.

Despite this, I’m willing to accept that a university with 20 Nobel prize winners is statistically different from a university with one, like UWA.  But is there really a difference between a university with one Nobel prize winner and two, or zero?  If Akshay wins tomorrow, then UWA will score some points in the 10% and move up the rankings, and be deemed a better university than it was before.

But does this make sense?

Akshay was at UWA nearly 20 years ago, and by all accounts he was prodigiously talented when he arrived at UWA and prodigiously talented when he left. It’s not clear to me that this reflects on UWA today in any particular way.

Nevertheless, a win for Akshay still might have some direct benefit for us in the publicity that follows a high profile win. Since Akshay’s day, the Maths department has dramatically shrunk due to being the main target of relentless rounds of budget cuts, staff sackings, more budget cuts and so on. Although we pride ourselves on being a member of Australia’s “Group of 8″ universites, we now have by far the weakest Mathematics course, with fewer than half the options of any of the others. Meanwhile, competitors like Monash are investing heavily in Maths with 15 new positions currently advertised. So if Akshay wins and the light of publicity shines on us, it might be a nice time to point out at the highest level that there aren’t any Top 50 universities with an emaciated Maths department.

So, one way or the other – good luck and  Go Akshay!

PS Another candidate “in the mix” is our old friend Harald Helfgott who has visited a couple of times and tried to teach us the mysterious properties of the sizes of triple products in groups. When we first invited him, we were told that he was of “Fields medal” quaility, but since he’s proven the ternary Goldbach conjecture, his odds are rapidly shortening. So good luck Harald!

Up to duality, there are two known families of finite (thick) generalised hexagons:

1. the Split Cayley hexagons of order $(q,q)$ related to the Dickson exceptional groups $G_2(q)$.
2. the Twisted Triality hexagons of order $(q,q^3)$ related to the Steinberg exceptional groups $\,^3D_4(q)$.

Since I will only consider generalised hexagons up to duality, I will assume that the order $(s,t)$ of a given one has $s\le t$. The parameter $s$ is one less than the number of points on a line, and $t+1$ is the number of lines on a point. To date, we do not know much about the possible values of these positive integers $s$ and $t$. Here is what we know:

• $s\le t^3$ and $t\le s^3$ (Haemers and Roos, 1981)
• $s^2+st+t^2$ divides $s^3(s^2t^2+st+1)$ (from the multiplicities of the eigenvalues of the point graph).
• There are two (up to duality) generalised hexagons with $s=2$ and they have $t\in\{2,8\}$ respectively (Cohen and Tits 1985).

We do not know that $s+1$ divides $t+1$, even though the known examples satisfy this simple divisibility relation.

Here is a cool number theoretic thing I’ve observed recently, but I can’t (quite yet) prove that it is true:

Claim: If $s,t$ are integers greater than 1, satisfying $s+1 \mid t+1$ and $s^2+st+t^2\mid s^3(s^2t^2+st+1)$, then apart from a small finite number of exceptions, we have $t=s$ or $t=s^3$.

I claim that the only exception is $(s,t)=(14,224)$.

We know that the known examples satisfy the extra relation $s+1\mid t+1$, and it is conceivable that such a simple looking relation holds for generalised hexagons and that it arises for some combinatorial reason. It’s simplicity is the only reason to believe that it might hold, but wouldn’t it be cool if every generalised hexagon satisfied it?

Last week I was at the Symmetries of Graphs and Networks IV conference at Rogla in Slovenia. The conference webpage is here.  At the same time was the annual  PhD summer school in discrete maths. As usual it was a very enjoyable and well organised conference. It was good to catch up with some of the regulars and meet a few new people as well

I was one of the invited speakers and spoke about some of the work that I have been doing recently with Luke Morgan on graph-restrictive permutation groups. The slides are available here.  The two relevant preprints are on the arxiv here and here.

Last week, I attended an excellent conference “Groups, Computation and Geometry” at the Pingree Park Conference Centre in the Colorado mountains. It was organised magnificently by Peter Brooksbank, Alexander Hulpke, Bill Kantor, Tim Penttila, and James Wilson. Pingree Park is at 9,000 feet (2,700 metres) and I did have headache on the last two days. My trip started badly with a case of gastro, but luckily I had arrived some three days before the conference, so by the second day of talks, I was feeling much better. My 50 minute talk had to be rescheduled to the Friday of the conference, because of this. Apart from your’s truly, the invited talks were given by Colva Roney-Dougal, Şükrü Yalçınkaya, Eric Moorhouse, and Simon Blackburn. Colva spoke about using pregroups to solve problems in combinatorial group theory, mainly word problems, and getting around in a Van Kampen diagram. Şükrü waved the flag for the computational group theorists and spoke on black-box recognition of classical groups. Eric waved the geometry flag and spoke on p-ranks of incidence matrices, and in particular, on ways to attack the “prime-order projective plane implies Desarguesian” conjecture. Simon’s talk carried the main theme of different types of discrete logarithm problems and their application to elliptic curve cryptography. All of these talks were amazing, and I followed each one with keen interest.

Bill Kantor was the master of the schedule, and I was one of only a few that enjoyed the format: only Bill knew who was speaking when, and the program for the subsequent day would be posted on a door each evening. No titles or abstracts, just the expectation that you would see the next talk and you would know then and there what it would be about. It’s good reasoning: if you wanted to know the title and abstract before hand, it might be because you were choosing whether to go or not, and with such a cosy conference, that is out of the question! The other argument, might be that you wanted to do some prior reading before the talk, but would you really? Read more…

Below is a guest blog post of Melissa Lee, who took part in a six-week summer vacation research project supported by the Australian Mathematical Sciences Institute.

Hi everyone! My name is Melissa Lee and I’ve recently started honours with John and Dr. Eric Swartz here at UWA. This past summer I have been working with John on an AMSI Vacation Research Scholarship. My AMSI VRS project was based on looking at structures embedded in an affine space by viewing them in terms of a game called SET. Read more…

I’m writing this post so that I can direct my students to it as I often have to go over the evolution of the concepts of ovoids of projective spaces and polar spaces, and to explain that they are (i) different, and (ii) connected. It’s  a bit dry, but it will serve as a reference for future posts.

## Ovoids of projective spaces

I will start with a result of Jacques Tits in 1962, though ovoids arose earlier in the work of Bose and Qvist. An ovoid of a projective space $\mathcal{S}$ is a set of points $\mathcal{O}$ such that for any point $P$ of $\mathcal{O}$, the union of the lines incident with $P$ that are tangent to $\mathcal{O}$ forms a hyperplane of $\mathcal{S}$. Tits showed that an ovoid of $PG(n,q)$ exists if and only if $n\le 3$. For $n\le 3$, we should be careful to stipulate that $q>2$, since the size of an ovoid of $PG(3,2)$ is just 5 and things are a bit messy to state. By a simple counting argument, the number of points of an ovoid of $PG(3,q)$ is $q^2+1$, and for a projective plane of order $q$, an ovoid has size $q+1$. For example, the elliptic quadric of $PG(3,q)$ is an example of an ovoid, and a non-singular conic is an example of an ovoid of $PG(2,q)$. For more on the elliptic quadric example, see this post.

In the planar case, we now call these objects ovals, and reserve the name ‘ovoid’ just for the 3-dimensional case. Now an ovoid and oval have the property that no 3 points are collinear. We have the following way to think of an ovoid in three different ways:

Let $\mathcal{O}$ be a set of points of $PG(3,q)$. The following are equivalent:

1. No 3 points of $\mathcal{O}$ are collinear and $|\mathcal{O}| = q^2+1$;
2. For any point $P$ of $\mathcal{O}$, the union of the lines incident with $P$ that are tangent to $\mathcal{O}$ span a plane of $PG(3,q)$;
3. At every point $P$ of $\mathcal{O}$, there is a unique plane that meets $\mathcal{O}$ only in $P$ (i.e., a tangent plane).

## Ovoids of polar spaces

Jef Thas defined ovoids of polar spaces in his seminal paper “Ovoidal translation planes” in 1972. It is a set of points such that every maximal totally isotropic subspace meets it in precisely one point. Alternatively, we could define it as a set of points no two collinear, with size the number of points divided by the number of points in a maximal. We have:

Let $\mathcal{O}$ be a set of points of a finite polar space $\mathcal{S}$, and let $\mu$ be the number of points of $\mathcal{S}$ divided by the number of points lying in a maximal totally istropic subspace. The following three are equivalent:

1. No 2 points of $\mathcal{O}$ are collinear and $|\mathcal{O}| = \mu$;
2. Every maximal totally isotropic subspace has exactly one point of $\mathcal{O}$ incident with it.
3. At every point $P$ of $\mathcal{O}$, there exists a maximal totally isotropic subspace through $P$ tangent to $\mathcal{O}$.

So where did Thas’ definition come from? The most impressive result of his 1972 paper is the connection between ovoids of the rank 2 symplectic space and ovoids of the 3-dimensional projective spaces.

Theorem (Thas 1972): Let $q$ be an even prime power. Then an ovoid of $W(3,q)$ is also an ovoid of $PG(3,q)$. Conversely, if $\mathcal{O}$ is an ovoid of $PG(3,q)$, then we can define a null polarity $\rho$ defining a $W(3,q)$ such that $\mathcal{O}$ are absolute points for $\rho$.

This result is quite remarkable. It certainly is easier to study and enumerate ovoids of $W(3,q)$, than in $PG(3,q)$. The null polarity hinted at by the theorem is truly beautiful and was first known to Segre (1959): given a point $P$ of $\mathcal{O}$, we define $P^\rho$ to be the unique tangent plane at $P$; for each secant plane $\pi$ of $\mathcal{O}$ we define $\pi^\rho$ to be the nucleus of the oval carved out (i.e., the nucleus of $\mathcal{O}\cap\pi$).

The situation for $q$ odd was already classified independently by Barlotti and Panella (1965): the only ovoids in this case are quadrics. So the open problem on classifying ovoids of $PG(3,q)$ boils down to the even case.

Another reason for defining ovoids this way is that it encapsulates natural examples. A non-degenerate hyperplane section of the Hermitian polar space $H(3,q^2)$ is an ovoid, and a non-degenerate hyperplane section of minus type of the parabolic quadric $Q(4,q)$ is an ovoid. Plus, we can think of the non-singular conic as a rank 1 polar space $Q(2,q)$, and hence an ‘ovoid’ of $Q(2,q)$ coincides trivially with an oval of $PG(2,q)$.

In a future post, I’ll talk about the open problems on ovoids of polar spaces.