Congratulations are due to Michael who is currently in a purple patch riding a wave of recognition and achievement over the last two or three weeks. (I had look up why it’s called a purple patch and not a green patch or a chartreuse patch etc.)

First, we heard that his latest ARC Discovery Grant application (along with Li and Gabriel) has been successful for a project involving symmetries of directed graphs, about which much less is known than for undirected graphs.

Secondly, he won a UWA Research Award in the “mid-career” category – these are internal awards across the whole university designed to recognise excellence in research; Gabriel also won the same award, but in the “early-career” category.

Then to top things off, we heard yesterday that his promotion to “something” was granted. Unfortunately it is not entirely clear what “something” should be. UWA used to use the British system of “lecturer/senior lecturer/associate professor/professor”, with only a handful of people in latter category. In other words, “Professor” is quite prestigious and only a few people will reach that level.

Then a few years ago, we decided to change to the US system of “assistant professor / associate professor / professor”. I’m not entirely sure why this change was made, but I think that it was partially to allow UWA to attract strong American academics (looking for jobs due to the hiring freeze in the US during the GFC) for whom dropping down to an “associate professor” would be seen as a backward step, and generally to align ourselves more with the US nomenclature than with the UK.

But then the local medical research funding body, the NHMRC, said that anyone who called themselves a professor would be evaluated as though they were an old-style prestigious professor, suddenly making some perfectly good research records look relatively mediocre.

So the edict has come down to, well, we don’t actually know what we’re meant to do. New jobs are to be advertised under the old titles, but holders of old jobs who chose to go to the new titles can continue to use the new titles or revert to the old titles depending on their preference or the day of the week or something. At least, I think I’ve got that right.

Anyway, there’s one thing that is clear. Congratulations to Michael on his promotion to Level D, his ARC Discovery Grant and his UWA Research Award. And to Gabriel for the latter two.

I’m in Singapore at the moment, spending the weekend watching my 11-year old daughter Amy in a gymnastics competition and the weekdays visiting Dave Roberson (currently at NUS) and Fengming Dong (currently at NTU).

Meeting up with Dave and thinking once again about the annoyingly resistant conjecture that the core of a cubelike graph is cubelike, reminded me of another “cubelike” question that is still unresolved.

First some terminology: a cubelike graph is a Cayley graph for the elementary abelian 2-group $Z_2^n$; the term “cubelike” arises for two reasons:

1. The $n$-dimensional cube is cubelike
2. If you talk, or write, anything about these graphs, you rapidly need something snappier for the phrase “a Cayley graph for an elementary abelian 2-group”

It has been known for decades that a cubelike graph cannot have chromatic number 3, and this was proved in a beautifully elegant fashion by Payan. A very ambitious conjecture that all cubelike graphs have chromatic number equal to a power of 2 can be disproved by exhibiting a cubelike graph on 16 vertices with chromatic number 7, and cubelike graphs on 64 vertices with chromatic number 6.

I simply cannot find any cubelike graph with chromatic number 5, and according to Brouwer’s website, this is unknown. Chris Godsil and I thought about this a few years ago, and someone said they had heard someone tell someone else that they had heard that perhaps some Russians had found such a graph on 128 vertices.

So, what’s the problem? Why not just construct all 128-vertex cubelike graphs and check their chromatic number?

The trouble with this is that the number of cubelike graphs grows very rapidly. There are only 1372 such graphs on 32 vertices, and we can find all the chromatic numbers. This jumps to 475499108 on 64 vertices (actually a handful less, this overcounts the number by about 10, due to some “accidental” isomorphisms), and although I don’t know the exact chromatic number of all of these, I can rule out enough of them as potential 5-chromatic graphs to complete the search.

But on 128 vertices, we have 1038397981840994509577948 graphs to work through (that’s just about $10^{30}$ and so unless we stumble on one somehow (perhaps someone knows who “the Russians” are) or make some theoretical advance, this problem is likely to remain unresolved for the time being.

In a colleague’s research grant proposal, under a heading about resources and equipment, he wrote something along the lines of:

All a mathematician needs is some paper, some pens, good access to online journals, and most importantly, a quiet place to work.

Of course, we also rely on good coffee, a buzzing environment of enthusiastic colleagues, natural light, and administrative support.

For other disciplines, lab equipment and technicians are extremely important, and not having the best equipment would severely cripple an experimental chemist, for example. So if you needed to find a way to stifle the progress of a group of mathematicians, what changes to their work environment would you make?

• You could take their blackboards/whiteboards away, but they might be just as happy with their endless supply of foolscap paper.
• You could take their paper away. This is difficult to do, as paper is easy to buy and very cheap. Mathematicians would bring their own paper, or just write on their desks and other flat surfaces.
• You could take their pens away. Again, pens are cheap and easy to buy, so it would be difficult to outlaw pens in the workplace. And anyway, where would you draw the line? Pencils, chalk, and crayons would have to be outlawed too.

Perhaps the best way is to create tension within the group, and somehow put in place a situation where the ambient noise in the workplace was disruptive and unpredictable. For those who wear headphones, they would need to be distracted by visible movement in their periphery. How can this be done effectively? Even more insidiously, we could create an environment that would increase the rate of infection due to colds, influenza, or other airborne viruses. Have you tried to solve a difficult mathematics problem whilst your head is congested and your joints feel like jelly?

So I leave the question to you: how can you create a work environment that not even a resource-minimalist mathematician can bear? To summarise:

1. it needs to organically create tension between colleagues,
2. it needs to be cost efficient,
3. it needs to be noisy,
4. it needs to have visual distractions,
5. it needs to foster airborne viruses.

I can only think of one solution to this problem. What is your solution?

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.