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I don’t usually post personal things on this maths blog, but today I’ll make an exception.

My father was born in Perth on 17 January 1914, which makes him 100 years old today (and still going strong). He’s had a pretty full life, and is mildly famous for his participation in The Great Escape, of which there are now only two survivors (the other being a sprightly 92-year old from Devon, UK).

Cheryl Praeger, Pablo Spiga and I have just uploaded to the arxiv our preprint  Finite primitive permutation groups and regular cycles of their elements.   Everyone recalls from their first course in group theory that if you write a permutation in cycle notation, for example (1,2)(3,4,5)(6,7,8,9,10), then its order is the lowest common multiple of it cycle lengths. As I have just demonstrated, not every permutation must have a cycle whose length is the same as the order of the element. We call a cycle of an element g whose length is the order of g, a regular cycle.  A natural question to ask is when can you guarantee that a permutation has a regular cycle? Or in particular, for which permutation groups do all elements have a regular cycle?

Clearly, the full symmetric group contains elements with no regular cycles, but what about other groups?  Siemons and Zalesskii showed that for any group G  between PSL(n,q) and PGL(n,q) other than for (n,q)=(2,2) or (2,3), then in any action of G, every element of G has a regular cycle, except G=PSL(4,2) acting on  8 points. The exceptions are due to isomorphisms with the symmetric or alternating groups.  They also later showed that for any finite simple group G,other than the alternating group, that admits a 2-transitive action, in any action of G every element has a regular cycle. This was later extended by Emmett and Zalesskii to any finite simple classical group not isomorphic to PSL(n,q).

With these results in mind, we started investigating elements in primitive permutation groups. The results of this work are in the preprint. First we prove that for $k\leq m/2$, every element of Sym(m) in its action on k-sets has a regular cycle if and only if m is less than the sum of the first k+1 primes. After further computation we were then willing to make the following conjecture:

Conjecture: Let $G\leqslant Sym(\Omega)$ be primitive such that some element has no regular cycle. Then there exist integers $k\geq1, r\geq1$ and $m\geq5$ such that
G preserves a product structure on $\Omega=\Delta^r$ with $|\Delta|=\binom{m}{k}$, and $Alt(m)^r \vartriangleleft G\leqslant Sym(m)\textrm{wr} Sym(r)$, where Sym(m) induces its k-set action on $\Delta$.

The general philosophy behind why such a result should be true is that most primitive groups are known to be small in terms of a function of the degree.  There is a large body of work bounding the orders of primitive permutation groups with the best results due to Maroti. The actions of Sym(m) on k-sets are the usual exceptions.

Our paper  goes about attempting to prove this conjecture. We make substantial progress and reduced its proof to having to deal with all the primitive actions of classical groups. Note that Emmett and Zalesskii only dealt with simple ones.  One consequence of our work is we showed that every automorphism of a finite simple group has a regular cycle in its action on the simple group.

Pablo and Simon Guest have subsequently gone on to prove the result for all actions of classical groups and so the conjecture is now a theorem.

Pablo gave a great talk about the conjecture and its subsequent proof  at the recent BIRS workshop on Permutation Groups in Banff which you can view here.

The Minister for Education (finally) announced the ARC Future Fellowships commencing in 2013 this morning, and simultaneously, the DECRA’s and Discovery Projects. Our research group missed out on a grant of the first and second kind, but we had success with two Discovery Projects:

1. Cheryl Praeger, Stephen Glasby, and Alice Niemeyer (“Finite linearly representable geometries & symmetries).
2. Gordon (“Real chromatic roots of graphs & matroids”)

I remember last summer when both grant applications were being written, and it was a lot of work for all involved, so it’s great to see that both have awarded to support cutting edge research.

Following on from my previous post on getting started with sage_matroids, let’s look at a few more of the built-in features.

One of the most fundamental classes of matroids is the class of graphic matroids. Given a graph $G$, the graphic matroid $M(G)$ has the edge set of $G$ as its elements, and the forests of $G$ as its independent sets. The circuits of $M(G)$, i.e. the minimal dependent sets, are then precisely the cycles of $G$, so sometimes $M(G)$ is called the cycle matroid of $G$.

The decision to go with Sage, rather than to either develop a stand-alone system or build on an existing stand-alone system, was largely driven by the fact that Sage provides both a standard programming language and a vast amount of inbuilt combinatorial functionality already. In particular, we can just piggyback on the very extensive “graphs” package to immediately create a graphic matroid using, of course, the canonical graph theory example.

sage: pm = Matroid(graphs.PetersenGraph())
sage: pm
Regular matroid of rank 9 on 15 elements with 2000 bases

Sage 5.12 has just been released, and this is the first release that automatically incorporates the sage_matroids package that was primarily written by Rudi Pendavingh and Stefan van Zwam. Over at The Matroid Union, Stefan has written an introductory post along with some examples of using the package, and I’m going to complement this with some additional examples. Unlike Stefan though, I’m not an expert in Python, so this post is really to show how easy it is to just get started, even if you don’t know all that much about Python.

I’ll assume that you’ve installed Sage 5.12 and you either have a notebook (the browser interface) or command line working; on the command line Sage prompts you for input with the “sage: ” or “…” when multiline input is incomplete.

I’m mostly interested in binary matroids, which are basically just sets of binary vectors; traditionally a binary matroid is represented by the columns of a binary matrix. Let’s start with an easy binary matroid, namely the Fano plane which is, of course, the “standard” starting example in matroid theory. We can build it in two steps by typing in the representing matrix row-by-row and then creating the matroid.

sage: mat = Matrix([[0,0,0,1,1,1,1],[0,1,1,0,0,1,1],[1,0,1,0,1,0,1]])
sage: bm
Binary matroid of rank 3 on 7 elements, type (3, 0)

Notice that I have to use the “full name” of the constructor for a binary matroid, but if I get tired of typing the whole thing, then I can just tell Sage once-and-for-all where to find it and then just use the “short name”.

sage: from sage.matroids.advanced import BinaryMatroid
sage: bm = BinaryMatroid(mat)

Either way, the  variable “bm” now reports that it represents a binary matroid of rank 3 with 7 elements and gives some additional information that it is of “type (3,0)” (of which more later).

Now I can ask Sage lots of things about this matroid, such as confirming that I entered the matrix correctly, by asking for the representing matrix: notice that the syntax is the common object-oriented style whereby the object “bm” is asked to run the method “representation” and the value returned by the method is then displayed (or assigned to a variable etc.)

sage: bm.representation()
[0 0 0 1 1 1 1]
[0 1 1 0 0 1 1]
[1 0 1 0 1 0 1]

So the elements of this matroid — the groundset — is the set of columns of the matrix. To refer to a particular subset of the columns, I’ll need to know the “names” of the elements.

sage: bm.groundset()
frozenset([0, 1, 2, 3, 4, 5, 6])

This tells me that the groundset is an immutable set (the “frozen” part) containing the elements $0$, $1$, $\ldots$, $6$. Mostly the end-user doesn’t need to know about immutable sets or why the groundset is immutable, so this is just a bit of the implementation showing and can safely be ignored. Or just use the alternative version of the command

sage: bm.groundset_list()
[0, 1, 2, 3, 4, 5, 6]

Now I know the groundset, I can find out all the usual “matroidy things” about the matroid, such as the rank of an arbitrary subset of the groundset.

sage: bm.rank([0,2,4])
3
sage: bm.rank([0,1,2])
2

The bases of the matroid are all the independent subsets of columns, and I can find out how many of these there are, and also which they are.

sage: bm.bases_count()
28
sage: bm.bases()
Iterator over a system of subsets

Ah, the first one is clear – there are 28 bases, but the second one didn’t give me what I expected, which was the list of bases, but rather an iterator. An iterator is a container object that supplies the objects it contains “one-by-one” (the container object can thereby represent a huge collection by not actually storing any of the objects but only producing them when called for). In this case though, we just want to see them all and so we use Python’s standard syntax for working with iterables.

sage: [b for b in bm.bases()]
[frozenset([0, 2, 3]),
frozenset([1, 2, 3]),
frozenset([0, 1, 3]),
frozenset([1, 3, 4]),
frozenset([2, 3, 4]),
frozenset([0, 2, 4]),
frozenset([1, 2, 4]),
frozenset([0, 1, 4]),
frozenset([0, 4, 5]),
frozenset([1, 4, 5]),
frozenset([3, 4, 5]),
frozenset([0, 3, 5]),
frozenset([2, 3, 5]),
frozenset([0, 2, 5]),
frozenset([1, 2, 5]),
frozenset([0, 1, 5]),
frozenset([1, 5, 6]),
frozenset([2, 5, 6]),
frozenset([3, 5, 6]),
frozenset([4, 5, 6]),
frozenset([0, 4, 6]),
frozenset([2, 4, 6]),
frozenset([3, 4, 6]),
frozenset([0, 3, 6]),
frozenset([1, 3, 6]),
frozenset([0, 2, 6]),
frozenset([1, 2, 6]),
frozenset([0, 1, 6])]
sage:

The notation is simple for mathematicians because it is so similar to set notation – the command can be essentially translated directly into mathematics as $\{b : b \in {\rm bases(bm)}\}$. Of course, listing the contents of an iterable object is such a common activity that you can just use

sage: list(bm.bases())

to get the same result.

The circuits of the matroid are the minimal dependent sets of columns, and unsurprisingly can be accessed by

sage: bm.circuits()
Iterator over a system of subsets
sage: len(bm.circuits())
14

What if we want to know the lines of the binary matroid, which are the circuits of size 3. Again the Python notation is straightforward for mathematicians:

sage: [c for c in bm.circuits() if len(c) == 3]
[frozenset([0, 1, 2]),
frozenset([0, 3, 4]),
frozenset([2, 4, 5]),
frozenset([1, 3, 5]),
frozenset([0, 5, 6]),
frozenset([1, 4, 6]),
frozenset([2, 3, 6])]

And there, we have the list of the lines of the Fano plane, exactly as expected. A frozen set is iterable, and so we can smarten up the output easily enough.

sage: [list(c) for c in bm.circuits() if len(c) == 3]
[[0, 1, 2], [0, 3, 4], [2, 4, 5], [1, 3, 5], [0, 5, 6], [1, 4, 6], [2, 3, 6]]

So there we have our first run through the absolute basics of using sage_matroids. Next time, I’ll move on to using – and building on – some of the more sophisticated built-in functions.

This is a continuation of my last post on this subject. As Gordon remarked in one of his posts, you may need to refresh your browser if some of the embedded gifs do not appear as they should.

### Dualities and isomorphisms of classical groups

Four of the five families of classical generalised quadrangles come in dual pairs: (i) $Q(4,q)$ and $W(3,q)$; (ii) $H(3,q^2)$ and $Q^-(5,q)$. Both can be demonstrated by the Klein correspondence. Recall from the last post that the Klein correspondence maps a line of $\mathsf{PG}(3,q)$ represented as the row space of

$M_{u,v}:=\begin{bmatrix}u_0&u_1&u_2&u_3\\v_0&v_1&v_2&v_3\end{bmatrix}$

to the point $(p_{01}:p_{02}:p_{03}:p_{12}:p_{31}:p_{23})$ where

$p_{ij}=\begin{vmatrix}u_i&u_j\\v_i&v_j\end{vmatrix}$.

Now consider the symplectic generalised quadrangle $W(3,q)$ defined by the bilinear alternating form

$B(x,y):=x_0y_1 - x_1y_0 + x_2y_3 - x_3y_2.$

A totally isotropic line $M_{u,v}$ must then satisfy

$0 = B(u,v) = p_{01} + p_{23}$.

Therefore, the lines of $W(3,q)$ are mapped to points of $Q^+(5,q)$ lying in the hyperplane $\pi:X_0+X_5=0$. Now the quadratic form defining $Q^+(5,q)$ is $Q(x)=x_0x_5+x_1x_4+x_2x_3$ and $\pi$ is the tangent hyperplane at the projective point $(1:0:0:0:0:1)$, which does not lie in the quadric. Hence the hyperplane $\pi$ is non-degenerate and so we see that $W(3,q)$ maps to points of $Q(4,q)$. That this mapping is bijective follows from noting that the number of lines of $W(3,q)$ is equal to the number of points of $Q(4,q)$ (namely, $(q+1)(q^2+1)$).

Hence $\mathsf{P\Gamma Sp}(4,q)\cong \mathsf{P\Gamma O}(5,q)$.

Now we will consider a more difficult situation which reveals that the generalised quadrangles $H(3,q^2)$ and $Q^-(5,q)$ are also formally dual to one another. Read more…

It’s been a long time since we posted anything here, but things have kept on happening, so I thought I’d just give a sketchy update.

One of the reasons behind the hiatus in blog posts is that there’s now only two months to go before we host 37ACCMCC, the annual Australasian Combinatorial Conference, here in Perth. I’m the director of the organising committee and it’s quite a juggling act trying to keep on top of everything. It’s a bit of a guessing game too, because we have to book things like the conference dinner and excursion, but with most people registering at the last minute, we have no real idea of numbers! Now I know what it’s like to be on the receiving end, I’m going to register early for all future conferences!

We have a poster now though, which you can see here (or as a PDF here); you can probably recognise the nice building from our blog’s banner image.