Complex Projective 4-Space

The abbreviation TFAE (the following are equivalent) is often used in the statement of various theorems. Of course, a completely synonymous phrase would be ‘any one of these implies the other n − 1′. This then admits a natural generalisation to ‘any k of these together imply the other n − k‘. We shall refer to these situations as (n,k)-TFAE.

It is obviously possible to construct artificial examples for any k, such as the following set of statements:

• a = 0
• b = 0
• …
• c = 0
• a + b + … + c = 0

However, these examples are completely boring, so we’ll just concentrate on more interesting examples instead. The case k = 1 is just regular TFAE, so the next thing to try is k = 2, i.e. ‘any two of these imply the others’. One such example is the new Mersenne conjecture, formulated by Bateman, Selfridge and Wagstaff:

Let p be an odd natural number (only interesting when p is prime). Then, any two of these implies the third:

(a) p = 2^k ± 1 or p = 4^k ± 3
(b) 2^p – 1 is a (Mersenne) prime
(c) (2^p + 1)/3 is a (Wagstaff) prime

Of course, this is only conjectural, although heuristically it is expected to be true (with the three properties being simultaneously satisfied only by 3, 5, 7, 13, 17, 19, 31, 61 and 127).

These cases of ‘any two conditions imply the third’, or (3,2)-TFAE, are reasonably common. If P and Q are two finite sets, and f is a function with domain P and codomain Q, then any two of the following statements imply the third:

• f is an injection.
• f is a surjection.
• P and Q are of equal cardinality.

Obviously, the case where all statements are true corresponds to f being a bijection. Equally obviously, this does not hold in general if P and Q are allowed to be infinite.

A more interesting example of (n,2)-TFAE, where there are four statements rather than three, is from Euclidean geometry. Suppose we have a pencil of four distinct lines passing through a common point, as shown above. Then, any two of the following statements together imply the other two:

• Lines A and C are perpendicular.
• A is an angular bisector of BD.
• C is an angular bisector of BD.
• (A, C; B, D) is a harmonic pencil.

I don’t have any other examples of (4,2)-TFAE, or indeed any examples of (n,k)-TFAE with k > 2. Using the ‘comments’ feature below, please post any non-trivial examples of (n,k)-TFAE (k > 1) you can find:

A particularly important function is Klein’s j-function. It is defined on the upper half-plane of the complex numbers, and is incredibly symmetrical. For example, it is periodic, and thus invariant under translations by integers:

It is also invariant under a larger group of Möbius transformations, namely those corresponding to matrices with integer entries. This group is called the modular group, or , since it is the special linear group of invertible matrices with integer entries, quotiented out by its centre of order 2.

We can think of the upper half-plane as Klein’s model of the hyperbolic plane, in which case the j-function is a function on the hyperbolic plane and the Möbius transformations correspond to orientation-preserving isometries. Hence, an equivalent definition of the modular group is the orientation-preserving symmetry group of a regular tiling of the hyperbolic plane: the Voronoi tessellation of the zeros of the j-function.

Recall that the group of proper orthochronous Lorentz transformations is isomorphic to . Consequently, we can think of the modular group as a discrete version of this.

Fundamental groups and the complement of the trefoil knot

A surprising isomorphism pertains to knot theory and homotopy. Let X be a path-connected topological space. Then, we consider closed paths in X, beginning and ending at some pre-specified point. Under concatenation of paths, these form a monoid. When we quotient out by the relation ‘two paths are equivalent if they can be continuously deformed into each other’, then we get the fundamental group of X.

The fundamental group of any simply-connected space is just the trivial group, as any two paths can be continuously deformed into each other. The once-punctured plane has the infinite cyclic group as its fundamental group, since a path is determined up to continuous deformation by the number of times it encircles the puncture. The fundamental group of the twice-punctured plane is the free group on two generators, best visualised by its Cayley graph:

Now embed the trefoil knot in and consider its complement. It is a remarkable fact that the fundamental group of the resulting space, when quotiented by its centre, yields a group isomorphic to the modular group!

A graph is vertex-transitive if its group of automorphisms acts transitively on the set of vertices. For example, the skeletons of uniform polyhedral are all vertex-transitive, as is the complete graph Kn. It’s obvious that a vertex-transitive graph must be regular, i.e. all vertices have the same degree.

Do all connected vertex-transitive graphs have a Hamiltonian cycle? Obviously not, since the second most basic connected vertex-transitive graph lacks one:

Is this the only counter-example? There are certainly no other obvious examples. In situations such as this one, it’s generally a good idea to employ the usual strategy of trying the Petersen graph. It does have a self-avoiding cycle of length 9, as shown in the embedding below, but not of length 10.

It transpires that only five counter-examples are known. In addition to K2 and the Petersen graph is the Coxeter graph, and graphs obtained by replacing each vertex of the Petersen and Coxeter graphs with triangles.

It is conjectured that these are the only counter-examples. Note that none of these are Cayley graphs (graphs whose vertices correspond to elements of a group, and edges correspond to multiplication by generators); indeed, the Petersen graph is the smallest vertex-transitive graph that is not a Cayley graph. Since all Cayley graphs are vertex-transitive, we obtain the following weaker conjecture:

Every Cayley graph contains a Hamiltonian cycle.

Consider the Cayley graph of the Rubik’s cube group, equipped with the twelve generators {U,U’,D,D’,L,L’,R,R’,F,F’,B,B’}. This has a vertex for each position of the Rubik’s cube, and an edge connecting two vertices if they differ by a quarter-turn. The existence of a Hamiltonian cycle is equivalent to the existence of a sequence of moves that visits every configuration exactly once before returning to the initial state.

A sequence of moves which eventually solves the Rubik’s cube irrespective of initial configuration is called a Devil’s algorithm. The Lovász conjecture, if true, would imply the following generalisation:

Every mechanical puzzle, the possible positions of which differ only in re-colouring the pieces, admits a Devil’s algorithm of length equal to the number of positions.

Indeed, this is actually equivalent to the Lovász conjecture on Cayley graphs, since every group can be expressed as the symmetries of some polytope in , and the polytope can be constrained in a casing so as to admit only motions corresponding to your favourite set of generators.

Some polyhedra have lots of symmetries. For example, consider the omnitruncated dodecahedron:

The interactive demonstration constructs it kaleidoscopically, by reflecting a fundamental region (the Schwarz triangle) in three mirrors. The same fundamental region, however, can result in lots of different polyhedra and tilings of Euclidean and hyperbolic space. For example, consider the following series:

These all share the same fundamental region, but with different reflection groups. To specify the omnitruncated dodecahedron, we need to include another piece of information: the angles between the mirrors. On the left of the following image is an annotated Schwarz triangle, giving the angles of 90°, 60° and 36° (these sum to more than 180°, since it is on the surface of a sphere).

If we want to describe tessellations of three-dimensional space (including polychora — four-dimensional polytopes — which can be regarded as tilings of the 3-sphere), we’ll need to have a Schwarz tetrahedron instead of a Schwarz triangle.

The second diagram is the dual of the Schwarz simplex, where each node corresponds to a mirror and each edge gives the reciprocal of the angle between the mirrors. The third diagram is a simplified version, where missing edges represent ‘2’ and unlabelled edges represent ‘3’. This is called a Coxeter-Dynkin diagram.

If we have a Coxeter-Dynkin diagram with n nodes, we can construct the symmetry group as follows:

• Start with the free group on n generators, one for each node;
• Quotient out by x² = 1 for each generator x, since these generators are reflections;
• Quotient out by (xy)^n = 1 if x and y are joined by an edge labelled n. If two nodes are not connected, we have n = 2, thus xyxy = 1. Since the generators are self-inverse, xyxy is the commutator of x and y. In other words, orthogonal mirrors correspond to reflections that commute with each other.

This gives a presentation of the group with ½(n² + n) relations on n generators.

Finite Coxeter groups

The finite reflection groups have been completely classified, and correspond to one of the following diagrams:

H3 is the familiar icosahedral symmetry, and H4 is the four-dimensional analogue. I2[n] is simply the reflection group of the regular n-gon (and therefore is isomorphic to the dihedral group D_2n). An and BCn are the symmetries of the regular simplex and hypercube, respectively. Dn is an index-2 subgroup of BCn corresponding to the symmetries that fix a 2-colouring of the vertices of the hypercube. F4 is the Coxeter group of the 24-cell (a special self-dual polytope in four dimensions).

This leaves three more exceptional cases, E6, E7 and E8. They’re fascinating objects, and really worth discussing, but I don’t really have time to discuss them at the moment. However, because it’s the most beautiful, I’ll use E8 as an example for presenting a Coxeter group:

The spider relation

Whilst E6, E7 and E8 are very interesting Coxeter groups, their exceptionality is rivalled only by the sporadic groups. Rather amazingly, some of the bigger sporadic groups can be obtained as quotients of infinite Coxeter groups. For example, start with the Coxeter group of the following diagram:

This is a huge infinite hyperbolic Coxeter group, G555. Nevertheless, a single extra relation, called the spider relation, causes it to collapse into a finite group, Y555:

• (a b1 c1 a b2 c2 a b3 c3)^10 = 1

As an aid to memory, here is a completely gratuitous photograph of a large spider:

Image courtesy of someone called John, shared under the Creative Commons licence.

The relevant nodes are highlighted in red in the Coxeter-Dynkin diagram. Ivanov and Norton proved a conjecture by Conway that Y555 is, in fact, the wreath product of the Monster group with the cyclic group C2. Consequently, it is a group of order 2m², where m is the (very large!) order of the Monster group.

This group, the bimonster, can also be constructed as Y554, Y544 or Y444. Similarly, the groups Y553, Y543 and Y443 are isomorphic to the direct product of the Monster group and C2. Hence, we have a presentation of the Monster group on 12 generators and 80 relations (78 for the Coxeter group, together with 1 spider relation and the relation Z = 1, where Z is the unique order-2 element in the centre of the group).

There are several items of news worth mentioning.

Firstly, I have been informed that the current status of the ‘bounded gaps between primes’ effort has now reached H = 6966, which is impressively low for these sieve methods. According to the status page, this is still an unconfirmed result. If we plot the results on a logarithmic plot, there appears to be an exponential decay.

Unfortunately, this trend is unlikely to continue, as major theoretical breakthroughs would be required in order to prove the twin prime conjecture (corresponding to the horizontal red line at H = 2).

Speaking of records, this year’s Senior Wrangler, Arran Fernandez, is suspected to be the youngest in history, winning Part II of the Mathematical Tripos at the age of just eighteen. Also, he is almost certainly the first Senior Wrangler from his college, Fitzwilliam.

Neither the collaborative ‘bounded gaps between primes’ project nor the Mathematical Tripos are actually competitions, but the same cannot be said of the 2013 Balkan Mathematical Olympiad, which is occurring in Agros, Cyprus at the moment. One of the contestants sent in a picture of an icosahedral Tarsia puzzle:

This is a welcome change from the more familiar planar variants. Continuing this sequence of smooth segues, the most common way to turn a dodecahedron into an icosahedron (or vice-versa) is by projective duality. There is a more incremental process, however, which consists of applying eight successive wye-delta transforms. If we insist that they occur in antipodal pairs to maintain central symmetry, a total of seven polyhedra can be formed, of which two are the dodecahedron and icosahedron. These polyhedra are double covers of the Petersen family of graphs, depicted in the following diagram by David Eppstein:

In the diagram, rotation by 180° corresponds to taking the projective dual of the corresponding polyhedron. In the middle of the diagram is the graph of the projectivisation of a centrally symmetric self-dual polyhedron, which can be regarded as precisely halfway between the icosahedron and dodecahedron.

Other polyhedra in this family include the hexakis hexagonal prism and its dual.

The partition numbers (sequence A000041) count the number of distinct ways to partition n identical objects. For example, p(5) = 7, as there are seven distinct ways to partition 5 objects:

This visualisation, where partitions are graphs whose connected components are path graphs, leads to an immediately equivalent formulation as the number of distinct graphs on n vertices that do not contain the claw graph or the triangle graph as a minor.

Unlike (for example) the number of permutations, the partition numbers lack a simple closed form. Rademacher’s formula is massively complicated:

In practice, one would never use this closed form, since there are cleaner ways of computing the partition numbers. For instance, it is not too difficult to find a recursive formula for the number p(n,k) of partitions of n into at most k parts, namely:

• p(n, 1) = 1 for all ;
• p(0, k) = 1 for all ;
• p(n, k) = p(n, k − 1) + p(n − k, k) if k is not larger than n;
• p(n, k) = p(n, k − 1) otherwise.

Then, we have p(n) = p(n, n).

Generating function

Equally straightforward to determine is the generating function for the partition numbers. It is a product of the geometric series over all k, which simplifies to the reciprocal of the following product:

It is a remarkable theorem, due to Euler, that the exponents are given by generalised pentagonal numbers. In other words, this product is expressible as a simple infinite sum:

This results in a recurrence relation for the number of partitions, involving pentagonal numbers, which is much more memory-efficient than the recurrence relation involving the two-argument generalisation p(n, k).

Odd parts and distinct parts

Similar generating functions can be found for the number of partitions into odd parts, and for the number of partitions into distinct parts. One may even notice that these actually give the same generating function! This suggests that there exists a nice bijection between the partitions of n into odd parts and the partitions of n into distinct parts. There are two well-known elegant bijections:

Proof 1: Using binary

Suppose we have a partition of n into odd parts, for example 40 = 7 + 7 + 5 + 5 + 5 + 3 + 3 + 1 + 1 + 1 + 1 + 1. For each odd integer k, we count the number of parts of size k, and express it as a sum of distinct powers of two (this representation is unique). Applying it to this example gives 40 = 14 + 10 + 5 + 6 + 4 + 1, which is a sum of distinct parts. Also, this process is clearly reversible in a unique way, so we have a bijection.

Proof 2: Using diagrams

We write each odd part as a L-shaped polyomino, and stack them together. For example, the partition 40 = 7 + 7 + 5 + 5 + 5 + 3 + 3 + 1 + 1 + 1 + 1 + 1 yields the following diagram:

We then read off the distinct partitions as V-shaped lines, like so:

Notice that this bijection gives 40 = 15 + 9 + 8 + 5 + 3, whereas the other method gave 40 = 14 + 10 + 6 + 5 + 4 + 1. In other words, the two different proofs are really different. More interestingly, we can compose one bijection with the inverse of the other to obtain a non-trivial natural permutation of the partitions into odd/distinct parts!

Ramanujan’s congruences

Srinivasa Ramanujan discovered three surprising congruences involving the partition function:

The first congruence can be explained by defining the rank of a partition as the difference between the largest part and the number of parts, and grouping the partitions into 5 equally large sets according to the residue class of the rank (modulo 5). This also works for the second congruence, by considering ranks modulo 7. Unfortunately, residue classes of ranks modulo 11 fails spectacularly for the third congruence.

Freeman Dyson conjectured in 1944 the existence of a more sophisticated combinatorial function, the crank, such that grouping partitions according to the residue class of the crank will give equally sized subsets. It took over forty years before such a function was discovered and proved to have the desired properties. Indeed, the crank of a function also proves the other two congruences.

Congruences exist for all other primes, except for 2 and 3. Hence, we know that there are infinitely many n such that p(n) is divisible by q, for any prime q other than 2 or 3. In fact, the case q = 2 can easily be dealt with, since its falsity would imply that for all sufficiently large n, p(n) is odd. We then choose a really large N (much greater than the last n such that p(n) is even), such that the pentagonal number recurrence avoids all of the even values of p(n), and such that there are an even number of terms in the recurrence. We can do this, since the gaps between pentagonal numbers become arbitrarily large. This means that p(N) is even, leading to a blatant contradiction.

This leaves just the case of q = 3, which is an unsolved problem. It is, of course, conjectured that there are infinitely many partition numbers divisible by 3, since the negation of this statement would be incredibly bizarre. Nevertheless, it remains to be proved.