Complex Projective 4-Space

This was inspired by a discussion with James Aaronson outside the Chapel. We were contemplating group theory and automorphisms (specifically, trying to determine whether there exists a group of size strictly larger than with only two automorphisms), and eventually considered the following question:

What happens if you begin with a group G, and iteratively apply the ‘automorphism group’ operator?

For a cyclic group of order n, where n has primitive roots, the automorphism group is the cyclic group of order φ(n). If a group is non-Abelian, its automorphism group is also non-Abelian (consider automorphisms arising from conjugation by elements which do not commute). Similarly, if a group is Abelian and non-cyclic, its automorphism group is non-Abelian. Hence, the groups that eventually reduce to the trivial group under repeated application of Aut are cyclic groups of order n such that {n, φ(n), φ(φ(n)), …, 1} all have primitive roots. James Aaronson completely characterised the integers with this property:

• , for all ;
• ;
• Primes p of the form ;
• 2p, where p is a prime of that form;
• 4, 5, 10, 11, 22, 23, 46, 47 and 94.

These form the sequence A117729.

The next case is C8, which has no primitive roots. Its automorphism group is the Klein 4-group, C2 × C2. Since it is the additive group of a vector space over F2, the automorphism group is the group of linear maps (2 × 2 invertible matrices over F2 with linearly independent columns). It is not too hard to show that this is isomorphic to S3 (symmetric group with 6 elements), and that the direct sum  has the general linear group GL(n, p) as its automorphism group.

Most symmetric groups (the exceptions being n = 2 and n = 6) are their own automorphism groups. The next few cases, such as D8 and Q8, are included in the diagram below:

As the size of the group increases, these become difficult to compute.

For simple non-Abelian groups, the automorphism group is complete (equal to its own automorphism group). This is proved here. More generally, for centreless groups, it is possible to define a limit of a sequence of iterated automorphism groups, known as an automorphism tower. It has been proved that this process eventually terminates at some ordinal (not necessarily finite).

This is discussed in further detail on Math Overflow.

A recurring general theme in mathematics is of a property being true of three objects when considered pairwise, but not true of the entire set. For example, earlier we considered non-trivial mutual friends, and how a simple graph is not sufficient to contain all information about the relationships between elements of a set. Here are a few more examples:

Coprimality

The integers 6, 10 and 15 are coprime, but any pair of them share a common factor. Consequently, the Diophantine equation cannot be solved by the Sam Cappleman-Lynes technique, nor can it be disproved by Fermat’s Last Theorem.

In the statements of Beal’s conjecture, Fermat-Catalan conjecture and the abc conjecture (proof still unverified at the time of writing, since precisely one person understands inter-universal Teichmüller theory.) it doesn’t matter whether we use coprimality or pairwise coprimality, since the relation between the variables causes the notions to coincide.

Probabilities

If an ordered triple of random variables (x, y, z) can take the values (0, 0, 0), (1, 1, 0), (1, 0, 1) and (0, 1, 1) with equal probability, then any pair of variables are independent, but the third is determined entirely by those two. A similar situation occurs in three-particle quantum entanglement, where the total parity of the electron spins is restricted by conservation laws.

Food incompatibility triad

George Hart (geometer, and co-creator of Vi Hart) proposed the following problem:

Find three foods, such that any two are compatible, but the three together are incompatible.

This has been fiercely debated in the comments section of an article in the Guardian column of Alex Bellos. There doesn’t seem to be any generally accepted solution (Feynman erroneously claimed that {milk, tea, lemon} was a solution, but this is broken by the incompatibility of milk and lemon), and little progress has occurred since George Hart closed the question on the basis of it being too subjective.

Until now, that is. John Howe, Cameron Ford et al claim the following solution:

1. Pancakes
2. Lemon liqueur
3. Beef

The first two constitute a crêpe suzette, and the other two combinations apparently exist on the Internet, whereas their union does not. The solution has apparently been sent to George Hart, and I am unaware whether a response has been received. This may finally put to rest an age-old puzzle believed to have originated with the philosopher Wilfrid Sellars.

Some people have recently been discussing Eric Weinstein*, who has proposed a 14-dimensional theory of everything known as Geometric Unity. The media attention is possibly as a result of the trivial difference between his surname and that of the patent office clerk responsible for formulating general relativity.

* Not to be confused with Eric Weisstein, who created Wolfram MathWorld.

Whilst these theories of everything are concerned with the local field equations and particles at the most minute of scales, the large-scale structure of the universe is also very interesting and equally unknown. Here we’ll examine a few suggested candidates for the shape of the universe:

Curvature of space

When you gaze off into the depths of space, it is natural to imagine space as being , an infinite Euclidean space. This is not, however, the only possibility. Space is locally curved (c.f. general relativity), and it has been suggested that it may in fact be globally curved. A constant positive Gaussian curvature would cause the universe to ‘curl up’ into a 3-sphere, whereas a negative Gaussian curvature corresponds to hyperbolic space.

Thanks to Gauss’s Theorem Egregium, it is possible to determine the local curvature by measuring distances in the propinquity of a point. Hence, it is theoretically possible to calculate the radius of the Earth by measuring distances between (four) cities. I tried this and computed a Cayley-Menger determinant; it differed from zero, proving that the Earth is not flat.

Cosmologists have measured the curvature of the universe to be within experimental error of zero, suggesting that (on a large scale) space is Euclidean. If space is indeed everywhere flat, this precludes the possibility of the universe being a 3-sphere. There are, however, several possible shapes, even with this restriction. Conway calls these platycosms.

Platycosms in two and three dimensions

In two dimensions, there are precisely two finite flat manifolds, the torus and Klein bottle. They differ in that one is an orientable surface, whereas the other is not. If we lived on a three-dimensional analogue of a Klein bottle, it would be possible to place a left-handed seashell on a spaceship, send it around the universe, and observe that it returns as a right-handed seashell.

There are ten finite three-dimensional platycosms (mentioned here), which are all feasible candidates for the shape of our universe. Other possibilities include an infinite universe (such as the Cartesian product of a circle and an infinite plane). I’m assuming that the universe doesn’t have a boundary, as this would be really weird.

Poincare dodecahedral space

Certain manifolds can be constructed by taking a simply-connected patch and identifying parts of its boundary. For example, a torus can be produced by identifying opposite edges of the unit square. A more complicated example is obtained by identifying opposite faces of a regular dodecahedron. Depending on how much you twist each face (by 36°, 108° or 180°), you can obtain different spaces (namely a Poincaré homology sphere, the Siefert-Weber space, and , respectively).

Based on data from the WMAP probe, some scientists conjecture that the shape of the universe is a Poincaré homology sphere. The evidence isn’t conclusive, however…

The Sierpinski triangle is a surprisingly ubiquitous mathematical object. Rather than describing what a Sierpinski triangle is, I may as well show you a picture of one.

Essentially, it consists of three identical copies of itself, scaled by a factor of ½. This means that any reasonable definition (e.g. Hausdorff) gives the dimension of the Sierpinski triangle as . We can ask more detailed quantitative questions  about the Sierpinski triangle:

Moment of inertia

This is quite an easy one, actually. If a Sierpinski triangle of mass m and side length l is rotated about an axis passing normally through its centre, dimensional analysis shows that the moment of inertia is given by for some dimensionless quantity k.

Using the recursive definition of the Sierpinski triangle and invoking the Huygens-Steiner theorem (you may know this as the parallel-axis theorem), we get a relation which can then be rearranged to give . So, the moment of inertia is simply .

From this, it is trivial to find the average squared distance between two points: , also by Huygens-Steiner.

Mean distance

Finding the mean (unsquared) distance between two randomly-chosen points is much more non-trivial, and the result is due to Andreas Hanz. It transpires that the mean distance is , and the variance of the distance is . These results were determined by taking the limit (as the number of discs tends to infinity) of the (normalised) mean distance between two randomly chosen configurations in Tower of Hanoi puzzle.

Sierpinski tetrahedron

This is an obvious generalisation of the Sierpinski triangle. It is, however, easier to construct: Let x,y,z be real numbers (which are not dyadic rationals) in the interval [0,1]. Then, (x,y,z) is in the Sierpinski tetrahedron if , where that operator XORs each bit of the binary expansion of the numbers. From this construction, it is easy to deduce that the Hausdorff dimension is precisely 2.

I’m not sure exactly how to motivate a discussion of the Hoffman-Singleton graph, except perhaps by beginning with the following problem from the October 2011 edition of the Advanced Mentoring Scheme:

This problem seems quite innocuous at first, and one would not expect there to be particularly deep mathematics associated with it. Even more surprising is that there’s a very specific unsolved problem related to this.

You might want to attempt this AMS question; it’s not particularly difficult. We can very quickly characterise these graphs as having diameter 2 (any pair of vertices can be connected by a path of length at most 2) and girth 5 (the length of the shortest cycle is 5), as well as noticing that every pair of vertices is contained in a 5-cycle, and so on. Pretty soon, it should be obvious that the smallest such graph is the pentagon itself:

Petersen graph

One may then wonder whether there are other graphs with this property. Generally, a good idea is to try your favourite graph. If you haven’t yet reached the stage where you have a favourite graph, adopt the Petersen graph as your newly-acquired favourite graph. Here are a few drawings of your favourite graph:

The leftmost diagram is the bog-standard drawing of the Petersen graph, which is the most instantly recognisable. The second one looks completely different, but more careful analysis demonstrates their equivalence. Taken together, it becomes clear that we’re dealing with a graph of immense symmetry. We can actually think of the Petersen graph as the skeleton of a dodecahedron, but with opposite vertices identified with each other. The complete graph K6 can be obtained by applying the same process to the icosahedron.

The third diagram differs only trivially from the first (there is a continuous deformation of the plane which relates them), and shows that the Petersen graph can be drawn such that all edges are the same length.

All of these drawings contain crossings, since the graph is non-planar (indeed, it is one of the simplest graphs that is not linklessly-embeddable) by virtue of having both the K5 and K3,3 as minors. It can, however, be drawn on the surface of a torus without crossings; I’ll leave the proof as an exercise to the reader.

Anyway, the point is that the Petersen graph satisfies the conditions specified in the problem.

Hoffman-Singleton graph

We’ve shown that there exist graphs of degree 2 and 3 with these properties. The next graph has degree 7, and is known as the Hoffman-Singleton graph. Noam Elkies proves its existence and uniqueness here.

Although it is not apparent from the rendering above, this graph is also massively symmetric. Ed Pegg provides a very elegant construction of the graph, which I’ll repeat below.

1. Consider the 35 3-subsets of {1, 2, 3, 4, 5, 6, 7}. We’ll call these triads.
2. Label the vertices of a Fano plane with the numbers {1, 2, 3, 4, 5, 6, 7}. By applying even permutations to the labels, we can obtain a total of distinct labellings of the Fano plane. Ed refers to these as fanos.
3. Let the 50 vertices of the Hoffman-Singleton graph be associated with the 35 triads and 15 fanos. We connect two triads with an edge if they are disjoint; we connect a triad to a fano if the former appears as a line in the latter.

It should be evident, with a little thought, that this gives a 7-regular graph. In fact, Pegg’s construction makes it straightforward to show that the conditions of the AMS problem are satisfied:

• Property (i) is obvious from the fact that it’s 7-regular.
• Property (ii) can be verified by a very short case check. No two fanos are connected, so three acquaintances must either (a) all be triads, or (b) one fano and two triads. Case (a) is impossible, since we cannot find three disjoint subsets of size 3 in the 7-element set {1,2,3,4,5,6,7}. Case (b) is similarly impossible, as we can’t find two disjoint lines in the Fano plane.
• Property (iii) takes slightly longer to verify. We split this into four cases, namely two fanos (c), one fano and unconnected triad (d), two triads intersecting in a single point (e) and two triads intersecting in two points (f). For case (c), note that two distinct fanos cannot share two lines (lest they are identical), so can share at most one. It’s similarly impossible for two fanos to have no common lines, so they must share precisely one. The corresponding triad, and only that triad, is connected to both fanos. For case (d), the three non-collinear points in the Fano plane can be extended to a set of four non-collinear points (an oval), the complement of which is a line disjoint from the original triad. In case (e), we can trivially embed the lines in a single labelled Fano plane. Case (f) is even easier: take the complement of the union of the two triads; this gives a third triad disjoint from the other two.

Hence, we can confirm that the Hoffman-Singleton graph actually works. Hooray!

Ed Pegg actually went further than merely constructing the graph; he designed a playable card game based on it. The case-bash above is precisely the solution to Pegg’s Puzzle E, showing that any two non-connected vertices have a unique common neighbour.

Although Pegg’s construction separates the vertices into clusters of sizes 15 (the fanos) and 35 (the triads), rest assured that the graph is actually vertex-transitive and thus all vertices are equivalent.

Beyond this?

It is a theorem, the Hoffman-Singleton theorem, that any graph satisfying those three properties must have either degree 2, 3, 7 or 57. In the first three cases, existence and uniqueness have been confirmed (with the pentagon, Petersen graph and Hoffman-Singleton graph, respectively).

As for degree 57, it is unknown whether such a graph exists. I very much hope that there is a unique such graph (since it would be another exceptional object, and exceptional objects are beautiful things), and there’s probably a really clever construction. It would have 3250 vertices and 92625 edges.

Observe that the Hoffman-Singleton graph contains lots of Petersen graphs, which in turn contain lots of pentagons. Similarly, I imagine that a degree-57 Moore graph would feature a plethora of Hoffman-Singleton graphs embedded within it.

Two open problems about the distribution of the primes have been solved within about 24 hours of each other, namely the ternary Goldbach conjecture and a weakened form of the twin prime conjecture. Let’s look at these in approximate chronological order:

Ternary Goldbach conjecture

This states that every odd integer can be expressed as the sum of three primes. Hardy and Littlewood first proved this for sufficiently large integers, conditional on the truth of the Riemann Hypothesis. Later, in 1937, Vinogradov proved the result for sufficiently large numbers without requiring the Riemann Hypothesis; this became known as Vinogradov’s theorem.

An lower bound was established, which was gradually reduced to in 2002, still ahead of the computer searches (which have only probed up to about ). Very recently, this was further reduced to 5, as desired; see the paper by Helfgott.

We can corollary-snipe† at this point, and state that this trivially implies that every integer can be expressed as the sum of four primes. The full version of Goldbach’s conjecture is equivalent to every integer being expressible as the sum of three primes.

Twin primes conjecture

Twin primes are primes which differ from their closest prime neighbours by 2. For instance, {5, 7}, {59, 61} and {, } are examples of pairs of twin primes. It has been conjectured that there are infinitely many such pairs, although no proof is known (there was a purported proof in 2004, but it was fallacious).

It has now been proved that infinitely many pairs of primes have a distance of at most 70000000. Again, this can be corollary-sniped to deduce that there exists some  such that infinitely many pairs of primes are separated by precisely N. No explicit values of N are known, and the original conjecture states that 2 has this property.

Conclusion and footnotes

Together, these results provide another piece of evidence for the following conjecture:

There are infinitely many pairs of exciting proofs published within 70000000 milliseconds of each other.

† Corollary-sniping is a rather impolite and dishonourable practice in which one jumps on a big theorem proved by someone else, and proves one or more corollaries using it. For instance, if someone suddenly exclaimed “Hence Fermat’s Last Theorem!” just as Andrew Wiles proved the necessary cases of the Tanayama-Shimura conjecture, that would have been an epic case of corollary-sniping (if that happened, hopefully the prize would still have been awarded to Wiles). An actual instance was when Xia’s proof that particles can be projected to infinity in finite time under Newtonian gravity was famously corollary-sniped to deduce that the n-body problem is undecidable.