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Category: groups

Kasha-eating dragons

This semester I’m teaching a first course in representation theory. On campus, IRL! It’s a bit strange, using a big lecture room for a handful of students, everyone wearing masks, keeping distances, etc.



So far, this is their only course on campus, so it has primarily a social function. The breaks in between are infinitely more important than the lectures themselves. I’d guess breaks take up more than one third of the four hours scheduled.

At first, I hoped to make groups and their representations relevant by connecting to the crisis at hand, whence the the symmetries of Covid-19 post, and the Geometry of Viruses series of posts.

Not a great idea. I guess most of us are by now over-saturated with Corona-related news, and if students are allowed to come to campus just one afternoon per week, the last thing they want to hear about is, right, Covid.

So I need to change tactics. By now we’ve reached the computation of character tables, and googling around I found this MathOverflow-topic: Fun applications of representations of finite groups.

The highest rated answer, by Vladimir Dotsenko, suggests a problem attributed to Kirillov:

An example from Kirillov’s book on representation theory: write numbers 1,2,3,4,5,6 on the faces of a cube, and keep replacing (simultaneously) each number by the average of its neighbours. Describe (approximately) the numbers on the faces after many iterations.

A bit further down the list, the Lecture notes on representation theory by Vera Serganova are mentioned. They start off with a variation of Kirillov’s question (and an extension of it to the dodecahedron):

Hungry knights. There are n hungry knights at a round table. Each of them has a plate with certain amount of food. Instead of eating every minute each knight takes one half of his neighbors servings. They start at 10 in the evening. What can you tell about food distribution in the morning?

Breakfast at Mars. It is well known that marsians have four arms, a standard family has 6 persons and a breakfast table has a form of a cube with each person occupying a face on a cube. Do the analog of round table problem for the family of marsians.

Supper at Venus. They have five arms there, 12 persons in a family and sit on the faces of a dodecahedron (a regular polyhedron whose faces are pentagons).

Perhaps the nicest exposition of the problem (and its solution!) is in the paper Dragons eating kasha by Tanya Khovanova.

Suppose a four-armed dragon is sitting on every face of a cube. Each dragon has a bowl of kasha in front of him. Dragons are very greedy, so instead of eating their own kasha, they try to steal kasha from their neighbors. Every minute every dragon extends four arms to the four neighboring faces on the cube and tries to get the kasha from the bowls there. As four arms are fighting for every bowl of
kasha, each arm manages to steal one-fourth of what is in the bowl. Thus each
dragon steals one-fourth of the kasha of each of his neighbors, while at the same
time all of his own kasha is stolen. Given the initial amounts of kasha in every
bowl, what is the asymptotic behavior of the amounts of kasha?

I can give them quick hints to reach the solution:

  • the amounts of kasha on each face gives a vector in R6, which is an S4-representation,
  • calculate the character of this kasha-representation,
  • use the character table of S4 to decompose the representation into irreducibles,
  • identify each of the irreducible factors as instances in the kasha-representation,
  • check that the food-grabbing operation is an S4-morphism,
  • remember Schur’s lemma, and compute the scaling factors on each irreducible component,
  • conclude!

But, I can never explain it better than Khovanova’s treatment of the kasha-eating dragons problem.

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de Bruijn’s pentagrids

In a Rhombic tiling (aka a Penrose P3 tiling) we can identify five ribbons.

Opposite sides of a rhomb are parallel. We may form a ribbon by attaching rhombs along opposite sides. There are five directions taken by sides, so there are five families of ribbons that do not intersect, determined by the side directions.

Every ribbon determines a skeleton curve through the midpoints of opposite sides of the rhombs. If we straighten these skeleton curves we get a de Bruijn’s pentagrid.



A pentagrid is a grid of the plane consisting of five families of parallel lines, at angles multiples of 72o=360o/5, and with each family consisting of parallel lines with a spacing given by a number γi for 0i4, satisfying
γ0+γ1+γ2+γ3+γ4=0

The points of the plane in the j-th family of the pentagrid are
{xR2 | x.vj+γjZ}
where vj=ζj=(cos(2πj/5),sin(2πj/5)).

A pentagrid is regular if no point in R2 belongs to more than two of the five grids. For almost all choices γ0,,γ4 the corresponding pentagrid is regular. The pentagrid coordinates of a point xR2 are the five integers (K0(x),,K4(x)Z5 defined by
Kj(x)=x.vj+γj
with r the smallest integer greater of equal to r, and clearly the function Kj(x) is constant in regions between the j-th grid lines.



With these pentagrid-coordinates one can associate a vertex to any point xR2
V(x)=j=04Kj(x)vj
which is constant in regions between grid lines.

Here’s de Bruijn‘s result:

For x running over R2, the vertices V(x) coming from a pentagrid determine a Rhombic tiling of the plane. Even better, every Rhombic tiling comes from a pentagrid.



We’ll prove this for regular pentagrids (the singular case follows from a small deformation).

Any intersection point x0 of the pentagrid belongs to exactly two families of parallel lines, so we have two integers r and s with 0r<s4 and integers kr,ksZ such that x0 is determined by the equations {x.vr+γr=krx.vs+γs=ks In a small neighborhood of x0, V(x) takes the values of four vertices of a rhomb. These vertices are associated to the 5-tuples in Z5 given by (K0(x0),,K4(x0))+ϵ1(δ0r,,δ4r)+ϵ2(δ0s,,δ4s) with ϵ1,ϵ2{0,1}. So, the intersection points of the regular pentagrid lines correspond to rhombs, and the regions between the grid lines (which are called meshes) correspond to vertices, whose positions are given by V(x).

Observe that these four vertices give a thin rhombus if the angle between vr and vs is 144o and determine a thick rhombus if the angle is 72o.

Not all pentagrid coordinates (k0,,k4) occur in the tiling though. For xR2 we have
Kj(x)=x.vj+γj+λj(x)
with 0λj(x)<1. In a regular pentagrid at most two of the λj(x) can be zero and so we have 0<λ0(x)++λ4(x)<5 and we have assumed that γ0++γ4=0, giving us j=04Kj(x)=x.(j=04vj)+j=04γj+j=04λj(x)=j=04λj(x) The left hand side must be an integers and the right hand side a number strictly between 0 and 5. This defines the index of a vertex
ind(x)=j=04Kj(x){1,2,3,4}
Therefore, every vertex in the tiling may be represented as
k0v0++k4v4
with (k0,,k4)Z5 satisfying j=04kj{1,2,3,4}. If we move a point along the edges of a rhombus, we note that the index increases by 1 in the directions of v0,,v4 and decreases by 1 in the directions v0,,v4.



It follows that the index-values of the vertices of a thick rhombus are either 1 and 3 at the 72o angles and 2 at the 108o angles, or they are 2 and 4 at the 72o angles, and 3 at the 108o angles. For a thin rhombus the index-values must be either 1 and 3 at the 144o angles and 2 at the 36o angles, or 2 and 4 at the 144o angles and 3 at the 36o angles.

We still have to show that this gives a legal Rhombic tiling, that is, that the gluing restrictions of Penrose’s rhombs are satisfied.



We do this by orienting and colouring the edges of the thin and thick rhombus towards the vertex defined by the semi-circles on the rhombus-pieces. Edges connecting a point of index 3 to a point of index 2 are coloured red, edges connecting a point of index 1 to a point of index 2, or connecting a point of index 3 to one of index 4 are coloured blue.

We orient green edges pointing from index 2 to index 1, or from index 3 to index 4. As this orientation depends only on the index-values of the vertices, two rhombs sharing a common green edge also have the same orientation on that edge.

On each individual rhomb, knowing the green edges and their orientation forced the red edges and their orientation, but we still have to show that if two rhombs share a common red edge, this edge has the same orientation on both (note in the picture above that a red edge can both flow from 2 to 3 as well as from 3 to 2).

From the gluing conditions of Penrose’s rhombs we see that if PQ be a red edge, in common to two rhombs, and these rhombs have angle α resp. β in P, then α and β must be both less that 90o or both bigger than 90o. We translate this in terms of the pentragrid.



The two rhombs correspond to two intersection points A and B, and we may assume that they both lie on a line l of family 0 of parallel lines (other cases are treated similarly by cyclic permutation), and that A is an intersection with a family p-line and B with a family q-line, with p,q{1,2,3,4}. The interval AB crosses the unique edge common to the two rhombs determined by A and B, and we call the interval AB red if jKj(x) is 2 on one side of AB and 3 on the other side. The claim about the angles α and β above now translates to: if AB is red, then p+q is odd (in the picture above p=1 and q=2).

By a transformation we may assume that γ0=0 and that l is the Y-axis. For every yR we then get
{K1(0,y)=y.sin(2π/5)+γ1K2(0,y)=y.sin(4π/5)+γ2K3(0,y)=y.sin(4π/5)+γ3K4(0,y)=y.sin(2π/5)+γ4
and γ1+γ4 and γ2+γ3 are not integers (otherwise the pentagrid is not regular). If y runs from to + we find that
K1(0,y)+K4(0,y)γ1+γ4={01
with jumps from 0 to 1 at places where (γ1+γ4γ1)/sin(2π/5)Z and from 1 to 0 when γ4/sin(2π/5)Z. (A similar result holds replacing K1,K4,γ1,γ4,sin(2π/5) by K2,K3,γ2,γ3,sin(4π/5).

Because the points on l intersecting with 1-family and 4-family gridlines alternate (and similarly for 2- and 3-family gridlines) we know already that pq. If we assume that p+q is even, we have two possibilities, either {p,g}={1,3} or {2,4}. As γ0=0 we have γ1+γ2+γ3+γ4=0 and therefore
γ1+γ4+γ2+γ3=1
It then follows that K1(0,y)+K2(0,y)+K3(0,y)+K4(0,y)=1 or 3 between the points A and B. Therefore K0(y,0)++K4(0,y) is either 1 on the left side and 2 on the right side, or is 3 on the left side and 4 on the right side, so AB must be green, a contradiction. Therefore, p+q is odd, and the orientation of the red edge in both rhombs is the same. Done!

An alternative way to see this correspondence between regular pentagrids and Rhombic tilings as as follows. To every intersection of two gridlines we assign a rhombus, a thin one if they meet at an acute angle of 36o and a thich one if this angle is 72o, with the long diagonal dissecting the obtuse angle.



Do this for all intersections, surrounding a given mesh.



Attach the rhombs by translating them towards the mesh, and finally draw the colours of the edges.



Another time, we will connect this to the cut-and-project method, using the representation theory of D5.

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Borcherds’ favourite numbers

Whenever I visit someone’s YouTube or Twitter profile page, I hope to see an interesting banner image. Here’s the one from Richard Borcherds’ YouTube Channel.

Not too surprisingly for Borcherds, almost all of these numbers are related to the monster group or its moonshine.

Let’s try to decode them, in no particular order.

196884

John McKay’s observation 196884=1+196883 was the start of the whole ‘monstrous moonshine’ industry. Here, 1 and 196883 are the dimensions of the two smallest irreducible representations of the monster simple group, and 196884 is the first non-trivial coefficient in Klein’s j-function in number theory.

196884 is also the dimension of the space in which Robert Griess constructed the Monster, following Simon Norton’s lead that there should be an algebra structure on the monster-representation of that dimension. This algebra is now known as the Griess algebra.

Here’s a recent talk by Griess “My life and times with the sporadic simple groups” in which he tells about his construction of the monster (relevant part starting at 1:15:53 into the movie).

1729

1729 is the second (and most famous) taxicab number. A long time ago I did write a post about the classic Ramanujan-Hardy story the taxicab curve (note to self: try to tidy up the layout of some old posts!).

Recently, connections between Ramanujan’s observation and K3-surfaces were discovered. Emory University has an enticing press release about this: Mathematicians find ‘magic key’ to drive Ramanujan’s taxi-cab number. The paper itself is here.

“We’ve found that Ramanujan actually discovered a K3 surface more than 30 years before others started studying K3 surfaces and they were even named. It turns out that Ramanujan’s work anticipated deep structures that have become fundamental objects in arithmetic geometry, number theory and physics.”

Ken Ono

24

There’s no other number like 24 responsible for the existence of sporadic simple groups.

24 is the length of the binary Golay code, with isomorphism group the sporadic Mathieu group M24 and hence all of the other Mathieu-groups as subgroups.

24 is the dimension of the Leech lattice, with isomorphism group the Conway group Co0=.0 (dotto), giving us modulo its center the sporadic group Co1=.1 and the other Conway groups Co2=.2,Co3=.3, and all other sporadics of the second generation in the happy family as subquotients (McL,HS,Suz and HJ=J2)



24 is the central charge of the Monster vertex algebra constructed by Frenkel, Lepowski and Meurman. Most experts believe that the Monster’s reason of existence is that it is the symmetry group of this vertex algebra. John Conway was one among few others hoping for a nicer explanation, as he said in this interview with Alex Ryba.

24 is also an important number in monstrous moonshine, see for example the post the defining property of 24. There’s a lot more to say on this, but I’ll save it for another day.

60

60 is, of course, the order of the smallest non-Abelian simple group, A5, the rotation symmetry group of the icosahedron. A5 is the symmetry group of choice for most viruses but not the Corona-virus.

3264

3264 is the correct solution to Steiner’s conic problem asking for the number of conics in PC2 tangent to five given conics in general position.



Steiner himself claimed that there were 7776=65 such conics, but realised later that he was wrong. The correct number was first given by Ernest de Jonquières in 1859, but a rigorous proof had to await the advent of modern intersection theory.

Eisenbud and Harris wrote a book on intersection theory in algebraic geometry, freely available online: 3264 and all that.

248

248 is the dimension of the exceptional simple Lie group E8. E8 is also connected to the monster group.

If you take two Fischer involutions in the monster (elements of conjugacy class 2A) and multiply them, the resulting element surprisingly belongs to one of just 9 conjugacy classes:

1A,2A,2B,3A,3C,4A,4B,5A or 6A

The orders of these elements are exactly the dimensions of the fundamental root for the extended E8 Dynkin diagram.

This is yet another moonshine observation by John McKay and I wrote a couple of posts about it and about Duncan’s solution: the monster graph and McKay’s observation, and E8 from moonshine groups.

163

163 is a remarkable number because of the ‘modular miracle’
eπ163=262537412640768743.99999999999925
This is somewhat related to moonshine, or at least to Klein’s j-function, which by a result of Kronecker’s detects the classnumber of imaginary quadratic fields Q(D) and produces integers if the classnumber is one (as is the case for Q(163)).

The details are in the post the miracle of 163, or in the paper by John Stillwell, Modular Miracles, The American Mathematical Monthly, 108 (2001) 70-76.

Richard Borcherds, the math-vlogger, has an entertaining video about this story: MegaFavNumbers 262537412680768000

His description of the j-function (at 4:13 in the movie) is simply hilarious!

Borcherds connects 163 to the monster moonshine via the j-function, but there’s another one.

The monster group has 194 conjugacy classes and monstrous moonshine assigns a ‘moonshine function’ to each conjugacy class (the j-function is assigned to the identity element). However, these 194 functions are not linearly independent and the space spanned by them has dimension exactly 163.

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