We've gone from ALE to ADE...
In article ,
Jeffery wrote:
Lubos Motl wrote in message
...
Let us now label the Z_n subgroups of SU(2) as A_n, the symmetries of the
polygons as D_n, and the symmetries of the platonic polyhedra 4~4, 6~8,
12~20 as E_6, E_7, E_8, respectively. It follows from my notation that the
discrete subgroups of SU(2) admit an ADE classification much like the
simply laced Lie groups: they either belong to the infinite A or D
families, or they can be of the special types E_6, E_7, E_8. Note that
A,D,E are exactly the simply-laced groups i.e. those who only contain
simple links in their Dynkin diagrams. Each discrete group Gamma can act
on C^2 = R^4, and generate an ALE space C^2 / Gamma.
I thought the symmetry groups of the dodecahedron and iocosahedron
were the H groups, which explains why they don't extend beyond 4
dimensions. Am I thinking of something else?
Yes. You're thinking of the well-known and important correspondence
between "Coxeter diagrams" and "reflection groups" - that is, finite
subgroups of O(n) that are generated by reflections. Under this
correspondence, the rotation/reflection symmetry group of the
dodecahedron (or icosahedron) corresponds to the Coxeter diagram H_3:
5
o----o-----o
However, Motl is using a deeper and very fascinating correspondence
between "simply-laced Dynkin diagrams" and "finite subgroups of SU(2)".
This is called the McKay correspondence.
A simply-laced Dynkin diagram is just a connected Coxeter diagram
where all the edges are unlabelled. These are just the A, D, and E
Dynkin diagrams:
An, which has n dots like this:
o---o---o---o
Dn, which has n dots, where n 3:
o
/
o---o---o---o
\
o
E6, E7, and E8:
o o o
| | |
o--o--o--o--o o--o--o--o--o--o o--o--o--o--o--o---o
Under the McKay correspondence these correspond to the cyclic
groups, the double covers of the dihedral groups, and the double
covers of the rotational symmetry groups of the tetrahedron,
cube/octahedron, and dodecahedron/octahedron, respectively.
A quick explanation of the McKay correspondence can be found
on my website, in a page written by McKay:
http://math.ucr.edu/home/baez/ADE.html
A summary appeared as part of "week65":
.................................................. .........................
There is a cool relationship between the ADE diagrams and the
symmetry groups of the Platonic solids, called the McKay correspondence.
Briefly, this is what you do to get it. First, learn about SO(3) and
SU(2) from "week61" or somewhere. Then, take the symmetry group of
a Platonic solid, or more generally any finite subgroup G of SO(3).
Since SO(3) has SU(2) as a double cover, you can get a double cover of
G, say G~, sitting inside SU(2). For example, if G was the symmetry
group of the icosahedron, G~ would be the icosians (see "week24").
Since G~ is finite, it has finitely many irreducible representations.
One of these will be 2-dimensional, coming from the spin-1/2 representation
of SU(2). Draw a dot for each of the irreducible representations, but
draw the dot for this 2-dimensional representation with red ink.
Now, when you tensor this 2d rep with any other irreducible rep R,
you get a direct sum of irreducible reps; draw one line for the dot
from R to each other dot for each time that other irreducible rep
appears in this direct sum. What do you get? Well, you get an
"affine Dynkin diagram" of type A, D, or E, which is like the
usual Dynkin diagram but with an extra dot thrown in. And you get
all of them this way!
.................................................. .........................
I also gave a low-budget explanation in terms of "Egyptian fractions"\0
in "week182". It's a low-budget explanation because it exhibits a
pretty direct connection between Platonic solids and "simply-laced
affine Dynkin diagrams", but doesn't completely explain where this
connection is coming from.
.................................................. .........................
Also available at
http://math.ucr.edu/home/baez/week182.html
June 19, 2002
This Week's Finds in Mathematical Physics (Week 182)
John Baez
It's been a long time, but in the last Week's Finds I was telling you
about my adventures this spring in northern California, and I hadn't
quite gotten around to telling you about that cool conference on
"Nonabelian Hodge Theory" at the MSRI in Berkeley. I'll continue my
story about that now...
....but first, a little detour through the Nile valley!
Egyptians liked to write fractions as the sum of reciprocals of
integers. For example, instead of writing
5/6
those folks would write something like
1/2 + 1/3
Nobody is sure why, but one possibility is that they started with a
neat notation for 1/n, and then wanted to extend this to handle other
fractions, and couldn't think of anything better.
Of course they *could* have written m/n as
1/n + ........... + 1/n
|--------m terms------|
but they preferred to use as few terms as possible. This leads to some
tricky questions. For example: clearly every fraction of the form 4/n
can be written using 4 terms - but can you always make do with just 3?
Nobody knows! Alan Swett claims to have shown you only need 3 terms if
n is less than or equal to 10^14. For example:
4/8689 = 1/2175 + 1/1718250 + 1/14929874250.
For much more on this, see:
1) David Eppstein, Egyptian fractions,
http://www.ics.uci.edu/~eppstein/numth/egypt/
2) Alan Swett, The Erdos-Strauss conjecture,
http://math.uindy.edu/swett/esc.htm
Egyptian fraction problems have a spooky way of showing up in
different unrelated mathematical contexts... which have a spooky way
of turning out not to be unrelated after all!
For example, suppose we are trying to classify all the Platonic
solids. We're looking for ways to tile the surface of a sphere
with regular n-gons, with m meeting at each vertex. Suppose there
is a total of V vertices, E edges, and F faces. Since the Euler
characteristic of the sphere is 2, we have
V - E + F = 2.
Since each face has n edges but 2 faces meet along each edge,
we have
nF = 2E.
Since each vertex has m edges meeting it but each edge
meets 2 vertices, we also have
mV = 2E.
Putting these equations together we get
2E(1/n + 1/m - 1/2) = 2
or
1/n + 1/m = 1/2 + 1/E.
An Egyptian fractions problem! It's obvious that this can only
have solutions if 1/n + 1/m 1/2. And interestingly, all the
solutions of this inequality do indeed correspond to Platonic solids...
at least if n,m 2. Here they a
(n,m) = (3,3) tetrahedron
(n,m) = (3,4) octahedron
(n,m) = (4,3) cube
(n,m) = (3,5) icosahedron
(n,m) = (5,3) dodecahedron
The cases n = 1,2 don't give Platonic solids in the usual sense:
after all, most people don't like polygons to have just 1 or 2 edges.
Neither do the cases m = 1,2, since most people don't like polyhedra
to have just 1 or 2 faces meeting at a vertex!
One can argue about whether these are irrational prejudices. But it's
actually good to study *all* unordered pairs of natural numbers with
1/n + 1/m 1/2
since they correspond to *all* the isomorphism classes of finite subgroups
of the rotation group! The Platonic solids have their symmetry groups,
which don't change when we switch n and m. The solution (n,1)
corresponds to the cyclic group Z_n: the symmetries of a regular n-gon,
where you're not allowed to flip it over. The solution (n,2)
corresponds to the dihedral group D_n: the symmetries of a regular n-gon
where you *are* allowed to flip it over.
In some weird sense, maybe we should think of Z_n and D_n as the
symmetry groups of Platonic solids with only 1 or 2 faces. I'll
leave you to ponder the Platonic solids with only 1 or 2 vertices.
If you get stuck, look up the word "hosohedron"!
The story gets better if we also consider solutions of
1/n + 1/m = 1/2
which formally correspond to Platonic solids where the number
E of edges is infinite. In fact, these correspond to tilings
of the plane by regular polygons:
(n,m) = (3,6): tiling by triangles
(n,m) = (6,3): tiling by hexagons
(n,m) = (4,4): tiling by squares
Similarly, solutions of
1/n + 1/m 1/2
give tilings of the hyperbolic plane: for example, Escher used
(n,m) = (3,7) in some of his prints.
Let me try to arrange this information in a table, using lines
to separate the spherical, planar and hyperbolic regions:
n=1 n=2 n=3 n=4 n=5 n=6 n=7
m=1 Z_1 Z_2 Z_3 Z_4 Z_5 Z_6 Z_7
m=2 Z_2 D_2 D_3 D_4 D_5 D_6 D_7
------------=======
m=3 Z_3 D_3 tetrahedron cube dodecahedron | hexagonal |
| tiling |
---------=============------------
m=4 Z_4 D_4 octahedron | square |
| tiling |
| --------
m=5 Z_5 D_5 isosahedron ||
||
------------ | hyperbolic tilings
m=6 Z_6 D_6 | triangular |
| tiling |
| -------------
m=7 Z_7 D_7 ||
||
It's not very pretty in ASCII, but hopefully you get the idea!
Now, the same Egyptian fraction problem comes up when studying other
problems, too. For example, suppose you are trying to find a basis of
R^n consisting of unit vectors that are all at 90-degree or 120-degree
angles from each other. We can describe a problem like this by drawing
a bunch of dots, one for each vector, and connecting two dots with an
edge when they're supposed to be at a 120-degree angle from each other.
If two dots are not connected, they should be at right angles to one
another.
So, for example, this diagram tells us to find a basis for R^3
consisting of unit vectors all at 120 degree angles from each other:
o
/ \
/ \
o-----o
It's easy to see this is impossible, since three vectors all
at 120 degrees from each must lie in a plane - so they can't be
linearly independent. On the other hand, this diagram gives a
solvable problem:
o-----o-----o
You just pick two unit vectors at right angles to each other and
wiggle the third one around until it's at a 120-degree angle to both.
It's not hard.
So, the question is: which diagrams give solvable problems?
This is actually a very fun puzzle: it's very famous, but most books
manage to make it seem really boring and "technical", so you should
really spend some time thinking about it for yourself. I'll give away
the answer, but I won't say how you prove it's true.
First, it's easy to see that if a diagram consists of a bunch of separate
pieces, and you can solve the problem for each piece, you can solve
the problem for the whole diagram. So, it's sufficient to consider
the case of connected diagrams.
Second, a connected diagram can only give a solvable problem if it's
Y-shaped, like this:
o
|
o
|
o--o--o--o--o--o--o--o--o--o
Third, a diagram like this gives a solvable problem only if
1/k + 1/n + 1/m 1
where (k,n,m) are the numbers labelling the tips of the Y when we
number it like this:
3
|
2
|
4--3--2--1--2--3--4--5--6--7
So for example, this particular problem is not solvable because
1/4 + 1/3 + 1/7 1.
Now, it's easy to see what we can only get 1/k + 1/n + 1/m 1 if one
of the numbers is 1 or 2, so except for the boring solution (1,1,1) we
might as well assume one of the numbers is 2. By symmetry we can assume
this number is k. We are thus looking for pairs (n,m) with
1/2 + 1/n + 1/m 1
or in other words
1/n + 1/m 1/2.
This is the same problem as before! So the problem we're dealing
with now is very much like classifying Platonic solids!
Even better, these diagrams I've been drawing are called "Dynkin
diagrams", and we can use them to get certain incredibly important
finite-dimensional Lie algebras called "simply-laced simple Lie algebras".
For a taste of how this works, reread "week65" and some previous Weeks.
Similarly, we get certain *infinite-dimensional* Lie algebras
called "simply-laced affine Lie algebras" when
1/n + 1/m = 1/2,
and "simply-laced hyperbolic Kac-Moody algebras" when
1/n + 1/m 1/2.
So, our whole big table above translates into a table of Lie algebras!
Let me draw it with the standard names of these Lie algebras below their
diagrams. Unfortunately, I'll have to make it very small to fit
everything in. So, for example, I'll draw the so-called E8 Dynkin
diagram:
o
|
o--o--o--o--o--o--o
as this puny miserable thing:
o
oooooo
This is what we get:
n=1 n=2 n=3 n=4 n=5 n=6
o o o o o o
m=1 o oo ooo oooo ooooo ooooooo
A2 A3 A4 A5 A6 A7
o o o o o o
m=2 oo ooo oooo ooooo oooooo oooooooo
A3 D4 D5 D6 D7 D8
----------------==
| |
o o o o o | o |
m=3 ooo oooo ooooo oooooo ooooooo | ooooooooo |
| |
A4 D5 E6 E7 E8 | E8^1 |
| |
------------===========-----------------
| |
o o o | o | o o
m=4 oooo ooooo oooooo | ooooooo | oooooooo oooooooooo
| |
A5 D6 E7 | E7^1 |
| |
| ----------
||
o o o || o o o
m=5 ooooo oooooo ooooooo || oooooooo ooooooooo ooooooooooo
||
A6 D7 E8 ||
|| hyperbolic Kac-Moody algebras
--------- |
| |
o o | o | o o o
m=6 oooooo ooooooo | oooooooo | ooooooooo oooooooooo oooooooooooo
A7 D8 | |
| E8^1 |
| |
| ----------
||
||
This mysterious way that the same Egyptian fraction problem shows up
in classifying Platonic solids and simply-laced simple Lie algebras is
actually the tip an iceberg sometimes called the "McKay correspondence" -
though important aspects of it go back to the theory of Kleinian
singularities. I talked about the McKay correspondence in "week65",
so that's a good place to dig deeper, but you should really
look at some of the references in there, and also these two -
both of which explain the mysterious word "hosohedron":
3) H. S. M. Coxeter, Generators and relations for discrete groups,
Springer, Berlin, 1984.
4) Joris van Hoboken, Platonic solids, binary polyhedral groups,
Kleinian singularities and Lie algebras of type A,D,E, Master's
Thesis, University of Amsterdam, 2002, available at
http://www.science.uva.nl/research/m...riptiejoris.ps