Hello dear group.

I have recently started to read the paper "D-branes, Quivers and ALE
Instantons" by Michael Douglas (hep-th/9603167) in order to understand
how the low energy effective field theory of D-branes on orbifolds is
derived. My trouble is that I don't know anything about ALE spaces.
Already equations 2.1 on page 2 managed to confuse me.

I see that V there is the potential generated by single-unit-of-charge
monopoles located at x_i. A, then, would be the vector potential
appearing if we think of those monopoles as magnetic. However, it
would then be a connection on a non-trivial bundle on R^3 minus the x_i
(or that times the "t" axis). How, then, is the line element to be well
defined? At first I thought that's the reason we get R^4/Gamma rather
than just R^4, but then I noticed Gamme is Z_n (n being the # of
monopoles), so for n = 1 we get just R^4, even though the bundle is still
non-trivial.

Moreover, on page 3 I read t is, in fact, an angular coordinate! This is
entirely mysterious, since we should get R^4/Gamma, not
R^3 X S^1/Gamma.

I would appreciate an explanation of these points, as well as some
basic references on ALE spaces, preferably available online. Thx!

Best regards,
Squark

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"Squark" wrote in message ...
Hello dear group.

I have recently started to read the paper "D-branes, Quivers and ALE
Instantons" by Michael Douglas (hep-th/9603167) in order to understand
how the low energy effective field theory of D-branes on orbifolds is
derived. My trouble is that I don't know anything about ALE spaces.
Already equations 2.1 on page 2 managed to confuse me.

I see that V there is the potential generated by single-unit-of-charge
monopoles located at x_i. A, then, would be the vector potential
appearing if we think of those monopoles as magnetic. However, it
would then be a connection on a non-trivial bundle on R^3 minus the x_i
(or that times the "t" axis). How, then, is the line element to be well
defined? At first I thought that's the reason we get R^4/Gamma rather
than just R^4, but then I noticed Gamme is Z_n (n being the # of
monopoles), so for n = 1 we get just R^4, even though the bundle is still
non-trivial.

Moreover, on page 3 I read t is, in fact, an angular coordinate! This is
entirely mysterious, since we should get R^4/Gamma, not
R^3 X S^1/Gamma.

I would appreciate an explanation of these points, as well as some
basic references on ALE spaces, preferably available online. Thx!

Best regards,
Squark


You should read the paper by Cifford Johnson and Rob Myers
"ASPECTS OF TYPE IIB THEORY ON ALE SPACES" which is much clearer.

I'll try and get to your technical questions later.


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On 27 Dec 2003, Squark wrote:

I have recently started to read the paper "D-branes, Quivers and ALE
Instantons" by Michael Douglas (hep-th/9603167)

I suppose that you wanted to say "Michael Douglas and Greg Moore". First
of all, ALE stands for "Asymptotically Locally Euclidean". It means that
it is a geometry that has a center, and if you are very far from this
center, the geometry looks exactly like the flat Euclidean space. It
follows that at infinity, the metric is asymptotic to R^4 / Gamma. Let me
talk about 4D ALE spaces only, because this is what we usually mean
anyway.

Gamma is a discrete group that identifies some points of the R^4. It must
be a symmetry of this R^4. We want to preserve some supersymmetry, and
therefore the holonomy of the ALE space must be in one of the SU(2)
subgroups of SO(4). The ALE space is a Kahler (which for d=4 automatically
means hyperKahler) manifold - a noncompact version of the K3 manifold. In
fact, the local geometry near singularities of K3 manifolds can look
exactly as an ALE space.

The geometry of the ALE space near the core is difficult, smooth and
complicated, see the equation (2.1), and it guarantees that the full
holonomy group is exactly SU(2) - nothing bigger and nothing smaller.
However if you only consider parallel transport around curves that avoid
the core, you would conclude that the holonomy is the same like the
holonomy of R^4 / Gamma - which is the asymptotic metric. And the holonomy
of R^4 / Gamma is Gamma, of course. We see that Gamma must be a discrete
subgroup of SU(2).

The discrete subgroups of SU(2) can be easily classified. Let us rename
R^4 to C^2, because we said that the ALE space should be a Kahler manifold
anyway. The two complex coordinates in C^2 transform as "2" of SU(2).
SU(2)/Z_2 is isomorphic to SO(3), and therefore every subgroup of SU(2)
gives us a subgroup of SO(3) - just by forgetting the difference between
the identity and the 2.pi-rotation. OK, what are the finite subgroups of
SO(3)?

We can pick a Z_n group of rotations around the z-axis by 2.pi/n. Then we
can take the symmetries of a polygon - that also includes the pi-rotation
around the x-axis (that flips the orientation of the polygon). Finally, we
have the symmetry groups of Platonic polyhedra that are also subgroups of
SO(3). That's it. Note that in all cases, the corresponding symmetry of
SU(2) always acts on all four real coordinates of R^4 = C^2 nontrivially.

Now realize that the symmetry of a polyhedron is isomorphic to the
symmetry of the dual polyhedron; the dual polyhedron is constructed as the
envelope of vertices that are sitting in the centers of the faces of the
original polyhedron. The tetrahedron is self-dual, the cube is dual to the
octahedron, and the dodecahedron is dual to the icosahedron.

(4~4, 6~8, 12~20), "~" represents the duality.

Notice that the duality exchanges faces with vertices, while mapping edges
to edges, and therefore the Euler character #(vertices)-#(edges)+#(faces)
is unchanged - in fact, it is 2 for all these polyhedra.

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.

So far my labels "A,D,E" that resemble the simply-laced Lie groups might
seem as arbitrary as Kepler's isomorphism between the polyhedra and the
planets. The nice surprise is that our new map is totally canonical; the
idea is sometimes referred to as the McKay correspodence. String theory
is the theory that should give physical meaning to all nice ideas in
mathematics, so here it is: if you define type IIA string theory or
M-theory on the singular space C^2 / Gamma where Gamma is a discrete group
of SU(2) that is labeled by the ADE label, the physics of string theory
will be totally smooth and well-defined and there will be an enhanced
gauge symmetry living on the fixed point of Gamma - i.e. at the origin of
the space - and it will be exactly one of these A,D,E groups (A is
SU(rank-1), D is SO(2.rank), and E_rank are the exceptional groups).

The Cartan subalgebra of the group comes from the 3-form potential
integrated over the homologically nontrivial 2-cycles of the ALE - or
C^2/Gamma (so far). There are exactly "rank" of them. The D2-branes (or
M2-branes if we go from type IIA string theory to M-theory) are charged
under this 3-form potential, and they can wrap various cycles. The lattice
of integer homology becomes the root lattice. For the cycles that are
rational curves, the wrapped D2-branes give exactly the physical states
that carry some charges under the Cartan subalgebra (roots), and they
exactly transform as the rest of the gauge supermultiplets. If we work
with the singular orbifold C^2/Gamma, all these 2-cycles are shrunk to
zero size, and the wrapped membranes are massless. When we take all states
into account and compute their interactions, we conclude that they are
described by a supersymmetric A/D/E gauge theory localized near the
singularity.

So far I was talking about the singular case C^2 / Gamma only. We can
study the smooth geometry, too, and the Eguchi-Hanson metric is the first
step to describe it quantitatively. The physical interpretation of
"blowing up" the singularity is the Higgs mechanism that breaks the gauge
group into its Cartan subalgebra U(1)^{rank}. Not surprisingly, all the
non-Cartan generators are broken, and the corresponding W-bosons become
massive. The mass is exactly calculable from the nonzero area of the
2-cycles - i.e. from the mass of the membranes that wrap these cycles.

I see that V there is the potential generated by single-unit-of-charge
monopoles located at x_i. A, then, would be the vector potential
appearing if we think of those monopoles as magnetic.

Yes!

However, it would then be a connection on a non-trivial bundle on R^3
minus the x_i (or that times the "t" axis). How, then, is the line
element to be well defined?

It is not only well-defined, but it is explicitly defined by equation
(2.1) itself. If you worry that the metric behaves in a singular fashion
at the position of the source x_i, then you should not worry because the
term that contains the problematic A(x_i) is proportional to V^{-1}, and
as you can see from the definition of V (that diverges at x_i), V^{-1} (x_i)
vanishes. So any ambiguity or singularity of the solution near the sources
x_i is completely illusory, and if you use better coordinates, it is
straightforward to see that the geometry around the point x_i is
completely smooth. In fact, you might worry that the metric has some
problems not only at x_i, but at the whole "Dirac string". But it is not a
problem because the gauge transformation that relates the potentials A in
various patches is exactly the Kaluza-Klein U(1) transformation that
rotates the coordinate "t", and therefore the Dirac string only means
that the angular coordinate can't be defined globally, i.e. the geometry
is *not* a Cartesian product but rather a more complicated "bundle".

At first I thought that's the reason we get R^4/Gamma rather
than just R^4, but then I noticed Gamme is Z_n (n being the # of
monopoles), so for n = 1 we get just R^4, even though the bundle is still
non-trivial.

No, the coordinates x1,x2,x3,t are not the same coordinates in which the
action of Gamma is described by the simple linear action - which is not
hard to see because the four coordinates in R^4 enter on equal footing
while "t" seems sort of priviliged among x1,x2,x3, right? You must make a
coordinate redefinition in order to see that the geometry is R^4 / Gamma.

Moreover, on page 3 I read t is, in fact, an angular coordinate! This is
entirely mysterious, since we should get R^4/Gamma, not
R^3 X S^1/Gamma.

It is a gravitational instanton, and in any kind of instanton, you always
work with a Wick-rotated, Euclidean time coordinate. The whole manifold is
Euclidean, with signature ++++, and relativity dictates that you should
not be asking which direction is time and which direction is space because
they are related by coordinate transformations. Nevevertheless, "t"
originated from time, but in the gravitational instanton it is a periodic
variable, and it has all the features to be called an angular variable.

Your conclusion that the geometry is R^3 x S^1 / Gamma is entirely
incorrect. The very purpose of all these "subtleties" with the potential
of the magnetic monopoles - those that you've complained so much about -
is exactly to make the geometry being more than just a product, and in
fact, it converts it to something that looks asymptotically as R^4 / Gamma.

All the best
Lubos
__________________________________________________ ____________________________
E-mail: fax: +1-617/496-0110 Web: http://lumo.matfyz.cz/
phone: work: +1-617/496-8199 home: +1-617/868-4487
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ ^^^^^^^^^^^^^^^^^^^^^^^^^^^^
Superstring/M-theory is the language in which God wrote the world.

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?

Jeffery Winkler

httpP://www.geocities.com/jefferywinkler

"Lubos Motl" wrote in message
...
...So any ambiguity or singularity of the solution near the sources
x_i is completely illusory, and if you use better coordinates, it is
straightforward to see that the geometry around the point x_i is
completely smooth.

Am I right assuming the spacetime is the total space of the gauge
bundle for the corresponding configuration of magnetic monopoles,
completed suitably over at the locations of the monopoles
themselves? Here, "completed suitable" means completed via the
total space of the U(1) bundle over S^2 corresponding to the
relevant monopole charge. In particular, for location with a single
unit of charge, the fiber is S^3 rather than the generic U(1). I guess
it may be related to the mechanism of appearance of magnetic
monopoles via SU(2) - U(1) SSB. The S^3 is then identified with
a fiber of the SU(2) bundle. The later only seems to work for unit
charges so something about it is obscure.

Best regards,
Squark

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On 30 Dec 2003, Squark wrote:

Am I right assuming the spacetime is the total space of the gauge
bundle for the corresponding configuration of magnetic monopoles,
completed suitably over at the locations of the monopoles
themselves?

Someone will need to recheck us, but I think that what you say is
essentially right. The Eguchi-Hanson instanton is written as a S^1 fiber
bundle (the bundle is S^1 i.e. a copy of the U(1) group manifold - I hope
that this is what you meant by the fiber) over the base that is
essentially R^3. Well, the solution is more than just its topology, and
you must take all the data about the metric seriously.

Here, "completed suitable" means completed via the
total space of the U(1) bundle over S^2 corresponding to the
relevant monopole charge. In particular, for location with a single
unit of charge, the fiber is S^3 rather than the generic U(1).

I think that this is extremely confusing. Why is it S^3? On the contrary,
the fiber S^1 shrinks to a point at the locus of the magnetic monopole
because the squared radius of S^1 is proportional to V^{-1} which goes to
zero at x_i.

I guess it may be related to the mechanism of appearance of magnetic
monopoles via SU(2) - U(1) SSB. The S^3 is then identified with a
fiber of the SU(2) bundle. The later only seems to work for unit
charges so something about it is obscure.

OK, sorry, I don't understand this line of reasoning at all. I see no
SU(2) group to start with, no S^3 fiber, nothing from what you're talking
about. Are we talking about the same solution? Yes, every U(1) magnetic
monopole can be extended into a smooth solution of an SU(2) gauge theory,
but there is apparently no SU(2) theory here to start with, is there?
__________________________________________________ ____________________________
E-mail: fax: +1-617/496-0110 Web: http://lumo.matfyz.cz/
phone: work: +1-617/496-8199 home: +1-617/868-4487
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ ^^^^^^^^^^^^^^^^^^^^^^^^^^^^
Superstring/M-theory is the language in which God wrote the world.

Someone will need to recheck us, but I think that what you say is
essentially right. The Eguchi-Hanson instanton is written as a S^1 fiber
bundle (the bundle is S^1 i.e. a copy of the U(1) group manifold - I hope
that this is what you meant by the fiber) over the base that is
essentially R^3.

More precisely, it's a U(1) bundle over R^3\{0} + something else over 0.
I don't think this "something else" is S^1, so the whole space is not an S^1
bundle.

Well, the solution is more than just its topology, and
you must take all the data about the metric seriously.

If we consider the formula for the metric over R^3\{0} and set V = 1 for
a moment (without changing A), we get at each point the metric that makes
the fiber orthogonal to the horizontal subspace, the metric on the
horizontal subspace being the lift from R^3 and the metric on the fiber a constant
U(1)-invariant metric. Returning the V factor, the metric on the fiber is
rescaled in an R^3-location dependant metric and so is the metric we lift
to the horizontal subspaces.

I think that this is extremely confusing. Why is it S^3? On the contrary,
the fiber S^1 shrinks to a point at the locus of the magnetic monopole
because the squared radius of S^1 is proportional to V^{-1} which goes to
zero at x_i.

You are right. If we decompose R^3\{0} as S^2 x R using speherical
coordinates, the bundle + connection is a lift from S^2. The total space of
the bundle is S^3 x R and the metric decomposes in a compatible fashion
before the V-rescaling. The V-rescaling comfortably depends only on the
R factor. The distance to the origin is easily seen to be finite, so the
primary remaing question is how the metric on S^3 behaves when we
come closer to it. I believe that due to SO(3) symmetry the fiber over the
origin is either a point - and then the global topology is R^4 - or S^2 -
and then the global topology is S^2 x R^2. It seems that the S^3 option doesn't
lead to a smooth manifold.

Best regards,
Squark

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A nice picture of the topology of this pair of spaces (when the
fibre over 0 is a point) is to
treat R^4 as the quaternions and then map any q \in R^4 to
q i bar(q) were bar(q) is the quaternion conjugate. Its
easy to see (because bar(q i bar(q)) = - q i bar(q) )
that the image is the imaginary quaternions so this is a map
R^4 - R^3

To find the fibres of this map note that

(i) q i bar(q) = 0 if and only if q = 0

(ii) if q \neq 0 , q i bar(q) = p i bar(p) if and only if
q = p exp(i theta)

so the fibres are the orbits of the action of the
circle exp(i theta) acting on the right of R^4.

- Michael

On 2 Jan 2004, Squark wrote:

More precisely, it's a U(1) bundle over R^3\{0} + something else over 0.
I don't think this "something else" is S^1, so the whole space is not an S^1
bundle.

The fiber at the origin {0} degenerates into a point, as I've already
said, and the geometry around this point is that of the R^4 / Z_2
orbifold, as I explain below.

You are right. If we decompose R^3\{0} as S^2 x R using speherical
coordinates, the bundle + connection is a lift from S^2. The total space of
the bundle is S^3 x R ...

No, no, no. The total (ALE) space, as repeated 10 times in the paper and
elsewhere, is asymptotic to

R^4 / Z_N = R^+ x (S^3 / Z_N)

where S^3 / Z_N is the so-called lens space. Once again, the space R^4 /
Z_N can be written as a cone - namely the S^3/Z_N fibration over the
semi-infinite line R^+ that starts at the origin and runs to infinity. The
general Eguchi-Hanson solution is a deformed version of the same metric,
and the fiber is S^3/Z_N at infinity while it approaches S^3/Z_2 at the
chosen center of the multi-center solution.

Consider the simple single-center Eguchi-Hanson instanton, eqn. (2.1) from
the Douglas+Moore paper, with x_1=(0,0,0). Then V=1/|x|. Use the radial
coordinate y where |x|=C.y^2. Then the radial part of V.dx^2 is
|x|^{-1}.dx^2 = C.y^{-2}.(2.C.y.dy)^2 = dy^2 for C^3 = 1/4.

OK, so this is the standard radial direction "y" going from 0 to infinity.
What about the other three directions? The other directions are the two
angular coordinates in the 3-vector "x", and also "t". For a fixed value
of "y", what space do they span? You see that if "y" is small, V is large,
and the term "dx^2" in the metric has a huge coefficient V=1/|x|
= 1 / (C.y^2) in front of it. On the other hand, the term dx^2 can be
rewritten in spherical coordinates, and its angular part is
|x|^2.d Omega^2 = C^2.y^4.d Omega^2 where

d Omega^2 = d theta^2 + sin^2(theta).d phi^2.

You see that the |x|^2 coefficient partly wins, and the overall
coefficient in front of d Omega^2 goes like C.y^2 which is the same
coefficient like the coefficient V^{-1} in front of the first term in the
metric (2.1), namely (dt+A.dx)^2; "A" depends on |x| as 1/|x| trivially,
and the nontrivial dependence is only on the angular parts of "x". In
other words, all three angular parts of the metric scale like the same
factor C.y^2, which is OK for a fixed fiber whose size linearly increases
with "y". What is the topology of this fiber?

It is a nontrivial S^1 fiber over S^2. This S^2 is spanned by theta,phi -
the angular coordinates of "x" - while the fiber S^1 is parameterized by
"t". The whole space of this fiber bundle can be seen to be compact
(finite volume), and it still has a SU(2) isometry that acts in the
obvious way on "theta,phi" (like on a sphere) while transforming the
vector field "A" accordingly - which turns out to be a gauge
transformation which is equivalent to a position dependent shift of "t".

The only compact 3-dimensional spaces with a faithful action of an SU(2)
symmetry are S^3 and S^3/Z_2, and you can see that we deal with the
latter. Write S^3 i.e. the SU(2) group manifold as a S^1 fiber over
SU(2)/U(1), and then notice that we don't really have the double cover of
the sphere SU(2)/U(1) but rather S^2 itself, and therefore our geometry
has the Z_2 identification.

Therefore the solution has a S^3/Z_2 fiber everywhere at y=(0,infty). If
you consider the multi-center solution and write it using a radial
coordinate around one of these centers (x_i), the fiber will be S^3/Z_2
near the center, but it will become S^3/Z_N with "N" jumping by one every
time you swallow another center.

There also exists a smoothed out version of this solution, where the
vicinity of the singularity is replaced by a smooth geometry, and the
singular point at the origin of R^4/Z_2 is replaced by a S^2 - which
becomes the homologically nontrivial cycle of the geometry.

All the best
Lubos
__________________________________________________ ____________________________
E-mail: fax: +1-617/496-0110 Web: http://lumo.matfyz.cz/
phone: work: +1-617/496-8199 home: +1-617/868-4487
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ ^^^^^^^^^^^^^^^^^^^^^^^^^^^^
Superstring/M-theory is the language in which God wrote the world.

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