This Week's Finds in Mathematical Physics (Week 182)
ba...@math.removethis.ucr.andthis.edu
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! David Eppstein has shown you only need 3 terms if n is
less than or equal to 12,500. 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/
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 are:
(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 if and 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 stuff is intimately connected to the tale of incidence
geometry I've been telling in "week178", "week180" and "week181",
but I don't have the energy now to say how. Instead, I'll
just recommend two references - both of which explain the
mysterious word "hosohedron":
2) H. S. M. Coxeter, Generators and relations for discrete groups,
Springer, Berlin, 1984.
3) 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
www.science.uva.nl/research/math/examen/2002/scriptiejoris.ps
Okay. Now - back to that conference at the Mathematical Sciences
Research Institute! You can look at transparencies and watch videos
of the talks here:
4) MSRI streaming video archive, Spring 2002,
http://www.msri.org/publications/video/index04.html
If you like watching math talks, there's a lot to see here - not just
this one conference, but all the MSRI conferences! For example, right
after the nonabelian Hodge theory conference there was one on conformal
field theory and supersymmetry, featuring talks by bigshots like Richard
Borcherds, Dan Freed, Igor Frenkel, Victor Kac, and Jean-Bernard Zuber -
just to name a few. You can see talks by all these folks.
But anyway, let me start by telling you what nonabelian Hodge theory is....
Hmm. I guess I should *start* by telling you what *abelian* Hodge
theory is!
In its simplest form, Hodge theory talks about how differential forms on
a smooth manifold get extra interesting structure when the manifold has
extra interesting structure. To warm up, let me remind you about what
we can do when our manifold has *no* extra interesting structure.
Whenever we have a smooth manifold M there's an "exterior derivative"
operator d going from p-forms on M to (p+1)-forms on M. This is just a
generalization of grad, curl, div and all that. In particular it
satisfies
d^2 = 0,
so the space of "closed" p-forms:
{w: dw = 0}
is a subspace of the "exact" p-forms:
{w: w = du for some u}.
This makes it fun to look at the vector space of closed p-forms modulo
exact p-forms. This is called the "pth de Rham cohomology group of M",
or
H^p(M)
for short. It only depends on the topology of M; its size keeps track
of the number of p-dimensional holes in M. When M is compact, it
agrees with the cohomology computed in a bunch of other ways that
topologists like.
Fine. But now, suppose M has a Riemannian metric on it! Then we can
write down a version of the Laplacian for differential forms. A
function is a 0-form, so we're just generalizing the Laplacian you
already know and love. Differential forms whose Laplacian is zero are
called "harmonic". Every harmonic p-form is closed, but if M is compact
life is even better: the vector space of harmonic p-forms is isomorphic
to the pth de Rham cohomology of M.
This is great: it means the de Rham cohomology, which only depends on
the *topology* of M, can also be thought of as the space of solutions
of a *differential equation* on M! This gets topologists and analysts
talking to each other, and has all sorts of marvelous spinoffs and
generalizations.
Some people call this stuff "Hodge theory". But Hodge theory
goes further when M has more structure - most notably, when it's
a Kaehler manifold!
A Kaehler manifold is to the complex plane as a Riemannian manifold is
to the real line. More precisely, it's is a manifold whose tangent
spaces have been made into *complex* vector spaces and equipped with a
*complex* inner product. Of course the real part of the inner product
makes it into a Riemannian manifold. That lets us parallel transport
vectors, so we demand a compatibility condition: parallel transporting
a vector and then multiplying it by i is the same as multiplying it by i
and then parallel transporting it! This makes complex analysis work
well on Kaehler manifolds.
Now, if you've taken complex analysis, you may remember how people use
it to find solutions of Laplace's equation... like when they're studying
electrostatics, or the flow of fluids with no viscosity or vorticity -
an idealization that von Neumann mockingly called "dry water". On
the complex plane we can talk about "holomorphic" functions, which
satisfy the Cauchy-Riemann equation:
_ _
df/dz = 0 (note: df/dz = df/dx + i df/dy)
and also the complex conjugates of these, called "antiholomorphic"
functions, which satisfy
df/dz = 0 (note: df/dz = df/dx - i df/dy)
Both holomorphic and antiholomorphic functions are automatically harmonic,
so we can find solutions of Laplace's equation this way. But even better,
every harmonic function is a linear combination of a holomorphic and an
antiholomorphic one!
All this stuff works much more generally for p-forms on Kaehler
manifolds. To get going, let's think a bit more about the complex
plane. If we have any 1-form on the complex plane we can write it as a
linear combination of dx and dy, where x and y are the usual coordinates
on the plane. But things get nicer if we work with *complex-valued*
differential forms. Then we can form linear combinations like
dz = dx + idy
and
_
dz = dx - idy
and express any 1-form as a linear combination of *these* in a unique way.
We call these the (1,0) and (0,1) parts of our 1-form.
This means that if we have a function f, we can take its exterior
derivative of f and chop it into its (1,0) part and (0,1) part:
_
df = Df + Df
_
These guys D and D are called "Dolbeault operators". People usually
write them using nice curly lower-case d's like you see in a partial
derivative, but I can't do that here: I'm a prisoner of low technology!
Anyway, it turns out that
_
Df = 0
is just a slick way of writing Cauchy-Riemann equation, which says that
f is holomorphic. You should check this for yourself! Similarly,
Df = 0
says that f is antiholomorphic.
Now let me say how all this stuff generalizes to arbitrary Kaehler
manifolds. We can decompose any p-form on a Kaehler manifold into
its (i,j) parts where i+j = p. For example, a (1,2)-form
in 4 dimensions might look something like this in complex coordinates:
_ _ _ _
f dz_1 ^ dz_3 ^ dz_2 + g dz_2 ^ dz_3 ^ dz_4.
We have
_
d = D + D
_
where D maps (i,j)-forms to (i+1,j)-forms, while D maps (i,j)-forms
to (i,j+1)-forms. This allows us to take the de Rham cohomology
groups of our manifold M and write them as a direct sum of smaller
vector spaces, which I'll call
H^{i,j}(M)
for short.
So far I don't think I've used anything about the metric on M, so all
this would work whenever M is a so-called "complex manifold". But if we
really have a Kaehler manifold, and it's compact, we can say more: a
p-form is harmonic if and only if all its (i,j) parts are. This
means H^{i,j}(M) is isomorphic to the space of harmonic (i,j)-forms.
_
Alternatively, you can describe H^{i,j}(M) just in terms of D:
you just take the (i,j)-forms in here:
_
{w: Dw = 0}
modulo those in here:
_
{w: w = Du for some u}
This is called the "(i,j)th Dolbeault cohomology group of M".
That's Hodge theory in a nutshell. There's even *more* you can
do when M is a Kaehler manifold, but I'm getting a little tired,
so I'll just let you read about that here:
5) R. O. Well, Differential analysis on complex manifolds,
Springer, Berlin, 1980.
This is a really *great* book for learning about all sorts of good
geometry stuff, starting with differential forms and working on up
through Hodge theory, pseudodifferential operators, sheaves and so on.
But anyway, I've given you a little taste of Hodge theory.
The main thing to remember is that when your manifold is complex,
the cohomology becomes "bigraded": instead of just
H^p(M)
you get
H^{i,j}(M).
So now, what's nonabelian Hodge theory?
The basic idea is simple: instead of askng what extra structure the
*homology groups* get when M is a complex manifold, we ask what
extra structure the *homotopy type* of M gets when M is a complex
manifold. The homotopy type includes invariants like the homotopy
groups, but also more. How are these constrained by the fact that
M is complex?
Unfortunately, to describe the answer - even a little teeny part of the
answer - I need to turn up the math level a notch.
For starters we can consider the fundamental group pi_1(M). But this is
hard to relate to differential geometry, so we will immediately water it
down by picking an algebraic group G and looking at homomorphisms of
pi_1(M) into G. These are basically the same thing as flat G-bundles
over M, so it's easier to see how M being a complex manifold affects
things. We can even be sneaky and study this for all G at once by forming
a group PI_1(M) called the "proalgebraic completion" of pi_1(M).
This is a proalgebraic group (an inverse limit of algebraic groups)
which a contains pi_1(M) and has the property that any homomorphism from
pi_1(M) into an algebraic group G extends uniquely to a proalgebraic
group homomorphism from PI_1(M) to G.
It's nice to ask what extra structure PI_1(M) gets when M is
a complex manifold, because this question has a nice answer.
To get ready for how nice the answer is, first go back to plain old
abelian Hodge theory. Note that making the cohomology of M bigraded
gives an obvious way for the algebraic group C*, the nonzero complex
numbers, to act on the cohomology. The reason is that for each integer
there's a representation of C* where the number z acts as multiplication
by z^n, so gradings are just another way of talking about C* actions.
Since the cohomology of M is automatically graded, putting *another*
grading on it amounts to letting C* act on it.
So in plain old Hodge theory, the answer to "What extra structure
does the cohomology of M get when M is complex?" is:
"It gets an action of C*!"
And it turns out that in nonabelian Hodge theory, the answer to
"What extra structure does Pi_1(M) get when M is complex?" is:
"It gets an action of C*!"
This is incredibly cool, but the story goes a lot further. The
fundamental group is just the beginning; you can do something similar
for the higher homotopy groups - but it's a lot more subtle. In fact,
you can do something similar directly to the homotopy type of M! When M
is a compact complex manifold, there's a homotopy type called the
"schematization of M" whose fundamental group is PI_1(M) - and there's
an action of C* on this homotopy type!
I learned about most of this fancy stuff from an incredibly lucid
talk by Bertrand Toen. Unfortunately there seems to be no video of his
talk, since he gave it down the hill at U. C. Berkeley instead of at
the MSRI - and the handwritten notes at the MSRI website are rather
illegible. So you want to learn more about this, you should probably
start with this quick summary of abelian Hodge theory:
6) Tony Pantev, Review of abelian Hodge theory,
http://www.msri.org/publications/ln/msri/2002/introstacks/pantev/1/index.html
and then take the deep plunge into this paper:
7) Ludmil Katzarkov, Tony Pantev and Bertrand Toen, Schematic homotopy
types and non-abelian Hodge theory I: The Hodge decomposition,
available at math.AG/0107129.
There are a lot of model categories and n-categories lurking in
the background of this subject, as well as ideas that originated
in physics, like "Higgs bundles". For the brave reader I recommend
these papers:
8) Bertrand Toen, Toward a Galoisian interpretation of homotopy theory,
available as math.AT/0007157.
This answers the question: "the fundamental group is to covering
spaces as the whole homotopy type is to... what?" The fact that
it's in French probably makes it easier to understand.
9) Bertrand Toen and Gabriele Vezzosi, Algebraic geometry over model
categories (a general approach to derived algebraic geometry),
available as math.AG/0110109.
This is only for badass mathematicians who find algebraic geometry
and homotopy theory insufficiently mindblowing when taken separately.
Ever wondered what an affine scheme would be like if you replaced the
ground field by an E_infinity ring spectrum? Then this is for you.
-----------------------------------------------------------------------
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http://math.ucr.edu/home/baez/
For a table of contents of all the issues of This Week's Finds, try
http://math.ucr.edu/home/baez/twf.html
A simple jumping-off point to the old issues is available at
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If you just want the latest issue, go to
Squark
> ...
> 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.
But in your web page you write (n,m) = (6,4)! What's more, you
give a picture example, which seems more like (6,4) than (3,7).
However, in the picture at some of the points 4 bats (supposedly
hexagons) meet and at other points 3 bats meet*. This doesn't
seem to make sense because all of the hexagons' angles are
supposed to be equal, aren't they? Moreover, why would 3 meet at
a point if we're in the (6,4) case?
* The triples join with the tails and the quadruples with the
wings.
Best regards,
The Puzzled Squark
------------------------------------------------------------------
Write to me using the following e-mail:
Skvark_N...@excite.exe
(just spell the particle name correctly and use "com" rather than
"exe")
Jeffery
series?
In this post, you imply that the dodecahedron and icosahedron
correspond to the E_8 group, although elsewhere you say they
correspond to the H_3 group.
David Eppstein
ba...@math.removethis.ucr.andthis.edu wrote:
> 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! David Eppstein has shown you only need 3 terms if n is
> less than or equal to 12,500. 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/
Thanks for the credit and the pointer, but I think other people searched
earlier and farther on this problem than I have. As far as I know the
current record for searching is more like n=10^12, by Allan Swett.
--
David Eppstein UC Irvine Dept. of Information & Computer Science
epps...@ics.uci.edu http://www.ics.uci.edu/~eppstein/
joe
>|--------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?
Maybe this is an oversimplification:
2/n + 1/n + 1/n + ....... + 1/n
|-----(m-1) terms---------------|
or, when m = 4
2/n + 1/n + 1/n
Revolution
> As far as I know the
> current record for searching is more like n=10^12
To your knowledge, has ANY speculation ever been computer tested this
deep and then proven NOT to be true?
Gerry Myerson
In article <ac2f081f.02062...@posting.google.com>,
kl...@webspan.net (Revolution) wrote:
Back in the pre-computer days, someone noticed that up to n = 100
the irreducible factors of x^n - 1 had only 0, 1, and -1 as
coefficients. But it turned out that x^105 - 1 has a factor with
a coefficient 2.
More to the point, there are quite a few things in number theory
that are known to be false even though they check out way far.
E.g., that at least half of the primes up to n are of the form
4k + 3; that the number of primes up to n does not exceed the
logarithmic integral up to n; that the sum of the Mobius function
up to n does not exceed the square root of n in absolute value.
--
Gerry Myerson (ge...@mpce.mq.edi.ai) (i -> u for email)
Gerry Myerson
Eppstein <epps...@ics.uci.edu> wrote:
=> In article <aercjt$k7b$1...@glue.ucr.edu>,
=> ba...@math.removethis.ucr.andthis.edu wrote:
=>
=> > For example: clearly every fraction of the form 4/n can be written
=> > using 4 terms - but can you always make do with just 3? Nobody
=> > knows! David Eppstein has shown you only need 3 terms if n is less
=> > than or equal to 12,500. For example:
=> >
=> > 4/8689 = 1/2175 + 1/1718250 + 1/14929874250.
=> >
=> > For much more on this, see:
=> >
=> > 1) David Eppstein, Egyptian fractions,
=> > http://www.ics.uci.edu/~eppstein/numth/egypt/
=>
=> Thanks for the credit and the pointer, but I think other people
=> searched earlier and farther on this problem than I have. As far as
=> I know the current record for searching is more like n=10^12, by
=> Allan Swett.
At the URL given above, there's a link to Swett's page. There, Swett
is now claiming 10^14.
Jim Heckman
On 24-Jun-2002, jeffery...@hotmail.com (Jeffery) wrote:
> Is the dihedral group D_n the same D_n group as in the ABCDEFGHI
> series?
No. Somewhat confusingly, the dihedral groups D_n are the Weyl
groups of the I_n series. They are the symmetry groups of the 2-d
polygons. The D_n Weyl groups are index-2 subgroups of the
B_n/C_n Weyl groups, which are the symmetry groups of the
n-dimensional analogues of the cube and octahedron.
> In this post, you imply that the dodecahedron and icosahedron
> correspond to the E_8 group, although elsewhere you say they
> correspond to the H_3 group.
The symmetry group of the dodecahedron and icosahedron is the
H_3 Weyl group.
Note that the notation for the H and I series isn't quite
standardized. Some authors reverse the two, calling H what
other authors call I, and vice versa.
--
Jim Heckman
Mark Hudson
Hi,
"joe" <joecy...@yahoo.com> wrote in message news:<vAJR8.47173$db.7...@twister.tampabay.rr.com>...
I thought that Egyptian fractions had to have unit numerators, hence
2/n would not be permitted.
Mark.
John Baez
Readers of sci.physics.research may be interested to know
that I just visited Oz on his farm near Oxford. I taught
him some math and physics; he showed me the megaliths of
Avebury. No fireballs were thrown.
I am now in an internet cafe in London. I apologize for
taking a while to respond to some interesting articles;
my internet access will be somewhat limited until July 15th.
In article <939044f.02062...@posting.google.com>,
Squark <fii...@yahoo.com> wrote:
>ba...@math.removethis.ucr.andthis.edu wrote in message
>news:<aercjt$k7b$1...@glue.ucr.edu>...
>> 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.
>
>But in your web page you write (n,m) = (6,4)! What's more, you
>give a picture example, which seems more like (6,4) than (3,7).
Yes, on my webpage
http://math.ucr.edu/home/baez/week182.html
I switched to talking about the (6,4) tiling of the hyperbolic
plane, because the first picture I found by him was an example
of (6,4).
>However, in the picture at some of the points 4 bats (supposedly
>hexagons) meet and at other points 3 bats meet*. This doesn't
>seem to make sense because all of the hexagons' angles are
>supposed to be equal, aren't they?
Escher has subdivided each hexagon into 3 triangles as part of
creating this design. He's also done some other clever things...
like putting a picture of a black bat inside each triangle,
and leaving white space between the bats that creates pictures of
angels! It's a bit hard to see the math behind all this artistry.
However, if you look at the wingtips of a triple of bats, you'll
see they form the vertices of a hexagon. These hexagons tile the
hyperbolic plane, with 4 hexagons meeting at each vertex. So,
there is really a (6,4) tiling here.
Escher corresponded with Coxeter about these issues.
Lest one think this is unrelated to physics, don't forget
that we can mod out the hyperbolic plane by a discrete group
like the symmetry group of this picture and get a surface
with constant negative curvature. With a little gauge-fixing,
these surfaces can be thought of as points in the configuration
space for 3d general relativity with nonzero cosmological constant!
For more on related stuff, see Kirill Krasnov's paper "Holography
and Riemann Surfaces":
http://xxx.lanl.gov/abs/hep-th/0005106
There is in fact a complete dictionary relating Riemann surfaces,
hyperbolic 3-manifolds and 3d quantum gravity, which develops
further in more recent papers.
John Baez
In article <8a8c1f93.02062...@posting.google.com>,
Jeffery <jeffery...@hotmail.com> wrote:
>Is the dihedral group D_n the same D_n group as in the ABCDEFGHI
>series?
No; as Chris Hillman explained this is just a notational
coincidence - though "week182" hints at a subtle relation between
the two.
>In this post, you imply that the dodecahedron and icosahedron
>correspond to the E_8 group, although elsewhere you say they
>correspond to the H_3 group.
I should have been a bit clearer.
First of all, there's a more or less straightforward classification
of finite reflection groups in terms of ABCDEFGHI Coxeter diagrams.
Among the simplest examples of finite reflection groups are the
reflection/rotation symmetry groups of the Platonic solids:
The symmetry group of the tetrahedron is called A_3.
The symmetry group of the cube or octahedron is called B_3.
The symmetry group of the dodecahedron or icosahedron is called H_3.
The group A_3 is part of an infinite series of A_n groups
since the tetrahedron has analogues in every dimension n.
The group B_3 is part of an infinite series of B_n groups
since the cube and octahedron have analogues in every dimension n.
The group H_3 is not part of an infinite series of H_n groups
since the dodecahedron and icosahedron do not have analogues
in every dimension n...
... however, they do have analogues in dimension 4, whose symmetry
group is called H_4! The buck stops in 4 dimensions if you're
looking for regular polytopes with 5-fold symmetry.
For more on the hyperdodecahedron, the hypericosahedron, and
other Platonic delights, try:
http://math.ucr.edu/home/baez/platonic.html
Anyway, in "week182" I was describing a *different* and rather
*bizarre* relationship between Platonic solids and the Coxeter-Dynkin
diagrams of type E. As I mentioned, this extends to a relationship
between finite subgroups of SO(3) and Coxeter-Dynkin diagrams of
types A, D, and E. This is sometimes called the "McKay correspondence" -
though as Minhyong Kim recently pointed out here on sci.physics.research,
there are aspects involving singularity theory which were understood
before the work of McKay.
In "week182" I was trying to boil this stuff down to its simplest
essence, so that even ancient Egyptians could understand it.
Unfortunately I forgot to say that's what I was doing! I'll fix that.
I talked about this mysterious relationship between ADE and
Platonic solids here:
http://math.uc.edu/home/baez/week65.html
and McKay talks about it here:
http://math.ucr.edu/home/baez/ADE.html
The most detailed online explanation is probably this:
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/math/examen/2002/scriptiejoris.ps
joe
news:a254de04.02062...@posting.google.com...
I thought that if 1/n + 1/n was going to be permitted then we were not
dealing with Egyptian fractions. Sorry.
Ralph E. Frost
John Baez <ba...@galaxy.ucr.edu> wrote in message
news:aft1c1$41u$1...@glue.ucr.edu...
>
...
> http://math.ucr.edu/home/baez/ADE.html
>
> The most detailed online explanation is probably this:
>
> 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/math/examen/2002/scriptiejoris.ps
Can you please explain a bit about what a binary polyhedral group is and
give some examples of things that that would fit into that classification?
Ralph Frost