Also available at http://math.ucr.edu/home/baez/week181.html
May 1, 2002
This Week's Finds in Mathematical Physics (Week 181)
John Baez
At the beginning of April I went up to Mathematical Sciences Research
Institute in Berkeley to a conference on ncategories and nonabelian
Hodge theory, which I should tell you about sometime... but the very
first thing I did is take a detour and give a talk at the University
of California at Santa Cruz.
U. C. Santa Cruz has a beautiful campus, with enormous rolling grassy
fields and redwood groves. And indeed it must be pretty idyllic there,
because the main thing the students used to complain about was that
the courses *aren't graded*  which makes it harder for them to get
jobs when they leave this paradise. I think grades are being phased
in now. Too bad!
Anyway, I wound up talking a lot to Richard Montgomery, who teaches in
the math department and works on the gravitational 3body problem.
Except when one mass is much smaller than the other two  see my
discussion of Lagrange points in "week150"  this problem is still
packed with mysteries! Montgomery and other have turned their attention
to the case where all 3 masses are equal and found solutions with
amazing properties: for example, one where the total angular momentum is
zero and all 3 masses chase each other around a figure8shaped curve!
For details, see:
1) Alain Chenciner and Richard Montgomery, A remarkable periodic
solution of the threebody problem in the case of equal masses,
Ann. of Math. 152 (2000), 881901. Also available as math.DS/0011268.
For a more popular account see:
2) Richard Montgomery, A new solution to the threebody problem,
AMS Notices 48 (May 2001), 471481. Also available as
http://www.ams.org/notices/200105/feamontgomery.pdf
and for Java applets illustrating this and other solutions based on
computer simulations by Carles Simo, try this:
3) Bill Casselman, A new solution to the three body problem  and more,
at http://www.ams.org/newinmath/cover/orbits1.html
There are lots of other unsolved puzzles concerning point particles
interacting via Newtonian gravity. They're not very practical, but
they're lots of fun!
For example, can we find a periodic orbit where N particles move around
in the plane and trace out an arbitrary desired *braid*? For strongly
attractive potentials like 1/r^a where a is greater than or equal to 2,
the answer is "yes"  this is not hard to prove by variational methods.
However, the question remains largely open for gravity, where a = 1.
See:
4) Cristopher Moore, Braids in classical gravity, Phys. Rev. Lett. 70
(1993), 36753679. Also available at
http://www.santafe.edu/media/workingpapers/9207034.pdf
Cristopher Moore, The 3body (and nbody) problem,
http://www.santafe.edu/~moore/gallery.html
In fact, Moore first discovered the figure8 solution of the gravitational
3body problem in this paper, using computer calculations. His student
Michael Nauenberg made a movie of it, which you can find with many others
on Moore's website.
Also see:
5) Richard Montgomery, The Nbody problem, the braid group, and
actionminimizing periodic solutions, Nonlinearity 11 (1998), 363371.
http://count.ucsc.edu/~rmont/papers/NbdyBraids.pdf
There is also the issue of whether a particle can shoot off to infinity
in a finite amount of time. Of course this isn't possible in the real
world, but Newtonian physics has no "speed limit", and we're idealizing
our particles as points. So, if two or more of them get arbitrarily
close to each other, the potential energy they liberate could in
principle give another particle enough kinetic energy to zip off to
infinity! Then our solution becomes undefined after a finite amount
of time.
Zhihong Xia showed this can actually happen in the Nbody problem
for N = 5 or bigger:
6) Zhihong Xia, The existence of noncollision singularities in Newtonian
systems, Ann. Math. 135 (1992), 411468.
or for a more popular account:
7) Donald G. Saari and Zhihong Xia, Off to infinity in finite time,
AMS Notices (May 1995), 538546. Also available at
http://www.ams.org/notices/199505/saari2.pdf
As far as I know, the question is still open for N = 4. Another
question concerns how *likely* it is for our solution to become
undefined in a finite amount of time. If it's infinitely improbable, we
say we have "asymptotic completeness" for the Nbody problem. I seem to
recall that the Nbody problem has been shown asymptotically complete
for N = 3, but not higher N.
Now  back to my tale of Lie groups and geometry!
So far I've talked about how to any complex simple Lie group G we can
associate an "incidence geometry": a generalization of projective
geometry having G as its symmetry group. Each different type of
"figure" in this geometry corresponds to a dot in the Dynkin diagram
for G. For example, when G = SL(4,C) we have
ooo
points lines planes
For each dot, the space of all figures of the corresponding type is
called a "Grassmannian", and it's a manifold of the form G/P, where P
is a "maximal parabolic" subgroup of G.
More generally, any subset of dots in the Dynkin diagram corresponds to
a type of "flag". A flag is a collection of figures satisfying certain
incidence relations. For example, this subset:
xox
points lines planes
corresponds to the type of flag consisting of a point lying on a plane.
The space of all flags of a particular type is called a "flag manifold",
and it's a manifold of the form G/P, where P is a "parabolic" subgroup
of G.
I also said a bit about how we can quantize this entire story! This is
actually what got me interested in this whole business. In loop quantum
gravity we run around claiming that space is made of quantum triangles,
quantum tetrahedra and the like  see "week113" and "week134" if you
don't believe me. The whole theory emerges naturally from the way
Euclidean and Lorentzian geometry are related to representations of the
rotation and Lorentz groups, but it got me wondering how the story
would change if we changed the group to something fancier  as we might
in a theory that tried to unify gravity with other forces, for example.
So I started studying incidence geometry and group representations, and
wound up learning lots of math so beautiful that it has, so far, completely
sidetracked me from my original goal! I'll get back to it eventually....
Anyway, let me say more about this quantum aspect now. This is the
royal road to understanding representations of simple Lie groups. For
starters, fix a complex simple Lie group G and any parabolic subgroup
P. Since G and P are complex Lie groups, the flag manifold G/P is a
complex manifold. More precisely, it has a complex structure that is
invariant under the action of G.
On the other hand, we can write the flag manifold as K/L, where K is the
maximal compact subgroup of G, and L is the intersection of K and P 
L is called a "Levi subgroup". Since K is compact, we can take any
Riemannian metric on the flag manifold and average it with respect to
the action of K to get a Riemannian metric that is invariant under the
action of K.
So, the flag manifold has a complex structure and metric that are both
invariant under K!
If this doesn't thrill you, consider the simplest example:
G = SL(2,C)
K = SU(2)
P = {upper triangular matrices in SL(2,C)}
L = {diagonal matrices in SL(2,C)}
Here G/P = K/L is a 2sphere, the complex structure is the usual way of
thinking of this as the Riemann sphere, and the metric can be any
multiple of the usual round metric on the sphere. The complex structure
is invariant under all of G = SL(2,C). That's why SL(2,C) is the double
cover of the group of conformal transformations of the Riemann sphere!
The metric is only invariant under K = SU(2). That's why SU(2) is the
double cover of the group of rotations of the sphere!
All this stuff is wonderfully important in physics  especially since
SL(2,C) is also the double cover of the Lorentz group, and the Riemann
sphere is also the "heavenly sphere" upon which we see the distant
stars. I have already lavished attention on this network of ideas
in "week162"... but what we're engaged in now is generalizing it to
*arbitrary* complex simple Lie groups!
Now, a basic principle of geometry is that any two of the following
structures on a manifold determine the third *if* they satisfy a
certain compatibility condition:
complex structure J
Riemannian metric g
symplectic structure w
and in this case we get a "Kaehler manifold": a manifold with a complex
structure J and a complex inner product on the tangent vectors whose
real part is g and whose imaginary part is w.
Furthermore, one of the big facts of quantization is that while the
phase space of a classical system is a symplectic manifold, we can only
quantize it and get a Hilbert space if we equip it with some extra
structure... for example, by making it into a Kaehler manifold! Once
the phase space is a Kaehler manifold, we can look for a complex line
bundle over it with a connection whose curvature is the symplectic
structure. If this bundle exists, it's essentially unique, and we can
take the space of its holomorphic sections to be the Hilbert space of
states of the *quantum* version of our system. For details, try my
webpage on geometric quantization, or these books, listed in rough order
of increasing difficulty and depth:
8) John Baez, Geometric quantization,
http://math.ucr.edu/home/baez/quantization.html
9) J. Snyatycki, Geometric Quantization and Quantum Mechanics,
SpringerVerlag, New York, 1980.
10) Nicholas Woodhouse, Geometric Quantization, Oxford U. Press, Oxford,
1992.
11) Norman E. Hurt, Geometric Quantization in Action: Applications of
Harmonic Analysis in Quantum Statistical Mechanics and Quantum Field
Theory, Kluwer, Boston, 1983.
In the beautiful situation I'm discussing now, the math gods are kind:
the complex structure and metric on the flag manifold fit together to
make it into a Kaehler manifold, so we can quantize it and get a Hilbert
space. And since everything in sight is invariant under the group K,
our Hilbert space becomes a unitary representation of K. This rep turns
out to be irreducible... and we get all the unitary irreps of compact
simple Lie groups this way!
By easy abstract nonsense, the unitary irreps of K are also all the
finitedimensional irreps of G. So, we've just conquered a great deal
of territory in the land of group representations. You may have seen
other ways to get all the irreps of simple Lie groups: for example,
"heighestweight representations" or "geometric quantization of
coadjoint orbits". In fact, all these tricks are secretly just
different ways of talking about the same thing. It took me years
to learn this secret, but it's yours for free!
However, there are some small subtleties we shouldn't sweep under the
rug. We've seen that our flag manifold has a godgiven complex
structure, but it usually has *lots* of Kinvariant metrics, since we
could take *any* metric and average it with respect to the action of K.
So, there are lots of Kinvariant Kaehler structures on our flag manifold.
How many are there? Well, I said that we get a flag manifold from any
subset of the dots in the Dynkin diagram for G. It turns out that
Kinvariant Kaehler structure on this flag manifold correspond to ways
of labelling the dots in this subset with positive real numbers. And
we can geometrically quantize the flag manifold to get an irrep of G
precisely when these numbers are *integers*!
The simplest situation is when our flag manifold is a Grassmannian.
This corresponds to a single dot in the Dynkin diagram. If we label
this dot with the number 1, we get a socalled "fundamental
representation" of our group. I sketched in "week180" how to get all
the other irreps from these.
Now let me illustrate all this stuff by going through all the classical
series of simple Lie groups and seeing what we get.
A_n: Here are the Grassmannians for some of the A_n series, that is, the
groups SL(n+1,C). I've drawn the Dynkin diagrams with each dot labelled
by the corresponding type of geometrical figure and the dimension of the
Grassmannian of all figures of this type. We can think of these figures
as vector subspaces of C^{n+1}. We can also think of them as spaces of
one less dimension in CP^n. Either way, we are talking about *projective*
geometry:
A1 1d subspaces
SL(2,C) o
1
A2 1d subspaces 2d subspaces
SL(3,C) oo
2 2
A3 1d subspaces 2d subspaces 3d subspaces
SL(4,C) ooo
3 4 3
A4 1d subspaces 2d subspaces 3d subspaces 4d subspaces
SL(5,C) oooo
4 6 6 4
Recognize the numbers labelling the Dynkin diagram dots? It's a weird
modified version of Pascal's triangle  but can you figure out the pattern?
No? I claim you learned this table of numbers when you were in grade
school: just tilt your head 45 degrees and you'll recognize it!
Next, here's what we get from quantizing these Grassmannians. I've
labelled each dot by the name of the corresponding fundamental
representation and its dimension. All these reps are exterior powers
of the obvious rep of SL(n+1,C) on C^{n+1}. We call elements of the pth
exterior power "pvectors", or "multivectors" in general:
A1 vectors
SL(2,C) o
2
A2 vectors bivectors
SL(3,C) oo
3 3
A3 vectors bivectors 3vectors
SL(4,C) ooo
4 6 4
A4 vectors bivectors 3vectors 4vectors
SL(5,C) oooo
5 10 10 5
Here the numbers labelling the dots form Pascal's triangle! So we
see that Pascal's triangle is a quantized version of the multiplication
table. (That was the answer to the previous puzzle, by the way  our
triangle was just the multiplication table viewed from a funny angle.)
B_n: Next let's look at the B_n series. B_n is another name for the
complexified rotation group SO(2n+1,C), or if you prefer, its double
cover Spin(2n+1,C). A Grassmannian for this group is a space consisting
of all pdimensional "isotropic" subspaces of C^{2n+1}  that is,
subspaces on which a nondegenerate symmetric bilinear form vanishes.
As I explained in "week180", these Grassmannians show up when we study
relativity in odddimensional Minkowski spacetime, especially when we
complexify and compactify. Another way to put it is that this is all
about *conformal* geometry in odd dimensions! We've already seen that
conformal geometry in even dimensions is very different, and we'll get
to that later.
Here are the Grassmannians and their dimensions:
isotropic
B1 1d subspaces
Spin(3,C) o
1
isotropic isotropic
B2 1d subspaces 2d subspaces
Spin(5,C) o=======>=======o
3 3
isotropic isotropic isotropic
B3 1d subspaces 2d subspaces 3d subspaces
Spin(7,C) oo=======>=======o
5 7 6
isotropic isotropic isotropic isotropic
B4 1d subspaces 2d subspaces 3d subspaces 4d subspaces
Spin(9,C) ooo=======>======o
7 11 12 10
I'm sure these are wellknown, but James Dolan and I had a lot of
fun one evening working these out, using a lot of numerology that we
eventually justified by a method I'll explain later.
Here's a bigger chart of these dimensions:
B1 1
B2 3 3
B3 5 7 6
B4 7 11 12 10
B5 9 15 18 18 15
B6 11 19 24 26 25 21
B7 13 23 30 34 35 33 28
B8 15 27 36 42 45 45 42 36
I leave it as an easy puzzle to figure out the pattern, and a harder
puzzle to prove it's true. Don't be overly distracted by the symmetry
lurking in rows 2, 5, and 8  every third row has this symmetry, but
it's a bit of a red herring!
If we quantize these Grassmannians we get these fundamental reps of
Spin(2n+1,C):
B1 spinors
Spin(3,C) o
2
B2 vectors spinors
Spin(5,C) o=======>=======o
5 4
B3 vectors bivectors spinors
Spin(7,C) oo=======>=======o
7 21 8
B4 vectors bivectors 3vectors spinors
Spin(9,C) ooo=======>======o
9 36 84 16
As before, the dimension of the space of pvectors in qdimensional
space comes straight from Pascal's triangle: it's q choose p. But now
we also have spinor reps; the dimensions of these are powers of 2.
C_n: Next let's look at the Grassmannians for the C_n series, that is,
the symplectic groups Sp(2n,C). This is the only series of classical
groups I haven't touched yet! Just as the A_n series are symmetry
groups of projective geometry and the B_n and D_n series are symmetry
groups of conformal geometry, the C_n series are symmetry groups of
"projective symplectic" geometry. Unfortunately I don't know much
about this subject  at least not consciously. It should be important
in physics, but I'm not sure where!
Anyway, Sp(2n,C) is the group of linear transformations of C^{2n} that
preserve a symplectic form: that is, a nondegenerate *antisymmetric*
bilinear form. A Grassmannian for this group again consists of all
pdimensional isotropic subspaces of C^{2n}, where now a subspace is
"isotropic" if the symplectic form vanishes on it.
Here's a little table of these Grassmannians:
isotropic
C1 1d subspaces
Sp(2,C) o
1
isotropic isotropic
C2 1d subspaces 2d subspaces
Sp(4,C) o=======<=======o
3 3
isotropic isotropic isotropic
C3 1d subspaces 2d subspaces 3d subspaces
Sp(6,C) oo=======<=======o
5 7 6
isotropic isotropic isotropic isotropic
C4 1d subspaces 2d subspaces 3d subspaces 4d subspaces
Sp(8,C) ooo=======<======o
7 11 12 10
You'll notice the dimensions are the same as in the B_n case! That's
because their Dynkin diagrams are almost the same: for reasons I may
someday explain, dimensions of flag manifolds don't care which way the
little arrows on the Dynkin diagrams point, since they depend only on
the *reflection group* associated to this diagram (see "week62").
However, the dimensions of the fundamental representations are different
from the B_n case  and I don't even know what they are! The basic idea
is this: the space of pvectors is no longer an irrep for Sp(2n,C), but
contracting with the symplectic form maps pvectors to (p2)vectors,
and the kernel of this map is the pth fundamental rep of Sp(2n). Let's
call these guys "irreducible pvectors".
Oh heck, I can *guess* the dimensions of these guys from this... I guess
they're just the dimension of the pvectors minus the dimension of the
(p2)vectors. Here's a table of these guesses:
C1 vectors
Sp(2,C) o
2
irreducible
C2 vectors bivectors
Sp(4,C) o=======<=======o
4 5
irreducible irreducible
C3 vectors bivectors 3vectors
Sp(6,C) oo=======<=======o
6 14 14
irreducible irreducible irreducible
C4 vectors bivectors 3vectors 4vectors
Sp(8,C) ooo=======<======o
8 27 48 42
Maybe someone can tell if they're right.
D_n: Finally, D_n is another name for the complexified rotation group
SO(2n,C) or its double cover Spin(2n,C). The pth Grassmannian for
this group consists of all pdimensional isotropic subspaces of the
space C^{2n} equipped with a nondegenerate symmetric bilinear form 
*except* for the topdimensional Grassmannians, as I explained
last week. These consist of selfdual or antiselfdual subspaces.
Selfduality is the special feature of conformal geometry in *even*
dimensions!
Here are the Grassmannians and their dimensions:
1
o selfdual 2d subspaces
D2
Spin(4,C)
o antiselfdual 2d subspaces
1
3
o selfdual 3d subspaces
/
/
D3 /
Spin(6,C) 4 o isotropic 1d subspaces
\
\
\
o antiselfdual 3d subspaces
3
6
o selfdual 4d subspaces
/
isotropic /
D4 1d subspaces /
Spin(8,C) oo isotropic 2d subspaces
6 9\
\
\
o antiselfdual 4d subspaces
6
10
o selfdual 5d subspaces
/
/
D5 1d subspaces 2d subspaces /
Spin(10,C) ooo isotropic 3d subspaces
8 13 15\
\
\
o antiselfdual 5d subspaces
10
You'll notice that the numbers on the "fishtails" are triangular
numbers: 1, 3, 6, 10... I'll say more later about how to calculate
the rest of these numbers.
As explained last week, the fundamental reps of the D_n consist of
pvectors, except for those at the fishtails, which are left and
righthanded spinor reps:
2
o lefthanded spinors
D2
Spin(4,C)
o righthanded spinors
2
4
o lefthanded spinors
/
/
D3 /
Spin(6,C) 6 o vectors
\
\
\
o righthanded spinors
4
8
o lefthanded spinors
/
/
D4 vectors /
Spin(8,C) oo bivectors
8 28\
\
\
o righthanded spinors
8
16
o lefthanded spinors
/
/
D5 vectors bivectors /
Spin(10,C) ooo 3vectors
10 45 120\
\
\
o righthanded spinors
16
Again the dimension of the space of pvectors in qdimensional space
comes from Pascal's triangle, while the dimensions of the spinor reps
are powers of 2.
Let me conclude by listing the dimensions of Grassmannians for the
exceptional groups, as computed by James Dolan. I strongly doubt he's
the first to have computed these  at this stage we're mainly learning
and reinventing known stuff  but he did it using a nice trick I'd like
to mention. I was shocked at how unfamiliar these numbers were to me,
because all these Grassmannians should be definable using the octonions:

G2: 5 > 5

F4: 4348===>===4843
21



E6 1625292516
42



E7 334753504227
92



E8 7898106104998357
You can calculate dimensions of these and all the other Grassmannians
for simple Lie groups by the following easy trick. Given the Dynkin
diagram for G and a chosen dot in it, remove this dot to get one or more
Dynkin diagrams for groups G_i. Work out the dimension of the space
of maximal flags for G, and subtract all the dimensions of the spaces
of maximal flags for the G_i. Voila! You get the dimensions of the
Grassmannian corresponding to the ith dot.
The dimensions of maximal flag manifold for G is easy to compute, in
turn, because it's just dim(G)  dim(B), where B is the Borel. And
dimension of the Borel is just (dim(G) + dim(T))/2, where T is the
maximal torus, so that dim(T) is the number of dots in the Dynkin diagram.

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