
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 proved there exist 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, Cris Moore first discovered the figure8 solution of the gravitational 3body problem in his 1993 paper, using computer calculations. His student Michael Nauenberg made this movie of it, which you can find with many others on Moore's website:
5) Richard Montgomery, The Nbody problem, the braid group, and actionminimizing periodic solutions, Nonlinearity 11 (1998), 363371. Also available at 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 planesFor 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 planescorresponds 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:
A_{1} 1d subspaces SL(2,C) o 1 A_{2} 1d subspaces 2d subspaces SL(3,C) oo 2 2 A_{3} 1d subspaces 2d subspaces 3d subspaces SL(4,C) ooo 3 4 3 A_{4} 1d subspaces 2d subspaces 3d subspaces 4d subspaces SL(5,C) oooo 4 6 6 4Recognize 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:
A_{1} vectors SL(2,C) o 2 A_{2} vectors bivectors SL(3,C) oo 3 3 A_{3} vectors bivectors 3vectors SL(4,C) ooo 4 6 4 A_{4} vectors bivectors 3vectors 4vectors SL(5,C) oooo 5 10 10 5Here 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 B_{1} 1d subspaces Spin(3,C) o 1 isotropic isotropic B_{2} 1d subspaces 2d subspaces Spin(5,C) o=======>=======o 3 3 isotropic isotropic isotropic B_{3} 1d subspaces 2d subspaces 3d subspaces Spin(7,C) oo=======>=======o 5 7 6 isotropic isotropic isotropic isotropic B_{4} 1d subspaces 2d subspaces 3d subspaces 4d subspaces Spin(9,C) ooo=======>======o 7 11 12 10I'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:
B_{1} 1 B_{2} 3 3 B_{3} 5 7 6 B_{4} 7 11 12 10 B_{5} 9 15 18 18 15 B_{6} 11 19 24 26 25 21 B_{7} 13 23 30 34 35 33 28 B_{8} 15 27 36 42 45 45 42 36I 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):
B_{1} spinors Spin(3,C) o 2 B_{2} vectors spinors Spin(5,C) o=======>=======o 5 4 B_{3} vectors bivectors spinors Spin(7,C) oo=======>=======o 7 21 8 B_{4} vectors bivectors 3vectors spinors Spin(9,C) ooo=======>======o 9 36 84 16As 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 C_{1} 1d subspaces Sp(2,C) o 1 isotropic isotropic C_{2} 1d subspaces 2d subspaces Sp(4,C) o=======<=======o 3 3 isotropic isotropic isotropic C_{3} 1d subspaces 2d subspaces 3d subspaces Sp(6,C) oo=======<=======o 5 7 6 isotropic isotropic isotropic isotropic C_{4} 1d subspaces 2d subspaces 3d subspaces 4d subspaces Sp(8,C) ooo=======<======o 7 11 12 10You'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:
C_{1} vectors Sp(2,C) o 2 irreducible C_{2} vectors bivectors Sp(4,C) o=======<=======o 4 5 irreducible irreducible C_{3} vectors bivectors 3vectors Sp(6,C) oo=======<=======o 6 14 14 irreducible irreducible irreducible C_{4} vectors bivectors 3vectors 4vectors Sp(8,C) ooo=======<======o 8 27 48 42Maybe 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 D_{2} Spin(4,C) o antiselfdual 2d subspaces 1 3 o selfdual 3d subspaces / / D_{3} / Spin(6,C) 4 o isotropic 1d subspaces \ \ \ o antiselfdual 3d subspaces 3 6 o selfdual 4d subspaces / isotropic / D_{4} 1d subspaces / Spin(8,C) oo isotropic 2d subspaces 6 9\ \ \ o antiselfdual 4d subspaces 6 10 o selfdual 5d subspaces / / D_{5} 1d subspaces 2d subspaces / Spin(10,C) ooo isotropic 3d subspaces 8 13 15\ \ \ o antiselfdual 5d subspaces 10You'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 D_{2} Spin(4,C) o righthanded spinors 2 4 o lefthanded spinors / / D_{3} / Spin(6,C) 6 o vectors \ \ \ o righthanded spinors 4 8 o lefthanded spinors / / D_{4} vectors / Spin(8,C) oo bivectors 8 28\ \ \ o righthanded spinors 8 16 o lefthanded spinors / / D_{5} vectors bivectors / Spin(10,C) ooo 3vectors 10 45 120\ \ \ o righthanded spinors 16Again 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:
 G_{2} 5 > 5  F_{4} 4348===>===4843 21    E_{6} 1625292516 42    E_{7} 334753504227 92    E_{8} 7898106104998357You 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.
© 2002 John Baez
baez@math.removethis.ucr.andthis.edu
