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Okay! Here comes the climax of our story linking q-mathematics, Dynkin diagrams, incidence geometry and categorification! I'll be amazed if you follow what I'm saying, because doing so will require that you remember everything I've ever told you. But I can't help talking about this stuff, because it's so cool. So, please at least pretend to pay attention. I promise to ratchet things down a notch next week.
Last time I described a bunch of things you can get from a Dynkin diagram:
DYNKIN DIAGRAM
/ \
/ \
choose a field/ \
/ \
/ \
/ Weyl group \
SIMPLE ALGEBRAIC ----------> COXETER GROUP
GROUP |
| |
FLAG VARIETY COXETER COMPLEX
\ /
\ /
\ /
\ /
\ /
\ /
\ /
q-POLYNOMIAL
If we choose any field, our Dynkin diagram gives us a group. If we
choose the real or complex numbers we get the real and complex simple
Lie groups that physicists know and love, but it's also fun to use other
fields. These other fields give us "simple algebraic groups",
which are not manifolds but instead algebraic varieties.
It's especially fun to choose the field Fq - the finite field with q elements, where q is any power of a prime number. The reason this is so fun is that we can also get a group from a Dynkin diagram without choosing a field: the so-called Coxeter group! Amazingly, all sorts of formulas about this Coxeter group are special cases of formulas about simple algebraic groups over Fq. To specialize, we just set q = 1.
In other words: the theory of simple algebraic groups over the field with q elements is a "q-deformation" of the theory of Coxeter groups, where q = 1. Even better, this q-deformation is closely related to other q-deformations that show up in the theory of quantum groups.
This is strange and mysterious, because it seems to be saying that Coxeter groups are simple algebraic groups over the field with one element - but there is no field with one element! This mystery, and its relation to q-deformation, is what I find so tantalizing about the whole subject.
To see the mystery play itself out before us, we need to look at the incidence geometries having simple algebraic groups and Coxeter groups as their symmetries. In both cases, these incidence geometries have one type of geometrical figure for each dot in the Dynkin diagram, and one basic incidence relation for each edge.
In the incidence geometry whose symmetries are a simple algebraic group over the field F, the set of figures of a given type will be an algebraic variety, say X(F). In the incidence geometry whose symmetries are the corresponding Coxeter group, the set of figures of this type will be a finite set, say X. When Fq is the field with q elements, the number of points in X(Fq) is finite and given by some polynomial in q. But when we set q = 1, we get the number of points in the set X.
To make this more clear - perhaps too clear for comfort! - I would like to show you how to calculate all these polynomials X(Fq). It's actually best to start by counting, not the set of figures of a given type corresponding to a given dot in the Dynkin diagram, but the set of all "maximal flags". A maximal flag is a collection of figures, one of each type, all incident. We'll soon see that if we can count these, we can count anything we want.
When we work over the field Fq, the set of maximal flags is actually an algebraic variety, and the number of maximal flags is a polynomial in q. Last week I called this the "q-polynomial" of our Dynkin diagram, and described how to calculate it. In a minute I'll say what this polynomial is in a bunch of cases. But I can't resist a short digression, to explain why I like this polynomial so much!
I'm always running around trying to "categorify" everything in sight, replacing equations by isomorphisms, numbers by finite sets, and so on. The reason is that we've been unconsciously "decategorifying" mathematics for the last couple of millenia, which is an information- destroying process, and I want to undo that process. For example, whenever we see a finite set, we have a tendency to decategorify it by counting it and just remembering its number of elements. Then we prove fun equations relating these numbers. But nowadays we know how to work directly with the finite sets and talk about isomorphisms between them, instead of just equations between their numbers of elements. This gives useful extra information.
The stuff I'm talking about now is a great example. Since the q-polynomial counts the number of maximal flags, it's really a decategorification of the variety consisting of all maximal flags. But what this means is that the maximal flag variety is a categorification of the q-polynomial. Using this way of thinking, all sorts of identities involving q-polynomials correspond to isomorphisms between algebraic varieties!
Here are the q-polynomials of the classical series of Dynkin diagrams. For maximum effect, this table should be read along with similar tables in "week64" and "week181".
o-------o-------o-------o-------o
The Lie group is SL(n+1). The Coxeter group is the symmetry group of
the regular n-simplex. This consists of all permutations of the n+1
vertices of the simplex, so it has (n+1)! elements. The Coxeter complex
is obtained by barycentrically subdividing the surface of the n-simplex.
The q-polynomial is the "q-factorial"
[n+1]! = [1] [2] ... [n+1]
o-------o-------o-------o====>====o
The Lie group is Spin(2n+1). The Coxeter group is the symmetry group of
an n-dimensional cube. This group is the semidirect product of the
permutations of the n axes and the group (Z/2)n generated by the
reflections along these axes. Thus the size of this group is the
"double factorial"
(2n)!! = 2 4 ... 2nThe Coxeter complex is obtained by barycentrically subdividing the surface of the n-dimensional cube. The q-polynomial is the "q-double factorial":
[2n]!! = [2] [4] ... [2n]
o-------o-------o-------o====<====o
The Coxeter group is the symmetry group of an n-dimensional cross-polytope,
which is the obvious generalization of an octahedron to arbitrary
dimensions. This is the exact same group as the Coxeter group of Bn,
with the same Coxeter complex, so the q-polynomial is again the q-double
factorial:
[2n]!! = [2] [4] ... [2n]
o
/
/
o------o-------o
\
\
o
The Lie group is Spin(2n). Since the Dynkin diagram is not just a
straight line of dots, it turns out the Coxeter group is not the full
symmetry group of some polytope. Instead, it's half as big as the Weyl
group of Bn: it's the subgroup of the symmetries of the
n-dimensional cube generated by permutations of the coordinate axes and
reflections along pairs of coordinate axes. I like to call the number
of elements of this group the "half double factorial", and use
this notation for it:
(2n)?! = (2n)!! / 2
= 2 4 ... (2n-2) n
The Coxeter complex is obtained from that for Bn by gluing together
top-dimensional simplices in pairs in a certain way to get bigger
top-dimensional simplices. The q-polynomial is the "q-half double
factorial":
[2n]?! = [2n]!! / q^n + 1
= [2] [4] ... [2n-2] [n]
At this point I guess I should "throw a concrete life preserver to the student drowning in a sea of abstraction", as the cruel joke goes. So let's actually work out some examples of these q-polynomials. As I explained in "week184", this is easiest in base q. The reason is that in this base, a q-integer like
[5] = q4 + q3 + q2 + q + 1is written as just a string of ones like 111111, and such numbers are pathetically easy to multiply and divide. So we do calculations resembling the work of an idiot savant gone berserk:
A2: [1]! = 1 A3: [2]! = 1 x 11 = 11 A4: [3]! = 1 x 11 x 111 = 1221 A5: [4]! = 1 x 11 x 111 x 1111 = 1356531 B1,C1: [2]!! = 11 B2,C2: [4]!! = 11 x 1111 = 12221 B3,C3: [6]!! = 11 x 1111 x 111111 = 1357887531 D1: [2]?! = 11 / 11 = 1 D2: [4]?! = 11 x 1111 / 101 = 121 D3: [6]?! = 11 x 1111 x 111111 / 1001 = 1357887531/1001 = 1356531but the results pack a considerable whallop. For example, to count the number of points in the maximal flag variety of Spin(7), we note that this group is also called B3, so its q-polynomial is [6]!!. This is 1357887531 in base q, or in other words:
q9 + 3q8 + 5q7 + 7q6 + 8q5 + 8q4 + 7q3 + 5q2 + 3q + 1And this is the number of points of the maximal flag variety if we work over the field with q elements!
By the way, you may have noticed a curious coincidence in the above table:
[4]! = [6]?!This is a spinoff of the fact that A3 and D3 are isomorphic: their Dynkin diagrams are both just 3 dots in a row. In "week180" I explained how this underlies Penrose's theory of twistors.
There's a lot more we can do with these q-polynomials. Back in "week179" and "week180" I explained some "flag varieties" of which the maximal ones we're discussing now are special cases. If you give me a Dynkin diagram and a field, I will give you a simple algebraic group G. If you pick a subset of the dots in this diagram, I will give you a subgroup P of G, called a "parabolic subgroup". The quotient G/P is called a "flag variety". A point in this flag variety consists of a collection of geometrical figures of different types, one for each dot in our subset, all incident.
The bigger the set of dots is, the smaller P is, and the bigger and fancier the corresponding flags are. For example, if we use all the dots, P is called the "Borel subgroup", and G/P is the maximal flag variety. On the other hand, if we use none of the dots, G/P is the minimal flag variety - just a point. That's boring. But if we use just one dot, G/P is a so-called "Grassmannian". I listed these back in "week181", and they're really interesting.
For example, if you give me the Dynkin diagram called D4:
o
/
/
o-------o
\
\
o
I'll give you the group G = Spin(8,C), and I'll tell you it's the group
of conformal transformations of 6-dimensional complexified compactified
Minkowski spacetime. If you pick out the subset consisting of just the
dot in the middle:
o
/
/
o-------x
\
\
o
I'll tell you that G/P is the space of null lines in this spacetime.
And if you say "huh?", I'll tell you to reread "week181"!
Now, for any Dynkin diagram and any subset of dots, there's a q-polynomial with all sorts of cool properties. It works just like last week:
a) the coefficient of qi in this polynomial is the number of i-cells in the Bruhat decomposition of G/P. Here the "Bruhat decomposition" is a standard way of writing G/P as disjoint union of i-cells, that is, copies of Fi where F is our field and i is a natural number.
b) if the coefficient of qi in this polynomial is k, the (2i)th homology group of G/P defined over the complex numbers is Zk.
c) the degree of this polynomial is the dimension of G/P.
d) the value of this polynomial at q a prime power is the cardinality of G/P defined over the field Fq.
e) the value of this polynomial at q = -1 is the Euler characteristic of G/P defined over the real numbers.
f) the value of this polynomial at q = 1 is the Euler characteristic of G/P defined over the complex numbers.
If we take property a) as the defining one, all the rest fall out automagically. By the way, the relation between the homology groups in part b) and the cardinalities in part d) is a special case of the "Weil conjectures", proved by Deligne. For an introduction to these, try:
1) Robin Harshorne, Algebraic Geometry, Appendix C: The Weil conjectures, Springer-Verlag, Berlin, 1977.
But now for the cute part: how you calculate this q-polynomial. It's actually really easy! You just calculate the q-polynomial for the whole Dynkin diagram and divide by the q-polynomial you get for the diagram you get when you remove the dots in your subset!
So, suppose for example you got really interested in the space of null lines in 6d complexified compactified Minkowski spacetime:
o
/
/
o-------x
\
\
o
The whole diagram is D4, so its q-polynomial is [8]?! If we remove
the dots in our subset we're left with
o
o
o
that is, three copies of A1. I never told you how to calculate the
q-polynomial for a diagram with more than one piece, but you just
multiply the q-polynomials for the pieces, so you get [2]! x [2]! x [2]!
This means the q-polynomial for our space is
[8]?! 11 x 1111 x 111111 x 11111111 / 10001
-------------- = --------------------------------------
[2]! [2]! [2]! 11 x 11 x 11
1111 111111 11111111
= ---- x ------ x --------
11 11 10001
= 101 x 10101 x 1111
= 1020201 x 1111
= 1133443311
You'll notice how all these numbers are palindromic; that comes from
Poincare duality. We can read of all sorts of wonderful things from
the final answer, as listed above. For example, the Euler characteristic
of our space G/P is
1+1+3+3+4+4+3+3+1 = 24
The Dynkin diagram D4 is all about triality and the octonions, which are
important in superstring theory. The number 24 plays an important role
in bosonic string theory. Does this "coincidence" make
anything good happen? I don't know!
That's enough for now... I'll leave off with a quote that reminds me of these weird base q calculations.
"I don't know", said Alice, "I lost count."
"She can't do addition."
- Lewis Carroll, Through the Looking Glass.
© 2002 John Baez
baez@math.removethis.ucr.andthis.edu
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