
To really know a subject you've got to learn a bit of its history. If that subject is topology, you've got to read this:
1) I. M. James, editor, History of Topology, Elsevier, New York, 1999.
From a blowbyblow account of the heroic papers of Poincare to a detailed account by Peter May of the prehistory of stable homotopy theory... it's all very fascinating. You'll probably want to study some more of the subject by the time you're done!
In order to satisfy that craving, I want to tell you how to compute some homology groups. But we'll do it a strange way: using "qmathematics". I began talking about qmathematics last week, but now I want to dig deeper.
At first, it looks like there are two really different places where this qstuff shows up. One is when you do mathematics with qdeformed quantum groups replacing the Lie groups you know and love  this is important in string theory, knot theory, and loop quantum gravity. In this case it's best if q is a unit complex number, especially an nth root of unity:
q = exp(2πi/n)You'll notice that in string theory, knot theory and loop quantum gravity, loops play a big role. This is no coincidence; in a way, quantum groups are just a technical device for studying "loop groups", which are groups consisting of functions from a circle to some specified Lie group.
See, in quantum physics problems with a loop group as the symmetry group, these symmetries tend to hold only up to a phase. The precise way these phases work depends on the parameter q. Mathematically, this means that instead the loop group itself, the symmetries are really described by a slightly larger group that keeps track of these phases, called a "central extension" of the loop group. This has led people to spend huge amounts of energy studying representations of central extensions of loop groups  which turn out to be much more economically understood, in a rather subtle way, as representations of quantum groups. In all this work the parameter q plays a major role.
For more on this try these books:
2) Andrew Pressley and Graeme Segal, Loop Groups, Oxford U. Press, Oxford, 1986.
3) Vyjayanathi Chari and Andrew Pressley, A Guide to Quantum Groups, Cambridge U. Press, Cambridge, 1994.
4) Jürgen Fuchs, Affine Lie Algebras and Quantum Groups, Cambridge U. Press, Cambridge, 1992.
Taken together, they provide a pretty good view of what I'm talking about. In case you're wondering, an "affine Lie algebra" is the Lie algebra of a central extension of a loop group.
Mathematical physicists know all about this sort of qstuff. But qstuff also shows up when we do mathematics with finite fields. Here I don't mean "field" in the physics sense  I mean an algebraic gadget where you can add, subtract, multiply and divide! Physicists are happiest when their field is the real or complex numbers. But mathematicians also like fields with finitely many elements. If q is a power of a prime number, there is a unique field with q elements, called F_{q}. Even better, these are all the finite fields. If q is itself prime, F_{q} is just the integers mod q. We can get the other finite fields using a trick very much like how we get the complex numbers from the reals by throwing in the square root of minus one.
Don't be scared: this is already more than what you'll need to know about finite fields to understand what I'm going to say!
And here's what I'm going to say: lots of formulas for counting structures on finite sets have qversions that tell you how to count structures on projective spaces over F_{q}. Remember, a "projective space" is just the space of all line through the origin in a vector space. The basic idea is this mystical analogy:
q = 1 FINITE SETS q = a power of a prime number PROJECTIVE SPACES OVER F_{q}This analogy is so powerful that it really pays to think of finite sets as projective spaces over F_{1}, the "field with one element"  even though there is no such field!
Let's start with some examples. How many points does an nelement set have? Answer: the integer
nNext: how many lines through the origin does an ndimensional vector space over F_{q} have? Of course, these lines are points in the corresponding projective space. Answer: the qinteger
q^{n}  1 [n] =  q  1 = 1 + q + q^{2} + ... + q^{n1}Remember from last week that the basic idea behind "qarithmetic" was to replace integers by these qintegers.
To see why this answer is right, first note that we determine a line through the origin by picking any nonzero vector. There are q^{n}  1 of these. However, two vectors determine the same line if one is a nonzero multiple of the other, and there are q  1 nonzero elements of F_{q}. So, the actual number of lines through the origin is
q^{n}  1  = [n] q  1Here's another example. How many ways are there to order a set with n elements? Answer:
n! = 1 2 ... nNext: how many maximal flags are there in an ndimensional vector space over F_{q}? Answer: the qfactorial
[n]! = [1] [2] ... [n]Remember, a maximal flag is a line through the origin contained in a plane through the origin contained in ... and so on, up to the top dimension. As we've seen, there are [n] ways to choose a line L like this in our vector space V. The next step is to choose a plane containing L. This is the same as choosing a line in the quotient space V/L, which has one dimension less, so there are [n1] ways to do this. And so on, giving us [n]! maximal flags.
Here's yet another example. How many melement subsets does an nelement set have? Answer: the binomial coefficient
n!  m! (nm)!Next: how many mdimensional subspaces are there of an ndimensional space over F_{q}? Answer: the qbinomial coefficient
[n]!  [m]! [nm]!I'll leave this one as an exercise.
It goes on and on like this: all sorts of structures that can be defined for finite sets have analogues for the projective geometry of finite fields, and when we count these, the former tend to give us "ordinary mathematics", while the latter give us "qmathematics", which reduces to ordinary mathematics at q = 1.
Clearly this pattern is trying to tell us something; the question is what. As always, it pays to focus on the simplest case, since that's where everything starts. I said that the number of lines through the origin in an ndimensional vector space over the field with q elements is
q^{n}  1 [n] =  q  1 = 1 + q + q^{2} + ... + q^{n1}But now let's think about why the second equation here is true!
Of course this is just the formula for summing a geometric series, but we can also categorify this formula. In other words: we can think of [n] not as the mere number of lines through the origin in an ndimensional vector space over F_{q}, but as the actual set of such lines. To prove the second equation, we should thus find a nice way to write this set as 1 special line, together with q more lines, and then q^{2} more, and so on.
To do this, pick a maximal flag: a 1d subspace contained in a 2d subspace contained in a 3d subspace... and so on. There is one line through the origin contained in our 1d subspace  namely the subspace itself. There are q lines through the origin contained in the 2d subspace but not in the 1d subspace. There are q^{2} lines in the 3d subspace but not the 2d subspace. And so on. Voila!
Combinatorists call this a "bijective proof": a proof that two numbers are equal which actually establishes a bijection between the finite sets they count. It's an example of "categorification" because we've taken an equation and found the isomorphism that explains it  taking us from math in the set of natural numbers to math in the category of finite sets.
The cool part is, this proof works for all fields, not just finite ones. For example, over the real numbers we can use it to take the projective space RP^{n1} and chop it into pieces like this:
RP^{n1} = R^{0} + R^{1} + ... + R^{n1}Topologically speaking, we've just decomposed RP^{n1} as a union of open balls, or "cells". This makes it easy to calculate its Euler characteristic. Evendimensional cells contribute 1 to the Euler characteristic, while odddimensional cells contribute 1, so we get
RP^{n1} = (1)^{0} + (1)^{1} + ... + (1)^{n1} (1)^{n}  1 =  , (1)  1or in other words, 0 if n is even and 1 if n is odd. Here I'm using X to stand for the Euler characteristic of X.
You'll notice that the Euler characteristic is working here exactly like the cardinality did in the finite field case. That's no coincidence! The Euler characteristic and its evil twin the "homotopy cardinality" are both generalizations of cardinality, as I explained in "week147". If we use Schanuel's improved version of the Euler characteristic, which lets us chop up a space X and calculate X by summing the Euler characteristics of the pieces, we have R = 1, so
RP^{n1} = R^{0} + R^{1} + ... + R^{n1} = R^{0} + R^{1} + ... + R^{n1} = [n]where [n] is the qinteger where q = R = 1. So if you want to shock your friends, you can tell them that the real numbers are the field with 1 elements!
What about the complex numbers? Well, as spaces we have
C = R^{2}so we get
C = R^{2} = 1.This implies that the Euler characteristic of CP^{n1} is [n], where now q = C = 1. In other words, it's just n.
Now that we've gotten this wonderful new insight we can test it on fancier examples, like flag manifolds. I already showed you that the number of maximal flags in an ndimensional vector space over F_{q} is the qfactorial
[n]!
And if you look back, you'll see I gave a bijective proof. This means that if we work over the real or complex numbers, the same proof gives a cell decomposition of the manifold of maximal flags in R^{n} or C^{n}  the "flag manifold", for short. So we can just calculate some qfactorials:
[1]! = 1 [2]! = 1 + q [3]! = 1 + 2q + 2q^{2} + q^{3} [4]! = 1 + 3q + 5q^{2} + 6q^{3} + 5q^{4} + 3q^{5} + q^{6}and read off all sorts of fun stuff. For example, the flag manifold of R^{4} has a cell decomposition like
R^{0} + 3R + 5R^{2} + 6R^{3} + 5R^{4} + 3R^{5} + R^{6}meaning that there's 1 zerocell, 3 onecells, 5 twocells and so on. Similarly, the flag manifold of C^{4} has a cell decomposition like
C^{0} + 3C + 5C^{2} + 6C^{3} + 5C^{4} + 3C^{5} + C^{6}meaning that there's 1 zerocell, 3 twocells, 5 fourcells and so on. (Their dimensions are twice as big now, since C has dimension 2.)
In particular, the Euler characteristic of the flag manifold in n dimensions is just [n]!, where we set q = 1 in the real case and q = 1 in the complex case. But in the complex case we can say more!
Whenever you build a space from cells, you can compute its homology from a chain complex with one generator for each cell and a differential saying how the cells of dimension n are glued to the cells of dimension n1. But since the complex flag manifold is built from only evendimensional cells, the differential is zero in this case. This means you can read off its nth homology group by just counting the number of ncells! The homology group is just Z^{k}, where k is this number.
So for example, if some nasty guy demands that you calculate the 10th homology of the complex flag manifold in 4 dimensions, you just tell him "I know it's a free abelian group..." and calculate
[4]! = 1 (1 + q) (1 + q + q^{2}) (1 + q + q^{2} + q^{3}) = 1 + 3q + 5q^{2} + 6q^{3} + 5q^{4} + 3q^{5} + q^{6}You know the q^{5} term gives you the 10cells in this flag manifold, since the complex numbers have dimension 2. You see the coefficient of this term is 3, so you say "... and it's Z^{3}." He will then think you know algebraic topology, and go away.
The same sort of trick works for Grassmannians, too. The Grassmannian Gr(n,k) is the set of all kdimensional subspaces of an ndimensional vector space. This makes sense over any field. I already said that over the finite field F_{q}, the cardinality of this Grassmannian is the qbinomial coefficient
[n]! Gr(n,k) =  [k]! [nk]!The same formula gives the Euler characteristic of this Grassmannian over the real numbers if we set q = 1, and over the complex numbers if we set q = 1. Of course q = 1 just gives the ordinary binomial coefficients.
So, for example, the Euler characteristic of the manifold of 2dimensional subspaces of C^{4} is the same as the number of ways of choosing 2 elements from a 4element set! A nice example of the unity of mathematics.
Also, since complex Grassmannians are built from only evendimensional cells, we can read off their homology groups just like we did for complex flag manifolds. Let's work out the homology of Gr(4,2), for example. We start by working out the qbinomial coefficient:
[4]! 1 (1 + q) (1 + q + q^{2}) (1 + q + q^{2} + q^{3})  =  [2]! [2]! 1 (1 + q) 1 (1 + q) = 1 + q + 2q^{2} + q^{3} + q^{4}It's mildly surprising that this ratio works out to be a polynomial, but of course we know it must! Reading off the coefficients, we get:
the 0th homology group is Z the 2nd homology group is Z the 4th homology group is Z^{2} the 6th homology group is Z the 8th homology group is Zand while we're at it, we've learned this Grassmannian is 8dimensional as a real manifold  or 4dimensional as a complex manifold. Note how the nth homology group is the same as the (8n)th; this comes from Poincare duality.
On a lighter note: the best way to simplify this sort of expression
1 (1 + q) (1 + q + q^{2}) (1 + q + q^{2} + q^{3})  1 (1 + q) 1 (1 + q)is to use base q. Then it's just
1 x 11 x 111 x 1111 111 x 1111 123321  =  =  = 11211 1 x 11 x 1 x 11 11 11where I did the last step using long division. And of course the last quantity is
1 + q + 2q^{2} + q^{3} + q^{4}.
By the way, the cells we've been counting are called "Schubert cells".
I'll quit here for now, but actually this is just the tip of the iceberg. I've been talking how qfactorials are related to projective geometry, but as readers of "week178", "week180" and "week181" will know, there exists a generalization of projective geometry for any simple Lie group. In fact, for any simple Lie group G and any parabolic subgroup P there is a decomposition of G/P into Schubert cells, and these cells are counted by the coefficients of a certain polynomial in q. Using these you can massively generalize everything I just told you! I'll explain this stuff in future Weeks.
Addendum: Here's my reply to a request for clarification from my friend Squark:
Squark wrote: >John Baez wrote: >> If we use Schanuel's improved version of the Euler characteristic, which >> lets us chop up a space X and calculate X by summing the Euler >> characteristics of the pieces, we have R = 1, C = R^2, so we get >> >> C = R^2 = 1. >How does this Schanuel thingie work? R and C are both contractible, so >it has to be principally different from the usual Euler characteristic! Right. Schanuel's Euler characteristic is not homotopy invariant like the usual Euler characteristic, and it's only defined for nice spaces, like polyhedral sets. However, it has a great property to make up for these sins: whenever we can chop up a polyhedral set A into nice parts B and C, we have A = B + C We also have XxY = X x Y, and homeomorphic nice spaces have the same Schanuel Euler characteristic. One can check that for compact manifolds, the Schanuel Euler characteristic matches the usual one, so my strange calculations really do give the standard "right answers". Schanuel's Euler characteristic of a point is 1: * = 1 so the Schanuel Euler characteristic of the open interval must be 1: we have (0,1) = (0,1/2) + {1/2} + (1/2,1) so if (0,1) = x we have x = x + 1 + x so x = 1. This means that the Schanuel Euler characteristic of a halfopen interval is zero: (0,1] = (0,1) + {1} so (0,1] = (0,1) + {1} = 1 + 1 = 0 The Schanuel Euler characteristic of a circle is 0 as well, since we can chop it into two (or three, or more) halfopen intervals. The Schanuel Euler characteristic of an open square is 1: (0,1) x (0,1) = (0,1) x (0,1) = 1 x 1 = 1 and the SE characteristic of a closed square is 1 x 1 = 1. Now, just as a consistency check, write the closed square as the union of an open square and its boundary. The boundary is homeomorphic to a circle, so we should get 1 + 0 = 1. It works! A good reference on this stuff is: James Propp, Exponentiation and Euler measure, available at http://www.arXiv.org/abs/math.CO/0204009. Here you'll see that what I'm calling Schanuel's Euler characteristic goes back to work before Schanuel. Also, if you push it far enough, it gives a fascinating approach to dealing with "sets with negative numbers of elements"  for example, it gives a kind of combinatorial interpretation of identities like 2 choose 3 = 4 Schanuel was trying to categorify the integers: that's why he came up with this stuff. Also see "week147" for more!
© 2002 John Baez
baez@math.removethis.ucr.andthis.edu
