August 4, 2002

This Week's Finds in Mathematical Physics (Week 184)

John Baez

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 blow-by-blow 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 "q-mathematics". I began talking about q-mathematics last week, but now I want to dig deeper.

At first, it looks like there are two really different places where this q-stuff shows up. One is when you do mathematics with q-deformed 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 q-stuff. But q-stuff 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 Fq. Even better, these are all the finite fields. If q is itself prime, Fq 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 q-versions that tell you how to count structures on projective spaces over Fq. 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 Fq
This analogy is so powerful that it really pays to think of finite sets as projective spaces over F1, the "field with one element" - even though there is no such field!

Let's start with some examples. How many points does an n-element set have? Answer: the integer

Next: how many lines through the origin does an n-dimensional vector space over Fq have? Of course, these lines are points in the corresponding projective space. Answer: the q-integer
          qn - 1
[n]  =   --------   
           q - 1

     = 1 + q + q2 + ... + qn-1
Remember from last week that the basic idea behind "q-arithmetic" was to replace integers by these q-integers.

To see why this answer is right, first note that we determine a line through the origin by picking any nonzero vector. There are qn - 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 Fq. So, the actual number of lines through the origin is

 qn - 1
-------- = [n]
 q - 1
Here's another example. How many ways are there to order a set with n elements? Answer:
n! =    1  2  ...  n
Next: how many maximal flags are there in an n-dimensional vector space over Fq? Answer: the q-factorial
[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 [n-1] ways to do this. And so on, giving us [n]! maximal flags.

Here's yet another example. How many m-element subsets does an n-element set have? Answer: the binomial coefficient

m! (n-m)!
Next: how many m-dimensional subspaces are there of an n-dimensional space over Fq? Answer: the q-binomial coefficient
[m]! [n-m]!
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 "q-mathematics", 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 n-dimensional vector space over the field with q elements is

           qn - 1
  [n]  =   --------   
            q - 1

       = 1 + q + q2 + ... + qn-1
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 n-dimensional vector space over Fq, 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 q2 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 q2 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 RPn-1 and chop it into pieces like this:

    RPn-1 = R0 + R1 + ... + Rn-1 
Topologically speaking, we've just decomposed RPn-1 as a union of open balls, or "cells". This makes it easy to calculate its Euler characteristic. Even-dimensional cells contribute 1 to the Euler characteristic, while odd-dimensional cells contribute -1, so we get
   |RPn-1| = (-1)0 + (-1)1 + ... + (-1)n-1 

            (-1)n - 1
          = ----------- ,
             (-1) - 1              
or 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

       |RPn-1| = |R0 + R1 + ... + Rn-1|

              = |R|0 + |R|1 + ... + |R|n-1

              = [n]
where [n] is the q-integer 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 = R2
so we get
|C| = |R|2 = 1.
This implies that the Euler characteristic of CPn-1 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 n-dimensional vector space over Fq is the q-factorial


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 Rn or Cn - the "flag manifold", for short. So we can just calculate some q-factorials:

[1]! = 1 
[2]! = 1 + q
[3]! = 1 + 2q + 2q2 +  q3
[4]! = 1 + 3q + 5q2 + 6q3 + 5q4 + 3q5 + q6
and read off all sorts of fun stuff. For example, the flag manifold of R4 has a cell decomposition like
     R0 + 3R + 5R2 + 6R3 + 5R4 + 3R5 + R6
meaning that there's 1 zero-cell, 3 one-cells, 5 two-cells and so on. Similarly, the flag manifold of C4 has a cell decomposition like
     C0 + 3C + 5C2 + 6C3 + 5C4 + 3C5 + C6
meaning that there's 1 zero-cell, 3 two-cells, 5 four-cells 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 n-1. But since the complex flag manifold is built from only even-dimensional cells, the differential is zero in this case. This means you can read off its nth homology group by just counting the number of n-cells! The homology group is just Zk, 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 + q2) (1 + q + q2 + q3) 
     = 1 + 3q + 5q2 + 6q3 + 5q4 + 3q5 + q6
You know the q5 term gives you the 10-cells 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 Z3." 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 k-dimensional subspaces of an n-dimensional vector space. This makes sense over any field. I already said that over the finite field Fq, the cardinality of this Grassmannian is the q-binomial coefficient

|Gr(n,k)| =  -----------
             [k]! [n-k]!
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 2-dimensional subspaces of C4 is the same as the number of ways of choosing 2 elements from a 4-element set! A nice example of the unity of mathematics.

Also, since complex Grassmannians are built from only even-dimensional 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 q-binomial coefficient:

    [4]!       1 (1 + q) (1 + q + q2) (1 + q + q2 + q3)
 ---------- =  -------------------------------------------
 [2]! [2]!             1 (1 + q)      1 (1 + q)

            = 1 + q + 2q2 + q3 + q4

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 Z2
the 6th homology group is Z
the 8th homology group is Z
and while we're at it, we've learned this Grassmannian is 8-dimensional as a real manifold - or 4-dimensional as a complex manifold. Note how the nth homology group is the same as the (8-n)th; this comes from Poincare duality.

On a lighter note: the best way to simplify this sort of expression

  1 (1 + q) (1 + q + q2) (1 + q + q2 + q3)
          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           11 
where I did the last step using long division. And of course the last quantity is
1 + q + 2q2 + q3 + q4.

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 q-factorials 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 half-open
interval is zero:

(0,1] = (0,1) + {1}


|(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) half-open 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 S-E 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

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!

We should declare instead candidly that we dwell on mathematics and affirm its statements for the sake of its intellectual beauty, which betokens the reality of its conceptions and the truth of its assertions. For if this passion were extinct, we would cease to understand mathematics; its conceptions would dissolve and its proofs carry no conviction. - Michael Polyani

© 2002 John Baez