Over on MathOverflow, we’ve had a bunch of discussions about the class number formula. In particular, Keith Conrad pointed out a paper of Orde which gives a beautiful nonsense proof of the class number formula. This post is my attempt to understand why Orde’s argument works. Specifically, I am going to use an idea which I learned from Terry Tao’s blog: Arguments about divergent sums are often really arguments about the constant term in asymptotic expressions for smoothed sums.
This post assumes familiarity with the concepts and notations of a first course in algebraic number theory.
The class number formula
Let be a positive integer which is either (a) squarefree and congruent to or (b) of the form , where is squarefree and congruent to or . Let be or respectively. Let and respectively be the class number and the number of units of . Let be the Kronecker symbol . (This is the same as the Jacobi symbol when is odd and positive, but it is defined for all integers. See the link for details.)
There are two formulas which I wish to discuss, and both of which I’ve heard called the class number formula.
The Cesaro sum in can be, and often is, rewritten into a finite sum. I’ll discuss this below, but I’ve become convinced that the Cesaro sum is the most natural formulation.
Until this week, all the proofs I knew obtained the equation in a reasonably motivated way. They then proved that in some mysterious way: either by proving the functional equation for , or by manipulations with Gauss sums. This is particular odd when one considers that equation is simpler than , and (after a little rewriting) can be made into a purely rational expression, with no limits or transcendental quantities at all. The first thing that fascinated me about Orde’s paper is that he gets directly, without passing through . In this post, we will prove both and directly.
If , then , and . The character periodically repeats
For , the partial sums of are periodic, repeating
By definition, the Cesaro sum is the limit of the average of these partial sums. If we average over terms, the average is exactly , and in general the averages approach . Sure enough, this is .
In general, the sum in is painful to compute directly. There is an algorithm to do it: write and evaluate the integral by the standard algorithm for integrating rational functions. (Note that , so the integral converges.) For example, when , we have , and
The numerators of these rational functions are the Fekete polynomials.
The first sum quickly gets out of hand for naive hand computation. The second sum, on the other hand, is very tractable. Let’s take . Clearly, and it turns out that . The character periodically repeats
+ + + + - + - + + - - + + - - + - + - - - - 0
with partial sums
1 2 3 4 3 4 3 4 5 4 3 4 5 4 3 4 3 4 3 2 1 0 0.
The average of these numbers is , as desired.
As should be clear, the Cesaro sum in can be easily turned into a finite sum. The most direct translation of the above argument gives . This can be simplified in various ways, to give formulas like the one discussed here. However, doing so tends to introduce cases based on what is modulo which, to my mind, obscure the main result.
Ideals and Quadratic forms
For a positive integer, let be the number of ideals of with norm . By unique factorization into prime ideals, is a multiplicative function. A little thought about the prime power case reveals that
We now give a second description of . Let , , … be representatives for the classes in the class group of . Let’s consider how many ideals of have norm and have class . A fractional ideal has class if and only if it is of the form , for . The number of ways to express a given fractional idea in this form is . The fractional ideal is an honest ideal of if and only if . And . In short,
Choosing an identification of with , let be the quadratic form . So
For future reference, note that has determinant .
Equation is valid for . Inspired by it, we define
To summarize, for , we have
and we want to prove that
Orde has an intuition for that term involving negative definite quadratic forms, but it doesn’t help me. I’ll just leave it as a mystery for now.
The divergent sum “argument”
I should say at the outset that, since I am going to give a correct proof later, I have felt free to abuse divergent sums even more blatantly than Orde does. I gave a more faithful write up of his argument here.
Consider . On the one hand, this is
As any student of divergent sums knows, , so this is .
On the other hand, is
Interchanging summation, we have
Now, . (Notice that . Consistency!) So we obtain
Exactly what we wanted!
In order to turn this into an actual proof, we will need to introduce a cutoff function, which we will call . Examples of ‘s which we might use are or . We should have , we should make sure that “dies fast enough near “, that is “differentiable enough” and that has bounded variation. The terms in quotes are intended to be sloppy.
For any sequence , the -summation of the sum is . Note that, if we take to be for and otherwise, then -summation is ordinary summation. However, this choice of will not be smooth enough for me. Note also that taking gives Cesaro summation; this choice is smooth enough for my argument.
Orde does something like this. He computes bounds for the sum , where and at certain linked rates, and is rational. The reason he chooses this smoothing is that he wants to work only with finite sums and rational numbers, so that he can claim to have a fully rational proof of the rational relation . In pursuit of this goal, Orde also rewrites all limits into their definitions, and uses some clever tricks so that he only has to do this a few times. I, on the other hand, will be profligate with my limits and integrals. If we had some sort of formal theory of constant terms of asymptotic series, comparable to the theory of formal power series, then one might be able to turn this post into a rational proof.
So, let us consider .
On the one hand, this is
Using the Euler-Maclaurin formula, we can approximate this sum by the integral . Ordinarily, the main source of error in such an approximation comes from boundary terms, but in this case there are none. The other source of error is lack of smoothness of . Our (unstated) assumptions on get this error down to at least . If is smooth, we get all the way down to , for any positive constant ! (We will feel free to omit the phrase “as ” when giving bounds like this.)
We can change variables in the integral so that becomes the standard norm on . The determinant of this change of variables is , since . So
Switching to polar coordinates, this is
Now, on the other hand, we have
Interchanging summation, we get
Defining , we can rewrite this as
Using Euler-Maclaurin summation,
Define the function by
So . If we choose is smooth enough and fast enough decaying then is a valid smoothing function. UPDATE I earlier asserted that the choice worked; I’m not sure that was true.
Using our new notation, the -summation of is
Comparing the terms of and , we see that
Canceling the integral, we get equation (1).
Now, let’s compare the constant terms. We get
The second term is the -summation of .
Key Fact: If is periodic with average , then the -summation of is always the Cesaro sum, independent of the choice of .
Terry proves this for (see the discussion above Exercise 1). It is a good exercise to adjust this proof for the general case. So we have shown that
This is equation .