26
$\begingroup$

(Crossposted on math stack exchange: https://math.stackexchange.com/questions/4040249/relation-between-schanuels-theorem-and-class-number-equation)

It was recently brought to my attention that there is a striking similarity between the Class Number Formula and Schanuel's Theorem. See for yourself:

Notation: \begin{align*} &K, \text{ number field}\\ &d, \text{ degree of $K$ over $\mathbb{Q}$}\\ &h_K, \text{ class number of $K$}\\ &R_K, \text{ regulator of $K$}\\ &D_K, \text{ discriminant of $K$}\\ &\mu_K, \text{ number of roots of unity contained in $K$}\\ &r_1, \text{ number of real places of $K$}\\ &r_2, \text{ number of complex places of $K$}\\ &\zeta_K(s), \text{ Dedekind zeta function of $K$}\\ &H_K(P), \text{ height relative to $K$ on $\mathbb{P}^n$} \end{align*}

Theorem (Class Number Formula) $$\lim_{s\to 1}(s-1)\zeta_K(s)=\frac{h_K R_K 2^{r_1}(2\pi)^{r_2}}{\mu_K \sqrt{|D_K|}}$$

Theorem (Schanuel's Theorem) \begin{align*} \#\{P\in \mathbb{P}^n(K): H_K(P)\leq T\}=\frac{h_K R_K}{\mu_K\ \zeta_K(n+1)}\left(\frac{2^{r_1} (2\pi)^{r_2}}{\sqrt{|D_K|}}\right)^{n+1} (n+1)^{r_1+r_2-1}T^{n+1}+O(T^{n+1-1/d}) \end{align*}

Both of these results can be proven using a geometry of numbers argument. So one reason for the similarity might simply be that they end up being answers to similar counting problems. But this is not very satisfying, and it would be nice to have a more conceptual reason that they are so similar.

Main Question: Is there an intuitive (or deep) reason that the analytic class number formula is so similar to the coefficient of the leading term in Schanuel's Theorem?

Secondary Question: What other examples are there of values of $L$-functions showing up in main terms of asymptotics?

$\endgroup$
1
  • 3
    $\begingroup$ Wouldn't it be more comparative to write the class number formula in a form involving the number of ideals of norm up to $T$? $\endgroup$
    – user174996
    Mar 6 at 4:01
19
$\begingroup$

A spirit. A general approach that may interest you is the following: counting laws can be obtained by studying suitable generating functions, via Tauberian arguments: the rightmost pole (resp. residue) of the generating function gives the growth order (resp. leading constant) in the counting law.

In the case of Schanuel's theorem, the generating function is in fact linked to the Dedekind zeta function $\zeta_K$, showing how the residue of $\zeta_K$ naturally appears as part of the main term. With this point of view, it is not a surprise.

Let me give some details and references below.

Counting laws from zeta functions. In many counting problems, we introduce a generating function (Fourier series, Dirichlet series, zeta functions, etc.) as an analytic tool to study the underlying combinatorics. In the realm of counting rational points on algebraic varieties $V$ over a number field $K$, a suitable such generating function is the height zeta function defined by $$Z_V(s) := \sum_{x \in V(K)} h(x)^{-s}.$$ For $s$ large enough, this series is convergent (as a consequence for instance of Northcott's theorem, stating that the number of rational points of bounded height grows polynomially).

The spirit now is the same as for the prime number theorem: obtain enough analytic properties of $Z_V(s)$. Suppose you are able to prove that $Z_V(s)$ has meromorphic continuation to a certain right half-plane, say with a single pole at $s=a$ and then remains holomorphic up to $\Re(s)>a-\delta$ for a certain $\delta>0$. Then, the (Wiener-Ikehara) Tauberian theorem gives you straight away the analogue of Schanuel's theorem for the variety $V$: $$N_V(X) := \# \{x \in V(K) \ : \ h(x) \leq X\} = \underset{s=a}{\rm Res} \ Z_V(s) \cdot X^a + O(X^{a-\delta}).$$

You can of course obtain finer asymptotic information on $N_V(X)$ with finer analytic information on $Z_V(s)$ (and may pick extra $\log(X)$ factors in case there is a vertical growth).

The case of Schanuel's theorem. Now, how do we study $Z_V(s)$? This is done in a great paper of Franke-Manin-Tschinkel (1989): they relate $Z_V(s)$ to an explicit Eisenstein series (see their Proposition 3), hence deducing its nice analytic properties, and then they evaluate precisely the residue to obtain the leading coefficient (see their equation (2.12)). Without much surprise if you are acquainted with Eisenstein series: it is expressed in terms of the Dedekind zeta function. Hence, the leading coefficient in Schanuel's law is essentially the residue at $s=1$ of $\zeta_K$.

A nice introduction to these geometric methods can be found for instance in the short and recent lecture notes by Chambert-Loir.

Another expression for the constant. Note that you may also be willing to forget about the expressions in terms of class numbers and Dedekind zeta functions (which bears a lot of arithmetics) and look for a more geometric expression. This is provided for instance in Chambert-Loir-Tschinkel where the leading constant is rephrased essentially as a height zeta integral (see their Theorem 1.3.1, I try to be consistent with my notations above) $$\int_{V(\mathbb{A}_K)} h(x)^{-a} dx$$ for a certain suitably normalized Tamagawa measure $dx$. Recall that $a$ is the growth order in the counting law.

A glimpse to the second question. What I said above already gives you a whole bunch of analogous phenomena: zeta functions are generating functions, hence their residues naturally appear in counting laws for the underlying objects.

An example with a pretty different flavor is the following: you can consider the analogous zeta functions (of the conductor) for automorphic forms, use to trace formulas to study it and in particular see that the identity term (which is the most significant one) can be expressed by the Dedekind zeta function of the underlying field. Hence, you get counting (Weyl) laws displaying a leading constant expressed with the same residue $\zeta_K^\star(1)$.

Hope it helps!

$\endgroup$
1
  • 2
    $\begingroup$ This is a great answer; thanks for the references too! $\endgroup$ Mar 11 at 2:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.