# Don Zagier's “Zetafunktionen und quadratische Körper”

Do you know of a text--preferably in English--whose treatment of the class number formula is based on (or follows closely) the one expounded by Zagier in sections II.8 (binary quadratic forms) and II.9 ($L(1, \chi)$ and the class number) of the aforementioned book? Unfortunately, my command of the German language is not so good yet and, to add insult to injury, it seems to me that Springer Verlag is still to commission a translation into English (or one of the Romance languages) of this notable book.

Let me thank you in advance for your attentive consideration of this question of mine.

BOUNTY! Jan-Christoph Schlage-Puchta suggests chapter 6 of H. Davenport's "Multiplicative Number Theory" as an alternative text for this topic. Though, I must confess that the fact that Davenport dedicates only two or three lines of the chapter to Lagrange's main result on reduction of binary quadratic forms makes one feel uneasy right from the start. So, if you are proficient in German and wish to provide a translation of pages 64-68 of Zagier's book, I will really appreciate your help and award you the bounty I am herewith offering on this question.

• there is actually not much text on these pages; would something like page 64 be sufficient for your purpose? If it is, I can easily do the other four pages, but if you want a fully editable translation including formulas, it will be much more work. – Carlo Beenakker Mar 27 '17 at 6:49
• If you can sent me the relevant pages (or provide a link), I can give the translation a try; from what @CarloBeenakker provided, it doesn't seem too hard, despite not having any background in the field; I am a native German, so just give it a try if you like. – Manfred Weis Mar 27 '17 at 9:56

here is my "screenshot translation", hoping it is sufficient for your purpose:

please correct me if I have misrepresented any technical term.

• Mr. Beenakker: Thank you very much for taking the time to render these valuable pages into English for me. I will start studying 'em in a couple of hours: I suppose I can leave you a message here in case I have doubts regarding the translation... – José Hdz. Stgo. Mar 27 '17 at 15:58
• certainly, you're welcome. – Carlo Beenakker Mar 27 '17 at 16:03
• well done Carlo, I think I couldn't improve much on that translation. – Manfred Weis Mar 28 '17 at 4:43

Zagier's proofs are compact, so I don't think there's any way to reduce further. 4 pages can be a lengthy translation but here is enough to to get started.

Theorem 2 Let $f(x,y) = ax^2 + bxy + cy^2$ be a primitive quadratic form of discriminant $D$ (squarefree). There is a bijection between solutions to Pell's equation and the automorphism group of $f$

$$(t,u) \to \left( \begin{array}{cc} \frac{t - bu}{2} & - cu \\ au & \frac{t+bu}{2}\end{array} \right)$$

This bijection is a group isomorhism by the composition rule: $$(t_1, u_1) \circ(t_2, u_2) = \left( \frac{t_1 t_2 + D u_1 u_2}{2}, \frac{t_1 u_2 + t_2 u_1}{2}\right)$$ The group $U_f$ is finite if $D < 0$ and cyclic of order $w$

• $\mathbb{Z}/6\mathbb{Z}$ for $D = - 3$
• $\mathbb{Z}/4\mathbb{Z}$ for $D = - 4$
• $\mathbb{Z}/2\mathbb{Z}$ for $D < - 4$

For positive $D > 0$ the group of solutions $U_f \simeq \mathbb{Z} \times \mathbb{Z}/2\mathbb{Z}$

Proof Zagier says the positive discriminant case follows from Pell's equation (it does). If you have a solution to Pell's equation (for any $f(x,y) = n$) , both group laws are pretty clear and that they map to one another. There is a map (injective homomorphism) from $U_f$ either to $\mathbb{C}^\ast$ or $\mathbb{R}^\ast$ given by $$(t,u) \mapsto \frac{t + u \sqrt{D} }{2}$$ the case $D = - 3$ corresonds to $\mathbb{Z}[e^{2\pi i / 3}]$ and $D = - 4$ matches $\mathbb{Z}[i]$, with units $e^{2\pi i /3}$ or $i$. In all other cases $D < -4$ there are only the trivial units $\pm 1$. $\hspace{1.4in}\square$

Theorem 3 Let $D$ be a "fundamental discriminant" $n \neq 0$ be a whole number. Then the total number $R(n)$ of representation of $n$ by primitive forms of discriminant $D$ is given by $$R(n) = \sum_{m|n} \chi_D(m)$$ where $m$ is a divisor of $n$ and $\chi_D(m)$ is a character. In particlar, $R(n)$ and all $R(n,f)$ are finite.

Proof There are no imprimitive forms, so we can leave outthe word "primitive" in the theorem. Let $R^\ast(n)$ be the number of inequivalent primitive representations of $n$ by (any) forms of discriminant $D$. Obviously $$R(n) = \sum_{g \geq 1, g^2 | n} R^\ast (n/g^2)$$ since every representation is a multiple of a primitive one. The main step in the proof is the formula $$R^\ast(n) = \{ b \pmod {2n} \big| b^2 \equiv D \pmod{4n}\}$$ Then he proceeds to explain the notion of a group action and variants of the orbit stabilizer formula

Let $G$ be a group $X$ and $Y$ be two sets and $S \subset X \times Y$. Then we can study the diagonal group action $(x,y) \mapsto (gx,gy)$ and observe the quotient sets $X/G$ as well as $S/G$.

• $Y_x = \{ y \in Y | (x,y) \in S \}$
• $G_x = \{ g \in G | gx = x \}$
• $|S/G| = \sum_{x \in X/G} |Y_x/G_x|$
• $|S/G| = \sum_{y \in Y/G} |X_y/G_y|$

The choice of group action by Zagier is set of invertible $2 \times 2$ matrices acting on pairs $X \times Y =$ $$\big\{\text{quadratic forms }ax^2 + bxy+cy^2 \text{ with }b^2 - 4ac = D \} \times \{ \text{pairs of integers }x,y \in \mathbb{Z}\big\}$$

and $G = SL(2,\mathbb{Z})$ which is solutions to $ad-bc = 1$ for $a,b,c,d \in \mathbb{Z}$. Lastly the special set

• $S = S(n) =\{ (f,z) \in X \times Y : f(z) = n \}$

Then $X/G$ is the set of equivalence classes of discriminant $D$ , and by the group action formula: $$|S/G| = \sum_{[f]} R^\ast (n,f) = R^\ast(n)$$

Every element of $Y$ is equivalent to $z = (1,0)$ (Exercise) for this element:

• $G_z = \left\{ \left(\begin{array}{cc} 1 & r \\ 0 & 1 \end{array} \right) : r \in \mathbb{Z}\right\}$

• $X_z = \{ nx^2 + bxy + \frac{b^2 - D}{4n} y^2: b \in \mathbb{Z}, b^2 \equiv D \pmod {4n} \}$

• these are related to the substitions $b \mapsto b + 2nr$

Zagier observes that $R^\ast(n)$ is multiplicative so that enough we solve for the primes $$R^\ast(p^r) = \# \big\{ b \pmod{p^r} \big| b^2 \equiv D \pmod{p^r} \big\}$$

Then we can reconstruct $R(n)$ from the count of "primitive" representations of $n$, $R^\ast(n)$

• $R(p^r) = \sum_{0 \leq s < r/2}2 + \sum_{s=r/2} 1 = r+1 = \sum_{0 \leq i \leq r} \chi_D(p^i)$ if $(\frac{D}{p})=+1$

• $R(p^r) = \sum_{0 \leq s < r/2}0 + \sum_{s=r/2} 1 = 0 \text{ or }1 = \sum_{0 \leq i \leq r} \chi_D(p^i)$ if $(\frac{D}{p})=-1$

• $R(p^r) = \sum_{0 \leq s < (r-1)/2}0 + \sum_{(r-1)/2 \leq s \leq r/2} 1 = 1 = \sum_{0 \leq i \leq r} \chi_D(p^i)$ if $(\frac{D}{p})=+1$

and Theorem 3 checks in all cases. $\hspace{2in}\square$

Corollary The average number of representations of $n$ is exactly the value of the L-function $$\lim_{N \to \infty} \left( \frac{1}{N}\sum_{n=1}^N R(n) \right) = L(1, \chi_D)$$ The proof involves counting lattice points under a hyperbola. For a specific primitive $f$ the formula involves the fundamental unit.

Theorem 4 Let $\epsilon_0$ be fundamental unit of $f$.

$$\lim_{N \to \infty} \left( \frac{1}{N}\sum_{n=1}^N R(n,f) \right) =$$ $$\left\{ \begin{array}{cc} \frac{2\pi}{w\sqrt{|D|}} & \text{ if }D < 0 \\ \frac{\log \epsilon_0 }{w\sqrt{D}} & \text{ if }D > 0 \\ \end{array}\right.$$

• Mr. Mangual: Thank you very much for taking the time to translate into English these two results... I have already upvoted your answer: unfortunately, the site allows one to upvote answers just once. – José Hdz. Stgo. Mar 27 '17 at 16:14
• @JoséHdz.Stgo. I translated what I can from a library copy of the book. My hope is you can look at my answer and skim the German equations and follow. His Theorem 2 can be found many times on this site and on Math.SE. His Theorem 3 is rather unique and related to his proof that all primes $p = 4k+1$ can be written as $p= a^2 + b^2$ here. If you have any doubts let me know. – john mangual Mar 28 '17 at 0:37

The content of chapter II.9 is contained in many textbooks on analytic number theory. A favourite of mine is Davenport's multiplicative number theory. For binary quadratic forms things are more difficult, somehow, they fell out of fashion. So most books either do not treat them at all, or they specialize on them, but I don't know of any treatment similar to the one in Zagier's book.