In Zagier's chapter of The 1-2-3 of Modular Forms, he defines the Theta series of a quadratic form $Q$ as $$\Theta_Q(z) = \sum_{n=0}^\infty r(Q,n) q^n$$ where $q = e^{2\pi i z}$. He identifies orbits of quadratic forms under an action of $SL_2(\mathbb{Z})$ with ideal classes in $\mathbb{Q}(\sqrt{D})$, and then takes a homomorphism $\chi$ from the ideal class group of $\mathbb{Q}(\sqrt{D})$ to $\mathbb{C}^*$, and sets $$f_\chi = \frac{1}{w} \sum_{\mathcal{A}} \chi(\mathcal{A}) \Theta_\mathcal{A}(z)$$ where $w$ is a constant, the sum is over ideal classes in $\mathbb{Q}(\sqrt{D})$, and $\Theta_\mathcal{A}(z)$ is $\Theta_Q(z)$ where $\mathcal{A}$ and the orbit of $Q$ under the action of $SL_2(\mathbb{Z})$ are identified.

He then claims that $f_\chi$ is a modular form of weight $1$, but I can't tell why that is the case. I'm guessing that it must be of weight $1$ and character $\chi$, because I don't see how it makes sense otherwise, but the definition I'm familiar with for modular forms of weight $k$ and character $\chi$ is that they are modular forms $f$ of weight $k$ on $\Gamma_1(N)$ such that $$f\left(\frac{az+b}{cz+d}\right) = \chi(d) (cz+d)^k f(z)$$ for all $\begin{pmatrix} a & b \\ c & d \end{pmatrix} \in \Gamma_0(N)$. Is this the right definition?

Here is the relevant section of the text.

Also, at the end, he claims that the Euler product for the $L$-series of $f_\chi$ means that $f_\chi$ is a normalized eigenform, but I thought that the Euler product must be of the form $$\prod_{p} \frac{1}{1 - a_p p^{-s} + \chi(p) p^{k-1-2s}}$$ where the product is over primes $p$ and $a_p$ are the Fourier coefficients of $f_\chi$. I can see why the $L$-series has a product over prime ideals of the ring of integers of $\mathbb{Q}(\sqrt{D})$, but not why it has this form.

  • $\begingroup$ Just to make clear, $\chi$ is not a Dirichlet character, so it doesn't make sense for $f_\chi$ to have character $\chi$. $\endgroup$
    – Mathmo123
    Commented Jan 9, 2018 at 18:20
  • $\begingroup$ Yeah, that's what was confusing me. I'm not sure in what other sense $f_\chi$ could be a modular form, though. $\endgroup$
    – user519967
    Commented Jan 9, 2018 at 20:12
  • 4
    $\begingroup$ Don't delete posts after getting an answer. Like ever. We have grown to think that this is rude to the nice people who put in some effort answering you. Basically deleting the question robs the answerers all opportunity to get their reward in terms of peer appreciation. You do not own the posts. After you, click the post -button, the question becomes collective property (read the fine print in legalese), and should be deleted if and only if the community feels that way. $\endgroup$ Commented Jan 10, 2018 at 13:46

1 Answer 1


Suppose that we have a Hecke character $\chi$, in this case a character $\chi:{Cl}(K)\to \mathbb C^*$ for some imaginary quadratic field $K$.

Using this character Zagier is showing how to construct a modular eigenform $f_\chi$ with the property that $$L(s,f_\chi) = L(s,\chi).$$ This equality makes complete sense as an equality of complex functions, despite the fact that the Euler product on the right is over the primes of $\mathcal O_K$, whilst the Euler product on the left is over the rational primes.

If you're happy with the fact that each $\Theta_Q(z)$ is a modular form of weight $1$, then since $f_\chi$ is a finite sum of such forms, it will also be a weight $1$ form. The character of $f_\chi$ will depend on $\chi$, but cannot be $\chi$, since $\chi$ is not a Dirichlet character.

On the one hand, $$\begin{align}L(s, f_\chi) &= L\left(s,\frac1w \sum_{\mathcal A}\chi(\mathcal A)\Theta_{\mathcal A}(z)\right) \\&=\sum_{n=1}^{\infty}n^{-s}\sum_{\mathcal A}\chi(\mathcal A)r(\mathcal A, n).\end{align}$$

On the other hand, since $r(\mathcal A,n)$ is the number of ideals $\mathfrak a$ of $\mathcal O_K$ of norm $n$ belonging to the class $\mathcal A$, it follows that $$\begin{align}\sum_{n=1}^{\infty}n^{-s}\sum_{\mathcal A}\chi(\mathcal A)r(\mathcal A, n)&=\sum_{n=1}^\infty\sum_{\mathfrak a :N(\mathfrak a) = n}\chi(\mathfrak a)N(\mathfrak a)^{-s}\\&=\sum_{\mathfrak a}\chi(\mathfrak a)N(\mathfrak a)^{-s}=L(s,\chi)\end{align}$$

Since $L(s,\chi)$ has an Euler product, it follows that $L(s,f_\chi)$ does too.

We have $$\begin{align}L(s,\chi) &= \prod_{\mathfrak p\subset\mathcal O_K}(1-\chi(\mathfrak p)N(\mathfrak p)^{-s})^{-1}\\&=\prod_{p\text{ inert},\ \mathfrak p\mid p}(1-\chi(\mathfrak p)p^{-2s})^{-1}\prod_{p\text{ split},\ \mathfrak p\mid p}(1-\chi(\mathfrak p)p^{-s})^{-1}\prod_{p\text{ ramifies},\ \mathfrak p\mid p}(1-\chi(\mathfrak p)p^{-s})^{-1}.\end{align}$$

Evaluating each of these products individually allows us to view the L-function as an Euler product over $\mathbb Z$, which will be exactly of the form you expect.

  • $\begingroup$ Thanks a ton for the help. I'm still a bit confused about a few things, though. Which group is $f_\chi$ a modular form of weight $1$ for? Is the proof that it satisfies the modular transformation property something to do with the Poisson summation formula, or is there a simple explanation? Also, on page 43, he claims that $f_\chi$ is in $S_1(\Gamma_0(23), \epsilon_{-23})$. Is that just a coincidence for this particular $\chi$, or is there something more general? $\endgroup$
    – user519967
    Commented Jan 9, 2018 at 18:52
  • $\begingroup$ The level and character will depend on $\chi$. The level will be closely related to the discriminant of $\mathbb Q(\sqrt D)$ and there is an explicit formula for the character. I don't have Zagier's book in front of me right now, but another reference is Miyake, Modular Forms Thm 4.8.2. $\endgroup$
    – Mathmo123
    Commented Jan 10, 2018 at 15:59

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .