Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

There is a formula given in the book saying that $$ \dim{\mathcal{M}_1(\Gamma)}=\dim\mathcal{S}_1(\Gamma)+\frac{\varepsilon^{\text{reg}}_\infty}{2} $$

Where $\varepsilon^{\text{reg}}_\infty$ means the number of regular cusps of $X(\Gamma)$.

But I think it is a typo and the difference should be $\varepsilon^{\text{reg}}_\infty$.

But on the other hand, we have no explicit formula for $\dim{\mathcal{M}_1(\Gamma)}$, only knowing that it is greater or equal to $\varepsilon^{\text{reg}}_\infty/2$ (proved in the book).

So I wonder if indeed the difference is $\varepsilon^{\text{reg}}_\infty$, why don't the authors state the stronger consequence $\dim{\mathcal{M}_1(\Gamma)}\geq \varepsilon^{\text{reg}}_\infty$?

So anyone know the correct result?


The book I'm reading the GTM228, A First Course In Modular Forms [by Diamond and Shurman], and the conclusion is on page 91, theorem 3.6.1.

There is another reasons for me to think the difference is $\varepsilon^\text{reg}_\infty$ rather than half of it: is there any theorem ganrantee that $\varepsilon^\text{reg}_\infty$ is even? For dimensions are both integer.

share|cite|improve this question
I think you are right about $\epsilon_{\infty}^{reg}$. For greater than $\epsilon_{\infty}^{reg}/2$ part, can you give a reference? – user27126 Jan 12 '13 at 18:36
I'm pretty sure the fomula given in the book is correct. The dimension of the weight 1 Eisenstein series is half the dimension of the Eisenstein series of odd weight $k\ge3$. – David Loeffler Jan 12 '13 at 20:14
PS: Which book are we talking about, by the way? – David Loeffler Jan 12 '13 at 20:35
@Sanchez, I have updated my post. – hxhxhx88 Jan 13 '13 at 1:20
@DavidLoeffler, I have updated my post, and how is Eisenstein series involved here? – hxhxhx88 Jan 13 '13 at 1:21
up vote 0 down vote accepted

I'm going to put together into an answer some of the things I've said in the comments.

Firstly, there is a direct sum decomposition

$$M_k(\Gamma) = S_k(\Gamma) \oplus N_k(\Gamma)$$

where $N_k(\Gamma)$ is the dimension of the space of weight $k$ Eisenstein series, which are defined to be those modular forms $f$ such that $\langle f, g \rangle_{\Gamma} = 0$ for all $g \in S_k(\Gamma)$ where $\langle, \rangle_{\Gamma}$ denotes the Petersson product. So the difference between $\dim M_k(\Gamma)$ and $\dim S_k(\Gamma)$ is $\dim N_k(\Gamma)$.

The dimension of $N_k$ is given by $$ \dim N_k(\Gamma) = \begin{cases} \varepsilon_{\infty} & \text{if $k \ge 4$ even}\\ \varepsilon_{\infty}^{\text{reg}} & \text{if $k \ge 3$ odd}\\ \varepsilon_{\infty} - 1 & \text{if $k = 2$}\\ \varepsilon_{\infty}^{\text{reg}} / 2 & \text{if $k = 1$}. \end{cases}$$

Rather than running through the proof (for which it's easier to just look in D + S) I suggest the following might make the result more believable. (Warning: I'm doing this from memory, it's Sunday and all my books are in the office, so some formulae may be slightly wrong.)

Suppose $\Gamma = \Gamma_1(N)$. Then for any primitive Dirichlet characters $\chi_1, \chi_2$ of conductors $N_1, N_2$ such that $N_1 N_2 \mid N$ and $\chi_1(-1) \chi_2(-1) = (-1)^k$, then there is an Eisenstein series $E_{k, \chi_1, \chi_2}$ given by $$ \text{constant} + \sum_{n \ge 1} \left( \sum_{d \mid n} \chi_1(d) \chi_2(n/d) d^{k-1} \right) q^n $$ (the constant is either 0 or some value of a Dirichlet $L$-function). If $k = 2$ there's an additional constraint to the effect that $\chi_1$ and $\chi_2$ aren't both trivial. And the space $N_k(\Gamma_1(N))$ is spanned by series of the form $E_{k, \chi_1, \chi_2}(q^t)$ for $t$ dividing $N / (N_1 N_2)$.

Now, for $k$ an odd integer the set of pairs of Dirichlet characters $ (\chi_1, \chi_2)$ that can come up is independent of $k$; but for $k = 1$ there is an extra symmetry, because $E_{1, \chi_2, \chi_1} = E_{1, \chi_1, \chi_2}$, and that cuts the size of the resulting set in half. So that gives a reason why one should expect $$ \dim N_1(\Gamma) = 1/2 \dim N_{2k + 1}(\Gamma)$$ for any $k \ge 1$.

share|cite|improve this answer
Thank you soooo much! – hxhxhx88 Jan 13 '13 at 13:14

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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