What is the difference between the dimension of $\mathcal{M}_1(\Gamma)$ and $\mathcal{S}_1(\Gamma)$? 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?

updated:
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.
 A: 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 $k \ge 1$.
