Dimension of space of Eisenstein series $\mathcal{E}_k(\Gamma)$

Question

Let $\Gamma$ be a congruence subgroup of $\text{SL}_2(\mathbb{Z})$, and define the space of Eisenstein series for $\Gamma$ to be $\mathcal{E}_k(\Gamma):=\{f\in M_k(\Gamma)\colon \langle f,S_k(\Gamma)\rangle_\Gamma=0\}$, the orthogonal complement of $S_k(\Gamma)$ in $M_k(\Gamma)$ with respect to the petersson inner product. Then I want to show a dimension bound of $$ \dim\mathcal{E}_k(\Gamma)\leq\#\text{Cusp}(\Gamma), $$ by constructing an element in $\mathcal{E}_k(\Gamma)$ for each cusp in $\text{Cusp}(\Gamma)$. However, I don't see how I can use cusps to cook up new Eisenstein series. Any help would be appreciated.

Answer 1

$$SL_2(\Bbb{Z})i\infty = \bigcup_{j=1}^m \Gamma \alpha_j i\infty$$ where the union is disjoint and $\alpha_j\in SL_2(\Bbb{Z})$. The RHS are the cusps.

The map $$M_k(\Gamma) \to \Bbb{C}^m, f\to (f|_k \alpha_1 i\infty,\ldots,f|_k \alpha_m i\infty)$$ is $\Bbb{C}$-linear with kernel $S_k(\Gamma)$.

Thus $$\dim_\Bbb{C}(E_k(\Gamma))=\dim_\Bbb{C}(M_k(\Gamma)/S_k(\Gamma))\le m$$

For many $k$ and $\Gamma$ it is a strict inequality (not for the normal subgroups $\Gamma=\Gamma(n)$ and $k\ge 4$ even)