I found the following statement in the paper(proof of lemma 4.1) https://www.researchgate.net/publication/347965880_The_arithmetic_of_modular_grids
$\begin{align} dim\mathbb{C}_{\hat{\Gamma},\nu}=dim\mathcal{E}_k(\Gamma,\nu)+dim\mathcal{E}_{2-k}(\Gamma,\bar{\nu})\end{align}$
The author said that above statement can be derived from modular pairing.
I have a problem to prove above statement.
Although the paper deal with the general case, let me assume the simple case.
$\Gamma$ is a congruence subgroup of $SL_2(\mathbb{Z})$, $\hat{\Gamma}=SL_2(\mathbb{Z})$, $k\in\mathbb{Z}$, and $\nu$ is a trivial multiplier.
We have $M_k(\Gamma)=S_k(\Gamma)\oplus\mathcal{E}_k(\Gamma)$,
$\mathcal{E}_k(\Gamma)$ is the subspace of $M_k(\Gamma)$ which is orthogonal to $S_k(\Gamma)$ with respect to the petersson inner product.
If I understood correctly, $dim\mathbb{C}_{\hat{\Gamma},\nu}$=$|\Omega(\Gamma)|$, where $\Omega(\Gamma):=\Gamma\backslash\mathbb{Q}^{\star}$
So I have to prove that $|\Omega(\Gamma)|$=$dim\mathcal{E}_k(\Gamma)+dim\mathcal{E}_{2-k}(\Gamma)$.
I will briefly explain the modular pairing.
Let $f\in M_k^!(\Gamma),g\in M_{2-k}^!(\Gamma)$
Then, for each $\lambda\in\hat{\Gamma},$ we have
$\begin{align}f^{\lambda}(z)=\sum\limits_{n\in\mathbb{Z}}a^{\lambda}(n)q_{w_\lambda}^n ,g^{\lambda}(z)=\sum\limits_{n\in\mathbb{Z}}b^{\lambda}(n)q_{w_\lambda}^n\end{align}$
Here, $w_{\lambda}$ is a width of $\lambda\infty$, $f^{\lambda}:=f|_k\lambda,g^{\lambda}:=g|_{2-k}\lambda$, and $q_{w_\lambda}:=e^{2\pi i\frac{z}{w_{\lambda}}}$
Then the pairing $\{f,g\}_{\Gamma}$ is given as a sum over the set of cusps $\Omega(\Gamma)$ by
$\begin{align}\{f,g\}_{\Gamma}=\sum\limits_{\rho\in\Omega(\Gamma)}w_{\rho}\sum\limits_{n\in\mathbb{Z}}a^{\gamma_{\rho}}(n)b^{\gamma_{\rho}}(-n), \text{ here }\gamma_{\rho} \text { is a any element in }\hat{\Gamma} \text{ such that }\gamma_{\rho}\infty=\rho \end{align}$
We can check that above formula is well defined and $\{f,g\}_{\Gamma}=0$
I have two questions
(1):How to derive that $|\Omega(\Gamma)|$=$dim\mathcal{E}_k(\Gamma)+dim\mathcal{E}_{2-k}(\Gamma)$ from modular pairing?
(2):If $k$ is odd and $-1\in\Gamma$, then $M_k(\Gamma)=\{0\}$ so I think that we may assume that $-1\notin\Gamma$. However, even though we assume that $-1\notin\Gamma$, if we consider the case $k=3$, then we can check that $dim\mathcal{E}_3(\Gamma)=\epsilon_{\infty}^{reg}$ and $dim\mathcal{E}_{-1}(\Gamma)=0$ from dimension formula, where $\epsilon_{\infty}^{reg}$ is a number of regular cusps of $\Gamma$. So if above formula hold then we conclude that there is no irregular cusps for $\Gamma$. I feel that there are some mistakes on my argument.
Thank you