Let $G=\mathrm{SL}_2(\mathbb{R})$, $K=\mathrm{SO}(2)$, and $f$ be a holomorphic modular form of weight $k$ for $\Gamma$ a Fuchsian group of the first kind. In Borel's book, 'automorphic forms on $\mathrm{SL}_2(\mathbb{R})$', the lifted function $\tilde{f}$ is considered, where $$ \tilde{f}(g)=j(g,i)^{-k}f(gi), \quad g \in G. $$
In this book, a Lemma says that $f$ satisfies $$ C\tilde{f}=(\frac{k^2}{2}-k)\tilde{f}, $$ where $C$ is the Casimir operator $C=\frac{1}{2}H^2+EF+FE$, $H=\begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix}, E=\begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}, F=\begin{bmatrix} 0 & 0 \\ 1 & 0 \end{bmatrix}$. This implies that $\tilde{f}$ is an automorphic form for $\Gamma$.
My first question is
- Why the above lemma is true?
To ask the second question, let me define the polyharmonic maass forms. A polyharmonic maass form $f$ of weight $k$, depth $m$ for $\Gamma$ is a smooth function on $\mathbb{H}$ with the following properties:
(1). $f|_k \gamma (z) = f(z) \quad$ for $\gamma \in \Gamma$.
(2). $\Delta_k^m f =0 \quad$ where $\Delta_k=-y^2(\frac{\partial^2}{\partial x^2}+\frac{\partial^2}{\partial y^2})-iky(\frac{\partial}{\partial x} + i\frac{\partial}{\partial y})$.
(3). moderate growth at the cusps
J.C.Lagarias and R.C.Rhoades' paper 'Polyharmonic Maass forms for $\mathrm{PSL}(2,\mathbb{Z})$' says that the finite dimensionality of the space of polyharmonic maass forms follows from the finite dimensionality of the space of automorphic forms. Hence it seems that there is a natural embedding from polyharmonic maass forms to automorphic forms.
I guess that the same lift induce the natural linear embedding. If this true, then for a polyharmonic maass form $f$, there is a polynomial $P$ in $C$ such that
$$ P(C)\tilde{f}=0. $$
Is my suggestion true?
If yes, what polynomial $P$ satisfies the above condition?
