Fourier Expansion in Siegel Modular forms Let $f: \mathbb{H_n} \mapsto \mathbb{C} $ be a Siegel modular form of weight $k \in  \mathbb{Z}$ of genus $n\geq 2$ with respect to the Symplectic group $\Gamma_n:=Sp(n,\mathbb{Z})$. I am only reading its basics. A statement first. "Since f is holomorphic on the Siegel Upper Half Plane, $\mathbb{H_n} :=\{Z \in M_n(\mathbb{C}) | Z= Z^{t}, ImZ> 0\}$ and satisfies $f(Z+S) = f(Z)$ for every $S=S^{t} \in M_n(\mathbb{Z})$, $f$ has a Fourier Series expansion of the form $f(Z) = \sum_{T=T^{t} \geq 0\\T\,\,\text{half-integral}} a(T)e^{2\pi itr(TZ)}$." I can't find any explanation beyond this in many books. Can someone suggest an easy reference? 
In n=1 case, we have the exponential map $e^{2\pi i z}$ from $\mathbb{H}$ to $\mathbb{D}$ which is an analytic map that is locally bi-holomorphic and the value of $f$ is constant on the fibres of $w\in \mathbb{D}\setminus\{0\}$ under $q$ all of which put together induces a holomorphic map $f_{1}:\mathbb{D}\setminus\{0\} \mapsto \mathbb{C}$ such that $f_1\circ q = f$. This fact allows me to have a fourier expansion for $f_1$ around a neighbourhood of $0$ with no negative coefficient terms (due to the boundedness of $f_1$ in the neighbourhood of $0$). The question is what is the analogue of $q$ in higher genus case?
The Fourier expansion in Siegel modular form case suggests to use the map $e^{2\pi i Tr(Z)}$. But then, in this case, $f_1$ will be a holomorphic function from $\mathbb{D}$ to $\mathbb{C}$ whose fourier expansion will have no connection(at least that is what I believe!) to the positive semi-definite matrices $T$. So $q(Z) := e^{2\pi i Z}$ seems to be only remaining choice from $\mathbb{H_n}$. But this leads me to several complex variables and I am zero on this. Where does this map land? How do we justify the fourier expansion in Siegel case? Any easy references for the above without going too much into the theory of several complex variables would also be equally useful.        
 A: Not a full answer because there is still the problem of the domain of convergence of the obtained Fourier series (the difficulty being that $Im(Z)>0$ means $Im(Z)$ is positive definite)

For $n=2$ then $F(u,v,w)=f(\pmatrix{u& v \\ v & w})$ is a $\Bbb{Z}^3$ periodic  holomorphic function on some open $U$ of $\Bbb{C}^3$ closed under $\Bbb{R}_{>0}$ linear combinations and $\Bbb{R}^3$ translates so
$$F(u,v,w) = \sum_{(n,m,k) \in \Bbb{Z}^3} A_{n,m,k} e^{2i \pi (nu+mv+kw)}$$

(pick some $u_0,v_0,w_0 \in U$, do the Fourier expansion of the smooth function $x \in\Bbb{R}^3\mapsto F((u_0,v_0,w_0)+x)$ and extend the obtained Fourier series by analytic continuation)

Then note that $$nu+mv+kw = Tr(\pmatrix{n& m/2 \\ m/2 & k}\pmatrix{u& v \\ v & w})$$
Finally since $$a(T) = \int_0^1\int_0^1\int_0^1 f(Z+\pmatrix{x_1 & x_2\\ x_2&x_3})e^{-2i \pi Tr(T(Z+\pmatrix{x_1 & x_2\\ x_2&x_3}))} dx_1dx_2dx_3$$ doesn't depend on $Z$ 
then $a(T)  =0$ whenever $T$ is not positive semi-definite because otherwise taking any $Z \in \Bbb{H}_n$ with $Tr(Im(Z)^{1/2} \ T \ Im(Z)^{1/2})=Tr(T\ Im(Z)) < 0$ then $f(r Z)$ would have to be unbounded as $r \to \infty$ which is forbidden by the cusp conditions.
