Behavior of a modular form in the lower strip Let $f$ be an (elliptic) modular form of weight $k>0$, and consider the vertical strip $S_m=\{x+iy\in\mathbb{C}:|x|\le 1/2, y>m$}. For every $m\ll 1$, the fundamental domain for $SL_2(Z)$ is included in $S_m$. We know that modular forms are bounded on the fundamental domain $\mathcal{F}$ i.e
$$|f(z)|\le c_f\quad\forall z\in \mathcal{F}$$
I want to study what happens in the lower part of the strip, i.e. for $0<y\le m$ for $m$ small. Let $z$ be in the lower part and find a $\gamma\in SL_2(Z)$ such that $\gamma z\in \mathcal{F}$, then
$$|f(z)|=|j(\gamma,z)^{-k}f(\gamma z)|\le c_f |cz+d|^{-k}$$
where $(c,d)$ is the bottom row of $\gamma$. I want an upper bound for $|f(z)|$ and since $k>0$ I just need to majorize 
$$|cz+d|=\sqrt{(cx+d)^2+c^2 y^2}\ge |c|y$$
I observe that $c\not=0$ (otherwise $\gamma$ would be an horizontal translation, but there are no points of $\mathcal{F}$ with $y\le m$). Hence $|c|\ge 1$ and finally
$$|f(z)|\le c_f y^{-k}\quad\forall y\le m\ll 1$$

Can we show something similar for a (scalar) Siegel modular form (of arbitrary degree $n$), namely
  $$|F(Z)|\ll_{n}\det(Y)^{-k}$$
  for every $Y\not> m I_n$, for $m\ll 1$?

The idea would be exactly the same, but I cannot find a lower bound for
$|det(CZ+D)|$ because the $n\times n$ matrix $C$ needs not to be invertible: my suspicion is that $rank(C)$ is roughly equal to the number of eigenvalues of $Y$ which are $\le m$.
UPDATE:
Let $E_q^n(Z,s)=\sum_\gamma \det(Y)^s |\det(CZ+D)|^{-2s} \det(CZ+D)^{-q}$ be the usual Siegel-Eisenstein series: it is known to converge absolutely and uniformly in $Z$ for $2\Re(s)>n+1-q$, hence by letting $q=0$ and $s=k/2$ (assuming $k>n+1)$ we have $|\det(CZ+D)|^{-k}\ll\det(Y)^{-k/2}$ and therefore
$$|F(Z)|\ll\det(Y)^{-k/2}$$
Not quite what I need, but at least the claim is not totally unreasonable.
EDIT:
The series converges only locally uniformly so that does not help at all.
 A: I worked out the case $n=2$, and I assume it can be generalized to any degree.
Fix any $m>0$ such that the Siegel fundamental domain is contained in $\{Z:\Im(Z)>mI_2\}$. Take any $Z$ in the lower strip, i.e. $Y=\Im(z)\not>mI_2$. Let $\gamma\in\Gamma_2$ be a $4\times 4$ symplectic matrix for which $\Im(\gamma Z)>mI_2$ and let $(C|D)$ be the lower part of $\gamma$; which exists because there must be an element of $\Gamma_2$ mapping $Z$ to the Siegel fundamental domain (by very definition of the latter, since $Z$ is not already contained in it).


*

*If $C=0$ then $D$ must be in $GL_2(\mathbb{Z})$. Then $|\det(CZ+D)|=1$.

*If $rank(C)=1$, then we follow Maass' exposition (page 165-167, Siegel modular forms and Dirichlet series, https://www.springer.com/us/book/9783540055631) which leads to $$|\det(CZ+D)|^2=\det(C_1)^2 \det(Q^t Y Q)^2\prod_v(h_v^2+1)\ge \det(Q^t YQ)^2$$
since $C_1\in\mathbb{Z}_0$, where $Q$ is a primitive vector in $\mathbb{Z}^{2}$ (i.e. with coprime entries): now $Q^tQ=Q_1^2+Q_2^2\ge 1$ so
$$Q^t YQ\ge\frac{Q^t YQ}{Q^tQ}\ge\min_{v\not= 0}\frac{v^t Y v}{v^t v}=\lambda(Y)$$
and therefore $|\det(CZ+D)|^{2}\ge \lambda(Y)^2$, where $\lambda(Y)$ is its smallest eigenvalue.

*If $C$ has full rank, then
$$|\det(CZ+D)|=|\det(C)||\det(iY+(X+C^{-1}D))|\ge|\det(R+iY)|$$
and, since $Y=Y^t$, $W=R+iY$ is in its cartesian form (i.e. $2R=W+W^*$ and $2iY=W-W^*$, where $W^*=\overline{W^t}$). Therefore $|\det(W)|\ge\det(Y)$.


All in all this means that $|F(Z)|\ll\prod_j \lambda_j^{-k}$ for some eigenvalues $\lambda_j$ of $Y$ (eventually none).
EDIT: Eventually I managed to generalise this to any genus $n$. With the same notation, let $0<r\le n$ be the rank of $C$. Then as already stated $|\det(CZ+D)|^2\ge \det(Q^t YQ)^2$ where $Q$ is a $n\times r$ matrix with integer entries and maximal rank $r$. Therefore
$$|\det(CZ+D)|^2\ge \frac{\det(Q^t YQ)^2}{\det(Q^t Q)^2}\ge \left(\min_P \frac{\det(P^t Y P)}{\det(P^t P)}\right)^2=\prod_{j=1}^r \lambda_j^2$$
where $P$ varies on the set of real $n\times r$ matrices of rank $r$ and the last equality is a generalisation of the min-max theorem, and the eigenvalues of $Y$ are ordered as $0<\lambda_1\le\dots\le \lambda_n$.
Therefore $|F(Z)|\ll \prod_{j=1}^{r(Z)}\lambda_j^{-k}$.
