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\maketitle

\noindent {\bf Abstract.} We prove a spanning result for vector-valued Poincar\'e series on a
bounded symmetric domain.
We associate a sequence of  holomorphic automorphic forms to a submanifold of the domain.
When the domain is the unit ball in $\C ^n$,
we provide estimates for the norms of these automorphic forms and we find
asymptotics of the norms  (as the weight goes to infinity)
for a class of totally real submanifolds. We give an example of a CR submanifold
of the ball, for which the norms of the associated automorphic forms have
a different asymptotic behaviour.

\

\noindent {\bf MSC 2010:} 32N15, 53C99

\

\noindent {\bf Keywords:} holomorphic automorphic forms, Poincar\'e series, spanning set, domain,
canonical bundle, Bergman kernel, complex hyperbolic space, submanifold, asymptotics.

\section{Introduction}

Let $D$ be a bounded symmetric domain (in $\C^n$, for $n\ge 1$).
Suppose $\Gamma$ is a cocompact discrete
subgroup of $Aut(D)$. Let $k$ be a positive integer.
A holomorphic function $f:D\to\C$ is called a {\it holomorphic
automorphic form of weight $k$ for} $\Gamma$ if $f(\gamma z)J(\gamma , z)^k=f(z)$ for all $\gamma\in \Gamma$, $z\in D$.
Here $J(\gamma ,z)$ denotes the determinant of the Jacobi matrix of $\gamma$ at $z$.
Let $m$ be a positive integer and let $\rho:\Gamma\to GL(m,\C )$ be a unitary representation of $\Gamma$.
A holomorphic function $F:D\to\C^m$ is called a {\it holomorphic $\C ^m$-valued
automorphic form of weight $k$} (for the pair ($\Gamma$, $\rho$))
if $F(\gamma z)=\rho(\gamma)F(z)J(\gamma , z)^{-k}$ for all $\gamma\in \Gamma$, $z\in D$.
Holomorphic automorphic forms correspond to holomorphic sections of ${\mathcal{L}}^{\otimes k}$, where ${\mathcal{L}}$
is the canonical bundle on
$M=\Gamma\backslash D$, and $\C ^m$-valued holomorphic automorphic forms correspond
to holomorphic sections of $E_{\rho}\otimes {\mathcal{L}}^{\otimes k}$,
where $E_{\rho}\to M$ is the flat vector bundle defined by $\rho$.

The theory of automorphic forms is a vast subject that has strong interaction with
many areas of mathematics, including
representation theory, number theory, semiclassical analysis and quantization.
One connection
between automorphic forms and quantization is as follows: $M$ is a K\"ahler manifold,
${\mathcal{L}}$ is a quantum line bundle, $\frac{1}{k}$ is interpreted as $\hbar$ (the Planck constant),
and the space of holomorphic automorphic forms, with the Petersson inner product, is isomorphic to the Hilbert space
$H^0(M,{\mathcal{L}}^{\otimes k})$
used in quantization \cite{barron:12}. Berezin-Toeplitz quantization or K\"ahler quantization is usually studied for $\C$-valued observables.
Extending the theory to $\C^m$-valued functions on $D$ is a non-trivial task which is interesting
from the mathematical
point of view and physically meaningful (see e.g. work by S.T. Ali and M. Englis \cite{ali:07} on domains
in $\C^n$).

There are many different kinds of automorphic/modular forms, and generalizations. Vector-valued automorphic forms are
ubiquitous and go back to classical works of Borel, Selberg and others (see, for example \cite{borel:65}).
Applications include work of R. Borcherds on singular Howe correspondence,
work of S. Kudla on arithmetic cycles, physics-related work by T. Gannon, G. Mason and others. The space of Jacobi
forms is isomorphic to a space of vector-valued modular forms. Various kinds of
vector-valued forms for $G=SU(n,1)$ (i.e. when $D$ is the $n$-dimensional complex hyperbolic space, or, equivalently, the open unit ball in $\C^n$ with the complex
hyperbolic metric: $D=\B^n\simeq SU(n,1)/S(U(n)\times U(1))$)
have been studied, in particular, in recent papers by E. Freitag, G. van der Geer and others \cite{clery:13}, \cite{freitag:14},
in work by Kato including \cite{kato:84}, and in work by Kojima including \cite{kojima:97}.
It is well known that modular forms appear in generating functions for arithmetic or algebraic
objects in Calabi-Yau varieties.
It would be interesting to see if Picard modular forms could play a similar role.

Poincar\'e series is a standard and powerful
tool that is used in automorphic forms, spectral theory, complex analysis,
Teichm\"uller theory, algebraic geometry, and other areas.
In \cite{foth:07, foth:08} T. Barron (Foth) studied automorphic forms
for compact smooth $M=\Gamma \backslash D$  (in \cite{foth:08} $D=\B^n$),
constructing explicitly the automorphic form $f_p$ ($p\in D$) with the property $(g,f_p)=g(p)$
for any other holomorphic automorphic form $g$. Here $(.,.)$ denotes the Petersson inner product.
Such $f_p$ is constructed via Poincar\'e series and is related to the (weighted) Bergman kernel
and to the concept of a coherent state. Choosing a $q$-form on a $q$-dimensional submanifold, and
integrating $f_p$,
one can get automorphic forms associated to submanifolds of $D$. In this paper we extend this
framework to vector-valued holomorphic automorphic forms.
In Section \ref{secspan}
we prove that sufficiently many of these vector-valued Poincar\'e series
span the space of vector-valued
holomorphic automorphic forms  - Theorem \ref{thspan}.

There are somewhat different, but closely related results in literature:
it is known that $\C$-valued Poincar\'e series of polynomials in $z_1$,...,$z_n$
span the space of holomorphic automorphic forms
on a bounded symmetric domain (for sufficiently large weights)
\cite{bell:67, foth:07, wu:15}.
In David Bell's thesis \cite{bell:67} it is stated that similar results also hold
for vector-valued automorphic forms on classical domains and it is explained how the proofs for $\C$-valued case
can be extended to the vector-valued case.

To give general context to Section \ref{secsubman}, we observe that
associating an automorphic form or, more generally, a section of a vector bundle,
to a submanifold
of a K\"ahler manifold is an idea that is used in many contexts.
In particular, relative Poincar\'e series
can be associated to closed geodesics on a hyperbolic Riemann surface
\cite{katok:85, katok:87}. S. Katok and
T. Foth (Barron)  generalized this construction
from compact Riemann surfaces of genus $g\ge 1$
to ball quotients in \cite{foth:01, foth:04} (where they addressed the spanning problem),
and more recently
T.B. addressed the non-vanishing question in \cite{barron:18}.
In \cite{kudla:79, tong:83} the submanifold is a closed geodesic or, more generally, a totally
geodesic submanifold. To mention a somewhat different kind of such technique,
there is a way to associate a section of a line bundle
to a Bohr-Sommerfeld Lagrangian submanifold,
which is used in semiclassical analysis and symplectic geometry (see, in particular,
\cite{borthwick:95, burns:10, deber:06, gorod:01, jeffrey:92,
paoletti:08}). In
\cite{guillemin:15, ioos:18} sections of vector bundles are associated to  isotropic submanifolds.

Here, in Section \ref{secsubman},
we take advantage of the fact that the K\"ahler manifold is $M=\Gamma\backslash D$,
the holomorphic sections of the vector bundle on $M$ correspond to holomorphic vector-valued
functions on $D$,
and we associate an automorphic form to a submanifold of a fundamental domain of $\Gamma$ in $D$,
and not to a submanifold of $M$ (see Remark \ref{bohrs}).
In the case when the domain is the unit ball in $\C^n$, we provide asymptotic
(as the weight goes to infinity)
statements about the inner products  - Theorems \ref{thxy}, \ref{thxx}.
In particular,
in Theorem \ref{thxy}{\it (ii)}
we show that if two submanifolds are at a positive distance from each other, then
the inner product of the associated automorphic forms decreases rapidly as the weight goes
to infinity. In Theorem \ref{thxx}{\it (ii)} we show, in particular,
that for a class of totally real submanifolds,  as $k\to\infty$,
the square of the norm of $\Theta_X^{(j;k)}$
grows as a positive constant
times $k^{n-\frac{q}{2}}$, where
$X$ is such a submanifold,  $\Theta_X^{(j;k)}$ is one of the $m$ $\C^m$-valued Poincar\'e series
associated to $X$ ($j\in \{ 1,2,...,m\}$), and
$q$ is the real dimension of $X$.
We work out several examples. In Example \ref{3ball} we estimate the asymptotics
of the norms for a $3$-dimensional submanifold which is CR and not totally real, and we show
that the leading term in the asymptotics the square of the norm of the associated Poincar\'e
series is \underline{not}
$const \cdot k^{n-\frac{3}{2}}$.

This paper contains results from the Ph.D. thesis of N.A. \cite{alluhaibi:17}
written under the supervision of T.B.

{\bf Acknowledgments.} We are thankful to A. Dhillon, Y. Karshon,  M. Pinsonnault, E. Schippers,
A. Uribe and N. Yui for related discussions. We acknowledge the referee's efforts.

\section{Preliminaries}
\label{secprelim}

Let $D=G/K\subset \C^n$, for $n\ge 1$, be a bounded symmetric domain
($G=Aut(D)$ is a real semisimple
Lie group that acts transitively on $D$, $K$ a maximal compact subgroup of $G$).
Denote by $z_1$,...,$z_n$ the complex coordinates. Also denote
$z_j=x_j+iy_j$ ($x_j,y_j\in\R$, $1\le j\le n$) and
denote the Euclidean volume form by
$$
dV_e=dx_1\wedge dy_1\wedge ... \wedge dx_n\wedge dy_n=
\Bigl (\frac{i}{2}\Bigr ) ^n dz_1\wedge d\bar{z}_1\wedge ... \wedge dz_n\wedge d\bar{z}_n.
$$
Let $\K (.,.)$ be the Bergman kernel for $D$. It has the reproducing property:
$$
f(z)=\int\limits_D f(w) \K (z,w)dV_e (w),
$$
$z\in D$, for all functions $f$ that are holomorphic on $D$ and such that $\int\limits_D |f(z)|^2dV_e(z)<\infty$. Also
$\K (z,w)=\overline{\K (w,z)}$ for $z,w\in D$, and
\begin{equation}
\label{bktransf}
J(\gamma,z) \overline{J(\gamma,w)}\K (\gamma z,\gamma w)=\K (z,w)
\end{equation}
for $z,w\in D$, $\gamma \in G$,
where $J(\gamma,z)$ is the complex Jacobian of the transformation $D\to D$ at $z$ defined by $\gamma$.
The $(1,1)$-form
 $\omega =i\partial\bar{\partial}\log K(z,z)$ is a $G$-invariant K\"ahler form on $D$.

Let $k\in\N$ be a positive integer. It will be usually assumed that $k$ is sufficiently large.
The volume form $dV(z)=\K (z,z) dV_e(z)$ is $G$-invariant.
The reproducing kernel for the Hilbert space
of holomorphic functions on $D$ satisfying $\int\limits_D |f(z)|^2\K (z,z)^{-k}dV(z)<\infty$
is $c(D,k)\K (z,w)^k$, where $c(D,k)$ is a constant (note: the proof of this fact in
\cite{selberg:57} uses the assumption
that $Aut(D)$ acts transitively on $D$).
The reproducing property is: for any such function $f$
\begin{equation}
  \label{reprod}
f(z)=c(D,k)\int\limits_D f(w) \K (z,w)^k\K (w,w)^{-k}dV (w),
\end{equation}
$z\in D$.
The value of the constant is determined by
(1.6)\cite{selberg:57}:
\begin{equation}
\label{constcdk}
c(D,k)\int\limits_D\frac{\K (z,w)^k \K (w,z)^k}{\K (w,w)^k}dV(w)=\K (z,z)^k
\end{equation}
for any $z\in D$.

Let $\Gamma$ be a discrete subgroup of $G$ such that the quotient
$M=\Gamma \backslash D=\Gamma\backslash G/K$ is smooth and compact.
Let $m$ be a positive integer.
Let $\rho : \Gamma\to GL(m,\C )$ be a unitary representation of $\Gamma$.
\begin{defn} \cite{baily:73}
A function $f:D\to\C$ is called a {\it (holomorphic) $\Gamma$-automorphic
form of weight $k$} if $f$ is holomorphic and
\begin{equation}
  \label{autfs}
f(\gamma z)J(\gamma ,z)^k=f(z) \ \forall
\gamma  \in \Gamma, z\in D .
\end{equation}
\end{defn}
\begin{defn} \cite{selberg:57}
\label{vvafdef}
A {\it vector-valued automorphic form of weight $k$ for $(\rho, \Gamma )$}
is $F=\begin{pmatrix} F_1 \\ ... \\ F_m \end{pmatrix}$, where $F_j: D\to\C$, $j=1,...,m$, are holomorphic functions,
and
\begin{equation}
\label{autfvv}
J(\gamma,z)^kF(\gamma z)=\rho(\gamma)F(z) \  \forall
\gamma   \in \Gamma, z\in D .
\end{equation}
\end{defn}
Denote the space of holomorphic $\Gamma$-automorphic forms of weight $k$ on $D$
by $\A(\Gamma,k)$. Denote the space of $\C ^m$-valued
holomorphic $(\rho,\Gamma)$-automorphic forms of weight $k$ on $D$
by $\A(\Gamma,m,k,\rho)$.
\begin{remark}
In a more general case when $M$ is of finite volume and not compact the definitions should include an appropriate
condition at the cusps.
The condition "$M$ is smooth" can be relaxed to allow $\Gamma$ such as, for example,
$SL(2,\Z)\subset SL(2,\R)\simeq SU(1,1)$
or $SU(2,1)\cap SL(3,\Z [i])$.
\end{remark}

Define the inner product on the space $\A(\Gamma,m,k,\rho)$ as follows:
\begin{equation}
  \label{inprod}
( F,G) = \int\limits_{\Gamma \backslash D} F(z)^T\overline{ G(z)} \K (z,z)^{-k} dV
\end{equation}
for $F,G\in \A(\Gamma,m,k,\rho)$.
This is well-defined because the function $F(z)^T\overline{ G(z)} \K (z,z)^{-k}$ is $\Gamma$-invariant
(note: for that it is essential that $\rho$ is unitary).

Define the inner product on the space $\A(\Gamma,k)$ by
$$
( f,g) = \int\limits_{\Gamma \backslash D} f(z)\overline{ g(z)} \K (z,z)^{-k} dV
$$
for $f,g\in \A(\Gamma,k)$.

Denote by $K_M$ the canonical bundle on $M$ and by $K_D$ the canonical bundle on $D$.
\begin{remark} We have isomorphisms of Hilbert spaces:
  $\A(\Gamma,k)\cong H^0(M,K_M^{\otimes k})$,
  $\A(\Gamma,m,k,\rho)\cong H^0(M,E_{\rho}\otimes K_M^{\otimes k})$.
  In particular, a holomorphic function $f$ on $D$ satisfies (\ref{autfs}) if and only if
  $f(z)(dz_1\wedge ...\wedge dz_n)^{\otimes k}$ is a $\Gamma$-invariant holomorphic section of
  $K_D^{\otimes k}$ (and thus descends to a holomorphic section of $K_M^{\otimes k}$).
\end{remark}
\begin{remark}
  There are irreducible unitary representations of the fundamental group
  of a compact Riemann surface of genus $\ge 2$ for each $m\in\N$ (Proposition 2.1
  \cite{narasimhan:64}). The proof in \cite{narasimhan:64} provides explicit examples of such
  representations.
\end{remark}

\section{Poincar\'e series and a spanning result}
\label{secspan}

Let $D$ be a bounded symmetric domain, and let $\Gamma$ be a discrete subgroup of $Aut(D)$
such that the quotient $M=\Gamma\backslash D$ is smooth and compact.
Let $k$ and $m$ be positive integers, and let $\rho$ be an $m$-dimensional unitary representation of $\Gamma$.
This is the setting for this section.

For an integrable holomorphic function $F:D\to\C^m$ we define, formally, the Poincar\'e series
of weight $k$
$$
\Theta_F(z)=\sum_{\gamma\in\Gamma} \rho(\gamma ^{-1})F(\gamma z)J(\gamma , z)^k
$$
(here we omit $k$ from notation and write simply $\Theta_F$).
If the series converges uniformly on compact sets in $D$,
then $\Theta_F\in \A(\Gamma,m,k,\rho)$. Indeed,
since the convergence is uniform on compact sets,
it follows that $\Theta_F$ is holomorphic. To verify (\ref{autfvv}), we observe:
for $g\in \Gamma$, $z\in D$
$$
\Theta_F(g z)=\sum_{\gamma\in\Gamma} \rho(\gamma ^{-1})F(\gamma gz)J(\gamma , gz)^k=
\sum_{\gamma\in\Gamma} \rho(g(\gamma g) ^{-1})F(\gamma gz)\frac{J(\gamma g, z)^k}{J(g, z)^k}=
\rho(g) J(g, z)^{-k}\Theta_F(z).
$$
Choose $p\in D$.
In \cite{foth:07} the $\C$-valued Poincar\'e series
$$
\theta_p(z)=\sum_{\gamma\in\Gamma}\Bigl ( \K (\gamma z,p)J(\gamma ,z)\Bigr ) ^k \in \A(\Gamma,k)
$$
(convergent absolutely and uniformly on compact sets for sufficiently large $k$, and
having  the property
$(f,\theta_p)={\mbox{const}}(D,k)f(p)$
for any $f\in \A(\Gamma,k)$) were considered, and it was shown
that such Poincar\'e series for an appropriate number of points in general position
form a basis in $\A(\Gamma,k)$. Note that the property $(f,\theta_p)={\mbox{const}}(D,k)f(p)$ reflects
the fact that the Bergman kernel for $K_M^{\otimes k}$ is the Poincar\'e series of the Bergman kernel for
$K_D^{\otimes k}$ (Theorem 2 \cite{ma:15} or Theorem 1 \cite{lu:15}).

Let us now generalize the construction from  \cite{foth:07}
by associating to a point $p\in D$ $m$ vector-valued Poincar\'e series
\begin{equation}
\label{vvps}
\Theta_p^{(j;k)}(z)=c(D,k) \sum_{\gamma\in\Gamma}\rho(\gamma^{-1}) \T _p (\gamma z)  J(\gamma,z)^k, \ j=1,...,m
\end{equation}
where $\T _p (z)=\begin{pmatrix} (\T_p) _1 (z) \\ ... \\ (\T _p) _m (z) \end{pmatrix}$,
$(\T _p)_j(z)=\K ( z,p)^k$ and $(\T _p)_l (z)=0$ for $l\ne j$
(i.e. $\T _p (\gamma z)$ is the vector-function whose components, except for the $j$-th one, are zero,
and $(\T _p)_j(\gamma z)=  \K ( \gamma z,p)^k$).

We shall also use the notation $\hat{\Theta}^{(j;k)}(z,p)$ for the function
$$
\hat{\Theta}^{(j;k)}:D\times D\to \C^m
$$
$$
(z,p)\to \Theta_p^{(j;k)}(z).
$$
\begin{lemma}
\label{propconvscalar}
Let $p\in D$.
For $k\ge 2$
the series $\sum\limits_{\gamma\in\Gamma}(\K (\gamma z,p)J(\gamma,z))^k$ converges absolutely and
uniformly on compact sets of $D$.
\end{lemma}
The proof is in the Appendix.
\begin{proposition}
  \label{propconv}
  Let $j\in \{ 1,...,m\}$ and let $p\in D$.
Suppose $k$ is  sufficiently large.

\noindent (i) The series (\ref{vvps}) converges absolutely and uniformly on compact sets.

\noindent (ii) For each $H\in \A(\Gamma,m,k,\rho)$
$$
( H,\Theta_p^{(j;k)})=H_j(p).
$$
\end{proposition}
\begin{theorem}
\label{thspan}
For sufficiently large $k$,
for sufficiently many points $p_1,...,p_d$ in general position, the $\C$-linear span of
$\{ \Theta_{p_l}^{(j;k)} | 1\le l\le d; \  1\le j\le m\}$ is
$\A(\Gamma,m,k,\rho)$.
\end{theorem}
\noindent {\bf Proof of Proposition \ref{propconv}.} Proof of {\it (i)}.
For $1\le l\le m$
$$
\Bigl | \Bigl ( \rho(\gamma^{-1}) \T _p (\gamma z) J(\gamma,z)^k
\Bigr )_l \Bigr |\le
\sqrt{\Bigl (\rho(\gamma^{-1}) \T _p (\gamma z) J(\gamma,z)^k
\Bigr )^T \overline{\rho(\gamma^{-1}) \T _p (\gamma z) J(\gamma,z)^k} }=
$$
$$
|\K (\gamma z,p)J(\gamma,z)|^k.
$$
The statement now follows from Lemma \ref{propconvscalar}.

Proof of {\it (ii)}. Let $\F$ be a Dirichlet fundamental domain for $\Gamma$
(or a canonical fundamental domain \cite{selberg:68}).
Denote $w=\gamma z$ for $\gamma\in\Gamma$, $z\in \F$. By (\ref{autfvv})
 $H(z)^T=H(w)^T(\rho (\gamma) ^{-1})^T J(\gamma,z)^k$.
Using (\ref{bktransf}), (\ref{reprod}), (\ref{inprod}), (\ref{vvps}), we get:
$$
( H,\Theta_p^{(j;k)}) =c(D,k) \int\limits_{\F} \begin{pmatrix} H_1(z) & ... & H_m(z) \end{pmatrix}
\sum_{\gamma\in\Gamma}\overline{\rho(\gamma^{-1})} \overline{\T _p(\gamma z) } \overline{J(\gamma,z)^k}
\K (z,z)^{-k}dV(z)=
$$
$$
c(D,k)\sum_{\gamma\in\Gamma}\int\limits_{\gamma \F} H(w)^T \overline{\T _p(w)}   K(w,w)^{-k}dV(w)=
$$
$$
c(D,k)\sum_{\gamma\in\Gamma}\int\limits_{\gamma \F} H_j(w)\overline{\K (w,p)}^k K(w,w)^{-k}dV(w)=
$$
$$
c(D,k)\int\limits_{D} H_j(w) \K (p,w)^k K(w,w)^{-k}dV(w)=
H_j(p).
$$
$\Box$

\noindent {\bf Proof of Theorem \ref{thspan}}.
Let $k\in\N$.
The holomorphic vector bundle $W=E_{\rho}\otimes K_M^{\otimes k}$ is positive.
Using the notations similar to those in Chapters 2, 3 \cite{kobayashi:87},
denote by $P(W)$ the fibre bundle over $M$ whose fibre at $x$ is $\Pr (W_x)$
(i.e. $P(W)=(W-\{ \text{zero section} \})/\C^*$), denote by $\pi: P(W)\to M$ the projection,
and by $L(W)$ the tautological line bundle over $P(W)$
(i.e. the subbundle of $\pi^*W$ with the fiber $L(W)_\xi$
at $\xi\in P(W)$ being the complex line in $W_{\pi(\xi)}$ represented by $\xi$).
Also denote by $L(W^*)$ the tautological line bundle over $P(W^*)=(W^*-\{ \text{zero section} \})/\C^*$
and by $\hat{\pi}: P(W^*)\to M$ the projection.
We note that a section $s$ of $W$  produces a section $\tilde{s}$ of $(L(W^*))^*\to P(W^*)$.
Specifically, $\tilde{s}=h\circ s\circ \hat{\pi}$,
where $h$ is the holomorphic surjection $\hat{\pi}^*W\to (L(W^*))^*\simeq L(W)$ given, fiberwise,
by the quotient map $W_x\to W_x/ \ker f$ over $(x,[f])\in P(W^*)$, where $x\in M$, $f\in W_x^*$, $f\ne 0$.

Suppose $k$ is large enough, so that the line bundle $(L(W^*))^*\to P(W^*)$ is very ample.
Let $d=\dim H^0(P(W^*),(L(W^*))^*)$ and let $\tilde{p}_1$,...,$\tilde{p}_d$ be points
in $P(W^*)$ in general position (i.e. such that their images under the projective embedding given
by  $(L(W^*))^*$ are not on the same hyperplane in $\Pr(H^0(P(W^*),(L(W^*))^*))^*$).
Such $d$ points exist because the linear system is base point free.
Select a Dirichlet fundamental domain $\F$ for $\Gamma$ and for each $j\in \{ 1,...,d\}$ let
$p_j$ be the point in $\F$ that corresponds to $\hat{\pi}(\tilde{p}_j)$.

Now, to prove the statement of the theorem,
suppose $H\in \A(\Gamma,m,k,\rho)$ is not in the linear span of
$\{ \Theta_{p_l}^{(j;k)}\} _{\substack{1\le l\le d \\ 1\le j\le m}}$.
Then $H$ is in the orthogonal complement of this subspace and therefore
$$
( H,\Theta_{p_l}^{(j;k)})=0
$$
for all $l\in \{ 1,...,d\}$ and all $j\in \{ 1,...,m\}$.
By Proposition \ref{propconv}{\it (ii)} $H(p_1)=...=H(p_d)=0$.
Let $s$ be the section of $W$ corresponding to $H$.
This section vanishes at $p_1$,...,$p_d$.
Therefore $\tilde{s}(\tilde{p_1})=...=\tilde{s}(\tilde{p_d})=0$. Since $\tilde{p_1}$,...,$\tilde{p_d}$
are in general position, we conclude that $\tilde{s}\equiv 0$. It follows that
 $s=0$. Hence $H=0$.
$\Box$

\section{Automorphic forms and submanifolds}
\label{secsubman}

Let $D$ be a bounded symmetric domain, and let $\Gamma$ be a discrete subgroup of $Aut(D)$
such that the quotient $M=\Gamma\backslash D$ is smooth and compact.
Let $k$ and $m$ be positive integers, and let $\rho$ be an $m$-dimensional unitary representation of $\Gamma$.

\noindent Let $\Lambda$ be a  $q$-dimensional submanifold of $D$ ($q\ge 1$) such that
$\Lambda \subset \bar{B}(z_0,r_0) \subset D$, where $\bar{B}(z_0,r_0)$ is the closed ball centered at $z_0$
of radius $r_0$ with respect to the Euclidean metric,
for some $z_0\in D$, $r_0>0$.
Let $\nu$ be a nonzero volume form (a real $q$-form)  on $\Lambda$
such that $\int\limits_{\Lambda}\nu >0$.
Set
\begin{equation}
\label{setup1}
\Theta_{\Lambda}^{(j;k)}(z)=\int\limits_{\Lambda} \hat{\Theta}^{(j;k)}(z,p)\K (p,p)^{-\frac{k}{2}}\nu(p)
\end{equation}
for $j\in \{1,...,m\}$.
By a standard differentiation under the integral sign argument $\Theta_{\Lambda}^{(j;k)}$ is holomorphic.
Moreover,
$\Theta_{\Lambda}^{(j;k)}\in \A(\Gamma,m,k,\rho)$ and
\begin{equation}
\label{setup2}
(H,\Theta_{\Lambda}^{(j;k)})=\int\limits_{\Lambda} H_j(z)\K (z,z)^{-\frac{k}{2}}\nu(z)
\end{equation}
for any $H\in \A(\Gamma,m,k,\rho)$.
\begin{remark}
  The statement analogous to
  Proposition \ref{propconv}{\it (ii)}, but  written for
  the corresponding sections of $E_{\rho}\otimes K_M^{\otimes k}$,
  would mean that the section of  $E_{\rho}\otimes K_M^{\otimes k}$, corresponding  to
  $\Theta_{p}^{(j;k)}$,
  is  the $j$-th row of the Bergman kernel for this vector bundle,
  where the Bergman kernel is written as an $m\times m$  matrix.
  The general idea of ``integrating the Bergman kernel over a submanifold $\Lambda$''
  was used in \cite{borthwick:95} (to obtain sections of powers of a line bundle,
  with $\Lambda$ being a Bohr-Sommerfeld Lagrangian submanifold of a compact K\"ahler
  manifold) and it is used in a recent preprint \cite{ioos:18}
  (to obtain sections of certain vector bundles, with $\Lambda$ being
  an isotropic  Bohr-Sommerfeld submanifold of a compact symplectic manifold).
\end{remark}
In this section,
the domain $D$ will be the unit ball $\B^n\subset \C^n$ ($n\ge 1$), with its Bergman metric.
 Recall that
$SU(n,1)=\{ A\in SL(n+1,\C ) \ | \
A^T\sigma \bar{A}=\sigma \}$, where
$\sigma =\begin{pmatrix}
1_{n\times n} & 0 \cr 0 & -1\end{pmatrix}$. The ball
is a bounded realization of the Hermitian symmetric space $SU(n,1)/S(U(n)\times U(1))$
(note that for $n=1$ $D$ is the unit disc:
$D\cong SU(1,1)/U(1)\cong SL(2,\R)/SO(2)$). The group $SU(n,1)$ acts on
$\B^n$ is by fractional-linear transformations:
for $\gamma =(a_{jk})\in SU(n,1)$ the corresponding automorphism
$\B^n\to \B^n$ is

\noindent
$z=(z_1,...,z_n)\mapsto \Bigl ( \dfrac{a_{11}z_1+...+a_{1n}z_n+a_{1,n+1}}{a_{n+1,1}z_1+...
+a_{n+1,n}z_n+a_{n+1,n+1}} , ...,
\dfrac{a_{n1}z_1+...+a_{nn}z_n+a_{n,n+1}}{a_{n+1,1}z_1+...
+a_{n+1,n}z_n+a_{n+1,n+1}} \Bigr )$.

The complex Jacobian is
$J(\gamma,z)=1/(a_{n+1,1}z_1+...+a_{n+1,n}z_n+a_{n+1,n+1})^{n+1}$.
\begin{remark}
  Each element of the center of $SU(n,1)$ acts as the identity map on $\B^n$, and $Aut(\B^n)$ is
  isomorphic to $PU(n,1)$. We will represent automorphisms of the ball by matrices from $SU(n,1)$,
  and we will use the same letter to denote the matrix and the corresponding automorphism.
\end{remark}
We will denote by $0$ the point
$(0,...,0)\in \B^n$.
Also, for $z,w\in \B^n$ denote
$$
\langle z,w\rangle=z_1\bar{w}_1+...+z_n\bar{w}_n-1.
$$
The $SU(n,1)$-invariant K\"ahler form on $\B^n$ is, up to a positive constant factor,
$$
i\partial\bar{\partial}\log (-\langle z,z\rangle)=
\frac{i}{\langle z,z\rangle ^2}\Bigl [
  (\sum_{j=1}^n\bar{z}_jdz_j)\wedge (\sum_{l=1}^nz_ld\bar{z}_l)-\langle z,z\rangle
  \sum_{r=1}^ndz_r\wedge d\bar{z}_r\Bigr ].
$$
Denote by $\tau (z,w)$ the distance between $z$ and $w$ with respect
to the complex hyperbolic metric. Note that
\begin{equation}
  \label{coshdistf}
\cosh ^2\frac{\tau(z,w)}{2}=\frac{\langle z,w\rangle \langle w,z\rangle}{\langle z,z\rangle \langle w,w\rangle}
\end{equation}
(see e.g. \cite{goldman:99} 3.1.7).
It is a standard fact (see e.g. \cite{rudin:80} or \cite{range:86}) that for the ball
\begin{equation}
  \label{bergkball}
\K (z,w)=\frac{n!}{\pi^n}(-\langle z,w\rangle)^{-(n+1)}.
\end{equation}
\begin{lemma}
\label{lemconstcdk}
For $D=\B^n$ the constant $c(D,k)$ from Section \ref{secprelim} is $c(\B^n ,k)=\binom{(n+1)(k-1)+n}{n}$.
\end{lemma}
This follows from Theorem 2.2 \cite{zhu:05} with $\alpha=(n+1)(k-1)$ (the constant $c(D,k)$ comes out to be
$c_{\alpha}$ given by (2.2)\cite{zhu:05}). This also can be verified in another way, by a direct calculation (see
the Appendix).
\begin{remark}
\label{remconstball}
Applying the Stirling formula $N!\sim (\frac{N}{e})^N\sqrt{2\pi N}\Bigl (1+O(\frac{1}{N})\Bigr )$ as $N\to\infty$ \cite{debru:61},
we get:  $c(\B^n ,k)\sim \frac{(n+1)^n}{n!}k^n\Bigl (1+O(\frac{1}{k})\Bigr )$ as $k\to\infty$.
\end{remark}
Let $\Gamma$ be a discrete subgroup of $SU(n,1)$ such that the quotient
$M=\Gamma \backslash \B^n$ is smooth and compact.
Let $k$, $m$ be positive integers, and let $\rho$ be an $m$-dimensional unitary representation of $\Gamma$.
Denote by $\pi:\B^n\to M$ the covering map. Let $\F$ be a Dirichlet fundamental domain for $\Gamma$ \cite{parker:09}.
Suppose $X$ and $Y$ are submanifolds of $\B^n$ of dimensions $q_X>0$
and $q_Y>0$ respectively, such that $X=\pi^{-1}(X')\cap \F$, $X\cong X'$, and
$Y=\pi^{-1}(Y')\cap \F$, $Y\cong Y'$, where $X'$ and $Y'$ are submanifolds of $M$, and
$\cong$ stands for diffeomorphism.
Let $\nu_X$ be a nonzero volume form on $X$ (a real $q_X$-form) such that $\int_X \nu_X>0$
and let $\nu_Y$ is a nonzero volume form on $Y$ (a real $q_Y$-form) such that $\int_Y \nu_Y>0$.
Denote $\tilde{X}=\Gamma X$, $\tilde{Y}=\Gamma Y$.
Define the $q_X$-form $\nu_{\tilde{X}}$ on ${\tilde{X}}$ by
$\nu_{\tilde{X}}\Bigr |_{\gamma^{-1}(X)}=\gamma^*\nu_X$ for each $\gamma\in\Gamma$.
Define $\nu_{\tilde{Y}}$ the same way. Note that $\nu_{\tilde{X}}$, $\nu_{\tilde{Y}}$
are $\Gamma$-invariant.
Assume
$\int\limits_{\tilde{X}}|K(z,w)|^2\frac{\nu_{\tilde{X}}(w)}{K(w,w)}<\infty$ for all $z\in \F$,
$\int\limits_{\tilde{Y}}|K(z,w)|^2\frac{\nu_{\tilde{Y}}(w)}{K(w,w)}<\infty$ for all $z\in \F$
(the last condition is satisfied, for example, when $Y$ is a small ball and $\nu_{\tilde{Y}}=dV\Bigr | _{\tilde{Y}}$, because
 $K(.,w)$ is square-integrable on $\B^n$).

For two subsets $A$, $B$ of $\B^n$ we will denote
$$
{\mathrm{dist}}(A,B)=\inf \{ \tau(z,w) \ | z\in A, w\in B\}.
$$
Since $\tau$ is $\Gamma$-invariant, the same notation can be used for subsets $A$, $B$ of $M$.
\begin{remark}
Recall that if $(a_k)$, $(b_k)$ are two sequences of complex numbers, then notation $a_k\sim b_k$ as $k\to\infty$ means
$\lim\limits_{k\to\infty}\frac{a_k}{b_k}=1$.
\end{remark}
\begin{theorem}
\label{thxy} Let $r,j\in \{ 1,...,m\}$.

\noindent (i) Suppose ${\mathrm{dist}}(\tilde{X}-X,Y)>0$  and $r\ne j$.  Then
for any $l\in\N$ there is a constant $C=C(l;n,X,Y,\Gamma , \nu_X, \nu_Y)$
such that, as $k\to\infty$
$$
|(\Theta_X^{(r;k)},\Theta_Y^{(j;k)})|\le \frac{C}{k^l}.
$$
(ii) Suppose ${\mathrm{dist}}(\tilde{X},Y)>0$.
Then
for any $l\in\N$ there is a constant $C=C(l;n,X,Y,\Gamma , \nu_X, \nu_Y)$
such that, as $k\to\infty$
$$
|(\Theta_X^{(r;k)},\Theta_Y^{(j;k)})|\le \frac{C}{k^l}.
$$
\end{theorem}
\begin{remark}
  \label{remdist}
  If ${\mathrm{dist}}(X,\partial \F)>0$ or ${\mathrm{dist}}(Y,\partial \F)>0$,
  then ${\mathrm{dist}}(\tilde{X}-X,Y)>0$.
\end{remark}
\begin{remark}
  If  ${\mathrm{dist}}(\tilde{X},Y)>0$, then ${\mathrm{dist}}(X',Y')>0$.
\end{remark}
\begin{theorem}
\label{thxx} Suppose $Y\subset X$, ${\mathrm{dist}} (X,\partial\F)>0$, and $j\in \{1,...,m\}$.

\noindent (i) If $q_X\le n$, then
$$
(\Theta_X^{(j;k)},\Theta_Y^{(j;k)})\le {\mbox{const}}(n,X,Y,\nu_X,\nu_Y)k^{n-\frac{q_X}{2}}
$$
as $k\to\infty$.

\noindent (ii) If $X\subset \{ z\in\B^n | y_1=...=y_n=0\}$, then
$$
(\Theta_X^{(j;k)},\Theta_Y^{(j;k)})\sim C(n,X,Y,\nu_X,\nu_Y)k^{n-\frac{q_X}{2}}
$$
as $k\to\infty$, where $C(n,X,Y,\nu_X,\nu_Y)$ is a positive constant.
\end{theorem}
\begin{remark}
  \label{bohrs}
  Because of the assumptions in Theorem \ref{thxx}{\it (ii)}, in this part
  of the theorem the submanifolds $X$ and $Y$ are isotropic submanifolds
  of $\B^n$. In Theorem \ref{thxx}{\it (i)} and in Theorem \ref{thxy}
  the submanifolds are not necessarily isotropic. Note that $X$ and $Y$ are submanifolds of $\B^n$, the
  universal cover of $M$, and not of $M$. We do not require $X$ and $Y$ to satisfy a Bohr-Sommerfeld
  condition. In the usual procedure of associating
  a section of a line bundle, $L$, to a Lagrangian or isotropic
  submanifold, $\Lambda$, of  a K\"ahler manifold, the Bohr-Sommerfeld condition ensures the existence
  of a covariant constant nonvanishing section, $\varphi$, of $L^*\Bigr |_{\Lambda}$.
  Having such $\varphi$, from a holomorphic section $s$ of $L^{\otimes k}$, one obtains
  a function on $\Lambda$, $\varphi^{\otimes k}(s)$, which then can be integrated over $\Lambda$. This
  provides a linear functional on the space of holomorphic sections of $L^{\otimes k}$.
  We do not need such $\varphi$, since in (\ref{setup2}) we are already integrating a function.
\end{remark}
\noindent {\bf Proof of Theorem \ref{thxy}.}
Using (\ref{vvps}), (\ref{setup1}), (\ref{setup2}), we get:
$$
|(\Theta_X^{(r;k)},\Theta_Y^{(j;k)})|=|\int\limits_{Y}(\Theta_X^{(r;k)}(z))_j \K ( z,z)^{-\frac{k}{2}}\nu_{Y}(z)|=
$$
$$
|\int\limits_{Y}\int\limits_{X}(\hat{\Theta}^{(r;k)}(z,\zeta))_j \K ( \zeta,\zeta)^{-\frac{k}{2}} \nu_{X}(\zeta)
\K ( z,z)^{-\frac{k}{2}} \nu_{Y}(z)|=
$$
$$
c(\B^n,k)|\int\limits_{Y}\int\limits_{X}\sum_{\gamma\in\Gamma}\rho(\gamma^{-1})_{jr}\K (\gamma z,\zeta)^kJ(\gamma,z)^k
\K ( \zeta ,\zeta)^{-\frac{k}{2}} \nu_{X}(\zeta)
\K (z,z)^{-\frac{k}{2}} \nu_{Y}(z) |\le
$$
$$
c(\B^n,k)\int\limits_{Y}\int\limits_{X}\sum_{\gamma\in\Gamma}|\rho(\gamma^{-1})_{jr}||\K (\gamma z,\zeta)J(\gamma,z)|^k
\K ( \zeta ,\zeta)^{-\frac{k}{2}} \nu_{X}(\zeta)
\K (z,z)^{-\frac{k}{2}} \nu_{Y}(z) .
$$
Setting $\zeta=\gamma w$, we get, using (\ref{bktransf}):
$$
|(\Theta_X^{(r;k)},\Theta_Y^{(j;k)})|\le c(\B^n,k)\int\limits_{Y}\sum_{\gamma\in\Gamma}
\int\limits_{\gamma ^{-1} X}
|\rho(\gamma^{-1})_{jr}||\K (z,w)|^k \K (w,w)^{-\frac{k}{2}} \nu_{\tilde{X}}(w)
\K (z,z)^{-\frac{k}{2}} \nu_{Y}(z).
$$
Since $\rho(\gamma^{-1})$ is a unitary matrix, we have: $|\rho(\gamma^{-1})_{jr}|\le 1$.
Using (\ref{coshdistf}), (\ref{bergkball}), we get:
$$
|(\Theta_X^{(r;k)},\Theta_Y^{(j;k)})|\le
c(\B^n,k)\int\limits_{Y}\sum_{\gamma\in\Gamma}
\int\limits_{\gamma ^{-1} X}
|\K (z,w)|^k \K (w,w)^{-\frac{k}{2}} \nu_{\tilde{X}}(w)
\K (z,z)^{-\frac{k}{2}} \nu_{Y}(z)=
$$
$$
c(\B^n,k)\int\limits_{Y}
\int\limits_{\tilde{X}}
|\K (z,w)|^k \K (w,w)^{-\frac{k}{2}} \nu_{\tilde{X}}(w)
\K (z,z)^{-\frac{k}{2}} \nu_{Y}(z)=
$$
$$
c(\B^n ,k)\int\limits_{Y}\int\limits_{\tilde{X}}
|\K (z,w)|^k \K (w,w)^{-\frac{k}{2}+1} \frac{\nu_{\tilde{X}}(w)}{K(w,w)}
\K (z,z)^{-\frac{k}{2}+1}  \frac{\nu_{Y}(z)}{K(z,z)}=
$$
$$
c(\B^n ,k)  \int\limits_{Y}\int\limits_{\tilde{X}}
\Bigl ( \frac{\langle z,z\rangle\langle w,w\rangle }{\langle z,w\rangle  \langle w,z\rangle}
\Bigr )^{ (n+1)(\frac{k}{2}-1)  }
|K(z,w)|^2    \frac{\nu_{\tilde{X}}(w)}{K(w,w)} \frac{\nu_{Y}(z)}{K(z,z)}=
$$
$$
c(\B^n ,k)  \int\limits_{Y}\int\limits_{\tilde{X}}
\Bigl ( \cosh \frac{\tau(z,w)}{2}\Bigr ) ^{-(n+1)(k-2)}
|K(z,w)|^2   \frac{\nu_{\tilde{X}}(w)}{K(w,w)} \frac{\nu_{Y}(z)}{K(z,z)}   \le
$$
$$
c(\B^n ,k)  \Bigl ( \frac{1}{\cosh [\frac{1}{2}{\mathrm{dist}}(\tilde{X},Y)] }\Bigr )^ {(n+1)(k-2)}
\int\limits_{Y}\int\limits_{\tilde{X}} |K(z,w)|^2   \frac{\nu_{\tilde{X}}(w)}{K(w,w)} \frac{\nu_{Y}(z)}{K(z,z)}  .
$$
For $r\ne j$, since $\rho({\mathrm{id}})_{jr}=0$, the argument above can be modified:
$$
|(\Theta_X^{(r;k)},\Theta_Y^{(j;k)})|\le c(\B^n,k)\int\limits_{Y}\sum_{\gamma\in\Gamma, \gamma\ne id}
\int\limits_{\gamma ^{-1} X}
|\K (z,w)|^k \K (w,w)^{-\frac{k}{2}} \nu_{\tilde{X}}(w)
\K (z,z)^{-\frac{k}{2}} \nu_{Y}(z)=
$$
$$
c(\B^n,k)\int\limits_{Y}
\int\limits_{\tilde{X}-X}
|\K (z,w)|^k \K (w,w)^{-\frac{k}{2}} \nu_{\tilde{X}}(w)
\K (z,z)^{-\frac{k}{2}} \nu_{Y}(z)=
$$
$$
c(\B^n ,k)\int\limits_{Y}\int\limits_{\tilde{X}-X}
|\K (z,w)|^k \K (w,w)^{-\frac{k}{2}+1} \frac{\nu_{\tilde{X}}(w)}{K(w,w)}
\K (z,z)^{-\frac{k}{2}+1}  \frac{\nu_{Y}(z)}{K(z,z)}=
$$
$$
c(\B^n ,k)  \int\limits_{Y}\int\limits_{\tilde{X}-X}
\Bigl ( \frac{\langle z,z\rangle\langle w,w\rangle }{\langle z,w\rangle  \langle w,z\rangle}
\Bigr )^{ (n+1)(\frac{k}{2}-1)  }
|K(z,w)|^2    \frac{\nu_{\tilde{X}}(w)}{K(w,w)} \frac{\nu_{Y}(z)}{K(z,z)}=
$$
$$
c(\B^n ,k)  \int\limits_{Y}\int\limits_{\tilde{X}-X}
\Bigl ( \cosh \frac{\tau(z,w)}{2}\Bigr ) ^{-(n+1)(k-2)}
|K(z,w)|^2   \frac{\nu_{\tilde{X}}(w)}{K(w,w)} \frac{\nu_{Y}(z)}{K(z,z)}   \le
$$
$$
c(\B^n ,k)  \Bigl ( \frac{1}{\cosh [\frac{1}{2}{\mathrm{dist}}(\tilde{X}-X,Y)] }\Bigr )^ {(n+1)(k-2)}
\int\limits_{Y}\int\limits_{\tilde{X}-X} |K(z,w)|^2   \frac{\nu_{\tilde{X}}(w)}{K(w,w)} \frac{\nu_{Y}(z)}{K(z,z)}  .
$$
Now,
$$
0<\int\limits_{Y}\int\limits_{\tilde{X}-X} |K(z,w)|^2 \frac{\nu_{\tilde{X}}(w)}{K(w,w)} \frac{\nu_{Y}(z)}{K(z,z)}\le
$$
$$
\int\limits_{Y}\int\limits_{\tilde{X}} |K(z,w)|^2 \frac{\nu_{\tilde{X}}(w)}{K(w,w)} \frac{\nu_{Y}(z)}{K(z,z)}  ={\mbox{const}} (\tilde{X},Y,\nu_{\tilde{X}},\nu_Y)
$$
Since $\cosh \frac{\varepsilon}{2}>1$ for $\varepsilon>0$, and with Remark \ref{remconstball},
the statements follow.  $\Box$

\noindent {\bf Proof of Theorem \ref{thxx}.}
Using (\ref{vvps}), (\ref{setup1}), (\ref{setup2}), we get:
$$
(\Theta_X^{(j;k)},\Theta_Y^{(j;k)})= \int\limits_{Y}(\Theta_X^{(j;k)}(z))_j \K ( z,z)^{-\frac{k}{2}}\nu_{Y}(z)=
$$
$$
\int\limits_{Y}\int\limits_{X}(\hat{\Theta}^{(j;k)}(z,\zeta))_j \K ( \zeta,\zeta)^{-\frac{k}{2}} \nu_{X}(\zeta)
\K ( z,z)^{-\frac{k}{2}} \nu_{Y}(z)=
$$
$$
c(\B^n ,k)\int\limits_{Y}\int\limits_{X}\sum_{\gamma\in\Gamma}\rho(\gamma^{-1})_{jj}(\K (\gamma z,\zeta)J(\gamma,z))^k
\K ( \zeta ,\zeta)^{-\frac{k}{2}} \nu_{X}(\zeta)
\K (z,z)^{-\frac{k}{2}} \nu_{Y}(z) =
$$
$$
I_1+I_2,
$$
where $I_1$ is the term with $\gamma={\mbox{id}}$ and $I_2$ is the rest. Thus,
$$
I_1=c(\B^n ,k)\int\limits_{Y}\int\limits_{X}\K (z,\zeta)^k
\K ( \zeta ,\zeta)^{-\frac{k}{2}} \nu_{X}(\zeta)
\K (z,z)^{-\frac{k}{2}} \nu_{Y}(z)
$$
and
$$
I_2=c(\B^n ,k)\int\limits_{Y}\int\limits_{X}
\sum_{\gamma\in\Gamma, \gamma\ne {\mbox{id}}}\rho(\gamma^{-1})_{jj}(\K (\gamma z,\zeta)J(\gamma,z))^k
\K ( \zeta ,\zeta)^{-\frac{k}{2}} \nu_{X}(\zeta)
\K (z,z)^{-\frac{k}{2}} \nu_{Y}(z).
$$
Since $\rho(\gamma^{-1})$ is a unitary matrix, $|\rho(\gamma^{-1})_{jj}|\le 1$.
Setting $\zeta=\gamma w$ and using (\ref{bktransf}), (\ref{coshdistf}), (\ref{bergkball}),
we get:
$$
|I_2|\le
c(\B^n ,k)\int\limits_{Y}\int\limits_{X}\sum_{\gamma\in\Gamma, \gamma\ne {\mbox{id}} }|\K (\gamma z,\zeta)J(\gamma,z)|^k
\K ( \zeta ,\zeta)^{-\frac{k}{2}} \nu_{X}(\zeta)
\K (z,z)^{-\frac{k}{2}} \nu_{Y}(z) =
$$
$$
 c(\B^n ,k)\int\limits_{Y}\sum_{\gamma\in\Gamma , \gamma\ne {\mbox{id}} } \ \int\limits_{\gamma ^{-1} X}
|\K (z,w)|^k \K (w,w)^{-\frac{k}{2}} \nu_{\tilde{X}}(w)
\K (z,z)^{-\frac{k}{2}} \nu_{Y}(z)=
$$
$$
c(\B^n ,k)\int\limits_{Y}\int\limits_{\tilde{X}-X}
|\K (z,w)|^k \K (w,w)^{-\frac{k}{2}} \nu_{\tilde{X}}(w)
\K (z,z)^{-\frac{k}{2}}  \nu_{Y}(z)=
$$
$$
c(\B^n ,k)
\int\limits_{Y}\int\limits_{\tilde{X}-X}
\Bigl ( \frac{\langle z,z\rangle\langle w,w\rangle }{\langle z,w\rangle  \langle w,z\rangle}
\Bigr )^{ (n+1)(\frac{k}{2}-1)  }|K(z,w)|^2  \frac{\nu_{\tilde{X}}(w)}{K(w,w)} \frac{\nu_{Y}(z)}{K(z,z)}=
$$
$$
c(\B^n ,k)
\int\limits_{Y}\int\limits_{\tilde{X}-X}
\Bigl ( \cosh \frac{\tau(z,w)}{2}\Bigr )  ^{ -(n+1)(k-2) }
|K(z,w)|^2  \frac{\nu_{\tilde{X}}(w)}{K(w,w)} \frac{\nu_{Y}(z)}{K(z,z)}\le
$$
$$
c(\B ^n,k)
\Bigl ( \cosh [\frac{1}{2}{\mathrm{dist}}(Y,\tilde{X}-X)]\Bigr )  ^{ -(n+1)(k-2) }
\int\limits_{Y}\int\limits_{\tilde{X}-X}
|K(z,w)|^2  \frac{\nu_{\tilde{X}}(w)}{K(w,w)} \frac{\nu_{Y}(z)}{K(z,z)}.
$$
Because of Remark \ref{remdist}, $\cosh [\frac{1}{2}{\mathrm{dist}}(Y,\tilde{X}-X)]>1$.
Also
$$
\int\limits_{Y}\int\limits_{\tilde{X}-X}
|K(z,w)|^2  \frac{\nu_{\tilde{X}}(w)}{K(w,w)} \frac{\nu_{Y}(z)}{K(z,z)}\le
\int\limits_{Y}\int\limits_{\tilde{X} }
|K(z,w)|^2  \frac{\nu_{\tilde{X}}(w)}{K(w,w)} \frac{\nu_{Y}(z)}{K(z,z)}<\infty.
$$
Using Remark \ref{remconstball},
we conclude that $I_2$ has the property: for any $l\in\N$ there is a constant
$C=C(l;n,X,Y,\Gamma , \nu_X,\nu_Y)$
such that
$$
|I_2|\le \frac{C}{k^l}
$$
as $k\to\infty$.

Now we consider
$$
I_1=
c(\B^n ,k)  \int_Y\int_X
 \frac{(\langle z,z\rangle\langle \zeta,\zeta\rangle)^{ \frac{(n+1)k}{2}  } }
{(-\langle z,\zeta\rangle )^{(n+1)k} }
\nu_{X}(\zeta)\nu_{Y}(z).
$$
We use Fubini's theorem to switch
to the integral over $Y\times X$ with respect to the product measure, then choose and fix
a sufficiently small $\delta>0$, and split $I_1$ into two parts:
$I_1^{(1)}$, where the integration is over the part of  $Y\times X$ where $\tau(z,\zeta)\le \delta$
and  $I_1^{(2)}$, where the integration is over the part of  $Y\times X$ where $\tau(z,\zeta)> \delta$.
Using (\ref{coshdistf}), we get:
$$
I_1^{(2)}=
c(\B^n ,k)
\iint\limits_{\substack{Y\times X \\ \tau(z,\zeta)> \delta} }
 \frac{(\langle z,z\rangle\langle \zeta,\zeta\rangle)^{ \frac{(n+1)k}{2}  } }
{(-\langle z,\zeta\rangle )^{(n+1)k} }
\nu_{Y}(z)\nu_{X}(\zeta),
$$
$$
|I_1^{(2)}|\le
c(\B^n ,k)
\iint\limits_{\substack{Y\times X \\ \tau(z,\zeta)> \delta} }
\Bigl ( \frac{\langle z,z\rangle\langle \zeta,\zeta\rangle }{\langle z,\zeta\rangle  \langle \zeta,z\rangle}
\Bigr )^{ \frac{(n+1)k}{2}  }\nu_{Y}(z)\nu_{X}(\zeta) =
$$
$$
c(\B^n ,k)
\iint\limits_{\substack{Y\times X \\ \tau(z,\zeta)> \delta} }
\Bigl (\cosh \frac{\tau (z,\zeta)}{2}\Bigr )^{-(n+1)k}\nu_{Y}(z)\nu_{X}(\zeta)\le
$$
$$
c(\B^n ,k)  \frac{1}{ \Bigl ( \cosh \frac{\delta}{2}\Bigr )^{(n+1)k} }
\iint\limits_{\substack{Y\times X \\ \tau(z,\zeta)> \delta} }\nu_{Y}(z)\nu_{X}(\zeta),
$$
therefore by Remark \ref{remconstball}, and since $\cosh \frac{\delta}{2}>1$,
it follows  that $I_1^{(2)}$ has the property:
for any $l\in\N$ there is a constant $C=C(l;n,X,Y,\delta , \nu_X, \nu_Y)$
such that
$$
|I_1^{(2)}|\le \frac{C}{k^l}
$$
as $k\to\infty$.

It remains to investigate the term
$$
I_1^{(1)}=
c(\B^n ,k)
\iint\limits_{\substack{Y\times X \\ \tau(z,\zeta)\le \delta} }
\frac{(\langle z,z\rangle\langle \zeta,\zeta\rangle) ^{ \frac{(n+1)k}{2}  }}
{(-\langle z,\zeta\rangle )
^{ (n+1)k } }\nu_{Y}(z)\nu_{X}(\zeta).
$$
To proceed with the proof of  $(i)$, we observe:
$$
|I_1^{(1)}|\le
c(\B^n ,k)
\iint\limits_{\substack{Y\times X \\ \tau(z,\zeta)\le \delta} }
\Bigl ( \frac{\langle z,z\rangle\langle \zeta,\zeta\rangle }{\langle z,\zeta\rangle  \langle \zeta,z\rangle}
\Bigr )^{ \frac{(n+1)k}{2}  }\nu_{Y}(z)\nu_{X}(\zeta).
$$
For the proof of $(ii)$: if $z\in Y$ and $\zeta\in X$, then $\langle z,\zeta\rangle =\langle \zeta ,z\rangle$ and
$$
I_1^{(1)}=
c(\B^n ,k)
\iint\limits_{\substack{Y\times X \\ \tau(z,\zeta)\le \delta} }
\Bigl ( \frac{\langle z,z\rangle\langle \zeta,\zeta\rangle }{\langle z,\zeta\rangle  \langle \zeta,z\rangle}
\Bigr )^{ \frac{(n+1)k}{2}  }\nu_{Y}(z)\nu_{X}(\zeta).
$$
Thus, to finish the proof of the theorem we need to treat the integral
$$
c(\B^n ,k)
\iint\limits_{\substack{Y\times X \\ \tau(z,\zeta)\le \delta} }
\Bigl ( \frac{\langle z,z\rangle\langle \zeta,\zeta\rangle }{\langle z,\zeta\rangle  \langle \zeta,z\rangle}
\Bigr )^{ \frac{(n+1)k}{2}  }\nu_{Y}(z)\nu_{X}(\zeta).
$$
Using (\ref{coshdistf}), we get:
$$
\iint\limits_{\substack{Y\times X \\ \tau(z,\zeta)\le \delta} }
\Bigl ( \frac{\langle z,z\rangle\langle \zeta,\zeta\rangle }{\langle z,\zeta\rangle  \langle \zeta,z\rangle}
\Bigr )^{ \frac{(n+1)k}{2}  }\nu_{Y}(z)\nu_{X}(\zeta)=
\iint\limits_{\substack{Y\times X\\ \tau(z,\zeta)\le \delta} }
\Bigl (\cosh \frac{\tau (z,\zeta)}{2}\Bigr )^{-(n+1)k}\nu_{Y}(z)\nu_{X}(\zeta)=
$$
$$
\iint\limits_{\substack{Y\times X \\ \tau(z,\zeta)\le \delta} }
e^{ -(n+1)k\ln \cosh \frac{\tau (z,\zeta)}{2} }\nu_{Y}(z)\nu_{X}(\zeta)=
\int\limits_{Y}\int\limits_{ \{ \zeta\in X|\tau(z,\zeta)\le \delta \}  }
e^{ -(n+1)k\ln \cosh \frac{\tau (z,\zeta)}{2} }\nu_{X}(\zeta)\nu_{Y}(z).
$$
Let $A_z\in SU(n,1)$, $z \in Y$, be a smooth family of automorphisms
$\B^n\to \B^n$ such that $A_z z=0$. Denote $\hat{X}=\cup_{z\in Y} A_z(X)$.
Let $\{ U_j\}$ be a finite cover of $\hat{X}$
by open subsets of $\B^n$ with smooth boundary, let $t_1^{(j)}$,...,$t_{q_X}^{(j)}$ be local
coordinates on $U_j\cap \hat{X}$, and let $\{ \psi^{(j)}\}$ be a partition of unity subordinate to the cover $\{ U_j\}$.

For a fixed $z\in Y$ consider the integral
$$
\int\limits_{ \{ \zeta\in X|\tau(z,\zeta)\le \delta \}  }
e^{ -(n+1)k\ln \cosh \frac{\tau (z,\zeta)}{2} }\nu_{X}(\zeta)=
\int\limits_{ \{ w\in A_z(X)|\tau(0,w)\le \delta \}  }
e^{ -(n+1)k\ln \cosh \frac{\tau (w,0)}{2} }[(A_z^{-1})^*\nu_{X}](w),
$$
where $w=A_z \zeta$. Note:
$\tau(0,w)=\tau(A_zz,A_z\zeta )=\tau(z,\zeta)$.
We have:  $(A_z^{-1})^*\nu_{X}\Bigr | _{U_j}=f^{(j)}(t)  dt_1^{(j)}\wedge ...\wedge dt_{q_X}^{(j)}$,
and the integral becomes
$$
\sum\limits_j
\int\limits_{ \{ w\in A_z(X)|\tau(w,0)\le \delta \} \cap U_j }
e^{ -(n+1)k\ln \cosh \frac{\tau (w,0)}{2} }\psi^{(j)}(t)f^{(j)}(t)  dt_1^{(j)}\wedge ...\wedge dt_{q_X}^{(j)}.
$$
Now we will work with the integral
\begin{equation}
\label{intlaplace}
\int\limits_{ \{ w\in A_z(X)|\tau(w,0)\le \delta \} \cap U_j }
e^{ -\frac{(n+1)k}{2}\ln \cosh^2 \frac{\tau (w,0)}{2} }
\psi^{(j)}(t)f^{(j)}(t)
dt_1^{(j)}...dt_{q_X}^{(j)}
\end{equation}
Apply the multivariable Laplace method.
If the point $w=0$ is in $U_j$ or on the boundary of $U_j$, then
the appropriate statement is, respectively,
Theorem 3 p. 495  or (5.15) p. 498 in \cite{wong:89}.
If the point  $w=0$ is not in $\overline{U_j}$,
then it follows that the contribution from the $j$-th
integral is rapidly decreasing as $k\to \infty$,
by an argument similar to the one that has already
been used earlier.

In order to use the Laplace method,
we need to show that the Hessian matrix $H_z$ of the function
$\ln \cosh ^2 \frac{\tau (w,0)}{2}=-\ln (-\langle w,w\rangle )$  at $w=0$
is positive definite. We have: for $l\in \{ 1,...,q_X\}$, $p\in \{1,...,q_X\}$
$$
\frac{\partial}{\partial t_p}( -\ln (-\langle w,w\rangle )=\frac{1}{-\langle w,w\rangle }
\sum\limits_{r=1}^n \Bigl ( w_r \frac{\partial \bar{w}_r}{\partial t_p}+\bar{w}_r \frac{\partial w_r}{\partial t_p}\Bigr )
$$
$$
\frac{\partial ^2}{\partial t_l\partial t_p}( -\ln (-\langle w,w\rangle )\Bigl | _{w=0}
=\sum\limits_{r=1}^n \Bigl ( \frac{\partial w_r}{\partial t_l} \frac{\partial \bar{w}_r}{\partial t_p}+
\frac{\partial \bar{w}_r}{\partial t_l} \frac{\partial w_r}{\partial t_p}\Bigr ).
$$
Therefore $H_{\zeta}=B_{\zeta}\bar{B}_{\zeta}^T+\bar{B}_{\zeta}B_{\zeta}^T$, where $B_{\zeta}$ is the $q_X\times n$
matrix $\begin{pmatrix} \frac{\partial w_1}{\partial t_1} & ... & \frac{\partial w_n}{\partial t_1} \\
... & & ... \\
\frac{\partial w_1}{\partial t_{q_X}} & ... & \frac{\partial w_n}{\partial t_{q_X}}
\end{pmatrix}$.
The matrix
$H_{\zeta}$ is symmetric. The matrices $H_{\zeta}$, $B_{\zeta}\bar{B}_{\zeta}^T$, $\bar{B}_{\zeta}B_{\zeta}^T$
are positive semidefinite, because for a vector $v\in \C^{q_X}$
$(B_{\zeta}\bar{B}_{\zeta}^Tv)^T\bar{v}=(\bar{B}_{\zeta}^T v)^T\overline{\bar{B}_{\zeta}^T v}$ and
$(\bar{B}_{\zeta}B_{\zeta}^Tv)^T\bar{v}=(B_{\zeta}^T v)^T\overline{B_{\zeta}^T v}$.
It remains to show: if $H_{\zeta}v=0$, then $v=0$. If  $H_{\zeta}v=0$, then
$\bar{v}^TH_{\zeta}v=0$, and it follows that
$B_{\zeta}^T v=0$. But $rk (B_{\zeta}^T)=q_X$, hence $\dim \ker B_{\zeta}^T=0$, therefore $v=0$.
Thus $H_{\zeta}$ is positive definite.

If the point $w=0$ is in $U_j$, then the integral (\ref{intlaplace}) is asymptotic, as $k\to\infty$, to
$\Bigl (\frac{4\pi}{(n+1)k}\Bigr )^{\frac{q_X}{2}}
 \psi^{(j)}f^{(j)} \Bigr| _{w=0}
(\det H_{\zeta})^{-\frac{1}{2}}$,
and if $0$ is on the boundary of $U_j$, then the integral (\ref{intlaplace}) is asymptotic to
$\frac{1}{2}\Bigl (\frac{4\pi}{(n+1)k}\Bigr )^{\frac{q_X}{2}}
 \psi^{(j)}f^{(j)}\Bigr| _{w=0}
(\det H_{\zeta})^{-\frac{1}{2}}$.
We conclude:
$$
c(\B^n ,k)
\int\limits_{Y}\int\limits_{ \{ \zeta\in X|\tau(z,\zeta)\le \delta \}  }
e^{ -(n+1)k\ln \cosh \frac{\tau (z,\zeta)}{2} }\nu_{X}(\zeta)\nu_{Y}(z)=
$$
$$
c(\B^n ,k)\int\limits_{Y} \sum\limits_j
\int\limits_{ \{ w\in A_z(X)|\tau(w,0)\le \delta \} \cap U_j }
e^{ -\frac{(n+1)k}{2}\ln \cosh^2 \frac{\tau (w,0)}{2}  }
\psi^{(j)}(t)f^{(j)}(t)  dt_1^{(j)}... dt_{q_X}^{(j)} \ \nu_{Y}(z)\sim
$$
$$
c(\B^n ,k)Ck^{-\frac{q_X}{2}}
$$
and the statements $(i)$, $(ii)$ now follow
from Remark \ref{remconstball}.
For the constant $C$ we have: $C>0$, because the number
$f^{(j)}\Bigr| _{w=0}(\det H_{\zeta})^{-\frac{1}{2}}$ is positive for each $j$,
the value of the function $\psi^{(j)}$ at the point $w=0$ is nonnegative for
each $j$, and there is $j_0$ such that the point $w=0$ is in $U_{j_0}$ and
$ \psi^{(j_0)}\Bigr| _{w=0}>0$.
$\Box$
\begin{remark}
The remainder in Theorem \ref{thxx}$(ii)$ is determined by $I_2$, $I_1^{(2)}$, the error term in the Laplace approximation
and the error in the Stirling formula.
\end{remark}
In the examples below, for specific $X$ and $Y$, we will work out the integral
$$
I_1=
c(\B^n ,k)
\int_Y\int_X
\frac{(\langle z,z\rangle\langle \zeta,\zeta\rangle) ^{ \frac{(n+1)k}{2}  }}
{(-\langle z,\zeta\rangle )
^{ (n+1)k } }\nu_{X}(\zeta)\nu_{Y}(z).
$$
This term appeared in the proof of Theorem \ref{thxx} as the term that determines
the behaviour of
$(\Theta_X^{(j;k)},\Theta_Y^{(j;k)})$ as $k\to\infty$.

\begin{example}
\label{exlinesegment}
Let $Y=X\subset \B^n$ be a ($1$-dimensional) line segment defined by
$z_1=te^{i\varphi}$, $-\alpha < t < \alpha$,
where $\alpha\in (0,1)$ and $\varphi\in [0,\frac{\pi}{2}]$ are fixed, $z_j=0$ for $j>1$, and let $\nu_X=dt$.
If $n=1$, then $X$ is a Lagrangian submanifold of $\B^1$. For arbitrary $n$ such $X$ is totally real.
We have:
$$
I_1=
c(\B^n ,k)
\int\limits_{-\alpha}^{\alpha}
\int\limits_{-\alpha}^{\alpha}
\Bigl ( \frac{(1-t^2)(1-T^2) }{ (1-tT)^2 }
\Bigr )^{ \frac{(n+1)k}{2}  } dt \ dT.
$$
Here $\zeta=Te^{i\varphi}$. For a fixed $T$
denote $f(t)=\Bigl ( \frac{(1-t^2)(1-T^2) }{ (1-tT)^2 }\Bigr ) ^{1/2}$. We have:
$$
f(T)=1, \  \frac{df}{dt}\Bigr | _{t=T}=0, \    \frac{d^2f}{dt^2}\Bigr | _{t=T}=-\frac{1}{(1-T^2)^2}<0.
$$
The function $f$ has a maximum at $t=T$.
Applying the $1$-dimensional Laplace approximation  ((1.5) p. 60 \cite{wong:89} or (5.1.21) \cite{bleistein:75})
we get: as $k\to\infty$
$$
\int\limits_{-\alpha}^{\alpha} f(t)^{ (n+1)k } dt\sim
\Bigl ( \frac{-2\pi}{ (n+1)kf''(T)  }  \Bigr ) ^{ \frac{1}{2}}=
\sqrt{\frac{2\pi}{(n+1)k}}(1-T^2),
$$
hence
$$
I_1\sim c(\B^n ,k) \sqrt{\frac{2\pi}{(n+1)k}}\int\limits_{-\alpha}^{\alpha}(1-T^2)dT
=c(\B^n ,k) \sqrt{\frac{2\pi}{(n+1)k}}2(\alpha-\frac{\alpha^3}{3})\sim
c(n)(\alpha-\frac{\alpha^3}{3}) k^{n-\frac{1}{2}},
$$
where $c(n)=\frac{(n+1)^{n-\frac{1}{2} } }{n!}2\sqrt{2\pi}$.
\end{example}
\begin{example}
\label{excircle}
Let $Y=X\subset \B^n$ ($n\ge 2$) be the circle of radius $0<\alpha <1$ in the $x_1x_2$-plane
centered at $(x_1,x_2)=(0,0)$:
$z_1=x_1=\alpha\cos \Theta$, $z_2=x_2=\alpha\sin \Theta$, $0\le \Theta <2\pi$,
$y_1=y_2=0$, $z_j=0$ for $j>2$.
Let $\nu_X=d\Theta$. For arbitrary $n\ge 2$ such $X$ is totally real.
We have:
$$
I_1=c(\B^n ,k)\int_X\int_X
\frac{[(1-x_1^2-x_2^2)(1-(Re(\zeta_1))^2-(Re(\zeta_2))^2)]^{\frac{(n+1)k}{2}}}
     {(1-x_1Re(\zeta_1)-x_2Re(\zeta_2))^{(n+1)k}}
 d\Theta \ d\varphi=
$$
$$
c(\B^n ,k)
\int\limits_{0}^{2\pi}
\int\limits_0^{2\pi }
\Bigl ( \frac{1-\alpha^2 }{ 1-\alpha^2\cos(\Theta-\varphi)}
\Bigr )^{ (n+1)k  } d\Theta \ d\varphi.
$$
Here $\zeta_1=\alpha\cos \varphi$, $0\le \varphi <2\pi$,
$\zeta_2=\alpha\sin \varphi$, $Im(\zeta_1)=Im(\zeta_2)=0$, $\zeta_j=0$ for $j>2$.
For a fixed $\varphi$
denote $f(\Theta)= \frac{1-\alpha^2 }{ 1-\alpha^2\cos(\Theta-\varphi)} $. We have:
$$
f(\varphi)=1, \  \frac{df}{d\Theta}\Bigr | _{\Theta=\varphi}=0, \
\frac{d^2f}{d\Theta^2}\Bigr | _{\Theta=\varphi}=-\frac{\alpha^2}{1-\alpha^2}<0,
$$
$f$ has a local maximum at $\Theta=\varphi$.
Applying the $1$-dimensional Laplace approximation ((1.5) p. 60 \cite{wong:89} or (5.1.21) \cite{bleistein:75})
we get:
$$
\int\limits_0^{2\pi} f(\Theta)^{ (n+1)k } d\Theta\sim
\Bigl ( \frac{-2\pi}{ (n+1)kf''(\varphi)  }  \Bigr ) ^{ \frac{1}{2}}=
\sqrt{\frac{2\pi}{(n+1)k}}\frac{\sqrt{1-\alpha^2}}{\alpha},
$$
hence
$$
I_1\sim c(\B^n ,k) \sqrt{\frac{2\pi}{(n+1)k}}\frac{\sqrt{1-\alpha^2}}{\alpha}
\int\limits_{0}^{2\pi}d\varphi
=c(\B^n ,k) \sqrt{\frac{2\pi}{(n+1)k}}\frac{\sqrt{1-\alpha^2}}{\alpha} 2\pi
\sim
k^{n-\frac{1}{2}}c(n)\frac{\sqrt{1-\alpha^2}}{\alpha},
$$
where $c(n)=\frac{(n+1)^{n-\frac{1}{2} } }{n!}2\pi\sqrt{2\pi}$.
\end{example}
\begin{example}
\label{exdisc}
Let $Y=X\subset \B^n$ ($n\ge 2$) be the disc of radius $\alpha\in (0,1)$ in the $x_1x_2$-plane
centered at $(x_1,x_2)=(0,0)$. Thus, $X$ is defined by
$x_1^2+x_2^2<\alpha^2$,   $y_1=y_2=0$,
$z_j=0$ for $j>2$. Let $\nu_X=dx_1\wedge dx_2$.
For arbitrary $n\ge 2$ such $X$ is totally real.  If $n=2$, then $X$ is a Lagrangian submanifold of $\B^2$.
$$
I_1=
c(\B^n ,k)
\int\limits_{X}\int\limits_X
\Bigl ( \frac{(1-x_1^2-x_2^2)(1-u_1^2-u_2^2)}{(1-x_1u_1-x_2u_2)^2} \Bigr ) ^{ \frac{(n+1)k}{2}  }dx_1dx_2du_1du_2=
$$
$$
c(\B^n ,k)
\int\limits_{X}\int\limits_X e^{\frac{(n+1)k}{2}\ln
\frac{(1-x_1^2-x_2^2)(1-u_1^2-u_2^2)}{(1-x_1u_1-x_2u_2)^2} }dx_1dx_2du_1du_2,
$$
where $u_1=Re(\zeta_1)$, $u_2=Re(\zeta_2)$. For fixed $u_1$, $u_2$ let
$f(x_1, x_2)=- \ln \frac{(1-x_1^2-x_2^2)(1-u_1^2-u_2^2)}{(1-x_1u_1-x_2u_2)^2}$.
We have: $f(u_1,u_2)=0$,
$$
\frac{\partial f}{\partial x_j}=2\Bigl ( \frac{x_j}{1-x_1^2-x_2^2}-\frac{u_j}{1-x_1u_1-x_2u_2}\Bigr ) , \ j=1,2
$$
$$
\frac{\partial f}{\partial x_1} \Bigr |_{(u_1,u_2)}=\frac{\partial f}{\partial x_2} \Bigr |_{(u_1,u_2)}=0
$$
$$
\frac{\partial ^2 f}{\partial x_j^2}=2\Bigl ( \frac{1-x_1^2-x_2^2+2x_j^2}{(1-x_1^2-x_2^2)^2}-
\frac{u_j^2}{(1-x_1u_1-x_2u_2)^2} \Bigr ) , \ j=1,2
$$
$$
\frac{\partial ^2 f}{\partial x_1 \partial x_2}=2\Bigl ( \frac{2x_1x_2}{(1-x_1^2-x_2^2)^2}-
\frac{u_1u_2}{(1-x_1u_1-x_2u_2)^2} \Bigr )
$$
$$
\frac{\partial ^2 f}{\partial x_1^2} \Bigr |_{(u_1,u_2)}=2\frac{1-u_2^2}{(1-u_1^2-u_2^2)^2}>0
$$
$$
H(u_1,u_2)= \Bigl (
\frac{\partial ^2 f}{\partial x_1^2} \frac{\partial ^2 f}{\partial x_2^2}
-\Bigl ( \frac{\partial ^2 f}{\partial x_1 \partial x_2} \Bigr )^2\Bigr )\Bigr |_{(u_1,u_2)}=
\frac{4}{(1-u_1^2-u_2^2)^3}>0
$$
Using Laplace approximation in $\R^2$ (\cite{wong:89} p. 495 or Theorem 2 \cite{hsu:51}) we get:
for a fixed $\zeta$
$$
\int\limits_X e^{-\frac{(n+1)k}{2}f(x_1,x_2) }
dx_1dx_2 \sim
\frac{4\pi}{(n+1)k\sqrt{H(u_1,u_2)}}=\frac{2\pi}{(n+1)k}(1-u_1^2-u_2^2)^{\frac{3}{2}}
$$
and
$$
I_1\sim
c(\B^n ,k)\frac{2\pi}{(n+1)k}\int\limits_{X}(1-u_1^2-u_2^2)^{\frac{3}{2}}du_1du_2=
c(\B^n ,k)\frac{4\pi ^2}{5}\frac{1}{(n+1)k}(1-(1-\alpha^2)^{\frac{5}{2}})\sim
$$
$$
k^{n-1}\frac{4\pi ^2}{5} \frac{(n+1)^{n-1}}{n!}(1-(1-\alpha^2)^{\frac{5}{2}}).
$$
\end{example}
\begin{example}
  \label{exlinedisc}
  Let $\beta\in (0,1)$ and $\alpha\in (\beta,1)$ be fixed.
  Let $Y$ be the line segment in $\B^n$ ($n\ge 2$) defined by $-\beta <x_1<\beta$,
  $y_1=0$, $z_j=0$ for $j>1$, and let
$X$ be the disc defined by $x_1^2+x_2^2<\alpha ^2$,  $y_1=y_2=0$,
$z_j=0$ for $j>2$. Let $\nu_X=dx_1\wedge dx_2$ and $\nu_Y=dx_1$.
$$
I_1=
c(\B^n ,k)
\int\limits_{Y}\int\limits_X
\Bigl ( \frac{(1-x_1^2-x_2^2)(1-u_1^2)}{(1-x_1u_1)^2} \Bigr ) ^{ \frac{(n+1)k}{2}  }dx_1dx_2du_1=
$$
$$
c(\B^n ,k)
\int\limits_{Y}\int\limits_X e^{\frac{(n+1)k}{2}\ln
\frac{(1-x_1^2-x_2^2)(1-u_1^2)}{(1-x_1u_1)^2} }dx_1dx_2du_1,
$$
where $u_1=Re(\zeta_1)$. For a fixed $u_1$ let
$f(x_1, x_2)=- \ln \frac{(1-x_1^2-x_2^2)(1-u_1^2)}{(1-x_1u_1)^2}$.
We have: $f(u_1,0)=0$,
$$
\frac{\partial f}{\partial x_1}=2\Bigl ( \frac{x_1}{1-x_1^2-x_2^2}-\frac{u_1}{1-x_1u_1}\Bigr ) , \
\frac{\partial f}{\partial x_2}=\frac{2x_2}{1-x_1^2-x_2^2}
$$
$$
\frac{\partial f}{\partial x_1} \Bigr |_{(u_1,0)}=\frac{\partial f}{\partial x_2} \Bigr |_{(u_1,0)}=0
$$
$$
\frac{\partial ^2 f}{\partial x_1^2}=2\Bigl ( \frac{1+x_1^2-x_2^2}{(1-x_1^2-x_2^2)^2}-
\frac{u_1^2}{(1-x_1u_1)^2} \Bigr ) , \
\frac{\partial ^2 f}{\partial x_2^2}=2\frac{1-x_1^2+x_2^2}{(1-x_1^2-x_2^2)^2}
$$
$$
\frac{\partial ^2 f}{\partial x_1 \partial x_2}=\frac{-4x_1x_2}{(1-x_1^2-x_2^2)^2}
$$
$$
\frac{\partial ^2 f}{\partial x_1^2} \Bigr |_{(u_1,0)}=\frac{2}{(1-u_1^2)^2}>0
$$
$$
H(u_1,0)= \Bigl (
\frac{\partial ^2 f}{\partial x_1^2} \frac{\partial ^2 f}{\partial x_2^2}
-\Bigl ( \frac{\partial ^2 f}{\partial x_1 \partial x_2} \Bigr )^2\Bigr )\Bigr |_{(u_1,0)}=
\frac{4}{(1-u_1^2)^3}>0
$$
Using Laplace approximation in $\R^2$ (\cite{wong:89} p. 495 or Theorem 2 \cite{hsu:51}) we get:
for a fixed $\zeta$
$$
\int\limits_X e^{-\frac{(n+1)k}{2}f(x_1,x_2) }
dx_1dx_2 \sim
\frac{4\pi}{(n+1)k\sqrt{H(u_1,0)}}=\frac{2\pi}{(n+1)k}(1-u_1^2)^{\frac{3}{2}}
$$
and as $k\to\infty$
$$
I_1\sim
c(\B^n ,k)\frac{2\pi}{(n+1)k}\int_{-\beta}^{\beta}(1-u_1^2)^{\frac{3}{2}}du_1
\sim {\mathrm{const}}(n,\beta ) \ k^{n-1}.
$$
\end{example}

\begin{example}
\label{3ball}
\noindent Let $\alpha\in (0,1)$ be fixed. Define a submanifold of $\B^n$, $n\ge 2$, by
$$
X=\{ (x_1,y_1,...,x_n,y_n)\in \B^n \ | \  x_1^2+y_1^2+x_2^2<\alpha^2, x_2> 0, y_2=0, x_j=y_j=0  \
{\mathrm{for}} \ j>2\}.
$$
and set  $\nu_X=dx_1\wedge dy_1\wedge dx_2$.

As a remark, we point out that $X$ is a CR submanifold which is not totally real and not complex
(see subsection \ref{appsec3} of the Appendix).

We will now estimate $I_1$, with $X=Y$:
$$
I_1=c(\B^n ,k)
\int_X \int_X
\frac{(\langle z,z\rangle\langle \zeta,\zeta\rangle) ^{ \frac{(n+1)k}{2}  }}
{(-\langle z,\zeta\rangle )
^{ (n+1)k } }\nu_{X}(z)\nu_{X}(\zeta).
$$
We set all coordinates, except for $x_1$, $y_1$ and $x_2$, to be zero. We
will use the spherical coordinates:
$$
x_1=\rho \sin\Phi \cos \Theta
$$
$$
y_1=\rho \sin\Phi\sin\Theta
$$
$$
x_2=\rho \cos\Phi
$$
$$
0< \rho< \alpha, \ 0\le \Theta<2\pi, \ 0\le \Phi < \frac{\pi}{2}
$$
$$
u_1=r \sin\psi \cos \beta
$$
$$
v_1=r \sin\psi\sin\beta
$$
$$
u_2=r \cos\psi
$$
$$
0< r< \alpha, \ 0\le \beta<2\pi, \ 0\le \psi< \frac{\pi}{2},
$$
where $\zeta_1=u_1+iv_1$ ($u_1,v_1\in\R$), $Re(\zeta_2)=u_2$.
The integral becomes
$$
I_1=c(\B^n,k) \int_X\int_X
\frac{[(1-x_1^2-y_1^2-x_2^2)(1-u_1^2-v_1^2-u_2^2)]^{\frac{(n+1)k}{2}}}{(1-(x_1+iy_1)(u_1-iv_1)-x_2u_2)^{(n+1)k}}
dx_1 dy_1 dx_2 du_1 dv_1 du_2.
$$
For fixed $u_1$, $v_1$, $u_2$, the integral
$$
\int_X
\frac{(1-x_1^2-y_1^2-x_2^2)^{\frac{(n+1)k}{2}}}{(1-(x_1+iy_1)(u_1-iv_1)-x_2u_2)^{(n+1)k}}
dx_1 dy_1 dx_2=
$$
$$
\int_X \frac{(1-\rho^2)^{\frac{(n+1)k}{2}} \rho^2\sin\Phi }
{(1-\rho\sin\Phi(\cos\Theta+i\sin\Theta)(u_1-iv_1)-u_2\rho\cos\Phi )^{(n+1)k}}
d\rho  \ d\Phi  \ d\Theta =
$$
$$
\int_0^{\frac{\pi}{2}}\sin\Phi \int_0^{\alpha} (1-\rho^2)^{\frac{(n+1)k}{2}} \rho^2
\int_0^{2\pi} \frac{1}
    {(1-\rho\sin\Phi  \ e^{i\Theta}(u_1-iv_1)-u_2\rho\cos\Phi )^{(n+1)k}ie^{i\Theta}}
    $$
    $$
    d(e^{i\Theta})d\rho  \ d\Phi  =
2\pi \int_0^{\frac{\pi}{2}}\sin\Phi \int_0^{\alpha} (1-\rho^2)^{\frac{(n+1)k}{2}} \rho^2
\frac{1}{(1-u_2\rho\cos\Phi )^{(n+1)k}} d\rho  \ d\Phi .
$$
Apply the Laplace method (\cite{wong:89} Theorem 3 p. 495)  to the integral
$$
\int_0^{\alpha} (1-\rho^2)^{\frac{(n+1)k}{2}} \rho^2
\frac{1}{(1-u_2\rho\cos\Phi)^{(n+1)k}} d\rho=
\int_0^{\alpha} e^{-(n+1)kf(\rho)} \rho^2  d\rho,
$$
where
$$
f(\rho)=-\ln \frac{\sqrt{1-\rho^2}}
{1-u_2\rho\cos\Phi }.
$$
We have:
$$
\frac{df}{d\rho}=-\frac{\rho-u_2\cos \Phi}{(1-\rho^2)(1-u_2\rho\cos\Phi )},
$$
$$
\frac{d^2f}{d\rho^2}\Bigr |_{\rho=u_2\cos\Phi}=\frac{1}{(1-(u_2\cos\Phi )^2)^2}>0,
$$
and as $k\to\infty$
$$
\int_0^{\alpha} e^{-(n+1)kf(\rho)} \rho^2  d\rho \sim
e^{-(n+1)kf(u_2\cos\Phi)}\Bigl [
\sqrt{\frac{2\pi}{(n+1)kf''(u_2\cos\Phi)}}
(u_2\cos\Phi)^2 +c_1((n+1)k)^{-\frac{3}{2}}+
$$
$$
c_2((n+1)k)^{-\frac{5}{2}}+...\Bigr ]
$$
where the constants $c_1$, $c_2$,..., depend on $\Phi$ and $u_2$. So,
$$
\int_0^{\alpha} e^{-(n+1)kf(\rho)} \rho^2  d\rho \sim (1-(u_2\cos\Phi)^2)^{-\frac{(n+1)k}{2}}
\Bigl [
  \sqrt{\frac{2\pi}{(n+1)k}}(u_2\cos\Phi)^2(1-(u_2\cos\Phi)^2)+
  $$
  $$
   c_1((n+1)k)^{-\frac{3}{2}}+
c_2((n+1)k)^{-\frac{5}{2}}+...\Bigr ] .
$$
Therefore
$$
I_1\sim c(\B^n,k)2\pi  \int_X \int_0^{\frac{\pi}{2}}
 (1-(u_2\cos\Phi)^2)^{-\frac{(n+1)k}{2}}
\Bigl [
  \sqrt{\frac{2\pi}{(n+1)k}}(u_2\cos\Phi)^2(1-(u_2\cos\Phi)^2)+
  $$
  $$
   c_1((n+1)k)^{-\frac{3}{2}}+
c_2((n+1)k)^{-\frac{5}{2}}+...\Bigr ]
(1-u_1^2-v_1^2-u_2^2)^{\frac{(n+1)k}{2}}\sin\Phi
 \ d\Phi  \ du_1 dv_1 du_2=
$$
$$
c(\B^n,k)2\pi \int_0^{2\pi}d\beta
\int_0^{\frac{\pi}{2}} \int_0^{\alpha}
\int_0^{\frac{\pi}{2}}
(1-(r\cos\psi\cos\Phi)^2)^{-\frac{(n+1)k}{2}}
$$
$$
\Bigl [
  \sqrt{\frac{2\pi}{(n+1)k}}(r\cos\psi\cos\Phi)^2(1-(r\cos\psi\cos\Phi)^2)+
   c_1((n+1)k)^{-\frac{3}{2}}+
   c_2((n+1)k)^{-\frac{5}{2}}+...\Bigr ]
$$
$$
(1-r^2)^{\frac{(n+1)k}{2}}\sin\Phi \
 r^2\sin\psi \
d\Phi  \ dr \  d\psi=
$$
$$
c(\B^n,k)4\pi^2
\int_0^{\frac{\pi}{2}}
\int_0^{\frac{\pi}{2}} \int_0^{\alpha}
(1-(r\cos\psi\cos\Phi)^2)^{-\frac{(n+1)k}{2}}
$$
$$
\Bigl [
  \sqrt{\frac{2\pi}{(n+1)k}}(r\cos\psi\cos\Phi)^2(1-(r\cos\psi\cos\Phi)^2)+
   c_1((n+1)k)^{-\frac{3}{2}}+
   c_2((n+1)k)^{-\frac{5}{2}}+...\Bigr ]
$$
$$
(1-r^2)^{\frac{(n+1)k}{2}}\sin\Phi \
 r^2\sin\psi \
dr  \ d\Phi \   d\psi =
$$
$$
c(\B^n,k)4\pi^2  \sqrt{\frac{2\pi}{(n+1)k}}
\int_0^{\frac{\pi}{2}} (\cos\psi )^2\sin\psi
\int_0^{\frac{\pi}{2}} (\cos\Phi)^2\sin\Phi   \int_0^{\alpha}
(1-(r\cos\psi\cos\Phi)^2)^{-\frac{(n+1)k}{2}}
$$
$$
(1-(r\cos\psi\cos\Phi)^2)
(1-r^2)^{\frac{(n+1)k}{2}}
 r^4
dr  \ d\Phi \  d\psi +
$$
$$
c(\B^n,k)4\pi^2
\int_0^{\frac{\pi}{2}}\sin\psi
\int_0^{\frac{\pi}{2}} \sin\Phi \int_0^{\alpha}
(1-(r\cos\psi\cos\Phi)^2)^{-\frac{(n+1)k}{2}}
$$
$$
\Bigl [
   c_1((n+1)k)^{-\frac{3}{2}}+
   c_2((n+1)k)^{-\frac{5}{2}}+...\Bigr ]
(1-r^2)^{\frac{(n+1)k}{2}}
 r^2
dr \ d\Phi \ d\psi .
$$
For fixed values of $\Phi$, $\Psi$, apply the Laplace method
 (\cite{wong:89} Theorem 3 p. 495) to the integral
$$
\int_0^{\alpha}e^{-(n+1)kf(r)}
(1-(r\cos\psi\cos\Phi)^2)
r^4 dr=\frac{1}{2}
\int_{-\alpha}^{\alpha}e^{-(n+1)kf(r)}
(1-(r\cos\psi\cos\Phi)^2)
r^4 dr
$$
 where
$$
f(r)=-\frac{1}{2}\ln\frac{1-r^2}{1-(r\cos\psi\cos\Phi)^2}.
$$
We get:
$$
\frac{df}{dr}=\frac{r-r(\cos\Phi\cos\psi)^2}{(1-r^2)(1-(r\cos\Phi\cos\psi)^2)}
$$
$$
\frac{d^2f}{dr^2}\Bigr |_{r=0}=1-(\cos\Phi\cos\psi)^2)>0
$$
and as $k\to\infty$
$$
\int_{-\alpha}^{\alpha}e^{-(n+1)kf(r)}
(1-(r\cos\psi\cos\Phi)^2)
r^4 dr\sim
$$
$$
e^{-(n+1)kf(0)}\Bigl [ \sqrt{\frac{2\pi}{(n+1)kf''(0)}}
(1-(r\cos\psi\cos\Phi)^2)r^4\Bigr |_{r=0}+O(k^{-\frac{3}{2}})\Bigr ].
$$
We also have, for fixed values of $\Phi$, $\psi$:
$$
| \int_0^{\alpha}
(1-(r\cos\psi\cos\Phi)^2)^{-\frac{(n+1)k}{2}}
\Bigl [
   c_1((n+1)k)^{-\frac{3}{2}}+
   c_2((n+1)k)^{-\frac{5}{2}}+...\Bigr ]
(1-r^2)^{\frac{(n+1)k}{2}}
r^2dr|
$$
$$
\le {\mathrm{const}} \ k^{-3},
$$
since by the Laplace method, as above,
$$
\int_0^{\alpha}
(1-(r\cos\psi\cos\Phi)^2)^{-\frac{(n+1)k}{2}}
(1-r^2)^{\frac{(n+1)k}{2}}
r^2dr=\frac{1}{2}
\int_{-\alpha}^{\alpha}
e^{-(n+1)k\frac{1}{2}\ln \frac{1-(r\cos\psi\cos\Phi)^2}{1-r^2}}
r^2dr\sim
$$
$$
O(k^{-\frac{3}{2}}).
$$
We recall: $c(\B^n ,k)\sim \frac{(n+1)^n}{n!}k^n\Bigl (1+O(\frac{1}{k})\Bigr )$
(Remark \ref{remconstball}). Combining all this together, we conclude:
$$
|I_1|\le {\mathrm{const}} \ k^{n-2}
$$
It follows that
$$
(\Theta_X^{(j;k)},\Theta_X^{(j;k)})\le {\mathrm{const}} \ k^{n-2}.
$$
We have: ${\mathrm{dim}}_{\R}X=3$,
for $n>2$ we are in the setting of  part {\it (i)} of Theorem \ref{thxx},
the asymptotic inequality holds and it is a strict inequality,
and the asymptotic behaviour we observe here is different
from the asymptotics for totally real submanifolds in
Theorem \ref{thxx}{\it (ii)}, for which
$(\Theta_X^{(j;k)},\Theta_X^{(j;k)})$ would have been asymptotic
to $Ck^{n-\frac{{\mathrm{dim}}_{\R}X}{2}}$ with $C>0$.
\end{example}
\begin{remark}
  In  \cite{borthwick:95} it is shown that
  the square of the norm of the sections of powers of
  the line bundle associated to  a  Bohr-Sommerfeld Lagrangian submanifold
of a compact $n$-dimensional K\"ahler manifold
 grows as a constant times
 $k^{\frac{n}{2}}$. In \cite{ioos:18} an analogous result for
 a $q$-dimensional isotropic Bohr-Sommerfeld submanifold of a $2n$-dimensional
 symplectic manifold and associated sections of vector bundles gives
 the leading term of $const \ k^{n-\frac{q}{2}}$. Our
 Theorem \ref{thxx}{\it (ii})  gives the same leading term for norms of
 vector-valued automorphic forms associated to isotropic submanifolds of ball quotients.
In the Example \ref{3ball} we have a submanifold $X$
which is not isotropic
and $(\Theta_X^{(j;k)},\Theta_X^{(j;k)})$ does not have the same kind of asymptotics.
All this raises a general question
how the geometric properties of submanifolds are reflected in asymptotics
of the associated sections of vector bundles.
\end{remark}

\appendix{}
\section{}
\subsection{Proof of Lemma \ref{propconvscalar}.} First, we observe that this statement is contained
in the general framework of Chapters 5 and 7 of \cite{kollar:95}.
Now we will present an actual proof. This is a modification of the proof of Prop. 1, p. 44 \cite{baily:73},
which will use that for a fixed $p\in D$
$\int\limits _D|K(z,p)^2|dV_e(z)<\infty$.
Let $A$ be a nonempty compact subset of $D$. We will prove that the series
$\sum\limits_{\gamma\in\Gamma}\Bigl ( K(\gamma z,p) J(\gamma,z)\Bigr ) ^k$
converges absolutely and uniformly on $A$, for $k\ge2$.
There are a compact subset $B$ of $D$ and $\delta>0$ such that $A$ is contained in the interior of $B$,
and for any $a\in A$ there is a polydisc $P_a$ of Euclidean volume $\delta$, with center $a$, such that
$\bar{P}_a\subset B$. Let $m_0$ be the number of elements in $\{ g\in \Gamma | gB\cap B\ne \emptyset\}$.
First consider the case $k=2$. For $a\in A$
$$
|K(\gamma a,p)^2 J(\gamma, a)^2|\le \frac{1}{\delta}\int\limits_{P_a}|K(\gamma z,p)^2J(\gamma,z)^2|dV_e(z)=
\frac{1}{\delta}\int\limits_{\gamma P_a}|K(w,p)^2|dV_e(w)
$$
where $w=\gamma z$. We get:
$$
\sum\limits_{\gamma\in\Gamma}|K(\gamma a,p)^2J(\gamma, a)^2|\le
\frac{1}{\delta} \sum\limits_{\gamma\in\Gamma} \int\limits_{\gamma P_a}|K(w,p)|^2dV_e(w)\le
\frac{m_0}{\delta}\int\limits_D|K(w,p)|^2dV_e(w).
$$
The last inequality is justified by observing that if $\gamma P_a \cap \gamma' P_a \ne \emptyset$
for $\gamma,\gamma'\in \Gamma$,
then $\gamma ^{-1}\gamma '\in \{ g\in\Gamma| gB \cap B\ne \emptyset\}$, so each $w\in D$ is in at most
$m_0$ of the sets $\gamma P_a$, $\gamma\in \Gamma$. This settles the case $k=2$.
Therefore for $a\in A$ $|K(\gamma a,p)J(\gamma ,a)|<1$ for all but at most finitely many $\gamma \in \Gamma$.
When $|K(\gamma a,p)J(\gamma ,a)|<1$, $|K(\gamma a,p)J(\gamma ,a)|^k$ is a decreasing function of $k\ge 2$.
The desired statement follows.
$\Box$

\subsection{Proof of Lemma \ref{lemconstcdk}}
\noindent From (\ref{constcdk}) for $D=\B ^n$ with $z=0$ we get:
$$
c(\B^n ,k)\frac{n!}{\pi^n}\int\limits_{\B^n }(-\langle w,w\rangle )^{(n+1)(k-1)}(\frac{i}{2})^n
dw_1\wedge d\bar{w}_1\wedge ...\wedge dw_n\wedge d\bar{w}_n=1.
$$
The integral in the left hand side is equal to $\pi^n\frac{((n+1)(k-1))!}{((n+1)(k-1)+n)!}$, and the statement readily
follows. To calculate this integral apply a change of variables
$(w_1,\bar{w}_1)\to (R_1,\varphi_1)$, where $0\le R_1\le 1$, $0\le \varphi_1<2\pi$,
$w_1=R_1e^{i\varphi_1}\sqrt{1-|w_2|^2-...-|w_n|^2}$. We get:
$$
\int\limits_{\B^n }(1-|w_1|^2-...-|w_n|^2 )^{(n+1)(k-1)}(\frac{i}{2})^n
dw_1\wedge d\bar{w}_1\wedge ...\wedge dw_n\wedge d\bar{w}_n=
$$
$$
(\frac{i}{2})^{n-1} \int\limits_{\B^n }(1-|w_2|^2-...-|w_n|^2 )^{(n+1)(k-1)+1}(1-R_1^2)^{(n+1)(k-1)}R_1
dR_1 \wedge d\varphi_1\wedge dw_2\wedge d\bar{w}_2\wedge ...\wedge dw_n\wedge d\bar{w}_n,
$$
then apply the change of variables
$(w_2,\bar{w}_2)\to (R_2,\varphi_2)$, where $0\le R_2\le 1$, $0\le \varphi_2<2\pi$,
$w_2=R_2e^{i\varphi_2}\sqrt{1-|w_3|^2-...-|w_n|^2}$ to transform the integral into
$$
(\frac{i}{2})^{n-2} \int\limits_{\B^n }(1-|w_3|^2-...-|w_n|^2 )^{(n+1)(k-1)+2}(1-R_1^2)^{(n+1)(k-1)}R_1
(1-R_2^2)^{(n+1)(k-1)+1}R_2
dR_1\wedge d\varphi_1\wedge
$$
$$
dR_2 \wedge d\varphi_2\wedge
dw_3\wedge d\bar{w}_3\wedge ...\wedge dw_n\wedge d\bar{w}_n,
$$
and so on. At the end we get:
$$
(2\pi)^{n-1}\int\limits_0^1(1-R_1^2)^{(n+1)(k-1)}R_1dR_1\int\limits_0^1(1-R_2^2)^{(n+1)(k-1)+1}R_2dR_2
...
$$
$$
\int\limits_0^1(1-R_{n-1}^2)^{(n+1)(k-1)+{n-2}}R_{n-1}dR_{n-1}\int\limits_{|w_n|\le 1} (1-|w_n|^2)^{(n+1)(k-1)+(n-1)}
\frac{i}{2}dw_n\wedge d\bar{w}_n,
$$
and with $w_n=R_ne^{i\varphi_n}$, $0\le R_n\le 1$, $0\le \varphi_n<2\pi$, the last integral is
$$
2\pi\int\limits_0^1(1-R_n^2)^{(n+1)(k-1)+n-1}R_ndR_n.
$$
An elementary calculation now yields the answer.
$\Box$

\subsection{Note for Example \ref{3ball}}
\label{appsec3}
In this subsection we explain why the submanifold
$$
X=\{ (x_1,y_1,...,x_n,y_n)\in \B^n \ | \  x_1^2+y_1^2+x_2^2<\alpha^2, x_2> 0, y_2=0, x_j=y_j=0  \
{\mathrm{for}} \ j>2\}.
$$
of $\B^n$, where  $n\ge 2$ and  $\alpha\in (0,1)$, is
a CR submanifold, which is not totally real and not complex.
The relevant part of the complex hyperbolic metric (3.3)\cite{goldman:99} is,
up to a positive factor,
$$
\frac{1}{(1-x_1^2-y_1^2-x_2^2-y_2^2)^2}[
  (1-x_2^2-y_2^2)(dx_1^2+dy_1^2)+(1-x_1^2-y_1^2)(dx_2^2+dy_2^2)
  $$
  $$
  +2(x_1x_2+y_1y_2)(dx_1dx_2+dy_1dy_2)+2(-y_1x_2+x_1y_2)(dx_1dy_2-dy_1dx_2)].
$$
The complex structure $J:T\B^n\to T\B^n$ acts as follows:
$$
\frac{\partial}{\partial x_j}\mapsto -\frac{\partial}{\partial y_j}, \
\frac{\partial}{\partial y_j}\mapsto \frac{\partial}{\partial x_j}
$$
for $j=1,2$.

The distribution ${\mathcal{D}}=span\{ \frac{\partial}{\partial x_1} ,
\frac{\partial }{\partial y_1}\}$ is a holomorphic distribution on $X$. The
complementary orthogonal distribution
${\mathcal{D}}^{\perp}=span\{ \frac{\partial}{\partial x_2}-
\frac{x_1x_2}{1-x_2^2}\frac{\partial}{\partial x_1}
-\frac{x_2y_1}{1-x_2^2}\frac{\partial}{\partial y_1}\}$
is a totally real distribution.
Indeed, $J( \frac{\partial}{\partial x_2}-
\frac{x_1x_2}{1-x_2^2}\frac{\partial}{\partial x_1}
-\frac{x_2y_1}{1-x_2^2}\frac{\partial}{\partial y_1})$ is orthogonal
to $ \dfrac{\partial}{\partial x_1}$,  $\dfrac{\partial}{\partial y_1}$,
$\dfrac{\partial}{\partial x_2}$, and is, therefore, in the normal bundle of $X$.


TITLE: A possible proof of the Borsuk Ulam theorem without "Homology-Cohomology"
QUESTION [0 upvotes]: Assume that $n>1$.
The  configuration space of  $S^n$ is  defined as follows $$M_n=\{(x,y)\in S^n\times S^n\mid x \neq y\}$$
We  have  two  questions:

1.Is there a  continuous  function $f:M_n \to S^{n-1}$  with $f(y,x)=-f(x,y)$, for all $x,y \in $S^{n}$?
2.Is there  a  continuos  function $h: M_n \to \mathbb{R}^n$  such that $h(x,y)=-h(y,x)  $  and $h(x,-x)\neq 0$  for  all  $x,y \in S^n$

If the  answer to  either of these two questions is  "affirmative ", then we can provide an alternative  proof  for  the  Borsuk Ulam theorem, inductively. Because  an  equivalent  formulation of the  Borsuk Ulam theorem is that:

There is  no  an odd continuous  function $g:S^{n+1}\to S^n$

Assuming that the  answer to either of  the  above two questions is  affirmative, we give a proof  for  the above  equivalent  formulation of the  Borsuk Ulam theorem. The  proof  is  as follows:
Assume  that $g:S^{n+1}\to S^n$ is  an odd continuous  function. then $f(g(x),g(-x))$ ( or  $h(g(x),g(-x))$ ) is  an odd  continuous  function from $S^{n+1}$ to $S^{n-1}$ ( or to $\mathbb{R}^n \setminus \{0\}$). This  obviously  gives  a  contradiction by induction. Because this  situation leads to existence  of  an odd continuous  function from $S^n$ to $S^{n-1}$. Now we apply the induction argument.

REPLY [3 votes]: Suppose such a function $M_n \to S^{n-1}$ existed. Consider the composition $S^n \to M_n \to S^{n-1}$, where the first map sends $x$ to the pair $(x,-x)$. That composition is an odd continuous function $S^n \to S^{n-1}$, hence can not exist by Borsuk-Ulam.


TITLE: Stacks and sheaves
QUESTION [24 upvotes]: I'm a bit confused by the double role which sheaves play in the theory of stacks.
On the one hand, sheaves on a site are the obvious generalization of a sheaf on a topological space. On the other hand a sheaf on a site is (or better its associated category fibered in sets is) a very particular stack itself, so a generalization of a space. This is not completely confusing: more or less it amounts (I believe) to identifying a space X with the sheaf of continuos functions with values in X.
But now my question is the following. An equivalent condition for a fibered category to be a prestack is that for any two objects (over the same base object), the associated functor of arrows should be a sheaf. In particular this is true for a stack, so for any stack and any two objects in it we have a sheaf, and so a stack (over a comma category).
What is the meaning of this geometrically?
For instance take the stack $\mathcal{M}_{g,n}$.
Giving two objects in the stack (over the same base object) means giving two families $X$ and $Y$ of stable pointed curves over the same scheme $S$, and the associated functor of arrows maps every other scheme $f \colon T \rightarrow S$ to the set of morphism between $f^* X$ and $f^* Y$. How should I think of the associated stack as a space?
To avoid misunderstandings I give the defition of the functor of arrows. Let $\mathcal{F}$ be a fibered category over $\mathcal{C}$. Take $U \in \mathcal{C}$ and $\xi, \eta \in \mathcal{F}(U)$. Then there is a functor $F \colon \mathcal{C}/U \rightarrow Set$ defined as follows. For a map $f \colon T \rightarrow U$ we put $F(f) = Hom(f^* \xi, f^* \eta)$. The action on arrows requires some diagrams to be described, but it's really the only possible one.

REPLY [2 votes]: As far as the first part of your question goes, I have exactly the same confusion (and it will probably get worse once I start thinking about sheaves on stacks...).
There are two things that give me the illusion of some comprehension.

Thinking of sheaves as (generalized) spaces is probably not terrible, in the same way you think of a bundle just as its total space. It is indeed true that any sheaf (with values in a reasonable category I guess) is the sheaf of sections of its etale space. (Although I must confess I don't particularly like etale spaces and I'm aware that this is probably not the right picture, as we shouldn't think of sheaves on a site in this fashion, but hey).
If we think of fibred categories (+cleavage) as presheaves with values in the 2-category Cat, then there is a generalization of the sheaf condition which yields the notion of a stack. Of course, arrows being a sheaf is a consequence of this, so if you find the generalization of the sheaf condition more natural, then arrows-being-a-sheaf might be viewed as a formal consequence I guess.

Anyway it's just a thought.


TITLE: Simple inequality question $\sqrt{3} - 2 + \frac{\pi}{3} < \pi -2$
QUESTION [1 upvotes]: Can someone please explain how this conclusion is made:
$\sqrt{3} - 2 + \frac{\pi}{3} <  \pi -2$
because $\sqrt{3} < 2 < \frac{2\pi}{3}$
I think this one is quite simple, just that I cant really get my head around it.
Thank you!

REPLY [4 votes]: When working with inequalities, remember that you can operate on each side, just as you would an equation except that when you multiply or divide by a negative number, the inequality reverses, for example:  $$ -2x < 8 \iff \left(-\frac{1}{2}\right)\cdot (-2x) > \left(-\frac{1}{2}\right) \cdot 8 \iff x> -4$$
Note the reversal of the "direction"  of the inequality from $<$ to $>$ when I multiplied each side by $(-\frac{1}{2})$. So you can operate on your inequality just as you normally would with an equation, just remember to use caution when multiplying or dividing by a negative value!
You have
$$\sqrt{3} - 2 + \frac{\pi}{3} <  \pi -2$$
and you want to confirm it is true.
$$3 \cdot \sqrt{3} - 6 + \pi < 3\pi -6\tag{multiply through by 3}$$
$$3 \cdot \sqrt{3} -6+6 < 3\pi-\pi\tag{+ 6, $-\pi$ on each side}$$
$$ 3\sqrt{3} < 2\pi\tag{yes?}$$


TITLE: Why is the "root field" premise necessary in this theorem about field extensions?
QUESTION [0 upvotes]: In Charles Pinter's "A Book of Abstract Algebra," chapter 31, Theorem 8 (p316-17), he states:

Suppose $I \subseteq E \subseteq K$, where $E$ is a finite extension of $I$ and $K$ is a finite extension of $E$. If $K$ is the root field of some polynomial over $E$, then $K$ is also the root field of some polynomial over $I$.

His short proof is then:

Suppose $K$ is a root field of some polynomial over $E$, and let $K = I(a)$. If $p(x)$ is the minimum polynomial of $a$ over $I$, its coefficients are certainly in $E$ and it has a root $a$ in $K$, so by Theorem 7, all its roots are in $K$. Therefore, $K$ is the root field of $p(x)$ over $I$.

I think "certainly in $E$" is a typo that should read "certainly in $I$," but here's my question:
Why is it required as a premise that $K$ be a root field of some polynomial over $E$?
Since $K$ is a finite extension of $E$ which is a finite extension of $I$, $K$ must also be a finite extension of $I$ (by Ch29 Th2). Therefore it must also be a simple extension (by Ch31 Th2). So we can write $K=I(a)$ for some $a \in K$, and be certain a minimum polynomial of $a$ over $I$ exists (since finite and simple extensions are defined to be in terms of algebraic elements). And from here, Pinter's argument (starting at sentence 2) still seems to hold. Am I missing something?

REPLY [0 votes]: After reviewing this more, I believe the actual theorem from Pinter is wrong! Counterexample: Let $I=\mathbb{Q}$ and $E=K=\mathbb{Q}(\sqrt[4]{2})$.
I found the 2nd edition of the same book and discovered the theorem is different:

Suppose $I\subseteq E\subseteq K$ where $E$ is a finite extension of $I$ and $K$ is a finite extension of $E$. If $K$ is the root field of some polynomial over $I$, then $K$ is also the root field of some polynomial over $E$.

This makes much more sense now.


TITLE: Union of subgroups is a subgroup if and only if one subgroup is a subset of the other
QUESTION [8 upvotes]: Let $H$ and $K$ denote two subgroups of a group $G$.  Prove that the union $H \cup K$ is a subgroup of $G$ if and only if $H \subset K$ or $K \subset H$.

REPLY [15 votes]: I shall use multiplicative notation for the group law.
If $H$ or $K$ is the trivial subgroup $\{1\}$ consisting only of the multiplicative identity, the result is trivial.  So, take $H, K$ to be subgroups of $G$ such that $H,K \neq \{1\}$.
$\Leftarrow$
If WLOG $H \subset K$, then $H \cup K=K$.  Because $K$ is a subgroup of $G$, $H \cup K$ is a subgroup of $G$.
$\Rightarrow$
Suppose for sake of contradiction that we have neither $H \subset K$ nor $K \subset H$ but that $H \cup K$ is a subgroup of $G$.  This implies that there exists $h \in H$ such that $h \notin K$; likewise, $\exists k \in K$ such that $k \notin H$.  Because $H \cup K$ is a subgroup of $G$, $hk \in H \cup K$, which implies $hk \in H \vee hk \in K$.  Say, WLOG, $hk \in H$.  Well, since $h \in H$, $h^{-1} \in H$.  Thus, $h^{-1}hk=k \in H$, contradiction.


TITLE: Proving right cosets are equal or disjoint
QUESTION [4 upvotes]: Let G be a group, H is a subgroup of G.  Let $Hx = \{hx | h\in H\}$.  Show that given $a, b \in G$, then $Ha \cap Hb = \emptyset$ or $Ha=Hb$.
Suppose we have a $y \in Ha$.  This means that $y=ha$ for some $h \in H$.  Now we have two cases.
Suppose a=b.  Then $y=ha=hb$.  Since $hb \in Hb$, $y \in Hb$.  Thus, $Ha\subseteq Hb$ and $Hb\subseteq Ha$
Now suppose $a\neq b$, then I know from a lemma in the textbook that $ha\neq hb$.  Thus, $Ha \nsubseteq Hb$.  Thus, $Hb \nsubseteq Ha$.  Thus,  $Ha \cap Hb = \emptyset$

REPLY [3 votes]: Suppose
$k \in Ha \cap Hb; \tag 1$
then there must exist $h_1, h_2 \in H$ with
$k = h_1a; \; k = h_2b; \tag 2$
thus
$h_1a = h_2b, \tag 3$
whence
$a = h_1^{-1}h_2b, \tag 4$
which clearly implies
$a \in Hb, \tag 5$
and so
$Ha \subset Hb; \tag 6$
interchanging he roles of $a$ and $b$, we see that
$Hb \subset Ha, \tag 7$
so
$Ha = Hb. \tag 8$
We have shown that if $Ha$ and $Hb$ have at least one element in common, then they are the same.  Therefore
$Ha \cap Hb \ne \emptyset \Longrightarrow Ha = Hb; \tag 9$
and by contraposition,
$Ha \ne Hb \Longrightarrow Ha \cap Hb = \emptyset. \tag{10}$


TITLE: What is the most Most relevant statistic method to compare those samples?
QUESTION [1 upvotes]: One person is answering a survey. There are two type of questions :

4 questions about intellectual work affinity.
4 questions about manual work affinity.

For each of these question, the person gives an aswer from 1 (do not agree) to 5 (totally agree).
I want to know if there is a significant difference between answers about intellectual work and manual work, and eventually be able to say "this individual is more an intellectual / a manual" or "the difference is not significant.
I am thinking of Wilcoxon method, but I would like to have a second opinion as I'm really just learning statistics by myself.

REPLY [0 votes]: I saw your question just after it was posted. My first thoughts were
that a satisfactory test is not going to be easy to find. I have
thought about it during the day, and I still see serious difficulties,
but by not maybe I can explain my reservations more clearly. And perhaps
suggest one workable method for you to try.
1) I think it is going to be difficult to frame eight Likert-scale
questions that allow subjects to show their views on intellectual
and manual work activities. But let's suppose that is possible.
2) To evaluate one person's view, it would be difficult to make good
judgments based on a rank-based test such as a Wilcoxon test. While
rank-based tests do not assume normal data, they do assume that the
observations are on a continuous scale. For discrete data such as
yours there are likely to be ties, which make it problematic to
compute a P-value.
For large samples, the Wilcoxon test statistics are nearly normal.
For eight observations nomality would not apply, and some sort of
adjustment (different depending on text or software package) would be
used. Even so, if someone marked four 5's for the intellectual questions and
four 1's for manual work questions, results from the software packages
I tried show a small P-value, suggesting a rejection of the null
hypothesis that the two types of activity are equally preferred, and
giving a warning that the P-value is not 'exact' because of ties.
3) Suppose you assign scores 1 through 5 for answers 1 through 5 on the
intellectual questions and scores -1 through -5 for answers 1 through 5 on
the manual work questions. And then take the sum $S$ of the eight scores as
a test statistic. (You would have to ponder whether this kind
of scoring matches your intent in asking the questions.)  Then scores
could be as low as $-15$ and as high as $15$.

In a simulation of 100,000 subjects choosing eight possible scores totally
at random (one model for people with no preference for either type of
activity or, perhaps, just not taking your questions seriously), the
distribution of $S$ turned out to be very nearly normal with mean $0$
and SD about $4.$ So you might judge those with $S \ge 8$ as
intellectually inclined, those with $S \le -8$ as inclined toward
manual labor and those in the middle as not showing a significant result.
My simulation involved only 'subjects' with no particular views or
ones not paying much attention to the questions. Among such subjects
about 3% are (perhaps wrongly) classified as preferring manual work
and another 3% classified as preferring intellectual pursuits.
Among subjects with strong views one way or the other, you might find
much higher percentages with significant results. However, you have
to realize that you may not get very good information from only eight
questions $-$ no matter how carefully crafted.


TITLE: Classification of a system of two second order PDEs with two dependent and two independent variables
QUESTION [2 upvotes]: If we have a second order quasilinear PDE of the form
$A\frac{\partial^2 u}{\partial x^2}+B\frac{\partial^2 u}{\partial y^2}+2C\frac{\partial^2 u}{\partial x\partial y}+ lower\,\, order \,\, terms=0$
where $A,B,C$ are functions of $x,y,u$,
then the equation is called elliptic if $det=\begin{vmatrix}A &C \\C & B\end{vmatrix}>0$, parabolic if $det=0$ and hyperbolic if $det<0$.
Now what happens if we have a system of two coupled PDEs of the form
$A\frac{\partial^2 u}{\partial x^2}+B\frac{\partial^2 u}{\partial y^2}+2C\frac{\partial^2 u}{\partial x\partial y}+D\frac{\partial^2 v}{\partial x^2}+E\frac{\partial^2 v}{\partial y^2}+2F\frac{\partial^2 v}{\partial x\partial y}+ lower\,\, order \,\, terms=0 \\
G\frac{\partial^2 u}{\partial x^2}+H\frac{\partial^2 u}{\partial y^2}+2K\frac{\partial^2 u}{\partial x\partial y}+L\frac{\partial^2 v}{\partial x^2}+M\frac{\partial^2 v}{\partial y^2}+2N\frac{\partial^2 v}{\partial x\partial y}+ lower\,\, order \,\, terms=0$
with $A,B,C,...$ being functions of $x,y,u,v$.
Does it make sense to construct the determinant
$det=\begin{vmatrix}A &C & D&F\\C & B&F&E\\G &K & L&N\\K &H & N&M\end{vmatrix}$
and investigate its sign, or something like that?

REPLY [2 votes]: The classification of linear PDEs with constant coefficients is based on the principle that any linear transformation of coordinates should preserve the classification. That is, we define
$$
\xi = ax+by+c,
\quad
\eta = dx+ey+f.
$$
where $ae \neq bd$ so that lines of constant $\xi$ and $\eta$ are not parallel. Substituting in this change of coordinates, you will find a new system of PDEs in $\xi, \eta$ which have a new set of coefficients. A classification of the system should be the same whether the coefficients of the $x, y$ or $\xi, \eta$ system are used. However, the classes of your system will be more numerous than those of the classical second order PDE.
With the general approach out of the way, I suspect that your system will be classified by the sign of the following 4 determinants:
$$
\left|\begin{array} {ccc}
A & C \\
C & B
\end{array} \right|
$$
$$
\left|\begin{array} {ccc}
D & F \\
F & E
\end{array} \right|
$$
$$
\left|\begin{array} {ccc}
G & K \\
K & H
\end{array} \right|
$$
$$
\left|\begin{array} {ccc}
L & N \\
N & M
\end{array} \right|
$$
and should be classified by four words, based on the sign of each. You should verify that the signs of these determinants are invariant under the change of coordinates.
Edit:
For the case where the coefficients are not constant, we use exactly the same classification system pointwise. Defining the change of coordinates
$$
\xi = \xi(x, y),
\quad
\eta = \eta(x, y)
$$
which has a nonzero and finite Jacobian at all points, then local to any point $(x_0,y_0)$ we have that
$$
\xi(x, y) \sim \xi(x_0, y_0) + (x-x_0) \cdot \frac{\partial \xi} {\partial x}(x_0, y_0) + (y-y_0) \cdot \frac{\partial \xi} {\partial y}(x_0, y_0)
$$
$$
\eta(x, y) \sim \eta(x_0, y_0) + (x-x_0) \cdot \frac{\partial \eta} {\partial x}(x_0, y_0) + (y-y_0) \cdot \frac{\partial \eta} {\partial y}(x_0, y_0)
$$
which gives us the local linear transform around each point. The coefficients at each point are locally given by there value at each point e.g.
$$
A(x, y, u,v) \sim A(x_0, y_0, u(x_0, y_0),v(x_0, y_0)).
$$
Thus, from an understand of the linear constant coefficient case, the quasi-linear variable coefficient case can be investigated in a pointwise manner.


TITLE: Integrate indicator function by Lebesgue measure
QUESTION [0 upvotes]: So I'm struggling with the following integral:
$\int_{[-1,1]^d} \mathbb{1_{||x||^2 \leq1}} d\lambda(x)$.
I know it's supposed to be the volume of the $d$-dimensional unit sphere, but I've trouble writing it down exactly. This is the result I arrived at, after calculating the expectation of a uniformly distributed random variable, transformed by a function containing among others the indicator left in the integral above.
Thanks ahead.

REPLY [0 votes]: I think the integral is just equal to $\int...=\lambda(\{x\in [-1,1]^d\})= \frac{\pi^{d/2}}{\Gamma(\frac{d}{2}+1)} $, if we assume euclidean norm. Then the indicator can be ignored, as it doesn't change the domain of x.


TITLE: Show that the following set is open
QUESTION [0 upvotes]: Show that the following set is open. $$S = \{(v,w,x,y,z) \, \epsilon \,\mathbb R ^5 \,|\, x^2e^{v + w^{100}} > 2 \;and\; xy-z^2 < -1\} \subset \mathbb R^5$$
Hint: Use the following theorem. Let f be a function from $\mathbb R^n \, to \,\mathbb R^m$ Then, f conts $\iff f^{-1}(U) $ is open for each open set U in $\mathbb R^m$
My idea is to prove that each individual set is open since the finite intersection of open sets is open. Starting with the first, I can come up with a function $f : \mathbb R^5 \to \mathbb R$ given by $f(v,w,x,y,z) = x^2e^{v + w^{100}}-2$ which is continuous since it is made up of powers and exponentials using basic operations. So now I need to show  that $f^{-1}(U)$ is open for each open set U in $\mathbb R$, but how do I find a U that contains each open set in $\mathbb R?$

REPLY [1 votes]: Here:$$S=f^{-1}\left((2,\infty)\right)\cap g^{-1}\left((-\infty,-1)\right)$$
where $f:\mathbb R^5\to \mathbb R$ is prescribed by $(v,w,x,y,z)\mapsto x^2e^{v+w^{100}}$ and $g:\mathbb R^5\to \mathbb R$ is prescribed by $(v,w,x,y,z)\mapsto xy-z^2$.
Both functions are continuous and the sets $(2,\infty)$ and $(-\infty,-1)$ are open subsets of $\mathbb R$.
So $S$ is an intersection of open sets, hence is an open set itself.