TITLE: differentiable function with only positive/negative slope impies strictly increasing/decreasing
QUESTION [0 upvotes]: If $f:(a,b)\rightarrow\mathbf{R}$ is differentiable and
1.$f'(x)>0$ for all x $\in (a,b)$.
$\ \ $then $f$ is strictly increasing on (a,b).
2.Similarly, if $f'(x)<0$ for all x $\in (a,b)$, then $f$ is strictly decreasing on (a,b)

Proof.(1.)
Assume $f:(a,b)\rightarrow\mathbf{R}$ is differentiable
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ $ and $f'(x)>0$ for all x $\in (a,b)$
Show that $\forall x_1,x_2 \in (a,b),x_1<x_2\rightarrow f(x_1)<f(x_2)$
Since $\exists x_3 \in (a,b) s.t. \forall x \in (a,b),0<f'(x_3)\leq f'(x)$
We can conclude that
$\exists x_3 \in (a,b) s.t. \forall x_1,x_2 \in (a,b),x_1<x_2 \rightarrow f(x_1)<f(x_1)+f'(x_3)(x_2-x_1)\leq f(x_2)$
Therefore, $f$ is strictly increasing on $(a,b)$
Proof.(2.)
Similar to (1.)

Questions:
1.Is my proof right?
2.Any other ways to prove this?
Thanks:)

REPLY [2 votes]: Here's a "hint". Pick any $x_1, x_2 \in (a,b)$ such that $x_1 < x_2$. By the mean value theorem, there's a $c \in (x_1, x_2)$ such that
\begin{equation}
f'(c) = \dfrac{f(x_2) - f(x_1)}{x_2 - x_1}
\end{equation}
I'll leave it to you to write and finish up the proof properly.


TITLE: Wave Equation in 2D and 3D
QUESTION [0 upvotes]: I am new to wave mechanics in 2 and 3 dimensions and would like to know: Do you simply add $\partial \psi \over\partial y$ and/or $\partial \psi \over\partial z$ term without the other parts of the equation changing? By changing I mean specifically about the expression for the phase velocity. I get that the phase velocity for electromagnetic waves should always be $1\over\sqrt{\mu_0\varepsilon_0}$, and by isomorphism we should have the phase velocity expressions for other types of waves be the same in 1, 2, and 3 dimensions. Is this the correct way to think about it?

REPLY [0 votes]: The wave equation contains the second derivatives in respect to position. So, you have the Laplace operator applied to the function (your $\psi$ ) that expreses the displacement from equilibrium, in either 2D or 3D. Nothing changes for the derivative in respect to time.


TITLE: The decoupling limit of the linear sigma model
QUESTION [0 upvotes]: I'm reading Schwartz's Quantum field theory and the standard model. I'm stuck at page 566. He says that when we take the decoupling limit, that's keeping $F_{\pi}$ fixed while sending $m$ and $\lambda$ to infinity, projects out the non-linear sigma model. Now, I don't understand what all this means. Doing this will take the coefficients of all terms in the potential to infinity while not affecting the kinetic terms.
So, How did he get:
$$L=(1/2) (\partial_{\mu}\pi)^2$$ after taking the decoupling limit?

REPLY [1 votes]: Substitute $F_\pi$ = $\frac{2m}{\sqrt{\lambda}}$ in (28.12 ) so the second term in between the parenthesis that's multiplied by $1/F^2_\pi$ becomes $\frac{1}{2} (\partial\pi)^2 + other \; terms$. Apply L'Hospital's rule and ignore all the $\infty$'s. Finally the Lagrangian becomes ($decoupled \: \pi$ & $\sigma$ fields):
$\frac{1}{2} (\partial\pi)^2 + \frac{1}{2} (\partial\sigma)^2$


text:
\begin{document}

\maketitle

\noindent {\bf Abstract.}
For a very ample line bundle $L$ on a compact connected complex manifold $X$, with a real structure,
we discuss entanglement properties of certain
sequences of vectors in tensor products of spaces of holomorphic sections of powers of $L$.

\

\noindent {\bf Keywords:} compact K\"ahler manifold, holomorphic hermitian line bundle, Lagrangian submanifold,
entropy of entanglement.

\section{Introduction}

Let $L\to X$ be a very ample holomorphic hermitian line bundle on a compact connected complex manifold $X$.
Paper \cite{borthwick:95} discusses a certain sequence $(u_k)$  of holomorphic sections of $L^{\otimes k}$
associated to a Legendrian submanifold of the unit circle bundle $Z$ in $L^*$
(projection of this submanifold to $X$ is a Lagrangian submanifold of $X$ satisfying a Bohr-Sommerfeld
condition). The K\"ahler form on $X$ and the contact form on $Z$ are obtained from the hermitian metric on $L$.
It is customary to interpret $k$ as $\frac{1}{\hbar}$.

Associating semiclassical states to Lagrangian submanifolds of a symplectic manifold is an idea that arises
in many contexts.
A partial list of references where various versions of this appear
 includes \cite{bates:97, borthwick:95, burns:10, deber:06, gorod:01, jeffrey:92,
paoletti:08}
and three papers (co-)authored by the first author of this paper: \cite{alluhaibi:16, foth:02, foth:08}.

Similar ideas are applicable to isotropic submanifolds \cite{guillemin:15}.

When the states are associated to submanifolds of a product manifold, we address
the question of entanglement, that comes from quantum information theory. We work with pure states.
Entanglement measures,
such as the entropy of entanglement, are defined for vectors in a tensor product of two or more
Hilbert spaces. In our setting the Hilbert space comes from K\"ahler quantization. We find that when the Lagrangian
submanifold $\Lambda$ of $X\times X$ is $X$, embedded antidiagonally (for this $X$ needs to have a real structure),
the sequence of normalized vectors in spaces of holomorphic sections of the line bundles, which is associated
to $\Lambda$, is a sequence of maximally entangled vectors (Theorem \ref{entropyth}).
We start with a very ample line bundle $L\to X$ on a compact connected complex manifold $X$, with a real structure.
Presence of the real structure makes $\Lambda\cong X$ a Bohr-Sommerfeld submanifold of $X\times X$. We construct a sequence
$\{ u_{\Lambda}^{(k)}\}$, $k=1,2,3,...$ (a sequence of holomorphic sections of
powers of the quantum line bundle on $X\times X$) by
applying a standard procedure of associating states to a Bohr-Sommerfeld Lagrangian submanifold.
Since the space of holomorphic sections of the $k$-th power of the quantum line bundle on $X\times X$ is isomorphic to
$H^0(X,L^{\otimes k})\otimes H^0(X,L^{\otimes k})$ (Remark \ref{remisom}), it is natural to inquire about the entanglement properties
of $u_{\Lambda}^{(k)}$.
We show that the entropy of entanglement
$\nu (\frac{u_{\Lambda}^{(k)}}{||u_{\Lambda}^{(k)}||})=\ln {\mbox{dim}} H^0(X,L^{\otimes k})$,
which means that $\{ \frac{1}{||u_{\Lambda}^{(k)}||}u_{\Lambda}^{(k)} \}$ is a sequence of maximally entangled vectors.
For this sequence we also describe
how the entropy of entanglement is related to the minimum distance to a separable vector (Corollary \ref{cormindist}).
As an example, when $X=\Prj ^1$, with the Fubini-Study form,
the entropy of entanglement of the $k$-th vector is $\ln(k+1)$.This is in Section \ref{sphere}. In Section \ref{torus} we consider
the case when $X$ is a $2$-dimensional torus.

We point out that a similar line of thought can be applied to a submanifold $\Lambda'$ of $\Lambda$.
As an example, we calculated the entropy of entanglement
for the sequence of sections associated to a circle in $\Prj ^1$. These vectors are entangled but not maximally
entangled, for each fixed $k$ or as $k\to\infty$. This calculation is in Section \ref{sphere}.

Real structures appear often in mathematical physics. See \cite{doran:15} as one among many papers on string theory
where real structures play a substantial role, also see \cite[Chapter 2]{manin:88} for a discussion in the context of field theory.
In K\"ahler quantization, which is relevant to this paper, existence of a real structure
on the K\"ahler manifold (in the sense of the definition in Section \ref{sectantidiag}) can be translated,
intuitively, into having quantum states that are their own antiparticles, analogously to Majorana fermions.

There is extensive literature on representation-theoretic aspects of entanglement.
We note papers
\cite{huckleberry:13} and \cite{sawicki:11} where the point of view taken is that on the tensor product
of two Hilbert spaces (or of finitely many Hilbert spaces)
entanglement
should be studied
on the orbits of the action of the unitary group.

In this paper we consider the questions about separability and entanglement combined with concepts
from K\"ahler quantization.

\section{Entropy of entanglement for sequences of sections of line bundles}

\subsection{Preliminaries}
Let $(X,\omega )$ be a compact connected $n$-dimensional K\"ahler manifold ($n\ge 1$).
Assume that $\frac{\omega}{2\pi}$ is an integral K\"ahler form.
There is a  holomorphic hermitian line bundle $L\to X$ such that the curvature of the Chern connection
is equal to $-i \omega$. For a positive integer $k$
denote $\Hk =H^0(X,L^{\otimes k})$ (the space of holomorphic sections of $L^{\otimes k}$).
It is a finite-dimensional Hilbert space and  $d_k={\mbox{dim}} \ \Hk \sim {\mathrm{vol}} (X) k^n+O(k^{n-1})$,
as $k\to\infty$, with ${\mathrm{vol}} (X)=\int\limits_X\frac{c_1(L)^n}{n!}$
(this is a standard fact that can be found, for example, in \cite[Chapter 4]{ma:07}).
The inner product $h_{L_x}(.,.)$ on $L_x$, $x\in X$, induces the inner product $h_{L^*_x}(.,.)$ on $L^*_x$,
the inner product  $h_{L_x\otimes L_x}(.,.)$ on $L_x\otimes L_x$ and so on.
Every element $\xi$ of $L^*_x$ can be represented as $h_{L_x}(.,v)$ for some $v\in L_x$, with $v$ and $\xi$ of the same norm.
If $\eta\in L^*_x$ is $h_{L_x}(.,w)$ for $w\in L_x$, then $h_{L_x^*}(\xi,\eta)=h_{L_x}(w,v)$.
For $v,w,v',w'\in L_x$ $h_{L_x\otimes L_x}(v\otimes v',w\otimes w')=h_{L_x}(v,w)h_{L_x}(v',w')$.
The Hermitian metric
on $L$ provides an inner product on $\Hk$. Let $dV=\frac{\omega^n}{n!}$.
For $s_1,s_2\in \Hk$
$$
\langle s_1,s_2\rangle =\int\limits_X h_{L_x^{\otimes k}}(s_1(x),s_2(x))dV(x).
$$
$\Hk\otimes \Hk$ is a tensor product of Hilbert spaces,
with the inner product $\langle u\otimes u', w\otimes w'\rangle = \langle u, w\rangle\langle u',w'\rangle$
for $u,u',w,w'\in \Hk$ and extended by linearity.
\begin{definition}
A vector $v\in \Hk \otimes \Hk$ is called {\it separable} or {\it decomposable} if it is of the form
$v_1\otimes v_2$, $v_1, v_2\in \Hk$.
\end{definition}
Let $\varphi_1^{(k)}$,...,$\varphi_{d_k}^{(k)}$ be an orthonormal basis in $\Hk$. Then
$\varphi_j^{(k)}\otimes \varphi_l^{(k)}$, $j,l=1,...,d_k$, is an orthonormal basis in
$\Hk\otimes\Hk$.
Let $\pi_j$, $j=1,2$, denote the projections $X\times X\to X$ onto the first and second factor respectively.
The line bundle $L$ is ample and for a sufficiently large $k_0\in \N$ the line bundle
$L^{\otimes k}$ is
very ample for $k\ge k_0$.
We shall assume that $\omega$ is such that $L$ is very ample (this can be achieved by replacing
$\omega $ by   $\omega'=k_0\omega$).
\begin{remark}
\label{remisom}
$H^0(X\times X,\pi_1^*L^{\otimes k}\otimes \pi_2^*L^{\otimes k})\cong \Hk\otimes \Hk$ for all $k$ (see a proof in \cite{barron:17}).
\end{remark}
\begin{definition}
The {\it entropy of entanglement} for a vector $v$
in $\Hk\otimes \Hk$, of norm $1$,
is
$$
\nu(v)=-\sum\limits_{j=1}^{d_k}\lambda_j\ln \lambda_j,
$$
where  $\lambda_1$,...,$\lambda_{d_k}$
are the eigenvalues of $\tr_2(v\bar{v}^T)$, and $\tr _2: \ndom (\Hk \otimes \Hk) \to \ndom (\Hk )$ is defined
by $\tr _2(A\otimes B ) =\tr (B )A$ for $A,B \in \ndom (\Hk)$ and extended by linearity.
The convention $0\ln 0 =0$ is used.
\end{definition}
In the definition above
the minimum value of $\nu(v)$ is zero, and   $\nu(v)=0$
if and only if $v$ is separable.
The maximum value of $\nu(v)$ is $\ln d_k$. It is not hard to see that $\nu(v)=\ln d_k$ if and only if
$\lambda_1=...=\lambda_{d_k}=\frac{1}{d_k}$.
\begin{definition}
A vector $v\in \Hk\otimes \Hk$, of norm $1$,
is called {\it maximally entangled} if $\nu(v)=\ln d_k$.
\end{definition}
The value of $\nu(v)$ characterizes how entangled  (or how "not separable")
$v$ is.
\begin{definition}
The {\it Schmidt decomposition} for $v\in \Hk\otimes \Hk$ is a representation of $v$ as
$v=\sum\limits_{j=1}^{d_k}\alpha_j^{(k)}\psi_j^{(k)}\otimes\chi_j^{(k)}$, where $\{ \psi_j^{(k)} \}$ is an orthonormal basis
of $\Hk$, $\{ \chi_j^{(k)} \}$ is another orthonormal basis
in  $\Hk$, and $\alpha_1^{(k)}\ge ...\ge \alpha_{d_k}^{(k)}\ge 0$ are real numbers.
\end{definition}
Such a representation
always exists and
$$
\tr _2(v\bar{v}^T)=\begin{pmatrix} (\alpha_1^{(k)})^2 & & & \\
& (\alpha_2^{(k)})^2 & & \\
 & &  ... & \\
 & & & (\alpha_{d_k}^{(k)})^2\end{pmatrix}.
$$
All this is standard material,  - a reference is, for example, \cite[Sections 9.2, 15.3]{bengt:06}.
Note that if $v$ is of norm $1$ then, of course, $\sum\limits_{j=1}^{d_k}(\alpha_{j}^{(k)})^2=1$.

For $u,v\in \Hk\otimes \Hk$ denote by $D(v,u)$ the distance $D(v,u)=\sqrt{ \langle v-u,v-u\rangle}$.
\begin{lemma}
\label{closestsepv}
For $v\in \Hk\otimes \Hk$ the minimum value of $D(v,u)$, for $u$ separable, is $\sqrt{\sum\limits_{j=2}^{d_k}(\alpha_j^{(k)})^2}=D(v,u_s)$, where
$u_s=\alpha_1^{(k)} \psi_1^{(k)}\otimes\chi_1^{(k)}$.
\end{lemma}
\begin{remark}
Lemma \ref{closestsepv} seems to be known (see e.g. \cite{lockhart:02}). However, we have not seen the statement
in this exact form, and we provide a proof.
\end{remark}
\noindent {\bf Proof.}
Let $v=\sum\limits_{j=1}^{d_k}\alpha_j^{(k)}\psi_j^{(k)}\otimes\chi_j^{(k)}$ be the Schmidt decomposition for $v$
and let $u_1=\sum\limits_{j=1}^{d_k}a_j\psi_j^{(k)}$, $u_2=\sum\limits_{l=1}^{d_k}b_l\chi_l^{(k)}$ be two vectors in $\Hk$.
$$
D(v,u_1\otimes u_2)^2=\langle v-u_1\otimes u_2,v-u_1\otimes u_2\rangle =
\sum\limits_{j=1}^{d_k}(\alpha_j^{(k)})^2+\sum\limits_{j=1}^{d_k}\sum\limits_{l=1}^{d_k}|a_jb_l|^2-
\sum\limits_{j=1}^{d_k}\alpha_j^{(k)}(a_jb_j+\bar{a}_j\bar{b}_j),
$$
$$
D(v,u_s)^2=\sum\limits_{j=2}^{d_k}(\alpha_j^{(k)})^2,
$$
and it remains to show:
$$
D(v,u_1\otimes u_2)^2\ge D(v,u_s)^2
$$
or, equivalently,
$$
(\alpha_1^{(k)})^2+\sum\limits_{j=1}^{d_k}\sum\limits_{l=1}^{d_k}|a_jb_l|^2-
\sum\limits_{j=1}^{d_k}\alpha_j^{(k)}(a_jb_j+\bar{a}_j\bar{b}_j)\ge 0.
$$
Using that $\alpha_1^{(k)}\ge \alpha_j^{(k)}\ge 0$ and the Cauchy-Schwarz inequality, we get:
$$
(\alpha_1^{(k)})^2+\sum\limits_{j=1}^{d_k}|a_j|^2\sum\limits_{l=1}^{d_k}|b_l|^2-
\sum\limits_{j=1}^{d_k}\alpha_j^{(k)}(a_jb_j+\bar{a}_j\bar{b}_j)
$$
$$
\ge (\alpha_1^{(k)})^2+(a_1b_1+...+a_{d_k}b_{d_k})(\bar{a}_1\bar{b}_1+...+\bar{a}_{d_k}\bar{b}_{d_k})-
\sum\limits_{j=1}^{d_k}\alpha_1^{(k)}(a_jb_j+\bar{a}_j\bar{b}_j)
$$
$$
=|\alpha_1^{(k)}-(a_1b_1+...+a_{d_k}b_{d_k})|^2\ge 0.
$$
$\Box$
\begin{corollary}
\label{cormaxent}
If $v$ is maximally entangled then
$$
D(v,u_s^{(k)})=\sqrt{1-e^{-\nu(v)}}
$$
\end{corollary}
\noindent {\bf Proof.}
If $v$ is maximally entangled then
$D(v,u_s^{(k)})=\sqrt{\frac{d_k-1}{d_k}}$, and the statement follows.
$\Box$

\begin{definition}
The {\it coherent vector} $u_\xi^{(k)}\in \Hk$ (for $x\in X$, $0\ne \xi\in L_x^*$)
is the unique element of $\Hk$ with the property
\begin{equation}
\label{defcohpoint}
\langle s,u_\xi^{(k)}\rangle=\xi^{\otimes k}(s(x))
\end{equation}
for any $s\in\Hk$.
The {\it coherent state} associated to $u_\xi^{(k)}$ is $\pi (u_\xi^{(k)})$, where
$\pi :\Hk -\{ 0\}\to \Prj(\Hk)$ is the projectivization map.
\end{definition}
\begin{remark}
As it is clear from the equality (\ref{defcohpoint}),
multiplying $\xi$ by a complex number $\alpha\ne 0$ leads to changing   $u_\xi^{(k)}$ by a factor of $\bar{\alpha}^k$.
\end{remark}
This kind of coherent states originates, in somewhat different ways, from work of
Berezin and of Rawnsley   \cite{schlich:10}, \cite{rawnsley:77}.

We observe that $u_\xi^{(k)}=\sum\limits_{j=1}^{d_k}\overline{\xi^{\otimes k}(\varphi_j^{(k)}(x))}\varphi_j^{(k)}$
(because, if we represent this vector as $u_\xi^{(k)}=\sum\limits_{j=1}^{d_k} c_j\varphi_j^{(k)}$, where $c_j$ are  coefficients, then
for each $l\in \{1,...,d_k\}$
$\langle \varphi_l^{(k)},u_{\xi}^{(k)}\rangle = \bar{c}_l=
\xi^{\otimes k}(\varphi_l^{(k)}(x))$).

For $x,y\in X$, $0\ne \xi\in L_x^*$, $0\ne \eta\in L_y^*$ we will denote by
$u_{\xi,\eta}^{(k)}\in \Hk\otimes \Hk$ the element of $\Hk\otimes \Hk$
with the property
$$
\langle w_1\otimes w_2 ,u_{\xi,\eta}^{(k)}\rangle=
\xi^{\otimes k}(w_1(x))\eta^{\otimes k}(w_2 (y))
$$
for all $w_1,w_2\in\Hk$.
It is not hard to see that
$$
u_{\xi,\eta}^{(k)}=u_\xi^{(k)}\otimes u_\eta^{(k)}=\sum\limits_{j=1}^{d_k}\sum\limits_{l=1}^{d_k}
\overline{\xi^{\otimes k}(\varphi_j^{(k)}(x))\eta^{\otimes k}(\varphi_l^{(k)}(y))}
\varphi_j^{(k)}\otimes \varphi_l^{(k)}.
$$
We will consider $X\times X$ with the symplectic form $\Omega=\pi_1^*\omega +\pi_2^*\omega$.
Denote the Chern connection in ${\mathcal{L}}=\pi_1^*L\otimes \pi_2^*L \to X\times X$ by $\nabla$
(note that its curvature is $-i\Omega$).

Suppose  $\Lambda$ is a Lagrangian submanifold of $X\times X$ satisfying the Bohr-Sommerfeld condition
(i.e. $\nabla$ over $\Lambda$ is trivial \cite{burns:10, eliashberg:95}).
\begin{remark}
Equivalently, the Bohr-Sommerfeld condition may be stated as a requirement that ${\mathcal{L}}\Bigr | _{\Lambda}$
has a nonzero covariant constant section \cite{burns:10, jeffrey:92}.
\end{remark}
Let $\nabla^*$ be the hermitian connection in ${\mathcal{L}} ^*$ induced by $\nabla$.
Let $\tau$ be a nonvanishing covariant constant section of ${\mathcal{L}}^*\Bigr | _{\Lambda}$.
For $(x,y)\in \Lambda$ we will denote by
$u_{\tau (x,y)}^{(k)}$ the  element of $\Hk\otimes \Hk$
with the property
$$
\langle w_1\otimes w_2,u_{\tau (x,y)}^{(k)}\rangle=
\tau(x,y)^{\otimes k}(w_1(x)\otimes w_2(y))
$$
for all $w_1,w_2\in\Hk$.
We observe that
$$
u_{\tau (x,y)}^{(k)}=\sum\limits_{j=1}^{d_k}\sum\limits_{l=1}^{d_k}\overline{\tau(x,y)^{\otimes k}(\varphi_j^{(k)}(x)\otimes \varphi_l^{(k)}(y))}\varphi_j^{(k)}\otimes \varphi_l^{(k)}.
$$
Let $d\mu$ be an $n$-form on $\Lambda$, with $\int\limits_{\Lambda} d\mu >0$.
We will denote by $u_{\Lambda}^{(k)}$ the element of $\Hk \otimes \Hk$ with the property
$$
\langle w_1\otimes w_2,u_{\Lambda}^{(k)}\rangle = \int\limits_{\Lambda}
\tau(x,y)^{\otimes k}( w_1(x)\otimes w_2(y)) \ d\mu (x,y)
$$
for all $w_1,w_2\in \Hk$.  Clearly
$$
u_{\Lambda}^{(k)}=\sum\limits_{j=1}^{d_k}\sum\limits_{l=1}^{d_k}
\varphi_j^{(k)}\otimes \varphi_l^{(k)} \int\limits_{\Lambda}
\overline{\tau(x,y)^{\otimes k} (\varphi_j^{(k)}(x)\otimes \varphi_l^{(k)}(y))}d\mu(x,y).
$$

\subsection{$X$ antidiagonally embedded into $X\times X$}
\label{sectantidiag}
Now suppose $X$ has a real structure (i.e.
an antiholomorphic involution $\sigma:X\to X$). We require that, moreover, $\sigma^*\omega =-\omega$
and
\begin{equation}
\label{sigmareq}
h_{L_{\sigma(x)}^{\otimes k}}(s_1(\sigma(x)),s_2(\sigma(x)))=\overline{h_{L_x^{\otimes k}}(s_1(x),s_2(x))}
\end{equation}
for all $x\in X$, $k\in \N$, $s_1,s_2 \in\Hk$. This implies that
$\sigma$ extends to
a norm-preserving bundle map $L^*\to L^*$ that is antilinear on the fibers,
for any $x\in X$, $\sigma$  identifies $L_{\sigma(x)}^*$ with $\overline{L_x^*}\cong L_x$, and so
$\sigma^*L\cong L^*$.

Define $\Lambda=\{ (x,y)\in X\times X| \ y=\sigma(x)\}$ and
$\iota:X\overset{\cong}{\to} \Lambda$, $x\mapsto (x,\sigma(x))$.
We have: $\iota^*\pi_1^*L\cong L$, $\iota^*\pi_2^*L\cong L^*$,
$\nabla$ restricted to  ${\mathcal{L}}\Bigr |_{\Lambda}$
is trivial. To show that $\nabla$ is trivial over $\Lambda$,
choose a local holomorphic frame $e(x)$ for $L$ and observe that
over $\Lambda \cong X$ the connection form for $\nabla$ is
$$
\iota^* \partial \log \Bigl ( h_{L_x}(e(x),e(x))h_{L_{\sigma(x)}}(e(\sigma(x)), e(\sigma(x)) ) \Bigr ) \Bigr | _{\Lambda}
$$
$$
=
  \partial \log h_{L_x}(e(x),e(x))+\bar{\partial }\log  h_{L_x}(e(x),e(x))=d \log  h_{L_x}(e(x),e(x)).
$$
Since the connection form is exact, the connection is trivial.

Let $\tau$ be a (nonvanishing, covariant constant) section of  ${\mathcal{L}}^*\Bigr |_{\Lambda}$ defined by
\begin{equation}
\label{deftau}
\tau(x,\sigma(x))(s_1(x)\otimes s_2(\sigma(x)))=
\frac{1}{h_{L_x}(e(x),e(x))}h_{L_x}(s_1(x),e(x))h_{L_{\sigma(x)}}(s_2(\sigma(x)),e(\sigma(x))),
\end{equation}
for holomorphic sections $s_1,s_2$ of $L$,
where $e$ is a holomorphic section of $L$, locally nonvanishing. It is clear that $\tau$ is a well defined global section
of ${\mathcal{L}}^*\Bigr |_{\Lambda}\to\Lambda$ (it does not depend on the choice of $e$, because replacing $e(x)$
by $f(x)e(x)$, where $f(x)$ is a local non-vanishing holomorphic function, does not change the right hand side of (\ref{deftau})
due to (\ref{sigmareq})). Now let us explain why $\tau$ is covariant constant.
We have already seen that with respect to a local holomorphic frame $e(x)$ in $L$ the connection form in $\iota^*({\mathcal{L}}\Bigr |_{\Lambda})$
is $d \log  h_{L_x}(e(x),e(x))$. Therefore the connection form in $\iota^*({\mathcal{L}} ^*\Bigr |_{\Lambda})$
is $-d \log  h_{L_x}(e(x),e(x))$.
We get, using (\ref{sigmareq}) and (\ref{deftau}):
$$
\nabla^* \tau (x,\sigma(x))(e(x)\otimes e(\sigma(x))=d\Bigl ( \tau (x,\sigma(x))(e(x)\otimes e(\sigma(x))) \Bigr )
$$
$$
-
\tau (x,\sigma(x))(e(x)\otimes e(\sigma(x)))d \log  h_{L_x}(e(x),e(x))
$$
$$
=d h_{L_x}(e(x),e(x))- h_{L_x}(e(x),e(x))d \log  h_{L_x}(e(x),e(x))=0.
$$
Let $d\mu=\pi_1^* dV$.
For an $m$-dimensional submanifold $\Lambda'$ of $\Lambda$, with an $m$-form $d\mu'$
we have:
$$
u_{\Lambda'}^{(k)}=\sum\limits_{j=1}^{d_k}\sum\limits_{l=1}^{d_k}
\varphi_j^{(k)}\otimes \varphi_l^{(k)} \int\limits_{\Lambda'}
\overline{\tau(x,\sigma(x))^{\otimes k} (\varphi_j^{(k)}(x)\otimes \varphi_l^{(k)}(\sigma(x)))}d\mu'(x,\sigma(x)).
$$
When $\Lambda'=\Lambda$ and $d\mu'=d\mu$, then
$$
u_{\Lambda}^{(k)}=\sum\limits_{j=1}^{d_k}\sum\limits_{l=1}^{d_k}
\varphi_j^{(k)}\otimes \varphi_l^{(k)} \int\limits_{X}
\overline{\tau(x,\sigma(x))^{\otimes k} (\varphi_j^{(k)}(x)\otimes \varphi_l^{(k)}(\sigma(x)))}dV(x).
$$
\begin{remark}
Our way of associating semiclassical states to Bohr-Sommerfeld Lagrangian submanifolds
is the same as the one, for example, in \cite{burns:10}. Instead, one could consider $\Lambda$ as
$p(\tilde{\Lambda})$, where
$p:P\to X\times X$ is the unit circle bundle in ${\mathcal{L}}^*\to X\times X $
(a principal $U(1)$-bundle, also
a contact manifold),
and  $\tilde{\Lambda}$ is a Legendrian submanifold of $P$ \cite{borthwick:95}, \cite{eliashberg:95}.
Specifically, in Section \ref{sectantidiag} $\tilde{\Lambda}=\{ (x,\sigma(x),\tau(x,\sigma(x)))|\ x\in X\}$.
\end{remark}
\begin{remark}
It would be interesting to interpret either the entropy of entanglement or the minimum distance
to a separable vector as a number that characterizes how far $\Lambda$ is from being a product submanifold.
\end{remark}
\begin{theorem}
\label{entropyth}
$$
\nu \Bigl ( \frac{u_{\Lambda}^{(k)}}{||u_{\Lambda}^{(k)}||}\Bigr ) = \ln d_k
$$
\end{theorem}
\noindent {\bf Proof.}
Let's write $u_{\Lambda}^{(k)}$ as a vector in the basis $\varphi_j^{(k)}\otimes \varphi_l^{(k)}$ and let's find
$v_k=\frac{1}{||u_{\Lambda}^{(k)}||}u_{\Lambda}^{(k)}$ and $\rho_k=v_k\bar{v}_k^T$. We have:
$$
u_{\Lambda}^{(k)}=\sum\limits_{j=1}^{d_k}\sum\limits_{l=1}^{d_k}
\varphi_j^{(k)}\otimes \varphi_l^{(k)} \int\limits_{X}
\overline{\tau(x,\sigma(x))^{\otimes k} (\varphi_j^{(k)}(x)\otimes \varphi_l^{(k)}(\sigma(x)))}dV(x),
$$
$$
\tau(x,\sigma(x))^{\otimes k}(\varphi_j^{(k)}(x)\otimes \varphi_l^{(k)}(\sigma(x) ))
$$
$$
=\frac{1}{h_{L_x^{\otimes k}}(\varphi^{\otimes k}(x),\varphi^{\otimes k}(x))}h_{L_x^{\otimes k}}(\varphi_j^{(k)}(x),\varphi^{\otimes k}(x))
h_{L_{\sigma(x)}^{\otimes k}} ( \varphi_l^{(k)}(\sigma(x)),\varphi^{\otimes k}(\sigma(x))),
$$
where $\varphi$ is a holomorphic section of $L$ (locally nonvanishing). Because of (\ref{sigmareq})
$$
\tau(x,\sigma(x))^{\otimes k}(\varphi_j^{(k)}(x)\otimes \varphi_l^{(k)}(\sigma(x) ))
$$
$$
=\frac{1}{h_{L_x^{\otimes k}}(\varphi^{\otimes k}(x),\varphi^{\otimes k}(x))}h_{L_x^{\otimes k}}(\varphi_j^{(k)}(x),\varphi^{\otimes k}(x))
\overline{h_{L_x^{\otimes k}}(\varphi_l^{(k)}(x),\varphi^{\otimes k}(x))}=
h_{L_x^{\otimes k}}
(\varphi_j^{(k)}(x),\varphi_l^{(k)}(x)).
$$
The last equality can be obtained by setting $\varphi_j^{(k)}(x)=a(x)\varphi^{\otimes k}(x)$, $\varphi_l^{(k)}(x)=b(x)\varphi^{\otimes k}(x)$.
Hence
$$
u_{\Lambda}^{(k)}=\sum\limits_{j=1}^{d_k}\sum\limits_{l=1}^{d_k}
\varphi_j^{(k)}\otimes \varphi_l^{(k)} \langle \varphi_l^{(k)}, \varphi_j^{(k)}\rangle =
\sum\limits_{j=1}^{d_k}\varphi_j^{(k)}\otimes \varphi_j^{(k)}.
$$
Therefore in the basis $\varphi_1^{(k)}\otimes \varphi_1^{(k)}$,
$\varphi_1^{(k)}\otimes \varphi_2^{(k)}$,...,$\varphi_1^{(k)}\otimes \varphi_{d_k}^{(k)}$,
$\varphi_2^{(k)}\otimes \varphi_1^{(k)}$,
$\varphi_2^{(k)}\otimes \varphi_2^{(k)}$,...,$\varphi_2^{(k)}\otimes \varphi_{d_k}^{(k)}$,...,
$\varphi_{d_k}^{(k)}\otimes \varphi_1^{(k)}$,$\varphi_{d_k}^{(k)}\otimes \varphi_2^{(k)}$,...,
$\varphi_{d_k}^{(k)}\otimes \varphi_{d_k}^{(k)}$ in $\Hk\otimes \Hk$ the vector
$v_k$ is $\frac{1}{\sqrt{d_k}}\begin{pmatrix} e_1 \\ ... \\  e_{d_k}\end{pmatrix}$, where
$\begin{pmatrix} e_1 &  ... &  e_{d_k}\end{pmatrix}=I_{d_k}$ is the $d_k\times d_k$
identity matrix. Then $\rho_k$ is
$ \frac{1}{d_k} \sum\limits_{j=1}^{d_k}\sum\limits_{l=1}^{d_k}E_{jl}\otimes E_{jl}$, where
$E_{jl}$ is the $d_k\times d_k$ matrix with the $jl$-th entry $1$ and the other entries equal to zero.
From this we get: $\tr _2(\rho_k)= \frac{1}{d_k}\sum\limits_{j=1}^{d_k}E_{jj}=\frac{1}{d_k}I_{d_k}$.
It follows that $v_k$ is maximally entangled and $\nu (v_k)=\ln d_k$.
$\Box$.

Since the vectors $v_k=\frac{1}{||u_{\Lambda}^{(k)}||}u_{\Lambda}^{(k)}$ in Theorem \ref{entropyth} are maximally
entangled, Corollary \ref{cormaxent} applies:
\begin{corollary}
\label{cormindist}
The minimum distance from $v_k$ to a separable vector is  $\sqrt{\frac{d_k-1}{d_k}}=\sqrt{1-e^{-\nu(v_k)}}$.
\end{corollary}

\section{Calculations on the sphere and on the torus}

\subsection{}
\label{sphere}
Suppose $X=\Prj^n$ ($n\ge 1$),
with the Fubini-Study K\"ahler form, $L$ is the hyperplane bundle, with the standard choice
of a Hermitian metric. Denote by $\zeta_0,..., \zeta_n$ the homogeneous coordinates on $X$.
The space $\Hk$ is usually identified with
the space of homogeneous polynomials of degree $k$ in $\zeta_0,..., \zeta_n$.
The antiholomorphic involution $X\to X$,
$\sigma: \zeta=[\zeta_0:...:\zeta_n]\mapsto \sigma(\zeta)=[\bar{\zeta}_0:...:\bar{\zeta}_n]$, provides
an antilinear map that sends the fiber $L^*_{\zeta}$ of the tautological bundle
(i.e. the line $l$ through $0$ in $\C^{n+1}$ spanned by $\begin{pmatrix} \zeta_0 \\ ...\\ \zeta_n\end{pmatrix} $)
to $\bar{l}=\{ w\in \C^{n+1}| \ \bar{w}\in l\}$.
Theorem \ref{entropyth} says that for the Lagrangian states associated to $\Prj^n$ antidiagonally embedded
into  $\Prj^n\times \Prj^n$ the entropy of entanglement is $\ln {\mbox{dim}} \Hk =\ln \binom{n+k}{n}$.

Everything can be written very explicitly for $n=1$. On $\Prj ^1$ we use the homogeneous coordinates $[\zeta_0:\zeta_1]$.
The space $\Hk$ of holomorphic sections of $L^{\otimes k}$, $k\in\N$,  is  the space of homogeneous degree $k$ polynomials
in $\zeta_0$, $\zeta_1$.
On the affine chart $\U= \{ [\zeta_0:\zeta_1]\in \Prj ^1 | \zeta_1\ne 0\}\cong \C$ homogeneous degree $k$ polynomials
in $\zeta_0$, $\zeta_1$ become polynomials in $z=\zeta_0/\zeta_1$ of degree $\le k$, and
$\Hk$
is identified with the space of polynomials in complex variable $z$ of degree $\le k$, with the inner product
$$
\langle \phi,\psi\rangle = \frac{i}{2\pi }\int\limits_{\C }\frac{\phi(z)\overline{\psi(z)}}{(1+|z|^2)^{k+2}}dzd\bar{z}
$$
(\cite{marche:15} 4.1.1. or \cite{bloch:03} 3.1). $\Prj^1\cong \U \cup \{ [1:0]\}$.
An orthonormal basis in $\Hk$ is
$$
\varphi_j^{(k)}=\sqrt{\frac{(k+1)!}{j!(k-j)!}}z^j, \ j=0,...,k
$$
(this is easily verified by a calculation in polar coordinates on $\R^2$). Note that
$\overline{\varphi_j^{(k)}(\sigma(z))}=\overline{\varphi_j^{(k)}(\bar{z})}=\varphi_j^{(k)}(z)$.
Denote ${\mathbf{z}}=[z:1]\in \U$.
The Hermitian metric on $L^*$ is given by
$$
h_{L_{\mathbf{z}}^*}\Bigl (\begin{pmatrix} z \\  1\end{pmatrix}, \begin{pmatrix} z \\  1\end{pmatrix} \Bigr ) =1+|z|^2,
$$
which implies that for $f,g\in\Hk$
$$
h_{L_{\mathbf{z}}^{\otimes k}}(f(z),g(z))=\frac{f(z)\overline{g(z)}}{(1+|z|^2)^k}.
$$
In particular,
$$
h_{L_{\mathbf{z}}^{\otimes k}}(\varphi_j^{(k)},\varphi_l^{(k)})=
\frac{(k+1)!}{\sqrt{j!(k-j)!l!(k-l)!}}\frac{z^j\bar{z}^l}{(1+|z|^2)^k}.
$$
The section $\tau$ is $\tau(z,\bar{z})=\frac{1}{1+|z|^2} \begin{pmatrix} z \\  1\end{pmatrix} \otimes
\begin{pmatrix} \bar{z} \\  1\end{pmatrix}$. We have:
$$
\tau(z,\bar{z})^{\otimes k} (\varphi_j^{(k)}(z)\otimes \varphi_l^{(k)}(\sigma(z))=
\frac{1}{(1+|z|^2)^k}\varphi_j^{(k)}(z) \varphi_l^{(k)}(\bar{z})=
h_{L_{\mathbf{z}}^{\otimes k}}(\varphi_j^{(k)}(z),\varphi_l^{(k)}(z)).
$$
$\tau$ is covariant constant, as it is shown in Section \ref{sectantidiag}.
$$
u_{\tau ({\mathbf{z}},{\mathbf{\bar{z}}})}^{(k)}( {\mathbf{w}},{\mathbf{\bar{w}}})
=\sum\limits_{j=0}^{k}\sum\limits_{l=0}^{k} \frac{
\overline{\varphi_j^{(k)}(z) \varphi_l^{(k)}(\bar{z})} }
{(1+|z|^2)^k}
\varphi_j^{(k)}(w)\otimes \varphi_l^{(k)}(\bar{w}),
$$
and
$$
u_{\Lambda}^{(k)}=\sum\limits_{j=0}^{k}\sum\limits_{l=0}^{k}
\varphi_j^{(k)}\otimes \varphi_l^{(k)} \int\limits_{\C}
\frac{
\overline{\varphi_j^{(k)}(z) \varphi_l^{(k)}(\bar{z})} }
{(1+|z|^2)^k} \frac{i}{2\pi}\frac{dz d\bar{z}}{(1+|z|^2)^2}
$$
$$
=\sum\limits_{j=0}^{k}\sum\limits_{l=0}^{k}
\varphi_j^{(k)}\otimes \varphi_l^{(k)} \int\limits_{\C}
\frac{\varphi_l^{(k)}(z)\overline{\varphi_j^{(k)}(z)} }{(1+|z|^2)^k }
\frac{i}{2\pi}\frac{dz d\bar{z}}{(1+|z|^2)^2}
$$
$$
=\sum\limits_{j=0}^{k}\sum\limits_{l=0}^{k}
\varphi_j^{(k)}\otimes \varphi_l^{(k)}
\langle \varphi_l^{(k)}, \varphi_j^{(k)}\rangle=
\sum\limits_{j=0}^{k}
\varphi_j^{(k)}\otimes \varphi_j^{(k)}.
$$
Therefore $u_{\Lambda}^{(k)}$ in the basis $\varphi_0^{(k)}\otimes \varphi_0^{(k)}$,
$\varphi_0^{(k)}\otimes \varphi_1^{(k)}$,...,$\varphi_0^{(k)}\otimes \varphi_k^{(k)}$,
$\varphi_1^{(k)}\otimes \varphi_0^{(k)}$,
$\varphi_1^{(k)}\otimes \varphi_1^{(k)}$,...,$\varphi_1^{(k)}\otimes \varphi_k^{(k)}$,...,
$\varphi_k^{(k)}\otimes \varphi_0^{(k)}$,$\varphi_k^{(k)}\otimes \varphi_1^{(k)}$,...,$\varphi_k^{(k)}\otimes \varphi_k^{(k)}$ in $\Hk\otimes \Hk$
is $\begin{pmatrix} e_1 \\ e_2 \\... \\  e_{k+1}\end{pmatrix}$, where
$\begin{pmatrix} e_1 & e_2 & ... &  e_{k+1}\end{pmatrix}=I_{k+1}$ is the $(k+1)\times(k+1)$
identity matrix. We get: $v_k=\frac{1}{||u_{\Lambda}^{(k)}||}u_{\Lambda}^{(k)}=
\frac{1}{\sqrt{k+1}}\begin{pmatrix} e_1 \\ e_2 \\... \\  e_{k+1}\end{pmatrix}$,
$v_k\bar{v}_k^T$ is
$ \frac{1}{k+1} \sum\limits_{j=1}^{k+1}\sum\limits_{l=1}^{k+1}E_{jl}\otimes E_{jl}$,
$\tr _2(v_k\bar{v}_k^T)= \frac{1}{k+1}\sum\limits_{j=1}^{k+1}E_{jj}=\frac{1}{k+1}I_{k+1}$.
It follows that $v_k$ is maximally entangled and $\nu (v_k)=\ln (k+1)$.

The Schmidt decomposition of $v_k$ is $v_k=\sum\limits_{j=1}^{k+1} \frac{1}{\sqrt{k+1}}e_j\otimes e_j$,
and $u_s^{(k)}=\frac{1}{\sqrt{k+1}}e_1\otimes e_1$.
The distance
to a closest separable vector $D(v_k,u_s^{(k)})=\sqrt{\frac{k}{k+1}}=\sqrt{1-e^{-\nu(v_k)}}$.

\begin{remark}
In general, a closest separable vector is not unique.
Here, for example,
$D(v_k, \frac{1}{\sqrt{k+1}}e_2\otimes e_2)=\sqrt{\frac{k}{k+1}}$ too.
\end{remark}

Let us now calculate the entropy of entanglement
for the states associated to $\Lambda'=
\{ ({\mathbf{z}},{\mathbf{\bar{z}}})\in\Lambda \ | \  z=e^{i\Theta}, \ 0\le \Theta < 2\pi\}$ with $d\mu=d\Theta$.
$$
u_{\Lambda '}^{(k)}=\sum\limits_{j=0}^{k}\sum\limits_{l=0}^{k}
\varphi_j^{(k)}\otimes \varphi_l^{(k)} \int\limits_0^{2\pi}
\frac{
\overline{\varphi_j^{(k)}(z) \varphi_l^{(k)}(\bar{z})} }
{(1+|z|^2)^k} \Bigr |_{z=e^{i\Theta}} d\Theta=
\sum\limits_{j=0}^{k} \frac{\pi}{2^{k-1}}\frac{(k+1)!}{j!(k-j)!}
\varphi_j^{(k)}\otimes \varphi_j^{(k)}.
$$
Using the identity $\sum\limits_{j=0}^k\binom{k}{j} ^2=\binom{2k}{k}$, we get:
$$
v_k=\frac{1}{||u_{\Lambda'}^{(k)}||}u_{\Lambda'}^{(k)}=
\frac{k!}{\sqrt{(2k)!}}\begin{pmatrix} \binom{k}{0}e_1 \\ \binom{k}{1}e_2 \\... \\  \binom{k}{k}e_{k+1}\end{pmatrix},
$$
$$
v_k\bar{v}_k^T=
\frac{(k!)^2}{(2k)!} \sum\limits_{j=1}^{k+1}\sum\limits_{l=1}^{k+1}\binom{k}{j-1}\binom{k}{l-1}E_{jl}
\otimes E_{jl},
$$
$$
\tr _2(v_k\bar{v}_k^T)=
\frac{(k!)^2}{(2k)!} \sum\limits_{j=1}^{k+1}\binom{k}{j-1}^2E_{jj}=
\frac{(k!)^2}{(2k)!}\begin{pmatrix} \binom{k}{0} ^2 & & & \\
& \binom{k}{1} ^2 & & \\
 & &  ... & \\
 & & & \binom{k}{k} ^2\end{pmatrix} ,
$$
$$
\nu(v_k)=-\frac{(k!)^2}{(2k)!} \sum\limits_{j=0}^{k}\binom{k}{j} ^2\ln
\Bigl [ \frac{(k!)^2}{(2k)!}\binom{k}{j} ^2\Bigr ] .
$$
We conclude that the vectors $v_k$ are not maximally entangled. Also, they are clearly not separable.

\subsection{}
\label{torus}
Another source of examples of complex manifolds with real structures is
compact Riemann surfaces. In genus $1$ those are
elliptic curves with a real $j$-invariant (biholomorphic to
$\C/\Gamma$ where $\Gamma=\Z+w\Z$ with $w$ either on the boundary of the region
$\{ z\in \C| \ |z|\ge 1, |Re(z)|\le 1/2\}$ or $w=it$ with $t\ge 1$) \cite{alling:81}.
For genus $g\ge 2$ there is extensive literature, - see, for example, an account given in \cite{bujalance:01}.

Let $X$ be the torus $\C/\Gamma$ where $\Gamma=\Z+i\Z$.
Let $\mu\in \R$, $k\in \N$, we will assume $k\ge 3$. Recall that a $k$-th order
theta-function with characteristics $(\mu,0)$ for this lattice
is an entire function $f:\C\to \C$ such that
$$
f(z+m+in)=e^{2\pi i(-\frac{k}{2}(in^2+2nz)+m\mu)}  f(z)
$$
for all $m,n\in\Z$, $z\in\C$, the space $\Theta (k;\mu,0)$ of these theta functions is $k$-dimensional, and the functions
$$
\theta_j^{(k,\mu)}(z)=\sum\limits_{n\in\Z}e^{2\pi i [\frac{1}{2}ki (n+\frac{\mu+j}{k})^2+(n+\frac{\mu+j}{k})kz ]}
 \ , \ j=1,...,k
$$
form  a basis in $\Theta (k;\mu,0)$ \cite{baily:65}.

Let $L_{k,\mu}\to X$
be the hermitian holomorphic line bundle on $X$ such that $H^0(X,L_{k,\mu})\cong \Theta (k;\mu,0)$
under the usual identification between holomorphic sections of a line bundle on a torus
and theta-functions (see e.g. \cite{debarre:05}), with the Hermitian metric determined by the function
$h(z)=e^{\frac{k\pi}{4}(z-\bar{z})^2}$, so that the inner product on $H^0(X,L_{k,\mu})\cong \Theta (k;\mu,0)$ is
$$
\langle f,g\rangle = \int\limits_0^1\int\limits_0^1 f(z)\overline{g(z)}
e^{\frac{k\pi}{2}(z-\bar{z})^2}\ dx \ dy.
$$
A straightforward calculation shows that the basis $\{ \theta_j^{(k,\mu)}(z)\}$
is orthogonal.
Denote $\varphi_j^{(k,\mu)}=\frac{1}{||\theta_j^{(k,\mu)}||}\theta_j^{(k, \mu)}$, $j=1,...,k$.
Take $\sigma:z\mapsto -\bar{z}$ as the antiholomorphic involution. Note that
$\overline{\varphi_j^{(k)}(\sigma(z))}=\varphi_j^{(k)}(z)$.

Define a section $\tau_k$ of
$\iota^*((\pi_1^*L_{k,\mu} \otimes \pi_2^*L_{k,\mu})^*\Bigr | _{\Lambda})\cong L_{k,\mu}^*\otimes L_{k,\mu}$
by
$$
\tau_k(z,-\bar{z})(s_1(z)\otimes s_2(-\bar{z}))=
e^{\frac{k\pi}{2}(z-\bar{z})^2}s_1(z) s_2(-\bar{z})
$$
(note that the expression in the right hand side is $\Gamma$-invariant).
\begin{remark}
In this section we slightly change notations. Previously we defined a covariantly constant section $\tau$ of  ${\mathcal{L}}^*\Bigr | _{\Lambda}$
and used $\tau^{\otimes k}$ for dealing with $L^{\otimes k}$. In this section, because of theta characteristics,
the role of $L^{\otimes k}$ is played by $L_1^{\otimes k}\otimes L_2$ for certain line bundles $L_1$, $L_2$. For this reason
we introduce and use $\tau_k$ instead of $\tau$ and $\tau^{\otimes k}$.
\end{remark}
We have:
$$
\tau_k(z,-\bar{z}) (\varphi_j^{(k,\mu)}(z)\otimes \varphi_l^{(k,\mu)}(\sigma(z)))=
e^{\frac{k\pi}{2}(z-\bar{z})^2}\varphi_j^{(k,\mu)}(z) \varphi_l^{(k,\mu)}(-\bar{z})=
e^{\frac{k\pi}{2}(z-\bar{z})^2}\varphi_j^{(k,\mu)}(z)\overline { \varphi_l^{(k,\mu)}(z)}.
$$
$\tau_k$ is covariant constant (see Section \ref{sectantidiag}).
$$
u_{\Lambda}^{(k)}=\sum\limits_{j=1}^{k}\sum\limits_{l=1}^{k}
\varphi_j^{(k,\mu)}\otimes \varphi_l^{(k,\mu)} \int\limits_0^1\int\limits_0^1
\overline{\tau_k(z,-\bar{z}) (\varphi_j^{(k,\mu )}(z)\otimes \varphi_l^{(k,\mu )}(-\bar{z}))}dxdy
$$
$$
=\sum\limits_{j=1}^{k}\sum\limits_{l=1}^{k}
\varphi_j^{(k,\mu)}\otimes \varphi_l^{(k,\mu)} \int\limits_0^1\int\limits_0^1
\overline{\varphi_j^{(k,\mu)}(z)}\varphi_l^{(k,\mu)}(z)e^{ k\frac{\pi}{2}(z-\bar{z})^2 }dxdy=
\sum\limits_{j=1}^{k}\varphi_j^{(k,\mu)}\otimes \varphi_j^{(k,\mu)},
$$
therefore $u_{\Lambda}^{(k)}$ in the basis  $\varphi_1^{(k,\mu)}\otimes \varphi_1^{(k,\mu)}$,
$\varphi_1^{(k,\mu)}\otimes \varphi_2^{(k,\mu)}$, ..., $\varphi_1^{(k,\mu)}\otimes \varphi_k^{(k,\mu)}$, ... ,
$\varphi_k^{(k,\mu)}\otimes \varphi_1^{(k,\mu)}$, ... ,$\varphi_k^{(k,\mu)}\otimes \varphi_k^{(k,\mu)}$,
in $\Theta (k;\mu,0)\otimes \Theta (k;\mu,0)$ is $\begin{pmatrix} e_1 \\ e_2 \\... \\  e_{k}\end{pmatrix}$,
$$
v_k=\frac{1}{||u_{\Lambda}^{(k)}||}u_{\Lambda}^{(k)}=
\frac{1}{\sqrt{k}}\begin{pmatrix} e_1 \\ e_2 \\... \\  e_{k}\end{pmatrix}
$$
and by a procedure similar to the one used in the proof of Theorem \ref{entropyth} or in Section \ref{sphere},
we get:
$\tr _2(v_k\bar{v}_k^T)= \frac{1}{k}I_{k}$, vectors $v_k$ are maximally entangled, $\nu (v_k)=\ln k$.

The Schmidt decomposition of $v_k$ is $v_k=\sum\limits_{j=1}^{k} \frac{1}{\sqrt{k}}e_j\otimes e_j$,
and $u_s^{(k)}=\frac{1}{\sqrt{k}}e_1\otimes e_1$.
The distance from $v_k$
to a closest separable vector $D(v_k,u_s^{(k)})=\sqrt{\frac{k-1}{k}}=\sqrt{1-e^{-\nu(v_k)}}$.

{\bf Acknowledgments.} We are thankful to N. Johnston, Y. Karshon, J. Rosenberg, A. Uribe, for discussions, and to the referees and the editors
for useful suggestions that helped improve the exposition in the paper.

config: arxiv


TITLE: Mechanics based query pertaining to force and conservation of energy
QUESTION [1 upvotes]: A wedge of mass M lies on a smooth horizontal plane while a block of mass $\:m=M/2\:$ is in contact with the smooth inclined face of the wedge . The inclination of this face to the horizontal is $\:\theta=30^{\rm{o}}\:$. The system is released from rest with the block at a vertical height $\:h\:$ above the horizontal plane. When the block reaches the horizontal plane, find (i) the speed $\:v_{\rm{wedge}}\:$ of the wedge, and (ii) the distance $\:s\:$the wedge has travelled.

So for this question, for question 1,
I know the potential energy of the block is $\:V_{\rm{block}}=mgh\:$ and by the time it reaches the ground it get's converted to kinetic energy ie, $\:T_{\rm{block}}=1/2\,mv_{\rm{block}}^{2}\:$.
Hence velocity of the block would be $\:v_{\rm{block}}=\sqrt{2gh}\:$ and the wedge would obviously be moving at the same speed, How do I go about calculating the distance?

REPLY [1 votes]: 1. Wedge-Block System (no friction)
Since the two bodies execute plane translations we'll make use of the following symbols for the position, velocity and acceleration vectors :
\begin{align}
\mathbf{r}_{\mathrm{w}} & = \left(x_{\mathrm{w}},y_{\mathrm{w}} \right)= \text{position vector of a fixed point of the wedge}
\tag{01a}\\
\mathbf{r}_{\mathrm{b}} & = \left(x_{\mathrm{b}},\,y_{\mathrm{b}} \right)= \text{position vector of a fixed point of the block}
\tag{01b}\\
\mathbf{\dot {r}}_{\mathrm{w}} & = \left(\dot{x}_{\mathrm{w}},\dot{y}_{\mathrm{w}} \right)=\left(\dot{x}_{\mathrm{w}},0\right)= \text{velocity of the wedge}
\tag{01c}\\
\mathbf{\dot {r}}_{\mathrm{b}} & = \left(\dot{x}_{\mathrm{b}},\dot{y}_{\mathrm{b}} \right)=\text{velocity of the block}
\tag{01d}\\
\mathbf{\ddot {r}}_{\mathrm{w}} & = \left(\ddot{x}_{\mathrm{w}},\ddot{y}_{\mathrm{w}} \right)=\left(\ddot{x}_{\mathrm{w}},0\right)= \text{acceleration of the wedge}
\tag{01e}\\
\mathbf{\ddot {r}}_{\mathrm{b}} & = \left(\ddot{x}_{\mathrm{b}},\ddot{y}_{\mathrm{b}} \right)=\text{acceleration of the block}
\tag{01f}\\
\mathbf{\ddot {r}}_{\mathrm{bw}} & =\mathbf{\ddot {r}}_{\mathrm{b}}-\mathbf{\ddot {r}}_{\mathrm{w}}=\text{acceleration of the block relatively to the wedge}
\tag{01g}
\end{align}

For the forces in free-body diagrams, see Figure, we have :
\begin{align}
\mathbf{W} & = \left(0,-Mg\right)= \text{weight of the wedge}
\tag{02a}\\
\mathbf{B} & = \left(0,-mg\right)=  \text{weight of the block}
\tag{02b}\\
\mathbf{N} & = \left(N\sin\theta,N\cos\theta\right)=\text{force on block due to wedge}
\nonumber\\
& =-[\text{force on wedge due to block}]
\tag{02c}\\
\mathbf{T} & = \left(0,T\right)=\text{force on wedge due to table}
\tag{02d}
\end{align}
Note that under the no friction assumption the force $\:\mathbf{N}\:$ is normal to the inclination plane while the force $\:\mathbf{T}\:$ is normal to the table.
The equations of motion for the wedge and the block are respectively :
\begin{align}
\mathbf{T}+\mathbf{W}-\mathbf{N} & = M\mathbf{\ddot {r}}_{\mathrm{w}}
\tag{03a}\\
\mathbf{N}+\mathbf{B} & = m\mathbf{\ddot {r}}_{\mathrm{b}}
\tag{03b}
\end{align}
The $\:x$-component of eq.(03a) and the $\:x,y$-components of eq.(03b) give  the following scalar equations :
\begin{align}
-N\sin\theta & = M\ddot{x}_{\mathrm{w}}
\tag{04a}\\
N\sin\theta & = m\,\ddot{x}_{\mathrm{b}}
\tag{04b}\\
N\cos\theta - mg  & = m\,\ddot{y}_{\mathrm{b}}
\tag{04c}
\end{align}
that is a system of 3 linear equations with 4 unknowns : the three acceleration components $\:\ddot{x}_{\mathrm{w}},\ddot{x}_{\mathrm{b}},\ddot{y}_{\mathrm{b}} \:$ and the force magnitude $\:N$. A 4th equation comes from the geometry of the system and more precisely from the demand the block to be in contact with the wedge and to slide down the wedge at an angle $\:\theta$. This equation is
\begin{equation}
\tan\theta \left(\ddot{x}_{\mathrm{w}}-\ddot{x}_{\mathrm{b}}\right) =  \ddot{y}_{\mathrm{b}}
\tag{04d}
\end{equation}
as proved in the end.
Now, we solve this system as follows : From equations (04a),(04b)
\begin{equation}
\ddot{x}_{\mathrm{w}} = -\dfrac{m}{M}\,\ddot{x}_{\mathrm{b}}
\tag{05}
\end{equation}
and
\begin{equation}
N = \dfrac{m}{\sin\theta}\,\ddot{x}_{\mathrm{b}}
\tag{06}
\end{equation}
Inserting (05) in (04d) we have
\begin{equation}
\ddot{y}_{\mathrm{b}}= - \tan\theta\left(1+\dfrac{m}{M}\right)\ddot{x}_{\mathrm{b}}
\tag{07}
\end{equation}
while inserting (06) in (04c)
\begin{equation}
\ddot{y}_{\mathrm{b}}=\dfrac{\ddot{x}_{\mathrm{b}}}{\tan\theta}-g
\tag{08}
\end{equation}
Equating the RHSs of (07),(08)
\begin{equation}
\boxed{\ddot{x}_{\mathrm{b}}=\dfrac{\tan\theta}{1+\tan^{2}\theta \left(1+\dfrac{m}{M}\right)}\,g}
\tag{09}
\end{equation}
which inserted in turn in (07) gives
\begin{equation}
\boxed{\ddot{y}_{\mathrm{b}}= - \dfrac{\tan^{2}\theta \left(1+\dfrac{m}{M}\right)}{1+\tan^{2}\theta \left(1+\dfrac{m}{M}\right)}\,g}
\tag{10}
\end{equation}
so the acceleration vector of the block in the table system (laboratory) is the constant vector
\begin{equation}
\mathbf{\ddot {r}}_{\mathrm{b}} = \left(\ddot{x}_{\mathrm{b}},\ddot{y}_{\mathrm{b}} \right)= \dfrac{\tan\theta }{1+\tan^{2}\theta \left(1+\dfrac{m}{M}\right)}\,g \biggl[1, -\tan\theta \left(1+\dfrac{m}{M}\right) \biggr]
\tag{11}
\end{equation}
with magnitude
\begin{equation}
\Vert\mathbf{\ddot {r}}_{\mathrm{b}}\Vert = \dfrac{\tan\theta\sqrt{1+\tan^{2}\theta \left(1+\dfrac{m}{M}\right)^{2}} }{1+\tan^{2}\theta \left(1+\dfrac{m}{M}\right)}\,g
\tag{12}
\end{equation}
Note that  $\:[\:\theta \rightarrow 90^{\rm{o}}\:]\Longrightarrow [\:\mathbf{\ddot {r}}_{\mathrm{b}}\rightarrow \left(0,-g\right)\:]\:$, that is block free fall, as expected.
Now, the acceleration vector of the block relatively to the wedge system is (see Figure below)
\begin{equation}
\mathbf{\ddot {r}}_{\mathrm{bw}}=\mathbf{\ddot {r}}_{\mathrm{b}}-\mathbf{\ddot {r}}_{\mathrm{w}} = \left(\ddot{x}_{\mathrm{b}}-\ddot{x}_{\mathrm{w}},\ddot{y}_{\mathrm{b}} \right)=\left(1+\dfrac{m}{M}\right)\biggl(1, - \tan\theta\biggr)\ddot{x}_{\mathrm{b}}
\tag{13}
\end{equation}
and inserting (09) this is the constant vector
\begin{equation}
\mathbf{\ddot {r}}_{\mathrm{bw}}=\mathbf{\ddot {r}}_{\mathrm{b}}-\mathbf{\ddot {r}}_{\mathrm{w}} = \dfrac{\tan\theta \left(1+\dfrac{m}{M}\right)}{1+\tan^{2}\theta \left(1+\dfrac{m}{M}\right)}\,g\biggl(1,-\tan\theta \biggr)
\tag{14}
\end{equation}
with magnitude
\begin{equation}
\Vert\mathbf{\ddot {r}}_{\mathrm{bw}}\Vert = \Vert\mathbf{\ddot {r}}_{\mathrm{b}}-\mathbf{\ddot {r}}_{\mathrm{w}}\Vert = \dfrac{\sin\theta \left(1+\dfrac{m}{M}\right)}{\cos^{2}\theta +\sin^{2}\theta \left(1+\dfrac{m}{M}\right)}\,g
\tag{15}
\end{equation}

Since in the wedge system the block starting from rest travels rectilinearly a distance $\:s_{\mathrm{bw}}=h/\sin\theta\:$ to reach the table with constant acceleration $\:\Vert\mathbf{\ddot {r}}_{\mathrm{bw}}\Vert\:$ and in time $\:t$, then
\begin{equation}
s_{\mathrm{bw}}=\dfrac{1}{2}\Vert\mathbf{\ddot {r}}_{\mathrm{bw}}\Vert \,t^{2} \quad \Longrightarrow \quad t=\sqrt{\dfrac{2s_{\mathrm{bw}}}{\Vert\mathbf{\ddot {r}}_{\mathrm{bw}}\Vert}}
\tag{16}
\end{equation}
so
\begin{equation}
\boxed{t=\sqrt{\dfrac{1+\tan^{2}\theta \left(1+\dfrac{m}{M}\right)}{\tan^{2}\theta \left(1+\dfrac{m}{M}\right)}}\sqrt{\dfrac{2h}{g}}}
\tag{17}
\end{equation}
Note that for $\:\theta \rightarrow 90^{\rm{o}}\:$ we have, as expected, the time of the block free fall
\begin{equation}
\theta \rightarrow 90^{\rm{o}}\quad \Longrightarrow \quad t\rightarrow t_{\textrm{free fall}}= \sqrt{\dfrac{2h}{g}}
\tag{18}
\end{equation}
Now, the wedge starting from rest travels rectilinearly  with constant acceleration  $\:\ddot{x}_{\mathrm{w}}\:$ which using equations (05) and (09) is
\begin{equation}
\boxed{\ddot{x}_{\mathrm{w}}=-\dfrac{\left(\dfrac{m}{M}\right)\tan\theta}{1+\tan^{2}\theta \left(1+\dfrac{m}{M}\right)}\,g=-\dfrac{m\tan\theta}{M+\tan^{2}\theta \left(M+m\right)}\,g}
\tag{19}
\end{equation}
Since
\begin{equation}
v_{\textrm{wedge}}=\vert\ddot{x}_{\mathrm{w}}\vert\,t
\tag{20}
\end{equation}
using (17),(19) we have
\begin{equation}
\boxed{v_{\textrm{wedge}}=\dfrac{\left(\dfrac{m}{M}\right)}{\sqrt{{\left(1+\dfrac{m}{M}\right)\biggl[1+\tan^{2}\theta \left(1+\dfrac{m}{M}\right)\biggr]}}}\sqrt{2gh}}
\tag{21}
\end{equation}
The distance traveled by the wedge is
\begin{equation}
s=\dfrac{1}{2}\vert\ddot{x}_{\mathrm{w}}\vert\,t^{2}
\tag{22}
\end{equation}
so
\begin{equation}
\boxed{s=\dfrac{\left(\dfrac{m}{M}\right)}{\tan\theta \left(1+\dfrac{m}{M}\right)}\,h}
\tag{23}
\end{equation}
2. Numerical Values
For $\:\theta = 30^{\rm{o}}\:$ and  $\:m/M=1/2\:$ we have
\begin{align}
v_{\textrm{wedge}} & =\tfrac{\sqrt{2}}{3}\,\sqrt{gh} \qquad s =\tfrac{\sqrt{3}}{3}\,h \qquad t =\sqrt{6}\,\sqrt{h/g}
\tag{24a}\\
\ddot{x}_{\mathrm{b}}&=\tfrac{2\sqrt{3}}{9}\,g \qquad \ddot{y}_{\mathrm{b}}=-\tfrac{1}{3}\,g \qquad \Vert\mathbf{\ddot {r}}_{\mathrm{b}}\Vert = \tfrac{1}{3}\sqrt{\tfrac{7}{3}}\, g
\tag{24b}\\
\ddot{x}_{\mathrm{w}}&=\tfrac{\sqrt{3}}{9}\,g \qquad \ddot{y}_{\mathrm{w}}=0\qquad \Vert\mathbf{\ddot {r}}_{\mathrm{w}}\Vert = \tfrac{\sqrt{3}}{9}\,g \qquad \Vert\mathbf{\ddot {r}}_{\mathrm{bw}}\Vert=\tfrac{2}{3}\,g
\tag{24c}
\end{align}
3. Proof of equation (04d)

Suppose that the block is translated infinitesimally by the vector $\:\mathrm{d}\mathbf{r}_{\rm{b}}=\left(\mathrm{d}x_{\rm{b}},\mathrm{d}y_{\rm{b}}\right)\:$
as in Figure. In order for the wedge to be in touch with the block the inclination plane $\:\rm{AB}\:$ must be translated horizontally to $\:\rm{A'B'}\:$ by $\:\mathrm{d}x_{\rm{w}}\:$. But then, as shown in the Figure
\begin{equation}
\mathrm{d}y_{\rm{b}}=\tan\theta \left(\mathrm{d}x_{\rm{w}}-\mathrm{d}x_{\rm{b}}\right)
\tag{25}
\end{equation}
If these translations concern the motion of the system in infinitesimal time $\:\mathrm{d}t\:$ then
\begin{equation}
\dfrac{\mathrm{d}y_{\rm{b}}}{\mathrm{d}t}=\tan\theta \left(\dfrac{\mathrm{d}x_{\rm{w}}}{\mathrm{d}t}-\dfrac{\mathrm{d}x_{\rm{b}}}{\mathrm{d}t}\right) \quad \text{or} \quad  \dot{y}_{\mathrm{b}}=\tan\theta \left(\dot{x}_{\mathrm{w}}-\dot{x}_{\mathrm{b}}\right)
\tag{26}
\end{equation}
and differentiating with respect to $\:t\:$
\begin{equation}
  \ddot{y}_{\mathrm{b}}=\tan\theta \left(\ddot{x}_{\mathrm{w}}-\ddot{x}_{\mathrm{b}}\right)
\tag{27}
\end{equation}
4. Conservation of energy
In the following we give a proof of the conservation of energy. It's also a verification of the validity of above equations. So we must prove that
\begin{equation}
\dfrac{1}{2}Mv_{\textrm{wedge}}^{2}+\dfrac{1}{2}mv_{\textrm{block}}^{2}=mgh
\tag{28}
\end{equation}
where $\:v_{\textrm{wedge}}^{2}\:$ is given by squaring equation (21)
\begin{equation}
v_{\textrm{wedge}}^{2}=2\dfrac{\left(\dfrac{m}{M}\right)^{2}}{\left(1+\dfrac{m}{M}\right)\biggl[1+\tan^{2}\theta \left(1+\dfrac{m}{M}\right)\biggr]}\,gh
\tag{29}
\end{equation}
and $\:v_{\textrm{block}}^{2}\:$ using equations (12),(17)

\begin{equation}
v_{\textrm{block}}^{2}=\Vert\mathbf{\ddot {r}}_{\mathrm{b}}\Vert^{2} \cdot t^{2}= \dfrac{\tan^{2}\theta\biggl[1+\tan^{2}\theta \left(1+\dfrac{m}{M}\right)^{2}\biggr] }{\biggl[1+\tan^{2}\theta \left(1+\dfrac{m}{M}\right)\biggr]^{2}}\,g^{2} \cdot\dfrac{\biggl[1+\tan^{2}\theta \left(1+\dfrac{m}{M}\right)\biggr]}{\tan^{2}\theta \left(1+\dfrac{m}{M}\right)}\dfrac{2h}{g}
\tag{30}
\end{equation}

so
\begin{equation}
v_{\textrm{block}}^{2}=2\dfrac{\biggl[1+\tan^{2}\theta \left(1+\dfrac{m}{M}\right)^{2}\biggr] }{\left(1+\dfrac{m}{M}\right)\biggl[1+\tan^{2}\theta \left(1+\dfrac{m}{M}\right)\biggr]}gh
\tag{31}
\end{equation}
From (29)
\begin{equation}
\dfrac{1}{2}Mv_{\textrm{wedge}}^{2}=\dfrac{\dfrac{m}{M}}{\left(1+\dfrac{m}{M}\right)\biggl[1+\tan^{2}\theta \left(1+\dfrac{m}{M}\right)\biggr]}\,mgh =\alpha\cdot mgh
\tag{32}
\end{equation}
and from (31)
\begin{equation}
\dfrac{1}{2}mv_{\textrm{block}}^{2}=\dfrac{1+\tan^{2}\theta \left(1+\dfrac{m}{M}\right)^{2}}{\left(1+\dfrac{m}{M}\right)\biggl[1+\tan^{2}\theta \left(1+\dfrac{m}{M}\right)\biggr]}\,mgh =\beta\cdot mgh
\tag{33}
\end{equation}
where
\begin{align}
\alpha & = \dfrac{\dfrac{m}{M}}{\left(1+\dfrac{m}{M}\right)\biggl[1+\tan^{2}\theta \left(1+\dfrac{m}{M}\right)\biggr]}
\tag{34a}\\
\beta & = \dfrac{1+\tan^{2}\theta \left(1+\dfrac{m}{M}\right)^{2}}{\left(1+\dfrac{m}{M}\right)\biggl[1+\tan^{2}\theta \left(1+\dfrac{m}{M}\right)\biggr]}
\tag{34b}
\end{align}
But
\begin{equation}
\alpha + \beta = 1
\tag{35}
\end{equation}
so proving the conservation of energy, equation (28).


text:
\begin{document}

\date{ }
\title{Asymptotic Cram\'er's theorem and analysis on Wiener space}
\author{ Ciprian A. TUDOR \\
 Laboratoire Paul Painlev\'e, Universit\'e de Lille 1,\\
  F-59655 Villeneuve d'Ascq, France,\\
   email: tudor@math.univ-lille1.fr.}
\maketitle

\begin{abstract}
We prove an asymptotic Cram\'er's theorem, that is, if the sequence $(X_{n}+ Y_{n})_{n\geq 1}$ converges in law to the standard normal distribution and for every $n\geq 1$ the random variables $X_{n}$ and $Y_{n}$ are independent, then $(X_{n})_{n\geq 1}$ {\it and }  $(Y_{n}) _{n\geq 1}$ converge in law to a normal distribution. Then we compare this result with  recent criteria for the central convergence obtained in terms of Malliavin derivatives.
\end{abstract}

{\bf  2000 AMS Classification Numbers:} 60G15,  60H05, 60F05, 60H07.

{\bf Key words:} multiple stochastic integrals,  limit theorems, Malliavin calculus, Stein's method.

\section{Introduction}
The sum of two independent random variables with Gaussian distribution is a  Gaussian random variable.  A famous result by Harald Cram\'er \cite{Cramer} says that the converse implication is also true. Namely, if the law of  $X+Y$ is Gaussian and $X$ and $Y$ are independent random variables, then $X$ and $Y$ are Gaussian.
We study in this paper  the following problem: given two sequences of centered square integrable random variables $(X_{n})_{n\geq 1}$ and $(Y_{n})_{n\geq 1} $ such that $\mathbf{E}X_{n}^{2} \to _{n\to \infty } c_{1} $ and $\mathbf{E}Y_{n}^{2} \to _{n\to \infty} c_{2}$ with $c_{1}, c_{2}>0$ and $c_{1}+ c_{2}=1$ and assuming that for every $n\geq 1$, $X_{n}$ and $Y_{n}$ are independent and $X_{n}+ Y_{n}\to _{n\to \infty }N(0,1) $ in law, can we get the convergence of $X_{n} $ to the normal law $N(0, c_{1})$ {\it and } the convergence of $Y_{n}$ to the normal law $N(0, c_{2})$? We will say in this case that the central limit of the sum is decoupled. A partial answer has been given in \cite{PT}: here the authors proved that the central limit for the sum  implies the central limit for each term when the random variables $X_{n}$ and $Y_{n}$ lives in a Wiener chaos of fixed order. In this work we will  prove this result for a very general class of random variables.

Then we will try to understand this asymptotic Cram\'er's theorem from the  perspective of some   recent ideas from \cite{NoPe1} and \cite{NoPe2} related to the Stein's method on Wiener space and some older ideas from \cite{NUZ}, \cite{UZ}, \cite{UZ2} where the independence of random variables is characterized in terms of the Malliavin derivatives. Let $\left( \Omega, {\cal{F}},P\right)$ be a probability space and let $(W_{t})_{t\in [0,1]}$ be  a Wiener process on this space.  Recall that a result in \cite{NoPe1} says that a sequence of Malliavin differentiable (with respect to $W$)  random variables $X_{n}$ (defined on $\Omega$)  converges to the normal law $N(0,1)$ if and only if
\begin{equation*}
\mathbf{E}\left( f'_{z} (X_{n}) (1-\langle DX_{n} , D(-L) ^{-1} X_{n} \rangle ) \right) \to _{n \to \infty }0.
\end{equation*}
where we denoted by $D$ the Malliavin derivative with respect to $W$, by $L$ the Ornstein-Uhlenbeck generator  and by $f_{z} $  the solution of the Stein's equation (for fixed $z\in \mathbb{R}$)
\begin{equation}\label{intro1}
1_{(-\infty , z] }(x)-P(Z\leq z) = f'(x)-xf(x), \hskip0.5cm x\in \mathbb{R}.
\end{equation}
(Throughout this paper we denote by $\langle \cdot , \cdot \rangle $ the scalar product in $L^{2}([0,1])$.) In particular, if $\mathbf{E}\left( 1-\langle DX_{n}, D(-L)^{-1} X_{n}\rangle \right) ^{2}\to _{n\to \infty }0$ then $X_{n}$ converges to $N(0,1)$ as $n\to \infty$ by using Schwarz's inequality and the fact that $f'_{z}$ is bounded (actually it suffices to have \\
$\mathbf{E}\left| 1-\langle DX_{n}, D(-L)^{-1} X_{n}\rangle \right| \to _{n\to \infty }0$).

Let us describe the basic idea  to treat the convergence of sums of independent random variables to the normal law. Let $X_{n}, Y_{n}$ be two sequences as above (that means Malliavin differentiable with $\mathbf{E}X_{n}^{2}\to_{n \to \infty} c_{1}>0$, $\mathbf{E}Y_{n}^{2} \to_{n \to \infty}c_{2}>0$ and $c_{1}+c_{2}=1$). The fact that $X_{n}+ Y_{n} \to _{n\to \infty } N(0,1)$ (in law) implies that
 \begin{equation}\label{int1}
 \mathbf{E}\left( f'_{z} (X_{n}+ Y_{n}) (1-\langle D(X_{n}+ Y_{n} ), D(-L) ^{-1} (X_{n}+ Y_{n}) \rangle ) \right) \to _{n \to \infty }0.
 \end{equation}
 Suppose now that $X_{n}$ and $Y_{n}$ are independent for every $n$. A result by \"Ustunel and Zakai (\cite{UZ}, Theorem 3) says that in this case
 \begin{equation}\label{intro2}
 \mathbf{E}(\langle DX_{n}, D(-L) ^{-1}Y_{n}\rangle | X_{n} )=0  \mbox{ and } \mathbf{E}(\langle DY_{n}, D(-L) ^{-1}X_{n}\rangle | Y_{n} )=0 \mbox{ a.s }.
\end{equation}
The relation (\ref{intro2}) induces the idea that the summands containing $\langle DX_{n} , D(-L) ^{-1}Y_{n} \rangle $ and \\
$\langle DY_{n} , D(-L) ^{-1}X_{n} \rangle $  could be eliminated from (\ref{int1}). We will see that it is  not immediate and  that actually a stronger condition than the independence of $X_{n}$ and $Y_{n}$ is necessary in order to do this. Therefore our first step is to introduce some classes of independent random variables $X,Y$ such that the "mixed" terms $\langle DX , D(-L) ^{-1}Y \rangle $ and $\langle DY , D(-L) ^{-1}X \rangle $ vanish. A first class that we consider here is the class of so-called {\it strongly independent} random variables for which every multiple integral in the chaos decomposition of $X$ is independent of every multiple integral in the chaos decomposition of $Y$. We will see that  if $X$ and $Y$ are strongly independent, then
$\langle DX , D(-L) ^{-1}Y \rangle =\langle DY , D(-L) ^{-1}X \rangle =0 $  almost surely.
Another class we consider is the class of random variables $X,Y$ that are differentiable in the Malliavin sense and such that $X$ is independent of the couple $(Y,\langle DY, D(-L) ^{-1}Y\rangle )$ and $Y$ is independent of the couple $(X, \langle DX, D(-L) ^{-1} X\rangle )$. We will say in this case that the couple $(X,Y)$ belongs to the class ${\cal{A}}$. A couple of strongly independent random variables belongs to ${\cal{A}}$ and in this sense this class is an intermediary class between the independent and strongly independent random variables. For couples in ${\cal{A}}$ we will show that $\mathbf{E}(\langle DX, D(-L)^{-1}Y\rangle | X+Y)=\mathbf{E}(\langle DY, D(-L)^{-1}X\rangle | X+Y) =0$ almost surely and it is then again possible to cancel the mixed terms in (\ref{int1}).
 We will prove, by an elementary argument coming from the original Cram\'er's theorem and  without using Malliavin calculus, that for independent random variables the asymptotic Cram\'er's theorem  holds. But  for random variables in these classes (in the class ${\cal{A}}$ or strongly independent) we can give  further results by using the tools of the Malliavin calculus. Concretely, we will treat the following problem: suppose that  the sum $X_{n}+Y_{n}$ converges to the normal law in a strong sense, that is, the upper bound $\mathbf{E}\left( 1-\langle D(X_{n}+ Y_{n}), D(-L) ^{-1} (X_{n}+ Y_{n}) \rangle \right) ^{2} $ converges to zero as $n\to \infty$. We can interpret this by saying that the sum $X_{n}+ Y_{n}$ is  "close" to $N(0,1)$, not in the sense of the rate of convergence but in the sense that $X_{n}+Y_{n}$ belongs to a subset of the set of the sequences of random variables converging to $N(0,1)$.  Then can we obtain the strong convergence of $X_{n}$ and $Y_{n}$ to the normal law, that is $\mathbf{E}\left( c_{1}-G_{X_{n}}\right) ^{2} \to _{n\to \infty}0$ and $\mathbf{E}\left( c_{2}-G_{Y_{n}}\right) ^{2}\to _{n\to \infty} 0$, where $G_{X_{n}}$ is given by (\ref{gx})? We prove that this property is true for strongly independent random variables while for couples in the class ${\cal{A}}$ a supplementary assumption is necessary in order to ensure the strong convergence of $X_{n}$ and $Y_{n}$ from the  convergence of $X_{n}+Y_{n}$.

 The organization of the paper is as follows. Section 2 contains preliminaries on the stochastic calculus of variations. In Section 3 we prove the asymptotic Cram\'er's theorem by using an elementary argument while  Section 4 contains some thoughts on this theorem from the perspective of recent results  on central limit theorem obtained via Malliavin calculus.

  \section{Preliminaries}

 Let $(W_{t})_{t\in [0,1]}$ be a classical
Wiener process on a standard Wiener space $\left( \Omega
,{\mathcal{F}},\mathbf{P}\right) $. If $f\in
L^{2}([0,1]^{n})$ with $n\geq 1$ integer, we introduce the multiple Wiener-It\^{o} integral of $f$ with respect to $W$. The basic references are the monographs \cite{Mal} or \cite{N}.
Let $f\in {\mathcal{S}_{n}}$ be an elementary functions with $n$
variables  that can be written as \begin{equation*}
f=\sum_{i_{1},\ldots ,i_{n}}c_{i_{1},\ldots ,i_{n}}1_{A_{i_{1}}\times
\ldots \times A_{i_{n}}}
\end{equation*}
where the coefficients satisfy $c_{i_{1},\ldots ,i_{n}}=0$ if two indices $
i_{k}$ and $i_{l}$ are equal and the sets $A_{i}\in
{\mathcal{B}}([0,1])$ are pairwise disjoint. For  such a step function $f$ we
define

\begin{equation*}
I_{n}(f)=\sum_{i_{1},\ldots ,i_{n}}c_{i_{1},\ldots
i_{n}}W(A_{i_{1}})\ldots W(A_{i_{n}})
\end{equation*}
where we put $W(A)=\int_{0}^{1} 1_{A}(s)dW_{s}$. It can be seen that the application $
I_{n}$ constructed above from ${\mathcal{S}}_{n}$ to $L^{2}(\Omega
)$ is an
isometry on ${\mathcal{S}}_{n}$ , i.e.
\begin{equation}
\mathbf{E}\left[ I_{n}(f)I_{m}(g)\right] =n!\langle f,g\rangle
_{L^{2}([0,1]^{n})}\mbox{ if }m=n  \label{isom}
\end{equation}
and
\begin{equation*}
\mathbf{E}\left[ I_{n}(f)I_{m}(g)\right] =0\mbox{ if }m\not=n.
\end{equation*}

Since the set ${\mathcal{S}_{n}}$ is dense in $L^{2}([0,1]^{n})$ for every $n\geq 1$ the mapping $
I_{n}$ can be extended to an isometry from $L^{2}([0,1]^{n})$ to $
L^{2}(\Omega)$ and the above properties hold true for this
extension.
It also holds that
\begin{equation}\label{ftilde}
I_{n}(f) = I_{n}\big( \tilde{f}\big)
\end{equation}
where $\tilde{f} $ denotes the symmetrization of $f$ defined by $\tilde{f}
(x_{1}, \ldots , x_{n}) =\frac{1}{n!} \sum_{\sigma \in {\bf S} _{n}}
f(x_{\sigma (1) }, \ldots , x_{\sigma (n) } ) $.
We will need the general formula for calculating products of Wiener chaos
integrals of any orders $m,n$ for any symmetric integrands $f\in
L^{2}([0,1]^{\otimes m})$ and $g\in L^{2}([0,1]^{\otimes n})$; it is
\begin{equation}
I_{m}(f)I_{n}(g)=\sum_{\ell=0}^{p\wedge q}\ell!C_{m}^{\ell}C_{n}^{\ell}
I_{m+n-2\ell}(f\otimes_{\ell}g) \label{product}
\end{equation}
where the contraction $f\otimes_{\ell}g$ is  defined by
\begin{eqnarray}
&&  (f\otimes_{\ell} g) ( s_{1}, \ldots, s_{m-\ell}, t_{1}, \ldots, t_{n-\ell
})\nonumber\\
&&  =\int_{[0,T] ^{m+n-2\ell} } f( s_{1}, \ldots, s_{m-\ell}, u_{1},
\ldots,u_{\ell})g(t_{1}, \ldots, t_{n-\ell},u_{1}, \ldots,u_{\ell})
du_{1}\ldots du_{\ell} . \label{contra}
\end{eqnarray}
Note that the contraction $(f\otimes_{\ell} g) $ is an element of $L^{2}([0,1]^{m+n-2\ell})$ but it is not necessary symmetric. We will denote by $(f\tilde{\otimes }_{\ell} g)$ its symmetrization.

We recall that any square integrable random variable which is
measurable with respect to the $\sigma$-algebra generated by $W$ can
be expanded into an orthogonal sum of multiple stochastic integrals
\begin{equation}
\label{sum1} F=\sum_{n\geq0}I_{n}(f_{n})
\end{equation}
where $f_{n}\in L^{2}([0,1]^{n})$ are (uniquely determined)
symmetric functions and $I_{0}(f_{0})=\mathbf{E}\left[  F\right]  $.

We denote by $D$  the Malliavin  derivative operator that acts on smooth functionals of the form $F=g(W(\varphi _{1}), \ldots , W(\varphi_{n}))$ (here $g$ is a smooth function with compact support and $\varphi_{i} \in L^{2}([0,1])$ for $i=1,..,n$)
\begin{equation*}
DF=\sum_{i=1}^{n}\frac{\partial g}{\partial x_{i}}(W(\varphi _{1}), \ldots , W(\varphi_{n}))\varphi_{i}.
\end{equation*}
We can define the $i$ th Malliavin derivative $D^{(i)}$ is defined iteratively. The operator $D^{(i)}$ can be extended to the closure $\mathbb{D}^{p,2}$ of smooth functionals with respect to the norm
\begin{equation*}
\Vert F\Vert _{p,2}^{2} = \mathbf{E}F^{2}+ \sum_{i=1}^{p} \mathbf{E} \Vert D^{i} F\Vert ^{2} _{L^{2}([0,1] ^{i})}
\end{equation*}
 The adjoint of
$D$ is denoted by $\delta $ and is called the divergence (or
Skorohod) integral. Its domain $Dom(\delta)$  coincides with the class of stochastic processes $u\in L^{2}(\Omega \times [0,1])$ such that
\begin{equation*}
\left| \mathbf{E}\langle DF, u\rangle \right| \leq c\Vert F\Vert _{2}
\end{equation*}
for all $F\in \mathbb{D}^{1,2}$ and $\delta (u)$ is the element of $L^{2}(\Omega)$ characterized by the duality relationship
\begin{equation*}
\mathbf{E}(F\delta (u))= \mathbf{E}\langle DF, u\rangle.
\end{equation*}
For adapted integrands, the divergence integral coincides with
the classical It\^o integral.

Let $L$ be the Ornstein-Uhlenbeck operator defined on $Dom (L)= \mathbb{D}^{2,2}$
\begin{equation*}
LF=-\sum_{n\geq 0} nI_{n}(f_{n})
\end{equation*}
if $F$ is given by (\ref{sum1}). There exists a connection between $\delta, D $ and $L$ in the sense that a random variable $F$ belongs to the domain of $L$ if and only if $F\in \mathbb{D}^{1,2}$ and $DF \in Dom (\delta)$ and then $\delta DF=-LF$. Also we will need in the paper  the integration by parts formula
\begin{equation}
\label{ip}
F\delta(u)= \delta (Fu) + \langle DF, u\rangle
\end{equation}
whenever $F\in \mathbb{D}^{1,2}$, $u\in Dom(\delta)$ and $\mathbf{E}F^{2}\int_{0}^{1}u_{s}^{2}ds <\infty$.

\section{Asymptotic Cram\'er's theorem  }

We start by proving an asymptotic version of the Cram\'er's theorem \cite{Cramer}. A particular case (when the sequences $X_{n}$ and $Y_{n}$ are multiple integrals in a Wiener chaos of fixed order) has been proven in \cite{PT}, Corollary 1. Our proof is based on the Cram\'er's theorem (see \cite{Cramer}) and an idea from \cite{NOT}.

\begin{theorem}\label{cra}
Suppose that $(X_{n})_{n\geq 1}$ and $(Y_{n}) _{n\geq 1} $are two sequences of centered random variables in $L^{2}(\Omega)$ such that $\mathbf{E}X_{n} ^{2}\to _{n\to \infty }c_{1}>0$ and $\mathbf{E}Y_{n}^{2}\to _{n\to \infty} c_{2}>0$ with $c_{1}+ c_{2}=1$. Assume that for every $n\geq 1$, the random variables $X_{n}$ and $Y_{n}$ are independent. Then
\begin{equation*}
X_{n}+ Y_{n} \to N(0,1) \Leftrightarrow (X_{n}\to N(0, c_{1}) \mbox{ and } Y_{n} \to N(0, c_{2})).
\end{equation*}
 \end{theorem}
{\bf Proof: } One direction is  trivial. Let us assume that $X_{n} + Y_{n} \to _{n\to \infty} N(0,1)$. We will prove that $X_{n}\to _{n\to \infty} N(0, c_{1})$ and $Y_{n}\to _{n\to \infty} N(0, c_{2})$. Since $\mathbf{E}X_{n}^{2} \to_{n \to \infty}c_{1}$ and  $\mathbf{E}Y_{n}^{2}\to _{n\to \infty} c_{2}$ it follows that the sequence $(X_{n}, Y_{n})_{n\geq 1}$ is bounded in $L^{2}(\Omega)$. By Prohorov's theorem it suffices to prove that for any subsequence which converges  in distribution to some random vector $(F,G)$,  then we must have $F\sim N(0, c_{1}), G\sim N(0, c_{2}) $ and $F,G$ are independent.  Let us consider  such an arbitrary sequence $(X_{n_{k}}, Y_{n_{k}})$ which  converges in law  to $(F,G)$ as $k\to \infty$.  Because $X_{n_{k}}$ and $Y_{n_{k}}$ are independent for each $k$, it is clear that $F$ and $G$ are independent. Since $X_{n}+ Y_{n}\to _{n\to \infty} N(0,1)$ it follows that $F+G\sim N(0,1)$.

Cram\'er's theorem implies that $F\sim N(0, c_{1}) $ and $G\sim N(0, c_{2})$. \qed

\vskip0.3cm

This result can be extended to finite and even infinite sums of independent random variables.
\begin{prop}
Suppose that for every $n\geq 1$, $X^{n}= \sum_{k\geq 1} X^{n}_{k}$ where for every $n$ the random variables $X^{k}_{n}, k\geq 1$ are mutually independent  and the series is convergent for every $\omega$. Assume also that $X^{n}_{k}$ are centered for every $n,k \geq 1$ and $ \mathbf{E}(X^{n}_{k}) ^{2} \to _{n\to \infty} c_{k}>0$ for every $k\geq 1$. Suppose that $X^{n}$ converging in law to $N(0,1)$ as $n\to \infty$. Then for every $k\geq 1$ the sequence $X^{n}_{k}$ converges to the normal law as $n\to \infty$.
 \end{prop}
 {\bf Proof: } Since $X^{n} =X_{1}^{n}+ \sum_{k\geq 2} X_{k}^{n}$ and the two summands are independent, Theorem \ref{cra} implies that $X_{1}^{n}$ converges to the normal law. Inductively, the conclusion can be obtained. \qed

\begin{remark}
When, for every $n\geq 1$,  $X_{n}=I_{k_{1}}(f^{n}) $ and $Y_{n}=I_{k_{2}}(g^{n})$ are multiple stochastic integrals (possibly of different orders, that can also vary with $n$) we can go further by proving the following result. If $\mathbf{E} (X_{n} + Y_{n} ) ^{2} \to _{n\to \infty}1$ and $X_{n}+ Y_{n} $ converges in law to $N(0,1)$, if $\liminf _{n}\mathbf{E} X_{n}^{2} >0$ and $\liminf _{n}\mathbf{E} Y_{n}^{2} >0$
then
\begin{equation}
\label{rev1}
d_{Kol} (X_{n}, N(0, \mathbf{E}X_{n}^{2} )) \to _{n\to \infty }0 \mbox{ and } d_{Kol} Y_{n}, N(0, \mathbf{E}Y_{n}^{2} ) \to _{n\to \infty }0
\end{equation}
Here $d_{Kol} $ means the Kolmogorov distance (recall that the Kolmogorov distance between the law of two random variables $U$ and $V$ is given by $d_{Kol}(U,V)= \sup_{x\in \mathbb{R}}\left| P(U\leq x) -P(V\leq x)\right| $). That is, there is an asymptotic Cram\'er's theorem even if the variances of $X_{n}$ and $Y_{n}$ do not converge a priori. Relation (\ref{rev1}) can be proved as follows. First, recall the following bound when $X$ lives in a chaos of fixed order (see e.g. \cite{NoPe3})
\begin{equation}
\label{rev2}
d_{Kol} X, N(0, \mathbf{E}X^{2} ) \leq \frac{ \left( \left| \mathbf{E} X^{4} -3(\mathbf{E} X^{2})^{2} \right|\right)^{\frac{1}{2}} }{\mathbf{E}X^{2} } =: \frac{\left(| k_{4} (X) | \right) ^{\frac{1}{2}}}{\mathbf{E}X^{2} }
\end{equation}
where $k_{4}(X)$ is the fourth cumulant of $X$. It is immediate, by the definition of the cumulant, that $k_{4}(X+Y)=k_{4}(X)+ k_{4}(Y)$ if $X$ and $Y$ are independent. Moreover, it follows from \cite{NoPe3}, identity (3. 31) that $k_{4}(X)\geq 0$ if $X$ is a multiple integral. Hence, if $\mathbf{E} (X_{n} + Y_{n} ) ^{2} \to _{n\to \infty}1$ and $X_{n}+ Y_{n} $ converges in law to $N(0,1)$, then
$$k_{4}(X_{n})+ k_{4}(Y_{n})=k_{4}(X_{n}+ Y_{n}) =\mathbf{E}(X_{n}+Y_{n}) ^{4} -3 (\mathbf{E}(X_{n}+Y_{n})^{2})^{2} \to _{n\to \infty}0$$
and this implies that $k_{4}(X_{n})\to _{n\to \infty}0$ and $k_{4}(Y_{n}) \to_{n\to \infty}0$. The convergence (\ref{rev1}) is obtained by using (\ref{rev2}) and the hypothesis $\liminf _{n}\mathbf{E} X_{n}^{2} >0$ and $\liminf _{n}\mathbf{E} Y_{n}^{2} >0$.
\end{remark}

 \vskip0.3cm

\section{Decoupling central limit under strong independence}

Let us regard Theorem \ref{cra} from the perspective of the results in \cite{NoPe1}. In this part all random variables are centered. We  recall some facts related to the convergence of a sequence of random variables to the normal law in terms of the Malliavin calculus. For any random variable $X \in \mathbb{D}^{1,2}$ we  denote by
\begin{equation}\label{gx}
G_{X} := \langle DX, D(-L) ^{-1}X\rangle.
\end{equation}
The following result is a slight extension of Proposition 3.1 in \cite{NoPe1}. See also Theorem 3 in \cite{T1}.

\begin{prop}\label{t1}
Let $(X_{n})_{n\geq 1}$ be a sequence of square integrable random variables such that $\mathbf{E}X_{n}^{2} \to _{n\to \infty} c>0$. Then the following are equivalent:
\begin{description}
\item{1. }The sequence $(X_{n})_{n\geq 1}$ converges in law, an $n\to \infty$,  to the normal random variable $N(0,c)$, $c>0$.
\item{2. }For every $t\in \mathbb{R}$,
$\mathbf{E}\left( e^{itX_{n} } (c-G_{X_{n}})\right)\to _{n\to \infty }0.$
\item{3. }$
\mathbf{E}\left( (c-G_{X_{n}})
| X_{n}\right) \to _{n\to \infty} 0$ \mbox{ a.s. }.
\item{4. }For every $z\in \mathbb{R}$,
$\mathbf{E}\left(  f'_{z} (X_{n}) (c-G_{X_{n}})\right) \to_{n\to \infty }0.$
\end{description}
\end{prop}
{\bf Proof: } We follow the scheme $1. \Rightarrow  2. \Rightarrow 3.\Rightarrow 4. \Rightarrow 1$. The implications $1. \Rightarrow  2.$ and $3.\Rightarrow 4. \Rightarrow 1$ follow exactly as in \cite{T1}, Theorem 3. Concerning $2. \Rightarrow 3.$, set $F_{n}=c-G_{X_{n}}$ for every $n\geq 1$. The random variable $\mathbf{E}(F_{n}|X_{n})$ is the Radon-Nykodim derivative with respect to $P$ of the measure $Q_{n}(A)= \mathbf{E}(F_{n}1_{A})$, $A\in \sigma(X_{n})$. Relation 2. means that
$\mathbf{E}\left( e^{itX_{n}} \mathbf{E}(F_{n} / X_{n})\right)=\mathbf{E}_{Q_{n} } (e^{itX_{n}}) \to _{n\to \infty} 0$  and hence $\int_{\mathbb{R}}e^{ity}d(Q_{n}\circ X_{n}^{-1})(y)\to _{n\to \infty}0$. This implies that $Q_{n} (A)=\mathbf{E}(F_{n}1_{A}) \to _{n\to \infty} 0$  for any $A \in \sigma (X_{n})$ or $\mathbf{E}(F_{n}|X_{n}) \to _{n\to \infty}0.$
\vskip0.2cm
As  an immediate consequence we have (see also \cite{NoPe1}).

\begin{corollary}\label{np}
Suppose that $(X_{n})_{n\geq 1}$ is a sequence of random variables such that $\mathbf{E}X_{n}^{2} \to_{n \to \infty} c$. suppose that
\begin{equation}\label{s1}
\mathbf{E}(c-G_{X_{n}})^{2}\to _{n\to \infty }0.
\end{equation}
Then $X_{n} \to _{n\to \infty } N(0,c)$.
\end{corollary}

\begin{remark}\label{chafin}
In the case when the variables $X_{n}$ live in a fixed Wiener chaos, $X_{n}= I_{k}(f_{n})$, then the convergence in distribution of $X_{n}$ to the normal law is equivalent to (\ref{s1}), see \cite{NOT}.
\end{remark}
\vskip0.3cm

Assume that $(X_{n})_{n\ge 1}$ and $(Y_{n})_{n\geq 1}$ are two sequences of random variables such that: i) for every $n\geq 1$ the random variables $X_{n}$ and $Y_{n}$ are independent and ii) $X_{n}+ Y_{n}\to N(0,1)$ in distribution as $n\to \infty$.
 The quantity $G_{X_{n} + Y_{n}}$, which plays a central role, can be written as
\begin{equation*}
G_{X_{n} + Y_{n}}= G_{X_{n}}+ G_{Y_{n}}+ \langle DX_{n} , D(-L)^{-1}Y_{n}\rangle + \langle DY_{n} , D(-L)^{-1}X_{n}\rangle.
\end{equation*}
The force of the Cram\'er's theorem can be observed here: the fact that
\begin{equation*}
 \mathbf{E} \left( c_{1}-G_{X_{n}}+ c_{2}-G_{X_{n}} -\langle DX_{n} , D(-L)^{-1}Y_{n}\rangle - \langle DY_{n} , D(-L)^{-1}X_{n}\rangle| X_{n}+Y_{n}\right)
\end{equation*}
converges to zero implies that $\mathbf{E}(c_{1}-G_{X_{n}}| X_{n}) $ and $\mathbf{E}(c_{2}-G_{Y_{n}}| Y_{n}) $ both converge to zero. It is  not obvious to prove this  directly. Note also that the independence of $X_{n}$ and $Y_{n}$ does not guarantee a priori that the terms $\mathbf{E}( \langle DY_{n} , D(-L)^{-1}X_{n}\rangle |X_{n}+Y_{n})$ $\mathbf{E}( \langle DX_{n} , D(-L)^{-1}Y_{n}\rangle |X_{n}+Y_{n})$ vanish.
But the situation when these two terms vanish is also interesting and we will analyze this case in the sequel. We will see that it requires a slightly stronger assumption than the independence of $X_{n}$ and $Y_{n}$. We introduce the following concept.

\begin{definition} Two random variables $X=\sum_{n\geq 0} I_{n}(f_{n})$ and $Y=\sum_{m\geq 0} I_{m}(g_{m}) $ are called {\it strongly independent} if  for every $m,n\geq 0$, the random variables $I_{n}(f_{n})$ and $I_{m}(g_{m})$ are  independent.
\end{definition}

\begin{remark}\label{indep}
Let us recall the criterion for the independence of two multiple integrals given in \cite{UZ}:  Let $X'=I_{p}(f)$ and $Y'=I_{q}(g)$ where $f\in L^{2}([0,1] ^{p})$ and $g\in L^{2}([0,1] ^{q})$ ($p,q\geq 1$) are symmetric functions. Then $X'$ and $Y'$ are independent if and only if
\begin{equation*}
f\otimes _{1}g=0 \mbox{ almost everywhere on  }[0,1] ^{p+q-2}.
\end{equation*}
As a consequence two random variables $X$ and $Y$ as in Definition 1 are strongly independent if and only if for every $m,n\geq 1$,  $f_{n} \otimes _{1} g_{m}=0$ almost everywhere on $[0,1]^{m+n-2}$.

\end{remark}

Let us also note that the class of strongly independent random variables is strictly included in the class of independent random variables. Indeed, consider
\begin{equation*}
X_{1}=\sqrt{2} I_{1} (1_{[\frac{1}{2}, 1]}) \mbox{ and }Y_{1}= \sqrt{2} \int_{0}^{\frac{1}{2}} sign (W_{s}) dW_{s}.
\end{equation*}
Then $X_{1}$ and $Y_{1}$ are independent standard normal random variables. Define
\begin{equation*}
X=\frac{1}{\sqrt{2}}(X_{1}+ Y_{1} ) \mbox{ and } Y= \frac{1}{\sqrt{2}}(X_{1}- Y_{1} ).
\end{equation*}
Then $X, Y$ are also independent standard normal but they are not strongly independent because for example the chaoses of order one of $X$ and $Y$ are not independent (note that the random variable $ \int_{0}^{\frac{1}{2}} sign (W_{s}) dW_{s}$ has only even order chaos components).

\begin{lemma}\label{n11}
Assume that $X,Y\in \mathbb{D}^{1,2}$ and $X,Y$ are strongly independent. Then
\begin{equation*}
\langle DX, D(-L)^{-1}Y\rangle =0  \hskip0.5cm \mbox{ a.s }.
\end{equation*}
\end{lemma}
{\bf Proof: } Suppose first that $X=I_{n}(f)$ and $Y=I_{m}(g)$. Then, since $D_{\alpha } X=n I_{n-1}(f(\cdot , \alpha ))$ and $D_{\alpha}(-L)^{-1}Y= I_{m-1}(g(\cdot , \alpha))$,  using (\ref{product})
\begin{eqnarray*}
\langle DX, D(-L)^{-1}Y\rangle &=& n\int_{0}^{1} d\alpha I_{n-1}(f(\cdot, \alpha )) I_{m-1}(g(\cdot , \alpha ))\\
&=&m \sum_{k=0}^{(m-1)\wedge (n-1)} k! C_{m-1}^{k}C_{n-1}^{k} \int_{0}^{1}d\alpha I_{m+n-2-2k} (f(\cdot , \alpha )\otimes _{k} g(\cdot , \alpha ))\\&=&
\sum_{k=0}^{(m-1)\wedge (n-1)} k! C_{m-1}^{k}C_{n-1}^{k} I_{m+n-2-2k}(f\otimes _{k+1} g)
\end{eqnarray*}
and this is equal to zero from the characterization of the independence of two multiple multiple integrals (see Remark \ref{indep}). The extension to the general case is immediate since, if $X=\sum_{n}I_{n}(f_{n})$ and $Y=\sum_{m}I_{m}(g_{m})$,
\begin{equation*}
\langle DX, D(-L) ^{-1}Y\rangle =\sum_{m,n}\langle DI_{n}(f_{n}), D(-L)^{-1}I_{m}(g_{m})\rangle .
\end{equation*} \qed

\vskip0.3cm

In view of Lemma \ref{n11}, the Proposition \ref{t1} can be formulated for strongly independent random variables as follows: Suppose that $(X_{n})_{n\geq 1}$ and $(Y_{n})_{n\geq 1}$ are two sequences of centered strongly independent random variables such that $\mathbf{E}X_{n}^{2}\to _{n\to \infty }c_{1} $ and $\mathbf{E}Y_{n}^{2} \to _{n \to \infty }c_{2}$ where $c_{1}, c_{2}>0$ are such that $c_{1}+ c_{2} =1$. Then the following affirmations are equivalent:
 \begin{description}
 \item{1. }The sequence $(X_{n}+ Y_{n})_{n\geq 1}$ converges in law to a standard normal random variable as $n\to \infty$;
 \item{2. } For every $t\in \mathbb{R}$, $\mathbf{E}\left( e^{it(X_{n}+Y_{n}) }(c_{1}-G_{X_{n}}+ c_{2}-G_{X_{n}}) \right) \to _{n\to \infty }0.$;
 \item{3. } $\mathbf{E}\left( c_{1}-G_{X_{n}}+ c_{2}-G_{Y_{n}} | X_{n}+ Y_{n}\right) \to _{n\to \infty}0.$;
  \item{4. } For every $z\in \mathbb{R}$, $\mathbf{E}\left( f'_{z}(X_{n}+ Y_{n})(c_{1}-G_{X_{n}}+ c_{2}-G_{Y_{n}}) \right) \to_{n\to \infty }0$.
\end{description}

\vskip0.3cm

Let us assume now that the two sequences of Theorem \ref{cra} are strongly independent. We will also  assume that the convergence of the sum $X_{n}+ Y_{n}$ to $N(0,1)$ is strong in the sense that $\mathbf{E}\left( 1-G_{X_{n}+Y_{n}}\right) ^{2}$ converges to zero as $n\to \infty$. We can say, somehow,  that in this case the sum $X_{n}+Y_{n}$ is rather close to the normal law since the upper bound of the distance between it and $N(0,1)$ goes to zero. We will prove that this implies that the convergence of $X_{n}$ and $Y_{n}$ to the normal law is also strong.

\begin{remark}
The case of multiple stochastic integrals can be easily understood. Suppose that $X_{n}=I_{k}(f^{n})$ and $Y_{n}=I_{l}(g^{n}) $ where for every $n\geq 1$ the kernels $f^{n}, g^{n}$ are in $L^{2}([0,1] ^{k})$ and $L^{2}([0,1]^{l})$ respectively. Assume that $EX_{n}^{2}\to_{n \to \infty}c_{1}>0$ and $EY_{n}^{2}\to_{n \to \infty}c_{2}>0$ such that $c_{1}+c_{2}=1$. Then if $X_{n}+ Y_{n}\to_{n \to \infty} N(0,1)$ and $X_{n}, Y_{n}$ are independent (thus strongly independent) it follows that $X_{n}\to N(0,c_{1})$ and $Y_{n}\to N(0,c_{2})$ and by Remark \ref{chafin}, $E\left( c_{1}-G_{X_{n}}\right) ^{2} \to_{n \to \infty}0$ and $E\left( c_{2}-G_{Y_{n}}\right) ^{2} \to_{n \to \infty}0$, so the convergence of $X_{n}$ and $Y_{n}$  to the normal distribution is strong.
\end{remark}
\vskip0.2cm

We will also need the following lemma.

\begin{lemma}\label{n12}
Assume that $X,Y\in \mathbb{D}^{1,2}$ and $X,Y$ are strongly independent. Then  the random variables $G_{X}$ and $G_{Y}$ are strongly independent.
\end{lemma}
{\bf Proof: } Let us assume once again that $X=I_{n}(f)$ and $Y=I_{m}(g)$. The result can  easily be extended to the general case.
We have
\begin{equation*}
G_{X}=n \sum_{k=0}^{n-1} \left( C_{n-1}^{k} \right) ^{2} I_{2n-2-2k}(f\otimes _{k+1} f)
\end{equation*}
and
\begin{equation*}
G_{Y}= m\sum_{l=0}^{m-1} l! \left( C_{m-1}^{l}\right) ^{2} I_{2m -2-2l} (g\otimes _{l+1} g).
\end{equation*}
It suffices to show that for every $k=1,..,n-1$ and $l=1, .., m-1$ the random variables $I_{2n-2k}(f\otimes _{k} f) $ and $I_{2m-2l} (g\otimes _{l}g)$ are independent or equivalently
\begin{equation*}
(f\tilde{\otimes }_{k}f) \otimes _{1} (g\tilde{\otimes }_{l} g)= 0 \mbox{ a.e.  on } [0,1] ^{2m-2k+2m-2l-2}.
\end{equation*}
But since
\begin{eqnarray*}
&&(f\tilde{\otimes }_{k}f)(x_{1},..,x_{2n-2k})\\
 &&=\sum_{\sigma \in {\bf S}_{2n-2k} } \int_{[0,1] ^{k}} f(u_{1},..,u_{k}, x_{\sigma (1)},..,x_{\sigma (n-k)})f(u_{1},..,u_{k}, x_{\sigma (n-k+1)},..,x_{\sigma (2n-2k)})du_{1}..du_{k}
\end{eqnarray*}
and
\begin{eqnarray*}
&&(g\tilde{\otimes }_{l} g)(y_{1},..,y_{2m-2l})\\
&& \sum_{\rho \in {\bf S}_{2m-2l}} \int_{[0,1]^{l}}g(v_{1},..,v_{l}, y_{\rho(1)},..,y_{\rho (m-l) })g(v_{1},..,v_{l}, y_{\rho(m-l+1)},..,y_{\rho (2m-2l) })
dv_{1}..dv_{l}
\end{eqnarray*}
then $(f\tilde{\otimes }_{k}f) \otimes _{1} (g\tilde{\otimes }_{l} g)= 0 $ almost everywhere on $[0,1] ^{2m-2k+2m-2l-2}$ by using Fubini and the fact that $\int_{0}^{1} d\alpha f(u_{1},..,u_{k}, x_{1},..,x_{n-k-1}, \alpha ) g(v_{1},..,v_{l}, y_{1}, ..,y_{m-l+1}, \alpha  ) =0$ for almost all $u_{1},v_{i}, x_{i},y_{i}$.
The general case demands to prove that $(f_{n} \tilde{\otimes} _{k}f_{n'} ) \otimes _{1} (g_{m}\tilde{\otimes }_{l} g_{m'})=0$ almost everywhere for every $n,n',m,m'\geq 1$ and for every  $k=1,..,n\wedge n '$ and $l=1, .., m\wedge m'$ and this can be done similarly as above (note that the fact that $k,l\geq 1$ and the value zero is excluded is essential for the proof). \qed

\begin{prop}\label{p101}
Suppose that $(X_{n})_{n\geq 1}$ and $(Y_{n})_{n\geq 1}$ are two sequences of centered strongly independent random variables such that $\mathbf{E}X_{n}^{2}\to _{n\to \infty }c_{1} $ and $\mathbf{E}Y_{n}^{2} \to _{n \to \infty }c_{2}$ where $c_{1}, c_{2}>0$ are such that $c_{1}+ c_{2} =1$. Then $\mathbf{E}\left(1-G_{X_{n}+Y_{n}}\right) ^{2} \to _{n\to \infty }0 $ if and only if
\begin{equation*}
\mathbf{E}\left( c_{1}-G_{X_{n}}\right) ^{2} \to _{n\to \infty }0 \mbox{ and }  \mathbf{E}\left( c_{2}-G_{Y_{n}}\right) ^{2} \to _{n\to \infty }0.
\end{equation*}
\end{prop}
{\bf Proof: } By using Lemmas \ref{n11} and \ref{n12} we have
\begin{equation*}
\mathbf{E}\left(1-G_{X_{n}+Y_{n}}\right) ^{2}=\mathbf{E}\left( c_{1}-G_{X_{n}}\right) ^{2}+\mathbf{E}\left( c_{2}-G_{Y_{n}}\right) ^{2}
\end{equation*}
and the conclusion is immediate. \qed

\vskip0.3cm

We will study  next if the result in Proposition \ref{p101} can be obtained by relaxing the hypothesis on the strong independence of $X_{n}$ and $Y_{n}$ for every $n$. As we have seen, the strong independence of two variables $X$ and $Y$ implies that $\langle DX, D(-L)^{-1}Y\rangle = \langle DY, D(-L)^{-1}X\rangle =0$ a.s. But in order to eliminate the "mixed" terms we only need $\mathbf{E}( \langle DX, D(-L)^{-1}Y\rangle |X+Y) =\mathbf{E}( \langle DY, D(-L)^{-1}X\rangle | X+Y) =0$ a.s. We  therefore introduce an intermediary class between the class of independent random variables and the class of strongly independent random variables for which this property holds.

\begin{definition}
We will say that a couple $(X,Y)$  of two random variables  in the space $\mathbb{D}^{1,2}$ belongs to the class ${\cal{A}}$  if the vector $X$ is independent of the vector $(Y, G_{Y})$ and $Y$ is independent of the vector  $(X, G_{X})$.
\end{definition}

We will give now examples of random variables in ${\cal{A}}$. First we recall the following result from \cite{UZ}.
\begin{lemma}\label{uz}
Let $X\in \mathbb{D}^{1,2}$ and $Y,Z\in L^{2}(\Omega)$. Then $X $ is independent of the pair $(Y,Z)$ if and only if for every $\alpha , \beta \in \mathbb{R}$
\begin{equation*}
\mathbf{E}\left( \langle DX, D(-L)^{-1}e^{i(\alpha Z + \beta Y)}\rangle | X \right)= 0 \mbox{ a.s. }.
\end{equation*}
\end{lemma}

We show that a couple of strongly independent random variables is in the set ${\cal{A}}$. We consider first the case of multiple integrals.
\begin{lemma}
Suppose that $X=I_{p}(f)$ and $Y=I_{q}(g)$ where $f\in L^{2}([0,1] ^{p}) $ and $g\in L^{2}([0,1] ^{q} )$ ($p,q\geq 1$)  are symmetric functions. Assume that $X$ and $Y$ are independent. Then $(X, Y)$ belongs to the class ${\cal{A}}$.
\end{lemma}
{\bf Proof: }We will prove that $X$ is independent of the couple $(Y, G_{Y})$. Similarly it will follow that $Y$ is independent of $(X, G_{X})$. We prove that
\begin{equation*}
\langle DX, D(-L)^{-1}e^{i(\alpha Y + \beta G_{Y}) } \rangle =0 \mbox{ a.s. }
\end{equation*}
or, since $D(-L) ^{-1}= (I+L)^{-1}D $,
\begin{equation*}
\langle DX, (I+L)^{-1}De^{i(\alpha Y + \beta G_{Y}) } \rangle =0  \mbox{ a.s. }
\end{equation*}
Note that $De^{i(\alpha Y + \beta G_{Y}) }= e^{i(\alpha Y+ \beta G_{Y})} (i\alpha DY + i\beta DG_{Y})$. First we will show that
\begin{equation*}
\langle DX, e^{i(\alpha Y+ \beta G_{Y})}DY \rangle =0 \mbox{ a. s. }
\end{equation*}
Assume that the random variable $e^{i(\alpha Y+ \beta G_{Y})}$ admits the chaos expansion $e^{i(\alpha Y+ \beta G_{Y})}=\sum_{N\geq 0} I_{N}(h_{N})$ (in the sense that its real part and its imaginary part admit  such a decomposition). Then
\begin{eqnarray*}
e^{i(\alpha Y+ \beta G_{Y})}D_{\alpha }Y &=& q\sum_{N\geq 0} I_{N}(h_{N}) qI_{q-1}(g(\cdot , \alpha ))\\
&=& q\sum_{N\geq 0} \sum_{r=0} ^{N\wedge (q-1)} r! C_{q-1}^{r} C_{N}^{r} I_{N+q-1-2r} (h_{N}\otimes _{r} g(\cdot , \alpha ))
\end{eqnarray*}
and
\begin{eqnarray*}
(I+L)^{-1}e^{i(\alpha Y+ \beta G_{Y})}D_{\alpha }Y &=&q\sum_{N\geq 0} \sum_{r=0} ^{N\wedge (q-1)} r! C_{q-1}^{r} C_{N}^{r} (1+N+(q-1)-2r)^{-1}I_{N+q-1-2r} (h_{N}\otimes _{r} g(\cdot , \alpha )).
\end{eqnarray*}
Therefore
\begin{eqnarray*}
&& \langle DX, (I+L)^{-1}De^{i(\alpha Y + \beta G_{Y}) } \rangle \\
&=&pq\sum_{N\geq 0} \sum_{r=0} ^{N\wedge (q-1)} r! C_{q-1}^{r} C_{N}^{r} (1+N+(q-1)-2r)^{-1}\\
&&\times \sum_{a=0}^{(N+q-1-2r)\wedge (p-1)}I_{N+q-1-2r+p-2a}\int_{0}^{1}
\left( (h_{N}\tilde{ \otimes } _{r} g(\cdot , \alpha )\otimes _{a} f(\cdot , \alpha )\right) d\alpha.
\end{eqnarray*}
Above, $(h_{N}\tilde{ \otimes } _{r} g(\cdot , \alpha )$ means the symmetrization of the function $(t_{1}, \ldots , t_{N+q-1-2r}) \to (h_{N} \otimes  _{r} g(t_{1}, \ldots , t_{N+q-1-2r} , \alpha )$ for fixed $\alpha$. In other words the above symmetrization does not affect the variable $\alpha$.  By interchanging the order of integration to integrate first with respect to $\alpha$ we will obtain that the last quantity is zero. Similarly it will follow that $\langle DX, e^{i(\alpha Y+ \beta G_{Y})}DG_{Y}\rangle $ is almost surely zero.
\qed

\vskip0.2cm

We can extend the previous result  to the case of infinite chaos expansion.
\begin{lemma}
Assume that $X,Y$ are two strongly independent random variables in $\mathbb{D}^{1,2}$.
Then $(X, Y)$ belongs to ${\cal{A}}$.
\end{lemma}
{\bf Proof: }The proof follows the lines of the proof of Lemma 4. In order to check that
\begin{equation*}
\langle DX, (I+L)^{-1}De^{i(\alpha Y + \beta G_{Y}) } \rangle =0 \mbox{ a.s. }
\end{equation*}
we write
\begin{eqnarray*}
&& \langle DX, (I+L)^{-1}De^{i(\alpha Y + \beta G_{Y}) } \rangle \\
&=&\sum_{p,q\geq 1}pq\sum_{N\geq 0} \sum_{r=0} ^{N\wedge (q-1)} r! C_{q-1}^{r} C_{N}^{r} (1+N+(q-1)-2r)^{-1}\\
&&\times \sum_{a=0}^{(N+q-1-2r)\wedge (p-1)}I_{N+q-1-2r+p-2a}\int_{0}^{1}
\left( (h_{N}\tilde{\otimes } _{r} g(\cdot , \alpha ))\otimes _{a} f(\cdot , \alpha )\right) d\alpha
\end{eqnarray*}
and we can finish as in the proof of the previous lemma. \qed

\vskip0.3cm

An interesting property of the couples in ${\cal{A}}$ is that the conditional expectation given $X+Y$ of the mixed scalar products $\langle DX, D(-L)^{-1} Y\rangle$ and $\langle DY, D(-L)^{-1} X\rangle$ vanish.

\begin{lemma}\label{a1}
Assume that $(X,Y)$  belongs to the class ${\cal{A}}$. Then
\begin{equation*}
\mathbf{E}\left(   e^{it (X+Y)}  \langle DX, D(-L)^{-1} Y\rangle \right) =0 \mbox{ a.s. }
\end{equation*}
\end{lemma}
{\bf Proof: } We have
\begin{eqnarray*}
\mathbf{E}\left(   e^{it (X+Y)}  \langle DX, D(-L)^{-1} Y\rangle \right) &=& \mathbf{E} \frac{1}{it}\langle  De^{itX}, e^{itY}D(-L)^{-1}Y \rangle \\
&=& \frac{1}{it} \mathbf{E} \left(  e^{itX}  \delta (e^{itY} D(-L)^{-1}Y) \right)\\
&=&\frac{1}{it} \mathbf{E}\left( e^{itX} \left( e^{itY}\delta D(-L)^{-1}Y   -it  e^{itY}\langle DY, D(-L)^{-1}Y\rangle \right)\right) \\
&=&\frac{1}{it} \mathbf{E}\left( e^{itX} \left( e^{itY}Y   -it  e^{itY}G_{Y} \right)\right)
\end{eqnarray*}
where we used the fact that since $e^{itY}\in \mathbb{D}^{1,2}$ and $D(-L)^{-1}Y \in Dom (\delta)$, then $e^{itY}D(-L^{-1})Y \in Dom (\delta )$ and by (\ref{ip})
$\delta (e^{itY}(D(-L)^{-1}Y))= e^{itY}\delta (D(-L)^{-1}Y) -ite^{itY}\langle DY, D(-L)^{-1}Y\rangle$.
By using the fact that $(X,Y)$ belongs to the class ${\cal{A}}$  we obtain
\begin{eqnarray*}
\mathbf{E}\left(   e^{it (X+Y)}  \langle DX, D(-L)^{-1} Y\rangle \right) &=&\frac{1}{it} \mathbf{E}(e^{itX}) \left(   \mathbf{E}(e^{itY}Y ) -it\mathbf{E}( e^{itY}G_{Y}) \right).
\end{eqnarray*}
 Now, going in the converse direction
 \begin{equation*}
 \mathbf{E}(e^{itY}Y ) -it\mathbf{E}( e^{itY}G_{Y}) =\mathbf{E}\delta (e^{itY}(D(-L)^{-1}Y))=0.
\end{equation*} \qed

We are now answering the following question: let $(X_{n})_{n\geq 1}$ and $(Y_{n})_{n\geq 1}$ be two sequences of random variables such that for every $n\geq 1$ the couple $(X_{n},Y_{n})$ is  in the class ${\cal{A}}$. Suppose that the sum $X_{n}+ Y_{n}$ converges  to the normal law and is such that the upper bound from Stein's method is attained, in the sense that $\mathbf{E}(1-G_{X_{n}+Y_{n} }) ^{2}$ converges to zero. Could we then conclude that both $X_{n}$ and $Y_{n}$ converge in a strong sense to the normal laws $N(0,c_{1}) $ and  $N(0, c_{2})$ respectively? We will see that this is true in some particular case under a supplementary hypothesis on the sequences $X_{n}$ and $Y_{n}$.

\subsection{Wiener chaos stable random variables}

Let us denote another class of families of random variables where the central limit of the sum implies central limit for each component. The idea is to assume a property on the filtration generated by $X_{n}+ Y_{n}$. Let us denote by $J_{n}$ the orthogonal projection of $L^{2}(\Omega)$ on the $n$-th Wiener chaos. We recall the following definition (see \cite{NUZ}, \cite{UZ2}).
\begin{definition}
We will say that a sigma-algebra $\tau \subset {\cal{F}}$ is Wiener chaos stable if for every $n$, $J_{n}\left( L^{2}(\tau)\right) \subset L^{2}(\tau)$. In other words, if a random variable $F\in L^{2}(\tau)$ admits the chaos decomposition  $F=\sum_{n\geq 0} I_{n}(f_{n})$ then for every $n\geq 0$ the random variable $I_{n}(f_{n})$ is $\tau$ -measurable.
\end{definition}

\begin{remark}
The Wiener chaos stable property for sigma-algebras is equivalent to the $L^{-1}$- stable property. Recall that a sigma-algebra $\tau $ is $L^{-1}$ stable if  $L^{-1}(L_{0}^{2}(\tau))\subset L^{2}_{0}(\tau)$ where $L^{2}_{0}(\tau)$ is the set of $\tau$-measurable square integrable random variables with zero expectation. As a matter of fact, the sigma -algebra generated by \\
$I_{p}(f), \langle DI_{p}(f), h_{1}\rangle, \langle DI_{p}(f), h_{2} \rangle, \ldots, \langle D^{p-1} I_{p}(f),h_{i_{1}}\otimes ..\otimes h_{i_{p-1}}\rangle $, where $h_{i}, i\geq 1$ is a complete orthogonal sequence in $L^{2}[0,1]$, is Wiener stable (see \cite{NUZ}, \cite{UZ2}).
\end{remark}

\begin{theorem}
Suppose that for every $X_{n}=\sum_{n\geq 1} I_{k}(f_{k}^{n})$ and $Y_{n}= \sum_{l\geq 1} I_{l}(g_{l}^{n}) $ are such that $\mathbf{E}X_{n}^{2} \to _{n\to \infty} c_{1}$ and $\mathbf{E}Y_{n}^{2}\to _{n\to \infty} c_{2}$ (such that $c_{1}, c_{2}>0$ and $c_{1}+ c_{2}=1$). Assume that
 \begin{description}
 \item{i. }for every $n\geq 1$ the couple $(X_{n}, Y_{n})$ belongs to the class ${\cal{A}}$.
 \item{ii. } for every $n\geq 1$ the sigma-algebras $\sigma (X_{n}) $ and $\sigma (Y_{n}) $ are Wiener chaos stable.
 \end{description}
Assume also that $\mathbf{E}\left( 1-G_{X_{n}+Y_{n}} \right) ^{2} \to _{n\to \infty }0.$ Then
\begin{equation*}
\mathbf{E}\left( c_{1}-G_{X_{n}}\right) ^{2} \to _{n\to \infty }0 \mbox{ and  } \mathbf{E}\left( c_{2}-G_{Y_{n}}\right) ^{2} \to _{n\to \infty }0.
\end{equation*}

\end{theorem}
{\bf Proof: } We will show that under assumption ii., the random variable $G_{X_{n}}$ belongs to $\sigma (X_{n})$ for every $n\geq 1$. Since $X_{n}$ is $\sigma(X_{n})$ measurable and $\sigma (X_{n}) $ is Wiener chaos stable, we get that $I_{k}(f_{k}^{n})$ is $\sigma(X_{n})$ measurable for every $n,k$. Consequently, $I_{k}^{n}(f_{k}^{n})I_{l}^{n}(f_{l}^{n})$ is $\sigma(X_{n})$ measurable for every $n,k,l$ and by using the product formula we will have that
\begin{equation*}
I_{k+l-2r} \left( f^{n}_{k} \otimes _{r} f^{n} _{l} \right)
\end{equation*}
is $\sigma(X_{n})$ measurable for every $n,k,l\geq 1$ and $r=0,.., k\wedge l $. As a consequence we can easily obtain that $G_{X_{n}}$ is measurable with respect to $\sigma(X_{n})$ and similarly $G_{Y_{n}}$ is measurable with respect to $\sigma(Y_{n})$.
Assume now that $\mathbf{E}(1-G_{X_{n}+Y_{n}})^{2} \to _{n\to \infty} 0$. The asymptotic Cram\'er's Theorem \ref{cra} together with Proposition \ref{t1} imply that $\mathbf{E}(c_{1}-G_{X_{n}}| X_{n}) \to 0$ and $\mathbf{E}(c_{2}-G_{Y_{n}}| Y_{n}) \to 0$  a.s. and by the measurability of $G_{X_{n}}$ and $G_{Y_{n}}$ we obtain the conclusion. \qed

\subsection{Vectorial convergence of $X_{n}+ Y_{n}$ and $G_{X_{n}}+ G_{Y_{n}}$}

A second class of sequences of random variables for which the central limit can be broken in order to ensure the strong convergence of each summand  is inspired by \cite{NoPe2}.
\begin{theorem}
Assume that $\mathbf{E}X_{n}^{2} \to c_{1}$ and $\mathbf{E}Y_{n}^{2}\to c_{2}$ (such that $c_{1}, c_{2}>0$ and $c_{1}+ c_{2}=1$). Assume that for every $n\geq 1$ the couple of random variables $(X_{n}, Y_{n})$ belongs to ${\cal{A}}$. Suppose moreover that the vector
\begin{equation}\label{cond3}
\left( X_{n}+ Y_{n}, \frac{c_{1}-G_{X_{n}}+ c_{2}-G_{Y_{n}}}{ \mathbf{E} \left( (c_{1}-G_{X_{n}})^{2} + (c_{2}-G_{Y_{n}}) ^{2} \right) ^{\frac{1}{2}}} \right)
\end{equation}
converges as $n\to \infty$ to the vector $(N_{1}, N_{2})$ where $N_{1}, N_{2}$ are standard normal random variables with correlation $\rho$. Then if $X_{n}+ Y_{n} \to_{n \to \infty} N(0,1)$ implies that
\begin{equation*}
\mathbf{E}\left(c_{1}-G_{X_{n}}\right) ^{2} \to_{n \to \infty}0 \mbox{ and } \mathbf{E}\left(c_{2}-G_{Y_{n}}\right) ^{2} \to_{n \to \infty}0
\end{equation*}
\end{theorem}
{\bf Proof: }On one hand, we have that
\begin{equation}\label{u1}
\mathbf{E}\left( f'_{z} (X_{n}+ Y_{n})(c_{1}-G_{X_{n}}+ c_{2}- G_{Y_{n}})\right) \to _{n\to \infty}0.
\end{equation}
On the other hand, from the convergence of the vector (\ref{cond3}) we get
\begin{equation*}
\mathbf{E}\left( f'_{z} (X_{n}+ Y_{n})(c_{1}-G_{X_{n}}+ c_{2}- G_{Y_{n}})a_{n}^{-1} \right) \to f'_{z}(N_{1})N_{2}
\end{equation*}
where $a_{n}= \mathbf{E} \left( (c_{1}-G_{X_{n}})^{2} + (c_{2}-G_{Y_{n}}) ^{2}  \right) ^{\frac{1}{2}}$.
It follows from the proof of Theorem 3.1 in \cite{NoPe2} that we can  find a constant $c\in (0,1)$ such that
\begin{equation}\label{u2}
\left| \mathbf{E}\left( f'_{z} (X_{n}+ Y_{n})(c_{1}-G_{X_{n}}+ c_{2}- G_{Y_{n}})\right)\right| ^{2} \geq c  \mathbf{E} \left( (c_{1}-G_{X_{n}})^{2} + (c_{2}-G_{Y_{n}}) ^{2} \right) ^{2}.
\end{equation}
By combining the  relations (\ref{u1}) and (\ref{u2}), we obtain that
$ \mathbf{E} \left( (c_{1}-G_{X_{n}})^{2} + (c_{2}-G_{Y_{n}}) ^{2} \right) ^{2}\to _{n\to \infty} 0 $ and this gives the convergence of $X_{n}$ and $Y_{n}$ to
$N(0, c_{1})$ and $N(0, c_{2})$ respectively. \qed

\subsection{Random variables with independent chaos components}

In this part we prove that in the case when the chaotic  components  appearing in the decomposition of $X_{n}$ are mutually independent (and  the same is true for  $Y_{n}$) then the central limit of the sum implies the central limit of the summands (in a strong sense) under simple independence.
\begin{prop}
Assume that for every $n\geq 1$, $X_{n} =\sum_{k\geq 1} I_{k}(f^{n}_{k})$ and $Y_{n}= \sum_{l\geq 1}I_{l} (g^{n}_{l}) $ and
 \begin{equation*}
 \mathbf{E}X_{n}^{2} \to_{n \to \infty} c_{1}, \hskip0.5cm \mathbf{E} Y_{n}^{2} \to_{n \to \infty} c_{2}
 \end{equation*}
 with  $c_{1}, c_{2}>0$ and $c_{1}+ c_{2}=1$.
 Suppose that the following conditions are fulfilled
 \begin{description}
\item{i. } for every $n\geq 1$ the random variables $X_{n}$ and $Y_{n}$ are  independent
\item{ii. } for every $n\geq 1$, the random variables $(I_{k}(f^{n}_{k}) )_{k\geq 1}$ are pairwise independent the same holds for $(I_{k} (g^{n}_{l}))_{l\geq 1}$.
\end{description}
Then $X_{n}+Y_{n} \to N(0,1)$ implies
\begin{equation*}
\mathbf{E}\left( c_{1}-G_{X_{n}}\right) ^{2} \to_{n \to \infty}0 \mbox{ and } \mathbf{E}\left( c_{2}-G_{Y_{n}}\right) ^{2} \to_{n \to \infty}0.
\end{equation*}
\end{prop}
{\bf Proof: } The Theorem \ref{cra} implies that $X_{n}\to N(0, c_{1})$ and $Y_{n}\to N(0,1)$ in law. Corollary  1 and Assumption ii. gives that for every $k$ the sequence $I_{k}(f_{l}^{n})$ converges to a normal law as $k\to \infty$. Finally, we use  Remark  \ref{chafin} and Lemmas \ref{n11}, \ref{n12}. \qed

{\bf Acknowledgments. }We wish to thank the anonymous referee for valuable  comments on our manuscript.


TITLE: Inverse of a composite function from $\mathbb R$ to $\mathbb{R}^p$ to $\mathbb R$ again with a non-zero continuous gradient in a point
QUESTION [2 upvotes]: Let $D \subset \mathbb{R}^p$ and $f: D \to \mathbb{R}$ a continious function. Let $a$ be a point in $D$. Let  $\nabla f(a) \neq 0$ and let the gradient be continious at $a$.
To proof: there exists a function $c: [a, b] \to \mathbb{R}^p$ such that $f(c(t)) = t$ for all $t \in [a, b]$.
I know that an inverse function exists if the function is injective, and what partial derivatives are. I don't know how to do the chain rule in multiple variables however.

REPLY [1 votes]: Question's claim  as stated is not true:
For $p=1$,
let $\phi : R \to R$ be the well-known continuous function, but  differentiable no where.
Now Take $f(x):=x + x^2 \phi(x)$ then $f$ is continuous every where $Dom (f')= \{0\}$ and $f' (0) =1 >0$ so $f'$ is continuous. But there is no sub interval $(a,b)$ on which $f$ be invertible on. !


TITLE: Graphing $x^2+y^2=\lvert x \rvert + \lvert y \rvert$
QUESTION [2 upvotes]: I was given the following equation to graph.
$$x^2+y^2=\lvert x \rvert + \lvert y \rvert$$
Entering it in Desmos produces a four-leaf clover.
However, I'm wondering how one can deduce its shape logically. For instance, the equation $y=2(x-3)^2 + 6$ can be formed by transformations of $y=x^2$.
The left side of the equation would be part of a circle equation and the right side would be a rotated square. I don't see how combining the two would produce that graph, but maybe there's a different way? Any advice would be greatly appreciated.

REPLY [7 votes]: $x^2+y^2=x+y$ is a circle.
In $x^2+y^2=\lvert x \rvert + \lvert y \rvert$, the equation $x^2+y^2=x+y$ only holds when $x\ge0, y\ge0$, and so the circle gets cut into a petal.
The other three petals come from the considering the other quadrants.


TITLE: A query about tensor product of graded modules.
QUESTION [0 upvotes]: Let $A=\bigoplus_{n\geq 0}A_n$, $B=\bigoplus_{n\geq 0}B_n$ be two graded Noetherian rings, where $A_n\subset B_n$ and $A_0=B_0$ is a local ring. Suppose $B$ is a finitely generated $A$ module. Let $I$ be an ideal in $A_0$. Define $M=\bigoplus_{n\geq 0}M_n$ such that $M_0=I$ and $M_i=0$ for $i\geq 1$, with the property $am=0$ for all $a\in A_n$ with $n\geq 1$ and $m\in I$. Then is the following statement true
Thm: $M$ and $\bigoplus(\frac{B_n}{A_n}\otimes I)$ are finitely generated $A$ modules.
I think this result is true but I want to be sure. Any suggestion would be really helpful. Thanks in advance.

REPLY [1 votes]: To expand on my comment, if you're asking whether the modules $M$ and $B/A \otimes_A M$ are finitely generated, the answer is yes. Since $A$ is noetherian, so is $A_0$. In particular, $I$ is a finitely generated ideal in $A_0$, so it's easy to see that $M$ is a finitely generated module over $A$. Lastly, since $B/A$ is the quotient of a finitely generated module, it too is a finitely generated module, and so $B/A \otimes M$ is finitely generated as well.