text:
\begin{document}
\maketitle
\begin{abstract}
Let $\mathcal{H}$ be a right quaternionic Hilbert space and let $T$ be a bounded normal right quaternionic linear operator on $\mathcal{H}$. In this paper, we prove that there exists a unique spectral measure $E$ in $\mathcal{H}$ such that
$$T=\int_{\sigma_S(T)}\lambda dE_\lambda,$$
where $\sigma_S(T)$ denotes the spherical spectrum of $T$.
\end{abstract}

\section{Introduction}
Let $\mathcal{H}$ be a complex Hilbert space and let $\mathcal{B}(\mathcal{H})$ denotes the set of all bounded quaternionic right linear operators from $\mathcal{H}$ to $\mathcal{H}$. Let $T\in\mathcal{B}(\mathcal{H})$ be normal, it is well-known that there exists a unique spectral measure $E$ in $\mathcal{H}$ such that
\begin{align}
T=\int_{\sigma(T)}\lambda dE_\lambda,\label{6eq1}
\end{align}
where $\sigma(T)$ denotes the spectrum of $T$ (see, e.g., \cite[Theorem 3.15]{6C}). A large part of the theory of normal operators depends on this fact. \\

Several authors discussed the spectral theorem based on the S-spectrum \cite{6ACK,6CGK,6RS,6V}. Results related to the quaternionic spectral theorem can furthermore be found in \cite{6Gh,6RS2}.\\

The reader is encouraged to see the book of Colombo, Gantner and Kimsey \cite{6CGK} for a very good and detailed write-up of spectral theory based on the S-spectrum. In this book, the authors showed that if $T$ is a bounded normal right quaternionic linear operator on a right quaternionic Hilbert space $\mathcal{H}$, then there exists a unique spectral measure $E^i$ on $\Omega_i^+:=\sigma_S(T)\cap\mathbb{C}_i^+$ (see, \cite[Theorem 11.2.1]{6CGK}) such that
$$\langle  Tx,y\rangle=\int_{\Omega_i^+}\text{Re}(\lambda) d\langle E_\lambda^i x, y\rangle+\int_{\Omega_i^+}\text{Im}(\lambda) d\langle E_\lambda^i Jx, y\rangle\;\;\text{ for all }\;\; x,y\in\mathcal{H},$$
where $J$ is the imaginary component in the $T = A+JB$ decomposition (see, \cite[Theorem 9.3.5]{6CGK}) of the operator $T$.

In \cite{6Ghh}, the authors showed a version of the spectral theorem (see, \cite[Theorem 4.1]{6Ghh}). Then in \cite[Theorem 14.4.2]{6CGK}, the authors  rewrite \cite[Theorem 4.1]{6Ghh} in the terminology of spectral systems as follows.\\
Let $T = A + JB\in\mathcal{B}_R(\mathcal{H})$ be a normal operator. There exists a unique quaternionic spectral measure $E$ on the $\sigma$-algebra of axially symmetric Borel subsets of $\sigma_S(T)$ \big(the spherical spectrum of $T$\big), such that $(E, J)$ is a spectral system and such that
$$T=\int_{\sigma_S(T)}\lambda dE_J(\lambda).$$

In this paper, we give a canonical decomposition of a bounded normal right quaternionic linear operator on a right quaternionic Hilbert space using only the classical decomposition of bounded normal operators in the complex Hilbert spaces.\\

Before starting, we recall some basic definitions and results which are useful in the sequel.
\subsection*{Quaternions}
Let $\mathbb{H}$ be the set of all elements, called quaternions, of the form $q = a + b\mathrm{i} + c\mathrm{j} + d\mathrm{k}; a,b,c,d\in\mathbb{R}$, where $\mathrm{i},\mathrm{j}$ and $\mathrm{k}$ are quaternion units satisfying:
\begin{equation}
\label{6e1}
\mathrm{i}^2=\mathrm{j}^2=\mathrm{k}^2=\mathrm{i}\mathrm{j}\mathrm{k}=-1.
\end{equation}
Then $\mathbb{H}$ is a non-commutative $\mathbb{R}$-algebra with the addition defined same as in $\mathbb{C}$ and multiplication given by Equations (\ref{6e1}). For a given $q\in\mathbb{H}$, we define the real part of $q$, $\text{Re}(q):=a$ and the imaginary part of $q$, $\text{Im}(q):=b\mathrm{i} + c\mathrm{j} + d\mathrm{k}$. The conjugate and the modulus of $q$ are given respectively by
$$\bar{q}=a-b\mathrm{i}-c\mathrm{j}-d\mathrm{k},\;\vert q\vert=\sqrt{q\bar{q}}=\sqrt{a^2+b^2+c^2+d^2}.$$
The unit sphere of imaginary quaternions is given by
$$\mathbb{S}=\{q\in\mathbb{H}:q^2=-1\}.$$
Let $p$ and $q$ be two quaternions. $p$ and $q$  are said to be conjugated, if there is $s\in\mathbb{H}\setminus\{0\}$ such that $p=sqs^{-1}$. The set of all quaternions conjugated with $q$, is equal to the 2-sphere
$$[q] =\{\text{Re}(q)+\vert\text{Im}(q)\vert i : i \in \mathbb{S}\}=\text{Re}(q)+\vert\text{Im}(q)\vert\mathbb{S}.$$
For every $U\subseteq\mathbb{H}$, $[U]$ denotes the set of all quaternions conjugated with the elements of $U$;
$$[U]:=\{[q]:q\in U\}.$$
For every $i \in \mathbb{S}$, denote by $\mathbb{C}_i$ the real subalgebra of $\mathbb{H}$ generated by $i$; that is,
$$\mathbb{C}_i:=\{\alpha+\beta i\in\mathbb{H}: \alpha,\beta\in\mathbb{R}\}.$$
We say that $U\subseteq\mathbb{H}$ is axially symmetric if $[q] \subset U$ for every $q \in U$.\\

For a thorough treatment of the algebra of quaternions $\mathbb{H}$, the reader is referred, for instance, to \cite{6Gh}.
\subsection*{Quaternionc Hilbert spaces}
\begin{definition}
Let $(\mathcal{H},+)$ be an abelian group. $\mathcal{H}$ is a right quaternionic vector space if it is endowed with a right quaternionic multiplication $(\mathcal{H},\mathbb{H})\rightarrow \mathcal{H}$, $(x,q) \mapsto xq$ such that for all $x,y \in \mathcal{H}$ and all $p, q \in\mathbb{H}$, we have:
\begin{enumerate}
\item[(i)]  $(x+y)q = xq + yq$;
\item[(ii)]  $x(p + q) = xp + xq$;
\item[(iii)] $(xp)q = x(pq)$;
\item[(iv)]  $ x1=x$.
\end{enumerate}
\end{definition}
\begin{definition}
Let $\mathcal{H}$ be a right quaternionic vector space. A Hermitian quaternionic scalar product is a map $\langle\cdot,\cdot\rangle:\mathcal{H}\times \mathcal{H}\rightarrow \mathbb{H},(x,y)\mapsto \langle x,y\rangle$ that satisfies the following properties.
\begin{itemize}
\item[(i)] $\langle xp+yq,z\rangle=\langle x,z\rangle p+\langle y,z\rangle q$;
\item[(ii)] $\overline{\langle x,y\rangle}=\langle y,x\rangle$;
\item[(iii)] $\langle x,x\rangle\geq0$ and $\langle x,x\rangle=0\Leftrightarrow x=0$;
\end{itemize}
for all $p,q\in\mathbb{H}$ and all $x, y, z \in \mathcal{H}$.
\end{definition}
Suppose that $\mathcal{H}$ is equipped with such a Hermitean quaternionic scalar product. Then we can define the quaternionic norm $\Vert\cdot\Vert : \mathcal{H}\rightarrow \mathbb{R}^+$ by setting
$$\Vert x\Vert=\sqrt{\langle x,x\rangle}\;\;\;\text{ for every }\;\;\; x\in\mathcal{H}.$$

A right quaternionic vector space $\mathcal{H}$ is said to be a right quaternionic Hilbert space if it is equipped with a Hermitean quaternionic scalar product which is complete with respect to its natural distance $d(x,y):=\Vert x-y\Vert$ for all $x,y\in\mathcal{H}$.
\begin{definition}
Let $\mathcal{H}$ be a right quaternionic Hilbert space. A right linear operator on $\mathcal{H}$ is a map $T : \mathcal{H} \rightarrow \mathcal{H}$ such that:
$$ T(xq+y) = (Tx)q + Ty\;\;\text{ for all }\;\; x, y\in \mathcal{H}\;\;\text{ and all }\;\;q\in\mathbb{H}.$$
A right linear operator $T$ on $\mathcal{H}$ is called bounded if
$$\Vert T\Vert:=\sup\{\Vert Tx\Vert: x\in \mathcal{H}, \Vert x\Vert=1\}<\infty.$$
For all $r\in\mathbb{R}$ and all $x\in \mathcal{H}$. Define  $(Tr)x := (Tx)r$ and $rT := Tr$. The adjoint of $T$ will be denoted by $T^*$. The set of all right linear bounded operators on $\mathcal{H}$ is denoted by $\mathcal{B}_R(\mathcal{H})$. For $T\in\mathcal{B}_R(\mathcal{H})$, let $\mathcal{N}(T)$ denote the null space of $T$, and let $\mathcal{R}(T)$ denote the range of $T$.
\end{definition}
\subsection*{Spherical spectrum}
Colombo, Sabadini and Struppa \cite{6CSS} extended the definitions of the spectrum and resolvent in quaternionic Hilbert spaces as follows.
\begin{definition}
Let $\mathcal{H}$ be a right quaternionic Hilbert space and let $T \in \mathcal{B}_R(\mathcal{H})$. For $q \in\mathbb{H}$, we set
$$Q_q(T) := T^2 -2\text{Re}(q)T + \vert q\vert^2I.$$
Where $I$ is the identity operator on $\mathcal{H}$.
We define the spherical resolvent set $\rho_S(T)$ of $T$ as
$$\rho_S(T) := \{q \in\mathbb{H} : Q_q(T) \text{ is invertible in } \mathcal{B}_R(\mathcal{H})\},$$
and we define the spherical spectrum $\sigma_S(T)$ of $T$ as
$$\sigma_S(T):=\mathbb{H}\setminus\rho_S(T).$$
\end{definition}
\begin{proposition}
\label{6p2}
Let $T\in\mathcal{B}_R(\mathcal{H})$. Then $\sigma_S(T)$ is a nonempty compact and axially symmetric set.
\end{proposition}

For more details on the notion of spherical spectrum, the reader may consult, e.g., \cite{6BB,6CGK,6CSS,6Gh}.
\section{The Spectral Theorem and Consequences}
In the following, $\mathcal{H}$ will be a fixed right quaternionic Hilbert space and $\langle\cdot,\cdot\rangle$ denotes the inner product in $\mathcal{H}$.
\begin{lemma}
\label{6l1}
Let $i\in\mathbb{S}$. The application $\langle\cdot,\cdot\rangle_i:\mathcal{H}\times \mathcal{H}\rightarrow \mathbb{C}_i,(x,y)\mapsto\mathrm{Re}\big(\langle x,y\rangle\big)-\mathrm{Re}\big(\langle x,y\rangle i\big)i$ satisfies the following properties.
\begin{itemize}
\item[(i)] $\langle xp+yq,z\rangle_i=\langle x,z\rangle_ip+\langle y,z\rangle_iq$;
\item[(ii)] $\overline{\langle x,y\rangle_i}=\langle y,x\rangle_i$;
\item[(iii)] $\langle x,x\rangle_i\geq0$ and $\langle x,x\rangle_i=0\Leftrightarrow x=0$;
\end{itemize}
for all $p,q\in\mathbb{C}_i$ and all $x, y, z \in \mathcal{H}$.
\end{lemma}
\begin{proof}
Let $p=\alpha+\beta i\in\mathbb{C}_i$ with $\alpha,\beta\in\mathbb{R}$ and let $x, y\in \mathcal{H}$. Then\\

\noindent (i) $\langle xp,y\rangle_i=\langle x(\alpha+\beta i),y\rangle_i=\mathrm{Re}\big(\langle x(\alpha+\beta i),y\rangle\big)-\mathrm{Re}\big(\langle x(\alpha+\beta i),y\rangle i\big)i=\mathrm{Re}\big(\langle x,y\rangle\big)\alpha+\mathrm{Re}\big(\langle x,y\rangle i\big)\beta-\mathrm{Re}\big(\langle x,y\rangle i\big)i\alpha+\mathrm{Re}\big(\langle x,y )\rangle\big)i\beta=\langle x,y\rangle_ip.$\\

\noindent (ii) $\langle y,x\rangle_i=\mathrm{Re}\big(\langle y,x\rangle\big)-\mathrm{Re}\big(\langle y,x\rangle i\big)i=\mathrm{Re}\big(\overline{\langle x,y\rangle}\big)+\mathrm{Re}\big(\overline{i\langle x,y\rangle}\big)i=\mathrm{Re}\big(\langle x,y\rangle\big)+\mathrm{Re}\big(i\langle x,y\rangle\big)i=\mathrm{Re}\big(\langle x,y\rangle\big)+\mathrm{Re}\big(\langle x,y\rangle i\big)i=\overline{\langle x,y\rangle_i}.$\\

\noindent (iii) $\langle x,x\rangle_i=\langle x,x\rangle\geq0$ and $\langle x,x\rangle_i=0\Leftrightarrow \langle x,x\rangle=0 \Leftrightarrow x=0$.
\end{proof}
\begin{definition}
Let $i\in\mathbb{S}$. We denote the space $\mathcal{H}$ considered as a complex vector space over the complex field $\mathbb{C}_i$ by $\mathcal{H}_i$ with respect to the structure induced by the right quaternionic Hilbert space $\mathcal{H}$; its sum is the sum of $\mathcal{H}$, its complex scalar multiplication is the right scalar multiplication of $\mathcal{H}$ restricted to $\mathbb{C}_i$.
\end{definition}
\begin{corollary}
\label{6c1}
For any $i\in\mathbb{S}$, the map $\langle\cdot,\cdot\rangle_i$ is an inner product on $\mathcal{H}_i$. Moreover, $\mathcal{H}_i$ is a complex Hilbert space with respect to the norm $\Vert\cdot\Vert_i:=\sqrt{\langle\cdot,\cdot\rangle_i}$.
\end{corollary}
\begin{remark}
Let $x,y\in \mathcal{H}$ and let $i\in\mathbb{S}$. Then
\begin{itemize}
\item[(i)] If $\langle x,y\rangle\in\mathbb{C}_i$. Then $\langle x,y\rangle_i=\langle x,y\rangle$.
\item[(ii)] $\langle x,x\rangle_i=\langle x,x\rangle$, hence $\Vert \cdot\Vert_i=\Vert \cdot\Vert$ where $\Vert \cdot\Vert$ is the norm on $\mathcal{H}$ and then the topologies of $\mathcal{H}$ and $\mathcal{H}_i$ coincide.
\end{itemize}
\end{remark}
\begin{definition}
\label{6d1}
Let $\Omega$ be a nonempty axially symmetric Borel subset of $\mathbb{H}$ and let $\mathcal{A}_{\Omega}^s$ be the $\sigma$-algebra of axially symmetric Borel subsets of $\Omega$. A spectral measure in a right quaternionic Hilbert space $\mathcal{H}$ is a mapping $E:\mathcal{A}_{\Omega}^s\rightarrow \mathcal{B}_R(\mathcal{H})$ such that
\begin{itemize}
\item[(i)] $E(\omega)$ is a bounded orthogonal projection for every $\omega\in\mathcal{A}_{\Omega}^s$;
\item[(ii)] $E(\emptyset) = 0$ and $E(\Omega) = I$;
\item[(iii)] $E(\omega_1\cap\omega_2) = E(\omega_1)E(\omega_2)$ for every $\omega_1,\omega_2\in\mathcal{A}_{\Omega}^s$;
\item[(iv)] $E(\cup_{n}\omega_n) =\sum_{n} E(\omega_n)$ whenever $\{\omega_n\}$ is a countable collection of pairwise disjoint sets in $\mathcal{A}_{\Omega}^s$ (i.e., $E$ is countably additive).
\end{itemize}
\end{definition}

Let $T\in\mathcal{B}_R(\mathcal{H})$ and $i\in\mathbb{S}$. Let $T_i$ be the linear operator on $\mathcal{H}_i$ defined by $T_i(x):=Tx$ for all $x\in \mathcal{H}_i$. Clearly, $T_i\in\mathcal{B}(\mathcal{H}_i)$. In order to show the spectral theorem, we need the following lemma.
\begin{lemma}
\label{6l4}
Let $T\in\mathcal{B}_R(\mathcal{H})$ be normal and let $i\in\mathbb{S}$. Then $(T_i)^*=(T^*)_i$, and furthermore, if $T$ is normal, then so is $T_i$.
\end{lemma}
\begin{proof}
Let $x,y\in \mathcal{H}_i$. We have
\begin{align*}
\langle T_ix,y\rangle_i&=\mathrm{Re}\big(\langle T_ix,y\rangle\big)-\mathrm{Re}\big(\langle T_ix,y\rangle i\big)i=\mathrm{Re}\big(\langle Tx,y\rangle\big)-\mathrm{Re}\big(\langle Tx,y\rangle i\big)i\\
&=\mathrm{Re}\big(\langle x,T^*y\rangle\big)-\mathrm{Re}\big(\langle x,T^*y\rangle i\big)i=\langle x,T^*y\rangle_i=\langle x,(T^*)_iy\rangle_i.
\end{align*}
Therefore, $(T_i)^*=(T^*)_i$. Moreover, if $T$ is normal, then $(T_i)^*T_i=(T^*)_iT_i=(T^*T)_i=(TT^*)_i=T_i(T^*)_i=T_i(T_i)^*$. Hence $T_i$ is normal.
\end{proof}
Let $T\in\mathcal{B}_R(\mathcal{H})$ be normal and let $i\in\mathbb{S}$. By \cite[Theorem 3.15]{6C}, there exists a unique complex spectral measure $E^i$ of the spectral decomposition of $T_i$ \big($E^i:\mathcal{A}_{\sigma(T_i)}\rightarrow \mathcal{B}(\mathcal{H}_i)$ where $\mathcal{A}_{\sigma(T_i)}$ is the $\sigma$-algebra of Borel subsets of $\sigma(T_i)$\big) such that
$$T_i=\int_{\sigma(T_i)}\lambda dE^i_\lambda.$$
That is $\langle  T_ix,y\rangle_i=\int_{\sigma(T_i)}\lambda d\langle E^i_\lambda x, y\rangle_i\;$ for all $x,y\in \mathcal{H}_i$.\\

Let $\omega\in\mathcal{A}_{\sigma(T_i)}$, then $E^i(\omega)$ is linear on $\mathcal{H}_i$. Let $\mathcal{A}_{\sigma(T_i)}^s$ be  the $\sigma$-algebra of Borel symmetric (with respect to the real axis) subsets  of $\sigma(T_i)$, the following lemma tells us that $E^i(\omega)$ is actually right quaternionic linear on $\mathcal{H}$ for all $\omega\in\mathcal{A}_{\sigma(T_i)}^s$.
\begin{lemma}
\label{6l3}
Let $T\in\mathcal{B}_R(\mathcal{H})$ be normal and $i\in\mathbb{S}$. Let $E^i$ be the complex spectral measure of the spectral decomposition of $T_i$. Then
$$E^i(\omega)\in\mathcal{B}_R(\mathcal{H}) \;\;\text{ for all }\;\; \omega\in\mathcal{A}_{\sigma(T_i)}^s.$$
\end{lemma}
\begin{proof}
An analytic extension of $z\mapsto (I_iz-T_i)^{-1}x$ with $x\in\mathcal{H}_i$ is a holomorphic function $f$ defined on a set $\mathcal{D}(f)\supset\rho(T_i)$ \big(where $\rho(T_i)$ is the resolvent set of $T_i$\big) such that $(I_iz-T_i)f(z)=x$ for all $z\in \mathcal{D}(f)$. $T_i$ is a bounded spectral operator in $\mathcal{H}_i$, then the function $z\mapsto (I_iz-T_i)^{-1}x$ has a unique maximal analytic extension. The local resolvent $\rho(x)$ is the domain of the unique maximal analytic extension of $z\mapsto(I_iz-T_i)^{-1}x$, and the local spectrum $\sigma(x)$ is the complement of $\rho(x)$ in $\mathbb{C}_i$ \big(see, \cite[Chapter XV]{6Dun}\big).\\
Let $\omega\in\mathcal{A}_{\sigma(T_i)}^s$ and let $\overline{\omega}$ be the closure of $\omega$. By \cite[ Theorem XV.3.4]{6Dun}, we have
$$\mathcal{R}\big(E^i(\overline{\omega})\big)=\{x\in\mathcal{H}_i:\sigma(x)\subseteq\overline{\omega}\}.$$
Let $j\in\mathbb{S}$ such that $ij=-ji$, then $\{1,i,j,ij\}$ is a basis of $\mathbb{H}$. It is sufficient to show that $E^i(\omega)(xj)=E^i(\omega)(x)j$ for all $x\in\mathcal{H}$.\\
Let $f$ be the unique maximal analytic extension of $z\mapsto(I_iz-T_i)^{-1}x$ defined on $\rho(x)$. The mapping $g:z\mapsto f(\bar{z})j$ defined on $\rho(x)^*:=\{\bar{z}:z\in\rho(x)\}$ is holomorphic. Indeed,
$$\lim_{h\rightarrow 0}\big(f(\overline{z+h})j-f(\bar{z})j\big)\frac{1}{h}=\lim_{h\rightarrow 0}\big(f(\bar{z}+\bar{h})-f(\bar{z})\big)\frac{1}{\bar{h}}j=f^\prime(\bar{z})j.$$
Moreover, if $z\in\rho(x)^*$, then
\begin{align*}
(I_iz-T_i)g(z)&=(I_iz-T_i)(f(\bar{z})j)=f(\bar{z})jz-T(f(\bar{z})j)=f(\bar{z})\bar{z}j-T(f(\bar{z}))j\\
&=\big((I_i\bar{z}-T_i)f(\bar{z})\big)j=xj.
\end{align*}
Hence $z\mapsto f(\bar{z})j$ is an analytic extension of $z\mapsto(I_iz-T_i)^{-1}(xj)$ that is defined on $\rho(x)^*$. Consequently $\rho(x)^*\subseteq\rho(xj)$. Similar arguments show that $\rho(xj)^*\subseteq\rho(-x)=\rho(x)$. Altogether, we obtain $\rho(x)=\rho(xj)^*$, and so $\sigma(xj)=\sigma(x)^*$. Thus
\begin{align}
\mathcal{R}\big(E^i(\overline{\omega})\big)=\{x\in\mathcal{H}_i:\sigma(x)\subseteq\overline{\omega}\}=\{xj\in\mathcal{H}_i:\sigma(x)\subseteq\overline{\omega}\}=\mathcal{R}\big(E^i(\overline{\omega})\big)j.
\label{6eq3}
\end{align}
If we choose an increasing sequence of closed sets $\omega_n\in\mathcal{A}_{\sigma(T_i)}^s$ with $\cup \omega_n=\sigma(T_i)\setminus\overline{\omega}$, we therefore have $E^i\big(\sigma(T_i)\setminus\overline{\omega}\big)x=\lim_{n\rightarrow\infty}E^i(\omega_n)x$ for all $x\in\mathcal{H}$. Hence by (\ref{6eq3}), it follows that
\begin{align}
\mathcal{R}\big(E^i(\sigma(T_i)\setminus\overline{\omega})\big)=\mathcal{R}\big(E^i(\sigma(T_i)\setminus\overline{\omega})\big)j.
\label{6eq4}
\end{align}
Now observe that
$$\mathcal{A}:=\{\Lambda\in\mathcal{A}_{\sigma(T_i)}^s:\mathcal{R}\big(E^i(\Lambda)\big)=\mathcal{R}\big(E^i(\Lambda)\big)j\;\text{ and }\;\mathcal{R}\big(E^i(\sigma(T_i)\setminus\Lambda)\big)=\mathcal{R}\big(E^i(\sigma(T_i)\setminus\Lambda)\big)j\}$$
is a $\sigma$-algebra that contains closed sets. Hence $\mathcal{A}=\mathcal{A}_{\sigma(T_i)}^s.$ Therefore,
\begin{align}
\mathcal{R}\big(E^i(\omega)\big)=\mathcal{R}\big(E^i(\omega)\big)j,
\end{align}
and so
$$\mathcal{N}\big(E^i(\omega)\big)=\mathcal{R}\big(E^i(\sigma(T_i)\setminus\omega)\big)=\mathcal{R}\big(E^i(\sigma(T_i)\setminus\omega)\big)j=\mathcal{N}\big(E^i(\omega)\big)j.$$
Since $\mathcal{H}=\mathcal{R}\big(E^i(\omega)\big)\oplus\mathcal{N}\big(E^i(\omega)\big)$, $xj=E^i(\omega)(xj)+\big(I-E^i(\omega)\big)(xj)$ and  $xj=E^i(\omega)(x)j+\big(I-E^i(\omega)\big)(x)j$ for all $x\in\mathcal{H}$. It follows that $E^i(\omega)(xj)=E^i(\omega)(x)j$. This completes the proof.
\end{proof}
Let $T\in\mathcal{B}_R(\mathcal{H})$ be normal and let $i,j\in\mathbb{S}$. The following lemma gives a relation between the complex spectral measures $E^i$ and $E^j$ of the spectral decompositions of $T_i$ and $T_j$, respectively.
\begin{lemma}
\label{6l2}
Let $i,j\in\mathbb{S}$ and $s\in\mathbb{H}\setminus\{0\}$ such that $s^{-1}i s=j$. Let $T\in\mathcal{B}_R(\mathcal{H})$ be normal, and let $E^i$ and $E^j$ be the complex spectral measures of the spectral decompositions of $T_i$ and $T_j$, respectively. Then
$$E^i(\omega)=E^j\big(s^{-1}\omega s\big) \;\;\text{ for all }\;\; \omega\in\mathcal{A}_{\sigma(T_i)}^s,$$
where $s^{-1}\omega s:=\{s^{-1}q s:q\in\omega\}$.
\end{lemma}
\begin{proof}
Let $x\in \mathcal{H}$ and let $P\in\mathbb{R}[X,Y]$. Define
\begin{align}
p(T):=P(T,T^*)\;\;\text{ and }\;\;  p(\lambda):=P(\lambda,\bar{\lambda})\;\;\text{ for all }\lambda\in\mathbb{H}.\label{6eq2}
\end{align}
We have
$$\langle p(T_j)^*p(T_j) x,x\rangle_j=\int_{\sigma(T_j)}\vert p(\lambda)\vert^2 d\langle E^j_\lambda x, x\rangle_j=\int_{\sigma(T_j)}\vert p(\lambda)\vert^2 d\langle E^j_\lambda x, x\rangle.$$
By \cite[Proposition 3.8]{6BB}, it follows that $\sigma(T_j)=s^{-1}\sigma(T_i)s$, hence
$$\int_{\sigma(T_j)}\vert p(\lambda)\vert^2 d\langle E^j_\lambda x, x\rangle=\int_{\sigma(T_i)}\vert p(\lambda)\vert^2 d\langle E^j_{s^{-1}\lambda s} x, x\rangle.$$
On the other hand, we have
$$\langle p(T_i)^*p(T_i) x,x\rangle_i=\int_{\sigma(T_i)}\vert p(\lambda)\vert^2 d\langle E^i_\lambda x, x\rangle_i=\int_{\sigma(T_i)}\vert p(\lambda)\vert^2 d\langle E^i_\lambda x, x\rangle.$$
Since $\langle p(T)^*p(T) x,x\rangle\in\mathbb{R}$, $\langle p(T_i)^*p(T_i) x,x\rangle_i=\langle p(T_j)^*p(T_j) x,x\rangle_j$. Thus
$$\int_{\sigma(T_i)}\vert p(\lambda)\vert^2 d\langle E^j_{s^{-1}\lambda s} x, x\rangle=\int_{\sigma(T_i)}\vert p(\lambda)\vert^2 d\langle E^i_\lambda x, x\rangle.$$
Set
$$\mu_x(\omega):=\langle E^i(\omega) x, x\rangle\;\;\text{ and }\;\; \nu_x(\omega):=\langle E^j(s^{-1}\omega s) x, x\rangle\;\; \text{ for all }\;\;\omega\in\mathcal{A}_{\sigma(T_i)}.$$
Let $\mathcal{P}$ be the set of all polynomials $P$ in the real variables $\alpha,\beta$ and with real coefficients $\sigma(T_i):\;\lambda=\alpha+\beta i\mapsto P(\alpha,\beta)$. By the Stone-Weierstrass Theorem for real functions, $\mathcal{P}$ is dense in the Banach space $\big(\mathcal{C}(\sigma(T_i)),\Vert\cdot\Vert_\infty\big)$ \big(the set of all real-valued continuous functions on $\sigma(T_i)$\big). Since $\sigma(T_i)$ is a compact set in $\mathbb{C}_i$ and $\mu_x$ is a finite positive measure on $\mathcal{A}_{\sigma(T_i)}$, it follows that $\mathcal{P}$ is also dense in the normed space $\big(\mathcal{C}(\sigma(T_i)),\Vert\cdot\Vert_2\big)$, which in turn is a dense linear manifold of the Banach space $\big(L^2(\sigma(T_i),\mu_x), \Vert\cdot\Vert_2\big)$, and hence $\mathcal{P}$ is dense in $\left(L^2\big(\sigma(T_i),\mu_x\big), \Vert\cdot\Vert_2\right)$. The same conclusion holds for $\nu_x$. It follows that
$$\int_{\sigma(T_i)}\chi_\omega(\lambda) d\langle E^j_{s^{-1}\lambda s} x, x\rangle=\int_{\sigma(T_i)}\chi_\omega(\lambda) d\langle E^i_\lambda x, x\rangle\;\;\text{ for all }\;\; \omega\in\mathcal{A}_{\sigma(T_i)}.$$
That is $\langle E^j(s^{-1}\omega s) x, x\rangle=\langle E^i(\omega) x, x\rangle$ for all $\omega\in\mathcal{A}_{\sigma(T_i)}$.\\ Let $\omega\in\mathcal{A}_{\sigma(T_i)}^s$. By Lemma \ref{6l3}, $E^i(\omega)\in\mathcal{B}_R(\mathcal{H})$ and so $E^i(\omega)$ is $\mathbb{C}_j$-linear. Thus $\langle \big(E^i(\omega)-E^j(s^{-1}\omega s)\big) x, x\rangle_j=\langle \big(E^i(\omega)-E^j(s^{-1}\omega s)\big) x, x\rangle=0$ for all $x\in\mathcal{H}_j$, hence $E^i(\omega)-E^j(s^{-1}\omega s)=0$ (see, e.g., \cite[Theorem 2.14]{6C}). It follows that $E^i(\omega)=E^j(s^{-1}\omega s)$. This completes the proof.
\end{proof}
We are now ready to show the main result of this paper: the canonical decomposition of a bounded normal right quaternionic linear operator, the quaternionic analogue of \cite[Theorem 3.15]{6C}.
\begin{theorem}
\label{6t1}
Let $T\in\mathcal{B}_R(\mathcal{H})$ be normal and let $J$ be the imaginary component in the $T = A+JB$ decomposition of $T$. There exists a unique spectral measure $E$ in $\mathcal{H}$ such that
$$\langle  Tx,y\rangle=\int_{\sigma_S(T)}\text{Re}(\lambda) d\langle E_\lambda x, y\rangle+\int_{\sigma_S(T)}\vert\text{Im}(\lambda)\vert d\langle E_\lambda Jx, y\rangle\;\text{ for all }\; x,y\in\mathcal{H}.$$
Notation: $$T=\int_{\sigma_S(T)}\lambda dE_\lambda.$$
\end{theorem}
\begin{proof}
We shall split the proof into three parts.\\

\noindent (i) Let $i\in\mathbb{S}$ and let $E^i$ be the complex spectral measure of the spectral decomposition of $T_i$. We claim that the map $E:\mathcal{A}_{\sigma_S(T)}^s \rightarrow\mathcal{B}_R(\mathcal{H})$ defined by
$E\big(\Omega\big)=E^i\big(\Omega\cap\mathbb{C}_i\big)$ is a spectral measure in $\mathcal{H}$.\\
To prove this, first note that $\mathcal{A}_{\sigma(T_i)}^s=\mathcal{A}_{\sigma_S(T)}^s\cap\mathbb{C}_i:=\{\Omega\cap\mathbb{C}_i:\Omega\in\mathcal{A}_{\sigma_S(T)}^s\}$. Indeed, $\sigma(T_i)=\sigma_S(T)\cap\mathbb{C}_i$ \big(see, \cite[Proposition 3.8]{6BB}\big) and $\mathbb{C}_i$ is closed in $\mathbb{H}$. Then one can easily check that $E$ is a spectral measure in $\mathcal{H}$.\\
Now, let us prove that $E$ is independent of the choice of the imaginary unit $i\in\mathbb{S}$. Let $j\in\mathbb{S}$, $s\in\mathbb{H}\setminus\{0\}$ such that $s^{-1}i s=j$ and $E^j$ be the complex spectral measure of the spectral decomposition of $T_j$. We have
$$\Omega\cap\mathbb{C}_j=s^{-1}\big(\Omega\cap\mathbb{C}_i\big)s\;\;\text{ for all }\;\;\Omega\in\mathcal{A}_{\sigma_S(T)}^s.$$
Hence by Lemma \ref{6l2},
$$E\big(\Omega\big)=E^i\big(\Omega\cap\mathbb{C}_i\big)=E^j\big(s^{-1}\big(\Omega\cap\mathbb{C}_i\big)s\big)=E^j\big(\Omega\cap\mathbb{C}_j\big).$$

\noindent (ii) Let $A$, $B$, and $J$ be as in \cite[Theorem 9.3.5]{6CGK}. Let $x,y\in \mathcal{H}$ and let $i\in\mathbb{S}$ such that $\langle T x,y\rangle\in\mathbb{C}_i$. Then
\begin{align*}
\langle  A_ix,y\rangle_i=\int_{\sigma(T_i)}\text{Re}(\lambda) d\langle E^i_\lambda x, y\rangle_i\;\text{ and }\;
\langle  B_iJx,y\rangle_i=\int_{\sigma(T_i)}\vert\text{Im}(\lambda)\vert d\langle E^i_\lambda Jx, y\rangle_i.
\end{align*}
It follows that
\begin{align*}
\langle  Tx,y\rangle=\langle  (A+BJ)x,y\rangle_i=\int_{\sigma(T_i)}\text{Re}(\lambda) d\langle E^i_\lambda x, y\rangle_i+\int_{\sigma(T_i)}\vert\text{Im}(\lambda)\vert d\langle E^i_\lambda Jx, y\rangle_i.
\end{align*}
Let $\omega\in\mathcal{A}_{\sigma(T_i)}$, by the polarization identity, we have
\begin{align*}
\langle E^i(\omega) x, y\rangle_i=&\frac{1}{4}\Big(\langle E^i(\omega) (x+y), x+y\rangle_i-\langle E^i(\omega) (x-y), x-y\rangle_i\\
&+i\langle E^i(\omega) (x+yi), x+yi\rangle_i-i\langle E^i(\omega) (x-yi), x-yi\rangle_i\Big).
\end{align*}
Since $\langle E^i(\omega)z, z\rangle_i\in\mathbb{R}$ for all $z\in\mathbb{H}$, $\langle E^i(\omega) x, y\rangle_i\in\mathbb{C}_i$ and so $\langle E^i(\omega) x, y\rangle_i=\langle E^i(\omega) x, y\rangle$. Thus
$$\langle  Tx,y\rangle=\int_{\sigma(T_i)}\text{Re}(\lambda) d\langle E^i_\lambda x, y\rangle+\int_{\sigma(T_i)}\vert\text{Im}(\lambda)\vert d\langle E^i_\lambda Jx, y\rangle.$$
Let $E$ be the spectral measure constructed in part (i), it is easy to see that $E^i\big(\omega\big)=E\big([\omega]\big)$ for all $\omega\in\mathcal{A}_{\sigma(T_i)}^s$. Thus for any continuous real slice function $f$ on $\sigma_S(T)$, we have
$$\int_{\sigma(T_i)}f(\lambda) d\langle E^i_\lambda u, v\rangle=\int_{\sigma_S(T)}f(\lambda) d\langle E_\lambda u, v\rangle\;\text{ for all }\;u,v\in\mathcal{H}.$$
Therefore,
$$\langle  Tx,y\rangle=\int_{\sigma_S(T)}\text{Re}(\lambda) d\langle E_\lambda x, y\rangle+\int_{\sigma_S(T)}\vert\text{Im}(\lambda)\vert d\langle E_\lambda Jx, y\rangle.$$

\noindent (iii) Uniqueness of the spectral measure $E :\mathcal{A}_{\sigma_S(T)}^s\rightarrow\mathcal{B}_R(\mathcal{H})$ is proved as follows. For all $P\in\mathbb{R}[X,Y]$, let $p(T)$ be as in (\ref{6eq2}), then
$$\int_{\sigma(T_i)}\vert p(\lambda)\vert^2 d\langle E^i_\lambda x, x\rangle_i=\langle p(T)^*p(T) x,x\rangle=\int_{\sigma_S(T)}\vert p(\lambda)\vert^2 d\langle E_\lambda x, x\rangle.$$
Thus (see proof of Lemma \ref{6l2})
$$\int_{\sigma(T_i)}\chi_{\Omega}(\lambda) d\langle E^i_{\lambda } x, x\rangle=\int_{\sigma_S(T)}\chi_\Omega(\lambda) d\langle E_\lambda x, x\rangle\;\;\text{ for all }\;\; \Omega\in\mathcal{A}_{\sigma_S(T)}^s.$$
And so  $E^i(\Omega\cap\mathbb{C}_i)=E(\Omega)$ for all $\Omega\in\mathcal{A}_{\sigma_S(T)}^s$. Since $E^i$ is the spectral measure of the spectral decomposition of $T_i$, $E^i$ is unique then so is $E$.
\end{proof}
\begin{lemma}
Let $T=\int_{\sigma_S(T)}\lambda dE_\lambda$ be the spectral decomposition of a normal operator $T\in\mathcal{B}_R(\mathcal{H})$ as in Theorem $\ref{6t1}$. If $\Omega\in\mathcal{A}_{\sigma_S(T)}^s$, then
\begin{itemize}
\item[(i)] $E(\Omega)\neq 0$ whenever $\Omega$ is open in the relative topology of $\sigma_S(T)$.
\item[(ii)] $E(\Omega)T = TE(\Omega)$.
\item[(iii)] $\sigma_S(T\vert_{\mathcal{R}(E(\Omega))})\subseteq\overline{\Omega}$.
\end{itemize}
\end{lemma}
\begin{proof}
Let $i\in\mathbb{S}$ and let $E^i$ be the spectral measure of the spectral decomposition of $T_i$.\\

\noindent (i) We have $E\big(\Omega\big)=E^i\big(\Omega\cap\mathbb{C}_i\big)$. Since $\Omega$ is open in the relative topology of $\sigma_S(T)$, $\Omega\cap\mathbb{C}_i$ is open relative to the topology of $\sigma(T_i)$. Then by \cite[Theorem 3.15]{6C}, $E^i\big(\Omega\cap\mathbb{C}_i\big)\neq0$, and so $E\big(\Omega\big)\neq0$.\\

\noindent (ii) By \cite[Lemma 3.16]{6C} it follows that $E^i(\Omega\cap\mathbb{C}_i)T_i = T_iE^i(\Omega\cap\mathbb{C}_i)$. Then clearly $E(\Omega)T = TE(\Omega)$.\\

\noindent (iii) It follows from \cite[Proposition 3.8]{6BB},  that $\sigma_S(T\vert_{\mathcal{R}(E(\Omega))})=[\sigma(T_i\vert_{\mathcal{R}(E(\Omega))_i})]$ and by \cite[Lemma 3.16]{6C}, we get $\sigma(T_i\vert_{\mathcal{R}(E(\Omega))_i})=\sigma(T_i\vert_{\mathcal{R}(E^i(\Omega\cap\mathbb{C}_i))_i})\subseteq\overline{\Omega\cap\mathbb{C}_i}$. Thus $[\sigma(T_i\vert_{\mathcal{R}(E(\Omega))_i})]\subseteq[\overline{\Omega\cap\mathbb{C}_i}]$, that is $\sigma_S(T\vert_{\mathcal{R}(E(\Omega))})\subseteq\overline{\Omega}$.
\end{proof}
\begin{corollary}[Fuglede Theorem]
Let $T=\int_{\sigma_S(T)}\lambda dE_\lambda$ be the spectral decomposition of a normal operator $T\in\mathcal{B}_R(\mathcal{H})$. If $S\in\mathcal{B}_R(\mathcal{H})$ commutes with $T$, then $S$ commutes with $E(\Omega)$ for every $\Omega\in\mathcal{A}_{\sigma_S(T)}^s$.
\end{corollary}
\begin{proof}
If $S\in\mathcal{B}_R(\mathcal{H})$ commutes with $T$, then $S$ commutes with $T_i$, and by \cite[Theorem 3.17]{6C}, $S$ commutes with $E^i(\Omega\cap\mathbb{C}_i)$ for all $\Omega\in\mathcal{A}_{\sigma_S(T)}^s$. Thus $S$ commutes with $E(\Omega)$ for all $\Omega\in\mathcal{A}_{\sigma_S(T)}^s$.
\end{proof}
\begin{corollary}[Fuglede-Putnam Theorem]
Suppose $T_1,T_2\in\mathcal{B}_R(\mathcal{H})$ are normal operators. If there exists $S\in\mathcal{B}_R(\mathcal{H})$ such that $T_1S=ST_2$, then $T_1^*S=ST_2^*$.
\end{corollary}
\begin{proof}
By Lemma \ref{6l4}, $(T_1)_i$ and $(T_2)_i$ are normal. Since $(T_1)_iS_i=S_i(T_2)_i$, $(T_1^*)_iS_i=S_i(T_2^*)_i$ (see,  \cite[Corollary 3.20]{6C}). It follows that $T_1^*S=ST_2^*$.
\end{proof}


TITLE: How do I prove that $\lim_{(x,y) \to (0,0)} \frac{2y^2}{\sqrt{x^2+xy}}$ exists?
QUESTION [2 upvotes]: Now I have learnt that to prove a function of 2 variables exists we must have the both the repeated limits as equal, which is $\lim_{(x=0,y\to0)}f(x,y) = \lim_{(x\to0,y=0)}f(x,y)$ , now in this case $f(x,y)= \frac{2y^2}{\sqrt{x^2+xy}}$ so $\lim_{(x=0,y\to0)}f(x,y) = \lim_{(x=0,y\to0)} 2y^2/\infty$ which is not defined ,while for $\lim_{(x\to0,y=0)}f(x,y) = 0$ which is I think is enough to prove that this function( $\frac{2y^2}{\sqrt{x^2+xy}}$) does not exist as $(x,y) \to (0,0)$
Yet my book says it does exist , and I really can't find here where I go wrong , so I decided to ask here.
(P S : I guess i would be ridiculous to add this but at point $(0,y)$ my function is undefined ! doesn't that mean I simply don't need the point $(x,0)$ at all to prove it exists because its undefined and that straight off finishes the matter)

REPLY [1 votes]: Choose the curve $y=x^3-x$ s.t.  $$lim_{x\rightarrow 0}\frac{2(x^3-x)^2}{\sqrt{x^2+x(x^3-x)}}=lim_{x\rightarrow 0} 2(x^2-1)^2=2$$ hence limit is not unique.


text:
The goal of this section is to prove the following proposition. Use the notation of $\mathsection$\ref{SectionQStabBig}.
\begin{prop}\label{PropNormOfQStab}
Without loss of generality we may assume $H \lhd P$.
\end{prop}

The proof of Proposition~\ref{PropNormOfQStab} is by upward induction on $\dim \mathscr{B}$. It is clearly true for the starting case $\dim \mathscr{B} = 0$ because $H = 1$ in this case.

\subsection{Algebraic family associated with $\mathscr{B}$}\label{SubsectionAlgFamilyAssoWithB}
Mok's idea to prove the Ax type transcendence results is to use algebraic foliations. In our situation, we wish to construct a family $\mathscr{F}'$ of varieties in $\cX^+$ associated with $\mathscr{Z}$, such that $\mathscr{F}'$ is $\Gamma_0$-invariant for a suitable subgroup $\Gamma_0$ of $\Gamma$. This construction can be realized, for example, by using Hilbert schemes. Then $\bu(\mathscr{F}')$ is a foliation on $M$. Next we wish to improve the algebraicity of $\bu(\mathscr{F}')$ to make it into an algebraic subvariety. We present this process for our situation in this subsection.
We point out that it is \cite[$\mathsection$3]{MokAx-Schanuel-for} adapted to mixed Shimura varieties of Kuga type with some slight modifications.

Let $Y = \bu(\tilde{Z})^{\mathrm{Zar}} = \mathrm{pr}_M(\mathscr{Z})^{\mathrm{Zar}}$. Let $\cX^\vee$ be as in Proposition~\ref{PropositionRealizationOfX}, then $\cX^+$ is open (in the usual topology) semi-algebraic in $\cX^\vee$. Let $\bH$ be the Hilbert scheme of all subvarieties of $\cX^\vee \times Y$ with the same Hilbert polynomial as $\mathscr{B}$, and let $\mathbscr{B} \rightarrow \bH$ be the (modified) universal family, namely
\[
\mathbscr{B} = \{(\tilde{x},m,[W])\in \cX^+ \times Y \times \bH : (\tilde{x},m) \in W \} \hookrightarrow (\cX^+ \times Y) \times \bH \subseteq (\cX^+ \times M) \times \bH
\]
where $[W]$ means the point of $\bH$ representing $W$.

It is known that $\bH$ is proper. Denote by
\[
\psi \colon \mathbscr{B} \rightarrow \cX^+ \times M
\]
the natural projection. Then $\psi$ is proper since $\bH$ is proper.

The action of $\Gamma$ on $\cX^+ \times M$ induces an action of $\Gamma$ on $\mathbscr{B}$ by
\begin{equation}\label{EqActionGammaOnB}
\gamma(\tilde{x}, m, [W]) = (\gamma\tilde{x}, m, [\gamma W]).
\end{equation}

It is clear that $\psi$ is $\Gamma$-equivariant.

Recall that $\Delta = \mathrm{graph}(\cX^+ \rightarrow M) \subseteq \cX^+ \times M$. Define $\mathbscr{Z} \subseteq \mathbscr{B} \cap (\Delta \times \bH)$ to be
\begin{equation}\label{EqFiberDimIntersectionDim}
\mathbscr{Z} = \{(\tilde{\delta},[W]) \in \Delta \times \bH : \tilde{\delta} \in \Delta \cap W, ~\dim_{\tilde{\delta}} (\Delta \cap W)\ge \dim \mathscr{Z}\}.
\end{equation}
Then $\mathbscr{Z}$ is a closed complex analytic subset of $\mathbscr{B}$. Hence $\psi(\mathbscr{Z})$ is closed complex analytic in $\cX^+ \times M$ since $\psi$ is proper.

Note that $\Delta$ is $\Gamma$-invariant. So $\mathbscr{Z}$ is $\Gamma$-invariant for the action \eqref{EqActionGammaOnB}. Hence we can define $\Gamma \backslash \mathbscr{Z}$, which is naturally a complex analytic variety. Moreover $\psi(\mathbscr{Z})$ is also $\Gamma$-invariant since $\psi$ is $\Gamma$-equivariant. We thus obtain the following commutative diagram:
\begin{equation}\label{EqBigDiagramForNormality}
\xymatrix{
\mathbscr{Z} \ar[r]^-{\psi} \ar[d] & \psi(\mathbscr{Z}) \ar@{}[r]|{\subseteq} \ar[d] & \cX^+ \times M \ar[d]^{(\bu,\mathrm{id})} \\
 \Gamma \backslash \mathbscr{Z} \ar[r]^-{\bar{\psi}} & \Gamma \backslash \psi(\mathbscr{Z}) \ar@{}[r]|{\subseteq} & M  \times M.
}
\end{equation}
Then $\bar{\psi}$ is proper since $\psi$ is proper.

We are now ready to prove the following result.

\begin{prop}\label{PropAlgebraicityOfTheAlgebraicFoliation}

The subset $(\bu,\mathrm{id})(\psi(\mathbscr{Z})) = \bar{\psi}(\Gamma \backslash \mathbscr{Z})$ is a closed algebraic subvariety of $M  \times M$.
\end{prop}
\begin{proof}
We claim that $\psi(\mathbscr{Z}) \cap (\mathfrak{F} \times M)$ is definable in $\R_{\mathrm{an},\exp}$. Since
\[
\psi(\mathbscr{Z}) \cap (\mathfrak{F} \times M) = \psi(\mathbscr{Z} \cap (\mathfrak{F} \times M \times \bH)),
\]
it suffices to prove that $\mathbscr{Z} \cap (\mathfrak{F} \times M \times \bH)$ is definable. But
\begin{align*}
\mathbscr{Z} \cap (\mathfrak{F} \times M \times \bH) = \{(\tilde{x},m,[W]) \in \mathfrak{F} \times M \times \bH :  & ~ (\tilde{x},m) \in W,~\tilde{x} \in (\bu|_{\mathfrak{F}})^{-1}(m), \\
& ~\dim_{(\tilde{x},m)} (\Delta \cap (\mathfrak{F} \times M) \cap W)\ge \dim \mathscr{Z} \}.
\end{align*}
Hence $\mathbscr{Z} \cap (\mathfrak{F} \times M \times \bH)$ is definable since $\bu|_{\mathfrak{F}}$ is definable.

Now since $\psi(\mathbscr{Z})$ is $\Gamma$-invariant, we have that
\[
(\bu,\mathrm{id})(\psi(\mathbscr{Z})) = (\bu,\mathrm{id})\left( \psi(\mathbscr{Z}) \cap (\mathfrak{F} \times M) \right)
\]
is closed complex analytic and definable in $M  \times M$. Hence $(\bu,\mathrm{id})(\psi(\mathbscr{Z}))$ is closed algebraic by Peterzil-Starchenko's o-minimal Chow \cite[Theorem~4.5]{PeterzilComplex-analyti}; see also \cite[Theorem~2.2]{MokAx-Schanuel-for}.
\end{proof}

\begin{cor}\label{CorAlgebraicityOfTheAlgebraicFoliation}
We have $\bar{\psi}(\Gamma \backslash \mathbscr{Z}) = Y$ with $Y$ viewed as a subvariety of $M  \times M$ via the diagonal embedding.
\end{cor}
\begin{proof}
First, observe that $\psi(\mathbscr{Z}) \subseteq \Delta$. Hence $\bar{\psi}(\Gamma \backslash \mathbscr{Z}) = (\bu,\mathrm{id})(\psi(\mathbscr{Z})) \subseteq M$ with $M$ viewed as a subvariety of $M \times M$ via the diagonal embedding. Next, we have $\psi(\mathbscr{Z}) \subseteq \psi(\mathbscr{B}) \subseteq \cX^+ \times Y$ by definition of $\mathbscr{B}$. Hence $\bar{\psi}(\Gamma \backslash \mathbscr{Z}) = (\bu,\mathrm{id})(\psi(\mathbscr{Z})) \subseteq \mathrm{pr}_2^{-1}(Y) \subseteq M \times M$ for the projection to the second factor $\mathrm{pr}_2 \colon M \times M \rightarrow M$. So $\bar{\psi}(\Gamma \backslash \mathbscr{Z}) \subseteq M \cap \mathrm{pr}_2^{-1}(Y) = Y$ with $Y$ viewed as a subvariety of $M \times M$ via the diagonal embedding.

By assumption we have $\mathscr{Z} \subseteq \psi(\mathbscr{Z})$. Hence $(\bu,\mathrm{id})(\mathscr{Z}) \subseteq (\bu,\mathrm{id})(\psi(\mathbscr{Z})) = \bar{\psi}(\Gamma \backslash \mathbscr{Z})$. Applying the projection $\mathrm{pr}_1 \colon M \times M \rightarrow M$ to the first factor, we get
\begin{equation}\label{EqTildeZT0}
\bu(\tilde{Z}) = \bu(\mathrm{pr}_{\cX^+}(\mathscr{Z})) \subseteq \mathrm{pr}_1(\bar{\psi}(\Gamma \backslash \mathbscr{Z})).
\end{equation}
Taking Zariski closures of both sides, we get $Y \subseteq \mathrm{pr}_1(\bar{\psi}(\Gamma \backslash \mathbscr{Z}))$. But $\bar{\psi}(\Gamma \backslash \mathbscr{Z}) \subseteq Y$ with $Y$ viewed as a subvariety of $M \times M$ via the diagonal embedding. We are done.
\end{proof}

\subsection{The $\Q$-stabilizer and first steps towards the normality}\label{SubsectionFirstStepTowardsNormality}
In this subsection we relate $Y$ to the $\Q$-stabilizer of $\mathscr{B}$ via monodromy. We point out that it is \cite[$\mathsection$3]{MokAx-Schanuel-for} adapted to the universal abelian variety. However we need to carefully distinguish the Mumford-Tate group and the monodromy group in the mixed case, since the underlying group is no longer reductive (so that we do not have a decomposition $\cX^+ = \cX_1^+ \times \cX_2^+$ as in the pure case, and are thus not able to replace $\cX^+$ by an $N(\R)^+$-orbit).

Let $\mathbscr{Z}_0$ be the complex analytic irreducible component of $\mathbscr{Z}$ which contains $\mathscr{Z} \times [\mathscr{B}]$. By Corollary~\ref{CorAlgebraicityOfTheAlgebraicFoliation},  $\bar{\psi}(\Gamma\backslash\mathbscr{Z}_0) = Y$ since $Y$ is irreducible. Consider the following map
\begin{equation}\label{EqLongMapFundamentalGroups}
\pi_1(\Gamma\backslash\mathbscr{Z}_0) \xrightarrow{\bar{\psi}_*} \pi_1(Y) \rightarrow \Gamma.
\end{equation}
Denote by $\Gamma_0$ the image of this map, and by $N$ the identity component of the $\Q$-Zariski closure (in $P$) of $\Gamma_0$. Then it is clear that $\Gamma_0 \mathbscr{Z}_0 \subseteq \mathbscr{Z}_0$ for the action $\Gamma$ on $\mathbscr{B}$ defined by \eqref{EqActionGammaOnB}.
\begin{lemma}\label{LemmaNnormalinP}
We have  $N \lhd P$. Moreover $\tilde{Z}^{\biZar} = N(\R)^+\tilde{z}$ for any $\tilde{z} \in \tilde{Z}$.
\end{lemma}
\begin{proof}

Recall our assumption that $(P,\cX^+)$ is the smallest connected mixed Shimura subdatum of Kuga type such that $\tilde{Z} \subseteq \cX^+$. Therefore $M$ is the smallest connected mixed Shimura subvariety of Kuga type which contains $Y = \bu(\tilde{Z})^{\mathrm{Zar}}$.

Since $\bar{\psi}$ is proper, the image of $\bar{\psi}_*$ has finite index in $\pi_1(Y)$.

Hence $N$ equals the connected algebraic monodromy group of $Y$. By the last paragraph, we have $N \lhd P$ by Andr\'{e} \cite[$\mathsection$5, Theorem~1]{AndreMumford-Tate-gr}; see \cite[$\mathsection$3.3]{GaoTowards-the-And}.

Next observe that $\bu(\tilde{Z})^{\biZar} = Y^{\biZar}$. So $\tilde{Z}^{\biZar} = N(\R)^+\tilde{z}$ (for any $\tilde{z} \in \tilde{Z}$) by \cite[Theorem~8.1]{GaoTowards-the-And}. Thus we finish the proof.
\end{proof}

Denote by $\theta \colon \mathbscr{B} \rightarrow \bH$ the natural projection. Let $\mathbscr{F} = \theta^{-1}(\theta(\mathbscr{Z}_0)) = \{(\tilde{x},m,[W]) : [W] \in\theta(\mathbscr{Z}_0), ~ (\tilde{x},m) \in W\}$, then $\mathbscr{F} \subseteq \mathbscr{B}$ is the family of algebraic varieties parametrized by $\theta(\mathbscr{Z}_0) \subseteq \bH$, with the fiber over each $[W] \in \theta(\mathbscr{Z}_0)$ being $W$. Then we have
\[
\Gamma_0 \mathbscr{F} \subseteq \mathbscr{F}
\]
for the action $\Gamma$ on $\mathbscr{B}$ defined by \eqref{EqActionGammaOnB}; see below \eqref{EqLongMapFundamentalGroups}.

For any $[W] \in \theta(\mathbscr{Z}_0)$ and any complex analytic irreducible component $\mathscr{Z}'$ of its fiber $(\mathbscr{Z}_0)_{[W]}$, we have $\bu(\mathrm{pr}_{\cX^+}(\mathscr{Z}'))^{\mathrm{Zar}} \subseteq Y$ as $\mathscr{Z}' \subseteq \psi(\mathbscr{Z}_0)$.

Thus $\mathrm{pr}_{\cX^+}(\mathscr{Z}')^{\biZar} \subseteq \tilde{Z}^{\biZar}$ as $Y^{\biZar} = \bu(\tilde{Z})^{\biZar}$. Moreover for a very general $[W]\in \theta(\mathbscr{Z}_0)$, \textit{i.e.} outside a countable union of proper closed subvarieties, there exists a component $\mathscr{Z}'$ of $(\mathbscr{Z}_0)_{[W]}$ such that $\mathrm{pr}_{\cX^+}(\mathscr{Z}')^{\biZar} = \tilde{Z}^{\biZar}$.

Denote by
\[
\Gamma_{\mathbscr{F}} = \{\gamma \in \Gamma : \gamma W \subseteq W,~\forall [W] \in \theta(\mathbscr{Z}_0)\}.
\]

Then for a very general $[W] \in \theta(\mathbscr{Z}_0)$ (but may be different from above), we have
\begin{equation}\label{EqStabOfGeneralFiber}
\mathrm{Stab}_{\Gamma}(W) = \Gamma_{\mathbscr{F}}.
\end{equation}
Indeed, let $\bH_{\gamma} := \{[W] \in \theta(\mathbscr{Z}_0) : \gamma W \subseteq W\}$ for each $\gamma \in \Gamma$. Then for any $\gamma \not \in \Gamma_{\mathbscr{F}}$, we have that $\bH_{\gamma}$ is a proper subvariety of $\theta(\mathbscr{Z}_0)$. Thus it suffices to take $[W] \in \theta(\mathbscr{Z}_0) \setminus \bigcup_{\gamma \in \Gamma \setminus \Gamma_{\mathbscr{F}}} \bH_{\gamma}$.

The two paragraphs above together imply: for a very general $[W]\in \theta(\mathbscr{Z}_0)$, we have $\mathrm{Stab}_{\Gamma}(W) = \Gamma_{\mathbscr{F}}$ and there exists a component $\mathscr{Z}'$ of $(\mathbscr{Z}_0)_{[W]}$ such that $\mathrm{pr}_{\cX^+}(\mathscr{Z}')^{\biZar} = \tilde{Z}^{\biZar}$.

It is known that $\mathscr{B}$ is a fiber of $\mathbscr{F}$. We claim that it suffices to handle the case where $\mathscr{B}$ is a very general fiber as in the previous paragraph. Indeed, take $[W] \in \theta(\mathbscr{Z}_0)$ very general. Let $\mathscr{Z}'$ be an irreducible component of $(\mathbscr{Z}_0)_{[W]}$ with $\bu(\mathrm{pr}_{\cX^+}(\mathscr{Z}'))^{\mathrm{biZar}} = \tilde{Z}^{\biZar}$. We have $\dim \mathscr{Z}' \ge \dim \mathscr{Z}$ by definition of $\mathbscr{Z}$, and $\dim W = \dim \mathscr{B}$ by definition of the Hilbert scheme $\bH$. Denote by $\mathscr{B}' = (\mathscr{Z}')^{\Zar}$, then $\mathscr{B}' \subseteq W$. Then $\dim \mathscr{B} - \dim \mathscr{Z} \ge \dim \mathscr{B}' - \dim \mathscr{Z}'$ and $\dim\tilde{Z}^{\biZar} = \dim \mathrm{pr}_{\cX^+}(\mathscr{Z}')^{\biZar}$. So to prove the Ax-Schanuel Theorem (Theorem~\ref{ThmEquivalentFormOfAS}) for $(\mathscr{B},\mathscr{Z})$, it suffices to prove it for $(\mathscr{B}', \mathscr{Z}')$. If $\dim \mathscr{B}' < \dim W = \dim \mathscr{B}$, then Proposition~\ref{PropNormOfQStab} follows from induction hypothese and we are done. If $\dim \mathscr{B}' = \dim W$, then $\mathscr{B}' = W$ is a very general fiber of $\mathbscr{F}$ as we desire.

 By \eqref{EqStabOfGeneralFiber}, $(\Gamma_{\mathbscr{F}}^{\Zar})^{\circ}$ is the $\Q$-stabilizer of $\mathscr{B}$, which is the $H$ defined in \eqref{EqDefnQStabilizerOfB}. The ``Moreover'' part of Lemma~\ref{LemmaNnormalinP} implies $\mathscr{B} = N(\R)^+\tilde{z} \times \bu(N(\R)^+\tilde{z})$, and hence $H$ is a subgroup of $N$.

\begin{lemma}\label{LemmaHnormalinN}
We have $H \lhd N$.
\end{lemma}
\begin{proof} Since $\Gamma_0 \mathbscr{F} \subseteq \mathbscr{F}$, every $\gamma \in \Gamma_0$ sends a very general fiber $W$ of $\mathbscr{F}$ to a very general fiber $W'$ of $\mathbscr{F}$. As $\mathrm{Stab}_{\Gamma}(W') = \mathrm{Stab}_{\Gamma}(\gamma W) = \gamma \mathrm{Stab}_{\Gamma}(W) \gamma^{-1}$, by \eqref{EqStabOfGeneralFiber} we get $\Gamma_{\mathbscr{F}} = \gamma \Gamma_{\mathbscr{F}} \gamma^{-1}$ for all $\gamma \in \Gamma_0$. Hence the conclusion follows by taking Zariski closures.
\end{proof}

\subsection{Normality of the $\Q$-stabilizer}\label{SubsectionLastStepTowardsNormality}
The argument in this subsection is new compared to the proof of the pure Ax-Schanuel theorem.

\begin{proof}[Proof of Proposition~\ref{PropNormOfQStab}]
Recall the definable set $\Theta$ defined in \eqref{EquationXiSetsForFiberOfMSV}
\[
\Theta=\{p\in P(\R): \dim(p^{-1}\mathscr{B}\cap  (\mathfrak{F} \times M) \cap \Delta)=\dim \mathscr{Z}\},
\]
where $\mathfrak{F}$ is the fundamental set for $\bu \colon \cX^+ \rightarrow M$ defined in $\mathsection$\ref{SubsectionFundamentalSet}.
By Remark~\ref{RmkAssumptionASEnlarge}, we may and do assume that every positive dimensional semi-algebraic block in $\Theta$ is contained in a left coset of $\stab_{P(\R)}(\mathscr{B})$.

We have $H(\R)^+ \mathscr{B} \subseteq \mathscr{B}$ since $\mathscr{B}$ is algebraic in $\cX^+ \times M$ and the action of $P(\R)^+$ on $\cX^+ \times M$ is semi-algebraic.

Let $V_H = V \cap H$ and $V_N = V \cap N$. Let $G_H = H/V_H$, $G_N = N/V_N$ and $G = P/V$. Then $G_H \lhd G_N \lhd G$ by Lemma~\ref{LemmaNnormalinP} and Lemma~\ref{LemmaHnormalinN}. Hence $G_H \lhd G$ since any reductive group over $\Q$, in particular $G$, is an almost direct product of its center and almost simple normal subgroups.

Now we are left to prove the following two facts.
\begin{enumerate}
\item[(i)] $V_H$ is a $G$-submodule of $V$.
\item[(ii)] $G_H$ acts trivially on $V/V_H$.
\end{enumerate}

We start with (ii). By Lemma~\ref{LemmaHnormalinN}, we have $H\lhd N$. It follows that $G_H$ acts trivially on $V_N/V_H$. Hence it suffices to prove that $G_H$ acts trivially on $V/V_N$. But Lemma~\ref{LemmaNnormalinP} asserts that $N \lhd P$, and hence $G_N$ acts trivially on $V/V_N$. In particular $G_H$ acts trivially on $V/V_N$. Thus we have established (ii).

Now we prove (i). The ``Moreover'' part of Lemma~\ref{LemmaNnormalinP} implies that $\mathscr{B} \subseteq \tilde{F} \times F$, where $\tilde{F} = N(\R)^+\tilde{z}$ and $F = \bu(\tilde{F})$. Consider $\tilde{X} = \mathrm{pr}_{\cX^+}(\mathscr{B})$, then $\tilde{X} \subseteq \tilde{F}$. Applying the natural projection $\tilde{\pi} \colon (P,\cX^+) \rightarrow (G,\cX_G^+)$, we get $\tilde{X}_G \subseteq G_N(\R)^+\tilde{z}_G$. Here we denote by $\tilde{X}_G = \tilde{\pi}(\tilde{X})$ and $\tilde{z}_G = \tilde{\pi}(\tilde{z})$.

For convenience of readers, the rest of the proof of (i) is divided into 3 steps.

\noindent\boxed{\text{Step~I}} Any $\tilde{x}_G \in \cX_G^+$ gives a morphism $\C^\times = \S(\R) \rightarrow G(\R)$. If we endow $V(\R)$ with the complex structure determined by $V(\R) \cong \tilde{\pi}^{-1}(\tilde{x}_G)$, then a real subspace $V'$ of $V_{\R}$ is complex if and only if the following condition holds: $\tilde{x}_G(\sqrt{-1}) \cdot V' \subseteq V'$.
Denote for simplicity by $J_{\tilde{x}_G} = \tilde{x}_G(\sqrt{-1})$. Thus under this complex structure on $V(\R)$, we have
\begin{itemize}
\item the smallest complex subspace of $V(\R)$ containing $V_H(\R)$ is $V_H(\R) + J_{\tilde{x}_G} V_H(\R)$.
\item the largest complex subspace of $V(\R)$ contained in $V_H(\R)$ is $V_H(\R) \bigcap J_{\tilde{x}_G} V_H(\R)$.
\end{itemize}

Now each point in $\tilde{X}_G$ is of the form $g\tilde{z}_G$ for some $g \in G_N(\R)^+$.

We claim that $J_{\tilde{x}_G} V_H(\R)$ is independent of $\tilde{x}_G$ when $\tilde{x}_G$ varies in $G_N(\R)^+\tilde{z}_G$. For any $g\in G_N(\R)^+$, we have $J_{\tilde{x}_G}^{-1} g J_{\tilde{x}_G} \in G_N(\R)^+$ since $G_N \lhd G$. Hence $g J_{\tilde{x}_G} = J_{\tilde{x}_G} g'$ for some $g' \in G_N(\R)^+$. Note that $J_{g\tilde{x}_G} = gJ_{\tilde{x}_G} g^{-1}$. Thus
\[
J_{g\tilde{x}_G} V_H(\R) = gJ_{\tilde{x}_G} g^{-1} \cdot V_H(\R) \subseteq g J_{\tilde{x}_G} \cdot V_H(\R) = J_{\tilde{x}_G} g' \cdot V_H(\R) \subseteq J_{\tilde{x}_G} \cdot V_H(\R).
\]
Here we used the fact that $V_H$ is stable under $G_N$ (since $H \lhd N$). This proves the claim.

In particular, $J_{\tilde{x}_G} V_H(\R)$ is independent of $\tilde{x}_G$ for $\tilde{x}_G \in \tilde{X}_G$.

\noindent\boxed{\text{Step~II}} Next for any $\gamma_G \in \tilde{\pi}(\Theta) \cap \Gamma_G \supseteq \{\gamma_G \in \Gamma_G : \gamma_G \mathfrak{F}_G \cap \tilde{Z}_G \not= \emptyset\}$,\footnote{It is not hard to show this inclusion by using $\Theta \cap \Gamma \supseteq \{\gamma \in \Gamma: \gamma\mathfrak{F} \cap \tilde{Z} \not= \emptyset\}$. See \cite[proof of Lemma~10.2]{GaoTowards-the-And}.}  let $\tilde{Z}|_{\gamma_G \mathfrak{F}_G}^+$ be a complex analytic irreducible component of $\tilde{Z}|_{\gamma_G \mathfrak{F}_G} := \tilde{Z} \cap \tilde{\pi}^{-1}(\gamma_G \mathfrak{F}_G)$, and define the following definable set
\[
\Theta' = \{v \in V(\R) : \dim\big( (-v,\gamma_G^{-1}) \mathscr{B} \cap (\mathfrak{F} \times M) \cap \Delta \big) = \dim \mathscr{Z} \} \subseteq V(\R).
\]
Then $\Theta' \cap \Gamma \supseteq \{\gamma_V \in \Gamma_V : (\gamma_V,\gamma_G)\mathfrak{F} \cap \tilde{Z}|_{\gamma_G \mathfrak{F}_G}^+ \not= \emptyset\}$, and $(\Theta', \gamma_G^{-1}) \subseteq \Theta$.
Denote by $\Gamma_{V_H} = V_H(\Q) \cap \Gamma_V$.

Suppose $\{\gamma_V \in \Gamma_V : (\gamma_V,\gamma_G)\mathfrak{F} \cap \tilde{Z}|_{\gamma_G \mathfrak{F}_G}^+ \not= \emptyset\}$ is not contained in a finite union of $\Gamma_{V_H}$-cosets. Identify $\cX^+ \cong V(\R) \times \cX_G^+$, then $\mathfrak{F} = (-k,k)^{2g} \times \mathfrak{F}_G$ for some $k \ge 1$, where $g = \dim V$. Hence
\[
\{\gamma_V \in \Gamma_V : (\gamma_V,\gamma_G)\mathfrak{F} \cap \tilde{Z}|_{\gamma_G \mathfrak{F}_G}^+ \not= \emptyset\} = \{\gamma_V \in \Gamma_V :  \left( (\gamma_V + \gamma_G(-k,k)^{2g}) \times \gamma_G \mathfrak{F}_G \right) \cap \tilde{Z}|_{\gamma_G \mathfrak{F}_G}^+ \not= \emptyset\}.
\]
Let $p_V \colon \cX^+ \cong V(\R) \times \cX^+_G \rightarrow V(\R)$, then $p_V((\gamma_V + \gamma_G(-k,k)^{2g}) \times \gamma_G \mathfrak{F}_G)$ is contained in a Euclidean ball of radius at most $kH(\gamma_G)$. But $\tilde{Z}|_{\gamma_G \mathfrak{F}_G}^+$ is connected by choice.

Thus if the set above is not contained in a finite union of $\Gamma_{V_H}$-cosets, then we get for free that it contains $\ge T$ elements in $\Gamma_V \setminus \Gamma_{V_H}$ of height at most $H(\gamma_G)T$ (for all $T\gg 0$). \footnote{The crucial point is that the group $V(\R)$ is Euclidean. See \cite[pp.~218-219]{TsimermanAx-Schanuel-and}.}
Hence by Pila-Wilkie \cite[Theorem~3.6]{PilaO-minimality-an}, there exist two constants $c,\epsilon > 0$ with the following property: for each $T \gg 0$, $\Theta'$ contains a  semi-algebraic block $B'$ which is not in any coset of $V_H(\R)$ and which contains $\ge c T^{\epsilon}$ elements in $\Gamma_V$ outside $\Gamma_{V_H}$ of height at most $T$. Recall our assumption that every positive dimensional semi-algebraic block in $\Theta$ is contained in a left coset of $\stab_{P(\R)}(\mathscr{B})$. In particular $(B', \gamma_G^{-1}) \subseteq (\gamma_V' , \gamma_G^{-1}) \cdot \stab_{P(\R)}(\mathscr{B})$ for some $\gamma_V' \in B' \cap \Gamma_V$. Hence $(-\gamma_V' + \gamma_G B', 1) \subseteq \stab_{P(\R)}(\mathscr{B})$; the argument is similar to the end of Proposition~\ref{PropBignessOfQStabilizer} (right above Remark~\ref{RmkAssumptionASEnlarge}). But then
\[
\left( (-\gamma_V' + \gamma_G B') \cap \Gamma_V, 1 \right)\subseteq \stab_{P(\R)}(\mathscr{B}) \cap \Gamma \subseteq H(\Q).
\]
By letting $T \rightarrow \infty$ and varying $B'$ accordingly, we see that this inclusion cannot hold since each $B' \subseteq V(\R)$ thus obtained is not contained in any coset of $V_H(\R)$.

Thus $\{\gamma_V \in \Gamma_V : (\gamma_V,\gamma_G)\mathfrak{F} \cap \tilde{Z}|_{\gamma_G \mathfrak{F}_G}^+ \not= \emptyset\}$ is contained in a finite union of $\Gamma_{V_H}$-cosets. Hence for the map $\bar{\bu} \colon \cX^+ \rightarrow \Gamma_V \backslash \cX^+$,\footnote{The complex space $\Gamma_V \backslash \cX^+$ is a family of abelian varieties. In fact it is the pullback of the abelian scheme $M \rightarrow M_G$ under the uniformization $\cX_G^+ \rightarrow M_G$.} we have that $\bar{\bu}(\tilde{Z}|_{\gamma_G \mathfrak{F}_G}^+)$ is closed in $(\Gamma_V \backslash \cX^+)|_{\gamma_G \mathfrak{F}_G}$ in the usual topology.

In the following paragraph we consider the usual topology. As $\gamma_G\mathfrak{F}_G$ is an open subset of $\cX^+_G$ for each $\gamma_G \in \Gamma_G$, each $\tilde{Z}|_{\gamma_G\mathfrak{F}_G}^+$ is an open subset of $\tilde{Z}$. In particular there exists a $\tilde{Z}|_{\gamma_G\mathfrak{F}_G}^+$ as above such that $\dim \tilde{Z}|_{\gamma_G\mathfrak{F}_G}^+ = \dim \tilde{Z}$. Consider the closure $W$ of $\bar{\bu}(\tilde{Z}|_{\gamma_G\mathfrak{F}_G}^+)$ in $\Gamma_V \backslash \cX^+$. By the last paragraph, we have $W^{\circ} \subseteq \bar{\bu}(\tilde{Z}|_{\gamma_G\mathfrak{F}_G}^+) \subseteq W$ where $W^{\circ}$ is the interior of $W$. Now let $\tilde{Z}'$ be the complex analytic irreducible component of $\bar{\bu}^{-1}(W)$ that contains $\tilde{Z}|_{\gamma_G\mathfrak{F}_G}^+$. Then we have $\tilde{Z}' \subseteq \tilde{Z}$ and $\bar{\bu}(\tilde{Z}') = W$ is closed in $\Gamma_V\backslash \cX^+$.

Denote by $\mathscr{Z}' := \mathrm{graph}(\tilde{Z}' \rightarrow \bu(\tilde{Z}'))$. Then $\dim \mathscr{Z}' = \dim \mathscr{Z}$ because $\dim \tilde{Z}' = \dim \tilde{Z}$. By analytic continuation, we then have $(\mathscr{Z}')^{\Zar} = \mathscr{Z}^{\Zar} = \mathscr{B}$ and $(\tilde{Z}')^{\Zar} = \tilde{Z}^{\Zar} = \tilde{X}$.

\noindent\boxed{\text{Step~III}}
Now let
\[
\Gamma' = \im \left( \pi_1(\bar{\bu}(\tilde{Z}')) \rightarrow \pi_1(\Gamma_V \backslash \cX^+) = \Gamma_V \right) \subseteq \Gamma_V.
\]
Then $\Gamma'$ stabilizes $\tilde{Z}'$ for the action of $\Gamma_V$ on $\cX^+$. Recall that $P(\R)^+$ acts on $\cX^+ \times M$ via its action on the first factor. So $\Gamma'$ stabilizes $\mathscr{Z}'$. Hence $(\Gamma')^{\Zar}(\R) \subset V(\R)$ stabilizes $\mathscr{B} = (\mathscr{Z}')^{\Zar}$. On the other hand $H$ is defined to be the $\Q$-stabilizer of $\mathscr{B}$. So $(\Gamma')^{\Zar} \subseteq H \cap V = V_H$.

Let us take a closer look at this. The identification $\pi_1(\Gamma_V \backslash \cX^+) = \Gamma_V$ is realized via \eqref{EqiDelta}
\[
i_{\cX_G^+}^{-1} \colon \cX^+ \cong V(\R) \times \cX_G^+
\]
and hence $\Gamma_V \backslash \cX^+ \cong (\Gamma_V\backslash V(\R)) \times \cX_G^+$. We have $\tilde{Z}' \subseteq \tilde{X} = (\tilde{Z}')^{\Zar} \subseteq \tilde{F} = N(\R)^+\tilde{z}$. Since $V_H \lhd N$, we can take the quotient of $\tilde{F} = N(\R)^+\tilde{z}$ by $V_H(\R)$ and get a real manifold. Call this quotient $q$. Denote by $\Gamma_{V_N} = \Gamma_V \cap N(\Q)$ and $\Gamma_{V_H} = \Gamma_V \cap H(\Q)$. Then we obtain a commutative diagram
\[
\xymatrix{
V_N(\R) \times \tilde{F}_G \cong \tilde{F} \ar[r]^-{q} \ar[d]_{\bar{\bu}} & q(\tilde{F}) \cong (V_N/V_H)(\R) \times \tilde{F}_G \ar[d]  \\
(\Gamma_{V_N}\backslash V_N(\R)) \times \tilde{F}_G \cong \bar{\bu}(\tilde{F}) \ar[r]^-{[q]} &  \big( (\Gamma_{V_N}/\Gamma_{V_H})\backslash (V_N/V_H)(\R) \big) \times \tilde{F}_G.
}
\]
Here all the isomorphisms in the diagram are compatible with the $i_{\cX_G^+}^{-1}$ above. For now on, by abuse of notation we no longer write $i_{\cX_G^+}^{-1}(\cdot)$.

Since $\Gamma' \subseteq V_H(\Q)$, we have that $[q](\bar{\bu}(\tilde{Z}'))$ is contained in $\{t\} \times \tilde{F}_G$ for some $t \in (\Gamma_{V_N}/\Gamma_{V_H})\backslash (V_N/V_H)(\R)$. But then $q(\tilde{Z}') \subseteq \{v'\} \times \tilde{F}_G$ for some $v' \in (V_N/V_H)(\R)$. Thus there exists $v \in V(\R)$ such that
\begin{equation}\label{EqInclusionTildeZFamily}
\tilde{Z}' - (\{v\} \times \tilde{Z}'_G) \subseteq V_H(\R) \times \tilde{F}_G.
\end{equation}
But $i_{\cX_G^+}(\{v\} \times \tilde{Z}'_G)$ is complex analytic in $\cX^+$ by property (ii) of $i_{\cX_G^+}$ below \eqref{EqiDelta}. So the left hand side of \eqref{EqInclusionTildeZFamily} is complex analytic.

We have proven in Step~I that over each $\tilde{x}_G \in \tilde{F}_G$, the largest complex subspace of $V(\R)$ contained in $V_H(\R)$ is $V_H(\R) \bigcap J_{\tilde{x}_G} V_H(\R)$ and $J_{\tilde{x}_G} V_H(\R)$ is independent of the choice of $\tilde{x}_G$. Hence \eqref{EqInclusionTildeZFamily} implies
\[
\tilde{Z}' - (\{v\} \times \tilde{Z}'_G) \subseteq (V_H(\R) \bigcap J_{\tilde{x}_G} V_H(\R)) \times \tilde{F}_G
\]
since the left hand side is complex analytic. Taking the Zariski closures, we get
\[
\tilde{X} - (\{v\} \times \tilde{X}_G) \subseteq (V_H(\R) \bigcap J_{\tilde{x}_G} V_H(\R)) \times \tilde{F}_G.\footnote{We use the fact that any subset of $\cX^+$ is algebraic if and only if it is complex analytic and semi-algebraic.}
\]
But $V_H(\R)$ stabilizes $\tilde{X}$, so $V_H(\R) = V_H(\R) \bigcap J_{\tilde{x}_G} V_H(\R)$. But then $V_H(\R)$ is complex for the complex structure of $V(\R) \cong \cX^+_{\tilde{x}_G}$ for any $\tilde{x}_G \in \tilde{F}_G$. Since $\tilde{Z}$ is assumed to be Hodge generic in $(P,\cX^+)$, we may take $\tilde{x}_G$ such that $\mathrm{MT}(\tilde{x}_G) = G$. Then we see that $V_H$ is a $G$-module.
\end{proof}

config: arxiv


TITLE: How can I obtain the definite integral of the modified Bessel function of the second kind $K_{i \alpha}(z)$?
QUESTION [1 upvotes]: I am wondering how to compute the following integral:
$$\int_{x_0}^{\infty} K_{i\alpha}(x) dx$$
where $K_{i\alpha}(x)$ is the modified Bessel function of the second kind, also known as the Bessel function of imaginary argument. Has anyone a reference that performs this integral in this way? I looked for it in NIST Digital Library of Mathematical Functions and other sources, with no success.
In fact, Ref. 1 says that I should find some further relations in Ref. [A. Apelblat (1983) Table of Definite and Infinite Integrals. Physical Sciences Data, Vol. 13, Elsevier Scientific Publishing Co., Amsterdam.], but I have no access to this reference.

REPLY [1 votes]: Assuming  $t>0$ , Mathematica returns
$$\int_{t}^{\infty} K_{i\alpha}(x)\,dx=2^{-1+i \alpha } \,\Gamma (i \alpha -1)\, t^{1-i \alpha } \,
   _1F_2\left(\frac{1-i \alpha }{2};\frac{3-i
   \alpha }{2},1-i \alpha ;\frac{t^2}{4}\right)+$$ $$2^{-1-i \alpha }\,
   \Gamma (-i \alpha -1)\, t^{1+i \alpha } \, _1F_2\left(\frac{1+i
   \alpha }{2};\frac{3+i \alpha }{2},1+i
   \alpha ;\frac{t^2}{4}\right)+$$ $$\frac{\pi}{2} \,
   \text{sech}\left(\frac{\pi  \alpha }{2}\right)$$


TITLE: Lie Derivative of a metric along a vector field
QUESTION [0 upvotes]: I know that the lie derivative of a metric $g_{ab}$ along a vector field $X$ can be written as $L_X g_{ab} = \nabla_a X_b + ∇_b X_a$. Now my question is, how would you use this to actually compute the lie derivative?
Say for example, if $X = (X^1, X^2) = (-y, x)$
Apologies for being unable to insert the correct notation.

REPLY [2 votes]: Assume that $n=2$. First, you have to find the metric components $g_{ab}$ and the connection components $\Gamma^c_{ab}$. Then, you may use the following lowering index to find $X_a$
$$X_a=g_{ab}X^b$$For example
$$X_1=g_{11}X^1+g_{12}X^2$$
Then,
$$\nabla_aX_b=\frac{\partial X_b}{\partial x^a}-X_c\Gamma^c_{ab}$$ For example,
$$\nabla_1X_1=\frac{\partial X_1}{\partial x^1}-X_1\Gamma^1_{11}-X_2\Gamma^2_{11}$$


TITLE: Physical interpretation of $\frac{\partial \rho}{\partial t} + \frac{\partial j}{\partial x}=0$
QUESTION [0 upvotes]: What is the physical interpretation of $\frac{\partial \rho}{\partial t} + \frac{\partial j}{\partial x}=0$? Here $\rho$ is the probability density and $j$ is the probability current of a particle of mass $m$ moving in one-dimension.

REPLY [1 votes]: Roughly speaking, the one-dimensional equation of continuity says that if the probability current going in is different from the probability current going out, then the probability density will change.
So, for instance, if there is more (of whatever) going in than going out, the probability will increase at that position.


TITLE: bayesian posterior of truncated normal distribution with uniform prior
QUESTION [2 upvotes]: Let $N_T(\mu,\sigma)$ be a truncated normal distribution with support on $[0,1]$.
Draw $x \sim N_T(\mu,\sigma)$
(What I want to model is, I have a unknown quantity $\mu \in [0,1]$, but I only observe it with some noise.)
What is $p(\mu|x)$ with uniform prior on $[0,1]$?
$$\begin{align}
p(\mu|x) & = \frac{p(x|\mu) p(\mu)}{\int p(x|\mu) p(\mu) \mathrm{d} \mu}  \\
& \propto \frac{\phi(\frac{x - \mu}{\sigma})}{\sigma ( \Phi(\frac{1-\mu}{\sigma}) - \Phi(\frac{-\mu}{\sigma}) )} \mathbf{1}(0 \leq \mu \leq 1)
\end{align}$$
$\phi(\cdot)$ is the standard normal distribution pdf. $\Phi(\cdot)$ is the standard normal distribution cdf.
Is $p(\mu|x)$ also a truncated normal distribution with different parameter? Similar to here? If not, what is it?
If I compute that $\displaystyle \int p(x|\mu)p(\mu) \mathrm{d} \mu$, I don't think I would get a constant which is same as a truncated normal distribution. Because $\mu$ is a variable in this case.

REPLY [3 votes]: Given a uniform prior and (independent) observations from a Normal distribution then the resulting posterior is a truncated normal distribution. However, in this case the observations are drawn from a truncated prior which makes it more complicated.
First, you can 'ignore' the integral in the denominator since this is just a constant assuring that the posterior is a density. In general
$$p(\mu | x) \propto p(x|\mu)p(\mu).$$
As you have derived (note that $1/\sigma$ is a constant and is not considered):
$$p(\mu|x) \propto \frac{\phi\left(\frac{x-\mu}{\sigma}\right)}{\Phi\left(\frac{1-\mu}{\sigma}\right) - \Phi\left(\frac{-\mu}{\sigma}\right)}I_{\mu \in [0,1]}.$$
At first glance it looks like it is a truncated normal again, however $\mu$ is now variable instead of $x$, so comparing with the truncated normal density, this is no longer the case.


TITLE: What is the probability that no two consecutive boxes have blue balls.?
QUESTION [3 upvotes]: There are red and blue balls which can be filled in 5 boxes.All balls are similar except color. what is the probability that no two consecutive boxes have blue balls.

Assume:A ball can be either red or blue with equal probability
Approach:

I could find only the no of ways for favourable cases.

Let we hv chose only ...1 B and 4 R balls...., So possible combinations are-
(B,R,R,R,R), (R,B,R,R,R), (R,R,B,R,R), (R,R,R,B,R), (R,R,R,R,B). => 5ways
Now assume we hv choose ....2 B and 3 R balls.., So possible combinations are-
(B,R,B,R,R), (B,R,R,B,R), (B,R,R,R,B), (R,B,R,B,R), (R,B,R,R,B), (R,R,B,R,B) => 6ways
similarly for 3 Blue balls, we hv only 1 option
(B, R, B, R, B)...........................................................................................=> 1 ways
we cant select 4 and 5 blue balls bcoz in these case we found 2 blue balls
at consecutive place!!
Total 5+6+1= 12 Ans.
what are the no of  possible outcomes i was confused here.
Is it correct approach correct me if i am wrong.

REPLY [1 votes]: I'm pretty sure Conrado Costa is correct. I ran the following simulation in R:
for (i in 1:100000){
  B<-rbinom(5,1,.5)
  Z<-(B[1:4]==B[2:5])*(B[2:5]==1)
  if(i==1) S=0
  S=S+(sum(Z*1)>0)
}

I get $\frac{S}{1E5} = 0.59 \implies $ no consecutive blue box probability is $1-0.59 = 0.41 \approx \frac{13}{32}$
It's hard to see how you can only have nine favorable outcomes. Each of yours are valid, so we know that 12 is a lower bound, at least.


TITLE: Practical example: how to formally verify "file name" implementation from a spec?
QUESTION [2 upvotes]: Say you have an OS-independent specification for file names. The file names are case sensitive, and let's say can't be more than 200 characters, and can't have : or / in the name.
Now say you want to implement this specification in C on different operating systems (different versions of linux and darwin, not windows for now).
Let's say for argument's sake, it turns out on some operating system (operating system "A"), that you also can't use | in the file name. Well, you didn't capture this in your spec initially.
The question is, how do you figure this out before running into this case? In circuits (from my understanding), you can use "model checking" to automatically explore all possible states of the system, to figure out edge cases your spec doesn't handle. That makes sense for relatively simple state machines like logic circuits. But what about this case with file names?
How would you generate file names, and try to save the file on that operating system, so it can discover these implementation errors, without having to generate every possible string up to 200 characters in length?
How do you verify that the implementation of the spec on operating system "A" is valid? The model checker would have to somehow randomly construct a file name using the | character, and then hit an error. Or maybe I am misunderstanding model checking, and what this really is is automatic testing.
What I'm wondering is, how do you account for these types of cases in advance using formal methods? What are the general techniques or topics to start looking into?

REPLY [3 votes]: In general, the technique used is known as "fuzzing". Not all errors are equally likely. Let's consider two hypothetical errors:

System A incorrectly rejects a filename if it contains an | anywhere.
System A incorrectly rejects a filename if it contains a prime number of b characters.

Errors of the second type are much, much rarer, but this is not explained by Computer Science. It's a result from how humans construct software. Fuzzing tries to focus on typical human bugs. In this case, we can predict the following types of errors are more likely:

Very short paths and very long paths may have boundary errors.
Errors near the begin or end of the path are more common than errors at the end
Many characters can be organized into classes which are treated identically, such as lowercase characters. All characters in this class are treated equally, so whether a filename is valid does not depend on a substitution within the class.
Some characters are a priori known to be problematic and each should be treated as a class on its own. (Examples: <>/\&:")
Duplicating characters is more likely to cause bugs the closer they're together.

Fuzzing aims to generate a smaller set of strings by omitting irrelevant variations. We obviously generate an initial set of testcases of all possible lengths (1-200). This checks rule 1. We then expand this set by varying a few characters near both ends. We might try varying the 150th character of the 170 character testcase, but we won't bother with all 254 variations. Nor will we add a variation in the 151th character.


TITLE: How did they come up with joint PDF representation?
QUESTION [0 upvotes]: I am looking at this problem, and get it up to the calculation of the joint PDF, which is 1/12.  From that point, I do not get how they achieve the parallelogram.  Could someone please tell me?  Thank you.

REPLY [1 votes]: Found the answer in this video.
From the inequality $\theta - 1 \leq x \leq \theta + 1$, two equations are obtained:
$\theta \leq x + 1$, and
$\theta \geq x - 1$.
Plot them on the $\theta$ vs $x$ plane, we got the shown graph.

The conditional expectation $\mathbf E[\Theta\mid X=x]$ of the uniformly distributed $\Theta$ is just the midpoint of that uniform distribution.  This explain the plot of $\mathbf E[\Theta\mid X=x]$ inside the parallelogram.


TITLE: If $x=[x_1,...,x_n]$ is Multivariate normal, what is the $x_1,...,x_k$ that will maximise $P(x_1,...,x_k , x_{k+1},...x_n| \mu, \Sigma)$?
QUESTION [0 upvotes]: How would you compute the $x_1,...,x_k$ that will maximise $P(x_1,...,x_k ,x_{k+1},...x_n| \mu, \Sigma)$?
if x was 2D then I think the main eigenvector of the covariance matrix at fixed $x_1$ will give you the $x_2$ that maximises $P(x_1,x_2|\mu,\Sigma)$.
However, say for n=3, you know $x_1,x_2$ and you want to know what will maximise $P(x_1,x_2,x_3 | \mu ,\Sigma)$. With $x_1,x_2$ known, you can't be on the eigenvector of the Covariance matrix anymore.
How would you do this for general n?

REPLY [0 votes]: Since we are dealing with continuous distribution, it is meaningful to study the pdf $f(x_1,\dots,x_k|\mu,\Sigma,x_{k+1},\dots,x_n).$
Let us write the vectors $y_1=(x_1,\dots,x_k)^T$ and $y_2=(x_{k+1},\dots,x_n)^T$
For the multivariate normal,
$$\begin{pmatrix}y_1
\\y_2\end{pmatrix}\sim\mathcal{N}\left(\begin{pmatrix}\mu_1
\\\mu_2\end{pmatrix},\begin{bmatrix}\Sigma_{11} &\Sigma_{12}
\\\Sigma_{21} &\Sigma_{22}\end{bmatrix}\right)$$
where $\mu_1,$ $\mu_2$ are $k\times 1$ and $(n-k)\times 1$ respectively, with $\mu=(\mu_1^T,\mu_2^T)^T.$ Similarly, $\Sigma_{11}$ is $k\times k,$ $\Sigma_{12}=\Sigma_{21}^T$ is $k\times (n-k)$ and $\Sigma_{22}$ is $(n-k)\times (n-k).$
 The required conditional distribution can then be written as
$$f(y_1|\mu,\Sigma,y_2=\mathbf{a})=\mathcal{N}\left(\mu_1+\Sigma_{12}\Sigma_{22}^{-1}(\mathbf{a}-\mu_2),\Sigma_{11}-\Sigma_{12}\Sigma_{22}^{-1}\Sigma_{21}\right)$$
More on this can be found here and here.
By the property of multivariate normal distribution, it is now clear that given $(x_{k+1},\dots,x_n)=\mathbf{a}^T,$ the pdf is maximized at the mean $\mu_1+\Sigma_{12}\Sigma_{22}^{-1}(\mathbf{a}-\mu_2).$