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Zheyuan Wu
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\usepackage{graphicx}
\usepackage{tabularx}
\usepackage{colortbl}
% for drawing the graph
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% set the author, title, and email
\author[Zheyuan Wu]{Zheyuan Wu}
\title{Measure concentration in complex projective space and quantum entanglement}
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%\logo{}
\institute[]{Washington University in St. Louis}
\date{\today}
%\subject{}
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\begin{document}
@@ -97,43 +64,431 @@
\end{frame}
\begin{frame}{Table of Contents}
\hypersetup{linkcolor=black}
\tableofcontents
\tableofcontents
\end{frame}
\section{Motivation}
\begin{frame}{Light polarization and non-commutative probability}
\begin{figure}
\includegraphics[width=0.6\textwidth]{../latex/images/Filter_figure.png}
\end{figure}
\begin{itemize}
\item Light passing through a polarizer becomes polarized in the direction of that filter.
\item If two filters are placed with relative angle $\alpha$, the transmitted intensity decreases as $\alpha$ increases.
\item In particular, the transmitted intensity vanishes when $\alpha=\pi/2$.
\end{itemize}
\end{frame}
\section{Memes}
\begin{frame}{Polarization experiment}
\begin{frame}{Memes}
\vspace{0.5em}
Now consider three filters $F_1,F_2,F_3$ with directions
$$
\alpha_1,\alpha_2,\alpha_3.
$$
Testing them pairwise suggests introducing three $0$--$1$ random variables
$$
P_1,P_2,P_3,
$$
where $P_i=1$ means that the photon passes filter $F_i$.
\begin{figure}
\includegraphics[width=0.5\textwidth]{./images/strengthvisuals.jpg}
\end{figure}
Note that the count of the beams is actually less than before.
\vspace{0.5em}
If these were classical random variables on one probability space, they would satisfy a Bell-type inequality.
\end{frame}
\section{Decomposing the statements}
\begin{frame}{Decomposing the statements}
\begin{block}{Concentration of measure effect}
Let $\psi\in \mathcal{P}(A\otimes B)$ be a random pure state on $A\otimes B$.
If we define $\beta=\frac{1}{\ln(2)}\frac{d_A}{d_B}$, then we have
$$
\operatorname{Pr}[H(\psi_A) < \log_2(d_A)-\alpha-\beta] \leq \exp\left(-\frac{1}{8\pi^2\ln(2)}\frac{(d_Ad_B-1)\alpha^2}{(\log_2(d_A))^2}\right)
$$
where $d_B\geq d_A\geq 3$.
\begin{frame}{A classical Bell-type inequality}
\begin{block}{Bell-type inequality}
For any classical random variables $P_1,P_2,P_3\in\{0,1\}$,
$$
\operatorname{Prob}(P_1=1,P_3=0)
\leq
\operatorname{Prob}(P_1=1,P_2=0)
+
\operatorname{Prob}(P_2=1,P_3=0).
$$
\end{block}
\cite{Hayden_2006}
Recall that the von Neumann entropy is defined as $H(\psi_A)=-\operatorname{Tr}(\psi_A\log_2(\psi_A))$.
\vspace{0.5em}
\begin{proof}
The event $\{P_1=1,P_3=0\}$ splits into two disjoint cases according to whether $P_2=0$ or $P_2=1$:
$$
\{P_1=1,P_3=0\}
=
\{P_1=1,P_2=0,P_3=0\}
\sqcup
\{P_1=1,P_2=1,P_3=0\}.
$$
Therefore,
$$
\begin{aligned}
\operatorname{Prob}(P_1=1,P_3=0)
&=
\operatorname{Prob}(P_1=1,P_2=0,P_3=0) \\
&\quad+
\operatorname{Prob}(P_1=1,P_2=1,P_3=0) \\
&\leq
\operatorname{Prob}(P_1=1,P_2=0)
+
\operatorname{Prob}(P_2=1,P_3=0).
\end{aligned}
$$
\end{proof}
\end{frame}
\begin{frame}{What the system actually looks like}
\begin{frame}{Experimental law}
For unpolarized incoming light, the \textbf{observed transition law} for a pair of filters is
$$
\operatorname{Prob}(P_i=1,P_j=0)
=
\operatorname{Prob}(P_i=1)-\operatorname{Prob}(P_i=1,P_j=1).
$$
Using the polarization law,
$$
\operatorname{Prob}(P_i=1)=\frac12,
\qquad
\operatorname{Prob}(P_i=1,P_j=1)=\frac12\cos^2(\alpha_i-\alpha_j),
$$
hence
$$
\operatorname{Prob}(P_i=1,P_j=0)
=
\frac12-\frac12\cos^2(\alpha_i-\alpha_j)
=
\frac12\sin^2(\alpha_i-\alpha_j).
$$
\vspace{0.5em}
So the experimentally observed probabilities depend only on the angle difference $\alpha_i-\alpha_j$.
\end{frame}
\begin{frame}{Violation of the classical inequality}
Substituting the experimental law into the classical inequality gives
$$
\frac12\sin^2(\alpha_1-\alpha_3)
\leq
\frac12\sin^2(\alpha_1-\alpha_2)
+
\frac12\sin^2(\alpha_2-\alpha_3).
$$
Choose
$$
\alpha_1=0,\qquad
\alpha_2=\frac{\pi}{6},\qquad
\alpha_3=\frac{\pi}{3}.
$$
Then
$$
\begin{aligned}
\frac12\sin^2\!\left(-\frac{\pi}{3}\right)
&\leq
\frac12\sin^2\!\left(-\frac{\pi}{6}\right)
+
\frac12\sin^2\!\left(-\frac{\pi}{6}\right) \\
\frac38 &\leq \frac18+\frac18 \\
\frac38 &\leq \frac14,
\end{aligned}
$$
which is false.
\vspace{0.5em}
Therefore the pairwise polarization data cannot come from one classical probability model with random variables $P_1,P_2,P_3$.
\end{frame}
\begin{frame}{The quantum model of polarization}
The correct model uses a Hilbert space rather than classical events.
\begin{itemize}
\item A pure polarization state is a vector
$$
\psi=\alpha|0\rangle+\beta|1\rangle \in \mathbb{C}^2.
$$
\item A filter at angle $\alpha$ is represented by the orthogonal projection
$$
P_\alpha=
\begin{pmatrix}
\cos^2\alpha & \cos\alpha\sin\alpha \\
\cos\alpha\sin\alpha & \sin^2\alpha
\end{pmatrix}.
$$
\item For a pure state $\psi$, the probability of passing the filter is
$$
\langle P_\alpha\psi,\psi\rangle.
$$
\end{itemize}
\vspace{0.4em}
The key point is that sequential measurements are described by \emph{ordered products} of projections, and these need not commute.
\end{frame}
\begin{frame}{Recovering the observed law from the operator model}
Assume the incoming light is unpolarized, so its state is the density matrix
$$
\rho=\frac12 I.
$$
The probability of passing the first filter $P_{\alpha_i}$ is
$$
\operatorname{Prob}(P_i=1)
=
\operatorname{tr}(\rho P_{\alpha_i})
=
\frac12\operatorname{tr}(P_{\alpha_i})
=
\frac12.
$$
If the photon passes the first filter, the post-measurement state is
$$
\rho_i
=
\frac{P_{\alpha_i}\rho P_{\alpha_i}}{\operatorname{tr}(\rho P_{\alpha_i})}
=
P_{\alpha_i}.
$$
$$
P_\alpha=
\begin{pmatrix}
\cos^2\alpha & \cos\alpha\sin\alpha \\
\cos\alpha\sin\alpha & \sin^2\alpha
\end{pmatrix}.
$$
Therefore
$$
\operatorname{Prob}(P_j=1\mid P_i=1)
=
\operatorname{tr}(\rho_i P_{\alpha_j})
=
\operatorname{tr}(P_{\alpha_i}P_{\alpha_j})
=
\cos^2(\alpha_i-\alpha_j).
$$
\end{frame}
\begin{frame}{Recovering the observed law from the operator model (cont.)}
$$
\begin{aligned}
\operatorname{Prob}(P_i=1,P_j=0)
&=
\operatorname{Prob}(P_i=1)
\bigl(1-\operatorname{Prob}(P_j=1\mid P_i=1)\bigr) \\
&=
\frac12\bigl(1-\cos^2(\alpha_i-\alpha_j)\bigr) \\
&=
\frac12\sin^2(\alpha_i-\alpha_j).
\end{aligned}
$$
This matches the experiment exactly.
\end{frame}
\begin{frame}{Conclusion}
\begin{itemize}
\item The classical model predicts a Bell-type inequality for three $0$--$1$ random variables.
\item The polarization experiment violates that inequality.
\item The resolution is that the quantities measured are \emph{sequential probabilities}, not joint probabilities of classical random variables.
\item In quantum probability, events are modeled by projections on a Hilbert space, and measurement order matters.
\end{itemize}
\vspace{0.6em}
This is one of the basic motivations for passing from classical probability to non-commutative probability.
\end{frame}
\section{Concentration on Spheres and quantum states}
\begin{frame}{Quantum states: pure vs.\ mixed}
\begin{itemize}
\item A finite-dimensional quantum system is modeled by a complex Hilbert space (a complete inner product space)
$$
\mathcal H \cong \mathbb C^{n+1}.
$$
\item A \textbf{pure state} is represented by a unit vector
$$
\psi \in \mathcal H, \qquad \|\psi\|=1.
$$
\item A \textbf{mixed state} is represented by a density matrix
$$
\rho \geq 0, \qquad \operatorname{tr}(\rho)=1.
$$
\item Pure states describe maximal information; mixed states describe probabilistic mixtures or partial information.
\end{itemize}
\vspace{0.4em}
\begin{block}{Key distinction}
Pure states form a curved geometric space; mixed states form a convex set inside the space of matrices.
\end{block}
\end{frame}
\begin{frame}{Why pure states are not vectors}
\begin{itemize}
\item Two nonzero vectors that differ by a nonzero complex scalar represent the same physical state:
$$
\psi \sim \lambda \psi, \qquad \lambda \in \mathbb C^\times.
$$
\item In particular, multiplying by a phase $e^{i\theta}$ does not change any physical predictions.
\item Therefore the physical pure state is not a single vector, but the \emph{complex line} spanned by that vector.
\end{itemize}
\vspace{0.4em}
Hence the space of pure states is
$$
\mathbb P(\mathcal H)
=
(\mathcal H \setminus \{0\})/\mathbb C^\times.
$$
After choosing a basis $\mathcal H \cong \mathbb C^{n+1}$, this becomes
$$
\mathbb P(\mathcal H) \cong \mathbb C P^n.
$$
\end{frame}
\begin{frame}{Relation with the sphere}
\begin{itemize}
\item Every nonzero vector can be normalized, so each pure state has a representative on the unit sphere
$$
S^{2n+1} \subset \mathbb C^{n+1}.
$$
\item Two unit vectors represent the same pure state exactly when they differ by a phase:
$$
z \sim e^{i\theta} z.
$$
\item Therefore
$$
\mathbb C P^n = S^{2n+1}/S^1.
$$
\end{itemize}
\vspace{0.4em}
The quotient map
$$
p:S^{2n+1}\to \mathbb C P^n, \qquad p(z)=[z]=\{\lambda z : \lambda \in \mathbb C^\times\},
$$
is the \textbf{Hopf fibration}.
\end{frame}
\begin{frame}{How the metric descends to $\mathbb C P^n$}
\begin{itemize}
\item The sphere $S^{2n+1}$ inherits the round metric from the Euclidean metric on
$$
\mathbb C^{n+1} \cong \mathbb R^{2n+2}.
$$
\item The fibers of the Hopf map are circles
$$
p^{-1}([z]) = \{e^{i\theta}z : \theta \in \mathbb R\}.
$$
\item Tangent vectors split into:
\begin{itemize}
\item \textbf{vertical directions}: tangent to the $S^1$-fiber,
\item \textbf{horizontal directions}: orthogonal complement to the fiber.
\end{itemize}
\item The differential $dp$ identifies horizontal vectors on the sphere with tangent vectors on $\mathbb C P^n$.
\end{itemize}
\vspace{0.4em}
This allows the round metric on $S^{2n+1}$ to define a metric on $\mathbb C P^n$.
\end{frame}
\begin{frame}{The induced metric: Fubini--Study metric}
\begin{itemize}
\item The metric on $\mathbb C P^n$ obtained from the Hopf quotient is the
\textbf{Fubini--Study metric}.
\item So the geometric picture is:
$$
S^{2n+1}
\xrightarrow{\text{Hopf fibration}}
\mathbb C P^n
$$
$$
\text{round metric}
\rightsquigarrow
\text{Fubini--Study metric}.
$$
\item The normalized Riemannian volume measure induced by this metric gives the natural probability measure on pure states.
\end{itemize}
\vspace{0.5em}
\begin{block}{Proof roadmap}
To prove this carefully, one usually shows:
\begin{enumerate}
\item $p:S^{2n+1}\to \mathbb C P^n$ is a smooth surjective submersion,
\item the vertical space is the tangent space to the $S^1$-orbit,
\item horizontal lifts are well defined,
\item the quotient metric is exactly the Fubini--Study metric.
\end{enumerate}
\end{block}
\end{frame}
\begin{frame}{Maxwell-Boltzmann Distribution Law}
\begin{columns}[T]
\column{0.58\textwidth}
Consider the orthogonal projection
$$
\pi_{n,k}:S^n(\sqrt{n})\to \mathbb{R}^k.
$$
Its push-forward measure converges to the standard Gaussian:
$$
(\pi_{n,k})_*\sigma^n\to \gamma^k.
$$
\vspace{0.5em}
This explains why Gaussian behavior emerges from high-dimensional spheres and supports the proof strategy for Levy concentration.
\column{0.42\textwidth}
\begin{figure}
\includegraphics[width=\textwidth]{../latex/images/maxwell.png}
\end{figure}
\end{columns}
\end{frame}
\begin{frame}{Levy Concentration}
\begin{block}{Levy's theorem}
If $f:S^n\to \mathbb{R}$ is $1$-Lipschitz, then there exists a median $a_0$ such that
$$
\mu\{x\in S^n:|f(x)-a_0|\geq \epsilon\}
\leq
2\exp\left(-\frac{(n-1)\epsilon^2}{2}\right).
$$
\end{block}
\begin{itemize}
\item In high dimension, most Lipschitz observables are almost constant.
\item This is the geometric mechanism behind generic entanglement.
\end{itemize}
\end{frame}
\section{Main Result}
\begin{frame}{Generic Entanglement Theorem}
\begin{block}{Hayden--Leung--Winter}
Let $\psi\in \mathcal{P}(A\otimes B)$ be Haar-random and define
$$
\beta=\frac{1}{\ln(2)}\frac{d_A}{d_B}.
$$
For $d_B\geq d_A\geq 3$,
$$
\operatorname{Pr}[H(\psi_A)<\log_2(d_A)-\alpha-\beta]
\leq
\exp\left(
-\frac{1}{8\pi^2\ln(2)}
\frac{(d_Ad_B-1)\alpha^2}{(\log_2 d_A)^2}
\right).
$$
\end{block}
With overwhelming probability, a random pure state is almost maximally entangled.
\end{frame}
\begin{frame}{How the Entropy Observable Fits In}
\begin{figure}
\centering
\begin{tikzpicture}[node distance=30mm, thick,
@@ -142,224 +497,145 @@
towards_imp/.style={->,red},
mutual/.style={<->}
]
% define nodes
\node[main] (cp) {$\mathbb{C}P^{d_A d_B-1}$};
\node[main] (pa) [left of=cp] {$\mathcal{P}(A\otimes B)$};
\node[main] (sa) [below of=pa] {$S_A$};
\node[main] (rng) [right of=sa] {$[0,\infty)$};
\node[main] (sa) [below of=pa] {$\mathcal{S}(A)$};
\node[main] (rng) [right of=sa] {$[0,\log_2 d_A]$};
% draw edges
\draw[mutual] (cp) -- (pa);
\draw[towards] (pa) -- node[left] {$\operatorname{Tr}_B$} (sa);
\draw[towards_imp] (pa) -- node[above right] {$f$} (rng);
\draw[towards] (sa) -- node[above] {$H(\psi_A)$} (rng);
\draw[towards_imp] (pa) -- node[above right] {$\psi\mapsto H(\psi_A)$} (rng);
\draw[towards] (sa) -- node[above] {$H$} (rng);
\end{tikzpicture}
\end{figure}
\begin{itemize}
\item The red arrow is the concentration of measure effect. $f=H(\operatorname{Tr}_B(\psi))$.
\item $S_A$ denotes the mixed states on $A$
\item The red arrow is the observable to which concentration is applied.
\item The projective description is natural because global phase does not change the physical state.
\end{itemize}
\end{frame}
\section{Geometry of Quantum States}
\begin{frame}{Wait, but what is $\mathbb{C}P^n$ and where they are coming from?}
$\mathbb{C}P^n$ is the set of all complex lines in $\mathbb{C}^{n+1}$, or equivalently the space of equivalence classes of $n+1$ complex numbers up to a scalar multiple. \cite{Bengtsson_Życzkowski_2017}
One can find that every odd dimensional sphere $S^{2n+1}$ under the group action of $S^1$, denoted by $S^{2n+1}/S^1$, is a complex projective space $\mathbb{C}P^n$ (complex-dimensional). Recall Math 416.
\begin{figure}
\includegraphics[width=0.5\textwidth]{./images/stereographic.png}
\end{figure}
Detailed proof involves the Hopf fibration structures.
It's a natural projective Hilbert space.
\end{frame}
\begin{frame}{Some interesting claims about $\mathbb{C}P^n$}
..... The claim is that every physical system can be modelled by $\mathbb{C}P^n$ for some (possibly infinite) value of $n$, provided taht a definite correspondence between the system and the point of $\mathbb{C}P^n$ is set up. \cite{Bengtsson_Życzkowski_2017}
\end{frame}
\begin{frame}{Initial attempts for Levy's concentration lemma}
Consider the orthogonal projection from $\mathbb{R}^{n+1}$, the space where $S^n$ is embedded, to $\mathbb{R}^k$, we denote the restriction of the projection as $\pi_{n,k}:S^n(\sqrt{n})\to \mathbb{R}^k$. Note that $\pi_{n,k}$ is a 1-Lipschitz function (projection will never increase the distance between two points).
We denote the normalized Riemannian volume measure on $S^n(\sqrt{n})$ as $\sigma^n(\cdot)$, and $\sigma^n(S^n(\sqrt{n}))=1$.
\begin{block}{Gaussian measure}
We denote the Gaussian measure on $\mathbb{R}^k$ as $\gamma^k$.
\begin{frame}{Ingredients Behind the Tail Bound}
\begin{block}{Page-type lower bound}
$$
d\gamma^k(x)\coloneqq\frac{1}{\sqrt{2\pi}^k}\exp(-\frac{1}{2}\|x\|^2)dx
$$
$x\in \mathbb{R}^k$, $\|x\|^2=\sum_{i=1}^k x_i^2$ is the Euclidean norm, and $dx$ is the Lebesgue measure on $\mathbb{R}^k$.
\end{block}
Basically, you can consider the Gaussian measure as the normalized Lebesgue measure on $\mathbb{R}^k$ with standard deviation $1$.
\end{frame}
\begin{frame}{Maxwell-Boltzmann distribution law}
\begin{block}{Maxwell-Boltzmann distribution law}
For any natural number $k$,
$$
\frac{d(\pi_{n,k})_*\sigma^n(x)}{dx}\to \frac{d\gamma^k(x)}{dx}
$$
where $(\pi_{n,k})_*\sigma^n$ is the push-forward measure of $\sigma^n$ by $\pi_{n,k}$.
In other words,
$$
(\pi_{n,k})_*\sigma^n\to \gamma^k\text{ weakly as }n\to \infty
\mathbb{E}[H(\psi_A)]
\geq
\log_2(d_A)-\frac{1}{2\ln(2)}\frac{d_A}{d_B}.
$$
\end{block}
\end{frame}
\begin{frame}{Maxwell-Boltzmann distribution law}
It also has another name, the Projective limit theorem. \cite{romanvershyni}
If $X\sim \operatorname{Unif}(S^n(\sqrt{n}))$, then for any fixed unit vector $x$ we have $\langle X,x\rangle\to N(0,1)$ in distribution as $n\to \infty$.
\begin{figure}
\includegraphics[width=0.8\textwidth]{./images/maxwell.png}
\end{figure}
\end{frame}
\begin{frame}{Proof of Maxwell-Boltzmann distribution law I}
This part is from \cite{shioya2014metricmeasuregeometry}.
We denote the $n$ dimensional volume measure on $\mathbb{R}^k$ as $\operatorname{vol}_k$.
Observe that $\pi_{n,k}^{-1}(x),x\in \mathbb{R}^k$ is isometric to $S^{n-k}(\sqrt{n-\|x\|^2})$, that is, for any $x\in \mathbb{R}^k$, $\pi_{n,k}^{-1}(x)$ is a sphere with radius $\sqrt{n-\|x\|^2}$ (by the definition of $\pi_{n,k}$).
So,
$$
\begin{aligned}
\frac{d(\pi_{n,k})_*\sigma^n(x)}{dx}&=\frac{\operatorname{vol}_{n-k}(\pi_{n,k}^{-1}(x))}{\operatorname{vol}_k(S^n(\sqrt{n}))}\\
&=\frac{(n-\|x\|^2)^{\frac{n-k}{2}}}{\int_{\|x\|\leq \sqrt{n}}(n-\|x\|^2)^{\frac{n-k}{2}}dx}\\
\end{aligned}
$$
as $n\to \infty$.
note that $\lim_{n\to \infty}{(1-\frac{a}{n})^n}=e^{-a}$ for any $a>0$.
\end{frame}
\begin{frame}{Proof of Maxwell-Boltzmann distribution law II}
$(n-\|x\|^2)^{\frac{n-k}{2}}=\left(n(1-\frac{\|x\|^2}{n})\right)^{\frac{n-k}{2}}\to n^{\frac{n-k}{2}}\exp(-\frac{\|x\|^2}{2})$
So
$$
\begin{aligned}
\frac{(n-\|x\|^2)^{\frac{n-k}{2}}}{\int_{\|x\|\leq \sqrt{n}}(n-\|x\|^2)^{\frac{n-k}{2}}dx}&=\frac{e^{-\frac{\|x\|^2}{2}}}{\int_{x\in \mathbb{R}^k}e^{-\frac{\|x\|^2}{2}}dx}\\
&=\frac{1}{(2\pi)^{\frac{k}{2}}}e^{-\frac{\|x\|^2}{2}}\\
&=\frac{d\gamma^k(x)}{dx}
\end{aligned}
$$
\end{frame}
\begin{frame}{Levy's concentration lemma}
\begin{block}{Levy's concentration lemma}
Let $f:S^{n-1}\to \mathbb{R}$ be a $\eta$-Lipschitz function. Let $M_f$ denote the median of $f$ and $\langle f\rangle$ denote the mean of $f$. (Note this can be generalized to many other manifolds (spaces that locally resembles Euclidean space).)
Select a random point $x\in S^{n-1}$ with $n>2$ according to the uniform measure (Haar measure). Then the probability of observing a value of $f$ much different from the reference value is exponentially small.
\begin{block}{Lipschitz estimate}
$$
\operatorname{Pr}[|f(x)-M_f|>\epsilon]\leq \exp(-\frac{n\epsilon^2}{2\eta^2})
\|H(\psi_A)\|_{\mathrm{Lip}}
\leq
\sqrt{8}\,\log_2(d_A),
\qquad d_A\geq 3.
$$
$$
\operatorname{Pr}[|f(x)-\langle f\rangle|>\epsilon]\leq 2\exp(-\frac{(n-1)\epsilon^2}{2\eta^2})
$$
\end{block}
The Maxwell-Boltzmann distribution law will help us find the limit of measures on hemisphere $S^{n-1}$ under the series of functions $f_n:S^{n-1}(\sqrt{n})\to \mathbb{R}$.
Levy concentration plus these two estimates produces the exponential entropy tail bound.
\end{frame}
\begin{frame}{Majorana stellar representation of the quantum state}
\section{Geometry of State Space}
\begin{figure}
\centering
\begin{tikzpicture}[node distance=40mm, thick,
main/.style={draw, draw=white},
towards/.style={->},
towards_imp/.style={<->,red},
mutual/.style={<->}
]
\node[main] (cp) {$\mathbb{C}P^{n}$};
\node[main] (c) [below of=cp] {$\mathbb{C}^{n+1}$};
\node[main] (p) [right of=cp] {$\mathbb{P}^n$};
\node[main] (sym) [below of=p] {$\operatorname{Sym}_n(\mathbb{C}P^1)$};
% draw edges
\draw[towards] (c) -- (cp) node[midway, left] {$z\sim \lambda z$};
\draw[towards] (c) -- (p) node[midway, fill=white] {$w(z)=\sum_{i=0}^n Z_i z^i$};
\draw[towards_imp] (cp) -- (p) node[midway, above] {$w(z)\sim w(\lambda z)$};
\draw[mutual] (p) -- (sym) node[midway, right] {root of $w(z)$};
\end{tikzpicture}
\end{figure}
Basically, there is a bijection between the complex projective space $\mathbb{C}P^n$ and the set of roots of a polynomial of degree $n$.
We can use a symmetric group of permutation of $n$ complex numbers (or $S^2$) to represent the $\mathbb{C}P^n$, that is $\mathbb{C}P^n=S^2\times S^2\times \cdots \times S^2/S_n$.
\end{frame}
\section{Future Plans}
\begin{frame}{Future Plans}
\begin{frame}{Observable Diameter}
\begin{block}{Definition}
For a metric-measure space $X$ and $\kappa>0$,
$$
\obdiam_{\mathbb{R}}(X;-\kappa)
=
\sup_{f\in \operatorname{Lip}_1(X,\mathbb{R})}
\diameter(f_*\mu_X;1-\kappa).
$$
\end{block}
\begin{itemize}
\item The physical meaning of the mathematical structures, the correspondence, and the relationship between the measures, quantum states, and the geometry of topological spaces.
\item It asks how concentrated every $1$-Lipschitz real observable must be.
\item In the thesis, entropy is used as a concrete observable-diameter proxy.
\item Hopf fibration lets us compare $\mathbb{C}P^n$ with spheres.
\end{itemize}
\end{frame}
\begin{frame}{A Geometric Consequence}
\begin{block}{Projective-space estimate}
For $0<\kappa<1$,
$$
\obdiam(\mathbb{C}P^n(1);-\kappa)\leq O(\sqrt{n}).
$$
\end{block}
\begin{itemize}
\item First estimate observable diameter on spheres via Gaussian limits.
\item Then use the Hopf map $S^{2n+1}(1)\to \mathbb{C}P^n$.
\item This gives a geometric explanation for why many projective-space observables concentrate.
\end{itemize}
\end{frame}
\section{Numerical Section}
\begin{frame}{Entropy-Based Simulations}
\begin{itemize}
\item Sample Haar-random pure states in $\mathbb{C}^{d_A}\otimes\mathbb{C}^{d_B}$.
\item Compute reduced density matrices and entanglement entropy.
\item Measure shortest intervals containing mass $1-\kappa$ in the entropy distribution.
\item Compare concentration across:
\begin{itemize}
\item Fiber bundles
\item Fubini-Study metric
\item Space of entangled states
\item real spheres,
\item complex projective spaces,
\item symmetric states via Majorana stellar representation.
\end{itemize}
\item Riemannian geometry of $\mathbb{C}P^n$.
\end{itemize}
\end{frame}
\begin{frame}{What the Data Suggests}
\begin{columns}[T]
\column{0.5\textwidth}
\begin{figure}
\includegraphics[width=\textwidth]{../latex/images/entropy_vs_dim.png}
\end{figure}
\centering
Entropy vs.\ ambient dimension
\column{0.5\textwidth}
\begin{figure}
\includegraphics[width=\textwidth]{../latex/images/entropy_vs_dA.png}
\end{figure}
\centering
Entropy vs.\ subsystem dimension
\end{columns}
\vspace{0.6em}
As dimension increases, the entropy distribution concentrates near the maximal value.
\end{frame}
\section{Conclusion}
\begin{frame}{Conclusion and Outlook}
\begin{itemize}
\item Concentration of measure explains generic high entanglement in large bipartite systems.
\item Complex projective space provides the natural geometric setting for pure quantum states.
\item Observable diameter gives a way to phrase concentration geometrically.
\item Ongoing directions:
\begin{itemize}
\item Ricci curvature
\item Levy's Isoperimetric inequality
\item Lipschitz constants and Levi-Civita connection
\item Local operations and classical communication (LOCC)
\end{itemize}
\item The proof of the Page's formula.
\item Majorana stellar representation of the quantum state. And possibly the concentration of measure effect on that.
\item Relations to Gromov's works \cite{MGomolovs}
\begin{itemize}
\item Levy families
\item Observable diameters
\item sharper estimates for $\mathbb{C}P^n$,
\item deeper use of Fubini--Study geometry,
\item Majorana stellar representation for symmetric states.
\end{itemize}
\end{itemize}
\end{frame}
\section{References}
\begin{frame}[allowframebreaks]{References}
\nocite{*} % This will include all entries from the bibliography file
\nocite{*}
\bibliographystyle{apalike}
\bibliography{references}
\end{frame}
\begin{frame}
\begin{center}
Q\&A
\end{center}
\end{frame}
\end{document}

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