bugfix

presentation/ZheyuanWu_HonorThesis_Presentation.tex

@@ -551,7 +551,7 @@ Thus entanglement entropy is the von Neumann entropy of a subsystem, and it meas
     \begin{block}{Projective-space estimate from Gromov}
         For $0<\kappa<1$,
         $$
-        \obdiam(\mathbb{C}P^n(1);-\kappa)\leq O(\sqrt{n}).
+        \obdiam(\mathbb{C}P^n(1);-\kappa)\leq O\left(\frac{1}{\sqrt{n}}\right).
         $$
     \end{block}
 
@@ -562,7 +562,7 @@ Thus entanglement entropy is the von Neumann entropy of a subsystem, and it meas
     \end{itemize}
 
 
-    Hayden's work suggests that if the entropy function is a ``good'' proxy for the observable diameter, the difference of \textbf{the order of the growth} between $\obdiam(\mathbb{C}P^n(1);-\kappa)$ and $\obdiam(S^{2n+1}(1);-\kappa)$ should be smaller than $O(\sqrt{n})$.
+    Hayden's work suggests that if the entropy function is a ``good'' proxy for the observable diameter, the difference of \textbf{the order of the growth} between $\obdiam(\mathbb{C}P^n(1);-\kappa)$ and $\obdiam(S^{2n+1}(1);-\kappa)$ should be smaller than $O\left(\frac{1}{\sqrt{n}}\right)$.
 \end{frame}
 
 \begin{frame}{A conjecture}
@@ -570,7 +570,7 @@ Thus entanglement entropy is the von Neumann entropy of a subsystem, and it meas
 
         For $0<\kappa<1$,
         $$
-        \obdiam(\mathbb{C}P^n(1);-\kappa)= O(\sqrt{n}).
+        \obdiam(\mathbb{C}P^n(1);-\kappa)= O\left(\frac{1}{\sqrt{n}}\right).
         $$
     
     \end{block}