Correct beta vs beta_{-} in loose bound theorem

main.tex

@@ -248,6 +248,7 @@ the circular walls must be nested and non-intersecting.
 \subsection{Characteristic curves for pseudo-semistabilizers}
 
 \begin{lemma}[Numerical tests for left-wall pseudo-semistabilizers]
+\label{lem:pseudo_wall_numerical_tests}
 Let $v$ and $u$ be Chern characters with positive ranks and $\Delta(v),
 \Delta(u)\geq 0$. Let $P$ be a point on $\Theta_v^-$.
 
@@ -470,18 +471,18 @@ normal one. So $0 \leq \Delta(E)$ yields:
 \end{equation}
 
 \begin{theorem}[Bound on $r$ - Benjamin Schmidt]
-Given a Chern character $v$ such that $\beta_{-}(v)\in\QQ$, the rank $r$ of
-any semistabilizer $E$ of some $F \in \firsttilt\beta$ with $\chern(F)=v$ is
+Given a Chern character $v$ such that $\beta_-:=\beta_{-}(v)\in\QQ$, the rank $r$ of
+any semistabilizer $E$ of some $F \in \firsttilt{\beta_-}$ with $\chern(F)=v$ is
 bounded above by:
 
 \begin{equation*}
-	r \leq \frac{mn^2 \chern^\beta_1(v)^2}{\gcd(m,2n^2)}
+	r \leq \frac{mn^2 \chern^{\beta_-}_1(v)^2}{\gcd(m,2n^2)}
 \end{equation*}
 \end{theorem}
 
 \begin{proof}
 
-The restrictions on $\chern^\beta_0(E)$ and $\chern^\beta_2(E)$
+The restrictions on $\chern^{\beta_-}_0(E)$ and $\chern^{\beta_-}_2(E)$
 is best seen with the following graph:
 
 % TODO: hyperbola restriction graph (shaded)
@@ -489,18 +490,19 @@ is best seen with the following graph:
 var("m") # Initialize symbol for variety parameter
 \end{sagesilent}
 
-This is where the rationality of $\beta_{-}$ comes in. If $\beta_{-} = \frac{*}{n}$
-for some $*,n \in \ZZ$.
-Then $\chern^\beta_2(E) \in \frac{1}{\lcm(m,2n^2)}\ZZ$ where $m$ is the integer
+This is where the rationality of $\beta_{-}$ comes in. If
+$\beta_{-} = \frac{*}{n}$ for some $*,n \in \ZZ$. Then
+$\chern^{\beta_-}_2(E) \in \frac{1}{\lcm(m,2n^2)}\ZZ$ where $m$ is the integer
 which guarantees $\chern_2(E) \in \frac{1}{m}\ZZ$ (determined by the variety).
-In particular, since $\chern_2(E) > 0$ we must also have
-$\chern^\beta_2(E) \geq \frac{1}{\lcm(m,2n^2)}$, which then in turn gives a bound
-for the rank of $E$:
+In particular, since $\chern_2^{\beta_-}(E) > 0$ (by using 	$P=(\beta_-,0)$ in
+lemma \ref{lem:pseudo_wall_numerical_tests}) we must also have
+$\chern^{\beta_-}_2(E) \geq \frac{1}{\lcm(m,2n^2)}$, which then in turn gives a
+bound for the rank of $E$:
 
 \begin{align}
-	\chern_0(E) &= \chern^\beta_0(E) \\
-	&\leq \frac{\lcm(m,2n^2) \chern^\beta_1(E)^2}{2} \\
-	&\leq \frac{mn^2 \chern^\beta_1(F)^2}{\gcd(m,2n^2)}
+	\chern_0(E) &= \chern^{\beta_-}_0(E) \\
+	&\leq \frac{\lcm(m,2n^2) \chern^{\beta_-}_1(E)^2}{2} \\
+	&= \frac{mn^2 \chern^{\beta_-}_1(F)^2}{\gcd(m,2n^2)}
 \end{align}
 
 \end{proof}