Adjust stronger theorem to more general expressions

main.tex

@@ -1403,7 +1403,7 @@ proof of theorem \ref{thm:rmax_with_uniform_eps}:
 	$\epsilon_{v,q}$
 	]
 	\label{lemdfn:epsilon_q}
-	Suppose $d \in \frac{1}{2}\ZZ$ satisfies the condition in
+	Suppose $d \in \frac{1}{\lcm(m,2n^2)}\ZZ$ satisfies the condition in
 	eqn \ref{eqn:positive_rad_condition_in_terms_of_q_beta}.
 	That is:
 
@@ -1425,51 +1425,72 @@ proof of theorem \ref{thm:rmax_with_uniform_eps}:
 
 	\begin{equation*}
 		\epsilon_{v,q} \coloneqq
-		\frac{k_{q}}{2n^2}
+		\frac{k_{q}}{\lcm(m,2n^2)}
+	\end{equation*}
+	with $k_{v,q}$ being the least $k\in\ZZ_{>0}$ satisfying
+	\begin{equation*}
+		k \equiv -\aa\bb \frac{m}{\gcd(m,2n^2)}
+		\mod{\gcd\left(
+			\frac{n^2\gcd(m,2)}{\gcd(m,2n^2)},
+			\frac{mn\aa}{\gcd(m,2n^2)}
+		\right)}
 	\end{equation*}
-	with $k_{v,q}$ being the least $k\in\ZZ_{>0}$ satisfying $k \equiv -\aa\bb \mod n$
 	
 \end{lemmadfn}
 
+\vspace{10pt}
+
 \begin{proof}
 
 Consider the following:
 
 \begin{align}
-	\frac{ x }{ 2 }
+	\frac{ x }{ \lcm(m,2) }
 	- \frac{
 		(\aa r+2\bb)\aa
 	}{
 		2n^2
 	}
-	= \frac{ k }{ 2n^2 }
+	= \frac{ k }{ \lcm(m,2n^2) }
 	\quad \text{for some } x \in \ZZ
 	\span \span \span \span \span
 	\label{eqn:finding_better_eps_problem}
 \\ \Longleftrightarrow& &
 	- (\aa r+2\bb)\aa
+	\frac{\lcm(m,2n^2)}{2n^2}
 	&\equiv k &&
-	\mod n^2
-\\ \Longleftrightarrow& &
-	- \aa^2 r - 2\aa\bb
-	&\equiv k &&
-	\mod n^2
-\\  \Longrightarrow& &
-  \aa^2 \aa^{-1}\bb - 2\aa\bb
+	\mod \frac{\lcm(m,2n^2)}{\lcm(m,2)}
+	\nonumber
+\\ \Longrightarrow& &
+	- \bb\aa
+	\frac{\lcm(m,2n^2)}{2n^2}
 	&\equiv k &&
-	\mod n
-	\label{eqn:better_eps_problem_k_mod_gcd2n2_a2mn}
+	\nonumber
+\\ &&&
+	\mod \gcd\left(
+		\frac{\lcm(m,2n^2)}{\lcm(m,2)},
+		\frac{mn \aa}{\gcd(m,2n^2)}
+	\right)
+	\span \span \span
+	\nonumber
 \\ \Longleftrightarrow& &
-  -\aa\bb
+	- \bb\aa
+	\frac{m}{\gcd(m,2n^2)}
 	&\equiv k &&
-	\mod n
 	\label{eqn:better_eps_problem_k_mod_n}
+\\ &&&
+	\mod \gcd\left(
+		\frac{n^2\gcd(m,n)}{\gcd(m,2n^2)},
+		\frac{mn \aa}{\gcd(m,2n^2)}
+	\right)
+	\span \span \span
+	\nonumber
 \end{align}
 
-In our situation, we want to find the least $k$ satisfying 
+In our situation, we want to find the least $k>0$ satisfying 
 eqn \ref{eqn:finding_better_eps_problem}.
 Since such a $k$ must also satisfy eqn \ref{eqn:better_eps_problem_k_mod_n},
-we can pick the smallest $k_{q,1} \in \ZZ_{>0}$ which satisfies this new condition
+we can pick the smallest $k_{q,v} \in \ZZ_{>0}$ which satisfies this new condition
 (a computation only depending on $q$ and $\beta$, but not $r$).
 We are then guaranteed that $k_{v,q}$ is less than any $k$ satisfying eqn
 \ref{eqn:finding_better_eps_problem}, giving the first inequality in eqn
@@ -1495,6 +1516,7 @@ from plots_and_expressions import r_upper_bound1, r_upper_bound2
 	\bgroup
 	\def\kappa{k_{v,q}}
 	\def\psi{\chern_1^{\beta}(F)}
+	\renewcommand\Omega{{\lcm(m,2n^2)}}
 	\begin{align*}
 		\min
 		\left(
@@ -1507,7 +1529,7 @@ from plots_and_expressions import r_upper_bound1, r_upper_bound2
 	and $R = \chern_0(v)$
 
 	Furthermore, if $\aa \not= 0$ then
-	$r \equiv \aa^{-1}b_q (\mod n)$.
+	$r \equiv \aa^{-1}b_q \pmod{n}$.
 \end{theorem}
 
 
@@ -1571,7 +1593,7 @@ This example was chosen because the $n$ value is moderatly large, giving more
 possible values for $k_{v,q}$, in dfn/lemma \ref{lemdfn:epsilon_q}. This allows
 for a larger possible difference between the bounds given by theorems
 \ref{thm:rmax_with_uniform_eps} and \ref{thm:rmax_with_eps1}, with the bound
-from the second being up to $\sage{n}$ smaller, for any given $q$ value.
+from the second being up to $\sage{n}$ times smaller, for any given $q$ value.
 The (non-exclusive) upper bounds for $r\coloneqq\chern_0(u)$ of a tilt semistabilizer $u$ of $v$
 in terms of the first few smallest possible values for $q\coloneqq\chern_1^{\beta}(u)$ are as follows:
 
@@ -1612,6 +1634,7 @@ end}
 However the reduction in the overall bound on $r$ is not as drastic, since all
 possible values for $k_{v,q}$ in $\{1,2,\ldots,\sage{n}\}$ are iterated through
 cyclically as we consider successive possible values for $q$.
+And for each $q$ where $k_{v,q}=1$, both theorems give the same bound.
 Calculating the maximums over all values of $q$ yields
 $\sage{max(theorem2_bounds)}$ for theorem \ref{thm:rmax_with_uniform_eps}, and
 $\sage{max(theorem3_bounds)}$ for theorem \ref{thm:rmax_with_eps1}.