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Commit e4c7684c authored by Luke Naylor's avatar Luke Naylor
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Adjust stronger theorem to more general expressions

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...@@ -1403,7 +1403,7 @@ proof of theorem \ref{thm:rmax_with_uniform_eps}: ...@@ -1403,7 +1403,7 @@ proof of theorem \ref{thm:rmax_with_uniform_eps}:
$\epsilon_{v,q}$ $\epsilon_{v,q}$
] ]
\label{lemdfn:epsilon_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}. eqn \ref{eqn:positive_rad_condition_in_terms_of_q_beta}.
That is: That is:
...@@ -1425,51 +1425,72 @@ proof of theorem \ref{thm:rmax_with_uniform_eps}: ...@@ -1425,51 +1425,72 @@ proof of theorem \ref{thm:rmax_with_uniform_eps}:
\begin{equation*} \begin{equation*}
\epsilon_{v,q} \coloneqq \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*} \end{equation*}
with $k_{v,q}$ being the least $k\in\ZZ_{>0}$ satisfying $k \equiv -\aa\bb \mod n$
\end{lemmadfn} \end{lemmadfn}
\vspace{10pt}
\begin{proof} \begin{proof}
Consider the following: Consider the following:
\begin{align} \begin{align}
\frac{ x }{ 2 } \frac{ x }{ \lcm(m,2) }
- \frac{ - \frac{
(\aa r+2\bb)\aa (\aa r+2\bb)\aa
}{ }{
2n^2 2n^2
} }
= \frac{ k }{ 2n^2 } = \frac{ k }{ \lcm(m,2n^2) }
\quad \text{for some } x \in \ZZ \quad \text{for some } x \in \ZZ
\span \span \span \span \span \span \span \span \span \span
\label{eqn:finding_better_eps_problem} \label{eqn:finding_better_eps_problem}
\\ \Longleftrightarrow& & \\ \Longleftrightarrow& &
- (\aa r+2\bb)\aa - (\aa r+2\bb)\aa
\frac{\lcm(m,2n^2)}{2n^2}
&\equiv k && &\equiv k &&
\mod n^2 \mod \frac{\lcm(m,2n^2)}{\lcm(m,2)}
\\ \Longleftrightarrow& & \nonumber
- \aa^2 r - 2\aa\bb \\ \Longrightarrow& &
&\equiv k && - \bb\aa
\mod n^2 \frac{\lcm(m,2n^2)}{2n^2}
\\ \Longrightarrow& &
\aa^2 \aa^{-1}\bb - 2\aa\bb
&\equiv k && &\equiv k &&
\mod n \nonumber
\label{eqn:better_eps_problem_k_mod_gcd2n2_a2mn} \\ &&&
\mod \gcd\left(
\frac{\lcm(m,2n^2)}{\lcm(m,2)},
\frac{mn \aa}{\gcd(m,2n^2)}
\right)
\span \span \span
\nonumber
\\ \Longleftrightarrow& & \\ \Longleftrightarrow& &
-\aa\bb - \bb\aa
\frac{m}{\gcd(m,2n^2)}
&\equiv k && &\equiv k &&
\mod n
\label{eqn:better_eps_problem_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} \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}. eqn \ref{eqn:finding_better_eps_problem}.
Since such a $k$ must also satisfy eqn \ref{eqn:better_eps_problem_k_mod_n}, 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 $r$). (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 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 \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 ...@@ -1495,6 +1516,7 @@ from plots_and_expressions import r_upper_bound1, r_upper_bound2
\bgroup \bgroup
\def\kappa{k_{v,q}} \def\kappa{k_{v,q}}
\def\psi{\chern_1^{\beta}(F)} \def\psi{\chern_1^{\beta}(F)}
\renewcommand\Omega{{\lcm(m,2n^2)}}
\begin{align*} \begin{align*}
\min \min
\left( \left(
...@@ -1507,7 +1529,7 @@ from plots_and_expressions import r_upper_bound1, r_upper_bound2 ...@@ -1507,7 +1529,7 @@ from plots_and_expressions import r_upper_bound1, r_upper_bound2
and $R = \chern_0(v)$ and $R = \chern_0(v)$
Furthermore, if $\aa \not= 0$ then Furthermore, if $\aa \not= 0$ then
$r \equiv \aa^{-1}b_q (\mod n)$. $r \equiv \aa^{-1}b_q \pmod{n}$.
\end{theorem} \end{theorem}
...@@ -1571,7 +1593,7 @@ This example was chosen because the $n$ value is moderatly large, giving more ...@@ -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 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 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 \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 $v$ 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: in terms of the first few smallest possible values for $q\coloneqq\chern_1^{\beta}(u)$ are as follows:
...@@ -1612,6 +1634,7 @@ end} ...@@ -1612,6 +1634,7 @@ end}
However the reduction in the overall bound on $r$ is not as drastic, since all 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 possible values for $k_{v,q}$ in $\{1,2,\ldots,\sage{n}\}$ are iterated through
cyclically as we consider successive possible values for $q$. 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 Calculating the maximums over all values of $q$ yields
$\sage{max(theorem2_bounds)}$ for theorem \ref{thm:rmax_with_uniform_eps}, and $\sage{max(theorem2_bounds)}$ for theorem \ref{thm:rmax_with_uniform_eps}, and
$\sage{max(theorem3_bounds)}$ for theorem \ref{thm:rmax_with_eps1}. $\sage{max(theorem3_bounds)}$ for theorem \ref{thm:rmax_with_eps1}.
......
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