Newer
Older
\bgroup
\def\psi{\chern_1^{\beta}(F)}
\begin{equation}
\min\left(
\sage{bgmlv2_d_upperbound_exp_term},
\sage{bgmlv3_d_upperbound_exp_term_alt2}
\geq \epsilon := \frac{1}{\lcm(m,2n^2)}
\noindent
This is equivalent to:
\bgroup
\def\psi{\chern_1^{\beta}(F)}
\def\Delta{\lcm(m,2n^2)}
\label{eqn:thm-bound-for-r-impossible-cond-for-r}
r \leq
\egroup % end scope where epsilon redefined
Luke Naylor
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\begin{corrolary}[Bound on $r$ \#2]
\label{cor:direct_rmax_with_uniform_eps}
Let $v$ be a fixed Chern character and
$R:=\chern_0(v) \leq \frac{1}{2}\lcm(m,2n^2)\Delta(v)$.
Then the ranks of the pseudo-semistabilizers for $v$
are bounded above by the following expression.
\bgroup
\begin{equation}
\frac{1}{2} \lcm(m,2n^2)
\left(
\frac{\chern_1^{\beta}(v)}{2}
+ \frac{R}{\chern_1^{\beta}(v)\lcm(m,2n^2)}
\right)^2
\end{equation}
\egroup
\bgroup
\begin{equation}
\frac{1}{8}
\Delta(v) \lcm(m,2n^2)
+ \frac{1}{2} R
+ \frac{R^2}{ 2 \Delta(v) \lcm(m,2n^2) }
\end{equation}
\egroup
\end{corrolary}
%% TODO simplified expression for rmax by predicting which q gives rmax
%% refinements using specific values of q and beta
This bound can be refined a bit more by considering restrictions from the
possible values that $r$ take.
Furthermore, the proof of theorem \ref{thm:rmax_with_uniform_eps} uses the fact
that, given an element of $\frac{1}{2n^2}\ZZ$, the closest non-equal element of
$\frac{1}{m}\ZZ$ is at least $\frac{1}{\lcm(m,2n^2)}$ away. However this a
conservative estimate, and a larger gap can sometimes be guaranteed if we know
this value of $\frac{1}{2n^2}\ZZ$ explicitly.
The expressions that will follow will be a bit more complicated and have more
parts which depend on the values of $q$ and $\beta$, even their numerators
$\aa,\bb$ specifically. The upcoming theorem (TODO ref) is less useful as a
`clean' formula for a bound on the ranks of the pseudo-semistabilizers, but has a
purpose in the context of writing a computer program to find
pseudo-semistabilizers. Such a program would iterate through possible values of
$q$, then iterate through values of $r$ within the bounds (dependent on $q$),
which would then determine $c$, and then find the corresponding possible values
for $d$.
Firstly, we only need to consider $r$-values for which $c:=\chern_1(E)$ is
integral:
\begin{equation}
c =
\sage{c_in_terms_of_q.subs([q_value_expr,beta_value_expr])}
\in \ZZ
\end{equation}
\noindent
That is, $r \equiv -\aa^{-1}\bb$ mod $n$ ($\aa$ is coprime to
$n$, and so invertible mod $n$).
\begin{sagesilent}
rhs_numerator = (
positive_radius_condition
.rhs()
.subs([q_value_expr,beta_value_expr])
.factor()
.numerator()
)
\end{sagesilent}
\noindent
Let $\aa^{'}$ be an integer representative of $\aa^{-1}$ in $\ZZ/n\ZZ$.
Next, we seek to find a larger $\epsilon$ to use in place of $\epsilon_F$ in the
proof of theorem \ref{thm:rmax_with_uniform_eps}:
\begin{lemmadfn}[
Finding better alternatives to $\epsilon_F$:
Suppose $d \in \frac{1}{m}\ZZ$ satisfies the condition in
eqn \ref{eqn:positive_rad_condition_in_terms_of_q_beta}.
That is:
\begin{equation*}
\sage{positive_radius_condition.subs([q_value_expr,beta_value_expr]).factor()}
\end{equation*}
\noindent
Then we have:
\begin{equation*}
d - \frac{(\aa r + 2\bb)\aa}{2n^2}
\geq \epsilon_{q,2} \geq \epsilon_{q,1} > 0
\end{equation*}
Where $\epsilon_{q,1}$ and $\epsilon_{q,2}$ are defined as follows:
\end{equation*}
\begin{align*}
\text{where }
&k_{q,1} \text{ is the least }
k\in\ZZ_{>0}\: s.t.:\:
k \equiv -\aa\bb m \mod n
\\
&k_{q,2} \text{ is the least }
k\in\ZZ_{>0}\: s.t.:\:
k \equiv \aa\bb m (\aa\aa^{'}-2)
\mod n\gcd(2n,\aa^2 m)
\end{align*}
\end{lemmadfn}
It is worth noting that $\epsilon_{q,2}$ is potentially larger than
$\epsilon_{q,2}$
but calculating it involves a $\gcd$, a modulo reduction, and a modulo $n$
inverse, for each $q$ considered.
\begin{proof}
- \frac{
(\aa r+2\bb)\aa
}{
2n^2
}
= \frac{ k }{ 2mn^2 }
\quad \text{for some } x \in \ZZ
\span \span \span \span \span
\label{eqn:finding_better_eps_problem}
&\equiv k &&
\mod 2n^2
\\ &\Longleftrightarrow&
- \aa^2 m r - 2\aa\bb m
&\equiv k &&
\mod 2n^2
\\ &\Longrightarrow&
\aa^2 \aa^{'}\bb m - 2\aa\bb m
&\equiv k &&
\mod \gcd(2n^2, \aa^2 mn)
\label{eqn:better_eps_problem_k_mod_gcd2n2_a2mn}
\label{eqn:better_eps_problem_k_mod_n}
In our situation, we want to find the least $k$ 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
(a computation only depending on $q$ and $\beta$, but not $r$).
We are then guaranteed that the gap $\frac{k}{2mn^2}$ is at least
$\epsilon_{q,1}$.
Furthermore, $k$ also satisfies
eqn \ref{eqn:better_eps_problem_k_mod_gcd2n2_a2mn}
so we can also pick the smallest $k_{q,2} \in \ZZ_{>0}$ satisfying this condition,
which also guarantees that the gap $\frac{k}{2mn^2}$ is at least $\epsilon_{q,2}$.
\begin{theorem}[Bound on $r$ \#3]
\label{thm:rmax_with_eps1}
Let $v$ be a fixed Chern character, with $\frac{a_F}{n}=\beta:=\beta(v)$
rational and expressed in lowest terms.
Then the ranks $r$ of the pseudo-semistabilizers $u$ for $v$ with
$\chern_1^\beta(u) = q = \frac{b_q}{n}$
are bounded above by the following expression (with $i=1$ or $2$).
\begin{sagesilent}
eps_k_i_subs = Delta == (2*m*n^2)/delta
\end{sagesilent}
\bgroup
\def\delta{k_{q,i}}
\def\psi{\chern_1^{\beta}(F)}
\begin{align*}
\min
\left(
\sage{r_upper_bound1.rhs().subs(eps_k_i_subs)}, \:\:
\sage{r_upper_bound2.rhs().subs(eps_k_i_subs)}
\right)
\end{align*}
\egroup
Where $k_{q,i}$ is defined as in definition/lemma \ref{lemdfn:epsilon_q},
and $R = \chern_0(v)$
Furthermore, if $\aa \not= 0$ then
$r \equiv \aa^{-1}b_q (\mod n)$.
\minorheading{Irrational $\beta$}
\egroup % end scope where beta redefined to beta_{-}
\section{Appendix - SageMath code}
\usemintedstyle{tango}
\inputminted[
obeytabs=true,
tabsize=2,
breaklines=true,
breakbefore=./
]{python}{filtered_sage.txt}