From 885d60070f98d5cbf131ae091cd74cf0a8da4f94 Mon Sep 17 00:00:00 2001
From: Luke Naylor <l.naylor@sms.ed.ac.uk>
Date: Wed, 21 Jun 2023 12:34:16 +0100
Subject: [PATCH] Remove useage of m in refinement section
---
main.tex | 95 ++++++++++++++++++++++++++++----------------------------
1 file changed, 48 insertions(+), 47 deletions(-)
diff --git a/main.tex b/main.tex
index dd9bc22..8197ab1 100644
--- a/main.tex
+++ b/main.tex
@@ -458,6 +458,7 @@ Component-wise, this is:
\\
\chern^\beta_2(E) &= \chern_2(E) - \beta \chern_1(E) + \frac{\beta^2}{2} \chern_0(E)
\end{align*}
+where $\chern_i$ is the coefficient of $\ell^i$ in $\chern$.
% TODO I think this^ needs adjusting for general Surface with $\ell$
\end{dfn}
@@ -504,7 +505,7 @@ 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
+$\beta_{-} = \frac{a_v}{n}$ for some $a_v,n \in \ZZ$. 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 $\chern_2^{\beta_-}(E) > 0$ (by using $P=(\beta_-,0)$ in
@@ -590,8 +591,8 @@ corresponding $\chern_1^{\beta}(E)$ fail one of the inequalities (which is what
was implicitly happening before).
First, let us fix a Chern character for $F$,
-$\chern(F) = (R,C,D)$, and consider the possible Chern characters
-$\chern(E) = (r,c,d)$ of some semistabilizer $E$.
+$\chern(F) = (R,C\ell,D\ell^2)$, and consider the possible Chern characters
+$\chern(E) = (r,c\ell,d\ell^2)$ of some semistabilizer $E$.
\begin{sagesilent}
# Requires extra package:
@@ -1217,7 +1218,7 @@ for the bounds on $d$ in terms of $r$ is illustrated in figure
(\ref{fig:d_bounds_xmpl_gnrc_q}).
The question of whether there are pseudo-destabilizers of arbitrarily large
rank, in the context of the graph, comes down to whether there are points
-$(r,d) \in \ZZ \oplus \frac{1}{m} \ZZ$ (with large $r$)
+$(r,d) \in \ZZ \oplus \frac{1}{2} \ZZ$ (with large $r$)
% TODO have a proper definition for pseudo-destabilizers/walls
that fit above the yellow line (ensuring positive radius of wall) but below the
blue and green (ensuring $\Delta(E), \Delta(G) > 0$).
@@ -1253,7 +1254,11 @@ $(r,c,d)$ that satisfy all inequalities to give a pseudowall.
\subsubsection{All Semistabilizers Left of Vertical Wall for Rational Beta min}
-The strategy here is similar to what was shown in (sect \ref{sec:twisted-chern}),
+The strategy here is similar to what was shown in (sect
+\ref{sec:twisted-chern}).
+One specialization here is to use that $\ell:=c_1(H)$ generates $NS(X)$, so that
+in fact, any Chern character can be written as
+$\left(r,c\ell,\frac{e}{2}\ell^2\right)$ for $r,c,e\in\ZZ$.
% ref to Schmidt?
\begin{sagesilent}
@@ -1281,7 +1286,7 @@ radius of the pseudo-wall being positive
\begin{equation}
\label{eqn:positive_rad_condition_in_terms_of_q_beta}
- \frac{1}{m}\ZZ
+ \frac{1}{2}\ZZ
\ni
\qquad
\sage{positive_radius_condition.subs([q_value_expr,beta_value_expr]).factor()}
@@ -1293,19 +1298,19 @@ radius of the pseudo-wall being positive
\begin{sagesilent}
var("nu", domain="real") # placeholder for the specific values of 1/epsilon
-assymptote_gap_condition1 = (1/nu < bgmlv2_d_upperbound_exp_term)
-assymptote_gap_condition2 = (1/nu < bgmlv3_d_upperbound_exp_term_alt2)
+assymptote_gap_condition1 = (1/(2*n^2) < bgmlv2_d_upperbound_exp_term)
+assymptote_gap_condition2 = (1/(2*n^2) < bgmlv3_d_upperbound_exp_term_alt2)
r_upper_bound1 = (
assymptote_gap_condition1
- * r * nu
+ * r * 2*n^2
)
assert r_upper_bound1.lhs() == r
r_upper_bound2 = (
assymptote_gap_condition2
- * (r-R) * nu + R
+ * (r-R) * 2*n^2 + R
)
assert r_upper_bound2.lhs() == r
@@ -1313,13 +1318,12 @@ assert r_upper_bound2.lhs() == r
\begin{theorem}[Bound on $r$ \#1]
\label{thm:rmax_with_uniform_eps}
- Let $v = (R,C,D)$ be a fixed Chern character. Then the ranks of the
+ Let $v = (R,C\ell,D\ell^2)$ be a fixed Chern character. Then the ranks of the
pseudo-semistabilizers for $v$ with
$\chern_1^\beta = q$
are bounded above by the following expression.
\bgroup
- \def\nu{\lcm(m,2n^2)}
\def\psi{\chern_1^{\beta}(F)}
\begin{align*}
\min
@@ -1340,9 +1344,9 @@ assert r_upper_bound2.lhs() == r
\noindent
Both $d$ and the lower bound in
(eqn \ref{eqn:positive_rad_condition_in_terms_of_q_beta})
-are elements of $\frac{1}{\lcm(m,2n^2)}\ZZ$.
+are elements of $\frac{1}{2n^2}\ZZ$.
So, if any of the two upper bounds on $d$ come to within
-$\frac{1}{\lcm(m,2n^2)}$ of this lower bound, then there are no solutions for
+$\frac{1}{2n^2}$ of this lower bound, then there are no solutions for
$d$.
Hence any corresponding $r$ cannot be a rank of a
pseudo-semistabilizer for $v$.
@@ -1381,7 +1385,7 @@ bounds_too_tight_condition2 = (
\sage{bgmlv2_d_upperbound_exp_term},
\sage{bgmlv3_d_upperbound_exp_term_alt2}
\right)
- \geq \epsilon := \frac{1}{\lcm(m,2n^2)}
+ \geq \epsilon := \frac{1}{2n^2}
\end{equation}
\egroup
@@ -1390,7 +1394,6 @@ This is equivalent to:
\bgroup
\def\psi{\chern_1^{\beta}(F)}
-\def\nu{\lcm(m,2n^2)}
\begin{equation}
\label{eqn:thm-bound-for-r-impossible-cond-for-r}
r \leq
@@ -1410,27 +1413,27 @@ This is equivalent to:
\end{proof}
\begin{sagesilent}
-var("Delta", domain="real")
+var("Delta nu", domain="real")
q_sol = solve(r_upper_bound1.rhs() == r_upper_bound2.rhs(), q)[0].rhs()
r_upper_bound_all_q = (
r_upper_bound1.rhs()
.expand()
.subs(q==q_sol)
- .subs(psi**2 == Delta)
- .subs(1/psi**2 == 1/Delta)
+ .subs(psi**2 == Delta/nu^2)
+ .subs(1/psi**2 == nu^2/Delta)
)
\end{sagesilent}
\begin{corrolary}[Bound on $r$ \#2]
\label{cor:direct_rmax_with_uniform_eps}
Let $v$ be a fixed Chern character and
- $R:=\chern_0(v) \leq \frac{1}{2}\lcm(m,2n^2)\Delta(v)$.
+ $R:=\chern_0(v) \leq n^2\Delta(v)$.
Then the ranks of the pseudo-semistabilizers for $v$
are bounded above by the following expression.
\bgroup
\let\originalDelta\Delta
- \def\nu{\lcm(m,2n^2)}
+ \let\nu\ell
\renewcommand\Delta{{\originalDelta(v)}}
\begin{equation*}
\sage{r_upper_bound_all_q.expand()}
@@ -1441,9 +1444,7 @@ r_upper_bound_all_q = (
\begin{proof}
\bgroup
\def\psi{\chern_1^{\beta}(F)}
-\def\nu{\lcm(m,2n^2)}
\let\originalDelta\Delta
-\renewcommand\Delta{{\psi^2}}
The ranks of the pseudo-semistabilizers for $v$ are bounded above by the
maximum over $q\in [0, \chern_1^{\beta}(F)]$ of the expression in theorem
\ref{thm:rmax_with_uniform_eps}.
@@ -1459,12 +1460,12 @@ and evaluating on of the $f_i$ there.
The intersection exists, provided that
$f_1(\chern_1^{\beta}(F))>f_2(\chern_1^{\beta}(F))=R$,
or equivalently,
-$R \leq \frac{1}{2}\lcm(m,2n^2){\chern_1^{\beta}(F)}^2$.
+$R \leq n^2{\chern_1^{\beta}(F)}^2$.
Setting $f_1(q)=f_2(q)$ yields
$q=\sage{q_sol.expand()}$.
And evaluating $f_1$ at this $q$-value gives:
-$\sage{r_upper_bound_all_q.expand()}$.
-Finally, noting that $\originalDelta(v)=\psi^2$, we get the bound as
+$\sage{r_upper_bound_all_q.expand().subs([nu==1,Delta==psi^2])}$.
+Finally, noting that $\originalDelta(v)=\psi^2\ell^2$, we get the bound as
stated in the corollary.
\egroup
\end{proof}
@@ -1498,7 +1499,7 @@ These 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
+$\frac{1}{2}\ZZ$ is at least $\frac{1}{2n^2}$ 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.
@@ -1547,7 +1548,7 @@ proof of theorem \ref{thm:rmax_with_uniform_eps}:
$\epsilon_{q,1}$ and $\epsilon_{q,2}$
]
\label{lemdfn:epsilon_q}
- Suppose $d \in \frac{1}{m}\ZZ$ satisfies the condition in
+ Suppose $d \in \frac{1}{2}\ZZ$ satisfies the condition in
eqn \ref{eqn:positive_rad_condition_in_terms_of_q_beta}.
That is:
@@ -1567,27 +1568,27 @@ Where $\epsilon_{q,1}$ and $\epsilon_{q,2}$ are defined as follows:
\begin{equation*}
\epsilon_{q,1} :=
- \frac{k_{q,1}}{2mn^2}
+ \frac{k_{q,1}}{2n^2}
\qquad
\epsilon_{q,2} :=
- \frac{k_{q,2}}{2mn^2}
+ \frac{k_{q,2}}{2n^2}
\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 \equiv -\aa\bb \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)
+ k \equiv \aa\bb (\aa\aa^{'}-2)
+ \mod n\gcd(n,\aa^2)
\end{align*}
\end{lemmadfn}
It is worth noting that $\epsilon_{q,2}$ is potentially larger than
-$\epsilon_{q,2}$
+$\epsilon_{q,1}$
but calculating it involves a $\gcd$, a modulo reduction, and a modulo $n$
inverse, for each $q$ considered.
@@ -1596,31 +1597,31 @@ inverse, for each $q$ considered.
Consider the following:
\begin{align}
- \frac{ x }{ m }
+ \frac{ x }{ 2 }
- \frac{
(\aa r+2\bb)\aa
}{
2n^2
}
- = \frac{ k }{ 2mn^2 }
+ = \frac{ k }{ 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 m
+ - (\aa r+2\bb)\aa
&\equiv k &&
- \mod 2n^2
+ \mod n^2
\\ &\Longleftrightarrow&
- - \aa^2 m r - 2\aa\bb m
+ - \aa^2 r - 2\aa\bb
&\equiv k &&
- \mod 2n^2
+ \mod n^2
\\ &\Longrightarrow&
- \aa^2 \aa^{'}\bb m - 2\aa\bb m
+ \aa^2 \aa^{'}\bb - 2\aa\bb
&\equiv k &&
- \mod \gcd(2n^2, \aa^2 mn)
+ \mod \gcd(n^2, \aa^2 n)
\label{eqn:better_eps_problem_k_mod_gcd2n2_a2mn}
\\ &\Longrightarrow&
- -\aa\bb m
+ -\aa\bb
&\equiv k &&
\mod n
\label{eqn:better_eps_problem_k_mod_n}
@@ -1631,12 +1632,12 @@ 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
+We are then guaranteed that the gap $\frac{k}{2n^2}$ 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 $\epsilon_{q,2}$.
+which also guarantees that the gap $\frac{k}{2n^2}$ is at least $\epsilon_{q,2}$.
\end{proof}
@@ -1678,9 +1679,9 @@ eps_k_i_subs = nu == (2*m*n^2)/delta
Just like in examples \ref{exmpl:recurring-first} and
\ref{exmpl:recurring-second},
take $\ell=c_1(\mathcal{O}(1))$ as the standard polarization on $\PP^2$, so that
-$m=2$, $\beta_-=\sage{recurring.b}$, giving $n=\sage{recurring.b.denominator()}$
+$m=2$, $\beta=\sage{recurring.b}$, giving $n=\sage{recurring.b.denominator()}$
and $\chern_1^{\sage{recurring.b}}(F) = \sage{recurring.twisted.ch[1]}$.
-% TODO transcode notebook code
+%% TODO transcode notebook code
Using the above theorem \ref{thm:rmax_with_eps1},
TODO fill in values
\end{example}
--
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