Implement Jun28 feedback (minus big lemma)

tex/setting-and-problems.tex

@@ -39,9 +39,11 @@ affect the results.
 	be considered but are left out for now as they do not have a great impact on
 	the finiteness of pseudo-walls.
 	In the case of a principally polarised abelian surface, the main example in
-	this thesis, the Euler characteristic condition is vacuous and the extension
-	group condition eliminates possibities with lower rank, and often none at all
-	for small values of $\chern_0(v)$.
+	this thesis, the Euler characteristic condition is vacuous.
+    The extension group condition can be shown to only
+    be redundant among the other conditions for sufficiently large rank of $u$
+    (when $\chern_0(u) \geq \chern_0(v)/2$), so does not affect the finiteness
+    of pseudo-semistabilisers.
 \end{remark}
 
 Later, Chern characters will be written $(r,c\ell,d\ell^2)$ because operations
@@ -119,7 +121,7 @@ $d \in \frac{1}{\lcm(m,2)}\ZZ$.
 \section{Characteristic Curves for Pseudo-semistabilisers}
 
 These characteristic curves introduced in
-subsection \ref{subsec:charact-curves}
+Subsection \ref{subsec:charact-curves}
 are convenient tools to think about the
 numerical conditions that can be used to test for pseudo-semistabilisers, and
 for solutions to the problems
@@ -159,10 +161,12 @@ are equivalent to the following more numerical conditions:
 	\item $\chern_1^{\beta(P)}(u) < \chern_1^{\beta(P)}(v)$
 	\item $\Delta(v-u) \geq 0$
 	\item $\chern_2^{P}(u)>0$
+	\footnote{Here $\chern_2^{P} = \chern_2^{\alpha, \beta}$ where $P=(\alpha, \beta)$.}
 \end{enumerate}
 \end{lemma}
 
-\begin{proof}[Proof for $\chern_0(v)>0$ case.]
+\begin{proof}
+First, consider the case where $\chern_0(v)>0$.
 Let $u,v$ be Chern characters with
 $\Delta(u),\Delta(v) \geq 0$, and $v$ has positive rank.
 
@@ -186,9 +190,9 @@ ${
 (consequence 5).
 Notice that $0 < \chern_1^{\beta(P)}(u)$
 follows from consequences 1 and 2, this is why it is not included in consequence 5.
-If $u$ were to have 0 rank, it's tilt slope would be decreasing as $\beta$
+If $u$ were to have 0 rank, its tilt slope would be decreasing as $\beta$
 increases, contradicting supposition b. So $u$ must have strictly non-zero rank,
-and we can consider it's characteristic curves (or that of $-u$ in case of
+and we can consider its characteristic curves (or that of $-u$ in case of
 negative rank).
 $\nu_Q(v)=0$, and hence $\nu_Q(u)=0$ too. This means that $\Theta_{\pm u}$ must
 intersect $\Theta_v^-$ at $Q$. Considering the shapes of the hyperbolae alone,
@@ -250,14 +254,14 @@ right of $\Theta_v^-$ at $\alpha=\alpha(P)$, but crosses to the left side as
 $\alpha \to +\infty$, intersection at some point $Q$ above $P$.
 This implies that the characteristic curves for $u$ and $v$ are in the
 configuration illustrated in Fig \ref{fig:correct-hyperbol-intersection}.
-We then have $\nu(u)=\nu(v)$ along a circle to the left of $V_u$ reaching it's
+We then have $\nu(u)=\nu(v)$ along a circle to the left of $V_u$ reaching its
 apex at $Q$, and encircling $P$. This along with consequence 3 implies that $u$
 is a pseudo-semistabiliser at the point on the circle with $\beta=\beta(P)$.
-Therefore, it's also a pseudo-semistabiliser further along the circle at $Q$
+Therefore, it is also a pseudo-semistabiliser further along the circle at $Q$
 (supposition a).
 Finally, consequence 4 along with $P$ being to the left of $V_u$ implies
 $\nu_P(u) > 0$ giving supposition b.
-\end{proof}
+
 \begin{sagesilent}
 from rank_zero_case_curves import pseudo_semistab_char_curves_rank_zero
 \end{sagesilent}
@@ -268,7 +272,8 @@ from rank_zero_case_curves import pseudo_semistab_char_curves_rank_zero
 	with rank 0 destabilising `downwards'}
 	\label{fig:hyperbol-intersection-rank-zero}
 \end{figure}
-\begin{proof}[Proof for $\chern_0(v)=0$ case.]
+
+Now consider the case where $\chern_0(v)=0$ but $\chern_1(v)>0$.
 Let $u,v$ be Chern characters with
 $\Delta(u),\Delta(v) \geq 0$, and $\chern_0(v)=0$ but $\chern_1(v)>0$.
 So $\Theta_v^-$ is the vertical line at $\beta = \beta_{-}(v)$, and
@@ -406,7 +411,7 @@ Where $u$ destabilises $v$ going `down' $\Theta_v^{-}$ (in the same sense as in
 Problem \ref{problem:problem-statement-1}.
 \end{problem}
 
-This is a specialization of Problem \ref{problem:problem-statement-1}
+This is a specialisation of Problem \ref{problem:problem-statement-1}
 with the choice $P=(\beta_{-},0)$, the point where $\Theta_v^-$ meets the
 $\beta$-axis.
 This is because all circular walls left of $V_v$ intersect $\Theta_v^-$ (once).
@@ -449,6 +454,7 @@ problem using Lemma \ref{lem:pseudo_wall_numerical_tests}.
 		\item $0<\chern_1^{\beta(P)}(u)<\chern_1^{\beta(P)}(v)$
 			\label{item:chern1bound:lem:num_test_prob1}
 		\item $\chern_2^{P}(u)>0$
+		\footnote{Here $\chern_2^{P} = \chern_2^{\alpha, \beta}$ where $P=(\alpha, \beta)$.}
 			\label{item:radiuscond:lem:num_test_prob1}
 		\end{multicols}
 	\end{enumerate}
@@ -497,6 +503,8 @@ problem using Lemma \ref{lem:pseudo_wall_numerical_tests}.
 \end{theorem}
 
 \begin{proof}
-	This is a specialization of the previous Lemma, using $P=(\beta_{-},0)$.
+	This is a specialisation of the previous Theorem
+	\ref{lem:num_test_prob1},
+    using $P=(\beta_{-},0)$.
 
 \end{proof}