From 07372ddecd55004d96f5b07fa576886292f00f77 Mon Sep 17 00:00:00 2001
From: Luke Naylor <l.naylor@sms.ed.ac.uk>
Date: Sun, 11 Jun 2023 18:44:00 +0100
Subject: [PATCH] Comment on geometry of hyperbolae
---
main.tex | 13 +++++++++++--
1 file changed, 11 insertions(+), 2 deletions(-)
diff --git a/main.tex b/main.tex
index edfd643..a929a45 100644
--- a/main.tex
+++ b/main.tex
@@ -104,7 +104,7 @@ Characters}
\[
u = \left(r, c\ell, d \frac{1}{2} \ell^2\right)
\]
- which has the same slope as $v$: $\mu_{\sigma}(u) = \mu_{\sigma}(v)$.
+ which has the same tilt slope as $v$: $\mu_{\sigma}(u) = \mu_{\sigma}(v)$.
Note $u$ does not need to be a Chern character of an actual sub-object of some
object in the stability condition's heart with Chern character $v$.
@@ -196,8 +196,17 @@ $\alpha$-$\beta$-half-plane into regions where the signs objects of Chern charac
Secondly, it gives more of a fixed target for some $u=(r,c\ell,d\frac{1}{2}\ell^2)$ to
be a pseudo-semistabilizer of $v$, in the following sense:
If $(\alpha,\beta)$, is on the hyperbola $\chern_2^{\alpha, \beta}(v)=0$, then
-$\mu_{\alpha,\beta}(v)=0$ and for any $u$, $u$ is a pseudo-semistabilizer of $v$
+for any $u$, $u$ is a pseudo-semistabilizer of $v$
iff $\mu_{\alpha,\beta}(u)=0$, and hence $\chern_2^{\alpha, \beta}(u)=0$.
+In fact, this allows us to use the characteristic curves of some $v$ and $u$
+(with $\Delta(v)\geq 0$, $\Delta(u)\geq 0$ and positive ranks) to determine the
+location of the pseudo-wall where $u$ pseudo-semistabilizes $v$.
+%TODO ref forwards
+
+Commenting on the geometry of the hyperbola, it always has left and right
+branches, or degenerates to 2 lines. This is a consequence of $\Delta(v)\geq 0$.
+Furthermore the assymptotes are angled at $\pm 45^\circ$, crossing through the
+base of the first characteristic curve $\chern_1^{\alpha,\beta}=0$ (vertical line).
--
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