From 3dab119c2032729102d233fad3e519e6089e9d53 Mon Sep 17 00:00:00 2001
From: Luke Naylor <l.naylor@sms.ed.ac.uk>
Date: Sun, 23 Jul 2023 13:59:58 +0100
Subject: [PATCH] Complete proof for sanity check for pseudo-semistabs
---
main.tex | 18 ++++++++++++++++--
1 file changed, 16 insertions(+), 2 deletions(-)
diff --git a/main.tex b/main.tex
index 4c6899f..a80b160 100644
--- a/main.tex
+++ b/main.tex
@@ -235,8 +235,22 @@ $\bddderived(X)$, some other sources may have this extra restriction too.
Since all the objects in the sequence are in $\firsttilt\beta$, we have
$\chern_1^{\beta} \geq 0$ for each of them. Due to additivity
($\chern(F) = \chern(E) + \chern(G)$), we can deduce
- $0 \leq \chern(E) \leq \chern(F)$.
- % TODO unfinished
+ $0 \leq \chern_1^{\beta}(E) \leq \chern_1^{\beta}(F)$.
+
+
+ $E \hookrightarrow F \twoheadrightarrow G$ being a semistabilizing sequence
+ means $\nu_{\alpha,\beta}(E) = \nu_{\alpha,\beta}(F) = \nu_{\alpha,\beta}(F)$.
+ % MAYBE: justify this harder
+ But also, that this is an instance of $F$ being semistable, so $E$ must also
+ be semistable
+ (otherwise the destabilizing subobject would also destabilize $F$).
+ Similarly $G$ must also be semistable too.
+ $E$ and $G$ being semistable implies they also satisfy the Bogomolov
+ inequalities:
+ % TODO ref Bogomolov inequalities for tilt stability
+ $\Delta(E), \Delta(G) \geq 0$.
+ Expressing this in terms of Chern characters for $E$ and $F$ gives:
+ $\Delta(\chern(E)) \geq 0$ and $\Delta(\chern(F)-\chern(E)) \geq 0$.
\end{proof}
--
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