# Verdict: the induction in "ICs and standards span alike"

Origin: Toan, 2026-10-20 12:58, on `solve-maths/satake-two-levels/props/summand.html`, step 5.

| statement | status | steps |
|---|---|---|
| I. \(P(\mu)\) for all \(\mu\), by strong induction on \(h(\mu)\); hypothesis used only for \(\Delta_\mu\in\operatorname{Thick}\{\mathrm{IC}_\nu:\nu\le\mu\}\) | proved (in-run audit) | A1-1–A1-5 |
| II. \(\operatorname{Thick}\{\Delta_\mu\}=\operatorname{Thick}\{\mathrm{IC}_\mu\}\iff\mu\) minimal | proved (in-run audit; second inclusion under PH) | A1-7, A1-8, A2-8, A2-9 |

## Proposition I: the induction made explicit (proved, in-run audit)
**Assumes.** H1–H3 (filtration.html), \(a_\mu^*\mathrm{IC}_\mu\simeq\underline k_{\mathcal O_\mu}[\delta(\mu)]\).
**Statement.** For \(\mu\in\Lambda_S^+\) let \(P(\mu)\): \(\operatorname{Thick}\{\mathrm{IC}_\nu:\nu\le\mu\}=\operatorname{Thick}\{\Delta_\nu:\nu\le\mu\}\). Then \(P(\mu)\) holds for every \(\mu\), by strong induction on \(h(\mu)=\#\{\nu\le\mu\}\) with hypothesis \(P(\nu)\) for all \(\nu<\mu\).
(a) \(\mathrm{IC}_\mu\in\operatorname{Thick}\{\Delta_\nu:\nu\le\mu\}\) uses no hypothesis.
(b) \(\Delta_\mu\in\operatorname{Thick}\{\mathrm{IC}_\nu:\nu\le\mu\}\) uses it once, through \(\operatorname{Thick}\{\Delta_\nu:\nu<\mu\}\subseteq\operatorname{Thick}\{\mathrm{IC}_\nu:\nu<\mu\}\).
**Proof.**
1. \(h\) is finite (H1(a)) and strictly increasing; \(\{\nu<\mu\}=\bigcup_{\nu<\mu}\{\nu'\le\nu\}\) in any partial order [[A1-1]], [[A2-4]].
2. For every \(M\), \(i_{\mu*}i_\mu^*M\in\operatorname{Thick}\{\Delta_\nu:\nu<\mu\}\): standard generation on the boundary and (T2), no induction [[A1-2]].
3. The triangle \(\Delta_\mu[\delta(\mu)]\to\mathrm{IC}_\mu\to i_{\mu*}i_\mu^*\mathrm{IC}_\mu\): outer terms give the middle, so (a) [[A1-3]].
4. From \(P(\nu)\), \(\nu<\mu\), and step 1: \(\operatorname{Thick}\{\Delta_\nu:\nu<\mu\}=\operatorname{Thick}\{\mathrm{IC}_\nu:\nu<\mu\}\) — the page's "by induction" [[A1-4]].
5. Middle and right terms of the triangle lie in the IC span (step 4), so the left one does: (b) [[A1-5]].

## Proposition II: one orbit alone (proved, in-run audit)
**Assumes.** As I, plus the recollement axioms \(f^*f_*\simeq\mathrm{id}\), \(i^*j_!=0\) (H2′, part of Kuijper Def. 2.3.4); for (c) also PH.
**Statement.** (a) If \(\mu\) is minimal, \(\mathrm{IC}_\mu\simeq\Delta_\mu[\delta(\mu)]\). (b) If some \(\nu<\mu\) exists, \(\mathrm{IC}_\mu\notin\operatorname{Thick}\{\Delta_\mu\}\). (c) Under PH, also \(\Delta_\mu\notin\operatorname{Thick}\{\mathrm{IC}_\mu\}\). (d) Always \(\operatorname{Thick}(\mathcal C_{<\mu}\cup\{\Delta_\mu\})=\operatorname{Thick}(\mathcal C_{<\mu}\cup\{\mathrm{IC}_\mu\})\), \(\mathcal C_{<\mu}=\operatorname{Thick}\{\Delta_\nu:\nu<\mu\}\).
**Proof.**
1. \(a_\nu^*\) is exact with thick kernel; \(a_\nu^*\Delta_\mu=0\), \(a_\nu^*\mathrm{IC}_\mu\ne0\) (Deligne's formula, BBD 2.1.11) [[A1-7]], [[A2-8]].
2. Under PH the constituent subcategory \(\mathcal D_{\{\mu\}}\) is thick and contains \(\mathrm{IC}_\mu\) but not \(\Delta_\mu\); in \(\mathrm{PGL}_2\times\mathrm{PGL}_2/\mathrm{PGL}_2\), \(\mu=2\): \({}^pH^{-1}(\Delta_2[\delta])\simeq\mathrm{IC}_0\) [[A2-9]].
3. (d) is the triangle modulo \(\mathcal C_{<\mu}\) [[A1-8]].

## Gaps
- None for the question. Without PH, whether \(\Delta_\mu\in\operatorname{Thick}\{\mathrm{IC}_\mu\}\) is not decided (not needed).

## Routes ruled out
- Separating \(\Delta_\mu\) from \(\operatorname{Thick}\{\mathrm{IC}_\mu\}\) by \(a_\nu^*\): it kills \(\Delta_\mu\); constituents are used instead.
- \(K_0\) test: thick spans contain retracts whose classes escape the generated subgroup.

## Proposed edit (summand.html, step 5)
"…which equals \(\operatorname{Thick}\{\mathrm{IC}_\nu:\nu<\mu\}\) by induction, i.e. by \(P(\nu)\) for all \(\nu<\mu\), since \(\{\nu<\mu\}=\bigcup_{\nu<\mu}\{\nu'\le\nu\}\); this is needed only for \(\Delta_\mu\in\operatorname{Thick}\{\mathrm{IC}_\nu:\nu\le\mu\}\)." Also: filtration.html says part (a) is used "in step 3"; proposition-I.tex uses it in step 1.