AI review · 20 October 2026 edits applied

Four questions from Toan. Each was framed as a problem and solved by /solve-maths with two agents: a solver, then a fresh auditor. Milton reviewed the edits; all four are applied (table below).

Table 1. Q and question and answer
Q	question	answer
Q1	Why is $H^{2e}_c(U)\ne0$? Where is symmetry used?	$U\ni t^\lambda$ is nonempty of dimension $e$, and $H^{2e}_c$ counts its top components. The deduction uses no symmetry; only the inputs S1, S3, S4 do.
Q2	Does recollement need finite unions of orbits?	No. The open piece must be quasi-compact (cut out at a finite jet level). $\{0\}\subseteq t^m\mathcal O^n$ is one orbit with no recollement. Finiteness is for the induction.
Q3	Where is the induction used in step 5?	On the whole down-set $\{\nu\le\mu\}$, once: to turn the boundary term into IC objects. The one-orbit reading is false.
Q4	For $\mathbb A^n$, do standards still generate?	Yes, they no longer generate: they miss the stalk at $0$. The standards of the pieces of the placid levels do generate.

Edits

Footnote: a proof step, a remark or a status cell changes; no proposition statement changes. Systematic: a statement or a hypothesis changes.

On the pages, each edit carries a [T] mark (T = Toan) linking back to its question here.

Table 2. Q and where and change
Q	where	change	level
Q1	split.html step 6 · criterion.html step 6 · Prop. III PDF	lowest-degree proof replaces weights (kept as a remark); "the deduction uses no symmetry"	footnote
Q2	summand.html H2 · affine.html Table 1 · Prop. IV PDF	remark after H2; "H2 empty" → "H2 fails at $\{0\}$"	footnote
Q3	summand.html step 5 · its PDF	"by induction" spelled out	footnote
Q4	filtration.html, after (b)	remark: without H1(a), (b) fails for $\mathbb A^n$	footnote

No systematic change: the level form that Q2 and Q4 point to is already Proposition IV §4. Dropped: the step-3/step-1 note for filtration.html, since both numberings are correct (HTML vs PDF). Q4's remark cites the run, not Proposition IV, which does not state the zero-stalk span.

Q1 · Why is $H^{2e}_c(U)$ nonzero, and where is symmetry used?

Toan · 20 Oct, 05:40 UTC · split.html, step 6

$H^{2e}_c(U)$ is nonzero

1) why should $H^{2e}_c(U)$ be nonzero? 2) where have we used $X$ is symmetric in this proof? technically we have not, trying to rewrite this in our notes

(1) $U$ is nonempty because it contains $t^\lambda$, and it has dimension $e$. Top compactly supported cohomology has one line for each $e$-dimensional component. (2) You are right: the lemma's deduction uses no symmetry. Symmetry enters only in verifying three of its inputs.

Proposition. Let $F=\{g\in\mathrm{Gr}_{\le\lambda}:g^{-1}t^\lambda x_0\in L^+X\}$, $U=F\cap\mathrm{Gr}_\lambda$, $e=\dim U$, $d=\langle2\rho,\lambda\rangle$.

$\dim H^{2e}_c(U)=\#\{\text{components of }U\text{ of dimension }e\}\ge1$, and $H^{2e}_c(U)\hookrightarrow H^{2e-d}\big((V_\lambda\ast\mathrm{IC}_0)_{t^\lambda x_0}\big)$, as the page claims.
Without weights: the stalk is already nonzero in degree $-d$.
S1–S4 ⇒ H5 is formal. Symmetry is used only for the support bound (S1), connected stabilizers (S3) and $t^\lambda x_0\in\mathcal O_{\varphi(\lambda)}$ (S4).

Proof plan. Steps 1–3 · The top class Step 4 · A weight-free shortcut Steps 5–7 · Symmetry

The top class

$0\ne H^{2e}_c(U)\hookrightarrow H^{2e-d}$ of the stalk at $x_\mu=t^\lambda x_0$.

Locate the base point

$F$ is closed in $\mathrm{Gr}_{\le\lambda}$, hence projective. $U$ is open in $F$ and contains $t^\lambda$. The stalk is the cohomology of $F$:
$$(t^\lambda)^{-1}\,t^\lambda x_0=x_0\in L^+X,\qquad (V_\lambda\ast\mathrm{IC}_0)_{t^\lambda x_0}\simeq R\Gamma(F,V_\lambda|_F).$$
$L^+X$ is closed in $LX$ because $X$ is affine. The action map is proper on $\mathrm{Gr}_{\le\lambda}\tilde\times L^+X$. Proper base change at a finite jet level then gives the stalk formula, with no symmetry used.

Sources
Count top components

For any nonempty separated $U$ of finite type and dimension $e$:
$$H^{2e}_c(U)\simeq\bigoplus_{\text{components }C,\ \dim C=e}k(-e)\ne0.$$
Remove a closed set of dimension $\lt e$, which does not change $H^{2e}_c$. On the smooth part, the trace map gives one line per component. Neither irreducibility nor equidimensionality is needed.
Proof
- Let $Z$ be the union of the lower-dimensional components, the pairwise intersections and the singular locus. Then $\dim Z\le e-1$, so $H^{2e-1}_c(Z)=H^{2e}_c(Z)=0$ and $H^{2e}_c(U\setminus Z)\xrightarrow{\sim}H^{2e}_c(U)$.
- For a smooth irreducible $V$ of dimension $e$, $\mathrm{tr}:H^{2e}_c(V)\xrightarrow{\sim}k(-e)$, which is pure of weight $2e$.
Kill the connecting map

In the sequence for $U\subseteq F\supseteq Z=F\setminus U$, the connecting map is zero:
$$H^{2e-d-1}(Z,V_\lambda)\xrightarrow{\ \delta\ }H^{2e}_c(U)\longrightarrow H^{2e-d}(F,V_\lambda).$$
The source has weights $\le 2e-1$, because $V_\lambda$ has weight $\le d$ and $*$-restriction and proper pushforward preserve upper weight bounds (BBD 5.1.14). The target is pure of weight $2e$.

Sources

A weight-free shortcut

This is recommended for the notes: no $\overline{\mathbb F}_q$ model and no transfer to $\mathbb C$ are needed.

Read the lowest degree

$\mathcal H^{-d}(V_\lambda)\supseteq\underline k$, so at every $y$ in the support:
$$H^{-d}\big((V_\lambda\ast\mathrm{IC}_0)_y\big)\supseteq H^0(F_y,k)\ne0,\qquad\operatorname{supp}(V_\lambda\ast\mathrm{IC}_0)=\mathrm{Gr}_{\le\lambda}\cdot L^+X.$$
Deligne's formula gives $\mathcal H^{-d}\,\mathrm{IC}(\mathrm{Gr}_{\le\lambda})=j_*\underline k_{\mathrm{Gr}_\lambda}$, which contains the constant sheaf because $\mathrm{Gr}_\lambda$ is dense. The functor $H^0$ is left exact.

Sources

Symmetry

The deduction is formal. Symmetry sits only in the inputs.

Deduce H5 formally

S1–S4 ⇒ H5 uses only S1–S4, the closure order on orbits, the H2 triangle and additivity of $j^*$.

None of $\theta,\ \tau,\ S,\ K,\ \varphi$ appears in steps 1–5 and 7 of the lemma. Here $\lambda$ is only the name of the object $V_\lambda\ast\mathrm{IC}_0$.

Locate the symmetric inputs

Symmetry is used in exactly three places:

Table 3. Comparison
S1	support bound $\nu_b\le\varphi(\lambda)$ (Proposition III, Lemma 3.1, via $\tau$ and $\theta$ on roots)
S3	connected stabilizers from $Z_K(S)$ connected (Lemma 3.2). At a fixed $\lambda$ only "⇐" holds: $\mathrm{SL}_2/\mathrm{SO}_2$, $\lambda=0$.
S4	$t^\lambda x_0=x_{\varphi(\lambda)}$ (Chen–Yi Prop. 17(i) and injectivity of $\tau$), plus the support bound

Placidity and the closure order enter by citation (Chen–Yi Thm. 36, Thm. 18). The decomposition step, steps 1–4, and the stalk formula hold for any smooth affine $X$.

Sources

State the general lemma

Let $X$ be smooth and affine, and let $Y_\lambda=\mathrm{Gr}_{\le\lambda}\cdot L^+X$ have dense orbit $\mathcal O_{\psi(\lambda)}$. Assume:
$$\text{(Q0) }L^+X\text{ is one orbit};\quad\text{(Q1) }Y_\lambda\text{ is a finite union of orbits, placid};\quad\text{(Q2) the stabilizer at }\psi(\lambda)\text{ is connected}.$$
Then $\lambda$ satisfies H5 at $\psi(\lambda)$, and only there. The symmetric statement "at $\varphi(\lambda)$" is the identity $\psi(\lambda)=\varphi(\lambda)$, which is exactly the support bound.

Applied edit to split.html, step 6: lead with step 4 (degree $-d$) and keep the weight argument as a remark, adding "$\dim H^{2e}_c(U)$ = number of $e$-dimensional components". In "For a symmetric space", name the three uses: the support bound, $Z_K(S)$ connected, and $t^\lambda x_0=x_{\varphi(\lambda)}$.

Run: verdict · attempt tree · 2 agents, 16:36–17:03 UTC · proved, in-run audit (A1-8 repaired; citations corrected).

Q2 · Does recollement need finite unions of orbits?

Toan · 20 Oct, 08:47 UTC · summand.html, H2

Let $D\subseteq Z$ be closed finite unions of orbits

do you really need finiteness of orbit union in orbit closure to have recollement?

No. Recollement needs the closed pieces to be cut out at a finite jet level: the open piece must be quasi-compact. The number of orbits does not matter. Finiteness is used elsewhere, to make the orbit-by-orbit induction terminate.

Proposition. Let $D\subseteq Z\subseteq LX$ be closed and $L^+G$-stable. Let $Y_N$ be the levels of the placid presentation ($Y_N=t^{-N}\mathcal O^n$ for $\mathbb A^n$), $Z_N=Z\cap Y_N$, $D_N=D\cap Y_N$.

H2 holds for $(Z,D)$ iff $Y_N\setminus Z_N$ and $Z_N\setminus D_N$ are quasi-compact for every $N$. Equivalently, $Z_N$ and $D_N$ have finitely many irreducible components, each of finite codimension.
Being a "finite union of orbits" is neither sufficient nor necessary. For $\mathbb A^n$, $\{0\}\subseteq t^m\mathcal O^n$ is one orbit and has no recollement. The lattice $t^{m+1}\mathcal O^n$ contains infinitely many orbits and has one.
For symmetric $X$, every closed finite union of orbits satisfies (a), by Chen–Yi Thm. 36. So H2 on the pages holds for a reason that has nothing to do with finiteness.

Proof plan. Steps 1–3 · The criterion Step 4 · The counterexample Steps 5–6 · What finiteness is for

The criterion

Recollement ⇔ "cylinder": a closed set that is the preimage of a closed set at one jet level.

Define cylinders

For a closed $D$ in a closed cylinder $Z=\pi_M^{-1}(\overline Z)$:
$$D=\pi_M^{-1}(\overline D)\ \text{for some }M\iff Z\setminus D\ \text{quasi-compact}\iff D\ \text{has finitely many components of finite codimension}.$$
Use a finite subcover by basic opens $D(g_i)$, all defined at one level. In the other direction, the codimension of a component stabilizes exactly when the component is pulled back from one level.

Sources
Glue level-wise recollements

If $Z$ and $D$ are cylinders, the six functors preserve the constructible categories, and for every $M\in\mathcal C_Z$:
$$j_!j^*M\longrightarrow M\longrightarrow i_*i^*M\xrightarrow{+1},\qquad\mathcal C_D\xrightarrow{i_*}\mathcal C_Z\xrightarrow{j^*}\mathcal C_U\ \text{a recollement}.$$
At each finite level, BBD §1.4 (equivariantly, Bernstein–Lunts) gives the recollement. The squares with the level transitions are cartesian, so base change shows that $j_!$ and $i_*$ of a pulled-back object are again pulled back.

Sources
Read necessity off stalks

If the triangle exists in $\mathcal C_Z$ for $\underline k_Z$, then $D$ is a cylinder.

The third term has stalk support exactly $D$. Stalks of a pulled-back object are constant on the fibres of $\pi_M$, so $D$ is a union of fibres over a constructible set. The same argument applied to $r_{Z!}\underline k_Z$ forces $Z$ itself to be a cylinder.

The counterexample

A single orbit with no recollement.

Remove the origin

For $X=\mathbb A^n$, $Z=t^m\mathcal O^n$ and $D=\{0\}$, every fibre of $\pi_M$ meets $U=Z\setminus\{0\}$. Hence:
$$\ker\big(j^*:\mathcal C_Z\to\mathcal C_U\big)=0\ne\mathcal C_{\{0\}},\qquad\delta_0,\ j_!\underline k_U\notin\mathcal C.$$
$\{0\}$ is a closed finite union of orbits (just one), but it has infinite codimension. So H2 as written fails at $\{0\}$. Proposition IV's phrase "H2 is empty" should say so.

What finiteness is for

It is used for the induction, not for the recollement.

Terminate the dévissage

In filtration.html and Proposition I, "finite" supplies a maximal orbit, the inductions on $n(Z)$ and $h(\mu)$, and finite support H1(b). For $\mathbb A^n$, even with $\delta_0$ added:
$$\mathrm{IC}_0\notin\operatorname{Thick}\{\Delta_m\ (m\in\mathbb Z),\ \delta_0\}.$$
Objects reachable from finitely many orbit standards have nonzero stalks on only finitely many orbits, and this property is thick. $\mathrm{IC}_0$ is nonzero on infinitely many shells.

Sources
Check the symmetric case

For $X=G/G^\theta$, every orbit closure is a cylinder and every orbit is quasi-compact open in its closure. So closed finite unions of orbits satisfy step 1.

Chen–Yi v4, Thm. 36(i)–(ii) gives finitely presented embeddings, and App. A.1(2),(4),(5) says these are pulled back from a finite level. Prop. 17 shows every orbit is of this form.

Sources

Applied edit after H2: "Recollement itself needs only that $D$ and $Z$ be cut out at a finite jet level ($Z\setminus D$ quasi-compact). For symmetric $X$, closed finite unions of orbits are such (Chen–Yi Thm. 36). Finiteness is used for the induction (H1)." In affine.html / Proposition IV: "H2 is empty" → "H2 fails at $\{0\}$".

Run: verdict · attempt tree · 2 agents, 16:36–17:11 UTC · proved, in-run audit, for $\mathbb A^n$; general $X$ under the placid assumption (P).

Q3 · Where is the induction used?

Toan · 20 Oct, 12:58 UTC · summand.html, step 5

$\operatorname{Thick}\{\Delta_\nu:\nu\lt\mu\}$, which equals $\operatorname{Thick}\{\mathrm{IC}_\nu:\nu\lt\mu\}$ by induction.

we are inducting on $\mu$ here? but then where is induction being used? in the proof, we assume $\operatorname{Thick}\{\Delta_{\nu}: \nu<\mu\}=\operatorname{Thick}\{\mathrm{IC}_{\nu}:\nu<\mu\}$ and want to show $\operatorname{Thick}\{\Delta_{\nu}: \nu=\mu\}=\operatorname{Thick}\{\mathrm{IC}_{\nu}:\nu=\mu\}$?

Yes, it is an induction on $\mu$, but the statement carried along is the equality over the whole down-set $\{\nu\le\mu\}$. The hypothesis is used once, for the standard $\Delta_\mu$. The one-orbit version is false.

Proposition. Assume H1–H3. For $\mu\in\Lambda_S^+$ let

$$P(\mu):\qquad\operatorname{Thick}\{\mathrm{IC}_\nu:\nu\le\mu\}=\operatorname{Thick}\{\Delta_\nu:\nu\le\mu\}.$$

$P(\mu)$ holds for every $\mu$. The proof is a strong induction on $h(\mu)=\#\{\nu\le\mu\}$, with hypothesis $P(\nu)$ for all $\nu<\mu$.
$\mathrm{IC}_\mu\in\operatorname{Thick}\{\Delta_\nu:\nu\le\mu\}$ needs no hypothesis. $\Delta_\mu\in\operatorname{Thick}\{\mathrm{IC}_\nu:\nu\le\mu\}$ uses it exactly once.
$\operatorname{Thick}\{\Delta_\mu\}=\operatorname{Thick}\{\mathrm{IC}_\mu\}$ holds iff $\mu$ is minimal. It always holds modulo $\operatorname{Thick}\{\Delta_\nu:\nu<\mu\}$.

Proof plan. Steps 1–2 · Setup Steps 3–5 · Induction step Steps 6–7 · One orbit alone

Setup

What is assumed at $\mu$: $P(\nu)$ for all $\nu<\mu$, and the boundary is standard-generated.

Fix the induction

$h(\mu)$ is finite and strictly increasing, and the strict down-set is a union of down-sets:
$$\nu<\mu\ \Rightarrow\ h(\nu)\lt h(\mu),\qquad\{\nu<\mu\}=\bigcup_{\nu<\mu}\{\nu'\le\nu\}.$$
H1(a) gives finiteness. The union holds in any partial order.
Proof
- If $\nu<\mu$, then $\{\nu'\le\nu\}\subsetneq\{\nu'\le\mu\}$, because $\mu\notin\{\nu'\le\nu\}$ by antisymmetry.
- If $\nu'\le\nu<\mu$, then $\nu'<\mu$. Conversely, $\nu\in\{\nu'\le\nu\}$.
- Strong induction on $h\in\mathbb Z_{\ge1}$. The base case is a minimal $\mu$, and there may be several: $0$ and $1$ for $\mathrm{PGL}_2$.
Sources
Generate the boundary

For every $M\in\mathcal C$, with no induction:
$$i_{\mu*}i_\mu^*M\in\operatorname{Thick}\{\Delta_\nu:\nu<\mu\}.$$
Apply standard generation (filtration.html, part (a)) to $Z=\partial\mathcal O_\mu=\bigsqcup_{\nu<\mu}\mathcal O_\nu$, then push forward to $LX$ by (T2).
Proof
- By H1(a), $\partial\mathcal O_\mu$ is a closed finite union of orbits with $I(Z)=\{\nu<\mu\}$.
- $i_\mu^*M\in\mathcal C_Z=\operatorname{Thick}\{\Delta_{\nu,Z}\}$. Since $i_{\mu*}$ is exact and $i_{\mu*}\Delta_{\nu,Z}=\Delta_\nu$, (T2) applies.
Sources

Induction step

One triangle, read in both directions:

$$\Delta_\mu[\delta(\mu)]\longrightarrow\mathrm{IC}_\mu\longrightarrow i_{\mu*}i_\mu^*\mathrm{IC}_\mu\xrightarrow{+1}$$

Read the triangle forward

The outer terms give the middle one, with no hypothesis:
$$\mathrm{IC}_\mu\in\operatorname{Thick}\{\Delta_\nu:\nu\le\mu\}.$$
The left term is a shift of $\Delta_\mu$, and the right term is covered by step 2. Then use two out of three, (T1).

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Use the hypothesis once

This is the page's "equals … by induction":
$$\operatorname{Thick}\{\Delta_\nu:\nu<\mu\}=\operatorname{Thick}\Big(\bigcup_{\nu<\mu}\operatorname{Thick}\{\Delta_{\nu'}:\nu'\le\nu\}\Big)\overset{P(\nu)}{=}\operatorname{Thick}\{\mathrm{IC}_\nu:\nu<\mu\}.$$
Step 1, plus $P(\nu)$ for every $\nu<\mu$. Only the inclusion $\subseteq$ is new, since $\supseteq$ is step 3 applied at each $\nu$.
Read the triangle backward

The middle and right terms give the left one, which proves $P(\mu)$:
$$\Delta_\mu\in\operatorname{Thick}\{\mathrm{IC}_\nu:\nu\le\mu\}.$$
Step 2 puts the right term only in the standard span below $\mu$. Step 4 turns that span into the IC span. This is the only place the hypothesis enters.
Proof
- $\Delta_\mu[\delta(\mu)]\simeq\operatorname{fib}(\mathrm{IC}_\mu\to i_{\mu*}i_\mu^*\mathrm{IC}_\mu)$. Both terms lie in $\operatorname{Thick}\{\mathrm{IC}_\nu:\nu\le\mu\}$, so by (T1) $\Delta_\mu$ does too.
- The lower generators $\Delta_\nu$, $\nu<\mu$, are in that span by step 4. Together with step 3 this gives $P(\mu)$.
- A remark from the audit: under the perverse t-structure hypothesis PH, finite length gives this inclusion without any induction on $\mu$.
Sources

One orbit alone

The reading "assume $\nu<\mu$, show $\nu=\mu$" asks for $\operatorname{Thick}\{\Delta_\mu\}=\operatorname{Thick}\{\mathrm{IC}_\mu\}$. This fails unless $\mu$ is minimal.

Test with a smaller orbit

If some $\nu<\mu$ exists, then $\mathrm{IC}_\mu\notin\operatorname{Thick}\{\Delta_\mu\}$:
$$a_\nu^*\Delta_\mu=0,\qquad a_\nu^*\mathrm{IC}_\mu\ne0.$$
$a_\nu^*$ is exact, so its kernel is thick. The stalk of $\mathrm{IC}_\mu$ is nonzero on the whole closure, in degree $-\delta(\mu)$, by Deligne's formula.
Proof
- $a_\nu^*\Delta_\mu$ factors through $i^*j_!=0$. This is the recollement axiom H2′, which Kuijper's Def. 2.3.4 contains but H2 does not display.
- $\mathcal H^{-d}\mathrm{IC}(\overline U)=j_*\underline k_U$, whose stalks on $\overline U$ are nonzero because $U$ is dense.
- Example: $\mathrm{PGL}_2\times\mathrm{PGL}_2/\mathrm{PGL}_2$, $\mu=2$. The IC stalk at the vertex of the quadric cone is $k[2]$.
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Compare modulo the boundary

With $\mathcal C_{<\mu}=\operatorname{Thick}\{\Delta_\nu:\nu<\mu\}$, the top pair agrees relative to the lower span:
$$\operatorname{Thick}(\mathcal C_{<\mu}\cup\{\Delta_\mu\})=\operatorname{Thick}(\mathcal C_{<\mu}\cup\{\mathrm{IC}_\mu\}).$$
The triangle's right term lies in $\mathcal C_{<\mu}$. Under PH, the inclusion $\Delta_\mu\notin\operatorname{Thick}\{\mathrm{IC}_\mu\}$ also holds, since ${}^pH^{-1}(\Delta_2[\delta])\simeq\mathrm{IC}_0$ for $\mathrm{PGL}_2$. So neither one-orbit inclusion holds.

Sources

Applied edit to summand.html, step 5: "…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 $\Delta_\mu\in\operatorname{Thick}\{\mathrm{IC}_\nu:\nu\le\mu\}$."

Run: verdict · attempt tree · 2 agents, 16:36–17:00 UTC · proved, in-run audit (all steps hold; A1-7 needs H2′).

Q4 · For $\mathbb A^n$, do standards still generate?

Toan · 20 Oct, 13:43 UTC · filtration.html, main proposition

The standards of all orbits generate $\mathcal C$: $\mathcal C=\operatorname{Thick}_{\mathcal C}\{\Delta_\mu:\mu\in\Lambda_S^+\}$.

Are you saying that in the case of $A^n$ and $GL_n$, where orbit closure is not finite union of other orbits, this is no longer true? i.e. the standard objects no longer generate the category Intuition: I feel that once we have placid presentation, standard objects should always generate the category, i.e. finite orbit union in closure is too strong of a condition to impose

Yes: for $\mathbb A^n$ the orbit standards no longer generate. They miss everything with a nonzero stalk at $0$. The intuition is right once "standards" means the standards of the pieces of the placid presentation, rather than of the orbits.

Proposition. Let $X=\mathbb A^n$, $G=\mathrm{GL}_n$, with shells $X_m=\{\operatorname{ord}v=m\}$ and $\mathcal C$ the objects pulled back from a level $Y_{N,M}=t^{-N}\mathcal O^n/t^M\mathcal O^n$.

The orbit standards span exactly the objects with zero stalk at the origin: $$\operatorname{Thick}_{\mathcal C}\{\Delta_m:m\in\mathbb Z\}=\{F\in\mathcal C:F_0=0\}\ne\mathcal C.$$
One lattice repairs this: $\mathcal C=\operatorname{Thick}(\{\Delta_m\}\cup\{\mathrm{IC}_0\})$, where $\mathrm{IC}_0=\underline k_{\mathcal O^n}$.
In general the standards of the pieces of all levels generate. The orbit standards suffice iff every piece is a finite union of orbits, each of them a piece of some level. This is equivalent to H1(a) + H1(b) + "every orbit is visible".

Proof plan. Steps 1–2 · The obstruction Step 3 · The repair Steps 4–6 · The general form

The obstruction

An exact functor that kills every orbit standard but not $\mathrm{IC}_0$.

Test at the origin

The stalk at $0$ is exact on $\mathcal C$ and separates:
$$i_0^*\Delta_m=0\ \ (m\in\mathbb Z),\qquad i_0^*\underline k_{t^M\mathcal O^n}=k.$$
It is computed at any level as the stalk at the origin of $Y_{N,M}$. Its kernel is thick, and no skyscraper $\delta_0$ is used.
Proof
- $0\notin X_m$, so extension by zero has zero stalk there. Since $0\in t^M\mathcal O^n$, the lattice sheaf has stalk $k$.
- The kernel of an exact functor contains $0$ and is closed under shifts, cofibres and retracts.
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Identify the span

Conversely, every $F$ with $F_0=0$ is built from shell standards:
$$F\in\mathcal C^{(N,M)},\ F_0=0\ \Longrightarrow\ F\in\operatorname{Thick}\{\Delta_m:-N\le m\lt M\}.$$
Dévissage over the pieces of one level. On the tail $T_M=t^M\mathcal O^n$, a single stalk determines the object, so the tail term vanishes.
Proof
- Under constant coefficients on a piece, the stalk cohomology at any point equals the t-cohomology multiplicities. So $F_0=0$ forces $F|_{T_M}=0$ (audit repair A2-3).
- The dévissage gives $F\in\operatorname{Thick}\{a_{P!}a_P^*F\}$. Only the shell terms $a_{m!}a_m^*F\in\operatorname{Thick}\{\Delta_m\}$ remain (A2-5, which replaces A1's step through a non-piece open).
Sources

The repair

One object that is not an orbit standard suffices.

Add one lattice

The shell triangle moves between lattices:
$$\Delta_m\longrightarrow\underline k_{t^m\mathcal O^n}\longrightarrow\underline k_{t^{m+1}\mathcal O^n}\xrightarrow{+1},\qquad\mathcal C=\operatorname{Thick}(\{\Delta_m\}\cup\{\underline k_{t^{M_0}\mathcal O^n}\}).$$
Starting from one lattice, the triangle reaches all the others, and these are the tail standards of every level. The same holds after passing to $\operatorname{Ind}\mathcal C$: there, $\varinjlim_M\underline k_{t^M\mathcal O^n}$ has no maps from any $\Delta_m$.

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The general form

The intuition, stated for a placid presentation with finitely many pieces per level.

Generate by pieces

Theorem G. If each level has finitely many $L^+G$-stable pieces satisfying the level hypotheses (L0)–(L3) of Proposition IV (recollement and constant coefficients), then
$$\mathcal C=\operatorname{Thick}\{\Delta_P:\ P\ \text{a piece of some level}\}.$$
This is the proof of Proposition I run piece by piece: a maximal piece is open, the triangle peels it off, and the induction ends because a level has finitely many pieces. No Satake input is needed.

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Detect visible orbits

An orbit standard lies in $\mathcal C$ only when the orbit is itself a piece of some level. Orbit standards generate iff
$$(\star)\quad\text{every piece is a finite union of orbits, each a piece of some level.}$$
Stalks are constant on pieces. If $(\star)$ fails at an invisible orbit through $x$, the stalk $i_x^*$ separates, as in step 1.
Compare with H1

$(\star)\iff$ H1(a) (finite orbit closures) + H1(b) (finite support) + every orbit visible.

For $\mathbb A^n$, H1(a) fails at every shell, and visibility fails at $\{0\}$. For symmetric spaces, $(\star)$ holds under H1–H3. H1(a) is necessary for orbit standards to generate. That it is not sufficient is shown only for the bare axioms.

Applied edit to filtration.html, after (b): "Without H1(a) this fails: for $\mathbb A^n$ the orbit standards span only the objects with zero stalk at $0$. The standards of the pieces of each level, e.g. $\{\Delta_m\}\cup\{\mathrm{IC}_0\}$, still generate (Proposition IV, section 4; this run)."

Run: verdict · attempt tree · 2 agents, 16:36–17:08 UTC · proved, in-run audit (two steps repaired, hypothesis (L0) added).

Q	question	answer
Q1	Why is \(H^{2e}_c(U)\ne0\)? Where is symmetry used?	\(U\ni t^\lambda\) is nonempty of dimension \(e\), and \(H^{2e}_c\) counts its top components. The deduction uses no symmetry; only the inputs S1, S3, S4 do.
Q2	Does recollement need finite unions of orbits?	No. The open piece must be quasi-compact (cut out at a finite jet level). \(\{0\}\subseteq t^m\mathcal O^n\) is one orbit with no recollement. Finiteness is for the induction.
Q3	Where is the induction used in step 5?	On the whole down-set \(\{\nu\le\mu\}\), once: to turn the boundary term into IC objects. The one-orbit reading is false.
Q4	For \(\mathbb A^n\), do standards still generate?	Yes, they no longer generate: they miss the stalk at \(0\). The standards of the pieces of the placid levels do generate.

S1	support bound \(\nu_b\le\varphi(\lambda)\) (Proposition III, Lemma 3.1, via \(\tau\) and \(\theta\) on roots)
S3	connected stabilizers from \(Z_K(S)\) connected (Lemma 3.2). At a fixed \(\lambda\) only "⇐" holds: \(\mathrm{SL}_2/\mathrm{SO}_2\), \(\lambda=0\).
S4	\(t^\lambda x_0=x_{\varphi(\lambda)}\) (Chen–Yi Prop. 17(i) and injectivity of \(\tau\)), plus the support bound

AI review · 20 October 2026 edits applied

Edits

Q1 · Why is \(H^{2e}_c(U)\) nonzero, and where is symmetry used?

The top class

Locate the base point

Count top components

Kill the connecting map

A weight-free shortcut

Read the lowest degree

Symmetry

Deduce H5 formally

Locate the symmetric inputs

State the general lemma

Q2 · Does recollement need finite unions of orbits?

The criterion

Define cylinders

Glue level-wise recollements

Read necessity off stalks

The counterexample

Remove the origin

What finiteness is for

Terminate the dévissage

Check the symmetric case

Q3 · Where is the induction used?

Setup

Fix the induction

Generate the boundary

Induction step

Read the triangle forward

Use the hypothesis once

Read the triangle backward

One orbit alone

Test with a smaller orbit

Compare modulo the boundary

Q4 · For \(\mathbb A^n\), do standards still generate?

The obstruction

Test at the origin

Identify the span

The repair

Add one lattice

The general form

Generate by pieces

Detect visible orbits

Compare with H1