§5 · Agent 5 · 𝔸ⁿ: lattices · 2026-10-18T18:40:46Z

Computation 5.3 · $\mathrm{IC}_{\tau^k}\ast_I\mathrm{IC}_0=S_k[k]$ and the simple reflections

proved ★ key step

Each simple object of $\mathcal C_I$ is, up to shift, the convolution of $\mathrm{IC}0$ with the skyscraper at a length-zero element $\tau^k$, so it is its own one-term decomposition. The simple affine reflections act on simple objects by $S_k\mapsto S_k[1]\oplus S_k[-1]$ or $S{k-1}\oplus S_{k+1}$, and every splitting is proved by parity, without the decomposition theorem.

Computation 5.3.

Over $k=\mathbb C$ or $\overline{\mathbb F}q$, in the $I$-equivariant categories, with $\Delta_w=j{w!}\underline k$ unshifted and $\mathrm{IC}_w$ perverse:

(a) For every $k\in\mathbb Z$: $\mathrm{IC}{\tau^k}=\Delta{\tau^k}=\delta_{\tau^kI}$ and $\mathrm{IC}{\tau^k}\ast_IA\simeq(\tau^k)!A$ (push-forward along $x\mapsto\tau^kx$) for every $A\in\mathcal C_I$. In particular $$\mathrm{IC}_{\tau^k}\ast_I\mathrm{IC}_0\simeq\underline k_{\tau^k\mathcal O^n}=S_k[k],\qquad (\tau^k)_!S_l\simeq S_{k+l}[k].$$

(b) For $\lambda=\lambda_{m,j}$ and $k=k(m,j)$: $t^\lambda=\tau^ku$ with $u\in W$, $\ell(u)=\ell(t^\lambda)=j(n-j)$, and $\Delta_{t^\lambda}\ast_I\mathrm{IC}0\simeq\underline k{\tau^k\mathcal O^n}[-2j(n-j)]$ (Tate twist $(-j(n-j))$ over $\mathbb F_q$). For $w_{\max}=\tau^kw_0$, of length $N_0=n(n-1)/2$: $\mathrm{IC}{w{\max}}\ast_I\mathrm{IC}0\simeq\bigoplus_iS_k[k+N_0-2i]^{\oplus b{2i}(G/B)}$.

(c) Let $s_1,\dots,s_{n-1}$ be the finite simple reflections and $s_0$ the affine one ($n\ge2$), so $\tau s_i\tau^{-1}=s_{i-1}$ (indices mod $n$). Then $\mathrm{IC}{s_i}=\underline k{P_{s_i}/I}[1]$ and $$\mathrm{IC}_{s_i}\ast_IS_k\simeq\begin{cases}S_{k-1}\oplus S_{k+1}, & i\equiv-k\pmod n,\\ S_k[1]\oplus S_k[-1], & \text{otherwise.}\end{cases}$$ In particular $\mathrm{IC}{s_0}\ast_I\mathrm{IC}_0\simeq S{-1}\oplus S_1$ (perverse, two simple summands, neither equal to $\mathrm{IC}_0$).

Proof. $H$ denotes a finite-level group $H_\alpha$; $H^*_H(\mathrm{pt})=H^*_T(\mathrm{pt})$ is a polynomial ring on degree-2 classes, so $H^{\mathrm{odd}}_H(V)=0$ for a vector space $V$ with linear $H$-action (parity).

  1. (a) $\ell(\tau^k)=0$ (Lemma 5.1(b)), so $I\tau^kI/I$ is the point $\tau^kI$ and $\mathrm{IC}=\Delta=\delta$. The convolution space is $\tau^kI\times^ILX\simeq LX$, $[\tau^k,y]\leftrightarrow y$ (because $[\tau^ki,x]=[\tau^k,ix]$), and $m[\tau^k,y]=\tau^ky$. So $\delta_{\tau^k}\ast A=(\tau^k)!A$; with $A=\underline k{\mathcal O^n}$ this is $\underline k_{\tau^k\mathcal O^n}$, which equals $S_k[k]$ by Lemma 5.2(b).
  1. (b) By Lemma 5.1(b), $u:=\tau^{-k}t^\lambda\in W$; $\ell(t^\lambda)=\sum_{\alpha>0}|\langle\lambda,\alpha\rangle|=\#\{(a,b):a\le j<b\}=j(n-j)$, and $\ell(\tau^ku)=\ell(u)$. Since $\tau^k$ normalizes $I$, $It^\lambda I=\tau^kIuI$ and $\Delta_{t^\lambda}=\delta_{\tau^k}\ast\Delta_u$. As $IuI\subseteq K$, $[g,v]\mapsto(gI,gv)$ identifies $IuI\times^I\mathcal O^n$ with $IuI/I\times\mathcal O^n$ and $m$ with the second projection, so $\Delta_u\ast\mathrm{IC}0=\underline k{\mathcal O^n}\otimes R\Gamma_c(\mathbb A^{\ell(u)})=\underline k_{\mathcal O^n}[-2\ell(u)]$ (one cohomology sheaf, an equivariant rank-one local system on a vector space, hence constant). Apply step 1. For $w_{\max}$: $\overline{Iw_{\max}I}=\tau^kK$, so $\mathrm{IC}{w{\max}}=\delta_{\tau^k}\ast\underline k_{K/I}[N_0]$ and $\underline k_{K/I}[N_0]\ast\mathrm{IC}0=\underline k{\mathcal O^n}\otimes R\Gamma(G/B)[N_0]$. Its cohomology sheaves are constant in even degrees only, and every extension class between two of them lies in some $H^{\mathrm{odd}}_H(V)=0$; so the complex splits.
  1. (c), $i=0$, $k=0$. $s_0=t^{-1}E_{1n}-tE_{n1}+\sum_{1<i<n}E_{ii}$ fixes every $\Lambda_{0,j}$, $1\le j\le n-1$, and moves $\mathcal O^n$; so $P_{s_0}=I\sqcup Is_0I$ fixes $\tau\mathcal O^n=\Lambda_{0,n-1}$ and $\tau^{-1}\mathcal O^n=t^{-1}\Lambda_{0,1}$, and every $L=g\mathcal O^n$, $g\in P_{s_0}$, satisfies $\tau\mathcal O^n\subset L\subset\tau^{-1}\mathcal O^n$ with $\dim L/\tau\mathcal O^n=1$. As $P_{s_0}\cap K=I$, $gI\mapsto g\mathcal O^n$ embeds $P_{s_0}/I\simeq\mathbb P^1$ in the $\mathbb P^1$ of lines of the plane $\tau^{-1}\mathcal O^n/\tau\mathcal O^n$, hence bijectively. So $P_{s_0}\times^I\mathcal O^n\simeq E:=\{(L,v):v\in L\}$ with $m(L,v)=v$. At a level with $V=\tau^{-1}\mathcal O^n/t^M\mathcal O^n\supset Z=\tau\mathcal O^n/t^M\mathcal O^n$ (codimension 2), $E$ is the base change of the blow-up $\beta:\mathrm{Bl}0\mathbb A^2\to\mathbb A^2$ along the smooth map $V\to V/Z$. The adjunction $\underline k\to\beta\underline k$ has cone $\underline k_0[-2]$ (stalk $H^(\mathbb P^1)/H^0$), and the connecting map lies in $\operatorname{Hom}(\underline k_0,\underline k_{\mathbb A^2}[3])=H^{3-4}H(\mathrm{pt})=0$ (purity $i^!\underline k=\underline k_0[-4]$). So $m!\underline k_E=\underline k_{\tau^{-1}\mathcal O^n}\oplus\underline k_{\tau\mathcal O^n}[-2]$, and with $\mathrm{IC}_{s_0}\mathbin{\tilde\boxtimes}\mathrm{IC}_0=\underline k_E[1]$:
$$\mathrm{IC}_{s_0}\ast\mathrm{IC}_0=\underline k_{\tau^{-1}\mathcal O^n}[1]\oplus\underline k_{\tau\mathcal O^n}[-1]=S_{-1}\oplus S_1 .$$
  1. (c), $i\ne0$, $k=0$. $P_{s_i}\subseteq K$, so $P_{s_i}\times^I\mathcal O^n\simeq\mathbb P^1\times\mathcal O^n$ and $\mathrm{IC}{s_i}\ast\mathrm{IC}_0=\underline k{\mathcal O^n}\otimes H^*(\mathbb P^1)[1]$; the possible extension class lies in $H^3_H(V)=0$, so this is $S_0[1]\oplus S_0[-1]$.
  1. (c), general $k$. $P_{s_i}$ moves exactly the vertex class of $\Lambda_{0,i}$ ($\Lambda_{0,0}\sim\mathcal O^n$), and $\tau$ shifts the chain by one step (Lemma 5.1(a)), so $\tau s_i\tau^{-1}=s_{i-1}$ and $\tau^{-k}s_i\tau^k=s_{i+k}$. Conjugation by $\tau^k$ preserves $I$, so $\mathrm{IC}{s_i}\ast\delta{\tau^k}=\delta_{\tau^k}\ast\mathrm{IC}{s{i+k}}$. Hence $\mathrm{IC}{s_i}\ast S_k[k]=(\tau^k)!(\mathrm{IC}{s{i+k}}\ast\mathrm{IC}0)$, which is $(\tau^k)!(S_{-1}\oplus S_1)=S_{k-1}[k]\oplus S_{k+1}[k]$ if $i+k\equiv0$ and $(\tau^k)_!(S_0[1]\oplus S_0[-1])=S_k[k+1]\oplus S_k[k-1]$ otherwise.

What this shows. No step uses the decomposition theorem: (a) is an isomorphism of spaces, (b) and (c) split by parity. Every simple object $S_k$ is a shift of the single generator $\mathrm{IC}_{\tau^k}\ast\mathrm{IC}_0$, which is a one-term direct sum.

Builds on

  • Lemma 5.1 · Orbits of $I$ on $L\mathbb A^n$: closures $\tau^k\mathcal O^n$, $\tau\in\Omega$
  • Lemma 5.2 · Simple objects of $\mathcal C_I$: $S_k=\underline k_{\tau^k\mathcal O^n}[-k]$

Built on by

  • Theorem 5.5 · $\mathbb A^n$: $\mathcal M_I=\{S_k\}$, so $\mathcal C_I^{\mathrm{Iwa}}=\mathcal C_I$
  • Proposition 5.6 · (IG) for $\mathbb A^n$: $T_{\tau^k}\ast\varphi_0=\mathbf 1_{\tau^k\mathcal O^n}$
  • Audit 8.9 · Audit of $\mathcal C_I^{\rm Iwa}=\mathcal C_I$ for $\mathbb A^n$ (A5-3, A5-5, A2-7)
  • Summary 10.2 · Verdict: the criterion and the bridge hold; the decided rows of Table 1 agree; GL₂/T fails at both levels
Original record, as the agent wrote it at 18:40 UTC

$\mathrm{IC}_{\tau^k}\ast_I\mathrm{IC}_0=S_k[k]$ and the simple reflections

Proof. $H$ denotes a finite-level group $H_\alpha$; $H^*_H(\mathrm{pt})=H^*_T(\mathrm{pt})$ is a polynomial ring on degree-2 classes, so $H^{\mathrm{odd}}_H(V)=0$ for a vector space $V$ with linear $H$-action (parity).

  1. (a) $\ell(\tau^k)=0$ (Lemma 5.1(b)), so $I\tau^kI/I$ is the point $\tau^kI$ and $\mathrm{IC}=\Delta=\delta$. The convolution space is $\tau^kI\times^ILX\simeq LX$, $[\tau^k,y]\leftrightarrow y$ (because $[\tau^ki,x]=[\tau^k,ix]$), and $m[\tau^k,y]=\tau^ky$. So $\delta_{\tau^k}\ast A=(\tau^k)!A$; with $A=\underline k{\mathcal O^n}$ this is $\underline k_{\tau^k\mathcal O^n}$, which equals $S_k[k]$ by Lemma 5.2(b).
  1. (b) By Lemma 5.1(b), $u:=\tau^{-k}t^\lambda\in W$; $\ell(t^\lambda)=\sum_{\alpha>0}|\langle\lambda,\alpha\rangle|=\#\{(a,b):a\le j<b\}=j(n-j)$, and $\ell(\tau^ku)=\ell(u)$. Since $\tau^k$ normalizes $I$, $It^\lambda I=\tau^kIuI$ and $\Delta_{t^\lambda}=\delta_{\tau^k}\ast\Delta_u$. As $IuI\subseteq K$, $[g,v]\mapsto(gI,gv)$ identifies $IuI\times^I\mathcal O^n$ with $IuI/I\times\mathcal O^n$ and $m$ with the second projection, so $\Delta_u\ast\mathrm{IC}0=\underline k{\mathcal O^n}\otimes R\Gamma_c(\mathbb A^{\ell(u)})=\underline k_{\mathcal O^n}[-2\ell(u)]$ (one cohomology sheaf, an equivariant rank-one local system on a vector space, hence constant). Apply step 1. For $w_{\max}$: $\overline{Iw_{\max}I}=\tau^kK$, so $\mathrm{IC}{w{\max}}=\delta_{\tau^k}\ast\underline k_{K/I}[N_0]$ and $\underline k_{K/I}[N_0]\ast\mathrm{IC}0=\underline k{\mathcal O^n}\otimes R\Gamma(G/B)[N_0]$. Its cohomology sheaves are constant in even degrees only, and every extension class between two of them lies in some $H^{\mathrm{odd}}_H(V)=0$; so the complex splits.
  1. (c), $i=0$, $k=0$. $s_0=t^{-1}E_{1n}-tE_{n1}+\sum_{1<i<n}E_{ii}$ fixes every $\Lambda_{0,j}$, $1\le j\le n-1$, and moves $\mathcal O^n$; so $P_{s_0}=I\sqcup Is_0I$ fixes $\tau\mathcal O^n=\Lambda_{0,n-1}$ and $\tau^{-1}\mathcal O^n=t^{-1}\Lambda_{0,1}$, and every $L=g\mathcal O^n$, $g\in P_{s_0}$, satisfies $\tau\mathcal O^n\subset L\subset\tau^{-1}\mathcal O^n$ with $\dim L/\tau\mathcal O^n=1$. As $P_{s_0}\cap K=I$, $gI\mapsto g\mathcal O^n$ embeds $P_{s_0}/I\simeq\mathbb P^1$ in the $\mathbb P^1$ of lines of the plane $\tau^{-1}\mathcal O^n/\tau\mathcal O^n$, hence bijectively. So $P_{s_0}\times^I\mathcal O^n\simeq E:=\{(L,v):v\in L\}$ with $m(L,v)=v$. At a level with $V=\tau^{-1}\mathcal O^n/t^M\mathcal O^n\supset Z=\tau\mathcal O^n/t^M\mathcal O^n$ (codimension 2), $E$ is the base change of the blow-up $\beta:\mathrm{Bl}0\mathbb A^2\to\mathbb A^2$ along the smooth map $V\to V/Z$. The adjunction $\underline k\to\beta\underline k$ has cone $\underline k_0[-2]$ (stalk $H^(\mathbb P^1)/H^0$), and the connecting map lies in $\operatorname{Hom}(\underline k_0,\underline k_{\mathbb A^2}[3])=H^{3-4}H(\mathrm{pt})=0$ (purity $i^!\underline k=\underline k_0[-4]$). So $m!\underline k_E=\underline k_{\tau^{-1}\mathcal O^n}\oplus\underline k_{\tau\mathcal O^n}[-2]$, and with $\mathrm{IC}_{s_0}\mathbin{\tilde\boxtimes}\mathrm{IC}_0=\underline k_E[1]$:
$$\mathrm{IC}_{s_0}\ast\mathrm{IC}_0=\underline k_{\tau^{-1}\mathcal O^n}[1]\oplus\underline k_{\tau\mathcal O^n}[-1]=S_{-1}\oplus S_1 .$$
  1. (c), $i\ne0$, $k=0$. $P_{s_i}\subseteq K$, so $P_{s_i}\times^I\mathcal O^n\simeq\mathbb P^1\times\mathcal O^n$ and $\mathrm{IC}{s_i}\ast\mathrm{IC}_0=\underline k{\mathcal O^n}\otimes H^*(\mathbb P^1)[1]$; the possible extension class lies in $H^3_H(V)=0$, so this is $S_0[1]\oplus S_0[-1]$.
  1. (c), general $k$. $P_{s_i}$ moves exactly the vertex class of $\Lambda_{0,i}$ ($\Lambda_{0,0}\sim\mathcal O^n$), and $\tau$ shifts the chain by one step (Lemma 5.1(a)), so $\tau s_i\tau^{-1}=s_{i-1}$ and $\tau^{-k}s_i\tau^k=s_{i+k}$. Conjugation by $\tau^k$ preserves $I$, so $\mathrm{IC}{s_i}\ast\delta{\tau^k}=\delta_{\tau^k}\ast\mathrm{IC}{s{i+k}}$. Hence $\mathrm{IC}{s_i}\ast S_k[k]=(\tau^k)!(\mathrm{IC}{s{i+k}}\ast\mathrm{IC}0)$, which is $(\tau^k)!(S_{-1}\oplus S_1)=S_{k-1}[k]\oplus S_{k+1}[k]$ if $i+k\equiv0$ and $(\tau^k)_!(S_0[1]\oplus S_0[-1])=S_k[k+1]\oplus S_k[k-1]$ otherwise.

What this shows. No step uses the decomposition theorem: (a) is an isomorphism of spaces, (b) and (c) split by parity. Every simple object $S_k$ is a shift of the single generator $\mathrm{IC}_{\tau^k}\ast\mathrm{IC}_0$, which is a one-term direct sum.

Sources