Proof

Step 1 reduces the claim to one object per orbit: every standard \(\Delta_\mu\) lies in \(\mathcal C^{\mathrm{Sat}}\). Steps 2–6 prove this by induction on \(\mu\), assuming \(\Delta_\nu\in\mathcal C^{\mathrm{Sat}}\) for all \(\nu\lt\mu\); step 7 concludes.

Dots: orbits. Edges: closure relations. Green: standard generated. Orange ring: current orbit.

1. Purpose Reduce to standards

   

   Under H1–H3, the claim is equivalent to placing one object per orbit in the Satake subcategory:

   $$\mathcal C^{\mathrm{Sat}}=\mathcal C\quad\Longleftrightarrow\quad\Delta_\mu\in\mathcal C^{\mathrm{Sat}}\ \text{ for every }\mu\in\Lambda_S^+.$$
   Proof

   • Standard generation. The standards generate every closed finite union of orbits, and all of \(\mathcal C\): $$\text{(a)}\quad\mathcal C_Z=\operatorname{Thick}_{\mathcal C_Z}\{\Delta_{\nu,Z}:\nu\in I(Z)\},$$ $$\text{(b)}\quad\mathcal C=\operatorname{Thick}_{\mathcal C}\{\Delta_\mu:\mu\in\Lambda_S^+\}.$$ On one orbit, truncation in the t-structure of H3 builds every object from the constant sheaf. In \(Z\), an orbit maximal for \(\le\) is open, and the H2(b) triangle leaves an object on one orbit fewer; H1(b) passes to \(\mathcal C\). Full proof →
   • Reduction. \(\mathcal C^{\mathrm{Sat}}\) is thick, so if it contains every standard, (b) gives $$\mathcal C=\operatorname{Thick}_{\mathcal C}\{\Delta_\mu:\mu\in\Lambda_S^+\}\subseteq\mathcal C^{\mathrm{Sat}}\subseteq\mathcal C.$$ Conversely, if \(\mathcal C^{\mathrm{Sat}}=\mathcal C\), it contains every standard.
2. Assumption Pick a Satake object for μ

   

   Fix \(\mu\) and assume \(\Delta_\nu\in\mathcal C^{\mathrm{Sat}}\) for every \(\nu\lt\mu\). The induction step starts from one Satake object that lives on the closure of \(\mathcal O_\mu\) and does not vanish on \(\mathcal O_\mu\):

   $$V_\lambda\ast\mathrm{IC}_0\in\mathcal C^{\mathrm{Sat}},\qquad\operatorname{supp}(V_\lambda\ast\mathrm{IC}_0)\subseteq\overline{\mathcal O_\mu},\qquad a_\mu^*(V_\lambda\ast\mathrm{IC}_0)\ne0.$$

   By H5(a), the object is the middle term of the localization triangle of step 4, whose outer terms are its top part and the boundary part of step 3. By H5(b), a shift of \(\Delta_\mu\) is a direct summand of the top part (steps 5–6).

   Proof

   • H5 gives \(\lambda\in\Lambda_T^+\) whose Satake object lives on \(\overline{\mathcal O_\mu}\) by H5(a). By H5(b) its restriction to \(\mathcal O_\mu\) has \(N_\mu\gt0\) summands, so \(a_\mu^*(V_\lambda\ast\mathrm{IC}_0)\ne0\).
   • H4 puts \(V_\lambda\ast\mathrm{IC}_0\) in \(\mathcal C\); as a Satake object it lies in \(\mathcal C^{\mathrm{Sat}}\).
3. Method Generate the boundary part

   

   The boundary part lies in \(\mathcal C^{\mathrm{Sat}}\), by the induction hypothesis:

   $$i_{\mu*}i_\mu^*(V_\lambda\ast\mathrm{IC}_0)\in\mathcal C^{\mathrm{Sat}}.$$
   Proof

   • By H1, the boundary \(Z=\overline{\mathcal O_\mu}\setminus\mathcal O_\mu\) is a closed finite union of orbits whose orbits are exactly the smaller ones (filtration proof). Part (a) of step 1 gives $$i_\mu^*(V_\lambda\ast\mathrm{IC}_0)\in\operatorname{Thick}_{\mathcal C_Z}\{\Delta_{\nu,Z}:\nu\lt\mu\}.$$
   • The exact functor \(i_{\mu*}\) sends \(\Delta_{\nu,Z}\) to \(\Delta_\nu\) by H2(c). By (T2) and the induction hypothesis: $$i_{\mu*}i_\mu^*(V_\lambda\ast\mathrm{IC}_0)\in\operatorname{Thick}_{\mathcal C}\{\Delta_\nu:\nu\lt\mu\}\subseteq\mathcal C^{\mathrm{Sat}}.$$
4. Method Generate the top part

   

   Removing the boundary part leaves the top part, which therefore lies in \(\mathcal C^{\mathrm{Sat}}\) too:

   $$a_{\mu!}a_\mu^*(V_\lambda\ast\mathrm{IC}_0)\in\mathcal C^{\mathrm{Sat}}.$$
   Proof

   • By H5(a), \(V_\lambda\ast\mathrm{IC}_0\) lives on \(\overline{\mathcal O_\mu}\), in which \(\mathcal O_\mu\) is open. H2(b) there, pushed to \(LX\) by H2(c), gives a triangle: $$a_{\mu!}a_\mu^*(V_\lambda\ast\mathrm{IC}_0)\longrightarrow V_\lambda\ast\mathrm{IC}_0\longrightarrow i_{\mu*}i_\mu^*(V_\lambda\ast\mathrm{IC}_0)\xrightarrow{+1}.$$
   • The middle term lies in \(\mathcal C^{\mathrm{Sat}}\) by step 2 and the right term by step 3. The left term is the fiber of the right-hand map, a cofiber shifted by \([-1]\), so it lies in \(\mathcal C^{\mathrm{Sat}}\) as well.
5. Method Split the top part

   

   The top part is a finite, nonzero sum of shifted copies of \(\Delta_\mu\):

   $$a_{\mu!}a_\mu^*(V_\lambda\ast\mathrm{IC}_0)\simeq\bigoplus_{s=1}^{N_\mu}\Delta_\mu[d_s],\qquad0\lt N_\mu\lt\infty.$$
   Proof

   • H5(b) splits the restriction to the orbit: $$a_\mu^*(V_\lambda\ast\mathrm{IC}_0)\simeq\bigoplus_{s=1}^{N_\mu}\underline k_{\mathcal O_\mu}[d_s]\quad\text{in }\mathcal C_\mu.$$
   • The exact functor \(a_{\mu!}\) preserves finite sums and shifts, and \(a_{\mu!}\underline k_{\mathcal O_\mu}=\Delta_\mu\). How H5 is verified →
6. Result Extract the standard

   

   This completes the induction step: the standard of \(\mathcal O_\mu\) lies in \(\mathcal C^{\mathrm{Sat}}\).

   $$\Delta_\mu\in\mathcal C^{\mathrm{Sat}}.$$
   Proof

   • Since \(N_\mu\gt0\), \(\Delta_\mu[d_1]\) is a direct summand of the top part (step 5), which lies in \(\mathcal C^{\mathrm{Sat}}\) (step 4): $$\Delta_\mu[d_1]\xrightarrow{u}a_{\mu!}a_\mu^*(V_\lambda\ast\mathrm{IC}_0)\xrightarrow{v}\Delta_\mu[d_1],\qquad vu=\mathrm{id}.$$
   • A thick subcategory contains retracts and shifts, so it contains \(\Delta_\mu[d_1]\) and then \(\Delta_\mu\).
7. Conclusion Conclude by induction

   

   The induction reaches every orbit, so step 1 gives the claim and its consequence:

   $$\mathcal C^{\mathrm{Sat}}=\mathcal C,\qquad\Map_{\widehat{\mathcal C}}(V_\lambda\ast\mathrm{IC}_0,F)=0\ \ \forall\lambda\ \Longrightarrow\ F=0.$$
   Proof

   • By H1(a), each \(\mu\) has finitely many \(\nu\le\mu\), so induction on \(h(\mu)=\#\{\nu:\nu\le\mu\}\) reaches every orbit. A minimal orbit has empty boundary, and step 3 is then vacuous.
   • If \(F\in\widehat{\mathcal C}\) is orthogonal to every \(V_\lambda\ast\mathrm{IC}_0\), exactness of \(\Map_{\widehat{\mathcal C}}(-,F)\) extends this to shifts, cofibers and retracts, hence to \(\mathcal C^{\mathrm{Sat}}=\mathcal C\). Objects of \(\mathcal C\) are compact generators of \(\widehat{\mathcal C}\), so \(F=0\).