Inputs

• H1(a) · Closure order.

  Each orbit closure is the disjoint union of the orbits below it (Chen–Yi Theorem 18): $$\overline{\mathcal O_\mu}=\bigsqcup_{\nu\le\mu}\mathcal O_\nu.$$
• F1 · Support bound.

  The Satake object lives on the closure of its leading orbit (Proposition III, step 2): $$\operatorname{supp}(M_\lambda)\subseteq\overline{\mathcal O_{\varphi(\lambda)}}.$$
• F2 · Leading stalk.

  Its stalk at the translated base point is nonzero (Proposition III, step 6; no use of H3): $$(M_\lambda)_{t^\lambda x_0}\ne0.$$
• F3 · Supports see orbits.

  For a closed finite union of orbits \(Z\), an object supported on \(Z\) restricts to zero on every orbit outside \(Z\): $$\operatorname{supp}(M)\subseteq Z,\quad\mathcal O_\nu\not\subseteq Z\ \Longrightarrow\ a_\nu^*M=0.$$ This is \(j^*i_*=0\) for the closed embedding of \(Z\), since an \(L^+G\)-stable \(Z\) either contains \(\mathcal O_\nu\) or misses it.

Proof

Steps 1–3 collect three facts about orbits and supports; steps 4–5 prove (a) and (b); steps 6–7 conclude.

1. Result Turn orbit containment into order

   By H1(a), inclusions of orbits are inequalities of labels:

   $$\mathcal O_\nu\subseteq\overline{\mathcal O_\mu}\iff\overline{\mathcal O_\nu}\subseteq\overline{\mathcal O_\mu}\iff\nu\le\mu.$$
   Proof

   • If \(\nu\le\mu\), H1(a) lists \(\mathcal O_\nu\) inside \(\overline{\mathcal O_\mu}\), which is closed, so it contains \(\overline{\mathcal O_\nu}\).
   • If \(\mathcal O_\nu\subseteq\bigsqcup_{\nu'\le\mu}\mathcal O_{\nu'}\), the orbits are disjoint and \(\mathcal O_\nu\ne\varnothing\), so \(\nu=\nu'\le\mu\) for one \(\nu'\).
2. Method Detect orbits in supports

   By F3, a nonzero restriction places the orbit inside any support:

   $$\operatorname{supp}(M)\subseteq Z,\quad a_\nu^*M\ne0\ \Longrightarrow\ \mathcal O_\nu\subseteq Z.$$
   Proof

   • This is the contrapositive of F3.
3. Result Find a nonzero leading stalk

   The Satake object is nonzero on its leading orbit:

   $$a_{\varphi(\lambda)}^*M_\lambda\ne0.$$
   Proof

   • The stalk at \(t^\lambda x_0=x_{\varphi(\lambda)}\) factors through restriction to its orbit, \((M_\lambda)_{t^\lambda x_0}=(a_{\varphi(\lambda)}^*M_\lambda)_{t^\lambda x_0}\), and is nonzero by F2.
4. Method Bound φ(λ) above by H5(a)

   Part (a). The support contains the leading orbit, so it lies in \(\overline{\mathcal O_\mu}\) only if \(\varphi(\lambda)\le\mu\):

   $$\operatorname{supp}(M_\lambda)\subseteq\overline{\mathcal O_\mu}\iff\varphi(\lambda)\le\mu.$$
   Proof

   • ⇒. Apply step 2 to \(Z=\overline{\mathcal O_\mu}\) and \(\nu=\varphi(\lambda)\), using step 3: $$\mathcal O_{\varphi(\lambda)}\subseteq\overline{\mathcal O_\mu},\qquad\text{so }\varphi(\lambda)\le\mu\text{ by step 1}.$$
   • ⇐. F1 and step 1 give \(\operatorname{supp}(M_\lambda)\subseteq\overline{\mathcal O_{\varphi(\lambda)}}\subseteq\overline{\mathcal O_\mu}\).
5. Method Bound φ(λ) below by H5(b)

   Part (b). The support lies in the leading closure, so a nonzero restriction to \(\mathcal O_\mu\) needs \(\mu\le\varphi(\lambda)\):

   $$a_\mu^*M_\lambda\ne0\implies\mu\le\varphi(\lambda).$$
   Proof

   • Apply step 2 to \(Z=\overline{\mathcal O_{\varphi(\lambda)}}\), which contains the support by F1: $$\mathcal O_\mu\subseteq\overline{\mathcal O_{\varphi(\lambda)}},\qquad\text{so }\mu\le\varphi(\lambda)\text{ by step 1}.$$
6. Result Pin λ to the fibre

   Part (c). H5(a) and H5(b) give both bounds:

   $$\varphi(\lambda)\le\mu\le\varphi(\lambda),\qquad\text{so}\qquad\varphi(\lambda)=\mu.$$
   Proof

   • H5(a) is the left side of (a). H5(b) has \(N_\mu\gt0\) summands, so \(a_\mu^*M_\lambda\ne0\) and (b) applies.
   • Antisymmetry of \(\le\): \(\varphi(\lambda)-\mu\) and \(\mu-\varphi(\lambda)\) both lie in the pointed positive coroot cone, so both are \(0\).
   • Clause (a) is needed. In the group case with \(H=\mathrm{SL}_2\) (labels), \(\lambda=(1,0)\) has \(\varphi(\lambda)=(1,-1)\) and $$M_{(1,0)}\simeq\underline k_{\overline{\mathcal O_{(1,-1)}}}[2],\qquad a_0^*M_{(1,0)}\simeq\underline k_{\mathcal O_0}[2]\ne0,$$ so H5(b) holds at \(\mu=0\) although \(\varphi(\lambda)\ne0\); H5(a) fails there.
7. Conclusion H5 forces φ onto

   H5 gives a \(\lambda\) at every \(\mu\), and step 6 puts it in \(\varphi^{-1}(\mu)\):

   $$\mathrm{H5}\implies\varphi(\Lambda_T^+)=\Lambda_S^+.$$
   Proof

   • Converse. Under H3, every dominant lift of \(\mu\) satisfies H5 at \(\mu\) (top-restriction lemma); Proposition III combines both directions.
   • A failure. For \(\mathrm{SL}_2/\mathrm{SO}_2\), \(\varphi\) doubles, so no \(\lambda\) lies over an odd label and H5 fails there (examples).