The computation
The leading-term map sends a Hecke label to the orbit of the translated base point:
$$\varphi(\lambda)=\lambda-\theta(\lambda),\qquad t^\lambda x_0\in\mathcal O_{\varphi(\lambda)}.$$Two rank-one examples run the well-definedness argument with explicit numbers. In the first \(\varphi\) is onto, as H5 requires of a symmetric space; in the second it doubles, so H5 fails.
| Group case of \(\mathrm{SL}_2\) | \(\mathrm{SL}_2/\mathrm{SO}_2\) | |
|---|---|---|
| Involution | \(\theta(g_1,g_2)=(g_2,g_1)\) on \(\mathrm{SL}_2\times\mathrm{SL}_2\) | \(\theta(g)=(g^t)^{-1}\) on \(\mathrm{SL}_2\) |
| Tori | \(T=T_1\times T_1\), \(S=\{(s,s^{-1})\}\) | \(T=S=T_1\) |
| \(\theta\) on \(\Lambda_T\) | \((a,b)\mapsto(b,a)\) | \(n\mapsto-n\) |
| \(\varphi\) | \((a,b)\mapsto(a-b,\,b-a)\) | \(n\mapsto2n\) |
| \(\Lambda_T^+\) | \(a\ge0\ge b\) | \(n\ge0\) |
| \(\Lambda_S^+\) | \((c,-c)\), \(c\ge0\) | \(m\ge0\) |
| Image of \(\varphi\) | all of \(\Lambda_S^+\) | even \(m\) only |
| Every label lifts | yes | no |
Both examples have \(L=Z_G(S)=T\), so only the second case of the pairing formula occurs. Here \(T_1\) is the diagonal torus of \(\mathrm{SL}_2\), with coroot \(\alpha^\vee(s)=\operatorname{diag}(s,s^{-1})\) and \(\langle\alpha,\alpha^\vee\rangle=2\).
1 · The group case of SL₂
Take \(H=\mathrm{SL}_2\). The swap involution fixes the diagonal copy of \(H\), and \(X\) is \(H\) itself.
$$G=H\times H,\qquad\theta(g_1,g_2)=(g_2,g_1),$$ $$X=G/\operatorname{diag}(H)\xrightarrow{\sim}H,\qquad(g_1,g_2)\longmapsto g_1g_2^{-1}.$$Labels. Write cocharacters of \(T=T_1\times T_1\) as integer pairs. The swap exchanges the entries, so \(\varphi\) lands in the antidiagonal \(\Lambda_S\).
$$(a,b)=(a\alpha^\vee,b\alpha^\vee),\qquad\theta(a,b)=(b,a),\qquad\varphi(a,b)=(a-b,\,b-a)\in\Lambda_S.$$Positive roots. Since \(L=T\), the minimal θ-split parabolic is a Borel with \(B\cap\theta(B)=T\): upper triangular in the first factor, lower in the second. Then \(-\theta\) sends one positive root to the other.
$$B=B_1\times B_1^-,\qquad\Phi^+=\{(\alpha,0),(0,-\alpha)\},\qquad-\theta(\alpha,0)=(0,-\alpha).$$Dominance. A dominant pair has \(a\ge0\ge b\). The pairing of the output splits into the two nonnegative terms of the general formula:
$$\langle(\alpha,0),\varphi(a,b)\rangle=2(a-b)=\underbrace{2a}_{\langle(\alpha,0),\lambda\rangle}+\underbrace{(-2b)}_{\langle(0,-\alpha),\lambda\rangle}\ge0.$$A dominant input. Take \(\lambda=(2,-1)\), so \(\theta\lambda=(-1,2)\) and \(\varphi(\lambda)=(3,-3)\). The Cartan embedding returns exactly \(t^{\varphi(\lambda)}\):
$$t^\lambda=\bigl(\operatorname{diag}(t^2,t^{-2}),\,\operatorname{diag}(t^{-1},t)\bigr),$$ $$\tau(t^\lambda x_0)=\bigl(\operatorname{diag}(t^3,t^{-3}),\,\operatorname{diag}(t^{-3},t^3)\bigr)=t^{\varphi(\lambda)}.$$Under \(X\simeq H\) the translate is \(\operatorname{diag}(t^3,t^{-3})\). It lies in the Cartan double coset of \(3\alpha^\vee\), the orbit labelled \(c=3\).
$$t^\lambda x_0\longmapsto t^{2\alpha^\vee}t^{\alpha^\vee}=t^{3\alpha^\vee}\in L^+H\,t^{3\alpha^\vee}L^+H.$$A non-dominant input. For \(\lambda=(-1,0)\) the output \((-1,1)\) is not dominant, yet the translate \(\operatorname{diag}(t^{-1},t)\) lies in the orbit labelled \(c=1\). Dominance is what makes \(\varphi(\lambda)\) the orbit label.
Every label lifts. Every orbit label \(c\ge0\) has the dominant lift \((c,0)\). The fibre over \(c\) has \(c+1\) points; H5 needs only one of them.
$$\varphi(c,0)=(c,-c),\qquad\varphi^{-1}(c,-c)=\{(c+b,\,b):-c\le b\le0\}.$$Proposition II proves H1–H5 for every group case; its H5 step uses the same lift.
2 · The doubling map of SL₂/SO₂
Take \(G=\mathrm{SL}_2\) with the inverse-transpose involution. Its fixed subgroup \(\mathrm{SO}_2\) is connected, and \(\tau\) sends a coset to a symmetric matrix.
$$\theta(g)=(g^t)^{-1},\qquad G^\theta=\mathrm{SO}_2,\qquad\tau(g\,\mathrm{SO}_2)=gg^t.$$Labels. \(\theta\) inverts the whole diagonal torus, so \(S=T=T_1\) and \(\theta=-1\) on cocharacters. The leading-term map doubles:
$$\Lambda_S=\Lambda_T=\mathbb Z\alpha^\vee,\qquad\theta(n\alpha^\vee)=-n\alpha^\vee,\qquad\varphi(n\alpha^\vee)=2n\alpha^\vee.$$Positive roots. Again \(L=T\). The upper-triangular Borel satisfies \(B\cap\theta(B)=T\), and \(-\theta\alpha=\alpha\), so the pairing formula doubles too:
$$\langle\alpha,\varphi(\lambda)\rangle=\langle\alpha,\lambda\rangle+\langle-\theta\alpha,\lambda\rangle=2\langle\alpha,\lambda\rangle\ge0.$$The translate. The Cartan embedding squares the diagonal entries, so \(t^{n\alpha^\vee}x_0\) lies in the orbit labelled \(2n\):
$$\tau(t^{n\alpha^\vee}x_0)=t^{n\alpha^\vee}\bigl(t^{n\alpha^\vee}\bigr)^t=\operatorname{diag}(t^{2n},t^{-2n}),\qquad t^{n\alpha^\vee}x_0\in\mathcal O_{2n\alpha^\vee}.$$Odd labels have no lift. Odd labels are orbit labels too (Chen–Yi, Proposition 17), but \(\varphi\) misses them. This matrix gives a point of \(\mathcal O_{\alpha^\vee}\):
$$g=\frac12\begin{pmatrix}1+t&i(1-t)\\ i(t-1)/t&(1+t)/t\end{pmatrix}\in\mathrm{SL}_2(F),\qquad gg^t=\operatorname{diag}(t,t^{-1})=t^{\alpha^\vee},$$ $$\Lambda_S^+\setminus\varphi(\Lambda_T^+)=\{\alpha^\vee,3\alpha^\vee,5\alpha^\vee,\dots\}.$$This matches Chen–Yi, Lemma 90: \(1-\theta\) maps \(\Lambda_T\) onto \(\Lambda_S\) when the orbit codimensions are even. Here \(\mathcal O_0\) has codimension \(\langle\rho,\alpha^\vee\rangle=1\) in \(\overline{\mathcal O_{\alpha^\vee}}\).
Every \(\lambda\) satisfying H5 at \(\mu\) lifts \(\mu\), with or without H3 (Proposition III, step 6), so H5 fails at every odd label. H3 fails too, since \(Z_{G^\theta}(S)=\mathrm{SO}_2\cap T_1=\{\pm1\}\) is disconnected (Proposition III, step 4).