10 · Practice

Three questions

Feedback appears only after you answer. Each question uses only the notation of step (1) of the Proposition as set up on these pages (notation).

1 Compute before checking

In \(\mathrm{SL}_2/\mathrm{SO}_2\), which orbits meet \(D_{\le 1}\cdot x_0=\{g\cdot I\}=\{gg^t:\ g\text{ has poles of order}\le1\}\)? Think in the tree: \(\ell_I\) passes through \(v_0\), and the observer \(g^{-1}v_0\) is at distance 0 or 2.

The answer is \(\overline{\mathcal O}_2\). In the tree, the observer is either on \(\ell\) (index 0), one step off it (index 1), or two steps off it (index 2). In matrices, an explicit point of \(\mathcal O_1\) is

\[ g=\begin{pmatrix}\tfrac{t+1}{2} & \tfrac{i(t-1)}{2}\\[2pt] -\tfrac{i(t-1)}{2t} & \tfrac{t+1}{2t}\end{pmatrix},\qquad \det g=1,\quad gg^t=\operatorname{diag}(t,t^{-1}). \]

Exhibiting \(g\) checks one point. The structural reason: the image is \(K\)-stable and closed (because \(D_{\le1}/K\) is projective), and it contains \(\mathcal O_2\), so it contains \(\overline{\mathcal O}_2\).

2 Explain

Why does the symmetric case give \(a+2m\), while the group case gives \(a+m_1+m_2\)? Answer in one or two sentences, in terms of \(\tau\) or the tree.

In the symmetric case one element \(g\) acts on both sides: \(\tau(gx)=g\,\tau(x)\,\theta(g)^{-1}\). Equivalently, \(g\) moves both ends of the line, and the observer moves by \(2m\). In the group case \(g\) and \(h\) are independent and cost \(m_1\) and \(m_2\) separately.

Check your answer. Did it locate the doubling in the equivariance of \(\tau\), rather than in the orbit poset? The poset is the same in both examples.

3 Transfer

Let \(X=\mathrm{SL}_3/\mathrm{SO}_3\) with \(\theta(g)=(g^t)^{-1}\), so \(S=T\) and \(\theta\lambda=-\lambda\). Can the inductive step of the draft's Sat = Full argument, as written, produce \(\Delta\) on \(\mathcal O_{(1,0,-1)}\)? In other words, is \((1,0,-1)=r(\lambda)\) for some \(\lambda\in\Lambda_T^+\)?

Here \(-\theta\lambda=\lambda\) is dominant, so part (ii) applies with \(r(\lambda)=2\lambda\). The tops are exactly the indices in \(2\Lambda_T^+\), and \((1,0,-1)\) has odd entries. It does lie inside a bound, since \((1,0,-1)\le(2,0,-2)\), which is why C is wrong. This is the parity phenomenon of the running example (Part III, “What to notice” 3) in a new case.

Sources