All skips and the cone walls of the sortable element s1s2 in A3

The element $v=s_1{s}_2$ has $\mathrm{\ell }(v)=2$ and $S(v)=\{s_1,s_2\}$ [F5]. The greedy scan of $c^{\mathrm{\infty }}=s_1{s}_2{s}_3{s}_1{s}_2{s}_3\cdots$ reads the remainders $v\to s_2\to 1$: position $1$ has letter $s_1\in D_L(v)$ and is selected, position $2$ has letter $s_2\in D_L(s_2)$ and is selected, and position $3$ has letter $s_3\mathrm{\notin }{D}_L(1)$ — the scan stops because the remainder is already $1$ after two selections [F4]. Hence the sorting word is $s_1{s}_2$ at positions $1,2$ and the block sequence is the single subset $\{s_1,s_2\}$, which is weakly decreasing; so $v$ is $c$-sortable [F3].

The selected positions are $1,2$, so the leftmost unselected occurrences are $s_3$ at position $3$ and $s_1,s_2$ at positions $4,5$; each follows exactly $i=2$ selected letters, so all three skips occur in the $3$rd position of the sorting word [F3].

The associated reflections are $t_{s_1}=a_1{a}_2{s}_1{a}_2{a}_1=s_1{s}_2{s}_1{s}_2{s}_1=s_2$, $t_{s_2}=s_1{s}_2{s}_2{s}_2{s}_1=s_1{s}_2{s}_1$ and $t_{s_3}=s_1{s}_2{s}_3{s}_2{s}_1$ [F3]; the reductions use $(s_1{s}_2{)}^3=1$ for the first and $s_2^2=1$ for the second [F10]. The words $a_1{a}_2{s}_1=s_1{s}_2{s}_1$ and $a_1{a}_2{s}_3=s_1{s}_2{s}_3$ are reduced while $a_1{a}_2{s}_2=s_1{s}_2{s}_2=s_1$ is not [F5]; hence the skips of $s_1$ and $s_3$ are unforced and the skip of $s_2$ is forced [F3].

The skip roots are $C_c^{s_1}(v)=\rho (s_1{s}_2)e_{s_1}=e_{s_2}$, $C_c^{s_2}(v)=\rho (s_1{s}_2)e_{s_2}=-(e_{s_1}+e_{s_2})$ and $C_c^{s_3}(v)=\rho (s_1{s}_2)e_{s_3}=e_{s_1}+e_{s_2}+e_{s_3}$: the images are computed from [F1] and [F11] as $\rho (s_2)e_{s_1}=e_{s_1}+e_{s_2}$, $\rho (s_2)e_{s_2}=-e_{s_2}$, $\rho (s_2)e_{s_3}=e_{s_2}+e_{s_3}$, $\rho (s_1)e_{s_1}=-e_{s_1}$, $\rho (s_1)e_{s_2}=e_{s_1}+e_{s_2}$ and $\rho (s_1)e_{s_3}=e_{s_3}$, with the sign of the $s_2$-skip negative because that skip is forced [F3].

The three vectors $e_{s_2}$, $-(e_{s_1}+e_{s_2})$ and $e_{s_1}+e_{s_2}+e_{s_3}$ form a basis of $V$: in the basis $(e_{s_1},e_{s_2},e_{s_3})$ they are $(0,1,0)$, $(-1,-1,0)$ and $(1,1,1)$, and the last has a nonzero third coordinate while the first two are independent. By [F3] and step 4.1, ${\mathcal{A}}_c(v)=\{-(e_{s_1}+e_{s_2})\}$ and ${\mathcal{B}}_c(v)=\{e_{s_2},e_{s_1}+e_{s_2}+e_{s_3}\}$.

Cover reflections: the right descents of $v=s_1{s}_2$ are read off the products $v{s}_1=s_1{s}_2{s}_1$, $v{s}_2=s_1$, $v{s}_3=s_1{s}_2{s}_3$; their lengths are $3$, $1$ and $3$ [F5], so the only cover relation $v⋗vs$ has $s=s_2$ and cover reflection $t=v{s}_2{v}^{-1}=s_1{s}_2{s}_1$, with positive root $-\rho (s_1{s}_2)e_{s_2}=-C_c^{s_2}(v)=e_{s_1}+e_{s_2}$ [F6, F11, step 4.1]. This matches [F7]: the unique negative skip root of $v$ is $-(e_{s_1}+e_{s_2})$ and $\mathrm{C}\mathrm{o}\mathrm{v}(v)=\{s_1{s}_2{s}_1\}$.

Cone and a chamber check: by [F3] the cone is ${\mathrm{C}\mathrm{o}\mathrm{n}\mathrm{e}}_c(v)=\{x\in V:B(x,e_{s_2})\ge 0,B(x,e_{s_1}+e_{s_2})\le 0,B(x,e_{s_1}+e_{s_2}+e_{s_3})\ge 0\}$. For $x\in C$, i.e. $B(x,e_{s_i})\ge 0$ for $i=1,2,3$, the adjoint identity $B(\rho (v)x,\beta )=B(x,\rho (v{)}^{-1}\beta )$ [F9] and the inverse images $\rho (v{)}^{-1}{e}_{s_2}=e_{s_1}$, $\rho (v{)}^{-1}(e_{s_1}+e_{s_2})=-e_{s_2}$, $\rho (v{)}^{-1}(e_{s_1}+e_{s_2}+e_{s_3})=e_{s_3}$ [F11, step 4.1] give $B(\rho (v)x,e_{s_2})=B(x,e_{s_1})\ge 0$, $B(\rho (v)x,e_{s_1}+e_{s_2})=-B(x,e_{s_2})\le 0$ and $B(\rho (v)x,e_{s_1}+e_{s_2}+e_{s_3})=B(x,e_{s_3})\ge 0$; hence $\rho (v)C\subseteq {\mathrm{C}\mathrm{o}\mathrm{n}\mathrm{e}}_c(v)$, that is $vC\subseteq {\mathrm{C}\mathrm{o}\mathrm{n}\mathrm{e}}_c(v)$. This is the instance ${\pi }_c(v)=v$ of the cone criterion at $w=v$ [F8], consistent with $v$ being $c$-sortable [step 1.1].