In A3 the moved spaces meet in a line, while the root complexes have no common nonempty face

In $S_4$, $\alpha$ and $\beta$ are products of two disjoint transpositions but are not themselves transpositions, so each has reflection length $2$. The four-cycle $c=(1243)=(13)(14)(12)$ has reflection length $3$: no product of at most two transpositions is a 4-cycle, since a product of two is the identity, a 3-cycle, or two disjoint transpositions. The identities $c=(14)\alpha$ and $c=\beta (14)$ show that ${\alpha }^{-1}c$ and ${\beta }^{-1}c$ are reflections (the first is a conjugate of $(14)$), hence $\alpha ,\beta {\le }_{T}c$.

A reflection in this model is a transposition. For $t\in T$, $t{\le }_T\alpha$ exactly when $t\alpha$ is a reflection, because ${\mathrm{\ell }}_T(\alpha )=2$ and ${\mathrm{\ell }}_T(t)=1$. If $t=(12)$ or $(34)$, $t\alpha$ is the other factor transposition. Any other transposition connects the two pairs $\{1,2\}$ and $\{3,4\}$, and $t\alpha$ is a four-cycle. Thus the only reflections below $\alpha$ are $(12)$ and $(34)$. The same argument with the factor pairs $\{1,3\}$ and $\{2,4\}$ shows that the only reflections below $\beta$ are $(13)$ and $(24)$. By the defining length equality, a proper lower element has strictly smaller reflection length, so the two lower intervals are exactly those listed in (i), and their intersection is $\{1\}$.

The coordinate action gives $M(\alpha )=\mathrm{s}\mathrm{p}\mathrm{a}\mathrm{n}\{E_1-E_2,E_3-E_4\}$ and $M(\beta )=\mathrm{s}\mathrm{p}\mathrm{a}\mathrm{n}\{E_1-E_3,E_2-E_4\}$. By [F4]-[F5] and step 2.1, their positive-root sets are respectively $\{{\rho }_1,{\rho }_2\}$ and $\{{\rho }_4,{\rho }_5\}$. The reverse products for the ordered pairs are $R({\rho }_2)R({\rho }_1)=(34)(12)=\alpha {\le }_{T}c$ and $R({\rho }_5)R({\rho }_4)=(13)(24)=\beta {\le }_{T}c$, so both pairs are edges by [F4]. The vertex sets are disjoint, hence every face of $X(\alpha )$ and every face of $X(\beta )$ have common face $\mathrm{\varnothing }$. By [F6], each corresponding pair of face cones intersects in $c[\mathrm{\varnothing }]=\{0\}$; taking the finite unions of these face cones gives $c[X(\alpha )]\cap c[X(\beta )]=\{0\}$. Intersecting with the unit sphere also gives disjoint spherical realizations.

Write a vector of $M(\alpha )$ as $a(E_1-E_2)+b(E_3-E_4)=(a,-a,b,-b)$ and a vector of $M(\beta )$ as $d(E_1-E_3)+e(E_2-E_4)=(d,e,-d,-e)$. Equating coordinates gives $d=a$, $e=-a$, and $b=-a$, so the intersection is exactly $\mathbb{R}(E_1-E_2-E_3+E_4)$. Every nonzero vector on this line has four nonzero coordinates, whereas every root in [F1] has two; therefore the line contains no root. The standard norm of its displayed generator is $2$ and that of each root is $\sqrt{2}$. By step 2.1, the meet is $1$, whose moved space is $\{0\}$; thus the moved-space intersection is strictly larger and is not the moved space of any common lower bound. Finally $c=(14)\alpha =\beta (14)$ is the asserted common upper bound.