Bipartite Coxeter Elements and Ordered Root Complexes — Examples

Put $\theta =\pi \mathrm{/}5$ and $u=\cos \theta >0$, since $\theta$ is acute. As $3\theta =\pi -2\theta$, [F10] gives $4u^3-3u=-(2u^2-1)$, or $(u+1)(4u^2-2u-1)=0$. Thus $2u=(1+\sqrt{5})\mathrm{/}2=\phi$, verifying the stated exact value. The Gram matrix is $G=\left(\begin{array}{cc}1& -\phi \mathrm{/}2\\ -\phi \mathrm{/}2& 1\end{array}\right)$ with determinant $1-{\phi }^2\mathrm{/}4=(3-\phi )\mathrm{/}4>0$, so it is positive definite. By [F2], $C_V=R_1{R}_2$, which has exact order $5$ by [F1]; [F9] gives $c^5=1$, so $h=\mathrm{ord}(c)=5$. From $c=s_1{s}_2$ and [F9] we have $s_2=s_{1}c$ and $s_{1}c{s}_1=c^{-1}$, so every group word has one of the ten normal forms $c^q$ or $c^q{s}_1$, $0\le q<5$. The five rotations are distinct by the order of $c$, as are the five elements $c^q{s}_1$; their determinants under $\rho$ differ, so the two lists are disjoint. Thus $\mathrm{\mid }W\mathrm{\mid }=10$. In the ordered basis $({\alpha }_1,{\alpha }_2)$, the reflection matrices give $R_1=\left(\begin{array}{cc}-1& \phi \\ 0& 1\end{array}\right)$ and $R_2=\left(\begin{array}{cc}1& 0\\ \phi & -1\end{array}\right)$, hence $C_V=R_1{R}_2=\left(\begin{array}{cc}\phi & -\phi \\ \phi & -1\end{array}\right)$.

Direct application of the reflection formula gives ${\rho }_1=(1,0)$ and ${\rho }_2=(\phi ,1)$ in simple-root coordinates. Multiplication by $C_V$ gives ${\rho }_3=(\phi ,\phi )$, ${\rho }_4=(1,\phi )$, and ${\rho }_5=(0,1)$; a second application gives ${\rho }_6=(-1,0)=-{\rho }_1$ and ${\rho }_7=(-\phi ,-1)=-{\rho }_2$. Since ${\rho }_{i+2}=C_V{\rho }_i$, induction on each parity gives ${\rho }_{i+5}=-{\rho }_i$ for every $i\ge 1$. The first five vectors are distinct, and [F5] identifies ${\mathrm{\Phi }}_{+}=\{{\rho }_1,\dots ,{\rho }_5\}$, so they exhaust the positive roots. Their squared norms are $Q(a,b):=a^2+b^2-\phi ab$: $Q(1,0)=Q(\phi ,1)=Q(\phi ,\phi )=Q(1,\phi )=Q(0,1)=1$ using [F7]. This proves (i).

The inverse Gram matrix is $G^{-1}=\frac{4}{3-\phi }\left(\begin{array}{cc}1& \phi \mathrm{/}2\\ \phi \mathrm{/}2& 1\end{array}\right)$, so its columns give the displayed ${\beta }_1,{\beta }_2$. From ${\mu }_{i+2}=C_V{\mu }_i$ in [F3] and $(C_V-\mathrm{i}\mathrm{d}){\mu }_i=-2{\rho }_i$, which follows from [F5] and the definition of $\mu$ in [F3], we obtain ${\mu }_{i+2}={\mu }_i-2{\rho }_i$; taking $i=1,2,3$ gives the three displayed recursions.

For coefficient vectors $(a,b)$ and $(u,v)$, the pairing is $B((a,b),(u,v))=au+bv-\frac{\phi }{2}(av+bu)$. The columns of the root-coordinate matrix and the $\mu$-coordinate matrix are respectively $R=\left(\begin{array}{ccccc}1& \phi & \phi & 1& 0\\ 0& 1& \phi & \phi & 1\end{array}\right)$ and $M=\left(\begin{array}{ccccc}\frac{4}{d}& \frac{2\phi }{d}& \frac{4}{d}-2& \frac{2\phi }{d}-2\phi & \frac{4}{d}-2-2\phi \\ \frac{2\phi }{d}& \frac{4}{d}& \frac{2\phi }{d}& \frac{4}{d}-2& \frac{2\phi }{d}-2\phi \end{array}\right)$, with $d:=3-\phi$. Therefore the desired dot-product matrix is $M^{\mathsf{T}}GR$, which, using [F7], is the displayed matrix. For example, its $(3,1)$ entry is $({\beta }_1-2{\alpha }_1)\cdot {\alpha }_1=1-2=-1$, and its $(4,3)$ entry is $({\beta }_2-2{\rho }_2)\cdot {\rho }_3=\phi -2(\phi \mathrm{/}2)=0$. The displayed entries give the stated signs and the one-step subdiagonal zeros. Specifically, these signs agree with F8(b),(d), and for $n=2$ the zero immediately below the diagonal is F8(c) with $t=1$. Finally ${\mu }_{i+2}=C_V{\mu }_i$ and ${\rho }_{j+2}=C_V{\rho }_j$, so ${\mu }_{i+2}\cdot {\rho }_{j+2}={\mu }_i\cdot {\rho }_j$ by $B$-invariance of $C_V$. This proves (ii)-(iii).

The odd-$h$ clause of [F5] gives $w_0=c^{(5-1)\mathrm{/}2}a=c^2{s}_1=(s_1{s}_2{)}^2{s}_1$ and asserts the alternating word $(s_1{s}_2{)}^2{s}_1$ is reduced of Coxeter length $5$ with prefix roots ${\rho }_1,\dots ,{\rho }_5$. It is the unique longest element by [F6]. The coordinate matrix $A=2(I-B)$ is $\left(\begin{array}{cc}0& \phi \\ \phi & 0\end{array}\right)$, so $A(1,1)=\phi (1,1)$; by [F7] and the definition of $\phi$, $\lambda =\phi =2\cos (\pi \mathrm{/}5)$. Here $J=\{s_1\}$ and $K=\{s_2\}$, so the theorem's $u=p_1{\alpha }_1$ and $v=p_2{\alpha }_2$ span $V$ since $p_1,p_2>0$; hence its Coxeter plane $P=\mathrm{span}(u,v)$ is $V$. No Choice is used.

The Gram matrix of $({\sigma }_1,{\sigma }_2,{\sigma }_3)$ is $G=\left(\begin{array}{ccc}1& 0& -1\mathrm{/}2\\ 0& 1& -1\mathrm{/}2\\ -1\mathrm{/}2& -1\mathrm{/}2& 1\end{array}\right)$, with determinant $1\mathrm{/}2>0$; $B$ is positive definite as it is half the Euclidean form on $V_0$. Thus the map from the abstract simple-root basis to $({\sigma }_1,{\sigma }_2,{\sigma }_3)$ is an isometry. In this basis $C_V=\left(\begin{array}{ccc}0& 1& -1\\ 1& 0& -1\\ 1& 1& -1\end{array}\right)$. For any $i<j$, the reflection with normal $E_i-E_j$ sends $x$ to $x-(x_i-x_j)(E_i-E_j)$ and swaps coordinates $i,j$. Thus the generators act as the stated transpositions, $c=(1243)$, and its order is $4$.

The prefix convention gives ${\rho }_1={\sigma }_1$, ${\rho }_2={\sigma }_2$, and ${\rho }_3=R({\sigma }_1)R({\sigma }_2){\sigma }_3=E_1-E_4$. Applying $c$ to these roots gives $E_2-E_4$, $E_1-E_3$, $E_2-E_3$, respectively, which are ${\rho }_4,{\rho }_5,{\rho }_6$. The roots of this coordinate realization are exactly $\pm (E_i-E_j)$; the six displayed vectors are exactly the ones with $i<j$, and their listed simple-root coordinates are nonnegative. This verifies the order, positivity, and full positive-root set in (i), while the cyclic recursion gives ${\rho }_{i+3}=C_V{\rho }_i$ for all $i\ge 1$.

Inverting $G$ gives $G^{-1}=\left(\begin{array}{ccc}3\mathrm{/}2& 1\mathrm{/}2& 1\\ 1\mathrm{/}2& 3\mathrm{/}2& 1\\ 1& 1& 2\end{array}\right)$, so ${\beta }_1=(3\mathrm{/}2,1\mathrm{/}2,1)$, ${\beta }_2=(1\mathrm{/}2,3\mathrm{/}2,1)$, and ${\beta }_3=(1,1,2)$ in the simple-root basis. Since $B({\beta }_2,{\sigma }_1)=0$ and $B({\beta }_3,{\sigma }_1)=B({\beta }_3,{\sigma }_2)=0$, the prefix formula gives ${\mu }_1={\beta }_1$, ${\mu }_2={\beta }_2$, and ${\mu }_3={\beta }_3$; the cyclic recursion gives ${\mu }_4=(-1\mathrm{/}2,1\mathrm{/}2,1)$, ${\mu }_5=(1\mathrm{/}2,-1\mathrm{/}2,1)$, and ${\mu }_6=(-1,-1,0)$. Taking $B({\mu }_i,{\rho }_j)={\mu }_i^{\mathsf{T}}G{\rho }_j$ yields exactly the displayed matrix. Reading its entries proves the diagonal, sign, and two-subdiagonal zero assertions.

Since $c=(1243)$, $c^2=(14)(23)$ is the reverse permutation $(4,3,2,1)$ and has six inversions, the maximum possible in $S_4$. Thus $w_0=c^2$ and $\mathrm{\ell }(w_0)=6$. The six-letter word $c^2=(s_1{s}_2{s}_3{)}^2$ is therefore reduced. Applying $C_V^2$ to ${\rho }_1,\dots ,{\rho }_6$ gives $-{\rho }_2,-{\rho }_1,-{\rho }_3,-{\rho }_5,-{\rho }_4,-{\rho }_6$; the period-three prefix recursion then gives the twelve prefix roots of $c^4$, with its second half exactly $-{\mathrm{\Phi }}_{+}$.

The matrix in step 3.1 has $B({\mu }_2,{\rho }_1)=0$. Also, using $R({\rho }_1)=(12)$ and $R({\rho }_2)=(34)$, $R({\rho }_1)R({\rho }_2)c=(12)(34)(12)(34)(23)=(23)$, a nonidentity reflection. Hence its reflection length is $1=n-2$, so both sides of the stated equivalence hold for this pair.

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.