Affine Reflections, Coroot Translations, and Alcoves — Examples

The set $\mathrm{\Phi }=\{1,-1\}$ is finite, spans $\mathbb{R}$, omits $0$, and meets the line $\mathbb{R}\alpha$ only in its two signs. The reflection in $0$ is $s_{\alpha }(x)=-x$, which exchanges the two roots, and the Cartan integers are $\pm 2$. Thus $\mathrm{\Phi }$ is a reduced crystallographic root system; its Weyl group is $\{1,s_{\alpha }\}$ by [F7].

For $\alpha =1$, $H_{\alpha ,k}=\{k\}$, while $H_{-\alpha ,k}=\{-k\}$; their union is $\mathbb{Z}$. By [F10], alcoves are the connected components of $\mathbb{R}\setminus \mathbb{Z}$, exactly the open intervals $A_n=(n,n+1)$: each such interval is connected and contains no integer, and any connected set meeting two of them would contain an intervening integer.

By [F3] and [F6], ${\alpha }^{\vee }=2$ and the affine reflection is $r_{\alpha ,k}(x)=x-(x-k)2=2k-x$. Composing two such reflections gives $x\mapsto x+2(k-l)$, so all translations by $2m$ occur. Every word is a composition of maps $x\mapsto 2k-x$; pairing consecutive factors shows it is either an even translation or a reflection of the same form. Conversely, $r_{\alpha ,m}=(x\mapsto -x+2m)$ and $r_{\alpha ,m}{r}_{\alpha ,0}=(x\mapsto x+2m)$, so [F9] gives exactly $W_a=\{x\mapsto x+2m,x\mapsto -x+2m:m\in \mathbb{Z}\}$; its translation subgroup is $2\mathbb{Z}=Q^{\vee }$. Also $Q=\mathbb{Z}$ by [F8].

The item Highest-root dominance and the fundamental alcove gives the fundamental factor $A=(0,1)$, with facet walls $\{0\}$ and $\{1\}$. Their reflections are $s_0(x)=-x$ and $s_1(x)=2-x$, and $s_1{s}_0(x)=x+2$.

Every element has one of the two forms $x\mapsto x+2m$ or $x\mapsto -x+2m$. The first sends $A$ to $(2m,2m+1)$, while the second sends it to $(2m-1,2m)$. These intervals are all distinct and exhaust the $A_n$, so each alcove is the image of $A$ under exactly one element; hence the action is simply transitive and ${\mathrm{Stab}}_{W_a}(A)=\{1\}$.

The left and right facets of $A_0$ have types $0$ and $1$, respectively, and $s_0,s_1$ exchange $A_0$ with $A_{-1},A_1$. By step 2.2 every interval is uniquely $g(A_0)$; applying $g$ to these two adjacencies gives every adjacent pair, since the maps $x\mapsto x+2m$ and $x\mapsto -x+2m$ send their shared endpoints through all integers. Thus right multiplication by a facet generator moves the interval index by $+1$ or $-1$, and [F12] guarantees that the shared endpoint has the same type from either interval. By [F5], each move crosses exactly its shared wall.

The relations $s_0^2=s_1^2=1$ reduce every word to an alternating word. Every nonempty alternating word of even length $2q$ is a nonzero translation by $2q$ or $-2q$; every alternating word of odd length is a reflection with slope $-1$. Hence no nonempty reduced alternating word is the identity, the only defining relations are the two involution relations, and $(s_1{s}_0{)}^q=t_{2q}\mathrm{\ne }1$ for $q>0$. Thus $m_{01}=\mathrm{\infty }$ and $W_a\cong ⟨s_0,s_1\mid s_0^2=s_1^2=1⟩$.

If $g(A)=A_n$, the walls separating $A_0$ and $A_n$ are the integer points between their intervals, and there are exactly $\mathrm{\mid }n\mathrm{\mid }$ of them: for $n>0$ they are $1,\dots ,n$, for $n<0$ they are $n+1,\dots ,0$, and for $n=0$ there are none. A word of length $m$ gives a gallery of $m$ adjacent intervals by step 3.1, so it must cross each separating wall and $m\ge \mathrm{\mid }n\mathrm{\mid }$. For $q\ge 0$, the maps $t_{2q}=(s_1{s}_0{)}^q$ and $t_{-2q}=(s_0{s}_1{)}^q$ send $A_0$ to $A_{2q}$ and $A_{-2q}$; the maps $t_{2q}{s}_1$ and $t_{-2q}{s}_0$ send it to $A_{2q+1}$ and $A_{-(2q+1)}$, respectively. These words have lengths equal to the absolute values of their indices, so the lower bound is attained for every $n$ and $\mathrm{\ell }(g)=\mathrm{\mid }\mathrm{Sep}(A,g(A))\mathrm{\mid }=\mathrm{\mid }n\mathrm{\mid }$.

The standard root lattice is $Q=\mathbb{Z}$ and the standard coroot translation lattice is $Q^{\vee }=2\mathbb{Z}$ by step 1.3. Under the dual-normal convention, the walls are $2x=k$; the coroot of ${\alpha }^{\vee }$ is $\alpha$, so reflection in $k\mathrm{/}2$ is $x\mapsto k-x$, and composing the reflections at $0$ and $1\mathrm{/}2$ gives translation by $1$. By [F13], ${\mathrm{\Phi }}^{\vee }$ meets the affine lemma's root-system hypotheses; applying [F11] to it gives translation subgroup $Q^{\vee }({\mathrm{\Phi }}^{\vee })=Q=\mathbb{Z}$ by [F6, F8]. Thus the alternative convention has translations by $Q$, with the two cosets $2\mathbb{Z}$ and $1+2\mathbb{Z}$ showing $[Q:Q^{\vee }]=2$. All computations use finite root lists and explicit integer arithmetic; no axiom of choice is used.

In $A_2$, all six roots have squared norm $1$ and directions with angles that are multiples of $\pi \mathrm{/}3$. The set spans ${\mathbb{R}}^2$; if a root has direction $j\pi \mathrm{/}3$, its reflecting line is perpendicular to it and reflection sends a root direction $k\pi \mathrm{/}3$ to $(2j+3-k)\pi \mathrm{/}3$, so it permutes the six roots. For roots $\alpha ,\beta$, the Cartan number is $2B(\beta ,\alpha )$, an integer because the possible inner products are $1,-1,\frac{1}{2},-\frac{1}{2}$. The only roots on each root line are its two displayed signs, so the system is reduced. The positive roots for the chamber bounded by ${\alpha }_1,{\alpha }_2$ are ${\alpha }_1,{\alpha }_2,\theta$; the first two are simple, $\theta ={\alpha }_1+{\alpha }_2$ is the highest root, and each coroot is $2\alpha$.

In $B_2$, the roots span ${\mathbb{R}}^2$. Reflection with short-root normal $e_i$ changes the sign of coordinate $i$, while reflection with long-root normal $e_1-e_2$ swaps coordinates and reflection with normal $e_1+e_2$ sends $(x_1,x_2)$ to $(-x_2,-x_1)$; these signed coordinate maps preserve the displayed set. If $\alpha$ is short, ${\alpha }^{\vee }=2\alpha$ and $B(\beta ,{\alpha }^{\vee })\in \{0,\pm 2\}$; if $\alpha$ is long, ${\alpha }^{\vee }=\alpha$ and $B(\beta ,{\alpha }^{\vee })\in \{0,\pm 1,\pm 2\}$. Thus all Cartan numbers are integral, and the explicit root lines show reducedness. The positive roots are ${\alpha }_1,{\alpha }_2,{\alpha }_1+{\alpha }_2=e_1,{\alpha }_1+2{\alpha }_2=\theta$, so ${\alpha }_1,{\alpha }_2$ are simple and $\theta$ is highest. The coroot formula gives ${\alpha }_1^{\vee }={\alpha }_1$, ${\alpha }_2^{\vee }=2e_2$, and ${\theta }^{\vee }=\theta$.

For $A_2$, [F4] gives the region $B(x,{\alpha }_1)>0$, $B(x,{\alpha }_2)>0$, $B(x,\theta )<1$. Writing $x=(x_1,x_2)$, the vertices are the pairwise intersections of its three boundary lines: the two simple-root walls meet at $0$; $H_{{\alpha }_2,0}\cap H_{\theta ,1}$ gives $-\frac{1}{2}{x}_1+\frac{\sqrt{3}}{2}{x}_2=0$ and $x_1=1$, hence $v_1=(1,1\mathrm{/}\sqrt{3})$; $H_{{\alpha }_1,0}\cap H_{\theta ,1}$ gives $x_1=0$ and $x_2=2\mathrm{/}\sqrt{3}$, hence $v_2=(0,2\mathrm{/}\sqrt{3})$. By [F5], these three facets have types $1,2,0$. Since $B(0,\theta )=0$ and $B(v_1,\theta )=B(v_2,\theta )=1$, the origin is the unique such corner.

For $B_2$, [F4] gives $x_1-x_2>0$, $x_2>0$, and $x_1+x_2<1$. Intersecting the boundary pairs gives $0$ from $x_1=x_2=0$, $e_1=(1,0)$ from $x_2=0$ and $x_1+x_2=1$, and $(1\mathrm{/}2,1\mathrm{/}2)$ from $x_1=x_2$ and $x_1+x_2=1$. By [F5], the facets on $H_{{\alpha }_1,0},H_{{\alpha }_2,0},H_{\theta ,1}$ have types $1,2,0$.

The three $A_2$ facet-wall pairs meet at $0,v_1,v_2$. Their normals have pairwise inner products of absolute value $1\mathrm{/}2$, so the corresponding lines meet at acute angle $\pi \mathrm{/}3$. Translating the intersection to the origin turns each affine reflection pair into a pair of linear line reflections; their product rotates by $2\pi \mathrm{/}3$ or $-2\pi \mathrm{/}3$ and has order $3$. Since the coroots are twice the roots, the coroot pairing products are $B({\alpha }_1^{\vee },{\alpha }_2)B({\alpha }_2^{\vee },{\alpha }_1)=1$ and $B({\alpha }_i^{\vee },\theta )B({\theta }^{\vee },{\alpha }_i)=1$ for $i=1,2$. Thus [F6] gives $m_{01}=m_{02}=m_{12}=3$.

At $v_1$, the root pairings are $B(v_1,{\alpha }_2)=0$ and $B(v_1,{\alpha }_1)=B(v_1,\theta )=1$. Since the positive roots are exactly ${\alpha }_1,{\alpha }_2,\theta$, these give precisely the three walls $H_{{\alpha }_2,0},H_{{\alpha }_1,1},H_{\theta ,1}$. Their normal lines have the three directions modulo $\pi$ separated by $\pi \mathrm{/}3$, so the local arrangement has six sectors; [F7] identifies them with the six incident alcoves. The fundamental alcove has types $2$ and $0$ at this vertex, and [F7] gives the alternating six-panel cycle. Its boundary word, in a suitable orientation and starting point, is $({\sigma }_2{\sigma }_0{)}^3$; [F6] identifies the Coxeter generators with the facet generators, and $m_{20}=3$, so $(s_2{s}_0{)}^3=1$. The panel of the fundamental alcove along the ray from $v_1$ toward $v_2$ has type $0$; three positions later in the alternating cycle the opposite ray of the same wall $H_{\theta ,1}$ has type $2$. Thus the type belongs to a panel, not to the whole wall.

In $B_2$, $B({\alpha }_1,{\alpha }_2)=-1$, $B(\theta ,{\alpha }_1)=0$, and $B(\theta ,{\alpha }_2)=1$. Together with the coroots from 1.2, the products of Cartan pairings for pairs $(1,2),(0,1),(0,2)$ are respectively $(-1)(-2)=2$, $0$, and $(2)(1)=2$. The corresponding line angles are $\pi \mathrm{/}4,\pi \mathrm{/}2,\pi \mathrm{/}4$, respectively, so products of the intersecting affine reflections are rotations by $\pi \mathrm{/}2$ or $\pi$, with orders $4,2,4$. By [F6], the affine matrix therefore has $m_{12}=4$, $m_{01}=2$, and $m_{02}=4$. Both coordinate models and all their corner and endpoint calculations are finite and explicit, so no axiom of choice is used.

The $A_2$ set is finite, nonzero, and spans the sum-zero plane because $e_1-e_2$ and $e_2-e_3$ are independent. Each root reflection swaps two coordinates, so it preserves the set. Every root has squared length $2$, and the dot product of any two listed roots is an integer; hence $2B(\beta ,\alpha )\mathrm{/}B(\alpha ,\alpha )=B(\beta ,\alpha )\in \mathbb{Z}$. No root line contains any listed multiple other than the two signs. Thus ${\mathrm{\Phi }}_{A_2}$ is a reduced crystallographic root system.

The $B_2$ set is finite, nonzero, and spans ${\mathbb{R}}^2$. Reflection in a short root $\pm e_i$ changes the sign of one coordinate; reflection in a long root $\pm e_1\pm e_2$ swaps or swaps-and-negates the two coordinates. These maps preserve the displayed set. Short roots have squared length $1$ and long roots squared length $2$; all dot products of listed roots are integers, so $2B(\beta ,\alpha )\mathrm{/}B(\alpha ,\alpha )$ is integral for either possible denominator. Each root line contains only the two signs. Thus ${\mathrm{\Phi }}_{B_2}$ is a reduced crystallographic root system.

By [F2], the standard walls give translation lattice $Q^{\vee }$. The dual-normal walls are exactly the standard affine walls of the reduced crystallographic root system ${\mathrm{\Phi }}^{\vee }$ from [F9], so [F2] applied to ${\mathrm{\Phi }}^{\vee }$ gives translation lattice $Q^{\vee }({\mathrm{\Phi }}^{\vee })={\sum }_{\alpha \in \mathrm{\Phi }}\mathbb{Z}({\alpha }^{\vee }{)}^{\vee }=Q$ by [F3, F4].

Every $A_2$ root has squared length $2$, so ${\alpha }^{\vee }=\alpha$ and the root and coroot lattices agree.

In $B_2$, the roots $\pm e_i$ have coroots $\pm 2e_i$, while the roots $\pm e_1\pm e_2$ have the same coroots. Since $e_1+e_2=(e_1-e_2)+2e_2$ and $2e_1=2(e_1-e_2)+2e_2$, all these coroots lie in $\mathbb{Z}(e_1-e_2)+\mathbb{Z}(2e_2)$; conversely both displayed generators are coroots. The roots include $\pm e_1,\pm e_2$, and every other root is their integer combination. Therefore $Q=\mathbb{Z}{e}_1+\mathbb{Z}{e}_2$ and $Q^{\vee }=\mathbb{Z}(e_1-e_2)+\mathbb{Z}(2e_2)\subset Q$.

Relative to the basis $(e_1,e_2)$ of $Q$, the two displayed generators of $Q^{\vee }$ are the columns of $M=\left(\begin{array}{cc}1& 0\\ -1& 2\end{array}\right)$, so $\mathrm{\mid }\det M\mathrm{\mid }=2$. By [F5], $Q\mathrm{/}{Q}^{\vee }$ has order $2$, and by [F6] this says $[Q:Q^{\vee }]=2$.

Relative to $(e_1,e_2)$, the basis matrices of $Q$ and $Q^{\vee }$ are $I$ and $M=\left(\begin{array}{cc}1& 0\\ -1& 2\end{array}\right)$. By [F7], their basis-parallelogram areas are $\mathrm{\mid }\det I\mathrm{\mid }=1$ and $\mathrm{\mid }\det M\mathrm{\mid }=2$. For any $x\in E_{B_2}$, write its unique coordinates in the latter basis as $(u_1,u_2)$ and let $m_i=\lfloor u_i\rfloor$ by [F8], so $u_i-m_i\in [0,1)$. Then $x=(m_1(e_1-e_2)+2m_2{e}_2)+((u_1-m_1)(e_1-e_2)+2(u_2-m_2)e_2)$. The first term lies in $Q^{\vee }$ and the second in the half-open parallelogram $P=\{t_1(e_1-e_2)+2t_2{e}_2:0\le t_1,t_2<1\}$. Uniqueness of the integer parts makes this decomposition unique; hence the translates of $P$ by $Q^{\vee }$ partition the plane. Its covolume is twice that of $Q$.

In $B_2$, take the generating root-reflection pair with normals ${\alpha }_1=e_1-e_2$ and ${\alpha }_2=e_2$; their reflections swap coordinates and change one coordinate sign, so they generate the signed permutation reflection group of $B_2$. Their dual roots are ${\beta }_1={\alpha }_1^{\vee }=e_1-e_2$ and ${\beta }_2={\alpha }_2^{\vee }=2e_2$, so ${\mathrm{\Phi }}_{B_2}^{\vee }=\{\pm 2e_i,\pm e_1\pm e_2\}$ is $C_2$, and [F3, F4] give the stated lattice exchange. The normal pairs satisfy $\frac{B({\alpha }_1,{\alpha }_2)}{\mathrm{\parallel }{\alpha }_1\mathrm{\parallel }\mathrm{\parallel }{\alpha }_2\mathrm{\parallel }}=\frac{B({\beta }_1,{\beta }_2)}{\mathrm{\parallel }{\beta }_1\mathrm{\parallel }\mathrm{\parallel }{\beta }_2\mathrm{\parallel }}=-\frac{1}{\sqrt{2}}$. Thus both pairs of reflecting hyperplanes meet at acute angle $\pi \mathrm{/}4$; the product of the two reflections is a rotation through $\pi \mathrm{/}2$ and has order $4$. This proves the diagram statement. All coordinate lists are finite and explicit, and the only interval representatives use the unique integer part; no axiom of choice is used.

Suppose $\mathrm{\Phi }\mathrm{\ne }\mathrm{\varnothing }$, let $\mathrm{\Delta }=\{{\alpha }_1,\dots ,{\alpha }_r\}$ be a simple-root basis, and let ${\omega }_1^{\vee },\dots ,{\omega }_r^{\vee }$ be the dual basis for $B({\alpha }_i,{\omega }_j^{\vee })={\delta }_{ij}$. By [F10], every root is an integer combination of the simple roots, so the condition defining $P^{\vee }$ is equivalent to $B({\alpha }_i,\lambda )\in \mathbb{Z}$ for every $i$; hence $P^{\vee }={⨁}_i\mathbb{Z}{\omega }_i^{\vee }$. By [F1], the simple coroots form a real basis and generate $Q^{\vee }$. Their coordinate matrix in the ${\omega }_i^{\vee }$ basis has entries $C_{ij}=B({\alpha }_i,{\alpha }_j^{\vee })\in \mathbb{Z}$ and is invertible, since its columns are a real basis. Thus $Q^{\vee }$ corresponds to $C{\mathbb{Z}}^r\subseteq {\mathbb{Z}}^r$, and [F8] shows $P^{\vee }\mathrm{/}{Q}^{\vee }$ is finite of order $\mathrm{\mid }\det C\mathrm{\mid }$. It is abelian because both lattices are additive groups.

For the displayed $A_2$ set, the roots span ${\mathbb{R}}^2$, have squared norm $1$, and occupy the six directions that are multiples of $\pi \mathrm{/}3$. The mirror of a root reflection is perpendicular to its root, so if the root direction is $j\pi \mathrm{/}3$, reflection sends a root direction $k\pi \mathrm{/}3$ to $(2j+3-k)\pi \mathrm{/}3$ and preserves the six roots. The possible Cartan numbers are $2B(\beta ,\alpha )\in \{2,1,-1,-2\}$, and each root line contains only the two displayed signs. Thus [F9] verifies the reduced crystallographic root-system axioms. Choose ${\mathrm{\Phi }}^{+}=\{{\alpha }_1,{\alpha }_2,\theta \}$; these are exactly the roots positive on $\theta$, since $B(\theta ,{\alpha }_1)=B(\theta ,{\alpha }_2)=1\mathrm{/}2$ and $B(\theta ,\theta )=1$. Then ${\alpha }_1,{\alpha }_2$ are simple and $\theta$ is highest. The coroots are $2{\alpha }_1,2{\alpha }_2,2\theta$ by [F3]. The vectors $v_1=(1,1\mathrm{/}\sqrt{3})$ and $v_2=(0,2\mathrm{/}\sqrt{3})$ satisfy $B({\alpha }_i,v_j)={\delta }_{ij}$, so they are the dual coweight basis. In this basis the simple coroots have columns $(2,-1)$ and $(-1,2)$, giving the matrix $\left(\begin{array}{cc}2& -1\\ -1& 2\end{array}\right)$ of determinant $3$. By [F8], $P^{\vee }\mathrm{/}{Q}^{\vee }$ has order $3$. The class of $v_1$ is nonzero because solving this matrix equation for $(1,0)$ gives $(2\mathrm{/}3,1\mathrm{/}3)$, not an integer vector; also $3v_1=2{\alpha }_1^{\vee }+{\alpha }_2^{\vee }$. Hence $[v_1]$ generates the quotient.

The Weyl group preserves $P^{\vee }$, since $B(\alpha ,w\lambda )=B(w^{-1}\alpha ,\lambda )\in \mathbb{Z}$ for every root $\alpha$, and it preserves $Q^{\vee }$ because it permutes coroots. For a root reflection, $s_{\alpha }(\lambda )-\lambda =-B(\lambda ,{\alpha }^{\vee })\alpha =-B(\lambda ,\alpha ){\alpha }^{\vee }$. If $\lambda \in P^{\vee }$, the last coefficient is integral, so every root reflection, and hence all of $W$, acts trivially on $P^{\vee }\mathrm{/}{Q}^{\vee }$. By [F2], $\pi ((\lambda ,w)(\mu ,v))=\lambda +w\mu +Q^{\vee }=\lambda +\mu +Q^{\vee }=\pi (\lambda ,w)+\pi (\mu ,v)$, so $\pi :W_a^{\mathrm{e}\mathrm{x}\mathrm{t}}\to P^{\vee }\mathrm{/}{Q}^{\vee }$, $\pi (\lambda ,w)=\lambda +Q^{\vee }$, is a homomorphism; it is onto and has kernel $Q^{\vee }⋊W=W_a$ by [F4]. Thus $W_a⊴W_a^{\mathrm{e}\mathrm{x}\mathrm{t}}$ with the stated quotient. Translations by $P^{\vee }$ shift each wall level by the integer $B(\lambda ,\alpha )$, while $W$ permutes roots, so $W_a^{\mathrm{e}\mathrm{x}\mathrm{t}}$ preserves the wall arrangement and its alcoves.