Crystallographic Root Lattices and Weyl Group Interfaces — Examples

Put $z=\cos (\pi \mathrm{/}3)>0$ by [F15]. The double-angle and supplementary identities give $2z^2-1=-z$, hence $(2z-1)(z+1)=0$ and $z=1\mathrm{/}2$. Since $m(s,t)=3$, [F4] gives the Gram matrix $\left(\begin{array}{cc}1& -1\mathrm{/}2\\ -1\mathrm{/}2& 1\end{array}\right)$ in $(e_s,e_t)$. With $a_s=e_s$ and $a_t=e_t$, one has $a_s^{\vee }=2a_s$, $a_t^{\vee }=2a_t$, $a_{ss}=a_{tt}=2$, $a_{st}=a_{ts}=-1$, and the displayed Cartan matrix. The form is positive definite because its quadratic form is $(x-y\mathrm{/}2{)}^2+3y^2\mathrm{/}4$. Thus the chosen scaling is crystallographic, and [F3] shows every crystallographic scaling has $c_s=c_t$, so this is unique up to a common positive multiple.

By bilinearity, the reflection formula [F5] defines linear maps and gives $r_{a_s}(a_s)=-a_s$, $r_{a_s}(a_t)=a_s+a_t$, $r_{a_t}(a_t)=-a_t$, $r_{a_t}(a_s)=a_s+a_t$, $r_{a_s}(a_s+a_t)=a_t$, and $r_{a_t}(a_s+a_t)=a_s$; by linearity the set $R=\{\pm a_s,\pm a_t,\pm (a_s+a_t)\}$ is stable under both reflections. By [F6], $\rho (s)=r_{a_s}$ and $\rho (t)=r_{a_t}$, since the reflection formula is unchanged by scaling its normal; by [F7], every $w\in W$ is a word in $s,t$, so ${\mathrm{\Phi }}_c\subseteq R$. Conversely, $a_s,a_t$ belong to ${\mathrm{\Phi }}_c$, while $-a_s=\rho (s)a_s$, $-a_t=\rho (t)a_t$, $a_s+a_t=\rho (s)a_t$, and $-(a_s+a_t)=\rho (t)\rho (s)a_s$. Thus ${\mathrm{\Phi }}_c=R$.

For $B_2$, set the roots ${\beta }_1={\epsilon }_1-{\epsilon }_2$ and ${\beta }_2={\epsilon }_2$ from the coordinate model [F9]. Their coroots are ${\beta }_1^{\vee }={\beta }_1$, ${\beta }_2^{\vee }=2{\epsilon }_2$. The full coordinate root set in [F9] contains ${\epsilon }_1,{\epsilon }_2$, so $Q(B_2)=\mathbb{Z}{\epsilon }_1+\mathbb{Z}{\epsilon }_2$. Its coroot set consists of $\pm ({\epsilon }_1\pm {\epsilon }_2)$ and $\pm 2{\epsilon }_i$; these span exactly $Q^{\vee }(B_2)=\mathbb{Z}{\beta }_1+\mathbb{Z}(2{\epsilon }_2)$, since both displayed generators are coroots and every listed coroot lies in their span. Thus [F12] gives $P(B_2)=\{(x,y):x-y\in \mathbb{Z},2y\in \mathbb{Z}\}=\mathbb{Z}{\epsilon }_1+\mathbb{Z}\frac{{\epsilon }_1+{\epsilon }_2}{2}$. The nontrivial coset is generated by $({\epsilon }_1+{\epsilon }_2)\mathrm{/}2$, of order two, so $[P(B_2):Q(B_2)]=2$.

For $C_2$, set the roots ${\gamma }_1={\epsilon }_1-{\epsilon }_2$ and ${\gamma }_2=2{\epsilon }_2$ from the coordinate model [F10]. Their coroots are ${\gamma }_1^{\vee }={\gamma }_1$ and ${\gamma }_2^{\vee }={\epsilon }_2$. The full root set in [F10] yields $Q(C_2)=\mathbb{Z}({\epsilon }_1-{\epsilon }_2)+\mathbb{Z}(2{\epsilon }_2)$: the two generators are roots and each other root is an integer combination of them. Its coroot set contains ${\epsilon }_1-{\epsilon }_2$ and ${\epsilon }_2$ and is contained in ${\mathbb{Z}}^2$, so $Q^{\vee }(C_2)=\mathbb{Z}({\epsilon }_1-{\epsilon }_2)+\mathbb{Z}{\epsilon }_2={\mathbb{Z}}^2$. Thus [F12] gives $P(C_2)=\{(x,y):x-y\in \mathbb{Z},y\in \mathbb{Z}\}={\mathbb{Z}}^2$. The quotient is generated by $[{\epsilon }_2]$, which has order two because $2{\epsilon }_2\in Q(C_2)$ and ${\epsilon }_2\mathrm{\notin }Q(C_2)$; hence $[P(C_2):Q(C_2)]=2$.

In each coordinate model, the reflection formula [F5] gives the maps $s_1(x,y)=(y,x)$ and $s_2(x,y)=(x,-y)$ for the displayed $B_2$ and $C_2$ root pairs; the second map is unchanged when its normal is $2{\epsilon }_2$ instead of ${\epsilon }_2$. Both maps are involutions. Their product sends $(x,y)\mapsto (-y,x)$; its square is $-I$ and its fourth power is $I$, so its order is four. Thus both standard systems realize the Coxeter diagram $I_2(4)$, giving the stated $B_2=C_2$ diagram coincidence.

Put ${\alpha }_1={\epsilon }_1-{\epsilon }_2$ and ${\alpha }_2={\epsilon }_2-{\epsilon }_3$. The linear map sending $a_s\mapsto {\alpha }_1\mathrm{/}\sqrt{2}$ and $a_t\mapsto {\alpha }_2\mathrm{/}\sqrt{2}$ is an isometry: the images have squared lengths $1,1$ and inner product $-1\mathrm{/}2$, matching their Gram matrix from [F4]. By step 1.2 it sends ${\mathrm{\Phi }}_c$ to $\frac{1}{\sqrt{2}}\{\pm {\alpha }_1,\pm {\alpha }_2,\pm ({\alpha }_1+{\alpha }_2)\}$, the standard $A_2$ coordinate root system of [F8]. For $v=(1,0,-1)$, its positive roots are ${\alpha }_1,{\alpha }_2,{\alpha }_1+{\alpha }_2$; thus [F11] makes ${\alpha }_1,{\alpha }_2$ its simple roots.

The two simple roots have squared length $1$, so [F13] gives their simple coroots $2a_s,2a_t$; hence $Q=\mathbb{Z}{a}_s+\mathbb{Z}{a}_t=\mathbb{Z}{\mathrm{\Phi }}_c$ and $Q^{\vee }=\mathbb{Z}(2a_s)+\mathbb{Z}(2a_t)=2Q$. Since every root in the standard $A_2$ coordinate set has squared length $2$, its coroot lattice is $Q_0$ by [F8,F11,F12,F13]. The isometry of step 2.1 sends $Q$ to $Q_0\mathrm{/}\sqrt{2}$ and $Q^{\vee }$ to $\sqrt{2}{Q}_0$, where $Q_0=\mathbb{Z}{\alpha }_1+\mathbb{Z}{\alpha }_2=\{x\in {\mathbb{Z}}^3:x_1+x_2+x_3=0\}$. By [F12], the image of $P$ is the lattice dual to $\sqrt{2}{Q}_0$, namely $P_0\mathrm{/}\sqrt{2}$, where $P_0=\{x\in E:(x,{\alpha }_1),(x,{\alpha }_2)\in \mathbb{Z}\}$.

For $x=(x_1,x_2,x_3)\in E$, the pairings defining $P_0$ are $x_1-x_2$ and $x_2-x_3$. Thus membership is equivalent to having integral coordinate differences. If those differences are integers, write $x_1=x_3+m$ and $x_2=x_3+n$ with $m,n\in \mathbb{Z}$; the sum-zero condition gives $3x_3=-m-n$, so all coordinates lie in $\frac{1}{3}\mathbb{Z}$. Conversely the displayed conditions make both pairings integral. Hence $P_0=\{x\in (\frac{1}{3}\mathbb{Z}{)}^3:{\sum }_i{x}_i=0,x_i-x_j\in \mathbb{Z}\text{ for all }i,j\}$.

The vectors ${\omega }_1=(2{\epsilon }_1-{\epsilon }_2-{\epsilon }_3)\mathrm{/}3$ and ${\omega }_2=({\epsilon }_1+{\epsilon }_2-2{\epsilon }_3)\mathrm{/}3$ satisfy $({\omega }_i,{\alpha }_j^{\vee })={\delta }_{ij}$, since ${\alpha }_j^{\vee }={\alpha }_j$ by [F13]. They are the fundamental weights by [F14] and form a basis of $P_0$: any $x\in P_0$ has integer pairings with the basis ${\alpha }_1^{\vee },{\alpha }_2^{\vee }$ and therefore is the corresponding integer linear combination of ${\omega }_1,{\omega }_2$. In this basis ${\alpha }_1=2{\omega }_1-{\omega }_2$ and ${\alpha }_2=-{\omega }_1+2{\omega }_2$; hence $[{\omega }_2]=2[{\omega }_1]$ and $3[{\omega }_1]=0$ in $P_0\mathrm{/}{Q}_0$. The quotient is nontrivial because ${\omega }_1\mathrm{\notin }{Q}_0$, so it is cyclic of order three. Therefore $P_0\mathrm{/}{Q}_0\cong \mathbb{Z}\mathrm{/}3$, $[P:Q]=3$, and $\det A=2\cdot 2-(-1)(-1)=3$.

The $A_2$ lattices satisfy $Q⊊P$ and $[P:Q]=3=\det A$, whereas both standard $B_2$ and $C_2$ coordinate systems have weight/root index two. No Choice is used: every step is a finite coordinate calculation on the displayed bases and finite root sets.

Put $u=st$. The relators in [F8] give $s^2=t^2=1$ and $u^4=1$; also $sus=u^{-1}$. Replacing $t$ by $su$ and moving each $s$ to the right with $su=u^{-1}s$, every word reduces to $u^k$ or $u^{k}s$, with $0\le k<4$. Every $W$-element is such a word by [F9], so $W$ has at most eight elements and is finite. In coordinates $x{e}_s+y{e}_t$, the form is $x^2-\sqrt{2}xy+y^2=(x-\frac{\sqrt{2}}{2}y{)}^2+\frac{1}{2}{y}^2$, so it is positive definite.

The single edge is a tree with label $4$. Rooting first at $s$ and then at $t$, [F3] gives $c=(1,\sqrt{2})$ and $c^{\mathrm{\prime }}=(\sqrt{2},1)$; both are crystallographic. Their scaled simple roots and coroots are $a_s=e_s$, $a_t=\sqrt{2}{e}_t$, $a_s^{\vee }=2e_s$, $a_t^{\vee }=a_t$ and $a_s^{\mathrm{\prime }}=\sqrt{2}{e}_s$, $a_t^{\mathrm{\prime }}=e_t$, $a_s^{\mathrm{\prime }\vee }=a_s^{\mathrm{\prime }}$, $a_t^{\mathrm{\prime }\vee }=2e_t$. Using [F1] and [F5] gives $A=\left(\begin{array}{cc}2& -1\\ -2& 2\end{array}\right)$ and $A^{\mathrm{\prime }}=\left(\begin{array}{cc}2& -2\\ -1& 2\end{array}\right)=A^T$.

For $c$, [F4] gives $r_s(a_s)=-a_s$, $r_s(a_t)=d:=a_t+2a_s$, $r_t(a_t)=-a_t$, and $r_t(a_s)=p:=a_s+a_t$. By linearity of the reflections, $r_s(p)=p$, $r_t(p)=a_s$, $r_s(d)=a_t$, and $r_t(d)=d$, so $R=\{\pm a_s,\pm a_t,\pm p,\pm d\}$ is stable under $r_s,r_t$. By [F6], these maps are linear and invariant under nonzero rescaling of their normals; since $a_s=c_s{e}_s$ and $a_t=c_t{e}_t$, $r_s=r_{a_s}$ and $r_t=r_{a_t}$. Then [F7] identifies them with $\rho (s),\rho (t)$, and [F9] gives ${\mathrm{\Phi }}_c\subseteq R$. For $w=r_s{r}_t$ one has $w(a_s)=p$, $w(a_t)=-d$, $w(p)=-a_s$, and $w(d)=a_t$, hence $w^2=-I$ on the basis $(a_s,a_t)$. The four positive listed vectors are $a_s,a_t$, $p=\rho (t)a_s$, and $d=\rho (s)a_t$; applying $w^2$ supplies their negatives. Therefore ${\mathrm{\Phi }}_c=R$.

Put ${\beta }_1={\epsilon }_1-{\epsilon }_2$, ${\beta }_2={\epsilon }_2$, ${\gamma }_1={\epsilon }_1-{\epsilon }_2$ and ${\gamma }_2=2{\epsilon }_2$. In $B_2$, choose $v=(2,1)$, which pairs nontrivially with every root in [F11]. Its positive roots are ${\epsilon }_2$, ${\epsilon }_1-{\epsilon }_2$, ${\epsilon }_1$ and ${\epsilon }_1+{\epsilon }_2$; the latter two are ${\beta }_1+{\beta }_2$ and ${\beta }_1+2{\beta }_2$, while ${\beta }_1,{\beta }_2$ are not sums of two listed positive roots. Thus they are simple by [F16]. In $C_2$, the same $v$ pairs nontrivially with every root in [F12] and gives positive roots ${\epsilon }_1-{\epsilon }_2$, $2{\epsilon }_2$, ${\epsilon }_1+{\epsilon }_2$ and $2{\epsilon }_1$; the latter two are ${\gamma }_1+{\gamma }_2$ and $2{\gamma }_1+{\gamma }_2$, while ${\gamma }_1,{\gamma }_2$ are not sums of two listed positive roots. Thus ${\gamma }_1,{\gamma }_2$ are simple by [F16]. The reflections in the first roots swap the coordinates and those in the second roots negate the second coordinate by [F6]; the second reflection is the same for normals ${\epsilon }_2$ and $2{\epsilon }_2$. Their product is a quarter-turn of order four. Therefore both coordinate root systems have Coxeter diagram $I_2(4)$.

In $A_2$, put ${\alpha }_1={\epsilon }_1-{\epsilon }_2$ and ${\alpha }_2={\epsilon }_2-{\epsilon }_3$. The vector $v=(1,0,-1)$ pairs nontrivially with every root in [F10], and its positive roots are ${\alpha }_1,{\alpha }_2,{\alpha }_1+{\alpha }_2$, so [F16] makes ${\alpha }_1,{\alpha }_2$ simple. All roots have squared length $2$, so their coroots equal the roots by [F15], and $Q_0=\mathbb{Z}{\alpha }_1+\mathbb{Z}{\alpha }_2=\{x\in {\mathbb{Z}}^3:{\sum }_i{x}_i=0\}$. By [F14] the dual weight lattice is $P_0=\{x\in E:(x,{\alpha }_1),(x,{\alpha }_2)\in \mathbb{Z}\}$. If $m=x_1-x_2$ and $n=x_2-x_3$, then $m,n\in \mathbb{Z}$ and $x=m{\omega }_1+n{\omega }_2$, where ${\omega }_1=(2{\epsilon }_1-{\epsilon }_2-{\epsilon }_3)\mathrm{/}3$ and ${\omega }_2=({\epsilon }_1+{\epsilon }_2-2{\epsilon }_3)\mathrm{/}3$; these vectors are in $P_0$, and every $x\in P_0$ has the same pairings with ${\alpha }_1,{\alpha }_2$ as $m{\omega }_1+n{\omega }_2$, so equality follows because ${\alpha }_1,{\alpha }_2$ span $E$. Thus they form a basis. In this basis ${\alpha }_1=2{\omega }_1-{\omega }_2$ and ${\alpha }_2=-{\omega }_1+2{\omega }_2$, so $P_0\mathrm{/}{Q}_0$ is generated by $[{\omega }_1]$, with $3[{\omega }_1]=0$ and $[{\omega }_1]\mathrm{\ne }0$ because ${\omega }_1\mathrm{\notin }{Q}_0$. Thus $[P(A_2):Q(A_2)]=3$.

For $c^{\mathrm{\prime }}$, [F4] gives $r_s(a_s^{\mathrm{\prime }})=-a_s^{\mathrm{\prime }}$, $r_s(a_t^{\mathrm{\prime }})=p^{\mathrm{\prime }}:=a_s^{\mathrm{\prime }}+a_t^{\mathrm{\prime }}$, $r_t(a_t^{\mathrm{\prime }})=-a_t^{\mathrm{\prime }}$, and $r_t(a_s^{\mathrm{\prime }})=d^{\mathrm{\prime }}:=a_s^{\mathrm{\prime }}+2a_t^{\mathrm{\prime }}$. The further images are $r_s(p^{\mathrm{\prime }})=a_t^{\mathrm{\prime }}$, $r_t(p^{\mathrm{\prime }})=p^{\mathrm{\prime }}$, $r_s(d^{\mathrm{\prime }})=d^{\mathrm{\prime }}$, and $r_t(d^{\mathrm{\prime }})=a_s^{\mathrm{\prime }}$, so $R^{\mathrm{\prime }}=\{\pm a_s^{\mathrm{\prime }},\pm a_t^{\mathrm{\prime }},\pm p^{\mathrm{\prime }},\pm d^{\mathrm{\prime }}\}$ is stable under both generators. The normal-scaling identity from step 1.3 applies to this scaling as well. For $w^{\mathrm{\prime }}=r_s{r}_t$, $w^{\mathrm{\prime }}(a_s^{\mathrm{\prime }})=d^{\mathrm{\prime }}$, $w^{\mathrm{\prime }}(a_t^{\mathrm{\prime }})=-p^{\mathrm{\prime }}$, $w^{\mathrm{\prime }}(p^{\mathrm{\prime }})=a_t^{\mathrm{\prime }}$ and $w^{\mathrm{\prime }}(d^{\mathrm{\prime }})=-a_s^{\mathrm{\prime }}$, hence $w^{\mathrm{\prime }2}=-I$. The positive listed vectors are generator roots or their images: $a_s^{\mathrm{\prime }},a_t^{\mathrm{\prime }}$ are generator roots, $p^{\mathrm{\prime }}=r_s(a_t^{\mathrm{\prime }})$, and $d^{\mathrm{\prime }}=r_t(a_s^{\mathrm{\prime }})$. Applying $w^{\mathrm{\prime }2}$ supplies their negatives. As in 1.3, ${\mathrm{\Phi }}_{c^{\mathrm{\prime }}}=R^{\mathrm{\prime }}$.

In $B_2$, the roots ${\epsilon }_1,{\epsilon }_2$ generate $Q(B_2)={\mathbb{Z}}^2$. The coroots of $\pm {\epsilon }_i$ are $\pm 2{\epsilon }_i$; the mixed roots are their own coroots. These coroots span exactly $Q^{\vee }(B_2)=\mathbb{Z}({\epsilon }_1-{\epsilon }_2)+\mathbb{Z}(2{\epsilon }_2)$: both displayed generators occur, and every other coroot is an integer combination of them. In $C_2$, the roots ${\epsilon }_1-{\epsilon }_2$ and $2{\epsilon }_2$ generate $Q(C_2)$, and the remaining roots lie in that span. Its coroot set contains ${\epsilon }_1-{\epsilon }_2,{\epsilon }_2$ and is contained in ${\mathbb{Z}}^2$, so $Q^{\vee }(C_2)={\mathbb{Z}}^2$. Hence $Q^{\vee }(B_2)=Q(C_2)$ and $Q(B_2)=Q^{\vee }(C_2)$.

By [F14], the dual of $Q^{\vee }(B_2)$ is $P(B_2)=\{(x,y):x-y\in \mathbb{Z},2y\in \mathbb{Z}\}$; writing $y=k\mathrm{/}2$, $x-y=m$ gives $P(B_2)=\mathbb{Z}{\epsilon }_1+\mathbb{Z}({\epsilon }_1+{\epsilon }_2)\mathrm{/}2$. Since $Q(B_2)={\mathbb{Z}}^2$, the quotient is generated by the nontrivial class of $({\epsilon }_1+{\epsilon }_2)\mathrm{/}2$; it is not in $Q(B_2)$ and its double lies in $Q(B_2)$, so it has order two. For $C_2$, $Q^{\vee }(C_2)={\mathbb{Z}}^2$, so $P(C_2)={\mathbb{Z}}^2$; the quotient by $Q(C_2)=\mathbb{Z}({\epsilon }_1-{\epsilon }_2)+\mathbb{Z}(2{\epsilon }_2)$ is generated by $[{\epsilon }_2]$, which is nonzero because ${\epsilon }_2\mathrm{\notin }Q(C_2)$ and has order two because $2{\epsilon }_2\in Q(C_2)$. With the simple systems of step 1.4, [F15] gives coroots ${\beta }_1^{\vee }={\beta }_1$, ${\beta }_2^{\vee }=2{\epsilon }_2$, ${\gamma }_1^{\vee }={\gamma }_1$, ${\gamma }_2^{\vee }={\epsilon }_2$. In the row-coroot convention [F17], $({\beta }_2,{\beta }_1^{\vee })=-1$, $({\beta }_1,{\beta }_2^{\vee })=-2$, $({\gamma }_2,{\gamma }_1^{\vee })=-2$ and $({\gamma }_1,{\gamma }_2^{\vee })=-1$, yielding Cartan matrices $\left(\begin{array}{cc}2& -1\\ -2& 2\end{array}\right)$ and $\left(\begin{array}{cc}2& -2\\ -1& 2\end{array}\right)$, both with determinant $2$.

The map $\psi$ in the Statement is an isometry: the images of $e_s,e_t$ have squared lengths $1,1$ and inner product $-1\mathrm{/}\sqrt{2}=-\sqrt{2}\mathrm{/}2$, which matches [F5]. It sends $a_s,a_t,p,d$ to $({\epsilon }_1-{\epsilon }_2)\mathrm{/}\sqrt{2}$, $2{\epsilon }_2\mathrm{/}\sqrt{2}$, $({\epsilon }_1+{\epsilon }_2)\mathrm{/}\sqrt{2}$, $2{\epsilon }_1\mathrm{/}\sqrt{2}$; it sends $a_s^{\mathrm{\prime }},a_t^{\mathrm{\prime }},p^{\mathrm{\prime }},d^{\mathrm{\prime }}$ to ${\epsilon }_1-{\epsilon }_2$, ${\epsilon }_2$, ${\epsilon }_1$, ${\epsilon }_1+{\epsilon }_2$. By [F11,F12], these are exactly $\frac{1}{\sqrt{2}}\mathrm{\Phi }(C_2)$ and $\mathrm{\Phi }(B_2)$; [F13] states that these coordinate root sets are reduced crystallographic systems. Positive scaling preserves those axioms: finiteness, spanning and reducedness are preserved, reflection normal lines are unchanged, and Cartan integers are unchanged by a common scalar. Thus both ${\mathrm{\Phi }}_c$ and ${\mathrm{\Phi }}_{c^{\mathrm{\prime }}}$ are reduced crystallographic root systems. The coroots of $\pm 2{\epsilon }_i$ in $C_2$ are $\pm {\epsilon }_i$, while mixed roots have squared length $2$ and are their own coroots; hence $\mathrm{\Phi }(C_2{)}^{\vee }=\mathrm{\Phi }(B_2)$. Since $(\lambda \mathrm{\Phi }{)}^{\vee }={\lambda }^{-1}{\mathrm{\Phi }}^{\vee }$ for $\lambda >0$ by [F15], $\psi ({\mathrm{\Phi }}_c^{\vee })=\sqrt{2}\mathrm{\Phi }(B_2)$ and $\psi ({\mathrm{\Phi }}_{c^{\mathrm{\prime }}})=\frac{1}{\sqrt{2}}\psi ({\mathrm{\Phi }}_c^{\vee })$.