Affine Coxeter Diagrams and Semidefinite Classification — Examples

(Definiteness is blockwise.) By [F1], if $v={\sum }_i{v}_i$ with $v_i\in V_i$, then $B(v,v)={\sum }_i{B}_i(v_i,v_i)$. If $B$ is positive semidefinite, taking $v$ supported in one block shows each $B_i$ is positive semidefinite; conversely, if every block is positive semidefinite, every term in the sum is nonnegative. The same one-block test proves that positive definiteness of $B$ implies positive definiteness of each $B_i$; conversely, if each $B_i$ is positive definite and $v\mathrm{\ne }0$, at least one $v_i\mathrm{\ne }0$, so the corresponding term is positive and all other terms are nonnegative.

(Radicals and coranks add.) Bilinearity and orthogonality give $B(v,w_i)=B_i(v_i,w_i)$ for every $w_i\in V_i$. Thus $v\in \mathrm{rad}(B)$ exactly when $v_i\in \mathrm{rad}(B_i)$ for every $i$, and the direct decomposition of $V$ makes $\mathrm{rad}(B)={⨁}_i\mathrm{rad}(B_i)$. The vectors $e_s$ for $s\in S_i$ are linearly independent by their coordinates and span $V_i$ by definition, so $V_i$ is finite-dimensional. Each radical is a linear subspace because its annihilation conditions are linear by bilinearity, hence it is finite-dimensional by the subspace dimension theorem. Applying the finite direct-sum dimension formula to these radical subspaces gives $\dim \mathrm{rad}(B)={\sum }_i\dim \mathrm{rad}(B_i)$. This also covers $k=0$: then $V=0$ and both sides are zero.

(Group and length factorization.) The factorwise group isomorphism and length-additivity assertion are precisely F1; their current supplier proof remains open for this run, so this citation is used provisionally and is recorded as an open obligation.

(The line reflection factors.) In $E={\mathbb{R}}^2$ define $r_1(x,y)=(-x,y)$, $r_2(x,y)=(2-x,y)$, $s_1(x,y)=(x,-y)$ and $s_2(x,y)=(x,2-y)$. The identity, composites and inverses of bijective distance-preserving maps are again bijective and distance-preserving, so $\mathrm{Isom}(E)$ is a group under composition; the four displayed maps are involutive isometries by the coordinate distance formula. Put $H_x:=⟨r_1,r_2⟩$ and $H_y:=⟨s_1,s_2⟩$. The maps in $H_x$ are exactly $(x,y)\mapsto (\epsilon x+2k,y)$ with $\epsilon \in \{1,-1\}$ and $k\in \mathbb{Z}$: composition sends parameters $(\epsilon ,k),(\eta ,l)$ to $(\epsilon \eta ,\epsilon l+k)$, the identity has parameters $(1,0)$, and the inverse of $(\epsilon ,k)$ has parameters $(\epsilon ,-\epsilon k)$, so these maps form a subgroup. The translation $r_2{r}_1$ is $x\mapsto x+2$, and its powers, followed by $r_1$, give every displayed map. By [F6], sending the two Coxeter generators $a,b$ of $W({\tilde{A}}_1)$ to $r_1,r_2$ defines a homomorphism to $H_x$: both images are involutions, and the infinity label supplies no further relator. Every word in $a,b$ reduces by $a^2=b^2=1$ to an alternating word, hence to $(ab{)}^k$ or $(ab{)}^{k}a$ for some $k\in \mathbb{Z}$. Their images are respectively $x\mapsto x-2k$ and $x\mapsto -x-2k$, which are pairwise distinct and exhaust the displayed maps, so the homomorphism is bijective and $H_x\cong W({\tilde{A}}_1)=D_{\mathrm{\infty }}$. The same coordinate calculation gives $H_y\cong D_{\mathrm{\infty }}$.

(Corank one.) Suppose first that $B$ is positive semidefinite with $\dim \mathrm{rad}(B)=1$. Step 1.1 makes every $B_i$ positive semidefinite, and step 1.2 says the nonnegative integers $d_i:=\dim \mathrm{rad}(B_i)$ sum to one; hence exactly one $d_i$ is one and all others are zero. For any positive-semidefinite block $C$ with zero radical, if $C(v,v)=0$, then for every $w$ and every real $t$, $0\le C(v+tw,v+tw)=2tC(v,w)+t^{2}C(w,w)$; if $C(v,w)\mathrm{\ne }0$, sufficiently small $t$ of the opposite sign makes the right-hand side negative, a contradiction. Therefore such $v$ is in the radical, so zero radical implies $C(v,v)>0$ for every nonzero $v$, i.e. $C$ is positive definite. Conversely, if exactly one block is positive semidefinite of radical dimension one and all others are positive definite, step 1.1 gives $B$ positive semidefinite and step 1.2 gives radical dimension one. The forms $B_i$ are symmetric bilinear and the component diagrams are connected by Fact F2. Thus the unique block is of affine form type by Fact F11; the other blocks are finite type by Fact F3.

(The two block examples.) For $m(s,t)=\mathrm{\infty }$, the defining entries of $B$ give $B_1((x,y),(x,y))=x^2-2xy+y^2=(x-y{)}^2$, so it is positive semidefinite with positive radical vector $e_s+e_t$ and radical $\mathbb{R}(e_s+e_t)$. A one-generator block has matrix $(1)$, hence is positive definite and has zero radical; with this block the full radical is spanned by $(1,1,0)$. With two $m=\mathrm{\infty }$ blocks, the orthogonal sum has radical spanned by $(1,1,0,0)$ and $(0,0,1,1)$ and has corank two by step 1.2. The group factorization from step 1.3 and the identification in step 1.4 give $W\cong D_{\mathrm{\infty }}\times D_{\mathrm{\infty }}$.

(The square group and its product map.) Let $G=⟨r_1,r_2,s_1,s_2⟩\le \mathrm{Isom}(E)$. The groups $H_x$ and $H_y$ from step 1.4 commute elementwise because they act on separate coordinates, their intersection is the identity because a map acting trivially on both coordinates is the identity, and they generate $G$. The map $\mu :H_x\times H_y\to G$, $\mu (h_x,h_y)=h_x{h}_y$, is a homomorphism by commutation, is surjective by generation, and is injective since $h_x{h}_y=1$ implies $h_x=h_y^{-1}\in H_x\cap H_y=\{1\}$. Thus it is an isomorphism by [F7], and step 1.4 identifies both factors with $D_{\mathrm{\infty }}$. Within either parallel pair the product is a nonzero translation by two units and has infinite order; across the pairs the reflections commute, and their product is a nonidentity involution because a vertical reflection moves some first coordinate while a horizontal reflection fixes every first coordinate. The products therefore give the disconnected diagram ${\tilde{A}}_1\perp {\tilde{A}}_1$.

(Grid and strict fundamental domain.) The orbit of $[0,1]$ under $H_x$ is $\{[2k,2k+1],[2k-1,2k]:k\in \mathbb{Z}\}$, exactly the unit intervals $[m,m+1]$; [F12] applied to each real $x$ gives $m\le x<m+1$, proving coverage of $\mathbb{R}$, and the interval interiors are pairwise disjoint. The corresponding statement holds for $H_y$, so the $G$-translates of $Q$ are exactly all grid squares, covering $E$ with pairwise disjoint interiors. To prove the stronger orbit assertion including boundary points, apply [F12] to $x\mathrm{/}2$ and put $n=\lfloor x\mathrm{/}2\rfloor$, $u=x-2n$; then $n\in \mathbb{Z}$ and $0\le u<2$. Its orbit under $H_x$ is $\{u+2j,-u+2j:j\in \mathbb{Z}\}$; if $0\le u\le 1$, its unique value in $[0,1]$ is $u$: among the values $u+2j$ only $j=0$ qualifies, and among $-u+2j$ the only possible additional values occur at $(u,j)=(0,0)$ or $(1,1)$ and equal that same point. If $1<u<2$, the unique value there is $2-u=-u+2$. Applying this independently to both coordinates shows every $G$-orbit meets $Q$ in exactly one point.

(The remaining geometric and factorwise conclusions.) The four side walls are $x=0$, $x=1$, $y=0$, and $y=1$; at each of the four vertices one vertical and one horizontal wall meet, so the interior angle is $\pi \mathrm{/}2$. Step 3.1 proves the precise closed-square tiling and orbit statement in (iv), including boundary points. Clauses (i)-(ii) give the positive-semidefinite corank-one factorization in (v); if one block has a vector of negative quadratic value, the same vector in $V$ shows that $B$ is indefinite, and if two blocks have nonzero radicals, step 1.2 gives radical dimension at least two. If $k=0$, then $V=0$ and $\dim \mathrm{rad}(B)=0$, so both sides of the equivalence in (ii) are false and the empty direct sums in (i) have value zero. All arguments use finite sums and explicit constructions, so no Choice is used.

The matrix in the statement gives $B(x_s{e}_s+x_t{e}_t,x_s{e}_s+x_t{e}_t)=(x_s-x_t{)}^2$. Also $B(x,e_s)=x_s-x_t$ and $B(x,e_t)=x_t-x_s$, so the radical is exactly $\mathbb{R}(e_s+e_t)$. Thus $B$ is positive semidefinite of corank one; the diagram is connected, so the pair is of affine form type by [F4].

In the ordered basis $(e_s,e_t)$, direct substitution in [F1] gives $[r_{e_s}]=\left(\begin{array}{cc}-1& 2\\ 0& 1\end{array}\right),[r_{e_t}]=\left(\begin{array}{cc}1& 0\\ 2& -1\end{array}\right)\mathrm{.}$ Both square to the identity. Their product is $[r_{e_s}{r}_{e_t}]=\left(\begin{array}{cc}3& -2\\ 2& -1\end{array}\right)=I+N,N=\left(\begin{array}{cc}2& -2\\ 2& -2\end{array}\right),N\mathrm{\ne }0,N^2=\left(\begin{array}{cc}2\cdot 2+(-2)\cdot 2& 2\cdot (-2)+(-2)\cdot (-2)\\ 2\cdot 2+(-2)\cdot 2& 2\cdot (-2)+(-2)\cdot (-2)\end{array}\right)=0.$ Since $m(s,t)=\mathrm{\infty }$, the presentation in [F3] has no relation beyond the two involutions, so the assignment $s\mapsto r_{e_s}$, $t\mapsto r_{e_t}$ defines a representation of $W$. For every integer $k$, $(I+N{)}^k=I+kN$ (use $(I+N{)}^{-1}=I-N$ for $k<0$), which is never the identity when $k\mathrm{\ne }0$. Thus the product is a nonidentity unipotent of infinite order and $W$ is infinite.

Here ${\delta }_s={\delta }_t=1$. The slice condition is $\phi (e_s)+\phi (e_t)=1$, so with $\alpha =\phi (e_s)$ its points are exactly $(\alpha ,1-\alpha )$ for $\alpha \in \mathbb{R}$. The vertices from [F5] are $v_s=(1,0)$ and $v_t=(0,1)$; the alcove inequalities give $0<\alpha <1$, and its closure is the Euclidean segment $[v_t,v_s]$. The two walls are its endpoints $\alpha =0$ and $\alpha =1$.

For the left dual action, $(w\cdot \phi )(v)=\phi (\rho (w{)}^{-1}v)$. Since each reflection is involutory, evaluating at $e_s$ gives $(s\cdot \phi )(e_s)=\phi (-e_s)=-\alpha$. Also $r_{e_t}(e_s)=e_s+2e_t$ by [F1], so $(t\cdot \phi )(e_s)=\alpha +2(1-\alpha )=2-\alpha$. Therefore $s\cdot \alpha =-\alpha$ and $t\cdot \alpha =2-\alpha$. Under the left-action convention, $(st)\cdot \alpha =s\cdot (t\cdot \alpha )=\alpha -2$, while $(ts)\cdot \alpha =t\cdot (s\cdot \alpha )=\alpha +2$.

If $m=m(s,t)<\mathrm{\infty }$, then $m\ge 2$ and $0<\pi \mathrm{/}m\le \pi \mathrm{/}2$. By [F7]-[F8] and $\cos 0=1$ from [F12], $0\le c:=\cos (\pi \mathrm{/}m)<1$. Therefore $\det \left(\begin{array}{cc}1& -c\\ -c& 1\end{array}\right)=1-c^2={\sin }^2(\pi \mathrm{/}m)>0,$ where the identity is [F9] and positivity follows from $0\le c<1$. Moreover the quadratic form is $(x_s-c{x}_t{)}^2+(1-c^2)x_t^2$, positive for every nonzero $(x_s,x_t)$; thus it has no radical. From [F3], $ts=(st{)}^{-1}$ and every word reduces to an alternating word, hence to $(st{)}^k$ or $(st{)}^{k}s$; the finite relator $(st{)}^m=1$ reduces $k$ modulo $m$, leaving at most $2m$ elements. At $m=2$ the generators commute; the presentation maps onto $A_1\times A_1$ by sending them to the two factors, and the normal-form bound gives at most four elements, so $W\cong A_1\times A_1$. For $m\ge 3$, let $X=\mathbb{Z}\mathrm{/}m\mathbb{Z}$ and define permutations $\sigma (j)=-j$ and $\tau (j)=1-j$. They are involutions and $(\sigma \tau )(j)=j-1$, which has exact order $m$. The $m$ maps $(\sigma \tau {)}^k$ are distinct translations, and the $m$ maps $(\sigma \tau {)}^k\sigma$ are distinct reflections. No reflection is a translation: equality of $j\mapsto -j-k$ and $j\mapsto j+\mathrm{\ell }$ at $j=0,1$ would imply $m\mid 2$, contrary to $m\ge 3$. By [F3], these permutations define a homomorphic image of $W$ with $2m$ elements. Together with the upper bound, this proves that $W$ is the finite dihedral group $I_2(m)$ of order $2m$, with the standard Coxeter notation [F17].

Compositions of $\alpha \mapsto -\alpha$ and $\alpha \mapsto 2-\alpha$ are precisely the maps $\alpha \mapsto \pm \alpha +2k$ for $k\in \mathbb{Z}$: the product $(ts{)}^k$ is translation by $2k$, and composing it with $s$ gives every orientation-reversing map of this form. The images of $(0,1)$ are all integer intervals: translations by $2k$ give $(2k,2k+1)$ and composing those translations with $\alpha \mapsto 2-\alpha$ gives $(2k+1,2k+2)$. A positive-orientation map stabilizes $(0,1)$ only when $k=0$; a negative-orientation map sends it to $(2k-1,2k)$, which cannot equal $(0,1)$ for integral $k$. Hence the action is simply transitive on these alcoves. This is the rank-one affine reflection group $W_a(A_1)$; its translation subgroup is $2\mathbb{Z}$ and its linear part is $\{\pm 1\}$, hence $W_a(A_1)\cong 2\mathbb{Z}⋊\{\pm 1\}$ as in Xiong's rank-one affine Weyl group example.

For $m(s,t)=4$, the finite diagram has one label-$4$ edge, whereas ${\tilde{A}}_1$ has one label-$\mathrm{\infty }$ edge by [F6]; these labelled graphs are distinct. Xiong's low-rank coincidence [F16] matches the two names $B_2$ and $C_2$ for this finite type. Their rank-two affine extensions have one additional label-$4$ edge by [F10], giving the three-vertex path $(4,4)={\tilde{B}}_2={\tilde{C}}_2$, not ${\tilde{A}}_1$.

The matrix entry $B(e_s,e_t)=-1$ for $m(s,t)=\mathrm{\infty }$ is the defining convention in [F1]. For any $\epsilon >0$, [F20] gives $\epsilon \mathrm{/}\pi >0$, so apply [F19] to choose $N\ge 1$ with $1\mathrm{/}N<\epsilon \mathrm{/}\pi$. If $m\ge N$, then $0<N\mathrm{/}m\le 1$, so $0<\pi \mathrm{/}m\le \pi \mathrm{/}N<\epsilon$. By the Lipschitz bound [F18] and $\cos 0=1$ [F12], $\mathrm{\mid }\cos (\pi \mathrm{/}m)-1\mathrm{\mid }\le \pi \mathrm{/}m<\epsilon$; hence ${\lim }_{m\to \mathrm{\infty }}\cos (\pi \mathrm{/}m)=1$, confirming that the matrix convention is the numerical limit. The infinite label still imposes no finite relator by [F3]. By [F15], Davis's cited Euclidean simplex criterion assumes every Coxeter label is finite, so the rank-one $\mathrm{\infty }$-edge case is established directly here. No choice principle is used.

For $x=(x_1,x_2,x_3)$, direct expansion gives $B(x,x)=\frac{1}{2}((x_1-x_2{)}^2+(x_2-x_3{)}^2+(x_3-x_1{)}^2)$. It is nonnegative and vanishes exactly on $\mathbb{R}(1,1,1)$; matrix multiplication also gives $B(1,1,1)=0$, so this is precisely the radical. Each two-coordinate principal form is $a^2-ab+b^2=(a-b\mathrm{/}2{)}^2+3b^2\mathrm{/}4$, positive definite; the one-coordinate forms are $(1)$ and the empty case is vacuous. Their determinants are $3\mathrm{/}4$, and the full determinant is zero.

The slice relation is ${\phi }_1+{\phi }_2+{\phi }_3=1$, and [F3] gives its coordinate vertices $v_i$ and triangular closure. Put $H:=\{z\in {\mathbb{R}}^3:{\sum }_i{z}_i=0\}$. Every class in $U=V\mathrm{/}\mathbb{R}(1,1,1)$ has a unique representative in $H$, obtained by subtracting the mean of its coordinates, so $H\to U$ is a linear isomorphism. The direction space $K_{\delta }$ consists of functionals with coordinate tuple $y=(\psi (e_{s_1}),\psi (e_{s_2}),\psi (e_{s_3}))\in H$; under the representative identification, $\psi (\bar{z})={\sum }_i{y}_i{z}_i$, so $y$ represents $\psi$ using the standard dot product on $H$. For $z\in H$, the matrix in [F1] gives $Bz=(3\mathrm{/}2)z$; hence $b^{\mathrm{♭}}(z)$ is represented by $(3\mathrm{/}2)z$, and $b^{\mathrm{♭}}{}^{-1}\psi$ is represented by $z=(2\mathrm{/}3)y$. Therefore $b^{\ast }(\psi ,\psi )=B(z,z)=(2\mathrm{/}3){\sum }_i{y}_i^2$. Each difference $v_i-v_j$ has tuple with one $1$, one $-1$ and one zero, so its squared length is $4\mathrm{/}3$. All three sides therefore have length $2\mathrm{/}\sqrt{3}$ in the prescribed Euclidean slice metric.