Finite Coxeter Invariants and Coinvariant Gradings — Examples

The Coxeter presentation maps onto $S_3$ by $s_1\mapsto (12)$ and $s_2\mapsto (23)$. Let $t=s_1{s}_2$; then $t^3=1$, $s_2=s_{1}t$, and $s_{1}t{s}_1=t^{-1}$. Every word is therefore one of $t^k$ or $s_1{t}^k$ for $k=0,1,2$, so $\mathrm{\mid }W\mathrm{\mid }\le 6$; the surjection onto $S_3$ gives $\mathrm{\mid }W\mathrm{\mid }\ge 6$ and hence $W\cong S_3$. The reflection in the $x$-axis has unit normal ${\alpha }_1=(0,1)$; the reflection in the line at angle $\pi \mathrm{/}3$ has unit normal ${\alpha }_2=(\sqrt{3}\mathrm{/}2,-1\mathrm{/}2)$, with $({\alpha }_1,{\alpha }_2)=-1\mathrm{/}2$. The Gram matrix of $({\alpha }_1,{\alpha }_2)$ is the $A_2$ Coxeter Gram matrix, so the map from the canonical simple-root basis to $({\alpha }_1,{\alpha }_2)$ is an isometry and conjugates the canonical representation to this model. The reflection actions are $s_1{\alpha }_1=-{\alpha }_1$, $s_1{\alpha }_2={\alpha }_1+{\alpha }_2$, $s_2{\alpha }_2=-{\alpha }_2$, and $s_2{\alpha }_1={\alpha }_1+{\alpha }_2$; hence $\{\pm {\alpha }_1,\pm {\alpha }_2,\pm ({\alpha }_1+{\alpha }_2)\}$ is exactly the root set and its positive elements are ${\alpha }_1,{\alpha }_2,{\alpha }_3$. The corresponding normalized forms are $y$, $\frac{\sqrt{3}}{2}x-\frac{1}{2}y$, and $\frac{\sqrt{3}}{2}x+\frac{1}{2}y$. Substitution $x=(u+z)\mathrm{/}2$, $y=(u-z)\mathrm{/}(2i)$ gives the three displayed scalar multiples in (1). Multiplying them gives $\mathrm{\Delta }=\frac{1}{8i}(u-z)(u-\zeta z)(u-{\zeta }^{2}z)=\frac{1}{8i}(u^3-z^3)$ because $1,\zeta ,{\zeta }^2$ are the distinct cube roots of unity; the product has degree $3=\mathrm{\mid }{\mathrm{\Phi }}_{+}\mathrm{\mid }$.

The reflection generator acts by $\sigma (u,z)=(z,u)$ and $R=s_2{s}_1$ acts by $R(u,z)=(\zeta u,{\zeta }^{-1}z)$. Since $s_2=R\sigma$, the pair $\sigma ,R$ generates $W$. Both $a=uz$ and $b=u^3+z^3$ are invariant. Let $f={\sum }_{p,q}{c}_{pq}{u}^p{z}^q$ be invariant. Rotation invariance and monomial independence imply $c_{pq}=0$ unless $3\mid (p-q)$; swap invariance gives $c_{pq}=c_{qp}$. Thus $f$ is a finite linear combination of $a^p$ and orbit sums $a^q(u^{3k}+z^{3k})$ with $k\ge 1$. Set $S_k=u^{3k}+z^{3k}$. Then $S_0=2$, $S_1=b$, and direct expansion gives $S_{k+1}=b{S}_k-a^3{S}_{k-1}$; induction shows each $S_k\in \mathbb{C}[a,b]$. Hence $\mathbb{C}[u,z{]}^W=\mathbb{C}[a,b]$.

The formal partial-derivative rules and the product rule give ${\mathrm{\partial }}_{u}a=z$, ${\mathrm{\partial }}_{z}a=u$, ${\mathrm{\partial }}_{u}b=3u^2$, and ${\mathrm{\partial }}_{z}b=3z^2$, so the Jacobian determinant is $\mathbf{J}=3(z^3-u^3)=-24i\mathrm{\Delta }$. It is not the zero polynomial. To prove $a,b$ algebraically independent, suppose a nonzero $H\in \mathbb{C}[Y_1,Y_2]$ of minimal total degree satisfies $H(a,b)=0$. Since the characteristic is zero, at least one partial derivative $H_j$ is nonzero. Expanding $H$ into monomials and applying the product rule gives the chain rule, so differentiating the relation with respect to $u,z$ gives $J^{\mathsf{T}}(H_1(a,b),H_2(a,b){)}^{\mathsf{T}}=0$. Multiply the displayed equation by the explicit adjugate $\left(\begin{array}{cc}3z^2& -3u^2\\ -u& z\end{array}\right)$ of $J^{\mathsf{T}}$. Its product with $J^{\mathsf{T}}$ is $3(z^3-u^3)I_2$, so the domain property and $3(z^3-u^3)\mathrm{\ne }0$ force both partial derivatives to vanish after substitution. A nonzero partial derivative is then a relation of strictly smaller total degree; if it were a nonzero constant it could not vanish, and otherwise this contradicts the minimal choice of $H$. Therefore $a,b$ are algebraically independent. They are homogeneous of degrees $2,3$ and generate $R$ by 2.1. Since $R_{+}=(a,b)R$, the coinvariant ideal is $(a,b)\mathbb{C}[u,z]$. Neither generator is in the ideal generated by the other: $a$ has degree $2<3$, while $b=u^3+z^3$ is not divisible by $uz$. Thus they form the basic family with degrees $2,3$ and exponents $1,2$. For anti-invariance, the swap sends $\mathrm{\Delta }$ to $-\mathrm{\Delta }$ and has determinant $-1$; the rotation sends $u^3-z^3$ to itself and has determinant $1$. Since these elements generate $W$, $w\cdot \mathrm{\Delta }=\det (w)\mathrm{\Delta }$ for every $w\in W$.

In $A=\mathbb{C}[u,z]\mathrm{/}(uz,u^3+z^3)$, all mixed monomials vanish, $z^3=-u^3$, and $u^4=z^4=0$, so $1,u,u^2,z,z^2,u^3$ span. The ideal is homogeneous: its degree-two part is spanned by $uz$, and its degree-three part by $u^{2}z,u{z}^2,u^3+z^3$. Thus $u^2,z^2$ are independent in degree two, and $u^3$ is nonzero in degree three because it is not in the span of those three degree-three relations; the degree-zero and degree-one classes are also independent because the ideal has no terms in those degrees. Hence the six displayed classes are a basis, $\mathrm{Hilb}(A,t)=1+2t+2t^2+t^3$, and $A_3$ is one-dimensional. Using $z^3=-u^3$ and the normalized scalar in step 1.1 gives $[\mathrm{\Delta }]=\frac{1}{8i}[u^3-z^3]=-\frac{i}{4}[u^3]\mathrm{\ne }0$, so it spans $A_3$ and carries the determinant character by step 3.1.

The degrees are $2,3$, so the exponents are $1,2$, their sum is $3=\mathrm{\mid }{\mathrm{\Phi }}_{+}\mathrm{\mid }$, and their product is $6=\mathrm{\mid }{S}_3\mathrm{\mid }$ by 1.1. The quotient basis in 4.1 has top degree $3$ and one-dimensional top component; the nonzero discriminant class is its generator. All constructions use explicit coordinates and finite polynomial identities, so no Choice is used. This proves the four clauses.

(Identification of the model.) Let $D\le \mathrm{O}(2)$ be the group generated by the reflection $\sigma$ in the $x$-axis and the rotation ${\rho }_{2\pi \mathrm{/}m}$ by angle $2\pi \mathrm{/}m$; it consists of the $m$ rotations by $2\pi k\mathrm{/}m$ and the $m$ reflections ${\rho }_{2\pi k\mathrm{/}m}\sigma$, so $\mathrm{\mid }D\mathrm{\mid }=2m$. The reflection $\tau$ in the line at angle $\pi \mathrm{/}m$ satisfies $\tau ={\rho }_{2\pi \mathrm{/}m}\sigma$, so $D=⟨\sigma ,\tau ⟩$ with $\sigma ,\tau$ involutions whose product has order $m$; hence the universal property of [F1] gives a homomorphism $\pi :W\to D$ with $\pi (s_1)=\sigma$, $\pi (s_2)=\tau$, and $\pi$ is onto. Conversely, put $t:=s_1{s}_2\in W$: then $t^m=1$, $s_2=s_{1}t$ and $s_{1}t{s}_1=s_2{s}_1=t^{-1}$, so every element of $W$ is a word in $s_1$ and $t$ and $W$ is a quotient of the group $⟨u,t\mid u^2=t^m=1,utu=t^{-1}⟩$, every element of which is $t^k$ or $u{t}^k$ with $k\in \{0,\dots ,m-1\}$; hence $\mathrm{\mid }W\mathrm{\mid }\le 2m$, and together with the surjection $\pi$ onto the group of order $2m$ this gives $\mathrm{\mid }W\mathrm{\mid }=2m$ and that $\pi$ is an isomorphism. Let $\phi :V\to {\mathbb{R}}^2$ be the linear isometry sending $e_1$ to the unit normal of the $x$-axis and $e_2$ to a unit normal of the line at angle $\pi \mathrm{/}m$ chosen with $(e_1,e_2)$-image equal to $-\cos (\pi \mathrm{/}m)$ (flipping a normal does not change its reflection); as in the reflection-model identification, $\phi \rho (s_i){\phi }^{-1}$ is the reflection of the model in the line fixed by $\sigma$ resp. $\tau$, so $\phi \rho {\phi }^{-1}=\pi$ on the generators and hence on all of $W$: the canonical representation of $W$ is conjugate to the model action, and the invariant rings correspond. Therefore $\mathbb{C}[V_{\mathbb{C}}{]}^W$ may be computed in the coordinates $u,z$, where the generators act by $u\leftrightarrow z$ and $u\mapsto \zeta u$, $z\mapsto {\zeta }^{-1}z$ with $\zeta =\exp (2\pi i\mathrm{/}m)$ [F2, F5].

(Generation of the invariants.) Let $f={\sum }_{p,q\ge 0}{c}_{pq}{u}^p{z}^q\in \mathbb{C}[u,z]$ be invariant. Applying the rotation gives $\sum c_{pq}{\zeta }^{p-q}{u}^p{z}^q=\sum c_{pq}{u}^p{z}^q$, so $c_{pq}({\zeta }^{p-q}-1)=0$ monomial by monomial; by [F5] ${\zeta }^k=1$ only when $m\mid k$, so every monomial of $f$ has $p\equiv q(\mathrm{m}\mathrm{o}\mathrm{d}m)$ [F3]. Applying the reflection $u\leftrightarrow z$ pairs the monomials $u^p{z}^q$ and $u^q{z}^p$ and forces $c_{pq}=c_{qp}$; together with the rotation condition this shows that the invariants are spanned by the elements $u^p{z}^p=(uz{)}^p$ and the sums $u^p{z}^q+u^q{z}^p$ with $p>q$ and $m\mid p-q$; writing $p=q+km$, $k\ge 1$, such a sum equals $(uz{)}^q(u^{km}+z^{km})$. The polynomials $S_k:=u^{km}+z^{km}$ satisfy $S_0=2$, $S_1=u^m+z^m=:b$ and, by expanding $(u^m+z^m)(u^{km}+z^{km})-(uz{)}^m(u^{(k-1)m}+z^{(k-1)m})=u^{(k+1)m}+z^{(k+1)m}$, the recurrence $S_{k+1}=b{S}_k-a^m{S}_{k-1}$ with $a:=uz$; induction gives $S_k\in \mathbb{C}[a,b]$ for every $k\ge 0$. Hence every invariant lies in $\mathbb{C}[a,b]$, and conversely $a$ and $b$ are invariant, so $\mathbb{C}[u,z{]}^W=\mathbb{C}[a,b]$.

(The coinvariant algebra.) In $B_0:=\mathbb{C}[u,z]\mathrm{/}(uz)$ every class corresponds uniquely to a pair $(p,q)$ with $p,q\in \mathbb{C}[t]$ and $p(0)=q(0)$, represented by $p(u)+q(z)-p(0)$, and monomials give the basis $1,u^p(p\ge 1),z^q(q\ge 1)$; the ring is a fibre product in which multiplication is componentwise, $(p,q)(p^{\mathrm{\prime }},q^{\mathrm{\prime }})=(p{p}^{\mathrm{\prime }},q{q}^{\mathrm{\prime }})$. Multiplication by $u^m+z^m$, i.e. by the pair $(t^m,t^m)$, preserves this decomposition and is injective because $\mathbb{C}[t]$ is a domain; its image consists of the pairs $(t^{m}p,t^{m}q)$ with $p(0)=q(0)$. Hence $0\to B_0\stackrel{\cdot (u^m+z^m)}{\to }{B}_0\to A\to 0$ is exact, $B_0$ has Hilbert series $1+{\sum }_{p\ge 1}{t}^p+{\sum }_{q\ge 1}{t}^q=(1+t)\mathrm{/}(1-t)$, and the multiplicativity of the Hilbert series in a short exact sequence of graded spaces gives $\mathrm{Hilb}(A,t)=(1-t^m)(1+t)\mathrm{/}(1-t)=(1+t)(1+t+\cdots +t^{m-1})$ [F3]. On the other hand the relations $uz=0$ and $z^m=-u^m$ show that the $2m$ displayed classes span: in $B_0$ the classes of $u^p$ and $z^q$ span, and $u^{m+k}=u^k(-z^m)=0$ for $k\ge 1$, likewise for $z$, so only $1,u,\dots ,u^{m-1},z,\dots ,z^{m-1},u^m$ remain, with $z^m\equiv -u^m$. Comparing with the Hilbert series degree by degree, these classes are linearly independent and form a basis, and ${\dim }_{\mathbb{C}}A=2m=\mathrm{\mid }W\mathrm{\mid }$ by step 1.1.

(Algebraic independence and the basic family.) The formal partials satisfy the product and power rules on monomials and hence, by linearity, on all polynomials; consequently ${\mathrm{\partial }}_u$ and ${\mathrm{\partial }}_z$ satisfy the two-variable chain rule for polynomial compositions. Suppose $0\mathrm{\ne }H\in \mathbb{C}[y_1,y_2]$ has minimal total degree among the nonzero polynomials with $H(a,b)=0$. Its partial derivatives $H_1,H_2$ are not both zero and have smaller total degree, and differentiating $H(a,b)=0$ with respect to $u$ and $z$ gives the two equations $H_1(a,b){\mathrm{\partial }}_{u}a+H_2(a,b){\mathrm{\partial }}_{u}b=0$ and $H_1(a,b){\mathrm{\partial }}_{z}a+H_2(a,b){\mathrm{\partial }}_{z}b=0$, i.e. $J^{\mathsf{T}}\cdot (H_1(a,b),H_2(a,b){)}^T=0$ for $J=\left(\begin{array}{cc}{\mathrm{\partial }}_{u}a& {\mathrm{\partial }}_{z}a\\ {\mathrm{\partial }}_{u}b& {\mathrm{\partial }}_{z}b\end{array}\right)=\left(\begin{array}{cc}z& u\\ m{u}^{m-1}& m{z}^{m-1}\end{array}\right)$. Multiplying these two equations by the explicit adjugate $\left(\begin{array}{cc}m{z}^{m-1}& -m{u}^{m-1}\\ -u& z\end{array}\right)$ of $J^{\mathsf{T}}$ gives $m(z^m-u^m)H_j(a,b)=0$ for $j=1,2$. Since $m(z^m-u^m)\mathrm{\ne }0$ and $\mathbb{C}[u,z]$ is a domain, both $H_1(a,b)$ and $H_2(a,b)$ vanish; a nonzero partial of $H$ is nonconstant (a nonzero constant cannot evaluate to $0$), so it is a nonzero polynomial of smaller total degree vanishing on $(a,b)$, contradicting minimality. Thus $a,b$ are algebraically independent [F4]. They are homogeneous of degrees $2$ and $m$, generate $R=\mathbb{C}[a,b]$, and are minimal: $a\mathrm{\notin }(b)$ because $\mathrm{deg}(b\cdot s)\ge m>2=\mathrm{deg}a$ for $s\mathrm{\ne }0$, and $b\mathrm{\notin }(a)$ because substituting $u=0$ gives $b(0,z)=z^m\mathrm{\ne }0=(uz\cdot s)(0,z)$. Since $R_{+}=(a,b)R$, the ideal is $I=S{R}_{+}=(a,b)S$, so $a,b$ are the basic invariants of the definition with degrees $2,m$ and exponents $1,m-1$.

(Noncrystallographic contrast in every rank.) In a crystallographic root system, a product of two root reflections has order in $\{1,2,3,4,6\}$: proportional roots give the identity; otherwise the product fixes the orthogonal complement of their plane and rotates that plane by twice the angle of the two reflecting lines, and [F6] gives exactly the angles with these orders. Suppose its Weyl group were abstractly $D_m=⟨r,s\mid r^m=s^2=1,srs=r^{-1}⟩$. The images of the root reflections form a generating set $U$ of involutions. Since the only possible nonidentity involution in $⟨r⟩$ is $r^{m\mathrm{/}2}$, some member lies outside this cyclic subgroup; relabel that member $s$. Every other member of $U$ is $r^{k}s$ or, for even $m$, $r^{m\mathrm{/}2}$. Products of the fixed root reflection $s$ with the other coset members give rotations $r^{-k}$ of orders in $\{1,2,3,4,6\}$. These rotations, together with the possible central involution, generate $⟨r⟩$: after replacing each coset generator by its product with $s$, all generators except $s$ are rotations, and $s$ merely inverts them. A cyclic group generated by elements of these orders has order dividing $\mathrm{lcm}(2,3,4,6)=12$. For $m\ge 3$ this leaves $m=3,4,6,12$. If $m=12$, the possible exponents $k$ of coset members are $0,2,3,4,6,8,9,10$, and every difference of two such exponents also gives a product of root reflections, so cannot have order $12$. If $3$ or $9$ occurs, none of $2,4,8,10$ can occur, as their differences are coprime to $12$; all rotations generated then have exponents divisible by $3$, including the central exponent $6$. If neither occurs, all exponents are even, again including $6$. Neither case generates the full rotation subgroup of order $12$, a contradiction. Thus an abstract Weyl-group isomorphism is possible only for $m=3,4,6$, proving the strongest asserted contrast without assuming the root-system rank is two. In particular, the prescribed plane action has no crystallographic root realization when $m\mathrm{\notin }\{3,4,6\}$. It also preserves no full lattice: if it did, the rotation would have an integer matrix and integer trace $2\cos (2\pi \mathrm{/}m)$; this trace is $-1,0,1$ for $m=3,4,6$, lies strictly between $0$ and $1$ for $m=5$, and strictly between $1$ and $2$ for $m>6$, by [F7], excluding every other $m\ge 3$. For $m=3,4,6$, step 3.1 gives degrees $2,3$, $2,4$, $2,6$, equal to one plus the spectral exponents in [F8].

Steps 1.1-4.1 prove all three clauses: the invariant algebra is the polynomial algebra $\mathbb{C}[a,b]$ on the algebraically independent basic invariants of degrees $2,m$, the coinvariant algebra has the displayed $2m$-element basis and Hilbert series with ${\dim }_{\mathbb{C}}A=2m=\mathrm{\mid }W\mathrm{\mid }$, and no reduced crystallographic root system realizes $I_2(m)$ for $m\mathrm{\notin }\{3,4,6\}$; the computation is a finite monomial and derivative calculation, uses no AC-scoped invariant-theory supplier, and the only choice made is the explicit order of the two colour classes, which does not enter the results.