The invariants and the coinvariant Hilbert series of $I_2(m)$: an explicit computation and the noncrystallographic contrast

(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.