A regular Coxeter eigenvector determines the basic degrees: the exponent-residue identification and the complete degree tables for all finite Coxeter types

Assume first that $W$ is irreducible and $n\ge 2$. Let $P={\mathrm{span}}_{\mathbb{R}}(u,v)$ be the Coxeter plane. Choose real vectors $\xi ,\eta$ forming a basis of $P$ so that $x=\xi +i\eta$ is an eigenvector of $C:={\rho }_{\mathbb{C}}(c)$ with eigenvalue ${\zeta }_h^{-1}$; this is possible because $C{\mathrm{\mid }}_P$ is rotation by $2\pi \mathrm{/}h$ [F2, F3]. If $x$ lay in a complexified reflecting hyperplane $H_{\alpha }{\otimes }_{\mathbb{R}}\mathbb{C}$, then its real and imaginary parts $\xi ,\eta$ would both lie in $H_{\alpha }$, so $P\subseteq H_{\alpha }$. But $H_{\alpha }\cap P$ is a line by [F3], so no reflecting hyperplane contains $P$. Therefore every factor of $\mathrm{\Delta }(x)$ is nonzero and $\mathbf{J}(x)\mathrm{\ne }0$ by [F4]. The matrix $(d{p}_i(x){)}_{i=1}^n$ is thus invertible, and $d{p}_1(x),\dots ,d{p}_n(x)$ form a basis of $V_{\mathbb{C}}^{\ast }$ [F4].

Let the connected components of a reducible diagram be $S_1,\dots ,S_k$, with corresponding spaces $V_{\mathbb{C},j}$ and groups $W_j$. By [F7], $W={\prod }_j{W}_j$ acts on $V_{\mathbb{C}}={⨁}_j{V}_{\mathbb{C},j}$ factorwise, and the coordinate polynomial algebra is $\mathbb{C}[V_{\mathbb{C}}]={⨂}_j\mathbb{C}[V_{\mathbb{C},j}]$. To compute invariants, expand any polynomial as a finite sum of block monomials with coefficients in the other blocks. Invariance under $W_j$ forces every coefficient in the $j$-th block to be $W_j$-invariant; doing this for each factor proves $\mathbb{C}[V_{\mathbb{C}}{]}^W={⨂}_j\mathbb{C}[V_{\mathbb{C},j}{]}^{W_j}$, and the reverse inclusion follows because each factor acts trivially on the other blocks [F7]. Each component invariant ring is polynomial on its homogeneous basic family, so the tensor product is polynomial on the union of these families. Its degrees are therefore the union of the component degree multisets, also as in [F5]. If $I_j$ is the ideal generated by positive-degree invariants in the $j$-th block, then the positive-degree ideal of ${⨂}_j\mathbb{C}[V_{\mathbb{C},j}{]}^{W_j}$ generates precisely $I={\sum }_j{I}_j\mathbb{C}[V_{\mathbb{C}}]$: every positive-degree pure tensor has at least one positive-degree factor, and each $I_j$ is generated by invariants of the full product. The tensor-product quotient is consequently $\mathbb{C}[V_{\mathbb{C}}]\mathrm{/}I\cong {⨂}_j(\mathbb{C}[V_{\mathbb{C},j}]\mathrm{/}{I}_j)$, which is the asserted tensor decomposition of coinvariant algebras. By [F5] the product of all degrees is $\mathrm{\mid }W\mathrm{\mid }={\prod }_j\mathrm{\mid }{W}_j\mathrm{\mid }$. There is no common Coxeter number for unequal component orders, and this clause uses no global $h$.

For each $i$ and all $y\in V_{\mathbb{C}}$, $p_i(Cy)=p_i(y)$ because $p_i$ is invariant [F1]. Differentiate this identity at $y=x$ in an arbitrary direction $v$ to get $d{p}_i(Cx)(Cv)=d{p}_i(x)(v)$, or $d{p}_i(Cx)\circ C=d{p}_i(x)$. Since $Cx={\zeta }_h^{-1}x$ and $p_i$ is homogeneous of degree $d_i$, $d{p}_i({\zeta }_h^{-1}x)={\zeta }_h^{-(d_i-1)}d{p}_i(x)$; hence $d{p}_i(x)\circ C={\zeta }_h^{d_i-1}d{p}_i(x)={\zeta }_h^{e_i}d{p}_i(x)$. These covectors form a basis by 1.1, so the eigenvalue multiset of the transpose action is $\{{\zeta }_h^{e_i}{\}}_{i=1}^n$. A matrix and its transpose have the same characteristic polynomial, so this is also the eigenvalue multiset of $C$, with multiplicity [F1, F2, F6, step 1.1, algebra]. By [F3], $C-I$ is invertible, so $1$ is not an eigenvalue. Thus each $e_i$ is congruent modulo $h$ to a unique residue in $\{1,\dots ,h-1\}$, and these residues, counted with multiplicity, are exactly the spectral exponents $r_1,\dots ,r_n$ of the statement [F2, F3, F6].

The matrix $C$ is real by [F2], so its nonreal eigenvalues pair as ${\zeta }_h^r,{\zeta }_h^{-r}={\zeta }_h^{h-r}$ with equal multiplicity; each such pair contributes $h$ to the sum of residues. The only real $h$-th roots of unity are $1$ and, when $h$ is even, $-1$; $1$ is excluded by 2.1, while each occurrence of $-1={\zeta }_h^{h\mathrm{/}2}$ contributes $h\mathrm{/}2$. Partitioning the full eigenvalue multiset into these pairs and occurrences of $-1$ therefore gives ${\sum }_i{r}_i=nh\mathrm{/}2$ [F2, F3, F6, algebra]. By [F4], ${\sum }_i{e}_i=\mathrm{\mid }{\mathrm{\Phi }}_{+}\mathrm{\mid }=nh\mathrm{/}2$ as well. Since each positive integer $e_i$ is congruent to a residue $r$ in $\{1,\dots ,h-1\}$, it has the form $e_i=r+h{q}_i$ with an integer $q_i\ge 0$; the residue multiset and exponent multiset have the same cardinality, and the equal sums force every $q_i=0$. Thus the exponent multiset is exactly the spectral-residue multiset, proving (1)'s identification and sum claim.

The rotation on $P$ has eigenvalues ${\zeta }_h$ and ${\zeta }_h^{-1}={\zeta }_h^{h-1}$, so the spectral residues $1$ and $h-1$ both occur, counted with multiplicity (when $h=2$ these are the same residue and the eigenvalue occurs with multiplicity two on $P$). By 3.1 the same residues occur among the $e_i$. Therefore ${\min }_i{e}_i=1$ and ${\max }_i{e}_i=h-1$, which gives ${\min }_i{d}_i=2$ and ${\max }_i{d}_i=h$.

Add $1$ to each spectral exponent from the two computed spectrum suppliers. The classical lists give $A_n:2,3,\dots ,n+1$, $B_n:2,4,\dots ,2n$, $D_n:2,4,\dots ,2n-2$ together with $n$, and $I_2(m):2,m$. The exceptional lists give $E_6:2,5,6,8,9,12$, $E_7:2,6,8,10,12,14,18$, $E_8:2,8,12,14,18,20,24,30$, $F_4:2,6,8,12$, $H_3:2,6,10$, and $H_4:2,12,20,30$ [F8, step 3.1]. The classification's coincidence conventions identify $A_2=I_2(3)$, $B_2=I_2(4)$, $G_2=I_2(6)$, $A_3=D_3$, and, with the extended notation, $A_1\times A_1=I_2(2)$; for the last case the degrees are the union $\{2,2\}$ [F7]. The product-degree formula ${\prod }_i{d}_i=\mathrm{\mid }W\mathrm{\mid }$ is [F5], so it holds for every listed irreducible type and for the reducible unions.

If $n=0$, then $W$ is trivial, the invariant and coinvariant algebras are $\mathbb{C}$, and all degree, exponent and product assertions are empty, with empty products equal to $1$ [F1, F5]. If $n=1$, the irreducible system is $A_1$: its generator acts by $x\mapsto -x$, so the invariant ring is $\mathbb{C}[x^2]$, the basic degree is $2$, $c=s$ has order $h=2$ and eigenvalue $-1={\zeta }_2$, the sole exponent/residue is $1$, $\mathrm{\mid }{\mathrm{\Phi }}_{+}\mathrm{\mid }=1=nh\mathrm{/}2$, and the coinvariant algebra is $\mathbb{C}[x]\mathrm{/}(x^2)$. Thus all claims hold directly; for reducible rank one factors the componentwise argument of 1.2 applies [F1, F5, F7, algebra]. The Axiom of Choice is used only for the basic-family and invariant-theory supplier conclusions in [F1] and [F5]; the plane, derivative, residue and tensor calculations above use no further choice [F1, F5, def-axiom-of-choice].