The total degree sum, the invariant Jacobian as the discriminant, anti-invariants, and the top coinvariant class

If $n=0$, then $W=\{1\}$, $S_0=R=A=\mathbb{C}$, and every family and product is empty; all clauses follow. If $n=1$, the Coxeter group is $W=\{1,s\}$, $T=\{s\}$ and $N=1$. For its single degree $d_1$, [F3] gives $\frac{1}{1-t^{d_1}}=\frac{1}{2}(\frac{1}{1-t}+\frac{1}{1+t})$. With $\delta =1-t$, the left side is $d_1^{-1}{\delta }^{-1}+(d_1-1)\mathrm{/}(2d_1)+O(\delta )$ and the right side is $\frac{1}{2}{\delta }^{-1}+\frac{1}{4}+O(\delta )$; comparing the ${\delta }^{-1}$ and constant coefficients gives $d_1=2$ and $d_1-1=1=N$. In rank one, write the unit positive-root form as $\mathrm{\Delta }=\epsilon x_1$ with $\epsilon =\pm 1$. Invariance under $x_1\mapsto -x_1$ gives $R=\mathbb{C}[x_1^2]$; the degree-two basic generator is $f_1=a{x}_1^2$ with $a\mathrm{\ne }0$, so $I=(x_1^2)$, $\mathbf{J}=2a{x}_1=(2a\mathrm{/}\epsilon )\mathrm{\Delta }$, and the anti-invariant polynomials are precisely the odd polynomials $\mathrm{\Delta }R$. The quotient has basis $1,[x_1]$, so its top component is the nonzero sign line spanned by $[\mathrm{\Delta }]$. This proves every clause for $n=1$. In the rest of the proof assume $n\ge 2$.

For $n\ge 2$, let $g\mathrm{\ne }1$ have fixed space of codimension one in $V_{\mathbb{C}}$. Because its matrix is real, the real and complex fixed spaces have the same codimension, so $H:={\mathrm{Fix}}_V(g)$ is a real hyperplane. If $H$ were not in the finite reflecting arrangement, a point $x\in H$ outside every arrangement hyperplane could be chosen: for each proper subspace cut out on $H$, choose a nonzero linear form vanishing there; their product is a nonzero polynomial on $H$, which cannot vanish everywhere over $\mathbb{R}$ (by induction on $\dim H$). It would lie in a chamber interior, whose point stabilizer is trivial by [F2], contrary to $gx=x$. Hence $H$ is an arrangement hyperplane. The same finite-union argument chooses $x\in H$ outside all other distinct arrangement hyperplanes; $x$ is nonzero. By [F2], $x\in w_0{C}_I$ for some $I⊊S$. Since $x$ lies on an arrangement hyperplane, $I$ is nonempty; the walls indexed by $I$ are distinct because their normals are the images of distinct basis vectors under the invertible map $\rho (w_0)$. They all pass through $x$, so $\mathrm{\mid }I\mathrm{\mid }=1$. For $I=\{s\}$, [F2] gives ${\mathrm{Stab}}_W(x)=w_0\{1,s\}{w}_0^{-1}$; since $g\mathrm{\ne }1$ fixes $x$, it is the reflection $w_{0}s{w}_0^{-1}\in T$. Conversely every element of $T$ has a fixed hyperplane by [F1]. Thus the nonidentity elements with fixed-space codimension one are exactly the reflections.

Assume $n\ge 2$ and put $\delta =1-t$. The left side of [F3] expands as ${\prod }_i(1-t^{d_i}{)}^{-1}=\frac{{\delta }^{-n}}{{\prod }_i{d}_i}(1+\frac{\delta }{2}{\sum }_i(d_i-1)+O({\delta }^2))$. In the normalized sum on its right, the identity contributes $\mathrm{\mid }W{\mathrm{\mid }}^{-1}{\delta }^{-n}$. Each of the $N$ reflections contributes $\mathrm{\mid }W{\mathrm{\mid }}^{-1}{\delta }^{-(n-1)}\mathrm{/}(1+t)=\frac{1}{2\mathrm{\mid }W\mathrm{\mid }}{\delta }^{-(n-1)}+O({\delta }^{-(n-2)})$. For every other element, [F5] and step 1.2 give at most $n-2$ eigenvalues equal to $1$ on $V_{\mathbb{C}}^{\ast }$: indeed $\mathrm{rank}(M^{-\mathsf{T}}-I)=\mathrm{rank}(M-I)$, so the fixed dimensions on a representation and its dual agree. Its Molien term is therefore $O({\delta }^{-(n-2)})$. Comparing the coefficients of ${\delta }^{-n}$ and ${\delta }^{-(n-1)}$ first gives ${\prod }_i{d}_i=\mathrm{\mid }W\mathrm{\mid }$ and then ${\sum }_i(d_i-1)=N$. The identity is a finite sum of rational functions, so these Laurent expansions compare coefficients without an infinite-limit interchange.

Write $M={\rho }_{\mathbb{C}}(w)$ in the chosen orthonormal coordinates. The tuple $F=(f_1,\dots ,f_n)$ satisfies $F(Mx)=F(x)$; differentiating gives $DF(Mx)M=DF(x)$ and hence $\mathbf{J}(Mx)=\det (M{)}^{-1}\mathbf{J}(x)=\det (M)\mathbf{J}(x)$, since $M$ is real orthogonal. If $x$ lies on the fixed hyperplane of $t_{\alpha }$, then $\mathbf{J}(x)=\mathbf{J}({\rho }_{\mathbb{C}}(t_{\alpha })x)=-\mathbf{J}(x)$, so $\mathbf{J}$ vanishes there. After taking ${\mathrm{\ell }}_{\alpha }$ as one coordinate, restriction to ${\mathrm{\ell }}_{\alpha }=0$ is the zero polynomial; thus ${\mathrm{\ell }}_{\alpha }$ divides $\mathbf{J}$. If ${\mathrm{\ell }}_{\alpha }$ and ${\mathrm{\ell }}_{\beta }$ are proportional, nondegeneracy of $B_{\mathbb{C}}$ gives $\alpha =c\beta$; since both roots are real with $B$-norm one, $c=\pm 1$, and positivity of both roots gives $c=1$, so $\alpha =\beta$. Thus the forms are pairwise nonproportional primes by [F6], and $\mathrm{\Delta }$ divides $\mathbf{J}$. Every determinant term of the Jacobian matrix has degree ${\sum }_i(d_i-1)$; [F4] makes this the degree of its nonzero determinant. Since $\mathrm{deg}\mathrm{\Delta }=N$, step 2.1 gives $\mathrm{deg}\mathbf{J}=\mathrm{deg}\mathrm{\Delta }$, so $\mathbf{J}=c\mathrm{\Delta }$ for a nonzero scalar $c$. This proves (2), including anti-invariance of $\mathrm{\Delta }$. If an orthonormal basis changes by $P$, its coordinate expressions satisfy ${\mathrm{\Delta }}_{\mathrm{n}\mathrm{e}\mathrm{w}}(y)={\mathrm{\Delta }}_{\mathrm{o}\mathrm{l}\mathrm{d}}(Py)$ and ${\mathbf{J}}_{\mathrm{n}\mathrm{e}\mathrm{w}}(y)=\det (P){\mathbf{J}}_{\mathrm{o}\mathrm{l}\mathrm{d}}(Py)$. Listing the fixed roots in a different order leaves the product unchanged; replacing ${\mathrm{\Phi }}_{+}$ by $-{\mathrm{\Phi }}_{+}$ multiplies it by $(-1{)}^N$.

If $p\in S_0^{\det }$, every reflection $t_{\alpha }$ acts by determinant $-1$, so $p$ vanishes on its fixed hyperplane and each ${\mathrm{\ell }}_{\alpha }$ divides $p$. The same pairwise-prime argument as in step 3.1 gives $p=\mathrm{\Delta }q$ for some $q\in S_0$. By step 3.1, $\mathrm{\Delta }$ is anti-invariant; applying any $w$ to $p=\mathrm{\Delta }q$ and cancelling $\mathrm{\Delta }$ in the domain $S_0$ then gives $w\cdot q=q$. Conversely, if $q\in R$, the product $\mathrm{\Delta }q$ is anti-invariant. The quotient $q$ is unique because $S_0$ is a domain and $\mathrm{\Delta }\mathrm{\ne }0$.

By [F3] the Hilbert series of $A$ is ${\prod }_i(1+t+\cdots +t^{d_i-1})$, whose top degree is ${\sum }_i(d_i-1)=N$ by step 2.1 and whose top coefficient is $1$, so $A_N$ is one-dimensional. For $f={\sum }_{\mathrm{\mid }a\mathrm{\mid }=N}{f}_a{x}^a$ and $g={\sum }_{\mathrm{\mid }a\mathrm{\mid }=N}{g}_a{x}^a$ in $(S_0{)}_N$, set $⟨f,g⟩:={\sum }_{\mathrm{\mid }a\mathrm{\mid }=N}a!f_a\overline{g_a}$ with $a!:={\prod }_j{a}_j!$; this is positive definite. If $p\in R_d$ is homogeneous with $d>0$ and $f\in (S_0{)}_{N-d}$, direct monomial expansion gives $⟨pf,\mathrm{\Delta }⟩=⟨f,\bar{p}(\mathrm{\partial })\mathrm{\Delta }⟩$, where $\bar{p}(\mathrm{\partial })$ conjugates the coefficients of $p$ and substitutes the formal partial derivatives for its variables. For a real orthogonal matrix $M$, the chain rule, first on linear symbols and then by products and linearity, gives $\bar{p}({\mathrm{\partial }}_x)(q\circ M)=(\bar{p}(M^{\mathsf{T}}\mathrm{\partial })q)\circ M$. Since $M^{\mathsf{T}}=M^{-1}\in W$ and real matrices preserve coefficientwise conjugation, $\bar{p}$ is invariant and $\bar{p}(M^{\mathsf{T}}\mathrm{\partial })=\bar{p}(\mathrm{\partial })$; thus this differential operator commutes with substitution by $M$. Since $\mathrm{\Delta }\circ M=\det (M)\mathrm{\Delta }$ by step 3.1, $\bar{p}(\mathrm{\partial })\mathrm{\Delta }$ is anti-invariant. If $d\le N$ it has degree $N-d<N$, so step 4.1 forces it to be zero; if $d>N$ the derivative is already zero. Every homogeneous element of $I_N$ is a sum of such products $pf$, hence $\mathrm{\Delta }$ is orthogonal to $I_N$. But $\mathrm{\Delta }$ is a nonzero real-coefficient polynomial, so $⟨\mathrm{\Delta },\mathrm{\Delta }⟩>0$; consequently $\mathrm{\Delta }\mathrm{\notin }{I}_N$ and $[\mathrm{\Delta }]\mathrm{\ne }0$ in $A_N$. It spans this one-dimensional component and has the determinant action by step 3.1.