Proof. Putting together the isomorphisms $S(V^*)\simeq S({E^*})\otimes {\mathcal {C}_N}$ as graded N-modules, $S({E^*})\simeq S({E^*})^H\otimes {\mathcal {C}_H}$ as bigraded H-modules (equivalently, as bigraded G-modules of N-invariants) and $S(V^*)^G=S({E^*})^H$ , we obtain the isomorphism $S(V^*)\simeq S(V^*)^G\otimes {\mathcal {C}_H}\otimes {\mathcal {C}_N}$ as graded G-modules. Letting $\pi :S(V^*)\rightarrow S(V^*)/{I_G^+}$ denote the canonical projection, we see that $\mathbb {C}\otimes {\mathcal {C}_H}\otimes {\mathcal {C}_N}$ must surject onto the image $S(V^*)/{I_G^+}\simeq {\mathcal {C}_G}$ , because $S(V^*)^G$ is generated as a $\mathbb {C}$ -algebra by the generators of the ideal ${I_G^+}$ . But this surjection ${\mathcal {C}_H}\otimes {\mathcal {C}_N}\rightarrow {\mathcal {C}_G}$ of graded G-modules must then be an isomorphism, because

$$ \begin{align*}\mathrm{dim}_{\mathbb{C}}({\mathcal{C}_H}\otimes{\mathcal{C}_N})=\mathrm{dim}_{\mathbb{C}}({\mathcal{C}_H})\cdot\mathrm{dim}_{\mathbb{C}}({\mathcal{C}_N})=|H|\cdot|N|=|G|=\mathrm{dim}_{\mathbb{C}}({\mathcal{C}_G}).\end{align*} $$

Because ${\mathcal {C}_H}\subset S({E^*})$ consists of N-invariants, we have $({\mathcal {C}_H}\otimes {\mathcal {C}_N})^N={\mathcal {C}_H}\otimes ({\mathcal {C}_N})^N={\mathcal {C}_H}\otimes \mathbb {C}$ , and therefore $({\mathcal {C}_G})^N\simeq {\mathcal {C}_H}$ as graded H-modules, as claimed.

Proof. Let $u_1,\dots ,u_m$ be a homogeneous basis for ${(U_M^N)^*}:=({\mathcal {C}_N}\otimes M^*)^N$ . We may assume that $\mathrm {deg}(u_i)=e_i^N(M)=\mathrm {deg}(\tilde {u}_i)$ for each $i=1,\dots ,m$ . Let $y_1,\dots ,y_m$ be a basis for $M^*$ , and write $\tilde {u}_i:=\sum _{j=1}^m\tilde {a}_{ij}\otimes y_j$ and $u_i=\sum _{j=1}^ma_{ij}\otimes y_j$ , where $\tilde {a}_{ij}\in S(V^*)$ and $a_{ij}\in {\mathcal {C}_N}$ and every nonzero $\tilde {a}_{ij}$ and $a_{ij}$ is homogeneous of degree $e_i^N(M)$ . Because the $\tilde {u}_i$ and the $u_i$ form bases for $(S(V^*)\otimes M^*)^N$ as a free $S(V^*)^N$ -module, there exists a matrix $[p_{ij}]\in \mathrm {GL}_m(S(V^*)^N)$ such that $[\tilde {a}_{ij}]=[p_{ij}]\cdot [a_{ij}]$ . Because the kernel of $\eta _N:S(V^*)\rightarrow {\mathcal {C}_N}$ is precisely the ideal generated by homogeneous N-invariants of positive degree, it follows that $[\eta _N(\tilde {a}_{ij})]=[p_{ij}(0)]\cdot [a_{ij}]$ , where $p_{ij}(0)$ denotes the evaluation at $0\in V$ . Because $\mathrm {deg}(\mathrm {det}[\tilde {a}_{ij}])=\sum _{i=1}^m e_i^N(M)=\mathrm {deg}(\mathrm {det}[a_{ij}])$ , it follows that $\mathrm {det}[p_{ij}]\in \mathbb {C}^\times $ , and therefore $[p_{ij}(0)]\in \mathrm {GL}_m(\mathbb {C})$ . Because $\eta _N\otimes 1$ is G-equivariant and $(\tilde {U}_M^N)^*$ is G-stable, $\eta _N\otimes 1:(\tilde {U}_M^N)^*\rightarrow {(U_M^N)^*}$ is a graded isomorphism of G-modules.

Proof. Applying Proposition 3.3, we see that

$$ \begin{align*}({\mathcal{C}_G}\otimes M^*)^G\simeq(({\mathcal{C}_H}\otimes{\mathcal{C}_N}\otimes M^*)^N)^H=({\mathcal{C}_H}\otimes({\mathcal{C}_N}\otimes M^*)^N)^H=({\mathcal{C}_H}\otimes {(U_M^N)^*})^H.\end{align*} $$

Let m be the rank of M, and let $y_1,\dots ,y_m$ be a basis of $M^*$ . Let $a_{ij}^N\in {\mathcal {C}_N}$ be $\mathbf {x}$ -homogeneous such that $u_i^N:=\sum _{j=1}^ma_{ij}^N\otimes y_j$ form a basis for ${(U_M^N)^*}$ with $\mathrm {deg}_{\mathbf {x}}(u_i^N)=e_i^N(M)$ . Letting ${\mathcal {E}_M^N}:=\{e_1^N(M),\dots ,e_m^N(M)\}$ denote the set of M-exponents of N, we see that

$$ \begin{align*}{(U_M^N)^*}\simeq\bigoplus_{e\in{\mathcal{E}_M^N}}{(U_M^N)^*_e}\end{align*} $$

as an H-module. Therefore, there exist $\mathbf {x}$ -homogeneous $a_{ij}^H\in {\mathcal {C}_H}$ such that $u_i^G:=\sum _{j=1}^ma_{ij}^H\otimes u_j^N$ form a basis for $({\mathcal {C}_H}\otimes {(U_M^N)^*})^H$ with $\mathrm {deg}_{\mathbf {x}}(u_i^G)=e_i^G(U_M^N)$ and such that $a_{ij}^H=0$ whenever $\mathrm {deg}_{\mathbf {x}}(u_j^N)\neq e_i^N(M)$ (in other words, the $[a_{ij}^H]$ may be chosen to be square-diagonal corresponding to the graded decomposition of ${(U_M^N)^*}$ ). But then the $u_i^G$ form an $\mathbf {x}$ -homogeneous basis for $({\mathcal {C}_G}\otimes M^*)^G\simeq ({\mathcal {C}_H}\otimes (U_M^N)^*)^H$ , and we see that

$$ \begin{align*}e_i^G(M)=\mathrm{deg}_{\mathbf{x}}(u_i^G)=e_i^G(U_M^N)+e_i^N(M).\\[-32pt]\end{align*} $$

Proof. The M-exponents $e_i^G(M)$ of G are obtained as the $\mathbf {x}$ -degrees of any $\mathbf {x}$ -homogeneous basis for $({\mathcal {C}_G}\otimes M^*)$ , and the M-exponents $e_i^H(M)$ of H are analogously obtained as the $\mathbf {N}$ -degrees of any (bi)homogeneous basis for $({\mathcal {C}_H}\otimes M^*)^H$ . Letting m denote the rank of M, the exponent $e_1^G(\bigwedge ^m M)$ is the $\mathbf {x}$ -degree of a basis element for $({\mathcal {C}_G}\otimes \bigwedge ^m M^*)^G$ , and the exponent $e_i^H(\bigwedge ^m M)$ is the $\mathbf {N}$ -degree of a basis element for $({\mathcal {C}_H}\otimes \bigwedge ^m M^*)^H$ . Because both M and ${\mathcal {C}_H}$ are N-invariant, and ${\mathcal {C}_G}\simeq {\mathcal {C}_H}\otimes {\mathcal {C}_N}$ as graded G-modules such that $({\mathcal {C}_G})^N\simeq {\mathcal {C}_H}$ by Proposition 3.3, we obtain

$$ \begin{align*}({\mathcal{C}_G}\otimes M^*)^G=(({\mathcal{C}_G}\otimes M^*)^N)^H= (({\mathcal{C}_G})^N\otimes M^*)^H\simeq ({\mathcal{C}_H}\otimes M^*)^H,\end{align*} $$

which establishes (i).

Applying (i) to $\bigwedge ^mM^*$ , we have that $({\mathcal {C}_G}\otimes {\textstyle \bigwedge ^m}M^*)^G\simeq ({\mathcal {C}_H}\otimes {\textstyle \bigwedge ^m}M^*)^H$ . By Lemma 3.9, the amenability of M as a G-module and the amenability of M as an H-module are respectively equivalent to

$$ \begin{align*}{\textstyle\bigwedge^m}(({\mathcal{C}_G}\otimes M^*)^G)\simeq({\mathcal{C}_G}\otimes{\textstyle\bigwedge^m}M^*)^G\quad\text{and}\quad{\textstyle\bigwedge^m}(({\mathcal{C}_H}\otimes M^*)^H)\simeq({\mathcal{C}_H}\otimes{\textstyle\bigwedge^m}M^*)^H,\end{align*} $$

which proves (ii).

Proof. Let m denote the rank of M. Define the amenability defects

$$ \begin{gather*} \gamma:=\sum_{i=1}^m e_i^G(M)-e_1^G({\textstyle\bigwedge^m} M);\qquad \nu:=\sum_{i=1}^me_i^N(M)-e_1^N({\textstyle\bigwedge^m}M);\qquad \text{and}\\ \eta:=\sum_{i=1}^m e_i^G(U_M^N)-e_1^G({\textstyle\bigwedge^m}U_M^N), \end{gather*} $$

so that M is amenable as a G-module if and only if $\gamma =0$ , M is amenable as an N-module if and only if $\nu =0$ and ${U_M^N}$ is amenable as a G-module if and only if $\eta =0$ . By Lemma 3.10, ${U_M^N}$ is also amenable as an H-module if and only if $\eta =0$ . By [Reference Orlik and Solomon16, Lemma 2.8] (see also Remark 3.8), in any case we have that $\gamma ,\nu ,\eta \geq 0$ .

Let $W_M^N$ be the dual of $(W_M^N)^*:=({\mathcal {C}_N}\otimes {\textstyle \bigwedge ^m}M^*)^N$ (instead of the more natural but cumbersome $U_{\bigwedge ^m\!\! M}^N$ in lieu of $W_M^N$ ). From Corollary 3.7, we know that

$$ \begin{align*}e_i^N(M)+e_i^G(U_M^N)=e_i^G(M)\qquad\text{and}\qquad e_1^N({\textstyle\bigwedge^m}M)+e_1^G(W_M^N)=e_1^G({\textstyle\bigwedge^m}M).\end{align*} $$

Summing the first equation over $i=1,\dots ,m$ and subtracting the second equation, we obtain

(10) $$ \begin{align}\nu+\eta+e_1^G({\textstyle\bigwedge^m}U_M^N)=\gamma+e_1^G(W_M^N).\end{align} $$

Consider now the natural inclusion of graded G-modules

$$ \begin{align*}S({E^*})\otimes{\textstyle\bigwedge^m}({\mathcal{C}_N}\otimes M^*)^N\hookrightarrow S({E^*})\otimes{\textstyle}({\mathcal{C}_N}\otimes{\textstyle\bigwedge^m}M^*)^N\end{align*} $$

in $(S(V^*)\otimes \bigwedge M^*)^N$ , which identifies $\bigwedge ^m(U_M^N)^*$ with $a\otimes (W_M^N)^*$ as graded G-modules for some $\mathbf {x}$ -homogeneous $a\in S({E^*})$ with $\mathrm {deg}_{\mathbf {x}}(a)=\nu $ (cf. Remark 3.8). From this it follows that $e_1^G(\bigwedge ^m U_M^N)+\nu \geq e_1^G(W_M^N)$ , which together with Equation (10) implies that $\eta \leq \gamma $ . Therefore, if M is amenable as a G-module, then $U_M^N$ is amenable. If M is amenable as an N-module, so that $\nu =0$ , then we see that $\bigwedge ^m{(U_M^N)^*}\simeq (W_M^N)^*$ as G-modules. Therefore, $e_1^G(\bigwedge ^m{U_M^N})= e_1^G(W_M^N)$ , and we obtain from Equation (10) that $\eta =\gamma $ in this case. Hence, if M is amenable as an N-module and ${U_M^N}$ is amenable, then M is amenable as a G-module.

Proof. We proceed as in the proof of Corollary 3.7: let $y_1,\dots ,y_m$ be a basis of $M^*$ . Let $a_{ij}^N\in {\mathcal {C}_N}$ be $\mathbf {x}$ -homogeneous such that $u_i^N:=\sum _{j=1}^ma_{ij}^N\otimes y_j$ form a basis for the $e_i^N(M)$ -homogeneous component of ${(U_M^N)^*}$ , and choose $\mathbf {x}$ -homogeneous $a_{ij}^H\in {\mathcal {C}_H}$ such that $u_i^G:=\sum _{j=1}^ma_{ij}^H\otimes u_j^N$ form a basis for $({\mathcal {C}_H}\otimes (U_M^N)^*_{e_i^N(M)})^H$ with $\mathrm {deg}_{\mathbf {x}}(u_i^G)=e_i^G(U_M^N)$ (where again $a_{ij}^H=0$ whenever $\mathrm {deg}_{\mathbf {x}}(u_j^N)\neq e_i^N(M)$ ). But then the $u_i^G$ form an $\mathbf {x}$ -homogeneous basis for $({\mathcal {C}_H}\otimes (U_M^N)^*)^H$ . Because ${U_M^N}$ consists of N-invariants, by Lemma 3.10 we have that $({\mathcal {C}_G}\otimes {(U_M^N)^*})^G\simeq ({\mathcal {C}_H}\otimes {(U_M^N)^*})^H$ . By Proposition 3.12, because M is amenable as a G-module, ${U_M^N}$ is amenable. Hence, by Theorem 2.7 we have that

$$ \begin{align*}(S(V^*)\otimes{\textstyle\bigwedge}{(U_M^N)^*})^G\simeq S(V^*)^G\otimes{\textstyle \bigwedge}({\mathcal{C}_G}\otimes{(U_M^N)^*})^G.\end{align*} $$

Because $({\mathcal {C}_G}\otimes {(U_M^N)^*})^G\simeq {\mathrm {span}}_{\mathbb {C}}\{u_1^G,\dots ,u_m^G\}$ and $u_i^G\in S(V^*)_{e_i^G({U_M^N})}\otimes (U_M^N)^*_{e_i^N(M)}$ , our result follows.

Proof. For $g\in G$ , let us write

$$ \begin{align*}\mathcal{P}^g_M(x,y,u)=\sum_{\ell,e,p\geq 0}\mathrm{tr}\bigl(g|S(V^*)_\ell\otimes({\textstyle\bigwedge^p}{(U_M^N)^*})_e\bigr)x^\ell y^e u^p,\end{align*} $$

so that $\mathcal {P}^G_M(x,y,u)=\frac {1}{|G|}\sum _{g\in G}\mathcal {P}^g_M(x,y,u)$ and

$$ \begin{align*}\mathcal{P}^g_M(x,y,u)=\left(\sum_{\ell\geq 0}\mathrm{tr}\bigl(g|S(V^*)_\ell\bigr)x^\ell\right)\left(\sum_{e,p\geq 0}\mathrm{tr}\bigl(g|({\textstyle\bigwedge^p}{(U_M^N)^*})_e\bigr)y^eu^p\right).\end{align*} $$

We know that $ \sum _{\ell \geq 0}\mathrm {tr}\bigl (g|S(V^*)_\ell \bigr )x^\ell =\mathrm {det}(1-xg^{-1}|V)^{-1}$ . On the other hand, because ${(U_M^N)^*}\simeq \bigoplus _{e\in {\mathcal {E}_M^N}}{(U_M^N)^*_e}$ , we have that $\bigwedge {(U_M^N)^*}\simeq \bigotimes _{e\in {\mathcal {E}_M^N}}\bigwedge {(U_M^N)^*_e}$ as bigraded G-modules. Hence, for each $g\in G$ ,

$$ \begin{align*}\sum_{e,p\geq 0}\mathrm{tr}\bigl(g|({\textstyle\bigwedge^p}{(U_M^N)^*})_e\bigr)y^eu^p=\prod_{e\in{\mathcal{E}_M^N}}\left(\sum_{p\geq 0}\mathrm{tr}\bigl(g|{\textstyle\bigwedge^p}{(U_M^N)^*_e}\bigr)y^{ep}u^p\right).\end{align*} $$

For each $e\in {\mathcal {E}_M^N}$ we have that $\sum _{p\geq 0}\mathrm {tr}\bigl (g|{\textstyle \bigwedge ^p}{(U_M^N)^*_e}\bigr )y^{ep}u^p=\mathrm {det}(1+y^eug^{-1}|{(U_M^N)_e})$ . Hence, for each $g\in G$ ,

$$ \begin{align*} \mathcal{P}^g_M(x,y,u)=\frac{\prod_{e\in{\mathcal{E}_M^N}}\mathrm{det}(1+y^eug^{-1}|{(U_M^N)_e})}{\mathrm{det}(1-xg^{-1}|V)}.\end{align*} $$

Our result follows after taking the average over $g\in G$ on each side.

Proof. The first equality is Corollary 3.7 applied to the G-module $M={V^\sigma }$ , and the last equality follows from the observations in Remark 3.1.

Let us establish the second equality. Let ${N}^\sigma _1,\dots ,{N}^\sigma _r$ denote a basis for ${(E^\sigma )^*}$ as a G-module. We will show that there exist $a_{ij}\in S({E^*})$ such that $u_i:=\sum _{j=1}^r a_{ij}\otimes {N}_j^\sigma $ form an $\mathbf {N}$ -homogeneous basis for $({\mathcal {C}_H}\otimes {(E^\sigma )^*})^H$ (so that each nonzero $a_{ij}$ is $\mathbf {N}$ -homogeneous of $\mathbf {N}$ -degree $e_i^H({E^\sigma })$ ) and, moreover, $a_{ij}=0$ whenever $d_j^N\neq d_i^N$ and each $a_{ij}\in S(\mathbf {N}_{d_i^N})$ , where as before $\mathbf {N}_d$ denotes the set of fundamental N-invariants of degree d. Because $({\mathcal {C}_G}\otimes {(E^\sigma )^*})^G\simeq ({\mathcal {C}_H}\otimes {(E^\sigma )^*})^H$ by Lemma 3.10, the existence of such $a_{ij}$ will establish our claim, because each nonzero $a_{ij}$ as above will then be $\mathbf {x}$ -homogeneous of $\mathbf {x}$ -degree $e_i^G({E^\sigma })= d_i^N\cdot e_i^H({E^\sigma })$ .

As in Remark 3.1, let us write ${\mathcal {D}_N}=\{d_1^N,\dots ,d_r^N\}$ for the set of degrees of N and ${E^*_d}$ for the graded component of ${E^*}$ spanned by the fundamental N-invariants of degree $d\in {\mathcal {D}_N}$ . To simplify notation, let us write ${E}_d$ for the graded dual of ${E^*_d}$ , instead of ${E}_{-d}$ , so that ${E}\simeq \bigoplus _{d\in {\mathcal {D}_N}}{E}_d$ , and similarly the Galois twist ${E^\sigma }\simeq \bigoplus _{d\in {\mathcal {D}_N}}{E}^\sigma _d$ . We saw in Remark 3.1 that H decomposes as a direct product $\unicode{x2A09} _{d\in {\mathcal {D}_N}}H_{(d)}$ , where each $H_{(d)}$ is a reflection group acting on ${E}_d$ and $H_{(d)}$ acts trivially on ${E}^\sigma _{d'}$ whenever $d\neq d'$ . We then see that $S({E^*})\simeq \bigotimes _{d\in {\mathcal {D}_N}} S({E^*_d})$ and each $S({E^*_d})\simeq S({E^*_d})^{H_{(d)}}\otimes \mathcal {C}_{H_{(d)}}$ as H-modules, so that in particular ${\mathcal {C}_H}\simeq \bigotimes _{d\in {\mathcal {D}_N}}\mathcal {C}_{H_{(d)}}$ as H-modules, where again $H_{(d)}$ acts trivially on $\mathcal {C}_{H_{(d')}}$ for $d\neq d'$ . It follows from the above observations that

so that we may indeed choose $a_{ij}\in \mathcal {C}_{H_{(d_i^N)}}\subset S({E}_{d_i^N}^*)$ such that the $\mathbf {N}$ -homogeneous $u_i=\sum _{j=1}^ra_{ij}\otimes {N}_j^\sigma \in (\mathcal {C}_{H_{(d_i^N)}}\otimes ({E}^\sigma _{d_i^N})^*)^{H_{(d_i^N)}}$ (i.e., with $a_{ij}=0$ whenever $d_i^N\neq d_j^N$ ) form a basis for $({\mathcal {C}_H}\otimes {(E^\sigma )^*})^H$ that is simultaneously $\mathbf {N}$ -homogeneous of $\mathbf {N}$ -degree $e_i^H({E^\sigma })$ and $\mathbf {x}$ -homogeneous of $\mathbf {x}$ -degree $e_i^G({E^\sigma })=d_i^N\cdot e_i^H({E^\sigma })$ .

Proof Proof of Theorem 1.4

By Corollaries 4.5 and 4.11, both sides are equal to $\lim _{x\to 1} \ |G|\cdot \mathcal {P}^G_{\sigma }\Bigl (x,x^t,qt(1-x)-1\Bigr )$ .

Proof. We compute:

$$ \begin{align*} &|G|\prod_{i=1}^r\frac{1+x^{e_i^G({U^N_\sigma}) }y^{e_i^N({V^\sigma})}u}{1-x^{d_i^G}}\Bigg|_{\substack{y=x^t \\ u=qt(1-x)-1\\ x\to 1}} \\&= |G|\lim_{x \to 1} \prod_{i=1}^r \left(\frac{x^{e_i^G({U^N_\sigma})+t e_i^N({V^\sigma}) }qt(1-x)}{1-x^{d_i^G}}+\frac{1-x^{e_i^G({U^N_\sigma})+t e_i^N({V^\sigma})}}{1-x^{d_i^G}}\right)\\ &=\prod_{i=1}^r \left(qt+e_i^N({V^\sigma}) t + e_i^G({U^N_\sigma})\right).\\[-44pt] \end{align*} $$

Proof. For each $n\in N$ ,

$$ \begin{align*}&\sum_{\ell,e,p\geq 0}\mathrm{tr}(ng|(S(V^*)_\ell\otimes({\textstyle\bigwedge^p}{(U^N_\sigma)^*})_e))x^\ell y^eu^p\\&\qquad\qquad\qquad =\left(\sum_{\ell\geq 0}\mathrm{tr}(ng|S(V^*)_\ell)x^\ell\right)\cdot\prod_{e\in{\mathcal{E}^N_\sigma}}\left(\sum_{p\geq 0}\mathrm{tr}(ng|{\textstyle\bigwedge^p}{(U^N_\sigma)^*_e})y^{ep}u^p\right),\end{align*} $$

and our result follows after taking the average over $n\in N$ .

Proof. We write $\mathcal {P}^{Ng}_\sigma (x,y,u)$ as in Lemma 4.6. First observe that, because ${U^N_\sigma }$ is N-invariant, for any $n\in N$ we have that

$$ \begin{align*}\prod_{e\in{\mathcal{E}^N_\sigma}}\mathrm{det}(1+uy^{e}(ng|_{{(U^N_\sigma)^*_e}}))=\prod_{i=1}^r(1+\epsilon_i^g({U^N_\sigma})uy^{e_i^N({V^\sigma})}),\end{align*} $$

independently of $n\in N$ .

Let $\mathcal {D}_N=\{d_1^N,\dots ,d_r^N\}$ denote the set of degrees for N, and let ${E}^*_d={\mathrm {span}}_{\mathbb {C}}\mathbf {N}_d$ for $d\in \mathcal {D}_N$ , where as before $\mathbf {N}_d$ denotes the set of fundamental N-invariants having degree d. Because $S({E}^*)\simeq \bigotimes _{d\in \mathcal {D}_N}S({E}^*_d),$ we have that

$$ \begin{align*}\sum_{\ell\geq 0}\mathrm{tr}\bigl(g|S({E}^*)_\ell\bigr)x^\ell = \prod_{d\in\mathcal{D}_N}\Bigl(\sum_{\ell\geq 0}\bigl(\mathrm{tr}(g|\mathrm{Sym}^\ell({E}^*_d))(x^{d})^\ell\bigr)\Bigr), \end{align*} $$

where $S({E}^*)_\ell :=S({E}^*)\cap \mathrm {Sym}^\ell (V^*)=\mathrm {Sym}^\ell (V^*)^N$ , and $\mathrm {Sym}^\ell (V^*)$ denotes the $\ell $ th symmetric power of $V^*$ . On the other hand,

$$ \begin{align*}\prod_{d\in\mathcal{D}_N}\Bigl(\sum_{\ell\geq 0}\mathrm{tr}\bigl(g|\mathrm{Sym}^\ell({E}^*_{d})\bigr)(x^{d})^\ell\Bigr)=\prod_{d\in\mathcal{D}_N}\frac{1}{\mathrm{det}(1-x^{d}(g|_{{E}^*_{d}}))}=\prod_{i=1}^r\frac{1}{1-\epsilon_i^g({E})x_i^{d_i^N}}.\end{align*} $$

Therefore,

$$ \begin{align*}\sum_{\ell\geq 0}\mathrm{tr}\bigl(g|S({E}^*)_\ell\bigr)x^\ell=\prod_{i=1}^r\frac{1}{1-\epsilon_i^g({E})x_i^{d_i^N}}.\end{align*} $$

Because, for $n\in N$ ,

$$ \begin{align*}\sum_{\ell\geq 0}\mathrm{tr}\bigl(ng|\mathrm{Sym}^\ell(V^*)\bigr)x^\ell = \frac{1}{\mathrm{det}(1-x(ng|_{V^*}))},\end{align*} $$

it remains to show that

$$ \begin{align*}\frac{1}{|N|}\sum_{n\in N}\Bigl(\sum_{\ell\geq 0}\mathrm{tr}\bigl(ng|\mathrm{Sym}^\ell(V^*)\bigr)x^\ell\Bigr)=\sum_{\ell\geq 0}\mathrm{tr}\bigl(g|S({E}^*)_\ell\bigr)x^\ell\end{align*} $$

or, equivalently, that for each $\ell \geq 0$ we have that

$$ \begin{align*}\frac{1}{|N|}\sum_{n\in N}\Bigl(\mathrm{tr}\bigl(ng|\mathrm{Sym}^\ell(V^*)\bigr)\Bigr)=\mathrm{tr}\bigl(g|S({E}^*)_\ell\bigr).\end{align*} $$

To see this, note that the operator on $\mathrm {Sym}^\ell (V^*)$ given by

$$ \begin{align*}\frac{1}{|N|}\sum_{n\in N}ng=g\cdot\left(\frac{1}{|N|}\sum_{n\in N}n\right)=g\circ\mathrm{pr}^N_\ell,\end{align*} $$

where $\mathrm {pr}^N_\ell =\frac {1}{|N|}\sum _{n\in N} n$ is the projection from $\mathrm {Sym}^\ell (V^*)$ onto its g-stable subspace $\mathrm {Sym}^\ell (V^*)^N=S(E^*)_\ell $ , whence $\mathrm {tr}\bigl ((g\circ \mathrm {pr}^N_\ell )|\mathrm {Sym}^\ell (V^*)\bigr )=\mathrm {tr}\bigl (g|S(E^*)_\ell \bigr )$ .

Proof. The first equality is a special case of [Reference Bonnafé, Lehrer and Michel7, Theorem 3.1] (where we note that the change in sign from the $-u$ in their notation to $+u$ in our notation is harmless), and the second equality follows directly from Proposition 4.8.

Proof. Let us agree to index the pairs $(\epsilon _i^g({E}),d_i^N)$ and $(\epsilon _i^g({U^N_\sigma }),e_i^N({V^\sigma }))$ in the multisets from Definition 4.7 such that $\epsilon _i^g({E})=1$ for $1\leq i \leq {\mathrm {fix}}_{{E}}(g)$ (if ${\mathrm {fix}}_{{E}}(g)\neq 0$ ) and $\epsilon _i^g({U^N_\sigma })=1$ for $1\leq i\leq {\mathrm {fix}}_{{U^N_\sigma }}(g)$ (if ${\mathrm {fix}}_{{U^N_\sigma }}(g)\neq 0$ ). By Proposition 4.8, the left-hand side of Equation (11) is

$$ \begin{align*}|N|\cdot\lim_{x\to 1}\ \displaystyle\prod_{i=1}^r \left(\frac{\epsilon_i^g({U^N_\sigma}) qt(1-x) x^{te_i^N({V^\sigma})}}{1-\epsilon_i^g({E}) x^{d_i^N}}+\frac{1-\epsilon_i^g({U^N_\sigma})x^{te_i^N({V^\sigma})}}{1-\epsilon_i^g({E})x^{d_i^N}}\right).\end{align*} $$

By [Reference Bonnafé, Lehrer and Michel7, Proposition 3.2], ${\mathrm {fix}}_{{U^N_\sigma }} (g) \geq {\mathrm {fix}}_{{E}} (g)$ . We will compute the above limit for the partial products ranging over $1\leq i \leq {\mathrm {fix}}_{{E}}(g)$ , ${\mathrm {fix}}_{{E}}(g)+1\leq i\leq {\mathrm {fix}}_{{U^N_\sigma }}(g)$ and ${\mathrm {fix}}_{{U^N_\sigma }}(g)+1\leq i \leq r$ separately.

Because $\epsilon _i^g({E})=1=\epsilon _i^g({U^N_\sigma })$ for $1\leq i \leq {\mathrm {fix}}_{{E}}(g)$ , we have that

$$ \begin{align*}\lim_{x\to 1}\prod_{i=1}^{{\mathrm{fix}}_{{E}}(g)}\frac{qt(1-x)x^{te_i^N({V^\sigma})}}{1-x^{d_i^N}}+\frac{1-x^{te_i^N({V^\sigma})}}{1-x^{d_i^N}}=\prod_{i=1}^{{\mathrm{fix}}_{{E}}(g)}\frac{qt+te_i^N({V^\sigma})}{d_i^N}.\end{align*} $$

Because $\epsilon _i^g({E})\neq 1\neq \epsilon _i^g({U^N_\sigma })$ for ${\mathrm {fix}}_{{U^N_\sigma }}(g)+1\leq i\leq r$ , we have that

$$ \begin{align*}\lim_{x\to 1}\prod_{i={\mathrm{fix}}_{{U^N_\sigma}}(g)+1}^r\left(\frac{1+\epsilon_i^g({U^N_\sigma}) \bigl(qt(1-x) -1\bigr)x^{te_i^N({V^\sigma})}}{1-\epsilon_i^g({E}) x^{d_i^N}}\right)=\prod_{i={\mathrm{fix}}_{{U^N_\sigma}}(g)+1}^r\frac{1-\epsilon_i^g({U^N_\sigma})}{1-\epsilon_i^g({E})}.\end{align*} $$

If the inequality ${\mathrm {fix}}_{{U^N_\sigma }}(g)>{\mathrm {fix}}_{{E}}(g)$ is strict, so that $\epsilon _i^g({E})\neq 1=\epsilon _i^g({U^N_\sigma })$ for ${\mathrm {fix}}_{{E}}(g)+1\leq i\leq {\mathrm {fix}}_{{U^N_\sigma }}(g)$ , then we see that for each such i the limit of the corresponding factor is

$$ \begin{align*}\lim_{x\to 1}\frac{1+\bigl(qt(1-x)-1\bigr)x^{te_i^N({V^\sigma})}}{1-\epsilon_i^g({E})x^{d_i^N}}=0.\end{align*} $$

Therefore, if ${\mathrm {fix}}_{{U^N_\sigma }}(g)>{\mathrm {fix}}_{{E}}(g)$ , then (cf. [Reference Bonnafé, Lehrer and Michel7, Theorem 3.3])

$$ \begin{align*}\lim_{x\to 1}|N|\cdot\mathcal{P}^{Ng}_\sigma\Bigl(x,x^t,qt(1-x)-1\Bigr)=0.\end{align*} $$

On the other hand, if ${\mathrm {fix}}_{{U^N_\sigma }}(g)={\mathrm {fix}}_{{E}}(g)$ , then (cf. [Reference Bonnafé, Lehrer and Michel7, Theorem 3.3])

$$ \begin{align*}\lim_{x\to 1}|N|\cdot\mathcal{P}^{Ng}_\sigma\Bigl(x,x^t,qt(1-x)-1\Bigr)=t^{{\mathrm{fix}}_{{E}}(g)}\!\prod_{i=1}^{{\mathrm{fix}}_{{E}}(g)}\!(q+e_i^N({V^\sigma}))\!\!\prod_{i={\mathrm{fix}}_{{E}}(g)+1}^r\!\!\frac{1-\epsilon_i^g({U^N_\sigma})}{1-\epsilon_i^g({E})}d_i^N.\end{align*} $$

In any case, we have shown that the left-hand side of Equation (11) is $t^{{\mathrm {fix}}_{{E}}(g)}\cdot P(q)$ for some $P(q)\in \mathbb {C}[q]$ . To conclude the proof, it suffices to compare the left- and right-hand sides of Equation (11) at $t=1$ . For this, we observe as in [Reference Bonnafé, Lehrer and Michel7, Theorem 3.3] that, as a consequence of Proposition 4.9 and the arguments of [Reference Orlik and Solomon16, Theorem 3.3] that are now standard,

$$ \begin{align*}\lim_{x\to 1} \ |N|\cdot\mathcal{P}^{Ng}_\sigma\Bigl(x,x,q(1-x)-1\Bigr)=&\sum_{n\in N}\left(\prod_{i=1}^r\frac{1+\bar{\lambda}_i(ng)^\sigma u}{1-\bar{\lambda}_i(ng)x\vphantom{\Big|}}\right)\Bigg|_{\substack{u=q(1-x)-1\\ x \to 1}}\\ =&\sum_{n \in N}\left(\prod\limits_{\bar{\lambda}_i(ng)\neq 1} \frac{1-\bar{\lambda}_i(ng)^\sigma}{1-\bar{\lambda}_i(ng)\vphantom{\Big|}} \right) q^{{\mathrm{fix}}_V (ng)}.\\[-40pt]\end{align*} $$

Proof. Let $g_1,\dots ,g_{|H|}\in G$ be a full set of coset representatives for $G/N=H$ . Because $\mathcal {P}^G_\sigma (x,y,u)=\frac {1}{|H|}\sum _{j=1}^{|H|}\mathcal {P}^{Ng_j}_\sigma (x,y,u)$ and ${\mathrm {fix}}_{{E}}(g_j)={\mathrm {fix}}_{{E}}(ng_j)$ for any $n\in N$ , it follows from Proposition 4.10 that

$$ \begin{gather*}\lim_{x\to 1}\ |G|\cdot\mathcal{P}^G_\sigma\Bigl(x,x^t,qt(1-x)-1\Bigr)=\sum_{j=1}^{|H|}\lim_{x\to 1}|N|\cdot\mathcal{P}^{Ng_j}_\sigma\Bigl(x,x^t,qt(1-x)-1\Bigr)=\\ \begin{aligned} =&\sum_{j=1}^{|H|}t^{{\mathrm{fix}}_{{E}}(g_j)}\sum_{n\in N}\left(\prod_{\bar{\lambda}_i(ng_j)\neq 1}\frac{1-\bar{\lambda}_i(ng_j)^\sigma}{1-\bar{\lambda}_i(ng_j)\vphantom{\Big|}}\right)q^{{\mathrm{fix}}_V(ng_j)}\\ =&\sum_{g\in G}\left(\prod_{\bar{\lambda}_i(g)\neq 1}\frac{1-\bar{\lambda}_i(g)^\sigma}{1-\bar{\lambda}_i(g)\vphantom{\Big|}}\right)q^{{\mathrm{fix}}_V(g)}t^{{\mathrm{fix}}_{{E}}(g)},\end{aligned}\end{gather*} $$

and our result follows after replacing g with $g^{-1}$ .

Proof. Note that $a\left \lceil \frac {s}{a} \right \rceil - s\in \{0,\dots ,a-1\}$ is congruent to $-s ({\text {mod }} a)$ . Letting $x^s$ denote a basis vector for $(V^{\otimes s})^*$ , we see that $c(x^s)=\zeta _a^sx^s$ and therefore $x^{a\left \lceil \frac {s}{a} \right \rceil - s}\otimes x^s$ is a basis vector for $(U_{V^{\otimes s}}^G)^*=({\mathcal {C}_G}\otimes (V^{\otimes s})^*)^G$ , because ${\mathcal {C}_G}={\mathrm {span}}_{\mathbb {C}}\{1,x,\dots ,x^{a-1}\}$ in this case.

Proof. For this choice of N, the fundamental N-invariants are ${N}_i = x_i^d$ for $1 \leq i \leq r$ , and we obtain the basis $u_i^N= x_i^{d\left \lceil \frac {s}{d}\right \rceil -s} \otimes x_i^\sigma $ for ${(U^N_\sigma )^*}$ as in Section 7.1.

Fix $g\in G$ and $\ell \in \{1,\dots ,r\}$ and suppose that $g(x_\ell )=\zeta _{ab}^k x_j$ for some $0\leq k<ab$ and $j\in \{1,\dots ,r\}$ . Then from the explicit descriptions of ${E^*}$ and ${(U^N_\sigma )^*}$ above we see that $g({N}_\ell )=\zeta _{ab}^{kd}{N}_j$ and $g(u_{\ell }^N)=\zeta _{ab}^{kd\left \lceil \frac {s}{d}\right \rceil }u_j^N$ . It follows that ${E}\simeq V^{\boxtimes d}$ and ${U^N_\sigma }\simeq V^{\boxtimes d\left \lceil \frac {s}{d}\right \rceil }$ , and therefore ${U^N_\sigma }\simeq {E}^{\boxtimes \left \lceil \frac {s}{d}\right \rceil }$ , as claimed.

Proof. By Lemma 3.6 (applied in the special case where $N=G$ ), it suffices to show that the $\tilde {u}_i^G$ form a basis for $(S(V^*)\otimes {(V^\sigma )^*})^G$ as a free $S(V^*)^G$ -module.

Because the group $G(ab,b,r)$ acts on $V^*$ by permutation of the $x_i$ and multiplication by $(ab)$ th roots of unity, it is clear that the $\tilde {u}_i^G\in S(V^*)\otimes {(V^\sigma )^*}$ are G-invariant. Let $u_i^G=\sum _{j=0}^ra_{ij}^G\otimes x_j^\sigma $ be a homogeneous basis for ${(U^G_\sigma )^*}$ , where $a_{ij}^G\in {\mathcal {C}_G}$ and $\mathrm {deg}(a_{ij}^G)=e_i^G({V^\sigma })$ whenever $a_{ij}^G\neq 0$ . Because $(S(V^*)\otimes {(V^\sigma )^*})^G\simeq S(V^*)^G\otimes {(U^G_\sigma )^*}$ , there exists a matrix $[p_{ij}^G]\in \mathrm {Mat}_{r\times r}(S(V^*)^G)$ such that $[p_{ij}^G]\cdot [a_{ij}^G]=[\tilde {a}_{ij}^G]$ , where

$$ \begin{align*}\tilde{a}_{ij}^G = \begin{cases} x_j^{abi-s} & \text{ for } 1 \leq i <r \\ (x_1\cdots x_r)^{\left\lceil \frac{s}{a}\right\rceil a} x_j^{-s} & \text{ for } i=r\end{cases}\end{align*} $$

or, equivalently, such that $\tilde {u}^G_i=\sum _{j=1}^rp_{ij}^G\cdot u_j^G$ . It is clear from the form of the $\tilde {u}_i^G$ that they are $\mathbb {C}$ -linearly independent, which implies that $\mathrm {det}(p_{ij}^G)\neq 0$ .

We claim that

(15) $$ \begin{align}\mathrm{det}(a_{ij}^G)=c\cdot (x_1\cdots x_r)^{\left\lceil \frac{s}{a}\right\rceil a-s} \prod_{1\leq i < j \leq r} (x_i^{ab}-x_j^{ab})\end{align} $$

for some $0\neq c\in \mathbb {C}$ . By Gutkin’s theorem [Reference Lehrer and Taylor14, Theorem 10.13],

$$ \begin{align*}\mathrm{det}(a_{ij}^G)=c\cdot \prod_{H\in\mathcal{R}_G} L_H^{C(H,{V^\sigma})}\end{align*} $$

for some $0\neq c\in \mathbb {C}$ , where $\mathcal {R}_G$ denotes the set of reflecting hyperplanes for G, $L_H\in V^*$ denotes a linear form defining H and $C(H,{V^\sigma })$ is defined as follows. Denoting by $\langle r_H\rangle = G_H<G$ the cyclic subgroup of G that stabilises H pointwise, decompose ${(V^\sigma )^*}\simeq \bigoplus _{i=1}^r\lambda ^{\otimes k_i}$ as a $G_H$ -module with $0\leq k_i<|G_H|$ for $1\leq i\leq r$ , where $\lambda $ denotes the standard representation of $G_H$ on the $G_H$ -stable complement of H in V, and define $C(H,{V^\sigma }):=\sum _{i=1}^r k_i$ . When $L_H=x_i-\zeta _{ab}^\ell x_j$ with $i\neq j$ and $0\leq \ell <ab-1$ , the cyclic generator $r_H$ has order $2$ and ${V^\sigma }\simeq \lambda \oplus \mathbb {C}^{\oplus (r-1)}$ as a $G_H$ -module, and therefore $C(H,{V^\sigma })=1$ in this case. If $a>1$ , then we have the additional reflecting hyperplanes defined by $L_H=x_i$ ; in this case, the cyclic generator $r_H$ has order a and ${(V^\sigma )^*}\simeq \lambda ^{\otimes \left (\left \lceil \frac {s}{a}\right \rceil a -s\right )}\oplus \mathbb {C}^{\oplus (r-1)}$ as a $G_H$ -module (cf. Lemma 7.1), and therefore $C(H,{V^\sigma })=\left \lceil \frac {s}{a}\right \rceil a -s$ . This concludes the proof of Equation (15).

We see by direct inspection that

$$ \begin{align*}\mathrm{deg}(\mathrm{det}(a_{ij}^G))=ab\cdot\binom{r}{2}+r\cdot\left(\left\lceil\frac{s}{a}\right\rceil a-s\right)=\mathrm{deg}(\mathrm{det}(\tilde{a}_{ij}^G)),\end{align*} $$

which implies that $\mathrm {deg}(\mathrm {det}(p_{ij}^G))=0$ , and because $\mathrm {det}(p_{ij}^G)\neq 0$ as we had already seen, it follows that $[p_{ij}^G]\in \mathrm {GL}_r(S(V^*)^G)$ , as we wanted to show.

Proof. Note that for this choice of G and N, with fundamental invariants defined as in Equation (14), we have ${N}_i={G}_i$ for $i=1,\dots ,r-1$ , and ${N}_r=(x_1\cdots x_r)^e$ while ${G}_r=(x_1\cdots x_r)^a$ . The matrix $\mathrm {diag}(\zeta _a,1,\dots ,1)\in G$ maps to the cyclic generator $c\in C_d\simeq H$ , acting trivially on ${\mathrm {span}}_{\mathbb {C}}\{{N}_1,\dots ,{N}_{r-1}\}$ and mapping $c:{N}_r\mapsto \zeta _d{N}_r$ , where $\zeta _d:=\zeta _a^{-e}$ is a primitive dth root of unity.

On the other hand, we see that the space $(\tilde{U}^N_\sigma)^* ={\mathrm {span}}\{\tilde {u}_1^N,\dots ,\tilde {u}_r^N\}$ constructed in Theorem 7.6 is G-stable and the $\tilde {u}_i^N$ form a homogeneous basis for $(S(V^*)\otimes {(V^\sigma )^*})^N$ as a free $S(V^*)^N$ -module, and therefore it is isomorphic to ${(U^N_\sigma )^*}$ as a graded G-module by Lemma 3.6. We see that $\mathrm {diag}(\zeta _a,1,\dots ,1)\in G$ acts trivially on ${\mathrm {span}}_{\mathbb {C}}\{\tilde {u}_1^N,\dots ,\tilde {u}_{r-1}^N\}$ and maps $c:\tilde {u}_r^N\mapsto \zeta _a^{-e\left \lceil \frac {s}{e}\right \rceil }\tilde {u}_r^N=\zeta _d^{\left \lceil \frac {s}{e}\right \rceil }\tilde {u}_r^N$ .