Cyclic fixed-space dimensions detect rational virtual characters

Let $C\le G$ be cyclic, and let $y\in R_{\mathbb{Q}}(C)$. By The rational representation ring $R_{\mathbb{Q}}(G)$ and rational-valued class functions, $y$ is an integral linear combination of characters of finite-dimensional $\mathbb{Q}$-representations of $C$, so it is enough to prove the next claim for one such character and then extend by linearity. Let $\rho :C\to \mathrm{GL}(V)$ be a finite-dimensional $\mathbb{Q}$-representation with character $y$, let $D\le C$, and let $d,d^{\mathrm{\prime }}$ be generators of $D$. Then $d^{\mathrm{\prime }}=d^u$ for some integer $u$ coprime to $\mathrm{\mid }D\mathrm{\mid }$. Because $d^{\mathrm{\mid }D\mathrm{\mid }}=1$, the matrix $\rho (d)$ has rational entries and satisfies $\rho (d{)}^{\mathrm{\mid }D\mathrm{\mid }}=I$, so over $\mathbb{C}$ it is diagonalizable with eigenvalues among the $\mathrm{\mid }D\mathrm{\mid }$-th roots of unity. Because the characteristic polynomial of $\rho (d)$ lies in $\mathbb{Q}[t]$, those eigenvalues occur with multiplicities stable under the Galois automorphism $\zeta \mapsto {\zeta }^u$ of the cyclotomic field. Thus the multisets of eigenvalues of $\rho (d)$ and $\rho (d^{\mathrm{\prime }})=\rho (d{)}^u$ agree, so $y(d^{\mathrm{\prime }})=\mathrm{tr}\rho (d^{\mathrm{\prime }})=\mathrm{tr}\rho (d)=y(d)$. Therefore every $y\in R_{\mathbb{Q}}(C)$ is constant on the set of generators of each subgroup of $C$. The rational representation ring $R_{\mathbb{Q}}(G)$ and rational-valued class functions

For each subgroup $D\le C$, choose a generator $d_D$ of $D$ when $D\mathrm{\ne }1$ and put $d_1=e$. Writing $b_D:=y(d_D)\in \mathbb{Q}$, step 1.1 gives $\mathrm{\mid }C\mathrm{\mid }y={\sum }_{D\le C}{b}_D{\tilde{\eta }}_D,$ where ${\tilde{\eta }}_D(c)=\mathrm{\mid }C\mathrm{\mid }$ when $⟨c⟩=D$ and ${\tilde{\eta }}_D(c)=0$ otherwise.

For each $D\le C$, one has ${\tilde{\eta }}_D={\mathrm{Ind}}_D^C{\eta }_D$. Indeed, if $c\in C$ generates $D$, then $c\in D$ and $C$ is abelian, so [F6] gives ${\mathrm{Ind}}_D^C{\eta }_D(c)=\mathrm{\mid }C\mathrm{\mid }$; if $⟨c⟩\mathrm{\ne }D$, then either $c\mathrm{\notin }D$ or $c$ does not generate $D$, and the induced value is $0$. Using [F4] inside $D$ and [F5] to induce further to $C$, each ${\tilde{\eta }}_D$ is therefore an integral linear combination of the permutation characters $U_E^C:={\mathrm{Ind}}_E^C{1}_E$ with $E\le D$. Thus every $y\in R_{\mathbb{Q}}(C)$ is a rational linear combination of the $U_E^C$.

Suppose now that $m_D(x)=0$ for every cyclic subgroup $D\le G$. Let $C\le G$ be cyclic. For each subgroup $E\le C$, the class functions ${\mathrm{Res}}_C^{G}x$ and $U_E^C$ are rational-valued, so [F1] makes their inner product symmetric. Using that symmetry and then Frobenius reciprocity [F2], we get $⟨{\mathrm{Res}}_C^{G}x,U_E^C{⟩}_C=⟨U_E^C,{\mathrm{Res}}_C^{G}x{⟩}_C=⟨1_E,{\mathrm{Res}}_E^{G}x{⟩}_E=⟨{\mathrm{Res}}_E^{G}x,1_E{⟩}_E=m_E(x)=0.$

By step 3.1, every $y\in R_{\mathbb{Q}}(C)$ is a rational linear combination of the class functions $U_E^C$, and both ${\mathrm{Res}}_C^{G}x$ and $y$ are rational-valued by The rational representation ring $R_{\mathbb{Q}}(G)$ and rational-valued class functions. Thus the $\mathbb{Q}$-bilinearity of [F1] on rational-valued class functions combines with step 4.1 to give $⟨{\mathrm{Res}}_C^{G}x,y{⟩}_C=0$ for every $y\in R_{\mathbb{Q}}(C)$. Applying [F2] again, we get $⟨{\mathrm{Ind}}_C^{G}y,x{⟩}_G=0$ for every cyclic subgroup $C\le G$ and every $y\in R_{\mathbb{Q}}(C)$.

By [F7], the rational virtual character $x$ is itself a rational linear combination of the induced characters ${\mathrm{Ind}}_C^{G}y$ from step 5.1. Linearity of [F1] in the first argument and step 5.1 therefore give $⟨x,x{⟩}_G=0$. The positive definiteness in [F1] forces $x=0$. Therefore the map $x\mapsto (m_C(x){)}_{(C)}$ is injective.

For an honest $\mathbb{Q}$-representation $V$, the equality $m_C({\chi }_V)=\dim V^C$ is exactly [F3].