The $r$-cycle virtual character filter

By the character form of the cancellation lemma [F2], ${\sum }_{x=1}^r(-1{)}^{r-x}{\chi }^{\nu }\circ {\chi }^{(x,1^{r-x})}={\mathrm{\Psi }}_{\nu }$; since the outer product is $\mathbb{Z}$-bilinear [F3] and ${\psi }_r={\sum }_{x=1}^r(-1{)}^{r-x}{\chi }^{(x,1^{r-x})}$ by [F1], the same sum is ${\chi }^{\nu }\circ {\psi }_r$. Hence ${\mathrm{\Psi }}_{\nu }={\chi }^{\nu }\circ {\psi }_r$ in $R(S_n)$.

We prove the evaluation ${\psi }_r=r\cdot 1_C$, where $C$ is the conjugacy class of the $r$-cycles of $S_r$ and $1_C$ is its indicator function. Let $\rho \in S_r$ be an $r$-cycle. For every $\lambda ⊢r$, orthonormality [F6] gives $⟨{\psi }_r,{\chi }^{\lambda }⟩={\sum }_{x=1}^r(-1{)}^{r-x}⟨{\chi }^{(x,1^{r-x})},{\chi }^{\lambda }⟩$, which equals $(-1{)}^{r-{\lambda }_1}$ if $\lambda =({\lambda }_1,1^{r-{\lambda }_1})$ is a hook and $0$ otherwise; by the full-cycle lemma [F5] this is exactly ${\chi }^{\lambda }(\rho )$. On the other hand, for $f:=r\cdot 1_C$ the definition of the standard inner product [F6], the integrality $\overline{{\chi }^{\lambda }(\sigma )}={\chi }^{\lambda }(\sigma )$ [F7] and the constancy of ${\chi }^{\lambda }$ on $C$ give $⟨f,{\chi }^{\lambda }⟩=\frac{1}{r!}r{\sum }_{\sigma \in C}{\chi }^{\lambda }(\sigma )=\frac{r\mathrm{\mid }C\mathrm{\mid }}{r!}{\chi }^{\lambda }(\rho )={\chi }^{\lambda }(\rho )$, because $C$ consists of the $r!\mathrm{/}r=(r-1)!$ cyclic orderings of the $r$ points up to rotation. Thus $⟨{\psi }_r-f,{\chi }^{\lambda }⟩=0$ for every $\lambda ⊢r$; the ${\chi }^{\lambda }$ exhaust the irreducible characters of $S_r$ and form an orthonormal basis of the class functions [F6], so ${\psi }_r-f=0$, that is, ${\psi }_r(\rho )=r$ for every $r$-cycle $\rho$ and ${\psi }_r(\tau )=0$ for every $\tau \in S_r$ that is not an $r$-cycle.

Let $\sigma \in S_n$ have cycle type with no part equal to $r$; we show ${\mathrm{\Psi }}_{\nu }(\sigma )=0$. By step 1.1, ${\mathrm{\Psi }}_{\nu }={\chi }^{\nu }\circ {\psi }_r$, and by [F3] this outer product is ${\mathrm{Ind}}_K^{S_n}({\chi }^{\nu }⊠{\psi }_r)$ for the two-block subgroup $K=\mathrm{Sym}(B_1)\times \mathrm{Sym}(B_2)$ with $B_1=\{0,\dots ,n-r-1\}$ and $B_2=\{n-r,\dots ,n-1\}$; applying Frobenius' formula [F4] to the honest characters of the finite expansion of ${\chi }^{\nu }⊠{\psi }_r$ and extending by linearity gives $({\chi }^{\nu }\circ {\psi }_r)(\sigma )=\frac{1}{\mathrm{\mid }K\mathrm{\mid }}{\sum }_{\begin{array}{c}x\in S_n\\ x^{-1}\sigma x\in K\end{array}}({\chi }^{\nu }⊠{\psi }_r)(x^{-1}\sigma x)\mathrm{.}$ For an index $x$ of the sum write $x^{-1}\sigma x=(u,v)$ with $u\in \mathrm{Sym}(B_1)$ and $v\in \mathrm{Sym}(B_2)$; then $({\chi }^{\nu }⊠{\psi }_r)(u,v)={\chi }^{\nu }(u){\psi }_r(v)$ by [F3]. If $r=1$, then $K$ fixes the single point of $B_2$, so no conjugate of $\sigma$ (which has no fixed point) lies in $K$ and the sum is empty; hence assume $r\ge 2$. If $v$ were an $r$-cycle on $B_2$, then the disjoint-cycle decomposition of $x^{-1}\sigma x$ would contain that $r$-cycle, so the cycle type of $x^{-1}\sigma x$ would have a part equal to $r$, contradicting [F8] and the hypothesis on $\sigma$. Therefore $v$ is not an $r$-cycle, so ${\psi }_r(v)=0$ by step 1.2 and every term of the sum vanishes; hence $({\chi }^{\nu }\circ {\psi }_r)(\sigma )=0$ and therefore ${\mathrm{\Psi }}_{\nu }(\sigma )=0$ by step 1.1.

Step 1.1 identifies ${\mathrm{\Psi }}_{\nu }$ with the outer product ${\chi }^{\nu }\circ {\psi }_r$, step 1.2 shows that ${\psi }_r$ equals $r$ on the class of $r$-cycles of $S_r$ and vanishes on every other class, and step 2.1 shows ${\mathrm{\Psi }}_{\nu }(\sigma )=0$ for every $\sigma \in S_n$ whose cycle type has no part equal to $r$. These are exactly the three assertions of the statement.