Padded irreducible characters are eventually given by character polynomials

Define $B_{\lambda }:={\sum }_{{\sum }_{r=1}^{m}r{k}_r=m}{\chi }^{\lambda }({\rho }_k){\prod }_{r=1}^m\left(\genfrac{}{}{0px}{}{X_r}{k_r}\right)$, where each $k_r\ge 0$, ${\rho }_k$ has $k_r$ parts equal to $r$, and $\left(\genfrac{}{}{0px}{}{X}{q}\right):=X(X-1)\cdots (X-q+1)\mathrm{/}q!$ with $\left(\genfrac{}{}{0px}{}{X}{0}\right)=1$. For $m=0$ use the empty tuple and $B_{\mathrm{\varnothing }}=1$. The sum is finite and has rational coefficients and weighted degree at most $m$, since each term has degree at most ${\sum }_{r}r{k}_r=m$.

Let ${\mathcal{E}}_{\lambda }$ be the finite set of partitions $\eta$ with ${\lambda }_i\ge {\eta }_i\ge {\lambda }_{i+1}$ for every $i$, padding both with zeros. For $n\ge m+{\lambda }_1$, a shape $\nu ⊢n$ has $\nu \mathrm{/}\lambda$ horizontal exactly when ${\nu }_i\ge {\lambda }_i\ge {\nu }_{i+1}$ for all $i$: column heights increase by at most one precisely under these interlacing inequalities. Writing $\nu =\eta [n]$, the inequalities for $i\ge 2$ and the upper bounds on ${\nu }_{i+1}$ say exactly $\eta \in {\mathcal{E}}_{\lambda }$; the remaining condition ${\nu }_1=n-\mathrm{\mid }\eta \mathrm{\mid }\ge {\lambda }_1$ holds because $\mathrm{\mid }\eta \mathrm{\mid }\le m$. Every such $\eta [n]$ is a partition, since $n-\mathrm{\mid }\eta \mathrm{\mid }\ge {\lambda }_1\ge {\eta }_1$. Therefore [F3] gives $M(\lambda {)}_n\cong {⨁}_{\eta \in {\mathcal{E}}_{\lambda }}{S}^{\eta [n]}$ on this whole range. The set contains $\lambda$, and every other member has $\mathrm{\mid }\eta \mathrm{\mid }<m$, because ${\eta }_i\le {\lambda }_i$ and equality of the total sizes would force equality in every coordinate.

For $\sigma \in S_n$, a subset $T$ is fixed setwise by $\sigma$ exactly when it is a union of cycles. The number of fixed $m$-subsets containing $k_r$ cycles of length $r$ for each $r$ is ${\prod }_r\left(\genfrac{}{}{0px}{}{X_r(\sigma )}{k_r}\right)$. On such a summand $S^{\lambda }(T)$ the restriction of $\sigma$ has cycle type ${\rho }_k$, so its trace is ${\chi }^{\lambda }({\rho }_k)$; moving summands have zero diagonal blocks. Thus ${\chi }_{M(\lambda {)}_n}(\sigma )=B_{\lambda }(X(\sigma ))$ for every $n$, by summing the fixed-block traces.

Induct on $m$. For $m=0$, step 1.2 has only the empty tail and step 2.1 gives $P_{\mathrm{\varnothing }}=1$. For $m>0$, assume that each smaller tail $\eta$ has its claimed polynomial $P_{\eta }$ and range. Define $P_{\lambda }:=B_{\lambda }-{\sum }_{\eta \in {\mathcal{E}}_{\lambda }\setminus \{\lambda \}}{P}_{\eta }$. This is a finite sum of rational polynomials of weighted degree at most $m$. For $n\ge m+{\lambda }_1$, every smaller tail in the sum satisfies $n\ge \mathrm{\mid }\eta \mathrm{\mid }+{\eta }_1$, since $\mathrm{\mid }\eta \mathrm{\mid }\le m$ and ${\eta }_1\le {\lambda }_1$. Subtracting their character identities from the direct-sum identity in step 1.2 and using step 2.1 proves ${\chi }^{\lambda [n]}(\sigma )=P_{\lambda }(X(\sigma ))$ throughout the required range. This establishes the induction.

The polynomial from step 3.1 has weighted degree at most $m$ and the claimed identity in every rank $n\ge m+{\lambda }_1$. Its uniqueness follows from the uniqueness of character polynomials agreeing in infinitely many ranks. All sums and choices in the argument are finite.