Hermite leading terms for normalized shifted characters

Exact removal of ones. For any partition $\tau$ with no ones and $q\ge 0$, repeated use of the exact $p_1^{\mathrm{\#}}$ identity in [F1], together with $p_1^{\mathrm{\#}}=n$, gives $p_{\tau \cup 1^q}^{\mathrm{\#}}=p_{\tau }^{\mathrm{\#}}{\prod }_{h=0}^{q-1}(n-\mathrm{\mid }\tau \mathrm{\mid }-h)$ on every $Y_n$ with $n\ge 1$. Dividing by the normalization gives ${\eta }_{\tau \cup 1^q}^{(n)}={\eta }_{\tau }^{(n)}{\prod }_{h=0}^{q-1}(1-(\mathrm{\mid }\tau \mathrm{\mid }+h)\mathrm{/}n)$. Thus every normalized term with ones is a finite polynomial in $1\mathrm{/}n$ times the corresponding observable without ones; its constant coefficient is one. These identities also hold below the partition size, because either $p_{\tau }^{\mathrm{\#}}=0$ or the product contains a zero factor.

Negative-degree remainders. Write a term as $c{n}^{a\mathrm{/}2}{p}_{\nu }^{\mathrm{\#}}$ and assign it degree $a+\mathrm{\mid }\nu {\mathrm{\mid }}_1$. The algebra-filtration inequality in [F1] makes degrees subadditive under multiplication, and $n=p_1^{\mathrm{\#}}$ has degree two. Each ${\eta }_{\nu }$ has degree zero. If a term has negative degree, dividing by the normalization rewrites it as $c^{\mathrm{\prime }}{n}^{-j\mathrm{/}2}{\eta }_{\nu }$ with an integer $j\ge 1$. Removing its ones by step 1.1 produces finitely many terms $c^{\mathrm{\prime }\mathrm{\prime }}{n}^{-j^{\mathrm{\prime }}\mathrm{/}2}{\eta }_{\tau }$ with $j^{\mathrm{\prime }}\ge j\ge 1$ and no ones in $\tau$. This rule is algebraic and exact, rather than a pointwise bound on the observables.

A single cycle size. Fix $k\ge 2$ and put $f_m={\eta }_{(k^m)}$, with $f_0=1$ and $f_1={\eta }_k$. Divide the single-cycle multiplication formula of [F1] by ${\sqrt{k}}^{m+1}{n}^{k(m+1)\mathrm{/}2}$. It gives $f_m{\eta }_k=f_{m+1}+m{\eta }_{(k^{m-1},1^k)}+r_m$, where $r_m$ has negative degree. Step 1.1 replaces the middle observable by $f_{m-1}$ plus negative-degree terms, so $f_m{\eta }_k=f_{m+1}+m{f}_{m-1}+r_m^{\mathrm{\prime }}$ with $r_m^{\mathrm{\prime }}$ of negative degree. Comparing with the recurrence [F2] proves by induction $f_m=H_m({\eta }_k)+E_m$, where $E_0=E_1=0$ and $E_{m+1}={\eta }_k{E}_m-m{E}_{m-1}-r_m^{\mathrm{\prime }}$ has negative degree by step 2.1. In particular the contraction coefficient is $m$; no extra power of $n$ remains.

Combining distinct sizes. Group the parts of $\rho$ into the blocks $(k^{m_k(\rho )})$ with distinct $k$. Successive applications of the distinct-size rule in [F1], with normalization denominators multiplying exactly, give ${\prod }_k{\eta }_{(k^{m_k})}={\eta }_{\rho }+(\text{negative-degree terms})$. Substituting step 3.1 and expanding the finite product, every correction includes a negative-degree $E_m$ and other factors of degree at most zero. Therefore ${\prod }_k{H}_{m_k}({\eta }_k)={\eta }_{\rho }+R_{\rho }$ with $R_{\rho }$ of negative degree. Step 2.1 writes it exactly as a finite sum ${\sum }_{\sigma ,j}{c}_{\sigma ,j}{n}^{-j\mathrm{/}2}{\eta }_{\sigma }$ with $j\ge 1$ and $m_1(\sigma )=0$. The support-union bound in [F1] shows that every partition in a product of cycle observables has size at most the sum of the cycle sizes; every Hermite monomial has that sum at most $\mathrm{\mid }\rho \mathrm{\mid }$. Removing ones only decreases size, so $\mathrm{\mid }\sigma \mathrm{\mid }\le \mathrm{\mid }\rho \mathrm{\mid }$, and hence $\mathrm{\mid }\sigma {\mathrm{\mid }}_1-j<\mathrm{\mid }\rho {\mathrm{\mid }}_1$. All constants are independent of $n$.

Expectations. The finite remainder expansion in step 4.1 and [F3] give $\mathrm{\mid }\mathbb{E}[R_{\rho }^{(n)}]\mathrm{\mid }=O(n^{-1\mathrm{/}2})$: each term has $j\ge 1$ and uniformly bounded expectation, indeed zero for nonempty $\sigma$ without ones. For nonempty $\rho$ its own expectation is zero by [F3], so the Hermite-product expectation is $O(n^{-1\mathrm{/}2})$. For $\rho =\mathrm{\varnothing }$ both empty products equal one and the remainder vanishes. This proves the exact expansion and all stated consequences.