The profile-moment generators in the shifted-character basis

Expansion: by [F1], ${\tilde{p}}_k=[u^k]\{B(u{)}^k\}+\text{(terms of weight }\le k-1\text{)}$. Every product appearing in $[u^k]\{B(u{)}^k\}$ has the form $p_{j_1-1}^{\mathrm{\#}}\cdots p_{j_i-1}^{\mathrm{\#}}$ with $j_1+\cdots +j_i=k$ and total weight $j_1+\cdots +j_i=k$, and by the top-term rule of [F2] its weight-$k$ component is $p_{(j_1-1)\cup \cdots \cup (j_i-1)}^{\mathrm{\#}}$ with coefficient $1$; in particular the term with a single factor ($i=1$, $j_1=k$) contributes $k{p}_{k-1}^{\mathrm{\#}}$, so $p_{k-1}^{\mathrm{\#}}$ occurs in the top weight component of ${\tilde{p}}_k$ with coefficient $k$. Hence the top weight component of ${\tilde{p}}_k$ is exactly the weight-$k$ component of $[u^k]\{B(u{)}^k\}$.

Multiplicativity of $L_k$: if $k$ is odd, $L_k=0$ and at least one of $L_{k^{\mathrm{\prime }}}$, $L_{k-k^{\mathrm{\prime }}}$ is zero, proving the identity. Suppose $k$ is even; let $f,g$ be weight-homogeneous of weights $k^{\mathrm{\prime }}$ and $k-k^{\mathrm{\prime }}$ and expand them in the basis $\{p_{\sigma }^{\mathrm{\#}}\}$, which is possible by The shifted character observables form a basis of $A$, with the Kerov weight filtration(i). Since the weight filtration has level $r$ spanned by the $p_{\sigma }^{\mathrm{\#}}$ with $\mathrm{wt}(p_{\sigma }^{\mathrm{\#}})\le r$ ([F2]), only $\sigma$ with $\mathrm{wt}(p_{\sigma }^{\mathrm{\#}})\le k^{\mathrm{\prime }}$ and $\tau$ with $\mathrm{wt}(p_{\tau }^{\mathrm{\#}})\le k-k^{\mathrm{\prime }}$ occur. The coefficient of $p_{(1^{k\mathrm{/}2})}^{\mathrm{\#}}$ in $fg$ is ${\sum }_{\sigma ,\tau }{f}_{\sigma }{g}_{\tau }{f}_{\sigma \tau }^{(1^{k\mathrm{/}2})}$; by [F3] and [F2], $f_{\sigma \tau }^{\rho }\mathrm{\ne }0$ forces ${\mathrm{deg}}_1(p_{\rho }^{\mathrm{\#}})\le {\mathrm{deg}}_1(p_{\sigma }^{\mathrm{\#}})+{\mathrm{deg}}_1(p_{\tau }^{\mathrm{\#}})\le \mathrm{wt}(p_{\sigma }^{\mathrm{\#}})+\mathrm{wt}(p_{\tau }^{\mathrm{\#}})\le k$, so a nonzero contribution to $\rho =(1^{k\mathrm{/}2})$, where ${\mathrm{deg}}_1(p_{\rho }^{\mathrm{\#}})=k$, forces equalities ${\mathrm{deg}}_1(p_{\sigma }^{\mathrm{\#}})=\mathrm{wt}(p_{\sigma }^{\mathrm{\#}})=k^{\mathrm{\prime }}$ and ${\mathrm{deg}}_1(p_{\tau }^{\mathrm{\#}})=\mathrm{wt}(p_{\tau }^{\mathrm{\#}})=k-k^{\mathrm{\prime }}$; as ${\mathrm{deg}}_1\le \mathrm{wt}$ with equality only for columns, this forces $\sigma =(1^{k^{\mathrm{\prime }}\mathrm{/}2})$ and $\tau =(1^{(k-k^{\mathrm{\prime }})\mathrm{/}2)}$, so $k$ and $k^{\mathrm{\prime }}$ are even and the coefficient is $f_{\sigma }{g}_{\tau }\cdot 1$ (the top coefficient being $1$ by [F2]). Summing gives $L_k(fg)=L_{k^{\mathrm{\prime }}}(f)L_{k-k^{\mathrm{\prime }}}(g)$; when $k$ or $k^{\mathrm{\prime }}$ is odd, no such pair exists and both sides are $0$.

Values on the generators: expand $[u^{2m}]B(u{)}^{2m}$ as a finite sum of products $p_{j_1-1}^{\mathrm{\#}}\cdots p_{j_i-1}^{\mathrm{\#}}$ with $j_l\ge 2$ and ${\sum }_l{j}_l=2m$. By [F2], the only weight-$2m$ partition in such a product is $(j_1-1)\cup \cdots \cup (j_i-1)$, with coefficient one; lower-weight terms cannot contribute to $L_{2m}$. This partition is $(1^m)$ precisely when $i=m$ and all $j_l=2$. There are $\left(\genfrac{}{}{0px}{}{2m}{m}\right)$ ways to select the $m$ factors supplying $p_1^{\mathrm{\#}}{u}^2$ among the $2m$ factors of $B(u{)}^{2m}$. Consequently $L_{2m}({\tilde{p}}_{2m})=\left(\genfrac{}{}{0px}{}{2m}{m}\right)$ by step 1.1. For odd $k$, $L_k$ is zero by definition.

Conclusion: step 1.1 gives the stated expansion with its equivalent form and the description of the top weight component; step 2.1 gives multiplicativity of $L_k$; step 2.2 evaluates $L$ on the generators ${\tilde{p}}_k$ as the central binomial coefficients for even $k$ and $0$ for odd $k$. No choice principle is used: the imported IvOl statements are algebraic.