The omega involution conjugates Schur functions

By [F3], each finite-rank coefficient of $E(-t)H(t)-1$ is zero. By [F1] and [F2], these are projections of stable coefficients, so equality of every projection gives $E(-t)H(t)=1$ in $\mathrm{\Lambda }\mathrm{⟦}t\mathrm{⟧}$. For finite rank $N\ge 0$, write $H_N(t)={\sum }_{k\ge 0}{h}_k(x_1,\dots ,x_N)t^k$ and $p_r^{(N)}={\sum }_{i=1}^N{x}_i^r$. The finite identity [F3] gives $H_N(t)={\prod }_{i=1}^N(1-x_{i}t{)}^{-1}$. Differentiating this finite product and expanding each geometric series gives $t{H}_N^{\mathrm{\prime }}(t)\mathrm{/}{H}_N(t)={\sum }_{i=1}^N{x}_{i}t\mathrm{/}(1-x_{i}t)={\sum }_{r\ge 1}{p}_r^{(N)}{t}^r$. Each coefficient is stable by [F1], [F2], and [F4], so equality of all rank projections gives $t{H}^{\mathrm{\prime }}(t)\mathrm{/}H(t)={\sum }_{r\ge 1}{p}_r{t}^r$ in ${\mathrm{\Lambda }}_{\mathbb{Q}}\mathrm{⟦}t\mathrm{⟧}$, where $H(t)={\sum }_{k\ge 0}{h}_k{t}^k$ and $H(t)$ is invertible because its constant term is $1$.

Since $\mathrm{\Lambda }=\mathbb{Z}[e_1,e_2,\dots ]$ freely by [F2], replacing each polynomial generator $e_r$ by the stable element $h_r\in {\mathrm{\Lambda }}^r$ defines a unique unital graded $\mathbb{Z}$-algebra endomorphism $\omega$ of $\mathrm{\Lambda }$.

The coefficient of $t^n$ in the stable identity of step 1.1 gives ${\sum }_{i=0}^n(-1{)}^i{e}_i{h}_{n-i}=0$ for each $n\ge 1$. Applying $\omega$ and using $\omega (e_i)=h_i$ gives $\omega (h_n)+{\sum }_{i=1}^n(-1{)}^i{h}_i\omega (h_{n-i})=0$. Reversing the index in the original recurrence also gives $e_n+{\sum }_{i=1}^n(-1{)}^i{h}_i{e}_{n-i}=0$. Starting with $\omega (h_0)=1=e_0$, induction on $n$ makes these last two sums identical after the leading term, so $\omega (h_n)=e_n$. Therefore ${\omega }^2(e_n)=e_n$ for every free generator; hence ${\omega }^2$ is the identity on $\mathrm{\Lambda }$ and $\omega$ is a ring automorphism. It sends $0$ to $0$ and $1$ to $1$.

For a partition $\lambda$, apply $\omega$ to the $h$ Jacobi–Trudi determinant in [F6]; for nonnegative indices step 2.1 gives $\omega (h_k)=e_k$, and for negative indices both are zero by [F6] and $\omega (0)=0$. Multiplicativity and additivity therefore give $\omega (s_{\lambda })=\det (e_{{\lambda }_i-i+j})$. Apply the dual formula in [F6] to ${\lambda }^{\mathrm{\prime }}$ with determinant size $r=\mathrm{\ell }(\lambda )$, which is allowed because $\mathrm{\ell }(({\lambda }^{\mathrm{\prime }}{)}^{\mathrm{\prime }})=\mathrm{\ell }(\lambda )$ and [F5] identifies $({\lambda }^{\mathrm{\prime }}{)}^{\mathrm{\prime }}=\lambda$. Thus this determinant is $s_{{\lambda }^{\mathrm{\prime }}}$, including at the minimal allowed size. The empty case gives $\omega (1)=1=s_{\mathrm{\varnothing }}$. For $\lambda =(1)$ the size-one determinant gives $\omega (s_{(1)})=e_1=h_1=s_{(1)}$, where $e_1=h_1$ is the coefficient of $t$ in the stable identity of step 1.1.

Extend $\omega$ to ${\mathrm{\Lambda }}_{\mathbb{Q}}$ and apply it coefficientwise to the logarithmic-derivative identity of step 1.1. By step 2.1, $\omega (H(t))=E(t):={\sum }_{k\ge 0}{e}_k{t}^k$, and a coefficientwise ring map commutes with formal differentiation and inverses of series with constant term $1$; hence $t{E}^{\mathrm{\prime }}(t)\mathrm{/}E(t)={\sum }_{r\ge 1}\omega (p_r)t^r$. At rank $N$, replacing $t$ by $-t$ in [F3] gives $E_N(t)={\prod }_{i=1}^N(1+x_{i}t)$, so $t{E}_N^{\mathrm{\prime }}(t)\mathrm{/}{E}_N(t)={\sum }_i{x}_{i}t\mathrm{/}(1+x_{i}t)={\sum }_{r\ge 1}(-1{)}^{r-1}{p}_r^{(N)}{t}^r$. Each coefficient is stable by [F1], [F2], and [F4]; therefore $t{E}^{\mathrm{\prime }}(t)\mathrm{/}E(t)={\sum }_{r\ge 1}(-1{)}^{r-1}{p}_r{t}^r$ in ${\mathrm{\Lambda }}_{\mathbb{Q}}\mathrm{⟦}t\mathrm{⟧}$. Comparing coefficients proves $\omega (p_r)=(-1{)}^{r-1}{p}_r$ for every $r\ge 1$, including the endpoint $r=1$.