[proofplan] We prove first that $\omega$ exchanges the complete homogeneous and elementary symmetric functions by applying $\omega$ to the generating-function identity relating $E(t)$ and $H(t)$. Since the elementary symmetric functions generate $\operatorname{Sym}$ as a $\mathbb{Q}$-algebra, this implies $\omega^2=\operatorname{id}_{\operatorname{Sym}}$. To prove preservation of the Hall [inner product](/page/Inner%20Product), we compute the action of $\omega$ on the power-sum basis and then check the inner product on basis elements. [/proofplan] [step:Show that $\omega$ sends each $h_n$ back to $e_n$] Define formal [power series](/page/Power%20Series) in $\operatorname{Sym}[[t]]$ by \begin{align*} E(t)&:=\sum_{n=0}^{\infty} e_n t^n,\\ H(t)&:=\sum_{n=0}^{\infty} h_n t^n, \end{align*} where $e_0=h_0=1$. The defining generating-function identity is \begin{align*} E(-t)H(t)=1. \end{align*} Since $\omega:\operatorname{Sym}\to\operatorname{Sym}$ is a $\mathbb{Q}$-algebra homomorphism, it extends coefficientwise to a $\mathbb{Q}[[t]]$-algebra homomorphism \begin{align*} \omega:\operatorname{Sym}[[t]]\to\operatorname{Sym}[[t]]. \end{align*} Applying $\omega$ to the identity $H(t)=E(-t)^{-1}$ gives \begin{align*} \omega(H(t))=\omega(E(-t))^{-1}=H(-t)^{-1}. \end{align*} Using again $E(t)H(-t)=1$, we obtain \begin{align*} \omega(H(t))=E(t). \end{align*} Comparing coefficients of $t^n$ gives \begin{align*} \omega(h_n)=e_n \end{align*} for every $n\in\mathbb{N}$. [/step] [step:Deduce that $\omega$ is an involution from the elementary generators] For every $n\in\mathbb{N}$, \begin{align*} \omega^2(e_n)=\omega(h_n)=e_n. \end{align*} The elements $e_1,e_2,\dots$ generate $\operatorname{Sym}$ as a $\mathbb{Q}$-algebra. Since $\omega^2:\operatorname{Sym}\to\operatorname{Sym}$ is a $\mathbb{Q}$-algebra homomorphism that fixes every algebra generator $e_n$, it fixes every polynomial expression in the $e_n$. Hence \begin{align*} \omega^2=\operatorname{id}_{\operatorname{Sym}}. \end{align*} [/step] [step:Compute the sign by which $\omega$ acts on each power-sum basis element] For each $r\in\mathbb{N}$, let $p_r\in\operatorname{Sym}$ denote the $r$-th power-sum symmetric function. The generating-function identities for the complete homogeneous and elementary symmetric functions are \begin{align*} H(t)&=\exp\left(\sum_{r=1}^{\infty}\frac{p_r}{r}t^r\right),\\ E(t)&=\exp\left(\sum_{r=1}^{\infty}(-1)^{r-1}\frac{p_r}{r}t^r\right). \end{align*} Applying $\omega$ coefficientwise to the first identity and using $\omega(H(t))=E(t)$ gives \begin{align*} \exp\left(\sum_{r=1}^{\infty}\frac{\omega(p_r)}{r}t^r\right) = \exp\left(\sum_{r=1}^{\infty}(-1)^{r-1}\frac{p_r}{r}t^r\right). \end{align*} Both sides have constant term $1$, so the formal logarithm is defined on both sides. Taking formal logarithms and comparing coefficients of $t^r$ gives \begin{align*} \omega(p_r)=(-1)^{r-1}p_r \end{align*} for every $r\in\mathbb{N}$. If $\lambda=(\lambda_1,\dots,\lambda_\ell)$ is a partition, define \begin{align*} p_\lambda:=p_{\lambda_1}\cdots p_{\lambda_\ell}, \qquad \varepsilon_\lambda:=\prod_{j=1}^{\ell}(-1)^{\lambda_j-1}=(-1)^{|\lambda|-\ell}. \end{align*} Since $\omega$ is a $\mathbb{Q}$-algebra homomorphism, \begin{align*} \omega(p_\lambda)=\prod_{j=1}^{\ell}\omega(p_{\lambda_j}) =\prod_{j=1}^{\ell}(-1)^{\lambda_j-1}p_{\lambda_j} =\varepsilon_\lambda p_\lambda. \end{align*} [guided] The purpose of this step is to move from the $e_n,h_n$ description of $\omega$ to the basis in which the Hall inner product is diagonal. For each $r\in\mathbb{N}$, let $p_r\in\operatorname{Sym}$ be the $r$-th power-sum symmetric function. We use the generating-function identities \begin{align*} H(t)&=\exp\left(\sum_{r=1}^{\infty}\frac{p_r}{r}t^r\right),\\ E(t)&=\exp\left(\sum_{r=1}^{\infty}(-1)^{r-1}\frac{p_r}{r}t^r\right). \end{align*} These identities are useful because the previous step showed that $\omega(H(t))=E(t)$. Applying $\omega$ coefficientwise to the first identity gives \begin{align*} \omega(H(t)) = \exp\left(\sum_{r=1}^{\infty}\frac{\omega(p_r)}{r}t^r\right). \end{align*} Since $\omega(H(t))=E(t)$, we therefore have \begin{align*} \exp\left(\sum_{r=1}^{\infty}\frac{\omega(p_r)}{r}t^r\right) = \exp\left(\sum_{r=1}^{\infty}(-1)^{r-1}\frac{p_r}{r}t^r\right). \end{align*} Both formal power series have constant term $1$, so the formal logarithm is valid on both sides. Taking logarithms gives \begin{align*} \sum_{r=1}^{\infty}\frac{\omega(p_r)}{r}t^r = \sum_{r=1}^{\infty}(-1)^{r-1}\frac{p_r}{r}t^r. \end{align*} Comparing the coefficient of $t^r$ and multiplying by $r$ yields \begin{align*} \omega(p_r)=(-1)^{r-1}p_r. \end{align*} Now let $\lambda=(\lambda_1,\dots,\lambda_\ell)$ be a partition. Define \begin{align*} p_\lambda:=p_{\lambda_1}\cdots p_{\lambda_\ell}, \qquad \varepsilon_\lambda:=\prod_{j=1}^{\ell}(-1)^{\lambda_j-1}=(-1)^{|\lambda|-\ell}. \end{align*} Because $\omega$ is an algebra homomorphism, it preserves products. Hence \begin{align*} \omega(p_\lambda) &=\omega(p_{\lambda_1}\cdots p_{\lambda_\ell})\\ &=\prod_{j=1}^{\ell}\omega(p_{\lambda_j})\\ &=\prod_{j=1}^{\ell}(-1)^{\lambda_j-1}p_{\lambda_j}\\ &=\varepsilon_\lambda p_\lambda. \end{align*} Thus each power-sum basis vector is an eigenvector of $\omega$, with eigenvalue $\varepsilon_\lambda\in\{1,-1\}$. [/guided] [/step] [step:Verify the Hall inner product on the power-sum basis] Let $\lambda$ and $\mu$ be partitions. Let $m_i(\lambda)$ denote the multiplicity of the part $i$ in $\lambda$, and define \begin{align*} z_\lambda:=\prod_{i=1}^{\infty} i^{m_i(\lambda)}m_i(\lambda)!. \end{align*} The Hall inner product is characterized on the power-sum basis by \begin{align*} \langle p_\lambda,p_\mu\rangle=z_\lambda\delta_{\lambda,\mu}, \end{align*} where $\delta_{\lambda,\mu}$ is $1$ if $\lambda=\mu$ and $0$ otherwise. Using the previous step and bilinearity of the Hall inner product, \begin{align*} \langle \omega(p_\lambda),\omega(p_\mu)\rangle &= \langle \varepsilon_\lambda p_\lambda,\varepsilon_\mu p_\mu\rangle\\ &= \varepsilon_\lambda\varepsilon_\mu\langle p_\lambda,p_\mu\rangle\\ &= \varepsilon_\lambda\varepsilon_\mu z_\lambda\delta_{\lambda,\mu}. \end{align*} If $\lambda\neq\mu$, then $\delta_{\lambda,\mu}=0$, so both $\langle \omega(p_\lambda),\omega(p_\mu)\rangle$ and $\langle p_\lambda,p_\mu\rangle$ are equal to $0$. If $\lambda=\mu$, then $\varepsilon_\lambda\varepsilon_\mu=\varepsilon_\lambda^2=1$, so \begin{align*} \langle \omega(p_\lambda),\omega(p_\lambda)\rangle = z_\lambda = \langle p_\lambda,p_\lambda\rangle. \end{align*} Therefore \begin{align*} \langle \omega(p_\lambda),\omega(p_\mu)\rangle = \langle p_\lambda,p_\mu\rangle \end{align*} for all partitions $\lambda$ and $\mu$. [/step] [step:Extend the equality from basis elements to all symmetric functions] The family $\{p_\lambda:\lambda \text{ is a partition}\}$ is a $\mathbb{Q}$-basis of $\operatorname{Sym}$. Thus every $f,g\in\operatorname{Sym}$ admit finite expansions \begin{align*} f=\sum_{\lambda} a_\lambda p_\lambda, \qquad g=\sum_{\mu} b_\mu p_\mu, \end{align*} with coefficients $a_\lambda,b_\mu\in\mathbb{Q}$. Using linearity of $\omega$ and bilinearity of the Hall inner product, \begin{align*} \langle \omega(f),\omega(g)\rangle &= \left\langle \sum_{\lambda}a_\lambda\omega(p_\lambda), \sum_{\mu}b_\mu\omega(p_\mu) \right\rangle\\ &= \sum_{\lambda,\mu}a_\lambda b_\mu \langle \omega(p_\lambda),\omega(p_\mu)\rangle\\ &= \sum_{\lambda,\mu}a_\lambda b_\mu \langle p_\lambda,p_\mu\rangle\\ &= \langle f,g\rangle. \end{align*} Together with $\omega^2=\operatorname{id}_{\operatorname{Sym}}$, this proves that $\omega$ is an isometric involution of $\operatorname{Sym}$. [/step]