[proofplan] The proof is an immediate filtration argument in the pseudodifferential calculus. First we use the formal adjoint formula, which says that $A^*$ is again of order $m$ and has principal symbol the complex conjugate of the principal symbol of $A$. Then the order-$m$ principal symbol of $A-A^*$ is $a_m-\overline{a_m}$, which vanishes because $a_m$ is real-valued in the quotient. Finally, the order filtration of $\Psi^m_{1,0}$ identifies vanishing principal symbol with membership in $\Psi^{m-1}_{1,0}$. [/proofplan] [step:Identify the principal symbol of the formal adjoint] Let \begin{align*} \sigma_m:\Psi^m_{1,0}(\mathbb{R}^n) \to S^m_{1,0}(\mathbb{R}^n \times \mathbb{R}^n)/S^{m-1}_{1,0}(\mathbb{R}^n \times \mathbb{R}^n) \end{align*} denote the principal symbol map. By the standard adjoint theorem for the $(1,0)$ pseudodifferential calculus, $A^*\in \Psi^m_{1,0}(\mathbb{R}^n)$ and \begin{align*} \sigma_m(A^*)=\overline{\sigma_m(A)}. \end{align*} Here the adjoint is the formal $L^2(\mathbb{R}^n,\mathcal{L}^n)$-adjoint, exactly the convention required by the adjoint formula. Since $\sigma_m(A)=a_m$, this gives \begin{align*} \sigma_m(A^*)=\overline{a_m}. \end{align*} (citing a result not yet in the wiki: adjoint of a pseudodifferential operator is pseudodifferential with conjugate principal symbol) [guided] Let \begin{align*} \sigma_m:\Psi^m_{1,0}(\mathbb{R}^n) \to S^m_{1,0}(\mathbb{R}^n \times \mathbb{R}^n)/S^{m-1}_{1,0}(\mathbb{R}^n \times \mathbb{R}^n) \end{align*} be the principal symbol map. The point of working with the quotient is that the principal symbol records only the order-$m$ part of the full symbol, while all terms of order at most $m-1$ are ignored. We now invoke the standard adjoint formula in the $(1,0)$ pseudodifferential calculus. Its hypotheses are satisfied because $A$ is assumed to belong to $\Psi^m_{1,0}(\mathbb{R}^n)$, and $A^*$ is taken to be the formal adjoint with respect to the $L^2(\mathbb{R}^n,\mathcal{L}^n)$ pairing. The theorem gives two conclusions: first, \begin{align*} A^* \in \Psi^m_{1,0}(\mathbb{R}^n), \end{align*} and second, \begin{align*} \sigma_m(A^*)=\overline{\sigma_m(A)}. \end{align*} Since the principal symbol class of $A$ is denoted by $a_m$, this becomes \begin{align*} \sigma_m(A^*)=\overline{a_m}. \end{align*} This is the only place where the adjoint calculus enters the argument: it tells us precisely what happens to the top-order symbol under passage to the formal adjoint. (citing a result not yet in the wiki: adjoint of a pseudodifferential operator is pseudodifferential with conjugate principal symbol) [/guided] [/step] [step:Compute the principal symbol of the skew-adjoint difference] Since both $A$ and $A^*$ belong to $\Psi^m_{1,0}(\mathbb{R}^n)$, their difference satisfies \begin{align*} A-A^* \in \Psi^m_{1,0}(\mathbb{R}^n). \end{align*} The principal symbol map is linear on the quotient, so \begin{align*} \sigma_m(A-A^*)=\sigma_m(A)-\sigma_m(A^*). \end{align*} Using $\sigma_m(A)=a_m$ and $\sigma_m(A^*)=\overline{a_m}$, we obtain \begin{align*} \sigma_m(A-A^*)=a_m-\overline{a_m}. \end{align*} By the hypothesis that $a_m$ is real-valued in $S^m_{1,0}/S^{m-1}_{1,0}$, we have \begin{align*} a_m-\overline{a_m}=0 \end{align*} in the quotient. Hence \begin{align*} \sigma_m(A-A^*)=0. \end{align*} [/step] [step:Use the order filtration to lower the operator order] The order filtration criterion for the pseudodifferential calculus says that if $B\in \Psi^m_{1,0}(\mathbb{R}^n)$ and $\sigma_m(B)=0$ in $S^m_{1,0}/S^{m-1}_{1,0}$, then \begin{align*} B\in \Psi^{m-1}_{1,0}(\mathbb{R}^n). \end{align*} Apply this criterion to the operator \begin{align*} B:=A-A^*. \end{align*} The previous step proved that $B\in \Psi^m_{1,0}(\mathbb{R}^n)$ and $\sigma_m(B)=0$. Therefore \begin{align*} A-A^* \in \Psi^{m-1}_{1,0}(\mathbb{R}^n). \end{align*} This is the desired conclusion. (citing a result not yet in the wiki: order filtration criterion for vanishing principal symbol) [/step]