[proofplan] We first invert the principal symbol of $A$ on $T^*M \setminus 0$, obtaining the leading symbol of a candidate inverse of order $-m$. Using the principal symbol exact sequence and the composition formula in the classical pseudodifferential calculus, we recursively add lower-order corrections so that the left error $I - BA$ has successively lower order. Asymptotic summation realizes the resulting formal inverse symbol as a genuine classical operator $B_L$ with $B_LA - I$ smoothing. Repeating the same construction on the right gives $B_R$ with $AB_R - I$ smoothing, and the two parametrices differ by a smoothing operator, which upgrades $B_L$ to a two-sided parametrix. [/proofplan]

[step:Invert the principal symbol to start the left inverse] Let $\pi: T^*M \setminus 0 \to M$ denote the cotangent projection restricted to the punctured cotangent bundle. Define \begin{align*} a_m: T^*M \setminus 0 \to \mathbb{C} \end{align*} by \begin{align*} a_m(x,\xi) = \sigma_m(A)(x,\xi). \end{align*} By ellipticity, $a_m(x,\xi) \neq 0$ for every $(x,\xi) \in T^*M \setminus 0$. Therefore the map \begin{align*} b_{-m}: T^*M \setminus 0 \to \mathbb{C} \end{align*} defined by \begin{align*} b_{-m}(x,\xi) = a_m(x,\xi)^{-1} \end{align*} is smooth and homogeneous of degree $-m$ in $\xi$. By the quantization part of the classical pseudodifferential calculus, choose an operator $B_0 \in \Psi^{-m}_{\mathrm{cl}}(M)$ whose principal symbol is $b_{-m}$. The composition formula for classical pseudodifferential operators gives \begin{align*} \sigma_0(B_0A) = \sigma_{-m}(B_0)\sigma_m(A) = b_{-m}a_m = 1. \end{align*} Hence the order-zero principal symbol of $I - B_0A$ vanishes. By the principal symbol exact sequence for classical pseudodifferential operators, citing a result not yet in the wiki: Principal Symbol Exact Sequence for Classical Pseudodifferential Operators, the operator $R_1 := I - B_0A$ lies in $\Psi^{-1}_{\mathrm{cl}}(M)$. [/step]

[step:Recursively remove each homogeneous term of the left error]Assume that for some integer $N \geq 1$ we have constructed an operator $B_{N-1} \in \Psi^{-m}_{\mathrm{cl}}(M)$ such that \begin{align*} R_N := I - B_{N-1}A \in \Psi^{-N}_{\mathrm{cl}}(M). \end{align*} Let \begin{align*} r_{-N}: T^*M \setminus 0 \to \mathbb{C} \end{align*} denote the principal symbol $\sigma_{-N}(R_N)$, homogeneous of degree $-N$ in $\xi$. Define \begin{align*} c_{-m-N}: T^*M \setminus 0 \to \mathbb{C} \end{align*} by \begin{align*} c_{-m-N}(x,\xi) = r_{-N}(x,\xi)a_m(x,\xi)^{-1}. \end{align*} Since $r_{-N}$ has degree $-N$ and $a_m^{-1}$ has degree $-m$, the function $c_{-m-N}$ is homogeneous of degree $-m-N$. Choose $C_N \in \Psi^{-m-N}_{\mathrm{cl}}(M)$ with principal symbol $c_{-m-N}$. Define \begin{align*} B_N := B_{N-1} + C_N. \end{align*} Then \begin{align*} I - B_NA = R_N - C_NA. \end{align*} The composition formula gives \begin{align*} \sigma_{-N}(C_NA) = \sigma_{-m-N}(C_N)\sigma_m(A) = c_{-m-N}a_m = r_{-N}. \end{align*} Thus the order $-N$ principal symbol of $R_N - C_NA$ vanishes. By the principal symbol exact sequence, $I - B_NA \in \Psi^{-N-1}_{\mathrm{cl}}(M)$.[/step]

[guided]The recursive step has one purpose: if the current left error has leading term of order $-N$, we add a correction to the inverse whose product with $A$ has exactly that same leading term. Assume that $B_{N-1} \in \Psi^{-m}_{\mathrm{cl}}(M)$ has already been chosen and that its left error \begin{align*} R_N := I - B_{N-1}A \end{align*} belongs to $\Psi^{-N}_{\mathrm{cl}}(M)$. The leading part of this error is its principal symbol \begin{align*} r_{-N} := \sigma_{-N}(R_N): T^*M \setminus 0 \to \mathbb{C}. \end{align*} This symbol is homogeneous of degree $-N$ in the covariable $\xi$. We want to find a correction $C_N$ such that $C_NA$ has principal symbol $r_{-N}$. Since $A$ has principal symbol \begin{align*} a_m := \sigma_m(A): T^*M \setminus 0 \to \mathbb{C} \end{align*} and $a_m$ is nonzero by ellipticity, the only possible leading symbol for $C_N$ is \begin{align*} c_{-m-N}: T^*M \setminus 0 \to \mathbb{C} \end{align*} defined by \begin{align*} c_{-m-N}(x,\xi) = r_{-N}(x,\xi)a_m(x,\xi)^{-1}. \end{align*} The degree check is essential: $r_{-N}$ has degree $-N$, while $a_m^{-1}$ has degree $-m$, so $c_{-m-N}$ has degree $-m-N$. Therefore it is the correct principal symbol for an operator of order $-m-N$. By the quantization theorem for classical symbols, citing a result not yet in the wiki: Quantization of Classical Symbols on a Compact Manifold, choose $C_N \in \Psi^{-m-N}_{\mathrm{cl}}(M)$ with \begin{align*} \sigma_{-m-N}(C_N) = c_{-m-N}. \end{align*} Set \begin{align*} B_N := B_{N-1} + C_N. \end{align*} Then the new error is \begin{align*} I - B_NA = I - B_{N-1}A - C_NA = R_N - C_NA. \end{align*} The composition formula for classical pseudodifferential operators, citing a result not yet in the wiki: Composition Formula for Classical Pseudodifferential Operators, gives \begin{align*} \sigma_{-N}(C_NA) = \sigma_{-m-N}(C_N)\sigma_m(A). \end{align*} Substituting the definitions gives \begin{align*} \sigma_{-N}(C_NA) = c_{-m-N}a_m = r_{-N}. \end{align*} Hence the principal symbol of $R_N - C_NA$ at order $-N$ is zero. The principal symbol exact sequence then lowers the order of the error by one: \begin{align*} I - B_NA \in \Psi^{-N-1}_{\mathrm{cl}}(M). \end{align*} This is the induction mechanism: each correction removes exactly one homogeneous layer of the error.[/guided]

[step:Sum the left inverse symbol asymptotically] The preceding recursion produces a sequence of correction operators \begin{align*} C_N \in \Psi^{-m-N}_{\mathrm{cl}}(M) \end{align*} for $N \geq 1$. By asymptotic summation for classical symbols, citing a result not yet in the wiki: Asymptotic Summation of Classical Symbols, there exists an operator $B_L \in \Psi^{-m}_{\mathrm{cl}}(M)$ such that \begin{align*} B_L - B_0 - \sum_{N=1}^{K-1} C_N \in \Psi^{-m-K}_{\mathrm{cl}}(M) \end{align*} for every integer $K \geq 1$. Fix $K \geq 1$. Since \begin{align*} I - B_{K-1}A \in \Psi^{-K}_{\mathrm{cl}}(M) \end{align*} and \begin{align*} B_L - B_{K-1} \in \Psi^{-m-K}_{\mathrm{cl}}(M), \end{align*} the composition formula gives \begin{align*} (B_L - B_{K-1})A \in \Psi^{-K}_{\mathrm{cl}}(M). \end{align*} Therefore \begin{align*} I - B_LA = I - B_{K-1}A - (B_L - B_{K-1})A \in \Psi^{-K}_{\mathrm{cl}}(M). \end{align*} Because this holds for every $K \geq 1$, we obtain \begin{align*} I - B_LA \in \bigcap_{K=1}^{\infty}\Psi^{-K}_{\mathrm{cl}}(M) = \Psi^{-\infty}(M). \end{align*} Thus \begin{align*} B_LA - I \in \Psi^{-\infty}(M). \end{align*} [/step]

[step:Construct a right inverse modulo smoothing operators] Apply the same construction to the right error. Start with an operator $D_0 \in \Psi^{-m}_{\mathrm{cl}}(M)$ with principal symbol $a_m^{-1}$. Since scalar principal symbols commute, the composition formula gives \begin{align*} \sigma_0(AD_0) = a_m a_m^{-1} = 1. \end{align*} Hence $I - AD_0 \in \Psi^{-1}_{\mathrm{cl}}(M)$. Inductively, suppose $D_{N-1} \in \Psi^{-m}_{\mathrm{cl}}(M)$ satisfies \begin{align*} S_N := I - AD_{N-1} \in \Psi^{-N}_{\mathrm{cl}}(M). \end{align*} Let \begin{align*} s_{-N}: T^*M \setminus 0 \to \mathbb{C} \end{align*} be $\sigma_{-N}(S_N)$. Define \begin{align*} d_{-m-N}: T^*M \setminus 0 \to \mathbb{C} \end{align*} by \begin{align*} d_{-m-N}(x,\xi) = a_m(x,\xi)^{-1}s_{-N}(x,\xi). \end{align*} Choose $D_N' \in \Psi^{-m-N}_{\mathrm{cl}}(M)$ with principal symbol $d_{-m-N}$, and set \begin{align*} D_N := D_{N-1} + D_N'. \end{align*} Then \begin{align*} I - AD_N = S_N - AD_N'. \end{align*} The composition formula gives \begin{align*} \sigma_{-N}(AD_N') = a_m d_{-m-N} = s_{-N}, \end{align*} so $I - AD_N \in \Psi^{-N-1}_{\mathrm{cl}}(M)$ by the principal symbol exact sequence. Asymptotic summation gives $B_R \in \Psi^{-m}_{\mathrm{cl}}(M)$ with \begin{align*} AB_R - I \in \Psi^{-\infty}(M). \end{align*} [/step]

[step:Show that the left parametrix is also a right parametrix] We have constructed operators $B_L, B_R \in \Psi^{-m}_{\mathrm{cl}}(M)$ such that \begin{align*} B_LA - I \in \Psi^{-\infty}(M) \end{align*} and \begin{align*} AB_R - I \in \Psi^{-\infty}(M). \end{align*} Compute \begin{align*} B_L - B_R = B_L(I - AB_R) + (B_LA - I)B_R. \end{align*} The class $\Psi^{-\infty}(M)$ is a two-sided ideal in the pseudodifferential calculus, citing a result not yet in the wiki: Smoothing Operators Form a Two-Sided Ideal in the Pseudodifferential Calculus. Therefore both terms on the right-hand side are smoothing, and hence \begin{align*} B_L - B_R \in \Psi^{-\infty}(M). \end{align*} Using this identity, \begin{align*} AB_L - I = A(B_L - B_R) + (AB_R - I). \end{align*} Again the smoothing ideal property implies $A(B_L - B_R) \in \Psi^{-\infty}(M)$, and therefore \begin{align*} AB_L - I \in \Psi^{-\infty}(M). \end{align*} [/step]

[step:Name the resulting operator as the parametrix] Set \begin{align*} B := B_L. \end{align*} From the left-parametrix construction, \begin{align*} BA - I = B_LA - I \in \Psi^{-\infty}(M). \end{align*} From the comparison with the right parametrix, \begin{align*} AB - I = AB_L - I \in \Psi^{-\infty}(M). \end{align*} Thus $B \in \Psi^{-m}_{\mathrm{cl}}(M)$ is a two-sided inverse for $A$ modulo smoothing operators. This is precisely a parametrix for the [elliptic operator](/page/Elliptic%20Operator) $A$. [/step]