[proofplan] We compare the same Schwartz kernel of $A$ in two coordinate systems. In $x$-coordinates the phase is $(x-x')\cdot\xi$; after writing the kernel in $y$-coordinates, the phase becomes $(\kappa^{-1}(y)-\kappa^{-1}(y'))\cdot\xi$. Near the diagonal, the first-order part of this phase changes the Fourier variable by the transpose Jacobian, while the coordinate density Jacobian is cancelled by the Fourier-variable Jacobian. The nonlinear part of the coordinate change and all off-diagonal Jacobian corrections enter only through the standard symbolic expansion and lower the symbol order by at least one, so the homogeneous degree-$m$ term transforms by the cotangent lift. [/proofplan] [step:Write the local kernel in the first coordinate chart] Let \begin{align*} X=x(U\cap V)\subset\mathbb R^n \end{align*} and \begin{align*} Y=y(U\cap V)\subset\mathbb R^n. \end{align*} The transition map is the diffeomorphism \begin{align*} \kappa:X\to Y. \end{align*} Its inverse is denoted \begin{align*} \lambda:Y\to X, \qquad \lambda=\kappa^{-1}. \end{align*} Since $a$ is a complete local symbol for $A$ in the $x$-chart and the quantisation convention is fixed, the local Schwartz kernel of $A$ modulo a smooth kernel has the oscillatory representation \begin{align*} K_x(x_0,x_1)=(2\pi)^{-n}\int_{\mathbb R^n}e^{i(x_0-x_1)\cdot\xi}a(x_0,\xi)\,d\mathcal L^n(\xi). \end{align*} Here \begin{align*} K_x:X\times X\to\mathcal D'(X\times X) \end{align*} denotes the coordinate expression of the kernel with respect to the coordinate density $d\mathcal L^n(x_1)$ in the integration variable, $\mathcal L^n$ denotes $n$-dimensional [Lebesgue measure](/page/Lebesgue%20Measure) on $\mathbb R^n$, and \begin{align*} a:X\times\mathbb R^n\to\mathbb C \end{align*} is classical of order $m$. The principal part $a_m$ is the homogeneous degree-$m$ component in the classical expansion of $a$ for $|\xi|\geq 1$. Replacing $a$ by another representative differing by an element of $S^{-\infty}$ changes no homogeneous component of finite degree, so $a_m$ is independent of the chosen complete representative. [/step] [step:Transform the kernel to the second coordinate chart] Let \begin{align*} J_\lambda:Y\to(0,\infty) \end{align*} be the absolute coordinate Jacobian \begin{align*} J_\lambda(y)=|\det d\lambda_y|. \end{align*} Since the local operator acts by integrating the kernel against the coordinate density in the second variable, define the $y$-coordinate kernel $K_y:Y\times Y\to\mathcal D'(Y\times Y)$ by \begin{align*} K_y(y_0,y_1)=K_x(\lambda(y_0),\lambda(y_1))J_\lambda(y_1) \end{align*} after the change of variables $x_1=\lambda(y_1)$. Therefore, modulo a smooth kernel, \begin{align*} K_y(y_0,y_1)=(2\pi)^{-n}\int_{\mathbb R^n}e^{i(\lambda(y_0)-\lambda(y_1))\cdot\xi}a(\lambda(y_0),\xi)J_\lambda(y_1)\,d\mathcal L^n(\xi). \end{align*} [guided] The kernel is not an ordinary scalar function under a change of coordinates in both variables, because the second variable is integrated against a density. If the operator is written in $x$-coordinates as \begin{align*} (Au)(x_0)=\int_X K_x(x_0,x_1)u(x_1)\,d\mathcal L^n(x_1), \end{align*} then writing $x_1=\lambda(y_1)$ transforms the measure by \begin{align*} d\mathcal L^n(x_1)=J_\lambda(y_1)\,d\mathcal L^n(y_1). \end{align*} Thus the $y$-coordinate kernel must be \begin{align*} K_y(y_0,y_1)=K_x(\lambda(y_0),\lambda(y_1))J_\lambda(y_1). \end{align*} Substituting the oscillatory formula for $K_x$ gives \begin{align*} K_y(y_0,y_1)=(2\pi)^{-n}\int_{\mathbb R^n}e^{i(\lambda(y_0)-\lambda(y_1))\cdot\xi}a(\lambda(y_0),\xi)J_\lambda(y_1)\,d\mathcal L^n(\xi). \end{align*} This is the exact place where the density bookkeeping enters. Without the factor $J_\lambda(y_1)$, the principal symbol transformation would appear to have an extra determinant factor after changing Fourier variables. That factor is cancelled at the diagonal in the next step. [/guided] [/step] [step:Linearize the phase at the diagonal and change the Fourier variable] Fix $y_0\in Y$, set \begin{align*} x_0=\lambda(y_0), \end{align*} and define the linear isomorphism \begin{align*} L_{y_0}:\mathbb R^n\to\mathbb R^n, \qquad L_{y_0}=d\lambda_{y_0}. \end{align*} Taylor expansion of $\lambda$ at $y_0$ gives \begin{align*} \lambda(y_0)-\lambda(y_1)=L_{y_0}(y_0-y_1)+R(y_0,y_1), \end{align*} where \begin{align*} R:Y\times Y\to\mathbb R^n \end{align*} is smooth and satisfies $R(y_0,y_1)=O(|y_0-y_1|^2)$ as $y_1\to y_0$. The linear part of the phase satisfies \begin{align*} L_{y_0}(y_0-y_1)\cdot\xi=(y_0-y_1)\cdot L_{y_0}^\top\xi. \end{align*} Set \begin{align*} \eta=L_{y_0}^\top\xi. \end{align*} Equivalently, \begin{align*} \xi=(L_{y_0}^{-1})^\top\eta=(d\kappa_{x_0})^\top\eta. \end{align*} The Lebesgue measure transforms by \begin{align*} d\mathcal L^n(\xi)=|\det L_{y_0}|^{-1}\,d\mathcal L^n(\eta). \end{align*} At the diagonal $y_1=y_0$, the density factor is \begin{align*} J_\lambda(y_0)=|\det L_{y_0}|. \end{align*} Hence the product of the coordinate-density factor and the Fourier-variable Jacobian is $1$ at principal order: \begin{align*} J_\lambda(y_0)|\det L_{y_0}|^{-1}=1. \end{align*} The leading amplitude after this linearized change is therefore \begin{align*} a\left(x_0,(d\kappa_{x_0})^\top\eta\right). \end{align*} [/step] [step:Show that nonlinear coordinate corrections lower the symbol order] We use the standard symbolic expansion for a nondegenerate change of phase in an oscillatory integral, with the result not yet cited in the wiki: Change of variables for oscillatory integral symbols. The phase \begin{align*} (y_0,y_1,\xi)\mapsto (\lambda(y_0)-\lambda(y_1))\cdot\xi \end{align*} has nondegenerate linear part at the diagonal because $d\lambda_{y_0}$ is invertible, and the amplitude \begin{align*} (y_0,y_1,\xi)\mapsto a(\lambda(y_0),\xi)J_\lambda(y_1) \end{align*} is smooth in $(y_0,y_1)$ and classical of order $m$ in $\xi$. Applied to this phase and amplitude, the theorem gives a complete $y$-symbol \begin{align*} \tilde b:Y\times\mathbb R^n\to\mathbb C \end{align*} such that the corresponding oscillatory kernel is \begin{align*} (2\pi)^{-n}\int_{\mathbb R^n}e^{i(y_0-y_1)\cdot\eta}\tilde b(y_0,\eta)\,d\mathcal L^n(\eta), \end{align*} and \begin{align*} \tilde b(y_0,\eta)-a(\lambda(y_0),(d\kappa_{\lambda(y_0)})^\top\eta)\in S^{m-1}(Y\times\mathbb R^n). \end{align*} The reason is that every term beyond the diagonal linearization contains either a derivative of $a$ with respect to the covariable or a factor vanishing to first order at $y_1=y_0$. Differentiating a symbol of order $m$ once in the covariable lowers its order to $m-1$, and multiplication by smooth coordinate coefficients does not increase symbol order. Thus the nonlinear part $R(y_0,y_1)$, the difference $J_\lambda(y_1)-J_\lambda(y_0)$, and all higher Taylor terms contribute only to $S^{m-1}$ or lower. [guided] The point of this step is to separate what can affect the top homogeneous term from what can only affect lower-order terms. After the previous step, the linearized phase already has the standard form \begin{align*} (y_0-y_1)\cdot\eta. \end{align*} The remaining difference between the exact phase and the linearized phase is the smooth remainder \begin{align*} R(y_0,y_1)\cdot\xi, \end{align*} where $R(y_0,y_1)=O(|y_0-y_1|^2)$ as $y_1\to y_0$. The symbolic expansion for nondegenerate changes of phase says that such a kernel can be rewritten in the standard left-quantised form with a complete symbol $\tilde b$. In the present notation, this gives \begin{align*} K_y(y_0,y_1)=(2\pi)^{-n}\int_{\mathbb R^n}e^{i(y_0-y_1)\cdot\eta}\tilde b(y_0,\eta)\,d\mathcal L^n(\eta) \end{align*} modulo a smooth kernel. The expansion is obtained by Taylor expanding the smooth coefficients at $y_1=y_0$ and converting powers of $y_0-y_1$ into derivatives with respect to $\eta$ by [integration by parts](/theorems/210) in the oscillatory integral. Why does this lower the order? A factor of $y_0-y_1$ paired with the exponential $e^{i(y_0-y_1)\cdot\eta}$ is converted into a derivative $\partial_{\eta_j}$ falling on the amplitude. If the amplitude is a symbol of order $m$, then one covariable derivative is a symbol of order $m-1$. The nonlinear phase remainder vanishes at least quadratically at the diagonal, so its first nonzero contributions contain at least one such covariable derivative. The off-diagonal density correction \begin{align*} J_\lambda(y_1)-J_\lambda(y_0) \end{align*} vanishes to first order at the diagonal, so it is handled in the same way and contributes only to $S^{m-1}$. Consequently the only contribution to the homogeneous degree-$m$ term comes from evaluating the transformed amplitude at the diagonal and using only the linear part of the phase. The Jacobian cancellation from the preceding step gives \begin{align*} \tilde b(y_0,\eta)=a(\lambda(y_0),(d\kappa_{\lambda(y_0)})^\top\eta)+r(y_0,\eta), \end{align*} where \begin{align*} r:Y\times\mathbb R^n\to\mathbb C \end{align*} is a symbol in $S^{m-1}(Y\times\mathbb R^n)$. [/guided] [/step] [step:Extract the homogeneous degree-$m$ term] The symbol $b$ is another complete local symbol for the same operator in the same $y$-quantisation convention. Therefore \begin{align*} b-\tilde b\in S^{-\infty}(Y\times\mathbb R^n). \end{align*} A smoothing symbol has no homogeneous component of degree $m$, and a symbol in $S^{m-1}$ has no homogeneous component of degree $m$. Taking homogeneous degree-$m$ parts in \begin{align*} \tilde b(y_0,\eta)=a(\lambda(y_0),(d\kappa_{\lambda(y_0)})^\top\eta)+r(y_0,\eta) \end{align*} therefore gives \begin{align*} b_m(y_0,\eta)=a_m(\lambda(y_0),(d\kappa_{\lambda(y_0)})^\top\eta) \end{align*} for every $y_0\in Y$ and every $\eta\neq0$. Substituting $y_0=\kappa(x_0)$ yields \begin{align*} b_m(\kappa(x_0),\eta)=a_m(x_0,(d\kappa_{x_0})^\top\eta). \end{align*} This is the asserted transformation law. [/step] [step:Patch the local principal parts on $T^*M\setminus 0$] Let $p\in U\cap V$, and let \begin{align*} \xi_p\in T_p^*M\setminus\{0\} \end{align*} be a nonzero covector. In the $x$-chart, write its coordinate representative as \begin{align*} \xi\in\mathbb R^n\setminus\{0\}. \end{align*} In the $y$-chart, write its coordinate representative as \begin{align*} \eta\in\mathbb R^n\setminus\{0\}. \end{align*} The coordinate representatives of the same covector satisfy \begin{align*} \xi=(d\kappa_{x(p)})^\top\eta. \end{align*} The transformation law just proved gives \begin{align*} b_m(y(p),\eta)=a_m(x(p),\xi). \end{align*} Thus the local prescriptions agree on overlaps. Define \begin{align*} \sigma_m(A):T^*M\setminus0\to\mathbb C \end{align*} by setting $\sigma_m(A)(p,\xi_p)$ equal to $a_m(x(p),\xi)$ in any coordinate chart. The overlap agreement proves that this definition is independent of the chart. Since each $a_m$ is smooth in the base variable and smooth homogeneous of degree $m$ in the nonzero covariable, the patched function $\sigma_m(A)$ is smooth on $T^*M\setminus0$ and homogeneous of degree $m$ on each cotangent fiber. This completes the proof. [/step]