This is a passage from Hormanders book, the analysis of linear partial differential operators volume 3.
If the adjoint of the symbol $a(x,\xi)$ is $b(x,\xi)=(2\pi)^{-n}\int e^{-i y \cdot \eta}\bar{a}(x-y,\xi - \eta)dyd\eta$. Since the quadratic form $(y,\eta) \rightarrow 2 y \cdot \eta$ has signature 0, determinant $(-1)^n$ and is its own dual, the fourier transform of $(2\pi)^{-n}e^{-iy\cdot \eta}$ is equal to $e^{i \hat{y} \cdot \hat{\eta}}$, where $\hat{y}, \hat{\eta}$ are the dual variables of $y,\eta$. Then $$b(x,\xi)=e^{iD_x \cdot D_{\xi}}\bar{a}(x,\xi)$$ In the sense that the fourier transform of b is equal to that of $\bar{a}$ multiplied by $e^{i\hat{x}\cdot\hat{\xi}}$.
My problems are the following
1)the fourier transform of $(2\pi)^{-n}e^{-iy\cdot \eta}$ is equal to $e^{i \hat{y} \cdot \hat{\eta}}$, is this a fourier transform with respect to both variables?
2)When it says the last part, the fourier tranform of b,is this applying a fourier transform to the first formula of the adjoint?
Thank you