I have seen various definitions of symbols used to define pseudo-differential operators. The class of symbols I am working with is
$S^d(U)=\{p(x,\xi)\in C^{\infty}\left (U\times \mathbb R^m ,M_{k\times l}(\mathbb C) \right): p$ has compact $x$-support in $U , \ \forall \ \alpha,\beta $ multi-indices $ \exists \ C_{\alpha,\beta}>0$ such that $\left | D^{\alpha}_xD^{\beta}_{\xi}p(x,\xi)\right |\leq C_{\alpha,\beta}(1+|\xi|)^{d-|\beta|} \}$
This defines a $\Psi DO$ of order $d$
$P: C^{\infty}_c (U)\rightarrow C^{\infty}_c(U) \\ f\longmapsto Pf$
defined by $Pf(x)=\int e^{ix\cdot\xi}p(x,\xi)\hat f(\xi) d\xi $
Let $\Psi^d(U)$ be the class of all pseudo-differential operators defined as above.
The leading symbol map is then given by
$\sigma_L :\Psi^d(U)\rightarrow \frac{S^d(U)}{S^{d-1}(U)} \\ P\mapsto [p(x,\xi)]$
Is there a canonical choice of representative for the leading symbol like in the case of differential operators of degree $d$, one can canonically assign a homogeneous polynomial of degree $d$.
https://mathoverflow.net/questions/75976/symbol-of-pseudodiff-operator/77437#77437
In the answer to this question there is a suggestion of defining the leading symbol as $\sigma_LP(x,\xi) = \lim_{\lambda \rightarrow \infty } \lambda^{-d}e^{-i\lambda \phi}P(e^{i\lambda \phi}) $ \ but I can't really see why this limit should exist.