I had a question about differential operators. I give some context. The operator \begin{align} x\partial_x \end{align} has as differential symbol the function $x\xi$ (replacing $\partial_x$ by $\xi$)
Another way to obtain the symbol $P(x,\xi)$ of an arbitrary differential operator $P(x,\partial_x)$ is by
\begin{align} P(x,\xi)=e^{-x\xi}P(x,\partial_x)e^{x\xi}(1) \end{align} where 1 is the function identically 1.
Given the above, with $P(x,\partial_x)=x\partial_x$, as
\begin{align} P(x,\xi)&=e^{-x\xi}x\partial_x e^{x\xi}(1)\\ &=e^{-x\xi}x\xi e^{x\xi}(1)\\ &=e^{-x\xi+x\xi}x\xi(1)\\ &=e^{0}x\xi(1)\\ &=x\xi \end{align}
Question 1. Does the operator $e^{\partial_x}$ have the symbol $e^{\xi}$? My attempt: \begin{align} e^{-x\xi}e^{\partial_x}e^{x\xi}(1)&=e^{-x\xi}\sum_{k=0}^{\infty}\frac{\partial_x^k}{k!}e^{x\xi}(1)\\ &=e^{-x\xi}\sum_{k=0}^{\infty}\frac{\xi^k}{k!}e^{x\xi}(1)\\ &=e^{-x\xi}e^{\xi}e^{x\xi}(1)\\ &=e^{\xi} \end{align}
Another way to calculate the symbol is with the Baker-Campbell formula (https://en.wikipedia.org/wiki/Baker–Campbell–Hausdorff_formula
My attempt:
\begin{align} e^{-x\xi}e^{\partial_x}e^{x\xi}(1)&=e^{-x\xi}e^{\partial_x+x\xi+\frac{1}{2}\xi}(1)\\ &=e^{(-x\xi+\partial_x+x\xi+\frac{1}{2}\xi)+\frac{1}{2}\xi}(1)\\ &=e^{\partial_x+\xi}(1)\\ &=\left.\sum_{k=0}^{\infty}\frac{(\xi+\partial_x)^k}{k!}(f)\right|_{f=1}\\ &=e^{\xi} \end{align}
Question 2. What should be the Lie group and Lie algebra to apply this formula? I don't know what they are. I guess that
\begin{align} \text{Lie algebra}=\left\{aI,b\partial_x:a,b\in\mathbb{R}\right\}\\ \text{Lie Group}=\left\{e^{aI}, e^{b\partial_x}:a,b\in\mathbb{R}\right\}? \end{align} Thanks.
