Connecting Differential Operators and Lie Brackets in ODEs

In ODE An Algebraic Approach 1, we observe a beautiful identity:

$$\begin{array}{}\text{(1)}& (D-\lambda )=e^{\lambda x}\circ D\circ e^{-\lambda x}\end{array}$$

In this essay, I would like to explain the connection between $(1)$$(1)$ and Lie bracket.

Firstly, we can prove this identity from the Lie bracket.

i.e.

$$\begin{array}{}\text{(2)}& [D,e^{\lambda x}]=\lambda e^{\lambda x}=D\circ e^{\lambda x}-e^{\lambda x}\circ D\end{array}$$

Thus

$$\begin{array}{}\text{(3)}& e^{\lambda x}\circ D+e^{\lambda x}\lambda =D\circ e^{\lambda x}⟹e^{\lambda x}(D+\lambda )=D\circ e^{\lambda x}\end{array}$$

Then take the inverse of $e^{\lambda x}$$e^{\lambda x}$

$$\begin{array}{}\text{(4)}& \begin{array}{c}\\ ⟹(D+\lambda )=e^{-\lambda x}\circ D\circ e^{\lambda x}⟹e^{\lambda x}\circ D\circ e^{-x}=(D-\lambda )\end{array}\end{array}$$

Conversely,

$$\begin{array}{}\text{(5)}& \frac{d}{dt}{|}_{t=0}(e^{tX}\circ Y\circ e^{-tX})=(X{e}^{tX}\circ Y\circ e^{-tX}+e^{tX}\circ Y\circ -X){|}_{t=0}=XY-YX=[X,Y]\end{array}$$

 

Cauchy integral formula and dual basis

Cauchy Integral Theorem is a well-known theorem in complex analysis.

i.e.

$$\begin{array}{}\text{(1)}& \frac{n!}{2\pi i}{\oint }_{\gamma }\frac{f(z)}{(z-a{)}^{n+1}}=f^{(n)}(a)\end{array}$$

It remind me that when I proved Taylor series, I used the fact that

$$\begin{array}{}\text{(2)}& \frac{1}{i!}\frac{d^i}{d{x}^i}{X}^j={\delta }_{ij}\end{array}$$

i.e. Using the fact that $X^j$$X^j$ form a basis of $\mathbb{R}[[X]]$$\mathbb R[[X]]$, then $\begin{array}{c}\frac{1}{i!}\frac{d^i}{d{x}^i}\end{array}$$\frac{1}{i!}\frac{d^i}{dx^i}\\$ as the dual basis of them will give the coordinate.

Similarly, we have

$$\begin{array}{}\text{(3)}& \begin{array}{c}{\oint }_{\gamma }{z}^n=\{\begin{array}{l}0,n\ne -1\\ 2\pi i,n=-1\end{array}\end{array}\end{array}$$

Thus

$$\begin{array}{}\text{(4)}& \frac{1}{2\pi i}{\oint }_{\gamma }\frac{(z-a{)}^i}{(z-a{)}^{j+1}}dz={\delta }_{ij}\end{array}$$

Therefore

$$\begin{array}{}\text{(5)}& \frac{1}{2\pi i}{\oint }_{\gamma }\frac{f(z)}{(z-a{)}^{j+1}}dz=\frac{f^{(n)}(a)}{n!}\end{array}$$