# Lie group of differential operators

I have the following three partial differential operators

$$A=y \frac{\partial}{\partial y}$$ $$B=y^{-1}(z\frac{\partial}{\partial z}+y\frac{\partial}{\partial y}+c-1)$$ $$C=y((1-z)\frac{\partial}{\partial z}-(a+b-c))$$

I found that the commentator relation of these operators are

$[A,B]=-B$ , $[A,C]=C$ and $$[B,C]=-z\frac{\partial}{\partial z}+(a+b-c)$$

Now to generate Lie group of these operators $[B,C]$ must be written as a linear combination of these operators

How could I write $[B,C]$ as a linear combination of these operators ?

If I add new operator $$D=z \frac{\partial }{\partial z}$$ Is the problem could be solved ??

• If you're going to add an operator, why not just add $[B,C]$? Mar 9, 2016 at 23:57
• Adding $[B,C]$ then It should be compute it's relation with the other operators $A,B,C$ and so the result may be could't written as a linear combination of the operators Mar 10, 2016 at 5:43
• "Adding $[B,C]$ then It should be compute it's relation with the other operators $A,B,C$" - No. Its relation to the other operators is $[B,C] = BC - CB$. It may turn out that $[B,C]$ is a linear combination of $A, B, C$, but there is no requirement that this has to be. My earlier comment was in response to your last question. There you suggest adding a different operator $D$ to $A, B, C$. But why would you choose that particular operator instead of defining $D = [B, C]$ to be the new add. Then your solution is obvious. Mar 10, 2016 at 15:46
"Adding $$[B,C]$$ then It should be compute it's relation with the other operators $$A,B,C$$" - No. Its relation to the other operators is $$[B,C]=BC−CB$$. It may turn out that $$[B,C]$$ is a linear combination of A,B,C, but there is no requirement that this has to be. My earlier comment was in response to your last question. There you suggest adding a different operator D to A,B,C. But why would you choose that particular operator instead of defining $$D=[B,C]$$ to be the new add. Then your solution is obvious.