MHB First Weyl Algebras .... A_1 ....

[Math Amateur]

I am reading Matej Bresar's book, "Introduction to Noncommutative Algebra" and am currently focussed on Chapter 1: Finite Dimensional Division Algebras ... ...

I need help with some remarks of Bresar on first Weyl Algebras ...

Bresar's remarks on Weyl Algebras are as follows:https://www.physicsforums.com/attachments/6236
View attachment 6237In the above text from Bresar we read the following: (see (1.4) above)

" ... ... It is straightforward to check that

$$[D, L] = I$$

the identity operator ..."Can someone please explain exactly why $$[D, L] = I$$ ... ... ?

(It looks more like $$[D, L] = 0$$ to me?)Help will be appreciated ...

Peter

 

[Opalg]

In the above text from Bresar we read the following: (see (1.4) above)

" ... ... It is straightforward to check that

$$[D, L] = I$$

the identity operator ..."Can someone please explain exactly why $$[D, L] = I$$ ... ... ?

(It looks more like $$[D, L] = 0$$ to me?)Help will be appreciated ...

Peter

Effectively, this is just the product rule from calculus, which tells you that $$\frac d{dx}\bigl(xf(x)\bigr) = f(x) + x\frac d{dx}\bigl(f(x)\bigr).$$ If $D$ denotes differentiation, $L$ is the operation of multiplication by $x$, and $I$ is the identity operator, then that equation can be written as $$DL\bigl(f(x)\bigr) = I\bigl(f(x)\bigr) + LD\bigl(f(x)\bigr).$$ When you hide the $\bigl(f(x)\bigr)$s and write this as an operator equation, it says $DL = I + LD$, in other words $[D,L] = DL - LD = I.$

 

[Math Amateur]

Effectively, this is just the product rule from calculus, which tells you that $$\frac d{dx}\bigl(xf(x)\bigr) = f(x) + x\frac d{dx}\bigl(f(x)\bigr).$$ If $D$ denotes differentiation, $L$ is the operation of multiplication by $x$, and $I$ is the identity operator, then that equation can be written as $$DL\bigl(f(x)\bigr) = I\bigl(f(x)\bigr) + LD\bigl(f(x)\bigr).$$ When you hide the $\bigl(f(x)\bigr)$s and write this as an operator equation, it says $DL = I + LD$, in other words $[D,L] = DL - LD = I.$

Thanks Opalg ...

Got the idea now ...

Thanks again,

Peter