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

  • #1
Math Amateur
Gold Member
MHB
3,988
48
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 6237


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

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

\(\displaystyle [D, L] = I\)

the identity operator ..."


Can someone please explain exactly why \(\displaystyle [D, L] = I\) ... ... ?

(It looks more like \(\displaystyle [D, L] = 0\) to me?)


Help will be appreciated ...

Peter
 

Answers and Replies

  • #2
Opalg
Gold Member
MHB
2,779
4,000
In the above text from Bresar we read the following: (see (1.4) above)

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

\(\displaystyle [D, L] = I\)

the identity operator ..."


Can someone please explain exactly why \(\displaystyle [D, L] = I\) ... ... ?

(It looks more like \(\displaystyle [D, L] = 0\) 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 $[D,L] = DL - LD = I.$
 
  • #3
Math Amateur
Gold Member
MHB
3,988
48
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
 

Suggested for: First Weyl Algebras .... A_1 ....

  • Last Post
Replies
4
Views
1K
Replies
8
Views
578
  • Last Post
Replies
2
Views
2K
Replies
8
Views
1K
Replies
2
Views
464
Replies
14
Views
798
  • Last Post
Replies
25
Views
915
Replies
1
Views
961
Replies
7
Views
369
Replies
6
Views
425
Top