Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

I First Weyl Algebras ... A_1 ...

  1. Nov 27, 2016 #1
    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:



    ?temp_hash=509e38ac20c62b45d30f637569fa41c6.png
    ?temp_hash=509e38ac20c62b45d30f637569fa41c6.png


    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/how ##[D, L] = I## ... ... ?

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


    Help will be appreciated ...

    Peter
     

    Attached Files:

  2. jcsd
  3. Nov 27, 2016 #2
    Apply the commutator ##[D,L]=DL-LD## to a test function ##f(\omega)##. What do you get for ##(DL)f(\omega)## and ##(LD)f(\omega)##?
     
  4. Nov 27, 2016 #3

    fresh_42

    Staff: Mentor

    You have to take the product rule of differentiation into account for ##D(L(f))##.
     
  5. Nov 27, 2016 #4
    Thanks for the help eys_physics and fresh_42 ... got the idea now ...

    Just needed the interpretation of DL as not a product but a composition of functions/operators ...which you provided ...

    By the way ... found it difficult to get a clear and consistent definition of "operator" from my various texts ...

    Thanks again,

    Peter
     
  6. Nov 27, 2016 #5

    fresh_42

    Staff: Mentor

    ... which is the product here. In case we deal with operators, the usual product is the matrix multiplication, resp. applying one operator after the other: ##(A\circ B)(v)=A(B(v))##. The successive application ##"\circ"## is usually written as a dot or left out. But it's right that one has to be careful, because for usual functions, say ##f,g : \mathbb{R} \rightarrow \mathbb{R}##, we have two (or three if you like) different multiplications:
    1. ##(f \cdot g) (x) = f(x) \cdot g(x)##
    2. ##(f \circ g) (x) = f(g(x))##
    3. and of course scalar multiplications ##(c\cdot f)(x)=c \cdot f(x)\, , \, c \in \mathbb{F}##
    The first one is differentiated with the product rule, the second with the chain rule (and the third by linearity of differentiation).

    Btw., the letters ##D## and ##L## for the operators above aren't chosen arbitrarily. ##D## stands for derivation (the result of a differentiation), and ##L## for left-multiplication (here with the function ##1(\omega)=\omega## the variable.)

    Not to confuse you, it's ##L(f)=1\cdot f## (multiplication #1 where the product rule for differentiation applies) and ##L(f)(\omega)=(1 \cdot f)(\omega)= 1(\omega)\cdot f(\omega)=\omega \cdot f(\omega)##. For the operators ##D## and ##L##, the product is #2, the successive application ##(D\circ L)(f) = D(L(f))## as linear operators on ##\mathbb{F}[\omega]## (and of course also #3 the scalar multiplication with elements of ##\mathbb{F}##).
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: First Weyl Algebras ... A_1 ...
  1. First Sylow Theorem (Replies: 1)

Loading...