# I First Weyl Algebras ... A_1 ...

1. Nov 27, 2016

### 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:

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

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2. Nov 27, 2016

### eys_physics

Apply the commutator $[D,L]=DL-LD$ to a test function $f(\omega)$. What do you get for $(DL)f(\omega)$ and $(LD)f(\omega)$?

3. Nov 27, 2016

### Staff: Mentor

You have to take the product rule of differentiation into account for $D(L(f))$.

4. Nov 27, 2016

### Math Amateur

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

5. Nov 27, 2016

### 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 $\mathbb{F}$).

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