9
$\begingroup$

Recall that the (first) Weyl algebra over $\mathbb{C}$ is the algebra generated by $x,y$ with the relation $yx-xy=1$. It can be realized as the algebra of polynomial differential operators in 1 variable, i.e. $\mathbb{C}[x]$ is a faithful representation, where $x$ acts by multiplication by $x$ and $y$ acts by $\frac{\partial}{\partial x}$.

In many places I've seen the q-Weyl algebra defined as the algebra generated by $x,y$ with relations yx-qxy=1, where q is some fixed nonzero scalar. This seems like a natural way to do it, and seems to be quite common.

Now in the notes Introduction to representation theory Etingof et al. define an algebra which they call the q-Weyl algebra. This is the $\mathbb{C}$-algebra generated by $x,x^{-1},y,y^{-1}$ with the relations $xx^{-1} = x^{-1}x = 1$, $yy^{-1}= y^{-1}y = 1$ and $xy=qyx$ where q is some fixed nonzero scalar.

My question then is: What is the reason for the name 'q-Weyl algebra' for the algebra defined by Etingof et al.?

$\endgroup$
2
  • $\begingroup$ What happens with the Weyl algebra if you replace $\partial/\partial x$ with a suitable $q$-differentiation? $\endgroup$ May 25, 2010 at 0:16
  • $\begingroup$ For another q-Weyl algebra, you can look at $A^-_q(n)$ from Hiyashi's "q-analog of Clifford and Weyl algebras" $\endgroup$
    – B. Bischof
    Apr 26, 2011 at 3:22

1 Answer 1

18
$\begingroup$

Suppose that $x$ and $y$ obey $xy-yx=h$, where $h$ is a central element. Set $X$ and $Y$ to be $e^x$ and $e^y$. For now, don't worry too much about what this exponentiation means. Then $XY=q YX$, where $q=e^h$.

If we interpret $X$ and $Y$ as operators on functions then $(Xf)(x)=e^x f(x)$ and $(Yf)(x) = f(x+h)$. You can check that $XY$ does indeed equal $q YX$. So the $q$-commutation relation can be seen as an exponentiation of the standard Weyl relation.

$\endgroup$
1
  • 6
    $\begingroup$ What I find interesting is that what we call the Weyl algebra is built from the canonical commutation relations (CCR) introduced by Heisenberg (thus also known as the Heisenberg commutation relations), whereas what Weyl really considered was the exponentiated form of these relations (replacing the Lie commutator with the group commutator) leading to what we call the $q$-Weyl algebra. $\endgroup$ May 25, 2010 at 4:25

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.