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Why is this algebra called the q-Weyl algebra?

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.?

MathOverflow Asked by GMRA on February 12, 2021

1 Answers

One Answer

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.

Correct answer by David E Speyer on February 12, 2021

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