# 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

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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