Gelfand-Kirillov dimension of the first Weyl algebra

Question

How can we compute the Gelfand-Kirillov dimension (GK for short) of the first Weyl algebra?

As we know we can look at the Weyl algebra as a generalized Weyl algebra in the following way:

Let $A=\mathbb F[h]$ be a polynomial algebra in the variable $h$. Consider its automorphism $\sigma$ defined by $\sigma(h)=h-1$ and the element $t=h$. Then the generalized Weyl algebra $A(\sigma, t)$ is isomorphic to the Weyl algebra $A_1(\mathbb F)$. And the Gelfand-Kirillov dimension (GK for short) of the generalized Weyl algebras are known in Zhao, Mo, and Zhang - Gelfand–Kirillov dimension of generalized Weyl algebras

I understood that GK($\mathbb F[h]$) is 1 and GK($A_1(\mathbb F)$) is 2, which means that $GK(A_1(\mathbb F))=GK(\mathbb F[h])+1$.

Now the question is that why $GK(\mathbb F[h])=1$? And why $GK(\mathbb F[h_1,\ldots,h_m])=m$?

Answer 1

Let $R=\mathbb{F}[h]$. That $GK(R)=1$ comes straight from the definition of Gelfand-Kirillov dimension. Let $R_n$ be the $\mathbb{F}$-subspace of $R$ spanned by monomials of degree at most $n$. Set $f(n)=\dim R_n = n+1$. Then $$GK(R) = \limsup(\log f(n)/\log n) = 1.$$ A similar argument applies if $R=\mathbb{F}[h_1,\dots,h_m]$. With $R_n$ as above, $f(n)=\dim R_n = \binom{n+m}{m}$ and $$GK(R) = \limsup(\log f(n)/\log n) = m.$$

Regarding the Weyl algebra $A=A_1(\mathbb{F})$, it helps to know that GK dimension is stable under taking associated graded rings. We have $A=\mathbb{F}\langle x,y : xy-yx-1\rangle$. Take the standard filtration on $A$ with $\deg(x)=\deg(y)=1$, so $gr(A)=\mathbb{F}[x,y]$. Thus, $GK(A)=GK(\mathbb{F}[x,y])=2$.

Good references on GK dimension are "Growth of algebras and Gelʹfand-Kirillov dimension" by Krause & Lenagan, as well as "Noncommutative noetherian rings" by McConnell & Robson.