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Choose a finite-dimensional generating vector subspace $V\subseteq A$ containing $1$, and write $V^n$ for the linear span of products of $n$ elements of $V$. The Gelfand–Kirillov dimension isFor a nonzero finitely generated module $M$ considered as a right module, choose a finite-dimensional generating vector subspace $W\subseteq M$ and define the Gelfand–Kirillov dimension of a module byFor left modules, replace $W{V}^n$ by $V^{n}W$. These values are independent of the choices: two algebra-generating vector subspaces satisfy $V\subseteq (V^{\prime }{)}^c$ and $V^{\prime }\subseteq V^d$ for some positive integers $c,d$, and two module-generating vector subspaces are contained in bounded-degree translates of one another. The corresponding growth bounds differ only by a constant rescaling and shift of $n$, which do not change the limit superior.

Let $S$ be a nonzero standard graded algebra, generated by $r$ elements of degree one. It is a homogeneous quotient ring of the polynomial ring $k[z_1,\dots ,z_r]$. By the Hilbert-Serre theorem, its Hilbert series is a rational function whose only possible pole is at $t=1$. After canceling, write it as $P(t)/(1-t{)}^d$ with $P(1)\ne 0$, where $0\le d\le r$. The cumulative dimensions are the coefficients of $P(t)/(1-t{)}^{d+1}$ and are eventually a polynomial in $n$ of degree $d$, with positive leading coefficient. With $V=S_0\oplus S_1$, these cumulative dimensions equal $\dim V^n$. ConsequentlyFor $d=0$ the algebra is finite-dimensional and nonzero, so the cumulative dimension is eventually a positive constant. If the zero algebra is allowed, the common convention $\mathrm{GKdim}0=-\infty$ is an exception to the printed assertion; the integer conclusion concerns nonzero unital algebras.

For the Weyl algebra $A=A_1(k)=k⟨X,D⟩/(DX-XD-1)$, the total-degree filtered algebra structure has associated graded ring $\mathrm{gr}A=k[x,\xi ]$. Its ordered monomials $X^i{D}^j$ form a basis. Every finitely generated module has a good filtration of a module, and its associated graded module is finitely generated over $k[x,\xi ]$. The Hilbert-Serre theorem therefore shows that its Gelfand–Kirillov dimension is an integer in $\{0,1,2\}$.

A nonzero module of Gelfand–Kirillov dimension zero here would have eventually constant cumulative dimension, hence be finite-dimensional. This is impossible in characteristic zero: taking the trace of the endomorphisms representing $DX-XD=1$ gives $0={\dim }_{k}M$. For right modules the right-action operators reverse composition and give the negative identity instead, with the same contradiction. ThusBoth occur. The regular module $A$ has $\dim V^n=\left(\genfrac{}{}{0ex}{}{n+2}{2}\right)$, hence Gelfand–Kirillov dimension two. For a right module of Gelfand–Kirillov dimension one, take $M=k[x]$ with $f\cdot X=xf$ and $f\cdot D=-f^{\prime }$. These actions satisfy the defining relation because $(f\cdot D)\cdot X-(f\cdot X)\cdot D=f$. The module is a cyclic module generated by $1$, and $\dim (1\cdot V^n)=n+1$. This also gives the lower bound for this particular Weyl algebra without invoking a general inequality for higher Weyl algebras.