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$\DeclareMathOperator\GKdim{GKdim}$Here I am trying to find the Gelfand–Kirillov dimension of the first Weyl algebra just by using the definition of the Gelfand–Kirillov dimension.

Let $A$ be an affine $\mathbb K$-algebra, and let $V\subseteq A$ be a finite dimensional $\mathbb K$ subspace (containing $1$) which generates $A$ as algebra. Let $V^n$ denote the subspace spanned by all products of $n$ elements from $V$. There is a chain of subspaces $$\mathbb K\subseteq V\subseteq V^2\subseteq\ldots\subseteq\cup_{n\geq0}V^n=A$$ and we define the Gelfand–Kirillov dimension of $A$, denoted $\GKdim(A)$, by $$\GKdim(A)=\lim\limits_{n\to\infty}\sup\frac{\log\dim(V^n)}{\log n}.$$

The first Weyl algebra $A_1=A_1(\mathbb K)$ is the ring of polynomials in 2 variables $x$ and $y$ with coefficients in $\mathbb K$, subject to the relation $yx-xy=1$.

Let us take $V=\mathbb K+\mathbb Kx+\mathbb Ky$. Hence it is the vector space generated by $1$, $x$ and y, and hence $\dim V=3$.

Let $V^2$ denote the subspace spanned by all products of 2 elements from $V$, i.e., generated by $\{ 1, x, y, x^2, xy, y^2\}$. And hence $\dim V^2=6$.

Now let $V^3$ denote the subspace spanned by all products of 3 elements from $V$, i.e. generated by $\{ 1, x, y, x^2, xy, y^2, x^3, x^2y, yx^2, yxy, y^3\}$. And hence $\dim V^3=11$.

But I think I am doing something wrong because I can see no recursive relation!?

I have to say that at the end we should get $\GKdim(A_1)=2$.

I will appreciate if someone can help me with this.

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    $\begingroup$ I would suggest to PBW order the monomials. Then you will see that $\dim V^3 = 10$ (you have $x^2y$ and $yx^2$ which have the same leading term). In general, $\dim V^n = (n+1)(n+2)/2$. $\endgroup$
    – Pavel Safronov
    Sep 3, 2020 at 18:21
  • $\begingroup$ Thank you very much for your comment! Yes, you are right. I was not seeing this point. Thanks! $\endgroup$
    – user513784
    Sep 3, 2020 at 19:16
  • $\begingroup$ If you're familiar with the fact that the associated graded with respect to this filtration is $k[x, y]$ with the usual grading then $\dim V^n - \dim V^{n-1} = n+1$ must be the number of monomials in $k[x, y]$ with total degree $n$ which gives Pavel's answer. $\endgroup$
    – Qiaochu Yuan
    Sep 3, 2020 at 21:57
  • $\begingroup$ @QiaochuYuan Thank you very much for your comment! $\endgroup$
    – user513784
    Sep 4, 2020 at 20:21

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