Bernstein growth dimension of a Weyl algebra module ($d(M)=\mathrm{deg}({\dim }_{\mathbb{C}}{F}_j{A}_n{M}_0)$)

For a nonzero finitely generated Weyl algebra module, choose a finite-dimensional generating subspace $M_0$. Under the Bernstein filtration, $\dim (F_j{A}_n{M}_0)$ is eventually polynomial because the graded module is finite over a polynomial ring. Its degree is independent of the generating subspace, since two generating filtrations bound each other after fixed shifts. It equals the Gelfand–Kirillov dimension of a module.

For a nonzero finitely generated module over the complex Weyl algebra $A_n$, its Bernstein growth dimension of a Weyl algebra module is at least $n$. The faithful finite-step action of a Weyl algebra gives $\dim F_j{A}_n\le (\dim M_j)(\dim M_{2j})$. Comparing degrees $2n$ and $2d(M)$ proves the inequality. The polynomial module has dimension $n$, so the bound is sharp. This is an algebraic growth inequality, distinct from probabilistic Bernstein inequalities.

For $M_j=F_j{A}_n{M}_0$ and $M_0\ne 0$, the action map in the display is injective. If $a\in F_j$ kills $M_j$, each commutator $[a,z]$ with a generator lies in $F_{j-1}$ and kills $M_{j-1}$. Induction makes all these commutators zero, so $a$ is scalar; since it kills $M_0$, it is zero. This finite-step faithfulness does not require the whole module to be finite-dimensional.