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For the order filtration of a Weyl algebra, a filtration of a module is good if it is increasing, exhaustive, zero in sufficiently negative degrees, compatible with multiplication, and has associated graded module finite over the associated graded ring. Equivalently it has the displayed finite shifted-generating description. Choose homogeneous generators of the associated graded module, lift them to $m_{\ell }$, and successively remove the top symbol of any element; the lower bound on the filtration of a module makes this process terminate. Conversely, the displayed description makes the symbols of the $m_{\ell }$ generate. Each $M_j$ is finite over the degree-zero polynomial ring $\mathbb{C}[x_1,\dots ,x_n]$; it need not be a finite-dimensional vector space over $\mathbb{C}$.

The algebraic set defined by the annihilator of a module of the associated graded module for a good filtration for the order filtration is the order characteristic variety. It lies in affine space with coordinates $x_1,\dots ,x_n,{\xi }_1,\dots ,{\xi }_n$ and is invariant under scaling the ${\xi }_i$. Its dimension is the dimension of a Weyl algebra module for the order filtration. The order filtration of a Weyl algebra assigns degrees zero to the $x_i$ and one to the ${\xi }_i$; a characteristic variety formed with the Bernstein filtration need not be the same algebraic set.

Work over $\mathbb{C}$, or another field of characteristic zero. The Weyl algebra $A_n$ is generated by $x_1,\dots ,x_n,{\partial }_1,\dots ,{\partial }_n$ with commuting $x$'s, commuting $\partial$'s and $[{\partial }_i,x_j]={\delta }_{ij}$. Its ordered monomial basis of a Weyl algebra consists of the monomials $x^{\alpha }{\partial }^{\beta }$. One way to verify independence is to apply a finite operator relation to the formal exponential $e^{\mathbf{t}\cdot \mathbf{x}}$: after canceling that exponential the relation becomes a polynomial identity $\sum c_{\alpha \beta }{\mathbf{x}}^{\alpha }{\mathbf{t}}^{\beta }=0$, so every coefficient vanishes. The relations already reorder every word into that basis.

Give both generators degree one. The Bernstein filtration isThus $\dim F_m{A}_n=\left(\genfrac{}{}{0ex}{}{m+2n}{2n}\right)$. Commutator terms have lower degree, accounting for the commutative symbol algebra. Since that algebra is Noetherian, leading-symbol generation followed by finite downward degree cancellation proves that $A_n$ is Noetherian on both sides.

A left finitely generated module $M$ has a good filtration, for example $M_m={\sum }_j{F}_m{A}_n{m}_j$ for a finite generating set of the module. More generally one allows fixed degree shifts of the generators. The associated graded module is finite over the symbol algebra. DefineThis is the characteristic variety of a Weyl algebra module for the Bernstein algebra filtration. It records where the module's leading symbols remain nonzero and makes noncommutative module growth accessible to commutative algebra.

For a fixed algebra filtration it is independent of the good filtration. Any two such filtrations differ by bounded shifts $M_i\subseteq {\tilde{M}}_{i+c}$ and conversely, since each finite set of generators occurs in bounded degree in the other filtration. If a degree-$d$ operator has symbol annihilating $\mathrm{gr}M$, its $r$th power maps $M_i$ into $M_{i+rd-r}$. For $r>2c$, bounded-shift comparison forces the power of that symbol to annihilate $\mathrm{gr}\tilde{M}$. Reversing the argument gives equal radicals of ideals of annihilators. This is independence of characteristic support from a good module filtration. Changing the algebra filtration itself, for example to the order filtration of a Weyl algebra with the $x_i$ in degree zero, need not preserve the same variety; the convention matters.

The Hilbert-Serre theorem gives an eventual cumulative polynomialHere $d$ is also the Gelfand–Kirillov dimension of a module, and $e(M)>0$ is an integer multiplicity of a filtered module for nonzero $M$. Finite generation bounds $d\le 2n$.

Here is a proof sketch of the essential lower bound, rather than an appeal to its name. For a nonzero $f\in F_m{A}_n$, choose a nonzero monomial of greatest total degree in its ordered expansion, say $x^{\alpha }{\partial }^{\beta }$. The operations $\mathrm{ad}{\partial }_i$ and $\mathrm{ad}(-x_i)$ differentiate this expansion with respect to $x_i$ and ${\partial }_i$. Apply them ${\alpha }_i$ and ${\beta }_i$ times. Lower-degree terms die, and every other monomial of the same degree also dies unless it has exactly those exponents. The result is a nonzero scalar multiple of the chosen coefficient. Characteristic zero is essential for the factorials. Expanding the repeated commutators therefore givesChoose $0\ne v\in M_0$. If $f$ acted as zero on all $M_m$, every $f{v}_{j}v$ would vanish, contrary to $v={\sum }_j{u}_{j}f{v}_{j}v$. Hence the linear action mapis injective. It follows that $\dim F_m{A}_n\le (\dim M_m)(\dim M_{2m})$. Comparing polynomial degrees gives $2n\le 2d$. ThusThis proves Bernstein inequality for Weyl algebra modules through the faithful finite-step action of a Weyl algebra.

The bound is sharp. The polynomial representation $\mathbb{C}[x_1,\dots ,x_n]\cong A_n/{\sum }_i{A}_n{\partial }_i$ has the zero section as its characteristic variety of a Weyl algebra module, namely $\xi =0$ and dimension $n$. It is a simple module: differentiate a nonzero polynomial sufficiently to obtain a nonzero constant, then multiply by the $x_i$ to generate the whole module. The module $A_n/{\sum }_i{A}_n{x}_i$ similarly has characteristic variety of a Weyl algebra module $x=0$, a fiber of the cotangent bundle. A nonzero free module has the whole cotangent bundle and dimension $2n$. No nonzero finite-dimensional module exists for $n\ge 1$, also seen by taking the trace of $[{\partial }_1,x_1]=I$.

Modules attaining dimension $n$ are holonomic Weyl algebra modules (with the zero module included conventionally). Every nonzero subquotient remains of dimension $n$ by the lower bound and monotonicity of growth. Compatible filtrations in a short exact sequence make the leading multiplicity of a filtered module additive. Since each nonzero factor has positive integer multiplicity of a filtered module, the number of factors in any strict submodule chain is bounded by $e(M)$. Together with Noetherianity, this proves finite composition length. It illustrates why characteristic variety of a Weyl algebra module and multiplicity of a filtered module provide useful structure beyond a list of presentations. In positive characteristic there are finite-dimensional Weyl modules, so the characteristic-zero inequality and this discussion must not be asserted without its field hypothesis.

The base ring must be specified in the algebraic definition. For a commutative $k$-algebra $R$, work in ${\mathrm{End}}_k(R)$ and let $m_f$ mean multiplication by $f\in R$. Set $D_k^{-1}(R)=0$ and define recursivelyThese are the operators of order at most $i$. The order of a nonzero differential operator relative to a base ring is the least $i$ for which it belongs to $D_k^i(R)$; the zero operator can be assigned order $-\infty$. An operator of order zero commutes with all multiplications and is itself multiplication by $P(1)$, so $D_k^0(R)=R$. The union $D_k(R)={\bigcup }_{i\ge 0}{D}_k^i(R)$ is the ring of finite-order operators, with composition as multiplication. The identityand induction on the sum of orders give $D_k^p{D}_k^q\subseteq D_k^{p+q}$. Sums have order at most the maximum of their orders, proving that this union is a ring.

For an arbitrary commutative ring, the absolute version uses $k=\mathbb{Z}$, so the ambient endomorphisms are additive. For the polynomial ring in this question, the intended Weyl-algebra identification is the relative one with $k=\mathbb{C}$. Absolute operators would also include nonzero derivations of the coefficient field $\mathbb{C}$: a derivation on a rational function field in a transcendence basis over $\mathbb{Q}$ extends to its algebraic closure in characteristic zero, and hence to $\mathbb{C}$. Extending it coefficientwise to $\mathbb{C}[x]$ gives an order-one additive operator which is not complex-linear. Such an operator cannot belong to the complex Weyl algebra. This explains why the linearity convention matters.

Now take $R=\mathbb{C}[x_1,\dots ,x_n]$. Coordinate multiplication has order zero, and each partial derivative ${\partial }_i$ has order one because $[{\partial }_i,m_f]=m_{{\partial }_{i}f}$. They satisfy the defining Weyl algebra relations, giving an algebra homomorphism $A_n\to D_{\mathbb{C}}(R)$. Reordering spans $A_n$ by $x^{\alpha }{\partial }^{\beta }$. If an ordered expression acted as zero, commutation with the coordinates according to a maximal derivative multiindex would give multiplication by $\beta !f_{\beta }(x)=0$; removing the top coefficients successively proves that every coefficient is zero. Thus the homomorphism is injective, as in the ordered monomial basis of a Weyl algebra proof.

For surjectivity, let $P$ have order at most $m$, and put ${\delta }_{i}P=[P,m_{x_i}]$. The operations ${\delta }_i$ commute, since the coordinate multiplications commute. A product of more than $m$ of them kills $P$. Define polynomial coefficientsCommuting $P$ through $m_{x^{\alpha }}$ repeatedly gives the finite identityIt follows by induction on $|\alpha |$ from $P{m}_{x_i}=m_{x_i}P+{\delta }_{i}P$ and Pascal's identity. Terms with $|\beta |>m$ vanish. Sincefor $\beta \le \alpha$, we obtain $P(x^{\alpha })={\sum }_{|\beta |\le m}{a}_{\beta }(x){\partial }^{\beta }(x^{\alpha })$. The monomials are a complex basis of $R$, so complex linearity yields equality of the operators on all of $R$. HenceFor a nonzero ordered expression ${\sum }_{\beta }{a}_{\beta }(x){\partial }^{\beta }$, its order is exactly ${\max }_{a_{\beta }\ne 0}|\beta |$. The upper bound follows from composition; the lower bound follows by taking that many coordinate commutators to extract a nonzero multiplication operator. This describes the order filtration of a Weyl algebra:

The associated graded ring is thereforeThe symbol of ${\partial }_i$ is ${\xi }_i$. In a product of ordered operators, moving derivatives past coefficients introduces terms of lower differential order. Thus top-order symbols multiply as ordinary polynomials, and the ordered basis proves there are no further relations. In particular $[D^p,D^q]\subseteq D^{p+q-1}$, with $D^{-1}=0$: the top-order symbols of the two products agree and cancel. Unlike the Bernstein filtration, the order filtration's degree-zero part is the entire polynomial ring and its pieces are infinite-dimensional over $\mathbb{C}$.

Form the Rees ring of a filtered algebraThe central element $h$ is a non-zero-divisor. This ring is generated by $h$, the $x_i$, and $y_i=h{\partial }_i$, withIts quotient modulo $h$ is the associated graded ring just computed. Explicitly the degree-$m$ component of $h\mathcal{A}$ is $D^{m-1}{h}^m$, so the degree-$m$ quotient is $D^m/D^{m-1}$. The quotient modulo $h-1$ recovers $A_n$.

Since $\mathcal{A}/h\mathcal{A}$ is commutative, any two lifts $F,G\in \mathcal{A}$ of symbols have $[F,G]\in h\mathcal{A}$. DefineDivision by $h$ is unique because $h$ is a non-zero-divisor. If $F$ is replaced by $F+hU$, the quotient changes by $[U,G]$, which is zero modulo $h$; changing $G$ works the same way. Thus the bracket is well defined. Antisymmetry and bilinearity follow from the commutator. The identity $[F,GH]=[F,G]H+G[F,H]$ proves the Leibniz rule. For the Jacobi identity, lift each inner bracket by $[G,H]/h$; the cyclic sum of the outer expressions is the commutator Jacobi identity divided by $h^2$, hence zero. These verifications prove that the graded polynomial ring is a Poisson algebra. This is the Poisson bracket from a central deformation parameter construction.

The generators give $\{{\xi }_i,x_j\}={\delta }_{ij}$ and all coordinate-coordinate and symbol-symbol brackets zero. The Leibniz rule therefore determines the answer on every polynomial:The sign here follows the chosen convention $[P,Q]=PQ-QP$: in particular $\{{\xi }_i,x_j\}={\delta }_{ij}$, rather than $\{x_i,{\xi }_j\}={\delta }_{ij}$. For homogeneous symbols the bracket has order degree $p+q-1$, and can equivalently be written $\{{\sigma }_p(P),{\sigma }_q(Q)\}={\sigma }_{p+q-1}([P,Q])$.

Use the Weyl algebra convention $[y_i,x_j]={\delta }_{ij}$, with $[x_i,x_j]=[y_i,y_j]=0$, where the brackets are commutators. Its order filtration of a Weyl algebra isThus $x_i$ has degree zero and $y_i$ degree one. By the ordered monomial basis of a Weyl algebra, the associated graded ring iswhere ${\xi }_i$ is the symbol of $y_i$.

The Rees ring of a filtered algebra is the graded ringwhere $t$ is a central indeterminate. Here it is generated by $x_i$, $Y_i=y_{i}t$, and $t$, with degrees $0,1,1$, respectively, and relationsIt interpolates between the associated graded ring and the original filtered ring:

A good filtration for the order filtration on $M$ is an increasing exhaustive family of complex vector subspaces $(M_j{)}_{j\in \mathbb{Z}}$, zero for sufficiently negative $j$, with $F_{i}R{M}_j\subseteq M_{i+j}$, such that the associated graded moduleis a finitely generated module over $A$. Equivalently, there are finitely many $m_1,\dots ,m_s$ and integer shifts $d_1,\dots ,d_s$ such that $M_j={\sum }_{\ell =1}^s(F_{j-d_{\ell }}R)m_{\ell }$. In particular, choose any finite generating family $m_1,\dots ,m_s$ of $M$ and putThis is a good filtration for the order filtration.

For $M\ne 0$, its dimension of a Weyl algebra module for the order filtration isHere support of a module, annihilator of a module, and Krull dimension are taken over the commutative polynomial ring $A$. The corresponding algebraic set is the order characteristic variety of a Weyl algebra module. This dimension is independent of the chosen good filtration for the order filtration; one may set $d(0)=-\infty$.

The pieces $F_{j}R$ generally have infinite complex dimension, since they contain all polynomials in the $x_i$. Thus a polynomial-growth definition based directly on ${\dim }_{\mathbb{C}}{M}_j$ would be inappropriate for this order filtration of a Weyl algebra.

Use the quotient filtration induced from the order filtration of a Weyl algebra. The key step is to identify the entire associated graded module of the left ideal $I$, rather than merely take symbols of its displayed generators.

For every member $x^{\alpha }{y}^{\beta }$ of the ordered monomial basis of a Weyl algebra, right multiplication by the pure $y$-monomials givesNo commuting past an $x_i$ is necessary. It follows that $I$ is exactly the complex linear span of the basis monomials satisfying eitherIntersecting this linear span with each filtration piece and then taking symbols therefore yields the monomial idealBy b,

To determine its Krull dimension, compute the radical of an ideal:Indeed, $K$ lies in the displayed intersection, an intersection of prime ideals. Conversely, $({\xi }_1{\xi }_2{)}^3\in K$ and $({\xi }_1{\xi }_3{)}^3\in K$, so both generators of that intersection lie in $\sqrt{K}$. The equality of the intersection with $({\xi }_1{\xi }_2,{\xi }_1{\xi }_3)$ follows by factoring out ${\xi }_1$ and reducing modulo $({\xi }_2,{\xi }_3)$.

The order characteristic variety of a Weyl algebra module in this case has two irreducible components:They are coordinate affine spaces of dimensions $2n-1$ and $2n-2$, respectively, in the $2n$ coordinates $x_1,\dots ,x_n,{\xi }_1,\dots ,{\xi }_n$. The Krull dimension is the larger of these component dimensions. ThusIn particular, two displayed generators do not force a drop of two in dimension: their common factor creates the component ${\xi }_1=0$ of dimension $2n-1$.

Let $h$ be central and a non-zero-divisor in an associative algebra, with commutative quotient modulo $h$. Dividing commutators by $h$ and reducing modulo $h$ is independent of the lifts, since changing a lift by $hc$ changes the quotient by a commutator, zero modulo $h$. Antisymmetry, the Leibniz rule and the Jacobi identity descend from the corresponding commutator identities. For the order filtration of a Weyl algebra, the convention is $\{{\xi }_i,x_j\}={\delta }_{ij}$, with the opposite sign for $\{x_i,{\xi }_j\}$.