Holomorphic Functions as a Module over the Weyl Algebra

Q: Why is H(U) a module at all — what has to be checked?

That z and d/dz preserve H(U) , and that they satisfy the single defining relation [∂, z] = 1 , which is the Leibniz rule. The universal property of the Weyl algebra then supplies the action of every operator. See defining an action from generators and relations .

Q: Does the answer depend on which open set U is?

Not for the statements on this page. The witness ( z) is entire, so it lies in every H(U) , and the polynomial relation it would produce extends from U to C by the identity theorem. Other features — the dimension of a particular solution space, for instance — depend on U very much.

Q: Is ( z) really not the solution of any differential equation?

It solves plenty. It satisfies h' = e^zh , whose coefficient is not a polynomial, and the non-linear equation hh'' - (h')^2 = hh' , which one verifies by substituting h' = e^zh and h'' = (e^z+e^2z)h . What it does not satisfy is any non-zero linear equation with polynomial coefficients.

Q: Where exactly does the proof need h to be non-vanishing?

At the cancellation step. From ( ∑ _i f_iF_i(e^z) )h = 0 one wants to conclude that the bracket vanishes. Since h = (something) is nowhere zero, that is legitimate; for a witness with zeros it would not be.

Q: How does this relate to holonomic modules?

A torsion element u generates a holonomic submodule A_1u , so the torsion part of H(U) is the union of its holonomic submodules. The module H(U) itself is not finitely generated, so it is not holonomic and has no dimension.

Q: What replaces H(U) when the solutions are not functions?

Distributions, hyperfunctions and microfunctions , in increasing order of generality. Each is a module over the same ring, and each is substituted into (5.7) in place of H(U) to define its own notion of solution. The delta distribution, for instance, generates the module P1 .

Overview

Up to this point the modules over the Weyl algebra in this collection have been algebraic objects: the polynomial ring, quotients of $A_n$ by left ideals, twists of these. The module of holomorphic functions is the first genuinely analytic one, and it is the module in which differential equations are actually solved. If $U ⊂ eq C$ is open and non-empty, $H(U)$ denotes the complex vector space of functions holomorphic on $U$ ; letting $z$ act by multiplication and $∂ = d/dz$ by differentiation makes it a left $A_1(C)$ -module.

The first observations are negative and easy. $H(U)$ is not simple, because the polynomials $C[z]$ form a proper non-zero submodule. It is not cyclic and not even finitely generated. What is not easy, and what this page proves, is that $H(U)$ is not a torsion module. A torsion element of $H(U)$ is precisely a function satisfying a non-trivial linear differential equation with polynomial coefficients, and such functions are abundant: $e^z$ , $sin z$ , $log z$ , $z^a$ , every algebraic function. The claim is that they do not exhaust $H(U)$ .

The witness is $h(z) = ( z)$ . Its derivatives have the shape $F_m(e^z)h(z)$ with $F_m = m$ , so an annihilating operator of order $k$ would produce a polynomial relation between $z$ and $e^z$ of degree exactly $k$ in the second variable. Since $e^z$ is not algebraic over $C(z)$ , no such relation exists. The argument is a clean template: a non-holonomy proof is a transcendence proof, and the algebra only supplies the reduction.

The interest of $H(U)$ is that it plays the role of the ambient space of solutions. For an operator $P$ , the solutions of $P(f) = 0$ in $H(U)$ are exactly the homomorphisms $A_1/A_1P → H(U)$ , so a module of functions converts differential equations into homological algebra. Enlarging $H(U)$ to distributions, hyperfunctions and microfunctions is what the next chapter does.

Definition

The module of holomorphic functionsCoutinho, Ch. 5 §3

Let $U ⊂ eq C$ be a non-empty open set and let $H(U)$ be the set of holomorphic functions $U \to ℂ$ , a complex vector space under pointwise operations. Write the generators of $A_1(C)$ as $z$ and $∂$ . Define

z ⋅ f = zf(z), ∂ ⋅ f = (df)/(dz).

Both operations map $H(U)$ into itself, and the Leibniz rule gives $∂(zf) = f + z,∂ f$ , that is $[∂, z] = 1$ on $H(U)$ . By the criterion for defining an action from generators and relations, this extends uniquely to a left $A_1(C)$ -module structure.

Torsion

Let $R$ be a ring and $M$ a left $R$ -module. An element $u ∈ M$ is a torsion element if $ann_R(u) = a ∈ R : au = 0$ is a non-zero left ideal. $M$ is a torsion module if every element of $M$ is torsion. For $M = H(U)$ and $R = A_1(C)$ , a torsion element is a function $f$ for which there exist polynomials $f_0, … ,f_k$ , not all zero, with

f_k(z),(d^k f)/(dz^k) + … + f_1(z),(df)/(dz) + f_0(z) f = 0 on U.

Such functions are called holonomic, or D-finite. See holonomic functions for the systematic theory.

Remark

The same recipe makes the smooth functions $C^ ∈ fty(V)$ on an open $V ⊂ eq R^n$ a module over $A_n(R)$ , and the holomorphic functions on an open $U ⊂ eq C^n$ a module over $A_n(C)$ . Nothing below is special to one variable except the explicit computations.

Core Concepts

Torsion means "satisfies an equation"

Over a commutative domain, torsion elements are the ones killed by a scalar and are usually the pathological part of a module. Here the dictionary is reversed. Being torsion over $A_1$ is a good property: it says the function is pinned down by finitely many pieces of data, since its derivatives satisfy a recursion. Non-torsion is the wild case. This is why the theory of holonomic modules, and eventually the whole machinery of automatic identity proving, sits on the torsion side.

The module of functions is a container for solution spaces

If $M = A_1/A_1P$ is the module of the equation $P(f) = 0$ , then a homomorphism $φ : M → H(U)$ is determined by $φ( 1) = f$ and is well defined exactly when $Pf = 0$ . So

Hom_A_1! (A_1/A_1P, H(U) ) ; ≅ ; , f ∈ H(U) : Pf = 0 ,.

(5.7)

Everything analytic — existence, dimension of the solution space, monodromy — enters the algebraic theory through the choice of the module on the right of (5.7). Choosing $H(U)$ gives classical solutions on $U$ ; choosing a bigger module gives distributions or hyperfunctions.

Why a transcendence input is unavoidable

To prove some function is not annihilated by any non-zero operator, one must rule out infinitely many operators at once. The only leverage is a structural statement about the derivatives. For $h = (e^z)$ the derivatives all lie in the $C[z]$ -module generated by $e^jzh$ , $j ≥ 0$ , and those generators are independent precisely because $e^z$ is transcendental over $C(z)$ . Algebra reduces the problem to that fact and cannot supply it.

Construction and Proof

The shape of the derivativesCoutinho (5.3.1)

Let $h(z) = ( z)$ . For every $m ≥ 0$ there is a monic polynomial $F_m ∈ C[y]$ of degree $m$ with $d^m h/dz^m = F_m(e^z)h(z)$ .

Proof

For $m = 0$ take $F_0 = 1$ . Since $h' = e^z h$ , the case $m = 1$ holds with $F_1(y) = y$ . Assume $h^(m) = F_m(e^z)h$ . Differentiating and using $(d)/(dz)F_m(e^z) = F_m'(e^z)e^z$ together with $h' = e^zh$ ,

h^(m+1) = F_m'(e^z)e^z h + F_m(e^z)e^z h = e^z (F_m'(e^z) + F_m(e^z) )h,

which is (5.8) for $m+1$ with $F_m+1(y) = y(F_m'(y) + F_m(y))$ . If $F_m$ is monic of degree $m$ then $yF_m'$ has degree $m$ and $yF_m$ is monic of degree $m+1$ , so $F_m+1$ is monic of degree $m+1$ . The induction is complete.

The exponential is not algebraic

There is no non-zero $G ∈ C[x,y]$ with $G(z,e^z) = 0$ for all $z$ in some non-empty open set.

Proof

Suppose $G(z,e^z)$ vanishes on a non-empty open $U$ . The function $z ↦ G(z,e^z)$ is entire, so by the identity theorem it vanishes on all of $ℂ$ . Write $G(x,y) = ∑ _j=0^d a_j(x)y^,j$ with $a_d ≠ 0$ . Restrict to real $z = t → + ∈ fty$ and divide by $e^dt$ :

a_d(t) + ∑ _j<d a_j(t),e^(j-d)t = 0.

Each term of the sum tends to $0$ , because a polynomial times $e^-ct$ with $c > 0$ tends to $0$ . Hence $a_d(t) → 0$ as $t → + ∈ fty$ , which for a polynomial forces $a_d = 0$ , a contradiction. (This is the fact Coutinho quotes from Hardy; the proof above is elementary and self-contained.)

$H(U)$ is not a torsion moduleCoutinho (5.3.2)

The entire function $h(z) = ( z)$ is not a torsion element of the $A_1(C)$ -module $H(U)$ , for any non-empty open $U ⊂ eq C$ . Consequently $H(U)$ is not a torsion module.

Proof

Suppose $P = ∑ _i=0^k f_i(z)∂^,i$ is non-zero with $Ph = 0$ ; discarding zero terms we may assume $f_k ≠ 0$ . By the first lemma, on $U$

0 = Ph = ( ∑ _i=0^k f_i(z)F_i(e^z) )h(z).

The function $h$ never vanishes, since it is an exponential, so the bracket vanishes identically on $U$ . That bracket is $G(z,e^z)$ with $G(x,y) = ∑ _i f_i(x)F_i(y)$ . Because $F_i = i$ , the coefficient of $y^k$ in $G$ is $f_k(x)$ , which is non-zero; hence $G ≠ 0$ . This contradicts the second lemma. So $ann(h) = 0$ and $h$ is not torsion.

Two points the source passes over

First, the relation $G(z,e^z) = 0$ is only obtained on $U$ , while non-algebraicity of $e^z$ is a statement about $ℂ$ ; the identity theorem bridges the gap, and it is worth stating because $U$ may be very small. Second, $h$ is entire, so it does belong to $H(U)$ for every $U$ — the conclusion is uniform in $U$ , which is not automatic for arguments of this kind.

Key Equations

The derivatives of $h(z) = ( z)$ have the closed form

(d^m h)/(dz^m) = F_m(e^z),h(z), F_m ∈ C[y], F_m = m,

(5.8)

where the polynomials $F_m$ are monic and obey the recursion

F_0 = 1, F_m+1(y) = y (F_m'(y) + F_m(y) ).

(5.9)

The first few are

F_1 = y, F_2 = y + y^2, F_3 = y + 3y^2 + y^3, F_4 = y + 7y^2 + 6y^3 + y^4.

If $P = ∑ _i=0^k f_i(z)∂^,i$ with $f_k ≠ 0$ annihilated $h$ , then dividing by the nowhere-zero function $h$ turns $Ph = 0$ into

G(z, e^z) = 0, G(x,y) = ∑ _i=0^k f_i(x)F_i(y),

(5.10)

and $G$ is a non-zero polynomial because the coefficient of $y^k$ in it is $f_k(x) ≠ 0$ .

Variable Definitions

$U$: a non-empty open subset of $ℂ$
$H(U)$: the holomorphic functions on $U$ , a left $A_1(C)$ -module
$z, ∂$: the generators of $A_1(C)$ , acting by multiplication and by $d/dz$
$h(z)$: the entire function $( z)$ , the witness for non-torsion
$F_m(y)$: the monic polynomial of degree $m$ with $h^(m) = F_m(e^z)h$
$P$: a differential operator $∑ _i f_i(z)∂^,i$ in $A_1(C)$ , of order $k$ when $f_k ≠ 0$
$G(x,y)$: the two-variable polynomial produced from $P$ and the $F_i$
$ann(f)$: the left ideal of operators annihilating $f$
$T$: the submodule of torsion (holonomic) elements of $H(U)$

Properties and Behaviour

$H(U)$ is not simple

$C[z] ⊂ eq H(U)$ is a submodule, non-zero and proper — proper because $e^z$ is holomorphic on $U$ and is not a polynomial. Moreover $C[z]$ is itself simple, so $H(U)$ has simple submodules without being semisimple in any useful sense.

The torsion elements form a submodule

Let $T ⊂ eq H(U)$ be the set of torsion elements. If $u$ is torsion, say $Pu = 0$ with $P ≠ 0$ , then $A_1u$ is a quotient of $A_1/A_1P$ and hence holonomic. For $u,v ∈ T$ the module $A_1u + A_1v$ is a quotient of $A_1u ⊕ A_1v$ and so is holonomic too, and every element of a holonomic module is torsion — an element with zero annihilator would generate a copy of $A_1$ , of dimension $2$ , inside a module of dimension $1$ . Hence $u + v$ and $P u$ are torsion, and $T$ is a submodule.

The same conclusion follows from the Ore condition: $A_1$ is a Noetherian domain, so any two non-zero left ideals meet non-trivially, and a common annihilator can always be produced.

$H(U)$ is not cyclicCoutinho, Ch. 5, Exercise 4.7

Suppose $H(U) = A_1 g$ for some $g$ . If $ann(g) ≠ 0$ then $H(U)$ is a quotient of $A_1/A_1P$ for a non-zero $P$ , hence holonomic, hence a torsion module — contradicting the proposition above, since $( z) ∈ H(U)$ . If $ann(g) = 0$ then $H(U) ≅ A_1$ as a left module; but $A_1$ is a domain, so it has no non-zero torsion elements, whereas $e^z ≠ 0$ is killed by $∂ - 1$ . Both cases are impossible.

$H(U)$ is not finitely generated

For distinct $λ ∈ C$ the submodules $A_1e^λ z = C[z]e^λ z$ are non-zero, and their sum inside $H(U)$ is direct because the functions $z^ke^λ z$ are linearly independent. So $H(U)$ contains an infinite direct sum of non-zero submodules and cannot satisfy the ascending chain condition. Since $A_1$ is Noetherian, a finitely generated module would be Noetherian; therefore $H(U)$ is not finitely generated. In particular it has no good filtration, no Hilbert polynomial and no dimension: statements such as Bernstein's inequality simply do not apply to it.

Solution spaces are finite dimensional

If $U$ is a simply connected domain and $P = ∑ _i ≤ k f_i ∂^,i$ has $f_k$ nowhere zero on $U$ , the classical existence and uniqueness theorem gives $_Cf ∈ H(U) : Pf = 0 = k$ . By (5.7) this is the statement that $Hom_A_1(A_1/A_1P, H(U))$ has dimension $k$ , the holonomic rank of the module. On a domain containing a zero of $f_k$ , or a non-simply-connected one, the dimension can drop — singular points and monodromy are exactly what obstructs it.

Examples and Special Cases

Torsion elements of $H(U)$ and annihilating operators. All are checked by direct differentiation.
Function	Domain $U$	Annihilating operator	Order
$e^λ z$	$ℂ$	$∂ - λ$	$1$
$(p(z))$ , $p$ a polynomial	$ℂ$	$∂ - p'(z)$	$1$
$sin z$ , $cos z$	$ℂ$	$∂^2 + 1$	$2$
$1/z$	$C ∖ 0$	$z∂ + 1$	$1$
$z^a$ , $a ∈ C$	$C ∖ (- ∈ fty,0]$	$z∂ - a$	$1$
$log z$	$C ∖ (- ∈ fty,0]$	$z∂^2 + ∂$	$2$
$( z)$	any $U$	none	—

Every algebraic function is torsion

If $G(z,f(z)) = 0$ with $G ≠ 0$ , then differentiating and solving for $f^{'}$ expresses $f^{'}$ as a rational function of $z$ and $f$ . Iterating, all derivatives of $f$ lie in the field $C(z)(f)$ , which is finite dimensional over $C(z)$ ; so $f, f^{'}, f^{''}, \dots$ are linearly dependent over $C(z)$ , and clearing denominators gives an annihilating operator. Hence $1-z^2$ , and every branch of an algebraic function, is a torsion element.

The exponential of a non-polynomial

$(p(z))$ is torsion for every polynomial $p$ , being killed by $∂ - p'$ . The function $( z)$ is the smallest natural perturbation of that pattern in which the logarithmic derivative $h'/h = e^z$ leaves the rational functions, and it is exactly at that point that torsion fails. The same phenomenon is what makes $(e^z)$ a plausible second example — Coutinho sets it as an exercise — and the derivative bookkeeping there again produces polynomials of growing degree in $e^z$ .

The submodule generated by a torsion element

For $u = e^z$ , $A_1u = C[z]e^z ≅ A_1/A_1(∂ - 1)$ , a simple holonomic module with $d = 1$ and $e = 1$ . For $u = z^a$ with $a ∉ Z$ , $A_1u ≅ A_1/A_1(z∂ - a)$ , again simple. These are the twists of $C[z]$ met on the isomorphism problem page — each is a copy of the polynomial module carrying a different exponential factor.

Worked Example

Ruling out every operator of order at most two by hand

Step 1 - compute the first four derivatives
With $h = (e^z)$ and $y = e^z$ : $h' = yh$ , so $F_1 = y$ . Then $h'' = (y + y^2)h$ , since differentiating $y h$ gives $yh + y ⋅ yh$ . Next $F_3 = y(F_2' + F_2) = y(1 + 2y + y + y^2) = y + 3y^2 + y^3$ , and $F_4 = y(F_3' + F_3) = y(1 + 6y + 3y^2 + y + 3y^2 + y^3) = y + 7y^2 + 6y^3 + y^4$ .

Direct check of $F_3$ : differentiating $(y+y^2)h$ gives $(y + 2y^2)h + (y+y^2)yh = (y + 3y^2 + y^3)h$ . It agrees.
Step 2 - write down the general order-two candidate
Let $P = f_0 + f_1∂ + f_2∂^2$ with $f_0,f_1,f_2 ∈ C[z]$ not all zero. Then

$Ph = (f_0 + f_1 y + f_2 (y + y^2) )h = (f_2 y^2 + (f_1 + f_2)y + f_0 )h, y = e^z.$
Step 3 - use that $h$ never vanishes
$Ph = 0$ forces $f_2(z)e^2z + (f_1(z)+f_2(z))e^z + f_0(z) = 0$ on $U$ , hence on $ℂ$ by the identity theorem.
Step 4 - peel off the coefficients
Divide by $e^2z$ and let $z = t → + ∈ fty$ along the reals: the last two terms tend to $0$ , so $f_2(t) → 0$ and therefore $f_2 = 0$ . The relation collapses to $f_1(z)e^z + f_0(z) = 0$ ; dividing by $e^z$ and repeating gives $f_1 = 0$ , and then $f_0 = 0$ .
Step 5 - see why order two was not special
The only inputs were $F_i = i$ — which makes the coefficient of the top power of $y$ equal to $f_k$ — and the growth comparison that kills a polynomial against an exponential. Both are available for every $k$ , which is exactly the general proof.

Result

No non-zero operator of order $≤ 2$ annihilates $( z)$ , and the same computation with $F_k$ monic of degree $k$ rules out every order. Contrast this with $(z^2)$ , which is annihilated by the first-order operator $∂ - 2z$ : the module $A_1 (z^2) ≅ A_1/A_1(∂ - 2z)$ is a simple holonomic module of multiplicity $1$ , sitting inside $H(C)$ .

Applications and Industry Use

In a mathematics topic, this section covers downstream use inside mathematics, computing and engineering rather than a manufactured product.

Solutions as homomorphisms. Identity (5.7) is the bridge between D-modules and analysis, and it is the definition used in the solutions functor. Replacing $H(U)$ by other modules — distributions, hyperfunctions, microfunctions — gives the generalised solution theories.
Holonomic functions and automatic proofs. The torsion elements of $H(U)$ are precisely the D-finite functions. Their closure under sums, products and integration is what makes creative telescoping and Zeilberger's method work, and the non-torsion examples mark the boundary of that machinery.
Special function libraries. Computer algebra systems represent a special function by an annihilating operator plus initial values. Such a representation exists exactly for torsion elements, so the proposition on this page is the statement that some perfectly ordinary functions cannot be stored that way.
Asymptotics and singularity analysis. For a holonomic function, the singularities of the solutions are confined to the zeros of the leading coefficient of the annihilating operator. That finite list is the starting point of asymptotic analysis; non-holonomic functions have no such control.

Computational Notes

Read this as the manufacturing section of the template: how the object is actually built by machine, at what cost, and where the computation stops being decidable.

Proving that a function is holonomic is a finite search; proving that it is not is a transcendence problem.

To certify holonomy, fix an order $k$ and a degree bound $d$ , write $P = ∑ _i ≤ k f_i(z)∂^,i$ with undetermined coefficients, expand $P f$ as a power series, and solve the resulting linear system over $ℚ$ for the coefficients of the $f_i$ . A solution is a candidate; it becomes a proof once it is verified symbolically or the ansatz is justified.
Closure properties do most of the practical work: sums, products, algebraic substitutions and integrals of holonomic functions are holonomic, with explicit bounds on order and degree, so holonomy is usually established by construction rather than by search.
To refute holonomy no algorithm is known in general. Each case is an argument about the derivatives, as here, and rests on a transcendence statement.
The dictionary between operators and recurrences on Taylor coefficients means all of this can be done on either side; implementations usually work with both.

Software: HolonomicFunctions.m for Mathematica, the ore_algebra package in SageMath, gfun for Maple, and the Dmodules package in Macaulay2 for the module-theoretic side. All of them assume holonomy as an input hypothesis when they compute; none of them decides it.

Limits of Validity

The witness must be entire, or the statement becomes $U$ -dependent. $( z)$ is entire, so it lies in every $H(U)$ and settles all $U$ at once. A function holomorphic only on part of the plane would prove the result only there.
Non-torsion is not the same as non-differentiability of any kind. $h = ( z)$ satisfies the first-order equation $h' = e^z h$ with an entire, non-polynomial coefficient, and it satisfies the algebraic differential equation $hh'' - (h')^2 = hh'$ . Only linear equations with polynomial coefficients are excluded.
The result says nothing about how big the non-torsion part is. It shows $T ≠ H(U)$ . In fact $T$ is a very thin subspace, but that requires a separate argument and is not proved here.
Finite generation fails, so the invariants of Chapters 9 to 11 are unavailable. Dimension, multiplicity and characteristic variety are defined only for finitely generated modules; $H(U)$ has none of them, and one works instead with its finitely generated submodules.
Deciding holonomy of a given function is not covered. The proof handles one specific function. There is no general algorithm that takes an analytic expression and decides whether it is holonomic.

Failure Modes and Common Mistakes

Confusing a torsion element with a torsion module

$H(U)$ is full of torsion elements — every exponential, every trigonometric function, every algebraic function — and this tempts one to call it a torsion module. It is not, and the difference is the entire content of the proposition. The correct summary is: $H(U)$ has a large torsion submodule $T$ and a non-zero quotient $H(U)/T$ .

Forgetting to require the leading coefficient non-zero

In the proof one writes $P = ∑ _i ≤ k f_i ∂^,i$ and needs $f_k ≠ 0$ ; otherwise the coefficient of $y^k$ in $G$ vanishes and $G$ might be zero. Since a non-zero operator has some highest index with non-zero coefficient, this costs nothing — but it must be said, because the conclusion $G ≠ 0$ is where the whole argument lives.

Assuming a closed form implies an equation

Elementary closed form and holonomy are unrelated notions. $( z)$ has a two-symbol closed form and is not holonomic; the Bessel functions have no elementary closed form and are holonomic. The right test is whether the derivatives span a finite-dimensional space over $C(z)$ .

Treating $h/h$ cancellation as automatic

The step from $( ∑ _i f_i F_i(e^z) )h = 0$ to $∑ _i f_i F_i(e^z) = 0$ uses that $h$ has no zeros. For a general witness function that step is illegal: a product can vanish because the other factor does on part of the domain. Choosing an exponential as the witness is what makes the division safe.

Historical Notes

The class of functions satisfying a linear differential equation with polynomial coefficients is classical: it is the setting of Fuchs, Frobenius and the nineteenth-century theory of special functions, where the equation, not the formula, is the definition of the function. The transcendence of $e^z$ over $C(z)$ used here is older still, and Coutinho refers the reader to Hardy's 1928 tract on the integration of functions of a single variable.

The module-theoretic reading — functions as a module over the ring of operators, solutions as homomorphisms — belongs to the 1960s and 1970s, in the work of Sato's school on algebraic analysis and of Bernstein on the algebraic side. The finiteness that holonomy encodes was made combinatorial by Stanley in 1980, who introduced D-finite series, and by Zeilberger in 1990, whose holonomic systems approach turned it into an algorithm.

Coutinho's use of $( z)$ as the witness is deliberately elementary: it needs only the chain rule and one transcendence fact, and it makes the point that the algebra can reduce an infinite family of questions to a single one about polynomials.

Comparison

Four modules over $A_1(C)$ , compared.
Module	Simple?	Torsion module?	Cyclic?	Finitely generated?
$C[z]$	yes	yes	yes, by $1$	yes
$A_1$	no	no	yes, by $1$	yes
$C(z)$	no	yes	no	no
$H(U)$	no	no	no	no

The rational function field is instructive as the intermediate case: every rational function satisfies a first-order equation, so it is a torsion module, but it is still not finitely generated, being the union of the localisations $C[z][1/f]$ . Torsion and finite generation are independent conditions, and holonomy is what one gets from both together.

Key Takeaways

Key points

$H(U)$ is a left $A_1(C)$ -module with $z$ acting by multiplication and $∂$ by differentiation.
Torsion elements of $H(U)$ are exactly the functions satisfying a non-trivial linear ODE with polynomial coefficients — the holonomic, or D-finite, functions.
$H(U)$ is not simple ( $C[z]$ is a submodule), not cyclic, and not finitely generated.
$H(U)$ is not a torsion module: $( z)$ has zero annihilator.
The proof writes $h^(m) = F_m(e^z)h$ with $F_m$ monic of degree $m$ , converts an annihilator of order $k$ into a non-zero polynomial relation between $z$ and $e^z$ , and appeals to the transcendence of $e^z$ .
By (5.7), solutions of $P(f) = 0$ in $H(U)$ are the homomorphisms $A_1/A_1P → H(U)$ ; this is how analysis enters the algebraic theory.

FAQs

Why is

H(U)

a module at all — what has to be checked?

That $z$ and $d/dz$ preserve $H(U)$ , and that they satisfy the single defining relation $[∂, z] = 1$ , which is the Leibniz rule. The universal property of the Weyl algebra then supplies the action of every operator. See defining an action from generators and relations.

Does the answer depend on which open set

U

is?

Not for the statements on this page. The witness $( z)$ is entire, so it lies in every $H(U)$ , and the polynomial relation it would produce extends from $U$ to $ℂ$ by the identity theorem. Other features — the dimension of a particular solution space, for instance — depend on $U$ very much.

( z)

really not the solution of any differential equation?

It solves plenty. It satisfies $h' = e^zh$ , whose coefficient is not a polynomial, and the non-linear equation $hh'' - (h')^2 = hh'$ , which one verifies by substituting $h' = e^zh$ and $h'' = (e^z+e^2z)h$ . What it does not satisfy is any non-zero linear equation with polynomial coefficients.

Where exactly does the proof need

h

to be non-vanishing?

At the cancellation step. From $( ∑ _i f_iF_i(e^z) )h = 0$ one wants to conclude that the bracket vanishes. Since $h = (something)$ is nowhere zero, that is legitimate; for a witness with zeros it would not be.

How does this relate to holonomic modules?

A torsion element $u$ generates a holonomic submodule $A_1u$ , so the torsion part of $H(U)$ is the union of its holonomic submodules. The module $H(U)$ itself is not finitely generated, so it is not holonomic and has no dimension.

What replaces

H(U)

when the solutions are not functions?

Distributions, hyperfunctions and microfunctions, in increasing order of generality. Each is a module over the same ring, and each is substituted into (5.7) in place of $H(U)$ to define its own notion of solution. The delta distribution, for instance, generates the module P1.

Is the module of smooth real functions any different?

It is much larger and behaves worse: it is not simple, not torsion, and it also contains functions vanishing on an open set without being zero, which no holomorphic module does. The failure of the identity theorem there is why the holomorphic setting is the one used for the sharp statements.

References

S. C. Coutinho, A Primer of Algebraic D-modules, London Mathematical Society Student Texts 33, Cambridge University Press, 1995 — Ch. 5 §3, results (5.3.1) and (5.3.2), and Exercises 4.7 to 4.12.
G. H. Hardy, The Integration of Functions of a Single Variable, 2nd edition, Cambridge University Press, 1928 — Ch. V §16, for the non-algebraicity of the exponential.
R. P. Stanley, Differentiably finite power series, European Journal of Combinatorics 1 (1980), 175–188.
D. Zeilberger, A holonomic systems approach to special functions identities, Journal of Computational and Applied Mathematics 32 (1990), 321–368.
J.-E. Björk, Rings of Differential Operators, North-Holland, 1979 — Ch. 1 and Ch. 5, for modules of analytic solutions.
M. Petkovšek, H. S. Wilf and D. Zeilberger, A = B, A K Peters, 1996 — for the algorithmic side of holonomic functions.
R. Hotta, K. Takeuchi and T. Tanisaki, D-modules, Perverse Sheaves, and Representation Theory, Birkhäuser, 2008 — Ch. 4, for solution complexes and the Riemann–Hilbert setting.
ISO 80000-2:2019, Quantities and units — Part 2: Mathematics, International Organization for Standardization.

AI Suggested Questions

Verify the recursion $F_m+1 = y(F_m' + F_m)$ by computing $F_5$ and checking it against a direct differentiation.
Show that $(e^z)$ and $(e^z)$ are not torsion elements, by tracking the pair of polynomials in $e^z$ that their derivatives produce.
Prove that the set of torsion elements of $H(U)$ is a submodule using the Ore condition, without invoking holonomy.
Compute the annihilator of $z^a z$ on the slit plane and identify the module it generates.
Show that $H(U)$ contains a submodule isomorphic to $C[z]$ for every choice of a nowhere-zero holonomic function.
Explain why the smooth functions on an interval fail every argument on this page that uses the identity theorem.
Determine the dimension of the solution space of $z∂ - a$ in $H(U)$ for $U$ a punctured disc, and relate the answer to monodromy.

Holomorphic Functions as a Module over the Weyl Algebra

Overview

Definition

Core Concepts

Torsion means "satisfies an equation"

The module of functions is a container for solution spaces

Why a transcendence input is unavoidable

Construction and Proof

Key Equations

Variable Definitions

Properties and Behaviour

Examples and Special Cases

Worked Example

Applications and Industry Use

Computational Notes

Limits of Validity

Failure Modes and Common Mistakes

Historical Notes

Comparison

Key Takeaways

Key points

FAQs

Related Engineering Topics

References

AI Suggested Questions

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