Does Peetre's theorem hold in complex analysis?

Question

Let $E, F$ be two smooth vector bundles over a smooth manifold $M$. Peetre's theorem states that any $\mathbb{R}$-linear morphism $D: \mathcal{E} \to \mathcal{F}$ of the sheaves of sections of $E$ and $F$ is a differential operator, i.e. there exist trivializing charts $U_i$ such that on $D|_{U_i}$ is a differential operator in the sense of real analysis.

Question 1: Does this result also hold for complex manifolds?

By this I mean that $M$ is a complex manifold, $E$ and $F$ are holomorphic vector bundles and $D$ is $\mathbb{C}$-linear. We then also restrict to sheaves of holomorphic sections. I don't think the proof of the real version can be adapted directly, since it relies on bump functions.

This is related to my m.SE question in which I showed that the result does not hold for the sheaf of rational functions on $\mathbb{C}$.

However, on any locally ringed space $(X, \mathcal{O})$ one can define differential operators of order $0$ as multiplication with a global section of $\mathcal{O}$ and differential operators of order $n$ inductively as $\mathbb{C}$-linear morphisms of sheaves $D: \mathcal{E} \to \mathcal{F}$ satisfying that $f D - D f$ is a differential operator of order $n-1$. General differential operators can then be defined as $\mathbb{C}$-linear morphisms of sheaves which locally are differential operators of finite order.
This gives the expected result for the differential operators on a Scheme (EGA IV Prop 16.8.8) (i. e. on $n$-dimensional affine space they are polynomials in the $n$ partial derivatives). By Peetre's theorem this definition also coincides with the classical notion of differential operators on a smooth manifold.

Question 2: If $\mathcal{E}$ and $\mathcal{F}$ are sheaves of sections of holomorphic vector bundles over a complex manifold, does the above definition of differential operators produce the classical differential operators between holomorphic vector bundles?

Answer 1

Yes, for both the second question.

Q2: Since it's a local question, it suffices to consider $M$ open in $\mathbb{C}^n$ and $\mathcal{E},\mathcal{F}$ one dimensional trivial complex bundles. So let $\mathcal{O}$ denote the ring of holomorphic functions on $M$, let $z_1,\ldots, z_n \in \mathcal{O}$ denote the standard coordinates on $\mathbb{C}^n$ restricted to $M$ and use multi-index notation $\mu=(\mu_1,\ldots,\mu_n)\in\mathbb{N}^n$ with $z^\mu=z_1^{\mu_1}\cdots z_n^{\mu_n}$, $\mu!=\mu_1!\cdots\mu_n!$ and $|\mu|=\mu_1+\ldots+\mu_n$. It's easy to show that an operator of the form $$ \sum\limits_{|\mu|\leq k}c_\mu \frac{\partial^{|\mu|}}{\partial z^\mu}: \mathcal{O} \to \mathcal{O}$$ with $c_\mu\in \mathcal{O}$, satisfies the algebraic definition of a $\mathbb{C}$-linear differential operator of rank at most $k$.

To show the converse, assume $D\colon \mathcal{O} \to \mathcal{O}$ is such an algebraic DO of rank at most $k$. We want to show $D = \sum\limits_{|\mu|\leq k}c_\mu \frac{\partial^{|\mu|}}{\partial z^\mu}$ for suitable coefficients $c_\mu \in \mathcal{O}$. Define them recursively as

$$ c_{\mu}= \cases{ D (1) \quad \text{ when } \mu=0 \\ \left(D - \sum\limits_{0\leq |\nu|< |\mu|} c_\nu \frac{\partial^{|\nu|}}{\partial z^\nu}\right)\left(\frac{z^\mu}{\mu!}\right) \quad \text{ when } 1\leq |\mu|\leq k. } $$ Its straightforward to checkt that $D$ and $\sum\limits_{|\mu|\leq k}c_\mu \frac{\partial^{|\mu|}}{\partial z^\mu}$ agree on polynomials in $z$ of degree at most $k$.

To show that they also agree on any other $u\in \mathcal{O}$, let $x\in M$ and use Taylor expansion around $x$ to write $u=p+r$ where $p$ is a polynomial in $z$ of degree at most $k$ and $r$ is an element of $I^{k+1}_x\subset \mathcal{O}$, the $k+1$st power of the vanishing ideal at $x$. Now use the fact that $D(I^{k+m})=I^m$, for any ideal $I\subset \mathcal{O}$ and any algebraic DO $D$ of rank at most $k$. So that $$ (D u)(x)=D(p+r)(x)=(D p)(x)= \left(\sum\limits_{|\mu|\leq k}c_\mu \frac{\partial^{|\mu|}}{\partial z^\mu}p\right)(x)=\sum\limits_{|\mu|\leq k}c_\mu \frac{\partial^{|\mu|}}{\partial z^\mu}(p+r)(x)=\left(\sum\limits_{|\mu|\leq k}c_\mu \frac{\partial^{|\mu|}}{\partial z^\mu} u\right)(x). $$ Since this holds for any $x$ in $M$ we conclude $D u =\sum\limits_{|\mu|\leq k}c_\mu \frac{\partial^{|\mu|}}{\partial z^\mu} u$.

Edit: After the clarification of the question what follows is not a correct answer.

Q1 should now be easy: since the topology of $M$ as complex manifold agrees with its topology as real manifold, by Peetres theorem a $\mathbb{C}$-linear local operator $D : C^\infty(M,\mathbb{C}) \to C^\infty(M,\mathbb{C})$ satisfies the algebraic definition of a DO with smooth $f$ (in the sense of your post $fD - Df$ being of lower rank). Since holomorphic $f$ are also smooth, the algebraic definition holds also for holomorphic $f$. Now if $D$ moreover preserves the subring of holomorphic functions it is a complex differential operator by Q2.

Answer 2

I think Peetre's theorem is false in the holomorphic category. Namely, one can construct a counter example (a differential operator of infinite order) already in the case of one complex variable. We want to exhibit a sheaf morphism $D\colon \mathcal{O} \to \mathcal{O}$, where $\mathcal{O}$ is the sheaf of holomorphic functions in one complex variable $z$ of the form $$ D[f](z) = \sum_{k=0}^\infty d_k \frac{f^{(k)}(z)}{k!} , $$ with infinitely many non-zero $d_k$. Since this is a purely local question, we can just assume that we are only dealing with functions $f$ defined on some neighborhood of $z=0 \in \mathbb{C}$.

First, recall that the Taylor coefficients of any holomorphic function $f(z)$ grow no faster than exponentially (as a consequence of the Cauchy integral formula), that is $$ \left|\frac{f^{(k)}(0)}{k!}\right| < \frac{C}{r^k} , $$ for some $C, r > 0$, where $r$ is at least as large as the radius of a disk that fits into the domain to which $f(z)$ has a unique analytic continuation. Since the same argument works also about any point of the domain of $f(z)$, we can always find $0 < R \le r$ such that $$ \left|\frac{f^{(k)}(z)}{k!}\right| < \frac{C}{R^k} , $$ uniformly on some (possibly small) neighborhood of $z=0$. The point is that such a neighborhood and corresponding constants $C, R > 0$ exist for any function $f(z)$ holomorphic at $z=0$.

Now, let $d_k$ be the coefficients of some non-polynomial entire function $d(z) = \sum_{k=0}^\infty d_k z^k$, meaning that infinitely many $d_k\ne 0$ and the sums $\sum_{k=0}^\infty |d_k|/R^k$ converge for any $R>0$. For example $d_k = 1/k!$. The question to answer is the following: given $f(z)$ holomorphic on some neighborhood of $z=0$, does $D[f](z)$ define a holomorphic function on some (possibly smaller) neighborhood of $z=0$? By combining the above estimates, we see that the answer is Yes, since the inequalities $$ \left|\sum_{k=0}^N d_k \frac{f^{(k)}(z)}{k!} \right| \le C \sum_{k=0}^\infty \frac{d_k}{R^k} < \infty $$ for arbitrarily large $N$ mean that the series defining $D[f](z)$ converges uniformly (and hence to a holomorphic function) on the same neighborhood of $z=0$ on which $|f^{(k)}(z)/k!| < C/R^k$ uniformly.

So $D$ maps germs of holomorphic functions to germs of holomorphic functions and, given how it was defined, it also clearly satisfies all the other properties of a morphism of sheaves. And yet, $D$ is not a differential operator of finite order. By construction, $D$ is a differential operator of infinite order with constant coefficients. But clearly, we can take the coefficients $d_k(z)$ as holomorphic functions, provided that they satisfy similar locally uniform bounds. So there are lots of possibilities for constructing sheaf morphisms that are not finite order differential operators.