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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 rank $0$ as multiplication with a global section of $\mathcal{O}$ and differential operators of rank $n$ 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 rank $n-1$. General differential operators can then be defined as $\mathbb{C}$-linear morphisms of sheaves which locally are differential operators of finite rank.
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?

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  • $\begingroup$ Welcome to Math Overflow. What is EGA? $\endgroup$ – Amir Sagiv Nov 9 at 19:32
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    $\begingroup$ Éléments de Géometrie Algébrique by Grothendieck $\endgroup$ – Carlos Esparza Nov 9 at 19:34
  • $\begingroup$ According to Wikipedia, Peetre's theorem needs a support-decreasing hypothesis (in the real differentiable case). $\endgroup$ – Qfwfq Nov 11 at 15:32
  • $\begingroup$ @Qfwfq on a (paracompact) manifold that's equivalent to saying that $D$ is a morphism if sheaves. But that's just a particularity if the real case, in general you should define differential operators as morphisms of sheaves (e.g. the support of an nonzero holomorphic function on a connected domain is the entire domain) $\endgroup$ – Carlos Esparza Nov 11 at 15:36
  • $\begingroup$ Ok, I see. - Assume we're on a space $(X,O_X)$ locally ringed in algebras over a char zero field $K$, and $E$ and $F$ are two (finite rank) locally free sheaves on $X$. We have: 1) the set $Hom_K (E,F)$ of $K$-linear maps of sheaves of $K$-vector spaces. 2) the set $Diff_K(E,F)$ of the $D\in Hom_K(E,F)$ locally satisfying the recursive $[D,M_f]\in Diff^{n-1}$ type relation. So, if I've understood well, smooth Peetre's theorem says the sets 1) and 2) are equal in this case? and your Q.1 was whether that was also true in the holomorphic category? $\endgroup$ – Qfwfq Nov 11 at 16:09
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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.

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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.

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    $\begingroup$ For Q1, I think the intended hypothesis on the operator is that its action is known only on holomorphic functions, $D\colon \mathcal{O} \to \mathcal{O}$, rather than all smooth functions. This is a strictly weaker hypothesis than is needed to apply the smooth Petree theorem in the way that you suggest. $\endgroup$ – Igor Khavkine Nov 11 at 12:05
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    $\begingroup$ The first paragraph of the question recalls the smooth Peetre theorem: $D$ is a morphism of sheaves of smooth functions $\iff$ $D$ is locally a differential operator. I presume that translating this statement to the complex world replaces "sheaves of smooth functions" by "sheaves of holomorphic functions" (meaning that $D$ depends only on the germ at any point and maps germs of holomorphic functions to germs of holomorphic functions). The notion of differential operator is the usual one, as you've discussed. Hence, Q1: is this translated statement an actual theorem? $\endgroup$ – Igor Khavkine Nov 11 at 13:42
  • $\begingroup$ @IgorKhavkine Yes, that is what I have in mind. I will update my question to be more precise $\endgroup$ – Carlos Esparza Nov 11 at 13:46
  • $\begingroup$ Is it obvious that $D(I^{m+k})=I^m$? $\endgroup$ – Qfwfq Nov 12 at 0:51
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    $\begingroup$ I should have written $D(I^{m+k})\subset I^m$. It's not obvious but follows from the equality $[f_1,[f_2,[\cdots[f_n,D]\cdots]]=\sum\limits_{s\subseteq \{1,2,\ldots,n\}} (-1)^{|s|}\left(\prod\limits_{i\in s}f_i\right)D\prod\limits_{k\in \{1,2,\ldots,n\}\setminus s}f_k$ which can be proven by induction. $\endgroup$ – Michael Bächtold Nov 12 at 13:12

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