I originally posted the question on math.stackexhange, but there doesn't seem to be an answer. I apalogize in advance for cross posting.

Let $E\rightarrow X$ be a holomorphic vector bundle over a compact Riemann surface with a holomorphic connection $\nabla:E\rightarrow E\otimes K$, where $K$ is the canonical bundle of $X$. Since the holomorphic connection is necessarily flat, its sheaf of local holomorphic sections $\mathcal{E}$ defines a (holonomic) $D$-module. Every holonomic D-module is locally cyclic, i.e. for any point $z_0$ there exists a neighborhood $U$ s.th. $\mathcal{E}(U)$ has a cyclic generator as a $D$-module (see e.g. Proposition 3.1.5. in Björk: Analytic $D$-Modules and Applications). Suppose we are given a coordinate $z$ on $U$ and identify $D(U)\cong D_1$, with $D_1=\mathbb{C}\left\lbrace z \right\rbrace \left\langle \partial_z \right\rangle$ (differential operators with coefficients in convergent power-series). So locally it holds $\mathcal{E}(U)\cong D_1/ I$, where $I$ is the ideal of differential operators annihilating the cyclic generator. This ideal is in general generated by two elements $P,Q$, with $P$ an operator of smallest possible degree in $I$ and furthermore $I/D_1P$ is of torsion type, i.e. for any $D\in I$ it holds $z^nD\in D_1P$ for some $n$ (Proposition 5.1.4 and Remark 5.1.5 in Björk: Analytic $D$-Modules and Applications). This implies for the dual $D$-module $\hom_{D_1}(D_1/I,\mathbb{C}\left\lbrace z\right\rbrace)=\left\lbrace f\in \mathbb{C}\left\lbrace z\right\rbrace \, \middle|\, Pf=Qf=0\right\rbrace=\left\lbrace f\in \mathbb{C}\left\lbrace z\right\rbrace \, \middle|\, Pf=0\right\rbrace$.

So far so good. Now on $U$ the holomorphic connection reads $\nabla|_U=\partial+A$ with $A$ some matrix of holomorphic functions.The dual bundle naturally comes with a holomorphic connection, too, which in local coordinates takes the form $\partial-A^T$. The whole discussion above shows that locally flat sections ($(\partial-A^T)Y=0$) are in one to one correspondence with solutions of $Pf=0$.

On the other hand there is Deligne's lemma of a cyclic vector. One way to formulate it, is to say that locally on a coordinate neighborhood $U$, for a vector bundle with holomorphic connection there exists $G\in \mathrm{GL}(n,\mathcal{O}(U))$, s.th. \begin{equation} \partial_z G-G A^T=\tilde{A}G \end{equation} with $\tilde{A}$ in companion form. Here $\partial-A^T$ is the local form of the holomorphic connection. But in general the non unit entries $a_i$ in $\tilde{A}$ are only meromorphic and $G$ might not be invertible as a holomorphic matrix.

It is clear that a system of linear differential equations $\partial_z Y=A^T Y$ with $A^T$ in companion form corresponds to a single $n$-th order scalar differential equation $Qf=0$. So from Deligne's cyclic vector lemma I get an $n$-th order scalar differential equation, but the corresponding differential operator might not be in $D_1$, but in $\mathbb{C}\left\lbrace z\right\rbrace [z^{-1}]\left\langle \partial_z\right\rangle$.

Q: Is there any relation between the differential operator I get from the discussion in the first paragraph applied to the dual bundle and the differential equation I get from Deligne's cyclic vector lemma?

I guess they are the same, maybe after imposing further constraints on the open $U$. It might very well be that the relation is obvious and just shows my lack of understanding.


I think I found the relation. The paper A Simple Algorithm for Cyclic Vectors by N. Katz seems crucial. The cyclic vector lemma is mostly stated for differential fields (see e.g. section 2 in Galois Theory of Linear Differential Equations), but theorem 1 in the paper by Katz actually doesn't need a field. To be more precise, let $R$ be a commutative, local ring with a derivation $\partial:R\rightarrow R$, an element $z\in R$ s.th. $\partial(z)=1$ and $(V,D,\mathbf{e})$ a freely, finitely generated $R$-module, where $D$ is an additive mapping $D:V\rightarrow V$ satisfying the usual Leibniz rule $D(rv)=\partial(r)v+rD(v)$ and $\mathbf{e}=(e_0,\dots e_{n-1})$ is an $R$-basis. Then under the assumption that $R$ is an $\mathbb{Z}\left[\frac{1}{(n-1)!}\right]$-algebra and $z$ is in the unique maximal ideal of $R$, there is a cyclic vector $c(\mathbf{e},z)$, i.e. $(c,Dc,\dots , D^{n-1}c)$ is an $R$-basis in $V$.

Since locally on a trivializing open $U$ with coordinate $z$, the sheaf of sections $\mathcal{E}(U)$ is a free $\mathbb{C}\left\lbrace z \right\rbrace$ module and $\left(\mathbb{C}\left\lbrace z \right\rbrace,\partial_z,z\right)$ is a local ring in which $(n-1)!$ is invertible one can apply the theorem to this situation.

Moreover the path trough the dual bundle seems unnecessary from this. Since $(c,\nabla_{\frac{\partial}{\partial z}}c,\dots, \nabla^{(n-1)}_{\frac{\partial}{\partial z}}c$) is a local frame, its corresponding matrix $C=\left(c\,\nabla_{\frac{\partial}{\partial z}}c\, \cdots\,\nabla^{(n-1)}_{\frac{\partial}{\partial z}}c\right)$ is invertible as a holomorphic matrix. Applying the gauge transformation $C^{-1}\left(\partial_z+A\right)C$ to the local form of the connection gives a connection in companion form and hence an $n$-th order scalar differential equation.

On the other hand, from the $D$-module point of view, it is clear that there is an equation \begin{equation} \nabla^{(n)}_{\frac{\partial}{\partial z}}c=\sum_{i=0}^{n-1}f_i(z)\nabla^{(i)}_{\frac{\partial}{\partial z}}c\end{equation}, thus $P\equiv \partial_z^n-\sum_{i=0}^{n-1}f_i \partial_z^i$ is in the ideal $I$ of $D_1$, killing the cyclic generator. In addition I think $I=D_1P$, since the leading coefficient in $P$ is just $1$ and $\mathcal{E}(U)$ is locally free $\mathbb{C}\left\lbrace z\right\rbrace $-module. Furthermore, $P$ is the differential operator one gets from the companion form of the connection.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.