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Let $K$ be a field of characteristic $0$, let $A = K[[t_1, \ldots, t_n]]$ be a power series ring over $K$, and let $V$ be a free $A$-module. Let $\nabla \colon V \rightarrow V \otimes_A \Omega^1_{A/K} = \bigoplus_{k=1}^n V \ dt_k$ be an integrable connection, so $\nabla^2 = 0$. I want to show that there always exist solutions $s \in V$ to the equation $\nabla(s) = 0$ subject to any initial condition $s(0) = \alpha$ for $\alpha \in V(0) = V \otimes_A A/\mathfrak{m}_A$.

More concretely, letting $e_1, \ldots, e_r$ be a basis for $V$, I have a system of $n \times r$ linear first-order homogeneous partial differential equations $$ \frac{\partial}{\partial t_k} s_i = \sum_j M^j_{ik} s_j $$ For some $M_{ik}^j \in A$. Or, writing $M_k$ for the matrix $(M_{i,k}^j)_{i,j}$, $$ \frac{\partial}{\partial t_k} s = M_k s $$

How can I show that integrability implies there are always solutions? Concretely, integrability is exactly the statement that for all $i,j,k, \ell$, we have $$ \frac{\partial}{\partial t_\ell} M_{k} - \frac{\partial}{\partial t_k} M_{\ell} = M_k M_\ell - M_\ell M_k $$

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  • $\begingroup$ What is $V(0)$? $\endgroup$ Commented Feb 24, 2019 at 17:49

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I hope you'll allow one or two slight adjustments to your question:

  • I think you are interested in $(t_1, \dots, t_n)$-adically continuous connections rather than arbitrary connections (for which the desired solution principle would not generally hold).

  • Judging from how you use "$r$," I think you intend $V$ to be a finite-rank free $A$-module (rather than an arbitrary free $A$-module).

You will find a very nice treatment of your question thus modified in Proposition 8.9 of Nicholas Katz's paper "Nilpotent connections and the monodromy theorem: Applications of a result of Turrittin."

Here is the statement of that proposition:

Let $K$ be a field of characteristic zero, $K[[t_1, \dots, t_n]]$ the ring of formal power series over $K$ in $n$ variables. Let $M$ be a finitely generated module over $K[[t_1, \dots, t_n]]$, given with an integrable connection $\nabla$ (for the continuous $K$-derivations of $K[[t_1, \dots, t_n]]$ to itself). Then $M^\nabla$, the $K$-space of horizontal elements of $M$, is finite-dimensional over $K$, and the pair $(M, \nabla)$ is isomorphic to the pair $(M^\nabla\otimes_K K[[t_1, \dots, t_n]], 1\otimes d)$ where $d$ denotes the ``identical'' connection on $K[[t_1, \dots, t_n]]$.

Note that in this proposition, one can even assume only that the module $M$ with integrable connection is finitely generated (rather than free of finite rank). As a consequence of fact that $M^\nabla \otimes_K K[[t_1, \dots, t_n]] \to M$ is an isomorphism, one can then conclude that $M$ is necessarily free.

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    $\begingroup$ Nice answer! I want to emphasize that this proposition is really saying that integrable connections are only really interesting as global objects---locally, they're all determined up to isomorphism by their rank. $\endgroup$ Commented Feb 24, 2019 at 17:58
  • $\begingroup$ Yes, I think both of those adjustments are right. Thanks for the reference! I'm surprised by the simplicity of the argument - it looks like Katz essentially just writes down a solution directly as a "matrix exponential". The condition of being $(t_1, \ldots, t_n)$-adically continuous should say exactly that the integrable connection is specified by differential equations $\frac{\partial}{\partial t_k} s = \sum_j M^j_{ik} s$, right? (i.e. some random integrable connection on the power series ring might not land in the span of the $dt_i$'s.) $\endgroup$
    – dorebell
    Commented Feb 25, 2019 at 0:07

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